Contents
Preface
2
1 The Wave Function
3
2 Time-Independent Schrödinger Equation
14
3 Formalism
62
4 Quantum Mechanics in Three Dimensions
87
5 Identical Particles
132
6 Time-Independent Perturbation Theory
154
7 The Variational Principle
196
8 The WKB Approximation
219
9 Time-Dependent Perturbation Theory
236
10 The Adiabatic Approximation
254
11 Scattering
268
12 Afterword
282
Appendix Linear Algebra
283
2nd Edition – 1st Edition Problem Correlation Grid
299
2
Preface
These are my own solutions to the problems in Introduction to Quantum Mechanics, 2nd ed. I have made every
effort to insure that they are clear and correct, but errors are bound to occur, and for this I apologize in advance.
I would like to thank the many people who pointed out mistakes in the solution manual for the first edition,
and encourage anyone who finds defects in this one to alert me (griffith@reed.edu). I’ll maintain a list of errata
on my web page (http://academic.reed.edu/physics/faculty/griffiths.html), and incorporate corrections in the
manual itself from time to time. I also thank my students at Reed and at Smith for many useful suggestions,
and above all Neelaksh Sadhoo, who did most of the typesetting.
At the end of the manual there is a grid that correlates the problem numbers in the second edition with
those in the first edition.
David Griffiths
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publisher.
CHAPTER 1. THE WAVE FUNCTION
3
Chapter 1
The Wave Function
Problem 1.1
(a)
j2 = 212 = 441.
j 2 =
=
1 2
1 2
j N (j) =
(14 ) + (152 ) + 3(162 ) + 2(222 ) + 2(242 ) + 5(252 )
N
14
1
6434
(196 + 225 + 768 + 968 + 1152 + 3125) =
= 459.571.
14
14
j
14
15
16
22
24
25
(b)
σ2 =
=
σ=
∆j = j − j
14 − 21 = −7
15 − 21 = −6
16 − 21 = −5
22 − 21 = 1
24 − 21 = 3
25 − 21 = 4
1 1 (∆j)2 N (j) =
(−7)2 + (−6)2 + (−5)2 · 3 + (1)2 · 2 + (3)2 · 2 + (4)2 · 5
N
14
1
260
(49 + 36 + 75 + 2 + 18 + 80) =
= 18.571.
14
14
√
18.571 = 4.309.
(c)
j 2 − j2 = 459.571 − 441 = 18.571.
[Agrees with (b).]
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4
CHAPTER 1. THE WAVE FUNCTION
Problem 1.2
(a)
h
1
1
x2 √ dx = √
2 hx
2 h
x2 =
0
h2
σ = x − x =
−
5
2
2
2
(b)
x+
P =1−
x−
h
h2
2 5/2 =
x
.
5
5
0
2
2h
4 2
h
=
h ⇒ σ = √ = 0.2981h.
3
45
3 5
√ x+
1
1 √
1
√
√ dx = 1 − √ (2 x) = 1 − √
x+ − x− .
2 hx
2 h
h
x−
x+ ≡ x + σ = 0.3333h + 0.2981h = 0.6315h;
√
P =1−
0.6315 +
√
x− ≡ x − σ = 0.3333h − 0.2981h = 0.0352h.
0.0352 = 0.393.
Problem 1.3
(a)
∞
Ae−λ(x−a) dx.
2
1=
Let u ≡ x − a, du = dx, u : −∞ → ∞.
−∞
∞
π
λ
e−λu du = A
2
1=A
−∞
(b)
x = A
∞
⇒ A=
xe−λ(x−a) dx = A
2
−∞
∞
=A
ue
−λu
2
x = A
∞
2
−∞
∞
−λu2
e
du = A 0 + a
π
λ
= a.
x2 e−λ(x−a) dx
−∞
∞
=A
2
2 −λu2
u e
∞
du + 2a
−∞
1
2λ
(u + a)e−λu du
−∞
=A
∞
du + a
−∞
2
λ
.
π
ue
−λu2
σ 2 = x2 − x2 = a2 +
π
λ
= a2 +
1
1
− a2 =
;
2λ
2λ
∞
du + a
−∞
π
+ 0 + a2
λ
2
−λu2
e
du
−∞
1
.
2λ
1
σ=√ .
2λ
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publisher.
CHAPTER 1. THE WAVE FUNCTION
5
(c)
ρ(x)
A
x
a
Problem 1.4
(a)
|A|2
1= 2
a
= |A|2
2
x dx +
0
|A|2
a
b
(b − x) dx = |A|
2
2
(b − a)
2
a
a b−a
b
+
= |A|2 ⇒ A =
3
3
3
(b)
1
a2
a
b (b − x)3 x3 1
−
+
3 0 (b − a)2
3
a
3
.
b
Ψ
A
a
b
x
(c) At x = a.
(d)
a
|Ψ|2 dx =
P =
0
|A|2
a2
a
x2 dx = |A|2
0
a
a
= .
3
b
P = 1 if b = a, P = 1/2 if b = 2a. (e)
a
b
1
1
2
x|Ψ|2 dx = |A|2 2
x3 dx +
x(b
−
x)
dx
a 0
(b − a)2 a
4 a
b 2
3 1
x x3
x4 1
2x
=
b
− 2b +
+
b a2
4 0 (b − a)2
2
3
4 a
2
3
=
a (b − a)2 + 2b4 − 8b4 /3 + b4 − 2a2 b2 + 8a3 b/3 − a4
2
4b(b − a)
4
3
b
2a + b
2 3
1
2 2
=
(b3 − 3a2 b + 2a3 ) =
−a b + a b =
.
2
2
4b(b − a)
3
3
4(b − a)
4
x =
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publisher.
6
CHAPTER 1. THE WAVE FUNCTION
Problem 1.5
(a)
∞
|Ψ| dx = 2|A|
2
1=
2
−2λx
e
2
dx = 2|A|
0
∞
e−2λx |A|2
=
;
−2λ 0
λ
A=
√
λ.
(b)
x =
x|Ψ| dx = |A|
2
x2 = 2|A|2
2
∞
∞
xe−2λ|x| dx = 0.
[Odd integrand.]
−∞
x2 e−2λx dx = 2λ
0
1
2
=
.
3
(2λ)
2λ2
(c)
σ 2 = x2 − x2 =
1
;
2λ2
√
1
σ=√ .
2λ
|Ψ(±σ)|2 = |A|2 e−2λσ = λe−2λ/
λ
2λ
√
= λe−
2
= 0.2431λ.
|Ψ| 2
.24λ
−σ
+σ
x
Probability outside:
∞
2
|Ψ|2 dx = 2|A|2
σ
∞
e−2λx dx = 2λ
σ
∞
√
e−2λx = e−2λσ = e− 2 = 0.2431.
−2λ σ
Problem 1.6
For integration by parts, the differentiation has to be with respect to the integration variable – in this case the
differentiation is with respect to t, but the integration variable is x. It’s true that
∂
∂x 2
∂
∂
(x|Ψ|2 ) =
|Ψ| + x |Ψ|2 = x |Ψ|2 ,
∂t
∂t
∂t
∂t
but this does not allow us to perform the integration:
a
b
∂
x |Ψ|2 dx =
∂t
a
b
b
∂
(x|Ψ|2 )dx = (x|Ψ|2 )a .
∂t
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CHAPTER 1. THE WAVE FUNCTION
Problem 1.7
From Eq. 1.33,
∂
∂t
dp
dt
∗ ∂Ψ
Ψ
∂x
= −i
∂
∂t
7
Ψ∗ ∂Ψ
∂x dx. But, noting that
∂2Ψ
∂x∂t
=
∂2Ψ
∂t∂x
and using Eqs. 1.23-1.24:
∂Ψ∗ ∂Ψ
∂Ψ
i ∂ 2 Ψ∗
i ∂ 2 Ψ
i
i
∗ ∂
∗ ∂Ψ
∗ ∂
=
+Ψ
= −
V
Ψ
+
Ψ
+
− VΨ
∂t ∂x
∂x ∂t
2m ∂x2
∂x
∂x 2m ∂x2
i
∂ 3 Ψ ∂ 2 Ψ∗ ∂Ψ
∂Ψ
∂
i
=
Ψ∗ 3 −
+
V Ψ∗
− Ψ∗ (V Ψ)
2
2m
∂x
∂x ∂x
∂x
∂x
The first term integrates to zero, using integration by parts twice, and the second term can be simplified to
∗ ∂Ψ
∗ ∂V
2 ∂V
V Ψ∗ ∂Ψ
∂x − Ψ V ∂x − Ψ ∂x Ψ = −|Ψ| ∂x . So
dp
= −i
dt
∂V
i
∂V
−|Ψ|2
dx = −
.
∂x
∂x
QED
Problem 1.8
∂ Ψ
Suppose Ψ satisfies the Schrödinger equation without V0 : i ∂Ψ
∂t = − 2m ∂x2 + V Ψ. We want to find the solution
2
2
∂ Ψ0
0
Ψ0 with V0 : i ∂Ψ
∂t = − 2m ∂x2 + (V + V0 )Ψ0 .
2
2
Claim: Ψ0 = Ψe−iV0 t/ .
2 2
∂Ψ −iV0 t/
∂ Ψ
−iV0 t/
0
Proof: i ∂Ψ
+ iΨ − iV0 e−iV0 t/ = − 2m
+ V0 Ψe−iV0 t/
∂t = i ∂t e
∂x2 + V Ψ e
2
∂ Ψ0
= − 2m
∂x2 + (V + V0 )Ψ0 .
2
QED
This has no effect on the expectation value of a dynamical variable, since the extra phase factor, being independent of x, cancels out in Eq. 1.36.
Problem 1.9
(a)
2
∞
1 = 2|A|
−2amx2 /
e
21
dx = 2|A|
0
2
π
= |A|2
(2am/)
π
;
2am
A=
2am
π
1/4
.
(b)
∂Ψ
= −iaΨ;
∂t
∂Ψ
2amx
=−
Ψ;
∂x
∂2Ψ
2am
=−
∂x2
∂Ψ
Ψ+x
∂x
2am
=−
2amx2
1−
Ψ.
∂ Ψ
Plug these into the Schrödinger equation, i ∂Ψ
∂t = − 2m ∂x2 + V Ψ:
2
2
2
2am
2amx2
V Ψ = i(−ia)Ψ +
−
1−
Ψ
2m
2amx2
= a − a 1 −
Ψ = 2a2 mx2 Ψ, so
V (x) = 2ma2 x2 .
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8
CHAPTER 1. THE WAVE FUNCTION
(c)
x =
∞
x|Ψ|2 dx = 0.
[Odd integrand.]
−∞
∞
x2 = 2|A|2
x2 e−2amx
2
/
dx = 2|A|2
0
p = m
dx
= 0.
dt
∂
i ∂x
2
∂2Ψ
Ψ∗ 2 dx
Ψ
Ψdx = −
∂x
2
2am
2amx
2am
2
∗
2
2
2
= −
Ψ −
1−
Ψ dx = 2am
|Ψ| dx −
x |Ψ| dx
2am 2
2am 1
= 2am 1 −
x = 2am 1 −
= 2am
= am.
4am
2
∗
p =
2
π
=
.
2am
4am
1
22 (2am/)
2
(d)
σx2 = x2 − x2 =
σx σp =
4am
√
=⇒ σx =
4am
;
4am
σp2 = p2 − p2 = am =⇒ σp =
√
am.
am = 2 . This is (just barely) consistent with the uncertainty principle.
Problem 1.10
From Math Tables: π = 3.141592653589793238462643 · · ·
(a)
P (0) = 0
P (5) = 3/25
P (1) = 2/25
P (6) = 3/25
In general, P (j) =
=
1
25 [0
(c) j 2 =
=
1
25 [0
P (3) = 5/25
P (8) = 2/25
1
25 [0
Median: 13 are ≤ 4, 12 are ≥ 5, so median is 4.
· 0 + 1 · 2 + 2 · 3 + 3 · 5 + 4 · 3 + 5 · 3 + 6 · 3 + 7 · 1 + 8 · 2 + 9 · 3]
+ 2 + 6 + 15 + 12 + 15 + 18 + 7 + 16 + 27] =
1
25 [0
P (4) = 3/25
P (9) = 3/25
N (j)
N .
(b) Most probable: 3.
Average: j =
P (2) = 3/25
P (7) = 1/25
118
25
= 4.72.
+ 12 · 2 + 22 · 3 + 32 · 5 + 42 · 3 + 52 · 3 + 62 · 3 + 72 · 1 + 82 · 2 + 92 · 3]
+ 2 + 12 + 45 + 48 + 75 + 108 + 49 + 128 + 243] =
σ 2 = j 2 − j2 = 28.4 − 4.722 = 28.4 − 22.2784 = 6.1216;
710
25
= 28.4.
√
σ = 6.1216 = 2.474.
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CHAPTER 1. THE WAVE FUNCTION
9
Problem 1.11
(a) Constant for 0 ≤ θ ≤ π, otherwise zero. In view of Eq. 1.16, the constant is 1/π.
1/π, if 0 ≤ θ ≤ π,
ρ(θ) =
0,
otherwise.
ρ(θ)
1/π
−π/2
π
0
θ
3π/2
(b)
θ =
θρ(θ) dθ =
1
θ =
π
π
1
π
1
θ dθ =
π
2
π
θdθ =
0
2
0
σ 2 = θ2 − θ2 =
1
π
π
π
θ2 =
2 0
2
[of course].
π
θ3 π2
.
=
3 0
3
π2
π2
π2
−
=
;
3
4
12
π
σ= √ .
2 3
(c)
1
sin θ =
π
cos θ =
1
π
π
sin θ dθ =
2
1
1
π
(− cos θ)|0 = (1 − (−1)) = .
π
π
π
cos θ dθ =
1
π
(sin θ)|0 = 0.
π
0
1
cos θ =
π
π
0
2
π
1
cos θ dθ =
π
2
0
π
(1/2)dθ =
0
1
.
2
[Because sin2 θ + cos2 θ = 1, and the integrals of sin2 and cos2 are equal (over suitable intervals), one can
replace them by 1/2 in such cases.]
Problem 1.12
(a) x = r cos θ ⇒ dx = −r sin θ dθ. The probability that the needle lies in range dθ is ρ(θ)dθ =
probability that it’s in the range dx is
ρ(x)dx =
1
π dθ,
so the
1 dx
dx
1
dx
= √
.
=
2
π r sin θ
π r 1 − (x/r)
π r2 − x2
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10
CHAPTER 1. THE WAVE FUNCTION
ρ(x)
-2r
√ 1
,
π r 2 −x2
∴ ρ(x) =
Total:
0,
r
√ 1
dx
−r π r 2 −x2
(b)
x =
1
π
2
π
x2 =
r
−r
0
x√
r
√
=
r2
2
π
-r
r
if − r < x < r,
otherwise.
r
0
√ 1
dx
r 2 −x2
1
dx = 0
− x2
2
π
=
x
2r
[Note: We want the magnitude of dx here.]
sin−1
x r
r 0
=
2
π
sin−1 (1) =
2
π
·
π
2
= 1.
[odd integrand, even interval].
x r
2
2
x2
2
r2
x 2
= 2 r sin−1 (1) = r .
dx =
r − x2 +
−
sin−1
π
2
2
r
π 2
2
r2 − x2
0
√
σ 2 = x2 − x2 = r2 /2 =⇒ σ = r/ 2.
To get x and x2 from Problem 1.11(c), use x = r cos θ, so x = rcos θ = 0, x2 = r2 cos2 θ = r2 /2.
Problem 1.13
Suppose the eye end lands a distance y up from a line (0 ≤ y < l), and let x be the projection along that same
direction (−l ≤ x < l). The needle crosses the line above if y + x ≥ l (i.e. x ≥ l − y), and it crosses the line
below if y + x < 0 (i.e. x < −y). So for a given value of y, the probability of crossing (using Problem 1.12) is
−y
P (y) =
−l
1
=
π
−1
sin
l
1
ρ(x)dx +
ρ(x)dx =
π
l−y
−y
−l
1
√
dx +
2
l − x2
l
l−y
1
√
dx
2
l − x2
x −y
l 1
−1 x =
− sin−1 (y/l) + 2 sin−1 (1) − sin−1 (1 − y/l)
+ sin
l −l
l l−y
π
sin−1 (y/l) sin−1 (1 − y/l)
−
.
π
π
Now, all values of y are equally likely, so ρ(y) = 1/l, and hence the probability of crossing is
1 l
l−y
1 l
−1 y
−1
P =
π − sin
π − 2 sin−1 (y/l) dy
− sin
dy =
πl 0
l
l
πl 0
=1−
=
l
1
2
2
2
πl − 2 y sin−1 (y/l) + l 1 − (y/l)2 = 1 − [l sin−1 (1) − l] = 1 − 1 + = .
πl
πl
π
π
0
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CHAPTER 1. THE WAVE FUNCTION
11
Problem 1.14
(a) Pab (t) =
b
a
|Ψ(x, t)2 dx,
∂|Ψ|2
∂
i
=
∂t
∂x 2m
∴
dPab
=−
dt
b
a
so
dPab
dt
=
b
∂
|Ψ|2 dx.
a ∂t
But (Eq. 1.25):
∂Ψ∗
∂
Ψ
−
Ψ
= − J(x, t).
∂x
∂x
∂t
∗ ∂Ψ
∂
b
J(x, t)dx = − [J(x, t)]|a = J(a, t) − J(b, t).
∂x
QED
Probability is dimensionless, so J has the dimensions 1/time, and units seconds−1 .
(b) Here Ψ(x, t) = f (x)e−iat , where f (x) ≡ Ae−amx
2
/
∗
df
−iat df iat
, so Ψ ∂Ψ
= f dx
,
∂x = f e
dx e
df
and Ψ∗ ∂Ψ
∂x = f dx too, so J(x, t) = 0.
Problem 1.15
(a) Eq. 1.24 now reads
∂Ψ∗
∂t
2
∗
i ∂ Ψ
i ∗ ∗
= − 2m
∂x2 + V Ψ , and Eq. 1.25 picks up an extra term:
∂
i
i
2Γ 2
|Ψ|2 = · · · + |Ψ|2 (V ∗ − V ) = · · · + |Ψ|2 (V0 + iΓ − V0 + iΓ) = · · · −
|Ψ| ,
∂t
2Γ ∞
2Γ
2
and Eq. 1.27 becomes dP
QED
dt = − −∞ |Ψ| dx = − P .
(b)
dP
2Γ
2Γ
= − dt =⇒ ln P = − t + constant =⇒ P (t) = P (0)e−2Γt/ , so τ =
.
P
2Γ
Problem 1.16
Use Eqs. [1.23] and [1.24], and integration by parts:
d
dt
∞
−∞
Ψ∗1 Ψ2
∞ ∗
∂Ψ1
∂
∗
∗ ∂Ψ2
(Ψ1 Ψ2 ) dx =
Ψ2 + Ψ1
dx
∂t
∂t
−∞ ∂t
−∞
∞ −i ∂ 2 Ψ∗1
i ∂ 2 Ψ2
i
i
∗
∗
Ψ
dx
+
+
Ψ
−
V
Ψ
V
Ψ
2
2
1
1
2m ∂x2
2m ∂x2
−∞
∞ 2 ∗
2
i
∂ Ψ1
∗ ∂ Ψ2
−
dx
Ψ
−
Ψ
2
1
2m −∞ ∂x2
∂x2
∞
∞
∞
∞
i ∂Ψ∗1 ∂Ψ∗1 ∂Ψ2
∂Ψ∗1 ∂Ψ2
∗ ∂Ψ2 −
−
+
Ψ2 dx − Ψ1
dx = 0. QED
2m ∂x
∂x −∞
−∞ ∂x ∂x
−∞ ∂x ∂x
−∞
dx =
=
=
=
∞
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12
CHAPTER 1. THE WAVE FUNCTION
Problem 1.17
(a)
1 = |A|
a
2 2
a −x
2
2
−a
a
a − 2a x + x dx = 2|A|
2
4
dx = 2|A|
2 2
4
2
0
2 1
16 5 2
= 2|A|2 a5 1 − +
=
a |A| , so A =
3 5
15
a
x5 a x − 2a
+
3
5 0
4
3
2x
15
.
16a5
(b)
x =
a
x|Ψ|2 dx = 0.
(Odd integrand.)
−a
(c)
p =
2
A
i
a
a2 − x2
−a
d 2
a − x2 dx = 0.
dx
(Odd integrand.)
−2x
Since we only know x at t = 0 we cannot calculate dx/dt directly.
(d)
a
2
x2 a2 − x2 dx = 2A2
−a
0
3
5
7 a
15
x
x
x
=
=2
a4
− 2a2
+
16a5
3
5
7 0
✚ 2 35 − 42 + 15
15a
a2 8
✚
=
=
· =
8
8 7
3·✁
5·7
✁
a
x2 = A2
a4 x2 − 2a2 x4 + x6 dx
15 7
a
8a5
1 2 1
− +
3 5 7
a2
.
7
(e)
p2 = −A2 2
=4·
a
−a
15 2
16a5
a2 − x2
d2 2
a − x2 dx = 2A2 2 2
2
dx
a
a2 − x2 dx
0
−2
a2 x −
a
5 2
152 2
x3 152 3 a3
a
=
=
−
· =
.
5
2
3 0
4a
3
4a
3
2 a2
(f )
σx =
x2 − x2 =
1 2
a
a = √ .
7
7
p2 − p2 =
5 2
=
2 a2
(g)
σp =
5
.
2a
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CHAPTER 1. THE WAVE FUNCTION
13
(h)
5
=
2a
a
σx σp = √ ·
7
5
=
14
10 > .
7 2
2
Problem 1.18
√
h
h2
>d ⇒ T <
.
3mkB d2
3mkB T
(a) Electrons (m = 9.1 × 10−31 kg):
T <
(6.6 × 10−34 )2
= 1.3 × 105 K.
3(9.1 × 10−31 )(1.4 × 10−23 )(3 × 10−10 )2
Sodium nuclei (m = 23mp = 23(1.7 × 10−27 ) = 3.9 × 10−26 kg):
T <
3(3.9 ×
(6.6 × 10−34 )2
= 3.0 K.
× 10−23 )(3 × 10−10 )2
10−26 )(1.4
(b) P V = N kB T ; volume occupied by one molecule (N = 1, V = d3 ) ⇒ d = (kB T /P )1/3 .
T <
h2
2mkB
P
kB T
2/3
⇒ T 5/3 <
h2 P 2/3
1
⇒T <
5/3
3m k
kB
B
h2
3m
3/5
P 2/5 .
For helium (m = 4mp = 6.8 × 10−27 kg) at 1 atm = 1.0 × 105 N/m2 :
1
T <
(1.4 × 10−23 )
(6.6 × 10−34 )2
3(6.8 × 10−27 )
3/5
(1.0 × 105 )2/5 = 2.8 K.
For hydrogen (m = 2mp = 3.4 × 10−27 kg) with d = 0.01 m:
T <
3(3.4 ×
(6.6 × 10−34 )2
= 3.1 × 10−14 K.
× 10−23 )(10−2 )2
10−27 )(1.4
At 3 K it is definitely in the classical regime.
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14
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
Chapter 2
Time-Independent Schrödinger
Equation
Problem 2.1
(a)
Ψ(x, t) = ψ(x)e−i(E0 +iΓ)t/ = ψ(x)eΓt/ e−iE0 t/ =⇒ |Ψ|2 = |ψ|2 e2Γt/ .
∞
−∞
|Ψ(x, t)| dx = e
2
2Γt/
∞
−∞
|ψ|2 dx.
The second term is independent of t, so if the product is to be 1 for all time, the first term (e2Γt/ ) must
also be constant, and hence Γ = 0. QED
2
∂ ψ
(b) If ψ satisfies Eq. 2.5, − 2m
dx2 + V ψ = Eψ, then (taking the complex conjugate and noting that V and
2
2
∗
∂ ψ
∗
∗
∗
E are real): − 2m
dx2 + V ψ = Eψ , so ψ also satisfies Eq. 2.5. Now, if ψ1 and ψ2 satisfy Eq. 2.5, so
too does any linear combination of them (ψ3 ≡ c1 ψ1 + c2 ψ2 ):
2 ∂ 2 ψ 3
∂ 2 ψ1
∂ 2 ψ2
2
−
+ V (c1 ψ1 + c2 ψ2 )
c
+
V
ψ
=
−
+
c
3
1
2
2m dx2
2m
dx2
∂x2
2 d 2 ψ 1
2 d 2 ψ2
= c1 −
+
c
−
+
V
ψ
+ V ψ2
1
2
2m dx2
2m dx2
2
= c1 (Eψ1 ) + c2 (Eψ2 ) = E(c1 ψ1 + c2 ψ2 ) = Eψ3 .
Thus, (ψ + ψ ∗ ) and i(ψ − ψ ∗ ) – both of which are real – satisfy Eq. 2.5. Conclusion: From any complex
solution, we can always construct two real solutions (of course, if ψ is already real, the second one will be
zero). In particular, since ψ = 12 [(ψ + ψ ∗ ) − i(i(ψ − ψ ∗ ))], ψ can be expressed as a linear combination of
two real solutions. QED
(c) If ψ(x) satisfies Eq. 2.5, then, changing variables x → −x and noting that ∂ 2 /∂(−x)2 = ∂ 2 /∂x2 ,
−
2 ∂ 2 ψ(−x)
+ V (−x)ψ(−x) = Eψ(−x);
2m dx2
so if V (−x) = V (x) then ψ(−x) also satisfies Eq. 2.5. It follows that ψ+ (x) ≡ ψ(x) + ψ(−x) (which is
even: ψ+ (−x) = ψ+ (x)) and ψ− (x) ≡ ψ(x) − ψ(−x) (which is odd: ψ− (−x) = −ψ− (x)) both satisfy Eq.
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CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
15
2.5. But ψ(x) = 12 (ψ+ (x) + ψ− (x)), so any solution can be expressed as a linear combination of even and
odd solutions. QED
Problem 2.2
2
Given ddxψ2 = 2m
2 [V (x) − E]ψ, if E < Vmin , then ψ and ψ always have the same sign: If ψ is positive(negative),
then ψ is also positive(negative). This means that ψ always curves away from the axis (see Figure). However,
it has got to go to zero as x → −∞ (else it would not be normalizable). At some point it’s got to depart from
zero (if it doesn’t, it’s going to be identically zero everywhere), in (say) the positive direction. At this point its
slope is positive, and increasing, so ψ gets bigger and bigger as x increases. It can’t ever “turn over” and head
back toward the axis, because that would requuire a negative second derivative—it always has to bend away
from the axis. By the same token, if it starts out heading negative, it just runs more and more negative. In
neither case is there any way for it to come back to zero, as it must (at x → ∞) in order to be normalizable.
QED
ψ
x
Problem 2.3
2
2
2
Equation 2.20 says ddxψ2 = − 2mE
= A + Bx;
2 ψ; Eq. 2.23 says ψ(0) = ψ(a) = 0. If E = 0, d ψ/dx = 0, so ψ(x) √
2
2
2
ψ(0) = A = 0 ⇒ ψ = Bx; ψ(a) = Ba = 0 ⇒ B = 0, so ψ = 0. If E < 0, d ψ/dx = κ ψ, with κ ≡ −2mE/
real, so ψ(x) = Aeκx + Be−κx . This time ψ(0) = A + B = 0 ⇒ B = −A, so ψ = A(eκx − e−κx ), while
ψ(a) = A eκa − eiκa = 0 ⇒ either A = 0, so ψ = 0, or else eκa = e−κa , so e2κa = 1, so 2κa = ln(1) = 0,
so κ = 0, and again ψ = 0. In all cases, then, the boundary conditions force ψ = 0, which is unacceptable
(non-normalizable).
Problem 2.4
nπ nπ
2 a
x dx.
Let y ≡
x, so
x|ψ|2 dx =
x sin2
a 0
a
a
2 a 2 nπ
2a
y sin 2y cos 2y
y2
=
y sin2 y dy = 2 2
−
−
a nπ
n π
4
4
8
0
x =
a
2a n2 π 2
cos 2nπ 1
= 2 2
−
+
= .
n π
4
8
8
2
a
dx =
dy;
nπ
nπ
y : 0 → nπ.
0
(Independent of n.)
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16
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
2 a 2 2 nπ 2 a 3 nπ 2 2
x sin
y sin y dy
x dx =
a 0
a
a nπ
0
3
nπ
2a2 y 3
y
1
y cos 2y
=
−
−
sin 2y −
(nπ)3 6
4
8
4
0
x2 =
=
1
2a2 (nπ)3
nπ cos(2nπ)
1
.
−
= a2
−
(nπ)3
6
4
3 2(nπ)2
p = m
dx
= 0.
dt
(Note : Eq. 1.33 is much faster than Eq. 1.35.)
2 2
d
d ψn
2
∗
p =
ψn
dx
ψn dx = −
i dx
dx2
2
nπ
2mEn
= (−2 ) − 2
ψn∗ ψn dx = 2mEn =
.
a
2
ψn∗
σx2
= x − x = a
2
2
2
1
1
1
−
−
2
3 2(nπ)
4
2
a2
=
4
1
2
−
3 (nπ)2
;
a
σx =
2
1
2
−
.
3 (nπ)2
2
(nπ)2
− 2.
3
2
The product σx σp is smallest for n = 1; in that case, σx σp = 2 π3 − 2 = (1.136)/2 > /2. σp2
= p − p =
2
2
nπ
a
;
σp =
nπ
.
a
∴ σx σp =
Problem 2.5
(a)
|Ψ|2 = Ψ2 Ψ = |A|2 (ψ1∗ + ψ2∗ )(ψ1 + ψ2 ) = |A|2 [ψ1∗ ψ1 + ψ1∗ ψ2 + ψ2∗ ψ1 + ψ2∗ ψ2 ].
1=
|Ψ| dx = |A|
2
2
√
[|ψ1 |2 + ψ1∗ ψ2 + ψ2∗ ψ1 + |ψ2 |2 ]dx = 2|A|2 ⇒ A = 1/ 2.
(b)
1 Ψ(x, t) = √ ψ1 e−iE1 t/ + ψ2 e−iE2 t/
2
1
=√
2
(but
En
= n2 ω)
π π 2π
2π
2
1
sin
x e−iωt + sin
x e−i4ωt = √ e−iωt sin
x + sin
x e−3iωt .
a
a
a
a
a
a
π 2π
1
2π
2 π
2
−3iωt
3iωt
|Ψ(x, t)| =
+e
+ sin
sin
x + sin
x sin
x e
x
a
a
a
a
a
π 2π
1
2π
2 π
2
=
sin
x + sin
x + 2 sin
x sin
x cos(3ωt) .
a
a
a
a
a
2
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CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
17
(c)
x =
x|Ψ(x, t)|2 dx
π 2π
1 a
2π
2 π
2
=
x sin
x + sin
x + 2 sin
x sin
x cos(3ωt) dx
a 0
a
a
a
a
a
0
a
0
π x dx =
x sin2
a
x sin 2π
cos 2π
x2
a x
a x
−
−
4
4π/a
8(π/a)2
a
a
2π
a2
2
=
x dx.
x sin
=
4
a
0
0
π π 2π
1 a
3π
x sin
x cos
x sin
x dx =
x − cos
x dx
a
a
2 0
a
a
π ax
π 1 a2
3π
a2
ax
3π
=
cos
x +
sin
x − 2 cos
x −
sin
x
2 π2
a
π
a
9π
a
3π
a
=
a2
1 a2
cos(π) − cos(0) − 2 cos(3π) − cos(0)
2
2 π
9π
∴ x =
a2
π2
1−
1
9
0
=−
8a2
.
9π 2
1 a2
a
a2
16a2
32
cos(3ωt) =
+
−
1 − 2 cos(3ωt) .
2
a 4
4
9π
2
9π
32 a = 0.3603(a/2);
9π 2 2
Amplitude:
=−
a
angular frequency: 3ω =
3π 2 .
2ma2
(d)
a 32 dx
8
p = m
=m
− 2 (−3ω) sin(3ωt) =
sin(3ωt).
dt
2
9π
3a
(e) You could get either E1 = π 2 2 /2ma2 or E2 = 2π 2 2 /ma2 , with equal probability P1 = P2 = 1/2.
So H =
1
5π 2 2
;
(E1 + E2 ) =
2
4ma2
it’s the average of E1 and E2 .
Problem 2.6
From Problem 2.5, we see that
Ψ(x, t) =
√1 e−iωt
a
|Ψ(x, t)|2 =
1
a
sin
2
sin
π
ax
π
ax
+ sin
+ sin2
2π
a x
2π
a x
e−3iωt eiφ ;
+ 2 sin
π
ax
sin
2π
a x
cos(3ωt − φ) ;
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18
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
and hence x =
a
2
1−
32
9π 2
cos(3ωt − φ) . This amounts physically to starting the clock at a different time
(i.e., shifting the t = 0 point).
If φ =
π
a
, so Ψ(x, 0) = A[ψ1 (x) + iψ2 (x)], then cos(3ωt − φ) = sin(3ωt); x starts at .
2
2
If φ = π, so Ψ(x, 0) = A[ψ1 (x) − ψ2 (x)], then cos(3ωt − φ) = − cos(3ωt); x starts at
a
32
1+ 2 .
2
9π
Problem 2.7
Ψ(x,0)
Aa/2
a/2
a
x
(a)
1=A
a/2
2
2
x dx + A
0
=
A2
3
a3
a3
+
8
8
a
(a − x) dx = A
2
2
a/2
=
√
A2 a3
2 3
⇒ A= √ .
12
a3
2
a/2
a
x3 (a − x)3 −
3 0
3
a/2
(b)
√ a/2
a
22 3
nπ
nπ
√
cn =
x dx +
x dx
x sin
(a − x) sin
aa a 0
a
a
a/2
√ 2
a/2
2 6
a
nπ
xa
nπ
= 2
sin
x −
cos
x a
nπ
a
nπ
a
0
2
a
a a
a
nπ
nπ
ax
nπ
+a −
−
sin
cos
x x −
cos
x nπ
a
nπ
a
nπ
a
a/2
a/2
√ 2
✟
✟
2
2
2
✟
✘✘ a ✟✟
2 6
a
a
a
nπ
nπ
✟✟
✘✘nπ
= 2
cos
cos
− ✘✘
− ✟cos nπ +
sin
✟
✘
a
nπ
2
2nπ
2
nπ
nπ
2
✟
✟✟
2
2
2
✟
✘
✟
✘
a
nπ
a ✟
a
✘✘nπ
+
sin
+ ✟cos
nπ − ✘✘
cos
✘
nπ
2
nπ
2nπ
2
✟
√
√
0,
n even,
2 6 a2
nπ
nπ
4 6
√
=
2
sin
sin
=
=
(n−1)/2 4 6
2
2
2
(−1)
(nπ)
2
(nπ)
2
a
(nπ)2 , n odd.
√
4 6
So Ψ(x, t) = 2
π
2 nπ
n2 π 2 2
(n−1)/2 1
−En t/
(−1)
sin
,
where
E
=
.
x
e
n
a n=1,3,5,...
n2
a
2ma2
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CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
19
(c)
16 · 6
= 0.9855.
π4
P1 = |c1 |2 =
(d)
H =
|cn |2 En =
96 π 2 2
π 4 2ma2
1
1
1
1
+ 2 + 2 + 2 + ···
1
3
5
7
=
482 π 2
62
.
=
2
2
π ma 8
ma2
π 2 /8
Problem 2.8
(a)
A, 0 < x < a/2;
Ψ(x, 0) =
0, otherwise.
a/2
dx = A2 (a/2) ⇒ A =
1 = A2
0
2
.
a
(b) From Eq. 2.37,
c1 = A
2
a
0
a/2
π a/2
π
π 2 a
2
2
x dx =
− cos
x − cos 0 = .
sin
= − cos
a
a
π
a
π
2
π
0
P1 = |c1 |2 = (2/π)2 = 0.4053.
Problem 2.9
ĤΨ(x, 0) = −
∗
Ψ(x, 0) ĤΨ(x, 0) dx = A
2
= A2
2
m
m
2
2 ∂ 2
2 ∂
2
[Ax(a − x)] = −A
(a − 2x) = A .
2
2m ∂x
2m ∂x
m
a
x(a − x) dx = A
2
x2
x3
a −
2
3
m
52
a3
a3
30 2 a3
−
= 5
=
2
3
a m 6
ma2
0
2
a
0
(same as Example 2.3).
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20
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
Problem 2.10
(a) Using Eqs. 2.47 and 2.59,
1
d
mω 1/4 − mω x2
e 2
−
+ mωx
dx
π
2mω
mω 1/4 mω mω 1/4
mω 2
mω 2
1
1
− −
=√
2mωxe− 2 x .
2x + mωx e− 2 x = √
2
2mω π
2mω π
mω 1/4
mω 2
1
d
(a+ )2 ψ0 =
2mω −
+ mωx xe− 2 x
2mω π
dx
mω 2 mω 1/4 2mω
mω 2
mω 1 mω 1/4 2
− 2 x
2
− 1 − x
=
=
2x + mωx e
x − 1 e− 2 x .
π
2
π
a+ ψ0 = √
Therefore, from Eq. 2.67,
1 mω 1/4
1
ψ2 = √ (a+ )2 ψ0 = √
2
2 π
(b)
mω 2
2mω 2
x − 1 e− 2 x .
ψ0
ψ1
(c) Since ψ0 and ψ2 are even, whereas ψ1 is odd,
we need to check is ψ2∗ ψ0 dx:
1
ψ2∗ ψ0 dx = √
2
=−
=−
ψ2
ψ0∗ ψ1 dx and
ψ2∗ ψ1 dx vanish automatically. The only one
mω 2
mω ∞ 2mω 2
x − 1 e− x dx
π −∞
∞
mω
2mω ∞ 2 − mω x2
− mω
x2
e
dx −
x e
dx
2π
−∞
−∞
mω
π
π
2mω −
= 0. 2π
mω
2mω mω
Problem 2.11
(a) Note that ψ0 is even, and ψ1 is odd. In either case |ψ|2 is even, so x = x|ψ|2 dx = 0. Therefore
p = mdx/dt = 0. (These results hold for any stationary state of the harmonic oscillator.)
√
2
2
From Eqs. 2.59 and 2.62, ψ0 = αe−ξ /2 , ψ1 = 2αξe−ξ /2 . So
n = 0:
x = α
2
2
∞
2 −ξ 2 /2
x e
−∞
dx = α
2
mω
3/2 ∞
2 −ξ 2
ξ e
−∞
1
dξ = √
π
mω
√
π
=
.
2
2mω
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CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
p2 =
ψ0
mω
=− √
π
d
i dx
∞
d2 −ξ2 /2
dξ
e
dξ 2
−∞
√
π √
mω
mω
2
−ξ 2 /2
ξ −1 e
dξ = − √
− π =
.
2
2
π
2
−∞
21
ψ0 dx = −2 α2
mω
∞
e−ξ
2
/2
n = 1:
x2 = 2α2
∞
x2 ξ 2 e−ξ dx = 2α2
2
−∞
p2 = −2 2α2
2mω
=− √
π
mω
∞
ξe−ξ
2
−∞
/2
mω
3/2 ∞
−∞
2
d2
ξe−ξ /2
dξ 2
ξ 4 e−ξ dξ = √
2
√
2 3 π
3
=
.
2mω
πmω 4
dξ
√ 2
π
3mω
2mω 3 √
π−3
ξ 4 − 3ξ 2 e−ξ dξ = − √
=
.
4
2
2
π
∞
−∞
(b) n = 0:
σx =
σx σ p =
x2 − x2 =
2mω
; σp = p2 − p2 =
2mω
mω
;
2
mω
= . (Right at the uncertainty limit.)
2
2
n = 1:
σx =
3
;
2mω
σp =
3mω
;
2
σx σp = 3
> .
2
2
(c)
T =
1
4 ω (n = 0)
1 2
p =
2m
T + V = H =
3
4 ω
(n = 1)
;
V =
1
2 ω (n = 0) = E0
3
2 ω
(n = 1) = E1
1
4 ω (n = 0)
1
mω 2 x2 =
2
3
4 ω
(n = 1)
.
, as expected.
Problem 2.12
From Eq. 2.69,
x=
so
mω
(a+ + a− ), p = i
(a+ − a− ),
2mω
2
x =
ψn∗ (a+ + a− )ψn dx.
2mω
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22
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
But (Eq. 2.66)
a+ ψn =
So
x =
p = m
x2 =
dx
= 0.
dt
2mω
√
n+1
2mω
x2 =
√
n + 1ψn+1 ,
ψn∗ ψn+1
dx +
√
a− ψn =
n
√
nψn−1 .
ψn∗ ψn−1 dx = 0 (by orthogonality).
(a+ + a− )2 =
a2 + a+ a− + a− a+ + a2− .
2mω
2mω +
ψn∗ a2+ + a+ a− + a− a+ + a2− ψn . But
2
a ψn
= a+
a+ a ψ = a
+ − n
+
a
a
ψ
=
a
+ n
−
−
a2− ψn
= a−
√
√
+ 1 n + 2ψn+2
= √ n√
= n nψn
√
= n + 1) n + 1ψn
√ √
= n n − 1ψn−2
√
√n + 1ψn+1
nψ
√ n−1
n + 1ψn+1
√
nψn−1
= (n + 1)(n + 2)ψn+2 .
= nψn .
=
(n + 1)ψn .
= (n − 1)nψn−2 .
So
x2 =
0+n
2mω
p2 = −
mω
mω 2
(a+ − a− )2 = −
a+ − a+ a− − a− a+ + a2− ⇒
2
2
|ψn |2 dx + (n + 1)
|ψn |2 dx + 0 =
mω
mω
p = −
[0 − n − (n + 1) + 0] =
(2n + 1) =
2
2
2
(2n + 1) =
2mω
n+
1 .
2 mω
1
n+
mω.
2
1
1
T = p /2m =
n+
ω .
2
2
2
σx =
x2 − x2 =
n+
;
mω
1
2
σp =
p2 − p2 =
n+
1√
mω;
2
σx σp =
n+
1
≥ .
2
2
Problem 2.13
(a)
1=
|Ψ(x, 0)| dx = |A|
2
2
9|ψ0 |2 + 12ψ0∗ ψ1 + 12ψ1∗ ψ0 + 16|ψ1 |2 dx
= |A|2 (9 + 0 + 0 + 16) = 25|A|2 ⇒ A = 1/5.
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CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
(b)
Ψ(x, t) =
23
1
1
3ψ0 (x)e−iE0 t/ + 4ψ1 (x)e−iE1 t/ =
3ψ0 (x)e−iωt/2 + 4ψ1 (x)e−3iωt/2 .
5
5
(Here ψ0 and ψ1 are given by Eqs. 2.59 and 2.62; E1 and E2 by Eq. 2.61.)
1 2
9ψ0 + 12ψ0 ψ1 eiωt/2 e−3iωt/2 + 12ψ0 ψ1 e−iωt/2 e3iωt/2 + 16ψ12
25
1 2
=
9ψ0 + 16ψ12 + 24ψ0 ψ1 cos(ωt) .
25
|Ψ(x, t)|2 =
(c)
x =
But
xψ02 dx =
1
9
25
xψ02 dx + 16
xψ12 dx + 24 cos(ωt)
xψ0 ψ1 dx .
xψ12 dx = 0 (see Problem 2.11 or 2.12), while
xψ0 ψ1 dx =
=
mω
π
2mω
2
− mω
2 x
xe
2 mω √
2 π2
π
'
2
− mω
2 x
xe
mω
1
2
(3
dx =
2 mω ∞ 2 − mω x2
x e dx
π
−∞
.
2mω
=
So
x =
24
25
cos(ωt);
2mω
p = m
mω
sin(ωt).
2
d
24
x = −
dt
25
(With ψ2 in place of ψ1 the frequency would be (E2 − E0 )/ = [(5/2)ω − (1/2)ω]/ = 2ω.)
Ehrenfest’s theorem says dp/dt = −∂V /∂x. Here
dp
24
=−
dt
25
so
mω
ω cos(ωt),
2
) ∂V *
24
−
= −mω 2 x = −mω 2
∂x
25
V =
∂V
1
mω 2 x2 ⇒
= mω 2 x,
2
∂x
24
cos(ωt) = −
2mω
25
mω
ω cos(ωt),
2
so Ehrenfest’s theorem holds.
(d) You could get E0 = 12 ω, with probability |c0 |2 = 9/25, or E1 = 32 ω, with probability |c1 |2 = 16/25.
Problem 2.14
The new allowed energies are En = (n + 12 )ω = 2(n + 12 )ω = ω, 3ω, 5ω, . . . . So the probability of
1
2
getting
2 ω is zero. The probability of getting ω (the new ground state energy) is P0 = |c0 | , where c0 =
Ψ(x, 0)ψ0 dx, with
Ψ(x, 0) = ψ0 (x) =
mω 1/4
π
2
− mω
2 x
e
,
ψ0 (x) =
m2ω
π
1/4
e−
m2ω 2
2 x
.
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24
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
So
mω
π
1/4
c0 = 2
∞
2
− 3mω
2 x
e
1/4
dx = 2
−∞
Therefore
mω √
2 π
π
'
2
3mω
1
2
(
= 21/4
2
.
3
2√
2 = 0.9428.
3
P0 =
Problem 2.15
ψ0 =
mω 1/4
π
e−ξ
2
/2
, so P = 2
mω
π
Classically allowed region extends out to:
2
P =√
π
∞
∞
e−ξ dx = 2
2
x0
1
2 2
2 mω x0
mω
mω
π
= E0 = 12 ω, or x0 =
∞
e−ξ dξ.
2
ξ0
mω ,
so ξ0 = 1.
√
2
e−ξ dξ = 2(1 − F ( 2)) (in notation of CRC Table) = 0.157.
1
Problem 2.16
−2(5−1)
−2(5−3)
4
n = 5: j = 1 ⇒ a3 = (1+1)(1+2)
a1 = − 43 a1 ; j = 3 ⇒ a5 = (3+1)(3+2)
a3 = − 15 a3 = 15
a1 ; j = 5 ⇒ a7 = 0. So
a1
4
4
3
5
3
5
H5 (ξ) = a1 ξ − 3 a1 ξ + 15 a1 ξ = 15 (15ξ − 20ξ + 4ξ ). By convention the coefficient of ξ 5 is 25 , so a1 = 15 · 8,
and H5 (ξ) = 120ξ − 160ξ 3 + 32ξ 5 (which agrees with Table 2.1).
n = 6: j = 0 ⇒ a2 =
−2(6−4)
(4+1)(4+2) a4
2
= − 15
a4 =
−2(6−0)
(0+1)(0+2) a0
8
− 15
a0 ; j =
−2(6−2)
(2+1)(2+2) a2 =
a0 − 6a0 ξ 2 + 4a0 ξ 4
= −6a0 ; j = 2 ⇒ a4 =
− 23 a2 = 4a0 ; j = 4 ⇒ a6 =
6 ⇒ a8 = 0. So H6 (ξ) =
−
8 6
15 ξ a0 .
The coefficient of ξ 6
8
is 26 , so 26 = − 15
a0 ⇒ a0 = −15 · 8 = −120. H6 (ξ) = −120 + 720ξ 2 − 480ξ 4 + 64ξ 6 .
Problem 2.17
(a)
2
d −ξ2
(e ) = −2ξe−ξ ;
dξ
d
dξ
d
dξ
3
4
d
dξ
2
e−ξ =
2
2
2
d
(−2ξe−ξ ) = (−2 + 4ξ 2 )e−ξ ;
dξ
e−ξ =
2
2
2
d
(−2 + 4ξ 2 )e−ξ = 8ξ + (−2 + 4ξ 2 )(−2ξ) e−ξ = (12ξ − 8ξ 3 )e−ξ ;
dξ
e−ξ =
2
2
2
d
(12ξ − 8ξ 3 )e−ξ = 12 − 24ξ 2 + (12ξ − 8ξ 3 )(−2ξ) e−ξ = (12 − 48ξ 2 + 16ξ 4 )e−ξ .
dξ
2
2
H3 (ξ) = −eξ
2
d
dξ
3
e−ξ = −12ξ + 8ξ 3 ; H4 (ξ) = eξ
2
2
d
dξ
4
e−ξ = 12 − 48ξ 2 + 16ξ 4 .
2
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CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
25
(b)
H5 = 2ξH4 − 8H3 = 2ξ(12 − 48ξ 2 + 16ξ 4 ) − 8(−12ξ + 8ξ 3 ) = 120ξ − 160ξ 3 + 32ξ 5 .
H6 = 2ξH5 − 10H4 = 2ξ(120ξ − 160ξ 3 + 32ξ 5 ) − 10(12 − 48ξ 2 + 16ξ 4 ) = −120 + 720ξ 2 − 480ξ 4 + 64ξ 6 .
(c)
dH5
= 120 − 480ξ 2 + 160ξ 4 = 10(12 − 48ξ 2 + 16ξ 4 ) = (2)(5)H4 . dξ
dH6
= 1440ξ − 1920ξ 3 + 384ξ 5 = 12(120ξ − 160ξ 3 + 32ξ 5 ) = (2)(6)H5 . dξ
(d)
2
d −z2 +2zξ
(e
) = (−2z + ξ)e−z +2zξ ; setting z = 0, H0 (ξ) = 2ξ.
dz
d
dz
2
(e−z
2
+2zξ
)=
=
d
dz
3
(e−z
2
+2zξ
2
d
(−2z + 2ξ)e−z +2zξ
dz
− 2 + (−2z + 2ξ)2 e−z
2
+2zξ
; setting z = 0, H1 (ξ) = −2 + 4ξ 2 .
2
d
− 2 + (−2z + 2ξ)2 e−z +2zξ
dz
2
= 2(−2z + 2ξ)(−2) + − 2 + (−2z + 2ξ)2 (−2z + 2ξ) e−z +2zξ ;
)=
setting z = 0, H2 (ξ) = −8ξ + (−2 + 4ξ 2 )(2ξ) = −12ξ + 8ξ 3 .
Problem 2.18
Aeikx + Be−ikx = A(cos kx + i sin kx) + B(cos kx − i sin kx) = (A + B) cos kx + i(A − B) sin kx
= C cos kx + D sin kx, with C = A + B; D = i(A − B).
C cos kx + D sin kx = C
eikx + e−ikx
2
+D
eikx − e−ikx
2i
= Aeikx + Be−ikx , with A =
=
1
1
(C − iD)eikx + (C + iD)e−ikx
2
2
1
1
(C − iD); B = (C + iD).
2
2
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26
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
Problem 2.19
k2
Equation 2.94 says Ψ = Aei(kx− 2m t) , so
k2
k2
k2
k2
i
∂Ψ
∂Ψ∗
i
J =
Ψ
− Ψ∗
=
|A|2 ei(kx− 2m t) (−ik)e−i(kx− 2m t) − e−i(kx− 2m t) (ik)ei(kx− 2m t)
2m
∂x
∂x
2m
=
i
k 2
|A|2 (−2ik) =
|A| .
2m
m
It flows in the positive (x) direction (as you would expect).
Problem 2.20
(a)
∞
∞
an inπx/a
bn inπx/a
− e−inπx/a +
+ e−inπx/a
e
e
2i
2
n=1
n=1
∞
∞
an
an
bn
bn
inπx/a
= b0 +
−
+
+
e
+
e−inπx/a .
2i
2
2i
2
n=1
n=1
f (x) = b0 +
Let
c0 ≡ b0 ; cn =
Then f (x) =
1
2
(−ian + bn ) , for n = 1, 2, 3, . . . ; cn ≡
∞
cn einπx/a .
1
2
(ia−n + b−n ) , for n = −1, −2, −3, . . . .
QED
n=−∞
(b)
a
−imπx/a
f (x)e
∞
dx =
−a
n=−∞
a
ei(n−m)πx/a dx =
−a
cn
a
ei(n−m)πx/a dx. But for n = m,
−a
a
ei(n−m)πx/a ei(n−m)π − e−i(n−m)π
(−1)n−m − (−1)n−m
=
=
= 0,
i(n − m)π/a −a
i(n − m)π/a
i(n − m)π/a
whereas for n = m,
a
ei(n−m)πx/a dx =
−a
a
dx = 2a.
−a
So all terms except n = m are zero, and
a
a
1
f (x)e−imπx/a = 2acm , so cn =
f (x)e−inπx/a dx. QED
2a
−a
−a
(c)
f (x) =
∞
n=−∞
π1
1 F (k)eikx ∆k,
F (k)eikx = √
2a
2π
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CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
where ∆k ≡
27
π
is the increment in k from n to (n + 1).
a
F (k) =
2 1
a
π 2a
a
−ikx
f (x)e
−a
1
dx = √
2π
a
f (x)e−ikx dx.
−a
(d) As a → ∞, k becomes a continuous variable,
1
f (x) = √
2π
∞
1
F (k)eikx dk; F (k) = √
2π
−∞
∞
f (x)eikx dx.
−∞
Problem 2.21
(a)
∞
1=
−∞
|Ψ(x, 0)| dx = 2|A|
2
2
∞
−2ax
e
−2ax ∞
2e
dx = 2|A|
0
(b)
A
φ(k) = √
2π
∞
A
e−a|x| e−ikx dx = √
2π
−∞
−2a 0
∞
−∞
=
√
|A|2
⇒ A = a.
a
e−a|x| (cos kx − i sin kx)dx.
The cosine integrand is even, and the sine is odd, so the latter vanishes and
∞
∞
A
A
−ax
φ(k) = 2 √
e
cos kx dx = √
e−ax eikx + e−ikx dx
2π 0
2π 0
∞
∞
A
A
e−(ik+a)x e(ik−a)x
= √
e(ik−a)x + e−(ik+a)x dx = √
+
−(ik + a) 0
2π 0
2π ik − a
a
A
−1
1
A −ik − a + ik − a
2a
= √
=
.
+
=√
−k 2 − a2
2π k 2 + a2
2π ik − a ik + a
2π
(c)
1
Ψ(x, t) = √ 2
2π
a3
2π
∞
−∞
k2
1
a3/2
ei(kx− 2m t) dk =
2
2
k +a
π
∞
−∞
k2
k2
1
ei(kx− 2m t) dk.
2
+a
(d) For large a, Ψ(x, 0) is a sharp narrow spike whereas φ(k) ∼
= 2/πa is broad and flat; position
is welldefined but momentum is ill-defined. For small a, Ψ(x, 0) is a broad and flat whereas φ(k) ∼
= ( 2a3 /π)/k 2
is a sharp narrow spike; position is ill-defined but momentum is well-defined.
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28
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
Problem 2.22
(a)
1 = |A|2
∞
−∞
π
;
2a
e−2ax dx = |A|2
2
A=
2a
π
1/4
.
(b)
∞
e−(ax
2
+bx)
∞
e−y
dx =
−∞
2
+(b2 /4a)
−∞
1
φ(k) = √ A
2π
∞
−ax2 −ikx
e
e
−∞
1
1
Ψ(x, t) = √
2π (2πa)1/4
1
=√
2π(2πa)1/4
√
1
4a
1
1 2
√ dy = √ eb /4a
a
a
1
dx = √
2π
∞
e−k
−∞
2
2a
π
1/4
∞
π b2 /4a
.
e
a
e−y dy =
2
−∞
2
π −k2 /4a
1
e
=
e−k /4a .
a
(2πa)1/4
/4a i(kx−k2 t/2m)
e dk
1
2
e−[( 4a +it/2m)k −ixk]
π
1
−x2 /4( 4a
+it/2m)
e
=
+ it/2m
2a
π
1/4
e−ax /(1+2iat/m)
.
1 + 2iat/m
2
(c)
2a e−ax /(1+iθ) e−ax /(1−iθ)
. The exponent is
π
(1 + iθ)(1 − iθ)
2
Let θ ≡ 2at/m. Then |Ψ| =
2
−
2
2a e−2ax /(1+θ )
√
.
π
1 + θ2
2
ax2
(1 − iθ + 1 + iθ)
ax2
−2ax2
; |Ψ|2 =
−
= −ax2
=
(1 + iθ) (1 − iθ)
(1 + iθ)(1 − iθ)
1 + θ2
a
, |Ψ|2 =
1 + θ2
Or, with w ≡
2
2 −2w2 x2
. As t increases, the graph of |Ψ|2 flattens out and broadens.
we
π
|Ψ| 2
t=0
|Ψ|2
x
x
t>0
(d)
x =
x =
2
∞
−∞
x|Ψ|2 dx = 0 (odd integrand); p = m
2
w
π
∞
2 −2w2 x2
x e
−∞
dx =
2
1
w
π 4w2
dx
= 0.
dt
π
1
=
.
2w2
4w2
p = −
2
2
∞
−∞
Ψ∗
d2 Ψ
dx.
dx2
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CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
Write Ψ = Be−bx , where B ≡
2
2a
π
1/4
√
29
1
a
.
and b ≡
1 + iθ
1 + iθ
2
d2 Ψ
d −bx2
= −2bB(1 − 2bx2 )e−bx .
=
B
−2bxe
2
dx
dx
Ψ∗
∗
2
d2 Ψ
a
a
2a
= −2b|B|2 (1 − 2bx2 )e−(b+b )x ; b + b∗ =
= 2w2 .
+
=
2
dx
1 + iθ 1 − iθ
1 + θ2
2a
1
√
=
π 1 + θ2
|B|2 =
p = 2b
2
2
= 2b2
But 1 −
2
w
π
2
w
π
∞
−∞
b
=1−
2w2
p2 = 2b2
2
d2 Ψ
w. So Ψ∗ 2 = −2b
π
dx
(1 − 2bx2 )e−2w
2
π
1
− 2b 2
2w2
4w
a
1 + iθ
a
= 2 a.
2b
x2
σx =
1
;
2w
dx
π
2w2
1 + θ2
2a
2 2
2
w(1 − 2bx2 )e−2w x .
π
b
= 2b2 1 −
.
2w2
=1−
(1 − iθ)
1 + iθ
a
=
= , so
2
2
2b
√
σp = a.
(e)
σx σp =
1 √
1 + θ2 =
1 + (2at/m)2 ≥ . a=
2w
2
2
2
Closest at t = 0, at which time it is right at the uncertainty limit.
Problem 2.23
(a)
(−2)3 − 3(−2)2 + 2(−2) − 1 = −8 − 12 − 4 − 1 = −25.
(b)
cos(3π) + 2 = −1 + 2 = 1.
(c)
0 (x = 2 is outside the domain of integration).
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30
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
Problem 2.24
If c > 0, y : −∞ → ∞.
If c < 0, y : ∞ → −∞.
1 ∞
1
∞
c −∞ f (y/c)δ(y)dy = c f (0) (c > 0); or
f (x)δ(cx)dx =
1 −∞
1 ∞
1
−∞
c ∞ f (y/c)δ(y)dy = − c −∞ f (y/c)δ(y)dy = − c f (0) (c < 0).
1
(a) Let y ≡ cx, so dx = dy.
c
In either case,
(b)
∞
f (x)
−∞
∞
1
f (x)δ(cx)dx =
f (0) =
|c|
−∞
∞
f (x)
−∞
1
1
δ(x)dx. So δ(cx) =
δ(x). |c|
|c|
∞
∞
df
dθ
−
dx = f θ
θdx (integration by parts)
dx
dx
−∞
−∞
∞
∞
df
= f (∞) −
f (x)δ(x)dx.
dx = f (∞) − f (∞) + f (0) = f (0) =
dx
0
−∞
So dθ/dx = δ(x). [Makes sense: The θ function is constant (so derivative is zero) except at x = 0, where
the derivative is infinite.]
Problem 2.25
√
ψ(x) =
mα −mα|x|/2
=
e
√
mα
2
e−mαx/ , (x ≥ 0),
2
emαx/ ,
(x ≤ 0).
x = 0 (odd integrand).
2 3
2
mα
2mα
4
2
x2 =
x2 |ψ|2 dx = 2 2
x2 e−2mαx/ dx = 2 2
=
; σx = √
.
2
2
2mα
2m α
2mα
−∞
0
mα −mαx/2
3
√
, (x ≥ 0) √
− e
2
2
dψ
mα 2
mα −θ(x)e−mαx/ + θ(−x)emαx/ .
=
=
dx
mα mαx/2
,
(x ≤ 0)
2 e
∞
√
3 ∞
mα
mα
−mαx/2
mαx/2
mαx/2
θ(x)e
−
δ(−x)e
+
θ(−x)e
2
2
√
3 2
mα
mα
=
−2δ(x) + 2 e−mα|x|/ .
d2 ψ
=
dx2
mα
−δ(x)e−mαx/ +
2
In the last step I used the fact that δ(−x) = δ(x) (Eq. 2.142), f (x)δ(x) = f (0)δ(x) (Eq. 2.112), and θ(−x) +
θ(x) = 1 (Eq. 2.143). Since dψ/dx is an odd function, p = 0.
√
3 ∞
√
∞
2
2
mα
mα
mα
d2 ψ
p2 = −2
ψ 2 dx = −2
e−mα|x|/ −2δ(x) + 2 e−mα|x|/ dx
dx
−∞
−∞
∞
mα 2
2
2
2
mα
mα mα
mα 2
=
2−2 2
1− 2
e−2mαx/ dx = 2
.
=
2mα
0
Evidently
mα
2 mα √ σp =
, so σx σp = √
= 2 > . 2
2
2mα c
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CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
31
Problem 2.26
1
Put f (x) = δ(x) into Eq. 2.102: F (k) = √
2π
1
∴ f (x) = δ(x) = √
2π
∞
−∞
1
1
√ eikx dk =
2π
2π
∞
1
δ(x)e−ikx dx = √ .
2π
−∞
∞
eikx dk. QED
−∞
Problem 2.27
V(x)
(a)
-a
a
x
(b) From Problem 2.1(c) the solutions are even or odd. Look first for even solutions:
−κx
(x < a),
Ae
ψ(x) = B(eκx + e−κx ) (−a < x < a),
κx
(x < −a).
Ae
Continuity at a : Ae−κa = B(eκa + e−κa ), or A = B(e2κa + 1).
Discontinuous derivative at a, ∆
dψ
2mα
= − 2 ψ(a) :
dx
2mα −κa
2mα
Ae
⇒ A + B(e2κa − 1) = 2 A; or
2
κ
2mα
2mα
2mα
2mα
2κa
2κa
2κa
−
1
=
B(e
−
1
⇒
e
−
1
+ 2 − 1.
B(e2κa − 1) = A
+
1)
−
1
=
e
2 κ
2 κ
2 κ
κ
−κAe−κa − B(κeκa − κe−κa ) = −
1=
2mα
2mα −2κa 2 κ
2 κ
−2κa
−2κa
−
1
+
e
=
1
+
e
− 1.
;
,
or
e
=
2 κ
2 κ
mα
mα
This is a transcendental equation for κ (and hence for E). I’ll solve it graphically: Let z ≡ 2κa, c ≡
so e−z = cz − 1. Plot both sides and look for intersections:
1
2
2amα ,
cz-1
e -z
1/c
z
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32
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
From the graph, noting that c and z are both positive, we see that there is one (and only one) solution
2
z2
(for even ψ). If α = 2ma
, so c = 1, the calculator gives z = 1.278, so κ2 = − 2mE
= (2a)
2 ⇒ E =
2
(1.278)2
2
2
− 8
ma2 = −0.204 ma2 .
Now look for odd solutions:
−κx
(x < a),
Ae
ψ(x) = B(eκx − e−κx ) (−a < x < a),
−Aeκx
(x < −a).
Continuity at a : Ae−κa = B(eκa − e−κa ), or A = B(e2κa − 1).
2mα −κa
Ae
⇒ B(e2κa + 1) = A
2
2mα
2mα
2mα
2κa
2κa
2κa
e
+ 1 = (e
− 1)
−1 =e
− 1 − 2 + 1,
2 κ
2 κ
κ
Discontinuity in ψ : −κAe−κa − B(κeκa + κe−κa ) = −
1=
2mα
−
1
,
2 κ
2mα
2 κ
2mα −2κa 2 κ
−2κa
−2κa
;
,
e
=
1
−
−
1
−
e
=
1
−
e
, or e−z = 1 − cz.
2 κ
2 κ
mα
mα
1
z
1/c
1/c
This time there may or may not be a solution. Both graphs have their y-intercepts at 1, but if c is too
large (α too small), there may be no intersection (solid line), whereas if c is smaller (dashed line) there
will be. (Note that z = 0 ⇒ κ = 0 is not a solution, since ψ is then non-normalizable.) The slope of e−z
(at z = 0) is −1; the slope of (1 − cz) is −c. So there is an odd solution ⇔ c < 1, or α > 2 /2ma.
Conclusion:
One bound state if α ≤ 2 /2ma; two if α > 2 /2ma.
ψ
ψ
-a
-a
a
Even
2
1
α=
⇒c= .
ma
2
a
x
x
Odd
Even: e−z = 12 z − 1 ⇒ z = 2.21772,
Odd: e−z = 1 − 12 z ⇒ z = 1.59362.
E = −0.615(2 /ma2 ); E = −0.317(2 /ma2 ).
α=
2
⇒ c = 2. Only even: e−z = 2z − 1 ⇒ z = 0.738835;
4ma
E = −0.0682(2 /ma2 ).
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CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
33
Problem 2.28
ikx
Ae + Be−ikx (x < −a)
ψ = Ceikx + De−ikx (−a < x < a) . Impose boundary conditions:
ikx
Fe
(x > a)
(1) Continuity at −a : Aeika + Beika = Ce−ika + Deika ⇒ βA + B = βC + D, where β ≡ e−2ika .
(2) Continuity at +a : Ceika + De−ika = F eika ⇒ F = C + βD.
−ika
(3) Discontinuity in ψ at −a : ik(Ce−ika − Deika ) − ik(Ae−ika − Beika ) = − 2mα
+ Beika )
2 (Ae
2
⇒ βC − D = β(γ + 1)A + B(γ − 1), where γ ≡ i2mα/ k.
ika
(4) Discontinuity in ψ at +a : ikF eika − ik(Ceika − De−ika ) = − 2mα
)
2 (F e
⇒ C − βD = (1 − γ)F.
To solve for C and D,
add (2) and (4) :
2C = F + (1 − γ)F ⇒ 2C = (2 − γ)F.
subtract (2) and (4) : 2βD = F − (1 − γ)F ⇒ 2D = (γ/β)F.
add (1) and (3) :
2βC = βA + B + β(γ + 1)A + B(γ − 1) ⇒ 2C = (γ + 2)A + (γ/β)B.
subtract (1) and (3) : 2D = βA + B − β(γ + 1)A − B(γ − 1) ⇒ 2D = −γβA + (2 − γ)B.
Equate the two expressions for 2C : (2 − γ)F = (γ + 2)A + (γ/β)B.
Equate the two expressions for 2D : (γ/β)F = −γβA + (2 − γ)B.
Solve these for F and B, in terms of A. Multiply the first by β(2 − γ), the second by γ, and subtract:
2
β(2 − γ)2 F = β(4 − γ 2 )A + γ(2 − γ)B ;
(γ /β)F = −βγ 2 A + γ(2 − γ)B .
4
F
⇒ β(2 − γ)2 − γ 2 /β F = β 4 − γ 2 + γ A = 4βA ⇒
=
.
2
A
(2 − γ) − γ 2 /β 2
Let g ≡ i/γ =
2 k
i
4g 2
F
; φ ≡ 4ka, so γ = , β 2 = e−iφ . Then:
=
.
2mα
g
A
(2g − i)2 + eiφ
Denominator: 4g 2 − 4ig − 1 + cos φ + i sin φ = (4g 2 − 1 + cos φ) + i(sin φ − 4g).
|Denominator|2 = (4g 2 − 1 + cos φ)2 + (sin φ − 4g)2
= 16g 4 + 1 + cos2 φ − 8g 2 − 2 cos φ + 8g 2 cos φ + sin2 φ − 8g sin φ + 16g 2
= 16g 4 + 8g 2 + 2 + (8g 2 − 2) cos φ − 8g sin φ.
2
F 2 k
8g 4
T = =
, where g ≡
and φ ≡ 4ka.
4
2
2
A
(8g + 4g + 1) + (4g − 1) cos φ − 4g sin φ
2mα
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34
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
Problem 2.29
−κx
(x > a)
Fe
In place of Eq. 2.151, we have: ψ(x) = D sin(lx) (0 < x < a) .
−ψ(−x) (x < 0)
Continuity of ψ : F e−κa = D sin(la); continuity of ψ : −F κe−κa = Dl cos(la).
Divide: − κ = l cot(la), or − κa = la cot(la) ⇒
z02 − z 2 = −z cot z, or − cot z = (z0 /z)2 − 1.
Wide, deep well: Intersections are at π, 2π, 3π, etc. Same as Eq. 2.157, but now for n even. This fills in the
rest of the states for the infinite square well.
Shallow, narrow well: If z0 < π/2, there is no odd bound state. The corresponding condition on V0 is
V0 <
π 2 2
⇒ no odd bound state.
8ma2
π
2π
z0
z
Problem 2.30
∞
a
∞
−2κx
|ψ| dx = 2 |D|
cos lx dx + |F |
e
dx
0
a
∞
a
−2κa
x
1 −2κx a sin 2la
1
2
2
2
2e
=
2
|D|
= 2 |D|
+ sin 2lx + |F | − e
+
+
|F
|
.
2 4l
2κ
2
4l
2κ
0
a
1=2
2
2
2
2
0
sin(2la) cos2 (la)
But F = Deκa cos la (Eq. 2.152), so 1 = |D|2 a +
+
.
2l
κ
Furthermore κ = l tan(la) (Eq. 2.154), so
2 sin la cos la cos3 la
cos la
1 = |D| a +
+
= |D|2 a +
(sin2 la + cos2 la)
2l
l sin la
l sin la
1
1
1
eκa cos la
= |D|2 a +
D= F =
,
.
= |D|2 a +
.
l tan la
κ
a + 1/κ
a + 1/κ
2
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CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
35
Problem 2.31
√
Equation 2.155 ⇒ z0 = a 2mV0 . We want α = area of potential = 2aV0 held constant as a → 0. Therefore
√
α
α
V0 = 2a
; z0 = a 2m 2a
= 1 mαa → 0. So z0 is small, and the intersection in Fig. 2.18 occurs at very small
z. Solve Eq. 2.156 for very small z, by expanding tan z:
tan z ∼
= z = (z0 /z)2 − 1 = (1/z) z02 − z 2 .
Now (from Eqs. 2.146, 2.148
and 2.155) z02 −z 2 = κ2 a2 , so z 2 = κa. But z02 −z 2 = z 4 1 ⇒ z ∼
= z0 , so κa ∼
= z02 .
1√
1
mα
∼
But we found that z0 = mαa here, so κa = 2 mαa, or κ = 2 . (At this point the a’s have canceled, and
we can go to the limit a → 0.)
√
mα
−2mE
m2 α 2
mα2
= 2 ⇒ −2mE =
.
E
=
−
(which agrees with Eq. 2.129).
2
22
V0
In Eq. 2.169, V0 E ⇒ T −1 ∼
sin2
= 1+ 4EV
0
V0 2a
and we can replace sin 9 by 9: T −1 ∼
= 1 + 4E
in agreement with Eq. 2.141.
2
2a
2
√
α
2a , so the argument of
1 + (2aV0 )2 2m2 E . But 2aV0 = α, so
2mV0 . But V0 =
2mV0 =
the sine is small,
2
mα
T −1 = 1 + 2
2E ,
Problem 2.32
Multiply Eq. 2.165 by sin la, Eq. 2.166 by
1
l
cos la, and add:
C sin2 la + D sin la cos la = F eika sin la
C cos2 la − D sin la cos la = ikl F eika cos la
Multiply Eq. 2.165 by cos la, Eq. 2.166 by
1
l
C = F eika sin la +
ik
cos la .
l
sin la, and subtract:
C sin la cos la + D cos2 la = F eika cos la
C sin la cos la − D sin2 la = ikl F eika sin la
D = F eika cos la −
ik
sin la .
l
Put these into Eq. 2.163:
ik
ik
cos la sin la + F eika cos la −
sin la cos la
l
l
ik
ik
cos2 la −
sin la cos la − sin2 la −
sin la cos la
l
l
ik
sin(2la) .
cos(2la) −
l
(1) Ae−ika + Beika = −F eika sin la +
= F eika
= F eika
Likewise, from Eq. 2.164:
−ika
(2) Ae
− Be
ika
ik
il ika
ik
sin la +
= − Fe
cos la cos la + cos la −
sin la sin la
k
l
l
il
ik
ik
= − F eika sin la cos la +
cos2 la + sin la cos la −
sin2 la
k
l
l
ik
il
il
cos(2la) = F eika cos(2la) − sin(2la) .
= − F eika sin(2la) +
k
l
k
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36
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
−ika
Add (1) and (2): 2Ae
F =
ika
= Fe
e−2ika A
l
k
−
k
l
sin(2la) , or:
(confirming Eq. 2.168). Now subtract (2) from (1):
2
2
cos(2la) − i sin(2la)
2kl (k + l )
2Beika = F eika i
2 cos(2la) − i
k
l
+
l
k
sin(2la) ⇒ B = i
sin(2la) 2
(l − k 2 )F (confirming Eq. 2.167).
2kl
2 2
A
sin(2la) 2
sin2 (2la) 2
T −1 = = cos(2la) − i
(k + l2 ) = cos2 (2la) +
(k + l2 )2 .
F
2kl
(2lk)2
But cos2 (2la) = 1 − sin2 (2la), so
(k 2 + l2 )2
−1
(2lk)2
T −1 = 1 + sin2 (2la)
1
(2kl)2
√
But k =
2mE
, l=
(k 2 − l2 )2
=
(2kl)2
4
∴T
−1
=1+
1
4
2 2
4
[k4 +2k2 l2 +l4 −4k2 l2 ]= (2kl)
2 [k −2k l +l ]=
(k2 −l2 )2
(2kl)2
(k 2 − l2 )2
sin2 (2la).
(2kl)2
.
2m(E + V0 )
2mV0
2a 2m(E + V0 ); k 2 − l2 = − 2 , and
; so (2la) =
2m 2 2
V0
2
2m 2
E(E +
2
V0 )
V02
=1+
sin2
4E(E + V0 )
=
V02
.
4E(E + V0 )
2a 2m(E + V0 ) , confirming Eq. 2.169.
Problem 2.33
E < V0 .
ikx
Ae + Be−ikx (x < −a)
ψ = Ceκx + De−κx (−a < x < a)
ikx
(x > a)
Fe
√
k=
2mE
; κ=
2m(V0 − E)
.
(1) Continuity of ψ at −a: Ae−ika + Beika = Ce−κa + Deκa .
(2) Continuity of ψ at −a: ik(Ae−ika − Beika ) = κ(Ce−κa − Deκa ).
κ −κa κ κa
⇒ 2Ae−ika = 1 − i
Ce
De .
+ 1+i
k
k
(3) Continuity of ψ at +a: Ceκa + De−κa = F eika .
(4) Continuity of ψ at +a: κ(Ceκa − De−κa ) = ikF eika .
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CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
⇒ 2Ce
κa
−ika
2Ae
=
ik
1+
κ
ika
Fe
;
−κa
2De
=
ik
1−
κ
F eika .
−2κa
e2κa
iκ
ik
iκ
ik
ika e
= 1−
1+
Fe
+ 1+
1−
F eika
k
κ
2
k
κ
2
F eika
κ
k
κ
k
=
1+i
−
+ 1 e−2κa + 1 + i
−
+ 1 e2κa
2
κ k
k
κ
F eika
(κ2 − k 2 ) 2κa
2 e−2κa + e2κa + i
e
− e−2κa
2
kκ
ex − e−x
ex + e−x
But sinh x ≡
, cosh x ≡
, so
2
2
F eika
(κ2 − k 2 )
=
4 cosh(2κa) + i
2 sinh(2κa)
2
kκ
(κ2 − k 2 )
= 2F eika cosh(2κa) + i
sinh(2κa) .
2kκ
=
T
T
−1
−1
37
.
2
A
(κ2 − k 2 )2
= = cosh2 (2κa) +
sinh2 (2κa). But cosh2 = 1 + sinh2 , so
F
(2κk)2
(κ2 − k 2 )2
V02
2
sinh
=1+ 1+
(2κa)
=
1
+
sinh2
(2κk)2
4E(V0 − E)
2a 2m(V0 − E) ,
where =
4κ k + k + κ − 2κ k
(κ + k )
=
=
(2κk)2
(2κk)2
2 2
4
4
2 2
2
2 2
2mE
2
+
2m(V0 −E)
2
2
2m(V0 −E)
4 2mE
2
2
=
V02
.
4E(V0 − E)
(You can also get this from Eq. 2.169 by switching the sign of V0 and using sin(iθ) = i sinh θ.)
ikx
Ae + Be−ikx (x < −a)
(−a < x < a)
E = V0 .
ψ = C + Dx
ikx
Fe
(x > a)
(In central region −
2 d 2 ψ
d2 ψ
+ V0 ψ = Eψ ⇒
= 0, so ψ = C + Dx.)
2
2m dx
dx2
(1) Continuous ψ at −a : Ae−ika + Beika = C − Da.
(2) Continuous ψ at +a : F eika = C + Da.
⇒ (2.5) 2Da = F eika − Ae−ika − Beika .
(3) Continuous ψ at −a : ik Ae−ika − Beika = D.
(4) Continuous ψ at +a : ikF eika = D.
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38
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
⇒ (4.5) Ae−2ika − B = F.
Use (4) to eliminate D in (2.5): Ae−2ika + B = F − 2aikF = (1 − 2iak)F , and add to (4.5):
2
A
2mE
−2ika
−1
2Ae
= 2F (1 − ika), so T = = 1 + (ka)2 = 1 + 2 a2 .
F
(You can also get this from Eq. 2.169 by changing the sign of V0 and taking the limit E → V0 , using sin 9 ∼
= 9.)
This case is identical to the one in the book, only with V0 → −V0 . So
E > V0 .
T −1 = 1 +
V02
sin2
4E(E − V0 )
2a 2m(E − V0 ) .
Problem 2.34
(a)
ψ=
Aeikx + Be−ikx (x < 0)
F e−κx
(x > 0)
√
where k =
2mE
; κ=
2m(V0 − E)
.
(1) Continuity of ψ : A + B = F.
(2) Continuity of ψ : ik(A − B) = −κF.
ik
ik
ik
⇒ A + B = − (A − B) ⇒ A 1 +
= −B 1 −
.
κ
κ
κ
2
B |(1 + ik/κ)|2
1 + (k/κ)2
R = =
=
= 1.
A
|(1 − ik/κ)|2
1 + (k/κ)2
Although the wave function penetrates into the barrier, it is eventually all reflected.
(b)
ψ=
Aeikx + Be−ikx (x < 0)
F eilx
(x > 0)
√
where k =
2mE
; l=
2m(E − V0 )
.
(1) Continuity of ψ : A + B = F.
(2) Continuity of ψ : ik(A − B) = ilF.
⇒A+B =
k
k
k
(A − B); A 1 −
= −B 1 +
.
l
l
l
2
B (1 − k/l)2
(k − l)2
(k − l)4
R = =
=
= 2
.
2
2
A
(1 + k/l)
(k + l)
(k − l2 )2
2m
Now k − l = 2 (E − E + V0 ) =
√
√
( E − E − V0 )4
R=
.
V02
2
2
2m
2
√
V0 ; k − l =
2m √
[ E − E − V0 ], so
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CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
39
(c)
vi
vi dt
vt dt
vt
From the diagram, T = Pt /Pi = |F |2 vt /|A|2 vi , where Pi is the probability of finding the incident particle
in the box corresponding to the time interval dt, and Pt is the probability of finding the transmitted
particle in the associated box to the right of the barrier.
2
√
vt
E − V0
E − V0 F √
But
=
. Alternatively, from Problem 2.19:
(from Eq. 2.98). So T =
vi
E A
E
Ji =
k 2
|A| ;
m
Jt =
l 2
|F | ;
m
T =
2
2
F l
F Jt
= = Ji
A k
A
E − V0
.
E
For E < V0 , of course, T = 0.
(d)
For E > V0 , F = A + B = A + A
k
l
k
l
−1
+1
=A
2k/l
2k
=
A.
k+l
+1
k
l
√ √
√
2
√
2
F l
l
4 E E − V0 ( E − E − V0 )2
4kl(k − l)2
2k
4kl
T = =
=
.
=
=
A k
k+l
k
(k + l)2
(k 2 − l2 )2
V02
T +R=
4kl
(k − l)2
4kl + k 2 − 2kl + l2
k 2 + 2kl + l2
(k + l)2
+
=
=
=
= 1. 2
2
2
2
(k + l)
(k + l)
(k + l)
(k + l)
(k + l)2
Problem 2.35
(a)
ψ(x) =
Aeikx + Be−ikx (x < 0)
F eilx
(x > 0)
√
where k ≡
Continuity of ψ ⇒ A + B = F
Continuity of ψ ⇒ ik(A − B) = ilF
k
A + B = (A − B);
l
k
A 1−
l
2mE
, l≡
2m(E + V0 )
.
=⇒
k
= −B 1 +
l
;
B
=−
A
1 − k/l
1 + k/l
.
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40
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
√ (2
2 2 ' √
B E + V0 − E
l−k
√
R= =
= √
A
l+k
E+V + E
'
=
1 + V0 /E − 1
1 + V0 /E + 1
(2
0
2
2 √
1+3−1
1
2−1
= √
=
= .
2
+
1
9
1+3+1
(b) The cliff is two-dimensional, and even if we pretend the car drops straight down, the potential as a function
of distance along the (crooked, but now one-dimensional) path is −mgx (with x the vertical coordinate),
as shown.
V(x)
x
-V0
(c) Here V0 /E = 12/4 = 3, the same as in part (a), so R = 1/9, and hence T = 8/9 = 0.8889.
Problem 2.36
Start with Eq. 2.22: ψ(x) = A sin kx + B cos kx. This time the boundary conditions are ψ(a) = ψ(−a) = 0:
A sin ka + B cos ka = 0; −A sin ka + B cos ka = 0.
Subtract : A sin ka = 0 ⇒ ka = jπ or A = 0,
Add :
B cos ka = 0 ⇒ ka = (j − 12 )π or B = 0,
(where j = 1, 2, 3, . . . ).
If B = 0 (so A = 0), k = jπ/a. In this case let n ≡ 2j (so n is an even
√ integer); then k = nπ/2a,
a
ψ = A sin(nπx/2a). Normalizing: 1 = |A|2 −a sin2 (nπx/2a) dx = |A|2 /2 ⇒ A = 2.
If A = 0 (so B = 0), k = (j − 12 )π/a. In this case let n ≡ 2j − 1 (n is an odd integer); again k = nπ/2a,
√
a
ψ = B cos(nπx/2a). Normalizing: 1 = |B|2 −a cos2 (nπx/2a)dx = |a|2 /2 ⇒ B = 2.
k
n π In either case Eq. 2.21 yields E = 2m
= 2m(2a)
2 (in agreement with Eq. 2.27 for a well of width 2a).
The substitution x → (x + a)/2 takes Eq. 2.28 to
2
nπx
n/2
(n even),
(−1)
a sin 2a
nπx nπ
2
2
nπ (x + a)
sin
=
sin
+
=
a
a
2
a
2a
2
(−1)(n−1)/2 2 cos nπx
(n odd).
a
2a
2 2
2
2 2
So (apart from normalization) we recover the results above. The graphs are the same as Figure 2.2, except that
some are upside down (different normalization).
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CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
cos(πx/2a)
sin(2πx/2a)
41
cos(3πx/2a)
Problem 2.37
Use the trig identity sin 3θ = 3 sin θ − 4 sin3 θ to write
3
sin
πx
a
3
= sin
4
πx
a
1
− sin
4
9
1
Normalize using Eq. 2.38: |A|
+
2 16 16
2a
So Ψ(x, 0) =
√1
10
3πx
. So (Eq. 2.28): Ψ(x, 0) = A
a
=
a 3
1
ψ1 (x) − ψ3 (x) .
2 4
4
5
4
a|A|2 = 1 ⇒ A = √ .
16
5a
[3ψ1 (x) − ψ3 (x)] , and hence (Eq. 2.17)
1 Ψ(x, t) = √
3ψ1 (x)e−iE1 t/ − ψ3 (x)e−iE3 t/ .
10
x =
|Ψ(x, t)|2 =
1
9ψ12 + ψ32 − 6ψ1 ψ3 cos
10
x|Ψ(x, t)|2 dx =
9
1
3
x1 + x3 − cos
10
10
5
a
0
E3 − E1
t
E3 − E1
t
; so
a
xψ1 (x)ψ3 (x)dx,
0
where xn = a/2 is the expectation value of x in the nth stationary state. The remaining integral is
πx
2πx
3πx
1 a
4πx
x cos
sin
dx =
− cos
dx
a
a
a 0
a
a
0
2
2
a
1
a
xa
2πx
a
xa
4πx 2πx
4πx
=
+
sin
−
−
sin
cos
cos
= 0.
a
2π
a
2π
a
4π
a
4π
a 2
a
a
x sin
0
Evidently then,
9 a
a
1 a
x =
+
= .
10 2
10 2
2
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42
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
Problem 2.38
π nπ 2
2
n2 π 2 2
; Ψ(x, 0) =
sin
x , ψn (x) =
sin
x .
2
2m(2a)
a
a
2a
2a
√ a+
√ a
n
nπ πx n
πx ,
π 2
2
cos
x sin
x dx =
−1
− cos
+1
dx.
sin
a 0
a
2a
2a 0
2
a
2
a
a
n
sin 2 − 1 πx
sin n2 + 1 πx
1
a
a
√
−
(for n = 2)
n
π
n
π
−
1
+
1
2a
2
a
2
a
0
sin n2 + 1 π
sin n2 + 1 π
sin n2 − 1 π
1
1
1
√
√
−
=
− n
n
n
n
2π
2π
2 −1
2 +1
2 −1
2 +1
n
√
0, √
if n is even
4 2 sin 2 + 1 π
=
.
4 2
2
± π(n2 −4) , if n is odd
π
(n − 4)
(a) New allowed energies: En =
cn =
=
=
=
√ a
√ a
2
2
1
1
2 π
c2 =
sin
x dx =
dx = √ . So the probability of getting En is
a 0
a
a 0 2
2
if n = 2
Pn = |cn |2 = π2 (n32
.
2 −4)2 , if n is odd
0,
otherwise
1
2,
Most probable: E2 =
π 2 2
(same as before).
2ma2
Probability: P2 = 1/2.
π 2 2
32
, with probability P1 =
= 0.36025.
8ma2
9π 2
2 2
a
d
π
(c) H = Ψ∗ HΨ dx = a2 0 sin πa x − 2m
dx2 sin a x dx, but this is exactly the same as before the wall
(b) Next most probable: E1 =
moved – for which we know the answer:
π 2 2
.
2ma2
Problem 2.39
(a) According to Eq. 2.36, the most general solution to the time-dependent Schrödinger equation for the
infinite square well is
∞
2 2
2
Ψ(x, t) =
cn ψn (x)e−i(n π /2ma )t .
n=1
2 2
2
2 2
2
2
n π n π 4ma
T =
= 2πn2 , so e−i(n π /2ma )(t+T ) = e−i(n π /2ma )t e−i2πn , and since n2 is
2 π
2ma2
2ma
2
an integer, e−i2πn = 1. Therefore Ψ(x, t + T ) = Ψ(x, t). QED
2 2
2 2
2
Now
(b) The classical revival time is the time it takes the particle to go down and back: Tc = 2a/v, with the
velocity given by
E=
1
mv 2 ⇒ v =
2
2E
⇒ Tc = a
m
2m
.
E
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CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
43
(c) The two revival times are equal if
4ma2
=a
π
2m
,
E
or
E=
π 2 2
E1
=
.
8ma2
4
Problem 2.40
(a) Let V0 ≡ 322 /ma2 . This is just like the odd bound states for the finite square well, since they are the
ones that go to zero at the origin. Referring to the solution to Problem 2.29, the wave function is
D sin lx,
l ≡ 2m(E + V0 )/ (0 < x < a),
√
ψ(x) =
F e−κx ,
κ ≡ −2mE/
(x > a),
and the boundary conditions at x = a yield
− cot z =
(z0 /z)2 − 1
2m(322 /ma2 )
2mV0
z0 =
a=
a = 8.
Referring to the figure (Problem 2.29), and noting that (5/2)π = 7.85 < z0 < 3π = 9.42, we see that there
are three bound states.
with
√
(b) Let
I1 ≡
a
a
|ψ| dx = |D|
2
0
I2 ≡
sin lx dx = |D|
2
2
2
0
∞
|ψ|2 dx = |F |2
a
∞
e−2κx dx = |F |2
a
a
x
a
1
1
− sin lx cos lx = |D|2
− sin lz cos la ;
2 2l
2 2l
0
−2κx ∞
−2κa
e
= |F |2 e
−
.
2κ 2κ
a
2
But continuity at x = a ⇒ F e−κa = D sin la, so I2 = |D|2 sin2κla .
Normalizing:
a
κ
1
sin2 la
1
− sin la cos la +
=
|D|2 κa − sin la cos la + sin2 la
2 2l
2κ
2κ
l
But (referring again to Problem 2.29) κ/l = − cot la, so
1
(1 + κa)
=
|D|2 κa + cot la sin la cos la + sin2 la = |D|2
.
2κ
2κ
1 = I1 + I2 = |D|2
So |D|2 = 2κ/(1 + κa), and the probability of finding the particle outside the well is
2κ sin2 la
sin2 la
=
.
1 + κa 2κ
1 + κa
We can express this interms of z ≡ la and z0 : κa = z02 − z 2 (page 80),
2
1
z2
1
z
sin2 la = sin2 z =
⇒P = 2
.
=
=
2
2
1 + (z0 /z) − 1
z0
1 + cot z
z0 (1 + z02 − z 2 )
P = I2 =
So far, this iscorrect for any bound state. In the present case z0 = 8 and z is the third solution
to − cot z = (8/z)2 − 1, which occurs somewhere in the interval 7.85 < z < 8. Mathematica gives
z = 7.9573 and P = 0.54204.
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44
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
Problem 2.41
(a) In the standard notation ξ ≡
mω/ x, α ≡ (mω/π)1/4 ,
Ψ(x, 0) = A(1 − 2ξ)2 e−ξ
2
/2
= A(1 − 4ξ + 4ξ 2 )e−ξ
2
/2
.
It can be expressed as a linear combination of the first three stationary states (Eq. 2.59 and 2.62, and
Problem 2.10):
√
2
2
2
α
ψ0 (x) = αe−ξ /2 ,
ψ1 (x) = 2 αξe−ξ /2 ,
ψ2 (x) = √ (2ξ 2 − 1)e−ξ /2 .
2
√
√ 2
2
1
−ξ
/2
So Ψ(x, 0) = c0 ψ0 + c1 ψ1 + c2 ψ2 = α(c0 + 2ξc1 + 2ξ c2 − √2 c2 )e
with (equating like powers)
√
√
⇒ c2 = 2 2A/α,
α√2c2 = 4A
√
α 2c1 = −4A
⇒ c1 = −2 2A/α,
√
√
α(c0 − c2 / 2) = A ⇒ c0 = (A/α) + c2 / 2 = (1 + 2)A/α = 3A/α.
Normalizing: 1 = |c0 |2 + |c1 |2 + |c2 |2 = (8 + 8 + 9)(A/α)2 = 25(A/α)2 ⇒ A = α/5.
√
√
3
2 2
2 2
c0 = , c1 = −
, c2 =
.
5
5
5
73
1
9 1
8 3
8 5
ω
2
H =
|cn | (n + )ω =
ω +
ω +
ω =
(9 + 24 + 40) =
ω.
2
25 2
25 2
25 2
50
50
(b)
√
√
√
√
3
2 2
2 2
−iωt/2 2 2
−3iωt/2 2 2
−5iωt/2
−iωt/2 3
−iωt
−2iωt
Ψ(x, t) = ψ0 e
.
ψ1 e
ψ2 e
ψ0 −
ψ1 e
ψ2 e
−
+
=e
+
5
5
5
5
5
5
To change the sign of the middle term we need e−iωT = −1 (then e−2iωT = 1); evidently ωT = π, or
T = π/ω.
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CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
45
Problem 2.42
Everything in Section 2.3.2 still applies, except that there is an additional boundary condition: ψ(0) = 0. This
eliminates all the even solutions (n = 0, 2, 4, . . . ), leaving only the odd solutions. So
En =
1
n+
2
ω, n = 1, 3, 5, . . . .
Problem 2.43
(a) Normalization is the same as before: A =
2a 1/4
.
π
(b) Equation 2.103 says
1
φ(k) = √
2π
2a
π
1/4 ∞
e−ax eilx e−ikx dx
2
−∞
1
1
Ψ(x, t) = √
2π (2πa)1/4
∞
[same as before, only k → k − l] =
2
1
e−(k−l) /4a .
(2πa)1/4
i(kx−k t/2m)
e−(k−l) /4a e
dk
−∞
1
l
t
−
k2 −(ix+
k
+i
2a ) ]
e−l2 /4a e [( 4a 2m )
2
2
1
π
1
=√
e(ix+l/2a) /[4(1/4a+it/2m)]
e−l /4a
1
t
+
i
2π (2πa)1/4
4a
2m
=
2a
π
1/4
2
2
2
1
e−l /4a ea(ix+l/2a) /(1+2iat/m) .
1 + 2iat/m
(c) Let θ ≡ 2at/m, as before: |Ψ| =
2
square brackets:
2
a
2
2a
1
√
e−l /2a e
π 1 + θ2
(ix+l/2a)2
(1+iθ)
+
(−ix+l/2a)2
(1−iθ)
. Expand the term in
2
2 1
l
l
[]=
(1 − iθ) ix +
+ (1 + iθ) −ix +
1 + θ2
2a
2a
1
ixl
ixl
l2
l2
2
2
=
−x
+
−x
+
−
+
+
1 + θ2
a
4a2
a
4a2
l2
l2
ixl
ixl
+iθ x2 −
− 2 + iθ −x2 −
+ 2
a
4a
a
4a
2
1
l
xl
xl θ2 l2
θ 2 l2
l2
1
=
−2x2 + 2 + 2θ
−2x2 + 2θ − 2 + 2 + 2
=
2
2
1+θ
2a
a
1+θ
a
2a
2a
2a
2
2
−2
θl
l
=
x−
+ 2.
1 + θ2
2a
2a
|Ψ(x, t)|2 =
2
π
2
a
− 2a (x−θl/2a)2 l2 /2a
e−l /2a e 1+θ2
e
=
2
1+θ
2 −2w2 (x−θl/2a)2
,
we
π
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46
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
θl
l
= x− m
where w ≡ a/(1 + θ2 ). The result is the same as before, except x → x − 2a
t , so |Ψ|2 has
the same (flattening Gaussian) shape – only this time the center moves at constant speed v = l/m.
(d)
x =
∞
=
Let y ≡ x − θl/2a = x − vt, so x = y + vt.
x|Ψ(x, t)|2 dx.
−∞
∞
2 −2w2 y2
dy = vt.
we
π
(y + vt)
−∞
(The first integral is trivially zero; the second is 1 by normalization.)
=
l
t;
m
x2 =
∞
p = m
(y + vt)2
−∞
1
x =
+
4w2
2
Ψ=
d2 Ψ
=
dx2
p2 =
=
=
=
=
=
σx2
dx
= l.
dt
lt
m
2 −2w2 y2
1
dy =
+ 0 + (vt)2 (the first integral is same as before).
we
π
4w2
2
.
p = −
2
2
∞
Ψ∗
−∞
d2 Ψ
dx.
dx2
1/4
l
2ia ix + 2a
2
2
1
dΨ
=
Ψ;
e−l /4a ea(ix+l/2a) /(1+iθ) , so
dx
(1 + iθ)
1 + iθ
2
2i2 a
−4a2 (ix + l/2a)
2ia(ix + l/2a) dΨ
2a
+
Ψ=
Ψ.
−
1 + iθ
dx
1 + iθ
(1 + iθ)2
1 + iθ
2
∞ l
4a2 2
(1 + iθ)
ix +
+
|Ψ|2 dx
(1 + iθ)2 −∞
2a
2a
2
∞ 4a2 2
il
(1 + iθ)
− y + vt −
+
|Ψ|2 dy
(1 + iθ)2 −∞
2a
2a
∞
∞
4a2 2
il
2
2
−
y |Ψ| dy − 2 vt −
y|Ψ|2 dy
(1 + iθ)2
2a
−∞
−∞
2
∞
il
(1 + iθ)
2
+ − vt −
+
|Ψ| dy
2a
2a
−∞
2
4a2 2
1
il
(1 + iθ)
− 2 + 0 − vt −
+
(1 + iθ)2
4w
2a
2a
2
4a2 2
1 + θ2
−il
(1 + iθ)
−
−
(1 + iθ) +
(1 + iθ)2
4a
2a
2a
a
l2
a2
l2
−(1 − iθ) + (1 + iθ) + 2 =
(1 + iθ) 1 +
= 2 (a + l2 ).
1 + iθ
a
1 + iθ
a
2a
π
√
1
= x − x =
+
4w2
2
2
lt
m
2
−
lt
m
2
=
1
1
⇒ σx =
;
4w2
2w
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CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
47
√
σp2 = p2 − p2 = 2 a + 2 l2 − 2 l2 = 2 a, so σp = a.
(e) σx and σp are same as before, so the uncertainty principle still holds.
Problem 2.44
√
Equation 2.22 ⇒ ψ(x) = A sin kx + B cos kx, 0 ≤ x ≤ a, with k = 2mE/2 .
Even solutions: ψ(x) = ψ(−x) = A sin(−kx) + B cos(−kx) = −A sin kx + B cos kx (−a ≤ x ≤ 0).
ψ continuous at 0 : B = B (no new condition).
Boundary
ψ discontinuous (Eq. 2.125 with sign of α switched): Ak + Ak = 2mα
2 B ⇒ B =
conditions
2 k
2 k
ψ → 0 at x = a : A sin(ka) + mα A cos(ka) = 0 ⇒ tan(ka) = − mα .
2 k
mα A.
2 k
ψ(x) = A sin kx +
cos kx (0 ≤ x ≤ a); ψ(−x) = ψ(x).
mα
π
2π
3π
ka
-h2k
mα
tan(ka)
From the graph, the allowed energies are slightly above
ka =
nπ
(n = 1, 3, 5, . . . ) so
2
En n2 π 2 2
(n = 1, 3, 5, . . . ).
2m(2a)2
These energies are somewhat higher than the corresponding energies for the infinite square well (Eq. 2.27, with
a → 2a). As α → 0, the straight line (−2 k/mα) gets steeper and steeper, and the intersections get closer to
nπ/2; the energies then reduce to those of the ordinary infinite well. As α → ∞, the straight line approaches
2 2 2
π horizontal, and the intersections are at nπ (n = 1, 2, 3, . . . ), so En → n2ma
– these are the allowed energies for
2
the infinite square well of width a. At this point the barrier is impenetrable, and we have two isolated infinite
square wells.
Odd solutions: ψ(x) = −ψ(−x) = −A sin(−kx) − B cos(−kx) = A sin(kx) − B cos(kx) (−a ≤ x ≤ 0).
ψ continuous at 0 : B = −B ⇒ B = 0.
ψ discontinuous: Ak − Ak = 2mα
Boundary conditions
2 (0) (no new condition).
ψ(a) = 0 ⇒ A sin(ka) = 0 ⇒ ka = nπ
2 (n = 2, 4, 6, . . . ).
ψ(x) = A sin(kx), (−a < x < a);
En =
n2 π 2 2
(n = 2, 4, 6, . . . ).
2m(2a)2
These are the exact (even n) energies (and wave functions) for the infinite square well (of width 2a). The point
is that the odd solutions (even n) are zero at the origin, so they never “feel” the delta function at all.
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48
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
Problem 2.45
d2 ψ1
2 d 2 ψ 1
2
ψ2
−
+ V ψ1 = Eψ1 ⇒ −
+ V ψ1 ψ2 = Eψ1 ψ2
2m dx2
2m
dx2
2
d2 ψ1
d2 ψ2
⇒−
= 0.
−
ψ
ψ2
1
2m
dx2
dx2
2 d 2 ψ 2
2
d2 ψ2
−
+ V ψ2 = Eψ2 ⇒ −
+ V ψ1 ψ2 = Eψ1 ψ2
ψ1
2m dx2
2m
dx2
d
dψ1
dψ2
d2 ψ1
d2 ψ2
d2 ψ1
d2 ψ2
dψ1 dψ2
dψ2 dψ1
But
−
= ψ2
− ψ1
. Since this is
ψ2
− ψ1
=
+ ψ2
− ψ1
2
2
2
dx
dx
dx
dx dx
dx
dx dx
dx
dx
dx2
dψ1
dψ2
zero, it follows that ψ2
− ψ1
= K (a constant). But ψ → 0 at ∞ so the constant must be zero. Thus
dx
dx
dψ1
dψ2
1 dψ1
1 dψ2
ψ2
= ψ1
, or
=
, so ln ψ1 = ln ψ2 + constant, or ψ1 = (constant)ψ2 . QED
dx
dx
ψ1 dx
ψ2 dx
Problem 2.46
−
2 d 2 ψ
d2 ψ
= Eψ (where x is measured around the circumference), or
= −k 2 ψ, with k ≡
2
2m dx
dx2
√
2mE
, so
ψ(x) = Aeikx + Be−ikx .
But ψ(x + L) = ψ(x), since x + L is the same point as x, so
Aeikx eikL + Be−ikx e−ikL = Aeikx + Be−ikx ,
and this is true for all x. In particular, for x = 0 :
(1) AeikL + Be−ikL = A + B. And for x =
π
:
2k
Aeiπ/2 eikL + Be−iπ/2 e−ikL = Aeiπ/2 + Be−iπ/2 , or iAeikL − iBe−ikL = iA − iB, so
(2) AeikL − Be−ikL = A − B. Add (1) and (2): 2AeikL = 2A.
Either A = 0, or else eikL = 1, in which case kL = 2nπ (n = 0, ±1, ±2, . . . ). But if A = 0, then Be−ikL = B,
leading to the same conclusion. So for every positive n there are two solutions: ψn+ (x) = Aei(2nπx/L) and
L
ψn− (x) = Be−i(2nπx/L) (n = 0 is ok too, but in that case there is just one solution). Normalizing: 0 |ψ± |2 dx =
√
1 ⇒ A = B = 1/ L. Any other solution (with the same energy) is a linear combination of these.
1
2n2 π 2 2
ψn± (x) = √ e±i(2nπx/L) ; En =
(n = 0, 1, 2, 3, . . . ).
mL2
L
The theorem fails because here ψ does not go to zero at ∞; x is restricted to a finite range, and we are unable
to determine the constant K (in Problem 2.45).
Problem 2.47
(a) (i) b = 0 ⇒ ordinary finite square well. Exponential decay outside; sinusoidal inside (cos for ψ1 , sin for
ψ2 ). No nodes for ψ1 , one node for ψ2 .
(ii) Ground state is even. Exponential decay outside, sinusoidal inside the wells, hyperbolic cosine in
barrier. First excited state is odd – hyperbolic sine in barrier. No nodes for ψ1 , one node for ψ2 .
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CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
ψ1
49
ψ2
-a
-a
a
a
x
ψ1
x
ψ2
-(b/2+a) -b/2
-(b/2+a) -b/2
b/2 b/2+a
x
b/2 b/2+a
x
(iii) For b a, same as (ii), but wave function very small in barrier region. Essentially two isolated finite
square wells; ψ1 and ψ2 are degenerate (in energy); they are even and odd linear combinations of the
ground states of the two separate wells.
ψ1
ψ
2
b/2
-b/2
-(b/2+a)
b/2+a x
-(b/2+a)
-b/2
b/2
b/2+a x
(b) From Eq. 2.157 we know that for b = 0 the energies fall slightly below
π 2 2
h
E1 + V0 ≈ 2m(2a)
π 2 2
2 = 4
.
where
h
≡
2 2
4π 2ma2
E2 + V0 ≈ 2m(2a)
2 = h
For b a, the width of each (isolated) well is a, so
E1 + V 0 ≈ E 2 + V 0 ≈
π 2 2
= h (again, slightly below this).
2ma2
2
Hence the graph (next page). [Incidentally, within each well, ddxψ2 = − 2m
2 (V0 + E)ψ, so the more curved
the wave function, the higher the energy. This is consistent with the graphs above.]
(c) In the (even) ground state the energy is lowest in configuration (i), with b → 0, so the electron tends to
draw the nuclei together, promoting bonding of the atoms. In the (odd) first excited state, by contrast,
the electron drives the nuclei apart.
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50
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
E+V0
h
E 2 +V0
E 1+V0
h/4
b
Problem 2.48
(a)
√
√ dΨ
2 3
a
2 3
1, (0 < x < a/2)
= √ 1 − 2θ x −
= √ ·
.
−1, (a/2 < x < a)
dx
2
a a
a a
(b)
√
√ d2 Ψ
2 3
4 3
a
a
√
√
=
−
−
2δ
x
−
δ
x
−
=
.
dx2
2
2
a a
a a
(c)
H = −
2
2m
√ √
62
a
2 · 3 · 2
4 3
2 32 ∗ a
√ Ψ
=
. Ψ∗ δ x −
− √
dx =
=
2
2
m·a·a
ma2
a a
ma a
√
3/a
Problem 2.49
(a)
∂Ψ mω a2
i
= −
−2iωe−2iωt +
− 2ax(−iω)e−iωt Ψ, so
∂t
2
2
m
i
∂Ψ
1
1
= − ma2 ω 2 e−2iωt + ω + maxω 2 e−iωt Ψ.
∂t
2
2
∂Ψ mω mω
= −
2x − 2ae−iωt Ψ = −
x − ae−iωt Ψ;
∂x
2
∂2Ψ
mω
mω mω 2
mω
−iωt ∂Ψ
x − ae−iωt
Ψ
−
x
−
ae
=
−
+
=
−
∂x2
∂x
2
Ψ.
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CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
−
51
1
2 ∂ 2 Ψ 1
2
mω mω 2
2
2 2
Ψ + mω 2 x2 Ψ
+
x
Ψ
=
−
x − ae−iωt
mω
−
+
2m ∂x2
2
2m
2
1
1
1
=
ω − mω 2 x2 − 2axe−iωt + a2 e−2iωt + mω 2 x2 Ψ
2
2
2
1
1
=
ω + maxω 2 e−iωt − mω 2 a2 e−2iωt Ψ
2
2
∂Ψ
= i
(comparing second line above). ∂t
(b)
|Ψ|2 =
=
=
2
2
iωt
−iωt
mω − mω
x2 + a2 (1+e2iωt )− it
+ x2 + a2 (1+e−2iωt )+ it
m −2axe
m −2axe
e 2
π
2
2
2
mω − mω
e 2 [2x +a +a cos(2ωt)−4ax cos(ωt)] . But a2 [1 + cos(2ωt)] = 2a2 cos2 ωt, so
π
2
2
2
mω − mω
e [x −2ax cos(ωt)+a cos (ωt)] =
π
mω − mω (x−a cos ωt)2
.
e π
The wave packet is a Gaussian of fixed shape, whose center oscillates back and forth sinusoidally, with
amplitude a and angular frequency ω.
(c) Note that this wave function is correctly normalized (compare Eq. 2.59). Let y ≡ x − a cos ωt :
x = x|Ψ|2 dx = (y + a cos ωt)|Ψ|2 dy = 0 + a cos ωt |Ψ|2 dy = a cos ωt.
p = m
−
dx
= −maω sin ωt.
dt
dp
= −maω 2 cos ωt.
dt
V =
1
dV
mω 2 x2 =⇒
= mω 2 x.
2
dx
dV
dp
= −mω 2 x = −mω 2 a cos ωt =
, so Ehrenfest’s theorem is satisfied.
dx
dt
Problem 2.50
(a)
(E + 12 mv 2 )
∂Ψ
mα ∂
= − 2
|x − vt| − i
Ψ;
∂t
∂t
∂
|x − vt| =
∂t
−v, if x − vt > 0
v, if x − vt < 0
.
We can write this in terms of the θ-function (Eq. 2.143):
∂
1, if z > 0
2θ(z) − 1 =
, so
|x − vt| = −v[2θ(x − vt) − 1].
−1, if z < 0
∂t
∂Ψ
i
=
∂t
mαv
1
2
i
[2θ(x − vt) − 1] + E + mv Ψ.
2
[]
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52
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
∂Ψ
mα ∂
imv
= − 2
|x − vt| +
Ψ
∂x
∂x
∂
|x − vt| = {1, if x > vt; −1, if x < vt} = 2θ(x − vt) − 1.
∂x
mα
imv
= − 2 [2θ(x − vt) − 1] +
Ψ.
∂2Ψ
=
∂x2
−
mα
imv
[2θ(x − vt) − 1] +
2
But (from Problem 2.24(b))
−
∂
∂x θ(x
2
Ψ−
2mα ∂
θ(x − vt) Ψ.
2
∂x
− vt) = δ(x − vt), so
2 ∂ 2 Ψ
− αδ(x − vt)Ψ
2m'∂x2
(
2
2
imv
mα
= −
+ αδ(x − vt) − αδ(x − vt) Ψ
− 2 [2θ(x − vt) − 1] +
2m
2 m2 α 2
m2 v 2
mv mα
2
=−
[2θ(x
−
vt)
−
1]
−
−
2i
[2θ(x
−
vt)
−
1]
Ψ
2m
4 2
2
1
mα2
1
mvα
∂Ψ
= − 2 + mv 2 + i
[2θ(x − vt) − 1] Ψ = i
(compare []). 2
2
∂t
(b)
|Ψ|2 =
mα −2mα|y|/2
e
(y ≡ x − vt).
2
Check normalization: 2
∞
mα
2
∞
e−2mαy/ dy =
0
2
2mα 2
= 1. 2 2mα
∂Ψ
, which we calculated above [].
∂t
−∞
imαv
1
1
2
=
[2θ(y) − 1] + E + mv |Ψ|2 dy = E + mv 2 .
2
2
H =
Ψ∗ HΨdx. But HΨ = i
(Note that [2θ(y) − 1] is an odd function of y.) Interpretation: The wave packet is dragged along (at speed
v) with the delta-function. The total energy is the energy it would have in a stationary delta-function
(E), plus kinetic energy due to the motion ( 12 mv 2 ).
Problem 2.51
(a) Figure at top of next page.
(b)
dψ0
= −Aa sech(ax) tanh(ax);
dx
d 2 ψ0
= −Aa2 − sech(ax) tanh2 (ax) + sech(ax) sech2 (ax) .
2
dx
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CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
53
V(x)
x
2 d 2 ψ0
2 a2
−
sech2 (ax)ψ0
2m dx2
m
2 a2
2
Aa2 − sech(ax) tanh2 (ax) + sech3 (ax) −
A sech3 (ax)
=
2m
m
2 a2 A =
− sech(ax) tanh2 (ax) + sech3 (ax) − 2 sech3 (ax)
2m
2 a2
=−
A sech(ax) tanh2 (ax) + sech2 (ax) .
2m
1
sinh2 θ + 1
sinh2 θ
But (tanh2 θ + sech2 θ) =
+
=
= 1, so
2
2
cosh θ cosh θ
cosh2 θ
Hψ0 = −
=−
2 a2
ψ0 . QED
2m
1 = |A|
2 a2
.
2m
∞
1
2
tanh(ax)
sech (ax)dx = |A|
= |A|2 =⇒ A =
a
a
−∞
−∞
2
Evidently E = −
∞
2
2
a
.
2
ψ(x)
x
(c)
dψk
A =
(ik − a tanh ax)ik − a2 sech2 ax eikx .
dx
ik + a
.
d2 ψk
A - =
ik (ik − a tanh ax)ik − a2 sech2 ax − a2 ik sech2 ax + 2a3 sech2 ax tanh ax eikx .
2
dx
ik + a
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54
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
2
2 a2
2 d 2 ψ k
A
− ik 2
2
2
−
+
V
ψ
=
−
iak
tanh
ax
−
a
sech
ax
+
−k
ik sech2 ax
k
2m dx2
ik + a
2m
2m
2 a3
2 a2
2
2
−
sech ax tanh ax −
sech ax(ik − a tanh ax) eikx
m
m
=
=
Aeikx 2
ik 3 − ak 2 tanh ax + ia2 k sech2 ax + ia2 k sech2 ax
ik + a 2m
−2a3 sech2 ax tanh ax − 2ia2 k sech2 ax + 2a3 sech2 ax tanh ax
Aeikx 2 2
2 k 2
k (ik − a tanh ax) =
ψk = Eψk . QED
ik + a 2m
2m
As x → +∞, tanh ax → +1, so ψk (x) → A
R = 0.
ik − a
ik + a
eikx , which represents a transmitted wave.
ik − a 2
ik − a
= −ik − a
T = = 1.
ik + a −ik + a
ik + a
Problem 2.52
(a) (1) From Eq. 2.133: F + G = A + B.
(2) From Eq. 2.135: F − G = (1 + 2iβ)A − (1 − 2iβ)B, where β = mα/2 k.
1
Subtract: 2G = −2iβA + 2(1 − iβ)B ⇒ B =
(iβA + G). Multiply (1) by (1 − 2iβ) and add:
1 − iβ
1
1
iβ 1
2(1 − iβ)F − 2iβG = 2A ⇒ F =
(A + iβG). S =
.
1 − iβ
1 − iβ 1 iβ
(b) For an even potential, V (−x) = V (x), scattering from the right is the same as scattering from the left, with
x ↔ −x, A ↔ G, B ↔ F (see Fig. 2.22): F = S11 G + S12 A, B = S21 G + S22 A. So S11 = S22 , S21 = S12 .
(Note that the delta-well S matrix in (a) has this property.) In the case of the finite square well, Eqs. 2.167
and 2.168 give
S21 =
cos 2la − i
(k2 +l2 )
2kl
−k
i (l 2kl
2
e−2ika
sin 2la
;
S11 =
2
)
sin 2la e−2ika
2
2
+l )
cos 2la − i (k 2kl
sin 2la
. So
' 2 2
(
−k )
sin 2la
1
i (l 2kl
S=
.
2
2 +l2 )
−k2 )
1
i (l 2kl
sin 2la
cos 2la − i (k 2kl
sin 2la
e−2ika
Problem 2.53
(a)
B = S11 A + S12 G ⇒ G =
1
S11
1
(B − S11 A) = M21 A + M22 B ⇒ M21 = −
, M22 =
.
S12
S12
S12
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CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
F = S21 A + S22 B = S21 A +
⇒ M11 = −
S22
(S11 S22 − S12 S21 )
S22
(B − S11 A) = −
A+
B = M11 A + M12 B.
S12
S12
S12
det S
S22
, M12 =
.
S12
S12
G = M21 A + M22 B ⇒ B =
M=
1
S12
− det(S) S22
.
−S11
1
Conversely:
1
M21
1
(G − M21 A) = S11 A + S12 G ⇒ S11 = −
; S12 =
.
M22
M22
M22
F = M11 A + M12 B = M11 A +
⇒ S21
55
M12
(M11 M22 − M12 M21 )
M12
(G − M21 A) =
A+
G = S21 A + S22 G.
M22
M22
M22
det M
M12
=
; S22 =
.
M22
M22
1
S=
M22
−M21 1
.
det(M) M12
[It happens that the time-reversal invariance of the Schrödinger equation, plus conservation of probability,
∗
∗
requires M22 = M11
, M21 = M12
, and det(M) = 1, but I won’t use this here. See Merzbacher’s Quantum
Mechanics. Similarly, for even potentials S11 = S22 , S12 = S21 (Problem 2.52).]
2
2
M21 2
1
, Tl = |S21 |2 = det(M) , Rr = |S22 |2 = M12 , Tr = |S12 |2 =
Rl = |S11 |2 = .
M22
M22
M22
|M22 |2
(b)
A
C
F
B
D
G
M2
M1
x
F
C
C
A
F
A
A
= M2
,
= M1
, so
= M 2 M1
=M
, with M = M2 M1 . QED
G
D
D
B
G
B
B
(c)
ψ(x) =
Aeikx + Be−ikx (x < a)
F eikx + Ge−ikx (x > a)
.
Continuity of ψ :
Aeika + Be−ika = F eika + Ge−ika
2mα
Aeika + Be−ika .
Discontinuity of ψ : ik F eika − Ge−ika − ik Aeika − Be−ika = − 2mα
2 ψ(a) = − 2
(1) F e2ika + G = Ae2ika + B.
2ika
(2) F e2ika − G = Ae2ika − B + i 2mα
+B .
2 k Ae
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56
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
Add (1) and (2):
2F e2ika = 2Ae2ika + i
mα
2mα
mα 2ika
Ae
A + i 2 e−2ika B = M11 A + M12 B.
+
B
⇒
F
=
1
+
i
2
2
k
k
k
So M11 = (1 + iβ); M12 = iβe−2ika ; β ≡
mα
.
2 k
Subtract (2) from (1):
2G = 2B − 2iβe2ika A − 2iβB ⇒ G = (1 − iβ)B − iβe2ika A = M21 A + M22 B.
So M21 = −iβe
2ika
; M22 = (1 − iβ).
M=
(1 + iβ) iβe−2ika
.
−iβe2ika (1 − iβ)
(d)
M1 =
(1 + iβ) iβe−2ika
(1 + iβ) iβe2ika
; to get M2 , just switch the sign of a: M2 =
.
−iβe2ika (1 − iβ)
−iβe−2ika (1 − iβ)
M = M2 M1 =
T = Tl = Tr =
[1 + 2iβ + β 2 (e4ika − 1)] 2iβ[cos 2ka + β sin 2ka]
.
−2iβ[cos 2ka + β sin 2ka] [1 − 2iβ + β 2 (e−4ika − 1)]
1
⇒
|M22 |2
T −1 = [1 + 2iβ + β 2 (e4ika − 1)][1 − 2iβ + β 2 (e−4ika − 1)]
= 1 − 2iβ + β 2 e−4ika − β 2 + 2iβ + 4β 2 + 2iβ 3 e−4ika − 2iβ 3 + β 2 e4ika
− β 2 − 2iβ 3 e4ika + 2iβ 3 + β 4 (1 − e4ika − e−4ika + 1)
= 1 + 2β 2 + β 2 (e4ika + e−4ika ) − 2iβ 3 (e4ika − e−4ika ) + 2β 4 − β 4 (e4ika + e−4ika )
= 1 + 2β 2 + 2β 2 cos 4ka − 2iβ 3 2i sin 4ka + 2β 4 − 2β 4 cos 4ka
= 1 + 2β 2 (1 + cos 4ka) + 4β 3 sin 4ka + 2β 4 (1 − cos 4ka)
= 1 + 4β 2 cos2 2ka + 8β 3 sin 2ka cos 2ka + 4β 4 sin2 2ka
T =
1
1 + 4β 2 (cos 2ka + β sin 2ka)2
Problem 2.54
I’ll just show the first two graphs, and the last two. Evidently K lies between 0.9999 and 1.0001.
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57
Problem 2.55
The correct values (in Eq. 2.72) are K = 2n + 1 (corresponding to En = (n + 12 )ω). I’ll start by “guessing”
2.9, 4.9, and 6.9, and tweaking the number until I’ve got 5 reliable significant digits. The results (see below)
are 3.0000, 5.0000, 7.0000. (The actual energies are these numbers multiplied by 12 ω.)
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58
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59
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60
Problem 2.56
The Schrödinger equation says − 2m
ψ = Eψ, or, with the correct energies (Eq. 2.27) and a = 1, ψ + (nπ)2 ψ =
0. I’ll start with a “guess” using 9 in place of π 2 (that is, I’ll use 9 for the ground state, 36 for the first excited
state, 81 for the next, and finally 144). Then I’ll tweak the parameter until the graph crosses the axis right
at x = 1. The results (see below) are, to five significant digits: 9.8696, 39.478, 88.826, 157.91. (The actual
energies are these numbers multiplied by 2 /2ma2 .)
2
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62
CHAPTER 3. FORMALISM
Chapter 3
Formalism
Problem 3.1
(a) All conditions are trivial except Eq. A.1: we need to show that the sum of two square-integrable functions
is itself square-integrable. Let h(x) = f (x) + g(x), so that |h|2 = (f + g)∗ (f + g) = |f |2 + |g|2 + f ∗ g + g ∗ f
and hence
∗
2
2
2
∗
∗
f g dx .
|h| dx = |f | dx + |g| dx + f g dx +
If f (x) and g(x)
are square-integrable, then the first two terms are finite, and (by Eq. 3.7) so too are the
last two. So |h|2 dx is finite. QED
The set of all normalized functions is certainly not a vector space: it doesn’t include 0, and the sum of
two normalized functions is not (in general) normalized—in fact, if f (x) is normalized, then the square
integral of 2f (x) is 4.
(b) Equation A.19 is trivial:
b
g|f =
∗
'
g(x) f (x) dx =
a
b
(∗
∗
f (x) g(x) dx
= f |g∗ .
a
Equation A.20 holds (see Eq. 3.9) subject to the understanding in footnote 6. As for Eq. A.21, this is
pretty obvious:
f |(b|g + c|h) = f (x)∗ (bg(x) + ch(x)) dx = b f ∗ g dx + c f ∗ h dx = bf |g + cf |h.
Problem 3.2
(a)
f |f =
1
x2ν dx =
0
1
1
1
x2ν+1 =
1 − 02ν+1 .
2ν + 1
2ν
+
1
0
Now 02ν+1 is finite (in fact, zero) provided (2ν + 1) > 0, which is to say, ν > − 12 . If (2ν + 1) < 0 the
integral definitely blows up. As for the critical case ν = − 12 , this must be handled separately:
1
1
f |f =
x−1 dx = ln x0 = ln 1 − ln 0 = 0 + ∞.
0
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CHAPTER 3. FORMALISM
63
So f (x) is in Hilbert space only for ν strictly greater than -1/2.
(b) For ν = 1/2, we know from (a) that f (x) is in Hilbert space: yes.
Since xf = x3/2 , we know from (a) that it is in Hilbert space: yes.
For df /dx = 12 x−1/2 , we know from (a) that it is not in Hilbert space: no.
[Moral: Simple operations, such as differenting (or multiplying by 1/x), can carry a function out of Hilbert
space.]
Problem 3.3
Suppose h|Q̂h = Q̂h|h for all functions h(x). Let h(x) = f (x) + cg(x) for some arbitrary constant c. Then
h|Q̂h = (f + cg)|Q̂(f + cg) = f |Q̂f + cf |Q̂g + c∗ g|Q̂f + |c|2 g|Q̂g;
Q̂h|h = Q̂(f + cg)|(f + cg) = Q̂f |f + cQ̂f |g + c∗ Q̂g|f + |c|2 Q̂g|g.
Equating the two and noting that f |Q̂f = Q̂f |f and g|Q̂g = Q̂g|g leaves
cf |Q̂g + c∗ g|Q̂f = cQ̂f |g + c∗ Q̂g|f .
In particlar, choosing c = 1:
f |Q̂g + g|Q̂f = Q̂f |g + Q̂g|f ,
whereas if c = i:
f |Q̂g − g|Q̂f = Q̂f |g − Q̂g|f .
Adding the last two equations:
f |Q̂g = Q̂f |g. QED
Problem 3.4
(a) f |(Ĥ + K̂)g = f |Ĥg + f |K̂g = Ĥf |g + K̂f |g = (Ĥ + K̂)f |g. (b) f |αQ̂g = αf |Q̂g; αQ̂f |g = α∗ Q̂f |g. Hermitian ⇔ α is real.
(c) f |Ĥ K̂g = Ĥf |K̂g = K̂ Ĥf |g, so Ĥ K̂ is hermitian ⇔ Ĥ K̂ = K̂ Ĥ, or [Ĥ, K̂] = 0.
(d) f |x̂g = f ∗ (xg) dx = (xf )∗ g dx = x̂f |g. 2
2 d 2
2
∗
∗d g
f |Ĥg = f −
+
V
g
dx
=
−
dx
+
f ∗ V g dx.
f
2m dx2
2m
dx2
Integrating by parts (twice):
∞
∞
∞
∞
∞ ∗
∞ 2 ∗
2
df dg
d f
df ∗ ∗d g
∗ dg ∗ dg f
dx = f
−
−
+
g dx.
dx = f
g
dx2
dx dx dx
dx dx dx2
−∞
−∞
−∞
−∞
−∞
−∞
But for functions f (x) and g(x) in Hilbert space the boundary terms vanish, so
∞
∞ 2 ∗
2
d f
∗d g
f
dx =
g dx, and hence (assuming that V (x) is real):
2
2
dx
−∞
−∞ dx
∗
∞
2 d 2 f
f |Ĥg =
−
+
V
f
g dx = Ĥf |g. 2m dx2
−∞
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64
CHAPTER 3. FORMALISM
Problem 3.5
f ∗ (xg) dx = (xf )∗ g dx = xf |g, so x† = x.
f |ig = f ∗ (ig) dx = (−if )∗ g dx = −if |g, so i† = −i.
(a) f |xg =
dg
f | =
dx
∞
−∞
∞
f
−
dx = f g dx
∗ dg
∗
−∞
∞
−∞
df
dx
∗
g dx = −xf |g, so
d
dx
†
=−
d
.
dx
1
1
(−ip + mωx). But p and x are hermitian, and i† = −i, so (a+ )† = √
(ip + mωx), or
2mω
2mω
(a+ )† = (a− ).
(b) a+ = √
(c) f |(Q̂R̂)g = Q̂† f |R̂g = R̂† Q̂† f |g = (Q̂R̂)† f |g, so (Q̂R̂)† = R̂† Q̂† . Problem 3.6
2π 2π ∗
2π
2π 2π 2 ∗
d2 g
df dg
d f
df ∗ ∗ dg ∗ dg dφ
=
f
g
dφ
=
f
−
−
g dφ.
+
2
dφ
dφ
dφ
dφ
dφ
dφ
dφ2
0
0
0
0
0
0
As in Example 3.1, for periodic functions (Eq. 3.26) the boundary terms vanish, and we conclude that f |Q̂g =
Q̂f |g, so Q̂ is hermitian: yes.
f |Q̂g =
2π
f∗
√
d2 f
= qf ⇒ f± (φ) = Ae± qφ .
2
dφ
√
√
The periodicity condition (Eq. 3.26) requires that q(2π) = 2nπi, or q = in, so the eigenvalues are
Q̂f = qf ⇒
q = −n2 , (n = 0, 1, 2, . . . ). The spectrum is doubly degenerate; for a given n there are two eigenfunctions
(the plus sign or the minus sign, in the exponent), except for the special case n = 0, which is not degenerate.
Problem 3.7
(a) Suppose Q̂f = qf and Q̂g = qg. Let h(x) = af (x) + bg(x), for arbitrary constants a and b. Then
Q̂h = Q̂(af + bg) = a(Q̂f ) + b(Q̂g) = a(qf ) + b(qg) = q(af + bg) = qh.
(b)
d2 f
d
d2
d x
d2 g
d2
= 2 (ex ) =
= 2 e−x =
(e ) = ex = f,
−e−x = e−x = g.
2
2
dx
dx
dx
dx
dx
dx
So both of them are eigenfunctions, with the same eigenvalue 1. The simplest orthogonal linear combinations are
sinh x =
1
1 x
e − e−x = (f − g)
2
2
and
cosh x =
1
1 x
e + e−x = (f + g).
2
2
(They are clearly orthogonal, since sinh x is odd while cosh x is even.)
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CHAPTER 3. FORMALISM
65
Problem 3.8
(a) The eigenvalues (Eq. 3.29) are 0, ±1, ±2, . . . , which are obviously real. For any two eigenfunctions,
f = Aq e−iqφ and g = Aq e−iq φ (Eq. 3.28), we have
2π
2π
i(q−q )φ A∗q Aq i(q−q )2π
e
f |g = A∗q Aq
eiqφ e−iq φ dφ = A∗q Aq
−1 .
e
=
i(q − q ) i(q − q )
0
0
i(q−q )2π
But q and q are integers, so e
nonzero). = 1, and hence f |g = 0 (provided q = q , so the denominator is
(b) In Problem 3.6 the eigenvalues are q = −n2 , with n = 0, 1, 2, . . . , which are obviously real. two eigenfunctions, f = Aq e±inφ and g = Aq e±in φ , we have
2π
2π
A∗q Aq ±i(n −n)2π
e±i(n −n)φ ∗
∓inφ ±in φ
∗
f |g = Aq Aq
e
e
dφ = Aq Aq
−1 =0
e
=
±i(n − n) ±i(n − n)
0
For any
0
(provided n = n ). But notice that for each eigenvalue (i.e. each value of n) there are two eigenfunctions
(one with the plus sign and one with the minus sign), and these are not orthogonal to one another.
Problem 3.9
(a) Infinite square well (Eq. 2.19).
(b) Delta-function barrier (Fig. 2.16), or the finite rectangular barrier (Prob. 2.33).
(c) Delta-function well (Eq. 2.114), or the finite square well (Eq. 2.145) or the sech2 potential (Prob. 2.51).
Problem 3.10
From Eq. 2.28, with n = 1:
p̂ ψ1 (x) =
d
i dx
π 2
sin
x =
a
a
i
π π 2π
π
cos
x = −i
cot
x ψ1 (x).
aa
a
a
a
Since p̂ ψ1 is not a (constant)
multiple of ψ1 , ψ1 is not an eigenfunction of p̂: no. It’s true that the magnitude
√
of the momentum, 2mE1 = π/a, is determinate, but the particle is just as likely to be found traveling to the
left (negative momentum) as to the right (positive momentum).
Problem 3.11
Ψ0 (x, t) =
mω
π
1/4
2
− mω
−iωt/2
2 x
e
e
;
1/4
∞
mω 2
1
mω
−iω/2
Φ(p, t) = √
e
e−ipx/ e− 2 x dx.
2π π
−∞
From Problem 2.22(b):
1/4
1
mω
Φ(p, t) = √
e−iωt/2
2π π
2
2π −p2 /2mω
1
=
e−p /2mω e−iωt/2 .
e
1/4
mω
(πmω)
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CHAPTER 3. FORMALISM
|Φ(p, t)|2 = √
√
2
1
p2
1
e−p /mω . Maximum classical momentum:
= E = ω =⇒ p = mω.
2m
2
πmω
So the probability it’s outside classical range is:
√
− mω
P =
−∞
√
mω
|Φ| dp +
2
∞
√
mω
|Φ| dp = 1 − 2
|Φ|2 dp.
Now
0
mω
√mω
2
1
|Φ| dp = √
e−p /mω dp.
πmω 0
√2
√
2
1
=√
e−z /2 dz = F ( 2) −
2π 0
2
0
√
2
Let z ≡
2
p, so dp =
mω
mω
dz.
2
1
, in CRC Table notation.
2
√
√ √
1
P = 1 − 2 (F ( 2) −
= 1 − 2F ( 2) + 1 = 2 1 − F ( 2) = 0.157.
2
To two digits: 0.16 (compare Prob. 2.15).
Problem 3.12
∞
1
From Eq. 3.55: Ψ(x, t) = √
eipx/ Φ(p, t)dp.
2π −∞
1
1
∗
−ip x/ ∗
√
x = Ψ xΨdx =
Φ (p , t)dp x √
e
e+ipx/ Φ(p, t)dp dx.
2π
2π
But xeipx/ = −i
x
d ipx/
, so (integrating by parts):
e
dp
ipx/
e
Φ dp =
1
So x =
2π
d ipx/
)Φ dp =
e
i dp
−ip x/
e
∗
eipx/ −
ipx/
Φ (p , t)e
∂
Φ(p, t) dp.
i ∂p
∂
−
Φ(p, t)
i ∂p
dp dp dx.
Do the x integral first, letting y ≡ x/:
1
1
e−ip x/ eipx/ dx =
ei(p−p )y dy = δ(p − p ), (Eq. 2.144), so
2π
2π
x =
Φ∗ (p , t)δ(p − p ) −
∂
Φ(p, t) dp dp =
i ∂p
Φ∗ (p, t) −
∂
Φ(p, t) dp.
i ∂p
QED
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CHAPTER 3. FORMALISM
67
Problem 3.13
(a) [AB, C] = ABC − CAB = ABC − ACB + ACB − CAB = A[B, C] + [A, C]B. (b) Introducing a test function g(x), as in Eq. 2.50:
d n
n
n dg
n dg
n−1
n dg
[x , p]g = x
g+x
−
(x g) = x
−
nx
= inxn−1 g.
i dx
i dx
i dx
i
dx
So, dropping the test function, [xn , p] = inxn−1 . dg
d
dg
df
dg
df
df
(c) [f, p]g = f
−
(f g) = f
−
g+f
= i g ⇒ [f, p] = i . i dx
i dx
i dx
i dx
dx
dx
dx
Problem 3.14
x,
p2
+V
2m
=
1 x, p2 + [x, V ];
2m
x, p2 = xp2 − p2 x = xp2 − pxp + pxp − p2 x = [x, p]p + p[x, p].
Using Eq. 2.51: x, p2 = ip + pi = 2ip. And [x, V ] = 0,
so x,
p2
+V
2m
=
1
ip
2ip =
.
2m
m
The generalized uncertainty principle (Eq. 3.62) says, in this case,
2 2
1 i
2
σx2 σH
≥
p =
p ⇒ σx σH ≥
|p|. QED
2i m
2m
2m
For stationary states σH = 0 and p = 0, so it just says 0 ≥ 0.
Problem 3.15
Suppose P̂ fn = λn fn and Q̂fn = µn fn (that is: fn (x) is an eigenfunction both of P̂ and of Q̂), and the set
/ {fn }
is complete, so that any function f (x) (in Hilbert space) can be expressed as a linear combination: f = cn fn .
Then
[P̂ , Q̂]f = (P̂ Q̂ − Q̂P̂ )
cn fn = P̂
cn λn fn =
cn µn λn fn −
cn µn fn − Q̂
cn λn µn fn = 0.
Since this is true for any function f , it follows that [P̂ , Q̂] = 0.
Problem 3.16
dΨ
i
a
= (iax − iax + p)Ψ =
dx
dΨ
a
=
Ψ
− x + x +
Let constant = −
i
−x + x + p Ψ.
a
ip
a
x2
ip
dx ⇒ ln Ψ =
−
+ xx +
x + constant.
a
2
a
x2 a
a
ip
+ B (B a new constant). Then ln Ψ = − (x − x)2 +
x + B.
2
2
Ψ = e− 2 (x−x) eipx/ eB = Ae−a(x−x)
a
2
2
/2 ipx/
e
, where A ≡ eB .
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68
CHAPTER 3. FORMALISM
Problem 3.17
(a) 1 commutes with everything, so
d
Ψ|Ψ = 0 (this is the conservation of normalization, which we origidt
nally proved in Eq. 1.27).
d
H = 0 (assuming H has no explicit time
dt
dependence); this is conservation of energy, in the sense of the comment following Eq. 2.40.
ip
p
dx
i
ip
(c) [H, x] = −
(Eq. 1.33).
(see Problem 3.14). So
=
−
=
m
dt
m
m
(b) Anything commutes with itself, so [H, H] = 0, and hence
0
0
1
1
p2
∂V
dV
dp
i
∂V
(d) [H, p] =
+ V, p = [V, p] = i
(Problem 3.13(c)). So
=
i
= −
.
2m
dx
dt
∂x
∂x
This is Ehrenfest’s theorem (Eq. 1.38).
Problem 3.18
1
1 Ψ(x, t) = √ ψ1 e−iE1 t/ + ψ2 e−E2 t/ . H 2 Ψ = √ (H 2 ψ1 )e−E1 t/ + (H 2 ψ2 )e−iEn t/ .
2
2
Hψ1 = E1 ψ1 ⇒ H 2 ψ1 = E1 Hψ1 = E12 ψ1 , and H 2 ψ2 = E22 ψ2 , so
1
ψ1 e−iE1 t/ + ψ2 e−iE2 t/ | E12 ψ1 e−iE1 t/ + E22 ψ2 e−iE2 t/ 2
1
= ψ1 |ψ1 eiE1 t/ E12 e−iE1 t/ + ψ1 |ψ2 eiE1 t/ E22 e−iE2 t/
2
1
+ ψ2 |ψ1 eiE2 t/ E12 e−iE1 t/ + ψ2 |ψ2 eiE2 t/ E22 e−iE2 t/ = E12 + E22 .
2
H 2 =
Similarly, H = 12 (E1 + E2 )
(Problem 2.5(e)).
1
1 2
1
E1 + E22 − (E1 + E2 )2 = 2E12 + 2E22 − E22 − E12 − 2E1 E2 − E22
2
4
4
1 2
1
1
= E1 − 2E1 E2 + E22 = (E2 − E1 )2 . σH = (E2 − E1 ).
4
4
2
2
σH
= H 2 − H2 =
1
ψ1 |x2 |ψ1 + ψ2 |x2 |ψ2 + ψ1 |x2 |ψ2 ei(E1 −E2 )t/ + ψ2 |x2 |ψ1 ei(E2 −E1 )t/ .
2
mπ
1 a 2
n+m
n−m
2 a 2
nπ
ψn |x2 |ψm =
x sin
x dx =
πx − cos
πx dx.
x sin
x cos
a 0
a
a
a 0
a
a
x2 =
a
2
Now
x cos
0
2
3 2
a
kπx
2a x
a
k
k
k
πx dx =
πx +
πx cos
− 2 sin
a
k2 π2
a
kπ
a
a
0
2a3
2a3
= 2 2 cos(kπ) = 2 2 (−1)k
k π
k π
(for k = nonzero integer).
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CHAPTER 3. FORMALISM
∴ ψn |x2 |ψm =
69
2a2
4nm
2a2 (−1)n−m
(−1)n+m
=
−
(−1)n+m 2
.
π 2 (n − m)2
(n + m)2
π2
(n − m2 )2
1
16a2
1
So ψ1 |x2 |ψ2 = ψ2 |x2 |ψ1 = − 2 . Meanwhile, from Problem 2.4, ψn |x2 |ψn = a2
.
−
9π
3 2(nπ)2
1
16a2 i(E2 −E1 )t/
1 2 1
1
1
Thus x2 =
e
+ e−i(E2 −E1 )t/ .
a
− 2 + a2
− 2 −
2
3 2π
3 8π
9π 2 E −E
2 cos( 2 1 t)
E2 − E 1
(4 − 1)π 2 2
3π 2 =
=
= 3ω [in the notation of Problem 2.5(b)].
2
2ma 2ma2
a2 2
32
a
5
32
x2 =
− 2 cos(3ωt) . From Problem 2.5(c), x =
−
1 − 2 cos(3ωt) .
2 3 8π 2
9π
2
9π
2
a2 4
64
64
32
5
2
2
2
So σx = x − x =
− 2 cos(3ωt) − 1 + 2 cos(3ωt) −
cos2 (3ωt) .
−
4 3 4π 2
9π
9π
9π 2
2
dx
a2 1
32
5
8
σx2 =
cos2 (3ωt) . And, from Problem 2.5(d):
− 2−
=
sin(3ωt).
2
4 3 4π
9π
dt
3ma
2
2 dx
2 2
Meanwhile, the energy-time uncertainty principle (Eq. 3.72) says σH
σx ≥
. Here
4
dt
2
2 2
2
1
1
32
32
1
5
5
2 2
2a
2
2 3
2
σH σx = (3ω)
−
cos (3ωt) = (ωa)
−
cos (3ωt) .
−
−
4
4 3 4π 2
9π 2
4
3 4π 2
9π 2
2
2
2 2 dx
2aω
8
8
2
=
.
=
sin (3ωt) =
(ωa)2 sin2 (3ωt), since
2
2
4
dt
2 3ma
3π
ma
π
So the uncertainty principle holds if
2
2
2
3
1
32
8
5
2
cos
(3ωt)
≥
sin2 (3ωt),
− 2−
4
3 4π
9π 2
3π 2
which is to say, if
2
2
2
1
5
32
4 8
32
2
2
−
≥
cos (3ωt) +
sin (3ωt) =
.
3 4π 2
9π 2
3 3π 2
9π 2
1
32 2
5
Evaluating both sides: − 2 = 0.20668;
= 0.12978. So it holds. (Whew!)
3 4π
9π 2
Problem 3.19
From Problem 2.43, we have:
x =
l
1
2at
1 2
dx
1 + θ2
l
1 2
=
t, so
= , σx2 =
, where θ =
; H =
p =
(a + l2 ).
2
m
dt
m
4w
4a
m
2m
2m
p2
We need H 2 (to get σH ). Now, H =
, so
2m
∞
∞
1
1
1
2
4
4
2
H =
p =
p |Φ(p, t)| dp, where (Eq. 3.54): Φ(p, t) = √
e−ipx/ Ψ(x, t) dx.
4m2
4m2 −∞
2π −∞
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70
CHAPTER 3. FORMALISM
From Problem 2.43: Ψ(x, t) =
2a
π
1/4
√
2
l
l2
1
e− 4a ea(ix+ 2a ) /(1+iθ) .
1 + iθ
1/4
∞
2
2
l
1
1
il
2a
√
e−l /4a
e−ipx/ ea(ix+ 2a ) /(1+iθ) dx. Let y ≡ x − .
π
2a
1
+
iθ
2π
−∞
1/4
∞
2
2
1
1
2a
√
=√
e−l /4a epl/2a
e−ipy/ e−ay /(1+iθ) dy.
1 + iθ
2π π
−∞
[See Prob. 2.22(a) for the integral.]
1/4
2
1
1
π(1 + iθ) − p2 (1+iθ)
2a
√
=√
e 4a2
e−l /4a epl/2a
a
1 + iθ
2π π
1/4
p2 (1+iθ)
pl
l2
1
1
=√
e− 4a e 2a e− 4a2 .
2aπ
So Φ(p, t) = √
|Φ(p, t)| = √
2
1
1 1 −l2 /2a pl/a −p2 /2a2
1
e
e
= √
e 2a
e
2aπ 2aπ
2
p
l2 − 2pl
+ 2
2
1
= √
e−(l−p/) /2a .
2aπ
∞
2
1
p
p4 = √
p4 e−(l−p/) /2a dp. Let − l ≡ z, so p = (z + l).
2aπ −∞
∞
2
1
= √
(z + l)4 e−z /2a dz. Only even powers of z survive:
5
2aπ
−∞
∞
√
2
4
(2a) √
4
3(2a)2 √
=√
2aπ + 6l2
2aπ + l4 2aπ
z 4 + 6z 2 l2 + l4 e−z /2a dz = √
4
2
2aπ −∞
2aπ
4
= 4 3a2 + 6al2 + l4 . ∴ H 2 =
3a2 + 6al2 + l4 .
4m2
2
σH
= H 2 − H2 =
2 2
σH
σx
4 a
2
4
2
2
4
2
2
4
2
2
+
6al
+
l
−
a
−
2al
−
l
+
4al
3a
=
2a
=
a + 2l2 .
4m2
4m2
2m2
2
2
4 l 2
a
2at
2at
4 a
2 1
=
1+ 2 1+
a + 2l
1+
=
2m2
4a
m
4m2
2l
m
2
2
4 l 2
2 l
2 dx
≥
=
=
, so it works.
2
4m
4 m
4
dt
Problem 3.20
dx dx
For Q = x, Eq. 3.72 says σH σx ≥ , so σx σH ≥
|p|, which is the Griffiths
. But p = m
2 dt
dt
2m
uncertainty principle of Problem 3.14.
Problem 3.21
P 2 |β = P (P |β) = P (α|β|α) = α|β(P |α) = α|β α|α |α = α|β|α = P |β.
1
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CHAPTER 3. FORMALISM
71
Since P 2 |β = P |β for any vector |β, P 2 = P . QED [Note: To say two operators are equal means that
they have the same effect on all vectors.]
If |γ is an eigenvector of P̂ with eigenvalue λ, then P̂ |γ = λ|γ, and it follows that P̂ 2 |γ = λP̂ |γ = λ2 |γ.
But P̂ 2 = P̂ , and |γ = 0, so λ2 = λ, and hence the eigenvalues of P̂ are 0 and 1. Any (complex) multiple of
|α is an eigenvector of P̂ , with eigenvalue 1; any vector orthogonal to |α is an eigenvector of P̂ , with eigenvalue
0.
Problem 3.22
(a) α| = −i1| − 22| + i3|;
β| = −i1| + 23|.
(b) α|β = (−i1| − 22| + i3|) (i|1 + 2|3) = (−i)(i)1|1 + (i)(2)3|3 = 1 + 2i.
β|α = (−i1| + 23|) (i|1 − 2|2 − i|3) = (−i)(i)1|1 + (2)(−i)3|3 = 1 − 2i = α|β∗ . (c)
A11 = 1|αβ|1 = (i)(−i) = 1; A12 = 1|αβ|2
A21 = 2|αβ|1 = (−2)(−i) = 2i; A22 = 2|αβ|2
A31 = 3|αβ|1 = (−i)(−i) = −1; A32 = 3|αβ|2
1 0 2i
A = 2i 0 −4 . No,
−1 0 −2i
= (i)(0) = 0; A13 = 1|αβ|3 = (i)(2) = 2i;
= (−2)(0) = 0; A23 = 2|αβ|3 = (−2)(2) = −4;
= (−i)(0) = 0; A33 = 3|αβ|3 = (−i)(2) = −2i.
it’s not hermitian.
Problem 3.23
Write the eigenvector as |ψ = c1 |1 + c2 |2, and call the eigenvalue E. The eigenvalue equation is
Ĥ|ψ = 9 (|11| − |22| + |12| + |21|) (c1 |1 + c2 |2) = 9 (c1 |1 + c1 |2 − c2 |2 + c2 |1)
= 9 [(c1 + c2 )|1 + (c1 − c2 )|2] = E|ψ = E(c1 |1 + c2 |2).
E
E
9(c1 + c2 ) = Ec1 ⇒ c2 =
− 1 c1 ; 9(c1 − c2 ) = Ec2 ⇒ c1 =
+ 1 c2 .
9
9
2
√
E
E
E
c2 =
−1
+ 1 c2 ⇒
− 1 = 1 ⇒ E = ± 2 9.
9
9
9
√
√
The eigenvectors are: c2 = (± 2 − 1)c1 ⇒ |ψ± = c1 |1 + (± 2 − 1)|2 .
1 1
The Hamiltonian matrix is
H=9
.
1 −1
Problem 3.24
|α =
n
cn |en ⇒ Q̂|α =
n
cn Q̂|en =
n
en |αqn |en =
'
n
(
qn |en en | |α ⇒ Q̂ =
qn |en en |. n
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72
CHAPTER 3. FORMALISM
Problem 3.25
1
1
1 dx = 2. So |e1 = √ .
2
−1
|e1 = 1; e1 |e1 =
1
|e2 = x; e1 |e2 = √
2
1
|e3 = x2 ; e1 |e3 = √
2
1
−1
x dx = 0; e2 |e2 =
1
x3 2
= . So |e2 =
3 −1
3
1
x2 dx =
−1
1
1 2
x2 dx = √ ; e2 |e3 =
23
−1
2
3
3
x.
2
1
x3 dx = 0.
−1
1 2
1
So (Problem A.4): |e3 = |e3 − √ |e1 = x2 − .
3
23
e3 |e3 =
|e3 =
1
x2 −
−1
1
3
2
45 2 1
x −
=
8
3
1
|e4 = x3 . e1 |e4 = √
2
5
2
e3 |e4 =
e4 |e4 =
5
|e4 =
2
dx =
1
−1
1
−1
1
2 x3
x 8
x5
2 4 2
− ·
+
. So
= − + =
5
3 3
9 −1
5 9 9
45
5 3 2 1
x −
.
2 2
2
1
x3 dx = 0;
−1
e2 |e4 =
3
2
1
x4 dx =
−1
3 2
· ;
2 5
3 5 1 3
x − x dx = 0. |e4 = |e4 − e2 |e4 |e2 = x3 −
2
2
2
3
x3 − x dx =
5
7 3 3
x − x =
2
5
3 2
2 5
3
3
x = x3 − x.
2
5
1
x7
2 12 18
2 · 3 x5
9 x3 8
= −
−
+
+
=
.
7
5 5
25 3 −1
7 25 75
7 · 25
7 5 3 3
x − x .
2 2
2
Problem 3.26
(a) Q = ψ|Q̂ψ = Q̂† ψ|ψ = −Q̂ψ|ψ = −(ψ|Q̂ψ)∗ = −Q∗ , so Q is imaginary. (b) From Problem 3.5(c) we know that (P̂ Q̂)† = Q̂† P̂ † , so if P̂ = P̂ † and Q̂ = Q̂† then
[P̂ , Q̂]† = (P̂ Q̂ − Q̂P̂ )† = Q̂† P̂ † − P̂ † Q̂† = Q̂P̂ − P̂ Q̂ = −[P̂ , Q̂]. If P̂ = −P̂ † and Q̂ = −Q̂† , then [P̂ , Q̂]† = Q̂† P̂ † − P̂ † Q̂† = (−Q̂)(−P̂ ) − (−P̂ )(−Q̂) = −[P̂ , Q̂].
So in either case the commutator is antihermitian.
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CHAPTER 3. FORMALISM
73
Problem 3.27
(a) ψ1 .
(b) b1 (with probability 9/25) or b2 (with probability 16/25).
(c) Right after the measurement of B:
• With probability 9/25 the particle is in state φ1 = (3ψ1 + 4ψ2 )/5; in that case the probability of
getting a1 is 9/25.
• With probability 16/25 the particle is in state φ2 = (4ψ1 − 3ψ2 )/5; in that case the probability of
getting a1 is 16/25.
So the total probability of getting a1 is
337
16 16
9 9
·
+
·
=
= 0.5392.
25 25 25 25
625
[Note: The measurment of B (even if we don’t know the outcome of that measurement) collapses the wave
function, and thereby alters the probabilities for the second measurment of A. If the graduate student
inadvertantly neglected to measure B, the second measurement of A would be certain to reproduce the
result a1 .]
Problem 3.28
Ψn (x, t) =
nπ 2
sin
x e−iEn t/ ,
a
a
with En =
n2 π 2 2
.
2ma2
∞
nπ 2 −iEn t/ a −ipx/
1
1
e−ipx/ Ψn (x, t) dx = √
e
sin
e
x dx
a
2π −∞
2π a
0
1
1 a i(nπ/a−p/)x
=√
e
− ei(−nπ/a−p/)x dx
e−iEn t/
2i 0
πa
a
1
1 ei(nπ/a−p/)x
ei(−nπ/a−p/)x =√
e−iEn t/
−
2i i(nπ/a − p/) i(−nπ/a − p/) 0
πa
Φn (p, t) = √
−1 −iEn t/ ei(nπ−pa/) − 1 e−i(nπ+pa/) − 1
= √
e
+
(nπ/a − p/)
(nπ/a + p/)
2 πa
n −ipa/
−1
−1 −iEn t/ (−1) e
(−1)n e−ipa/ − 1
= √
e
a+
a
(nπ − ap/)
(nπ + ap/)
2 πa
a −iEn t/
2nπ
1
n −ipa/
(−1)
=−
e
e
−
1
2 π
(nπ)2 − (ap/)2
=
ne−iEn t/
aπ
1 − (−1)n e−ipa/ .
2
2
(nπ) − (ap/)
Noting that
n −ipa/
1 − (−1) e
−ipa/2
=e
ipa/2
e
− (−1)n e−ipa/2 = 2e−ipa/2
cos(pa/2)
i sin(pa/2)
(n odd),
(n even),
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74
CHAPTER 3. FORMALISM
we have
|Φ1 (p, t)|2 =
4πa cos2 (pa/2)
,
[π 2 − (pa/)2 ]2
|Φ2 (p, t)|2 =
16πa
sin2 (pa/2)
.
[(2π)2 − (pa/)2 ]2
Mathematica has no trouble with the points p = ±nπ/a, where the denominator vanishes. The reason is that
the numerator is also zero there, and the function as a whole is finite—in fact, the graphs show no interesting
behavior at these points.
|Φ1 | 2
|Φ 2| 2
p
p2 =
∞
−∞
4n2
=
a2
p2 |Φn (p, t)|2 dp =
∞
−∞
4n2 πa
∞
p
p2
2
2
2
−∞ [(nπ) − (ap/) ]
2
2
x
4n
Tn (x) dx = 2 In ,
2
2
(1 − x )
a
where
Tn (x) ≡
cos2 (nπx/2),
sin2 (nπx/2),
cos2 (pa/2)
sin2 (pa/2)
if n is odd,
if n is even.
[let x ≡
dp
ap
]
nπ
The integral can be evaluated by partial fractions:
x2
1
1
1
1
1
−
=
+
+
(x2 − 1)2
4 (x − 1)2
(x + 1)2
(x − 1) (x + 1)
1
In =
4
∞
−∞
1
Tn (x) dx +
(x − 1)2
∞
−∞
1
Tn (x) dx +
(x + 1)2
∞
−∞
⇒
1
Tn (x) dx −
(x − 1)
∞
−∞
1
Tn (x) dx .
(x + 1)
For odd n:
∞
−∞
1
2 nπx
cos
dx =
(x ± 1)k
2
1
2 nπx
sin
dx =
(x ± 1)k
2
∞
−∞
∞
1
1
2 nπy
2 nπ
cos
sin
(y
∓
1)
dy
=
dy.
k
yk
2
2
−∞ y
For even n:
∞
−∞
In either case, then,
In =
1
2
∞
−∞
∞
−∞
∞
1
1
2 nπ
2 nπy
sin
sin
(y
∓
1)
dy
=
dy.
k
yk
2
2
−∞ y
1
nπ
2 nπy
sin
dy =
2
y
2
4
Therefore
4n2
4n2 nπ 2
p = 2 In = 2
=
a
a
4
2
nπ
a
∞
−∞
sin2 u
nπ 2
du
=
.
u2
4
2
(same as Problem 2.4).
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CHAPTER 3. FORMALISM
75
Problem 3.29
nλ
1
e
Ψ(x, 0) dx = √
ei(2π/λ−p/)x dx
2 nπλ −nλ
−∞
nλ
1
1
ei(2π/λ−p/)x ei2πn e−ipnλ/) − e−i2πn eipnλ/)
= √
= √
i(2π/λ − p/)
2 nπλ i(2π/λ − p/) −nλ
2 nπλ
1
Φ(p, 0) = √
2π
∞
−ipx/
λ sin(npλ/)
.
nπ (pλ − 2π)
=
|Ψ(x, 0)|2 =
1
2nλ
(−nλ < x < nλ);
|Ψ| 2
λ sin2 (npλ/)
.
nπ (pλ − 2π)2
|Φ| 2
x
nλ
-nλ
|Φ(p, 0)|2 =
2πh/λ
p
The width of the |Ψ|2 graph is wx = 2nλ. The |Φ|2 graph is a maximum at 2π/λ, and goes to zero on either
2π
1
2π
side at
1±
, so wp =
. As n → ∞, wx → ∞ and wp → 0; in this limit the particle has a
λ
2n
nλ
well-defined momentum, but a completely indeterminate position. In general,
2π
= 4π > /2,
nλ
so the uncertainty principle is satisfied (using the widths as a measure of uncertainty). If we try to check the
uncertainty principle more rigorously, using standard deviation as the measure, we get an uninformative result,
because
λ ∞ 2 sin2 (npλ/)
2
p =
p
dp = ∞.
nπ −∞ (pλ − 2π)2
wx wp = (2nλ)
(At large |p| the integrand is approximately (1/λ2 ) sin2 (npλ/), so the integral blows up.) Meanwhile p is
zero, so σp = ∞, and the uncertainty principle tells us nothing. The source of the problem is the discontinuity
in Ψ at the end points; here p̂ Ψ = −i dΨ/dx picks up a delta function, and Ψ|p̂2 Ψ = p̂ Ψ|p̂ Ψ → ∞ because
the integral of the square of the delta function blows up. In general, if you want σp to be finite, you cannot
allow discontinuities in Ψ.
Problem 3.30
(a)
1 = |A|
2
=
∞
−∞
1
dx = 2|A|2
(x2 + a2 )2
1
π
1
|A|2 tan−1 (∞) = 3 |A|2
2
a
a
2a
∞
0
⇒
∞
1
1
x
2 1
−1 x
dx = 2|A|
+ tan
(x2 + a2 )2
2a2 x2 + a2
a
a 0
A=a
2a
.
π
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76
CHAPTER 3. FORMALISM
(b)
x = A
∞
2
−∞
∞
x2 = 2A2
(a2
0
x
dx = 0.
+ x2 )2
x2
x2
a
√
dx.
[Let
y
≡
, x = a y, dx = √ dy.]
2
2
2
2
(a + x )
a
2 y
√
√
1/2
2
2
2a ( π/2)( π)
y
2a Γ(3/2)Γ(1/2)
=
= a2 .
dy
=
(1 + y)2
π
Γ(2)
π
1
∞
2a
π 0
σx = x2 − x2 = a.
2
=
(c)
px px 1
− i sin
, and sine is odd.]
dx. [But e−ipx/ = cos
2
+a
−∞
∞
2A
2A π −|p|a/ cos(px/)
a −|p|a/
√
= √
dx
=
.
=
e
e
2 + a2
x
2a
2π 0
2π
Φ(p, 0) = √
A
2π
∞
−∞
∞
e−ipx/
x2
|Φ(p, 0)|2 dp =
a
∞
e−2|p|a/ dp =
−∞
2a
∞
e−2pa/ = 1. −2a/ 0
(d)
a
p =
a
p = 2
∞
2
(e)
pe−2|p|a/ dp = 0.
−∞
∞
2 −2pa/
p e
0
2a
2
dp =
2a
3
=
2
.
2a2
σp =
p2 − p2 = √ .
2a
√ σx σp = a √
= 2 > . 2
2
2a
Problem 3.31
d
i
Equation 3.71 ⇒ xp = [H, xp]; Eq. 3.64 ⇒ [H, xp] = [H, x]p + x[H, p]; Problem 3.14 ⇒ [H, x] =
dt
ip
dV
−
; Problem 3.17(d) ⇒ [H, p] = i
. So
m
dx
d
i
i
p2
dV
dV
dV
xp =
− p2 + ix
= 2
− x
= 2T − x
. QED
dt
m
dx
2m
dx
dx
In a stationary state all expectation values (at least, for operators that do not depend explicitly on t) are
time-independent (see item 1 on p. 26), so dxp/dt = 0, and we are left with Eq. 3.97.
For the harmonic oscillator:
1
dV
dV
V = mω 2 x2 ⇒
= mω 2 x ⇒ x
= mω 2 x2 = 2V ⇒ 2T = 2V ⇒ T = V . QED
2
dx
dx
In Problem 2.11(c) we found that T = V = 14 ω (for n = 0); T = V = 34 ω (for n = 1). In Problem 2.12 we found that T = 12 n+ 12 ω, while x2 = (n+ 12 )/mω, so V = 12 mω 2 x2 = 12 (n+ 12 )ω,
and hence T = V for all stationary states. c
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CHAPTER 3. FORMALISM
77
Problem 3.32
1
Ψ(x, t) = √ ψ1 e−iE1 t/ + ψ2 e−iE2 t/ ;
2
Ψ(x, t)|Ψ(x, 0) = 0 ⇒
1 iE1 t/
ψ1 |ψ1 + eiE1 t/ ψ1 |ψ2 + eiE2 t/ ψ2 |ψ1 + eiE2 t/ ψ2 |ψ2 e
2
=
1 iE1 t/
+ eiE2 t/ = 0, or eiE2 t/ = −eiE1 t/ , so ei(E2 −E1 )t/ = −1 = eiπ .
e
2
Thus (E2 − E1 )t/ = π (orthogonality also at 3π, 5π, etc., but this is the first occurrence).
∴ ∆t ≡
t
1
. But ∆E = σH = (E2 − E1 ) (Problem 3.18). So ∆t ∆E = . =
π
E2 − E1
2
2
Problem 3.33
Equation 2.69: x =
(a+ + a− ), p = i
2mω
n|x|n =
=
n|p|n = i
mω
(a+ − a− ); Eq. 2.66 :
2
√
a+ |n = √n + 1 |n + 1,
a− |n = n |n − 1.
√
√
n|(a+ + a− )|n =
n + 1 n|n + 1 + n n|n − 1
2mω
2mω
√
√
√
√
n + 1 δn,n +1 + n δn,n −1 =
n δn ,n−1 + n δn,n −1 .
2mω
2mω
√
mω √
n δn ,n−1 − n δn,n −1 .
2
Noting that n and n run from zero to infinity, the matrices are:
√
1 √0 0 0 0
√0
1 0
2 √0 0 0
√
.
0
2 √0
3 √0 0 :
; P = i mω
X=
2mω
2
3
0
4
0
0
0
√
√
0 0 0
4 0
5
···
Squaring these matrices:
√
1·2 √0
0
0
1
0
0
2
·
3
0
0
3
0
√
√
1·2 0
X2 =
5
0
3
·
4
√0
2mω
0 √2 · 3 0
7
0
4·5
···
−1
0
0
−3
√
mω
1·2 0
P2 = −
√
2
2·3
0
√
√
0
0
0
0
√0 − 1 √
1 0 − 2 0
0
0
√
√
0
2
0
−
3
0
0
√
√
0
3
0
−
4
0
0
√
√
0
0
0
4
− 5
···
:
.
.
.
: ;
1·2 √0
0
0
2·3 √0
0
0
−5
0
3·4 √0
0
−7
0
4·5
···
.
: .
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78
CHAPTER 3. FORMALISM
So the Hamiltonian, in matrix form, is
1 2 mω 2 2
P +
X
2m
2
√
1·2 √0
0
0
−1
0
0
−3
0
2·3 √0
0
√
ω
.
1·2 0
=−
−5
0
3
·
4
0
:
√
4
0 √2 · 3 0
−7
0
4·5
···
√
1
1·2 √0
0
0
1
0
0
2·3 √0
0
3
0
0
√
ω
ω
.
0
1·2 0
+
5
0
3 · 4 √ 0 :
= 2
√
0
4
0
2·3 0
7
0
4·5
···
H=
It’s plainly diagonal, and the nonzero elements are Hnn = n +
1
2
0
3
0
0
0
0
5
0
0
0
0
7
..
.
.
ω, as they should be.
Problem 3.34
√
√
Evidently Ψ(x, t) = c0 ψ0 (x)e−iE0 t/ + c1 ψ1 (x)e−iE1 t/ , with |c0 |2 = |c1 |2 = 1/2, so c0 = eiθ0 / 2, c1 = eiθ1 / 2,
for some real θ0 , θ1 .
p = |c0 |2 ψ0 |pψ0 + |c1 |2 ψ1 |pψ1 + c∗0 c1 ei(E0 −E1 )t/ ψ0 |pψ1 + c∗1 c0 ei(E1 −E0 )t/ ψ1 |pψ0 .
But E1 − E0 = ( 32 ω) − ( 12 ω) = ω, and (Problem 2.11) ψ0 |pψ0 = ψ1 |pψ1 = 0, while (Eqs. 2.69 and 2.66)
ψ0 |pψ1 = i
mω
ψ0 |(a+ − a− )ψ1 = i
2
√
√
mω ψ0 | 2ψ2 − ψ0 | 1ψ0 = −i
2
mω
; ψ1 |pψ0 = i
2
mω
.
2
'
'
(
(
1 −iθ0 1 iθ1 −iωt
1 −iθ1 1 iθ0 iωt
mω
mω
√ e e
√ e e
p = √ e
−i
i
+√ e
2
2
2
2
2
2
i mω
mω
=
−e−i(ωt−θ1 +θ0 ) + ei(ωt−θ1 +θ0 ) = −
sin(ωt + θ0 − θ1 ).
2
2
2
The maximum is
mω/2; it occurs at t = 0 ⇔ sin(θ0 − θ1 ) = −1, or θ1 = θ0 + π/2. We might as well pick
θ0 = 0, θ1 = π/2; then
1 1
Ψ(x, t) = √ ψ0 e−iωt/2 + ψ1 eiπ/2 e−3iωt/2 = √ e−iωt/2 ψ0 + iψ1 e−iωt .
2
2
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CHAPTER 3. FORMALISM
79
Problem 3.35
(a) x = α|xα =
α|(a+ + a− )α =
2mω
(a− α|α + α|a− α) =
2mω
(α + α∗ ).
2mω
a2 + a+ a− + a− a+ + a2− . But a− a+ = [a− , a+ ] + a+ a− = 1 + a+ a− (Eq. 2.55).
2mω +
=
a2 + 2a+ a− + 1 + a2− .
2mω +
x2 =
α| a2+ + 2a+ a− + 1 + a2− α =
a2− α|α + 2a− α|a− α + α|α + α|a2− α
2mω
2mω
∗ 2
=
(α ) + 2(α∗ )α + 1 + α2 =
1 + (α + α∗ )2 .
2mω
2mω
x2 =
p = α|pα = i
p2 = −
mω
α|(a+ − a− )α = i
2
mω
(a− α|α − α|a− α) = −i
2
mω
(α − α∗ ).
2
mω 2
mω 2
a+ − a+ a− − a− a+ + a2− = −
a+ − 2a+ a− − 1 + a2− .
2
2
mω
mω
α| a2+ − 2a+ a− − 1 + a2− α = −
a2− α|α − 2a− α|a− α − α|α + α|a2− α
2
2
mω mω ∗ 2
=−
(α ) − 2(α∗ )α − 1 + α2 =
1 − (α − α∗ )2 .
2
2
p2 = −
(b)
1 + (α + α∗ )2 − (α + α∗ )2 =
;
2mω
2mω
mω
mω σp2 = p2 − p2 =
1 − (α − α∗ )2 + (α − α∗ )2 =
. σx σ p =
2
2
σx2 = x2 − x2 =
2mω
mω
= . QED
2
2
(c) Using Eq. 2.67 for ψn :
1
1
1
αn
cn = ψn |α = √ (a+ )n ψ0 |α = √ ψ0 |(a− )n α = √ αn ψ0 |α = √ c0 . n!
n!
n!
n!
(d) 1 =
∞
|cn |2 = |c0 |2
n=0
(e) |α(t) =
∞
n=0
∞
2
|α|2n
= |c0 |2 e|α|
n!
n=0
cn e−iEn t/ |n =
⇒
c0 = e−|α|
2
/2
.
∞
∞
αe−iωt
2
1
αn
√ e−|α| /2 e−i(n+ 2 )ωt |n = e−iωt/2
√
n!
n!
n=0
n=0
n
e−|α|
2
/2
|n.
Apart form the overall phase factor e−iωt/2 (which doesn’t affect its status as an eigenfunction of a− , or
its eigenvalue), |α(t) is the same as |α, but with eigenvalue α(t) = e−iωt α. (f ) Equation 2.58 says a− |ψ0 = 0, so yes, it is a coherent state, with eigenvalue α = 0.
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80
CHAPTER 3. FORMALISM
Problem 3.36
(a) Equation 3.60 becomes |z|2 = [Re(z)]2 + [Im(z)]2 =
2 2
σA
σB ≥
1
(z + z ∗ )
2
2
1
(f |g + g|f )
2
+
2
+
1
(z − z ∗ )
2i
1
(f |g − g|f )
2i
2
; Eq. 3.61 generalizes to
2
.
But f |g − g|f = [Â, B̂] (p. 111), and, by the same argument,
f |g + g|f = ÂB̂ − AB + B̂ Â − AB = ÂB̂ + B̂ Â − 2AB = D.
2 2
So σA
σB ≥
1
D2 + C2 . 4
(b) If B̂ = Â,
2 2
σA
σA
≥
then Ĉ = 0,
4
(1/4)4σA
=
4
σA
,
D̂ = 2 Â2 − A2 ;
2
D = 2 Â2 − A2 = 2σA
.
So Eq. 3.99 says
which is true, but not very informative.
Problem 3.37
First find the eigenvalues and eigenvectors of the Hamiltonian. The characteristic equation says
(a − E)
0
b 0
(c − E)
0 = (a − E)(c − E)(a − E) − b2 (c − E) = (c − E) (a − E)2 − b2 = 0,
b
0
(a − E)
Either E = c, or else (a − E)2 = b2 ⇒ E = a ± b. So the eigenvalues are
E1 = c,
E2 = a + b,
E3 = a − b.
To find the corresponding eigenvectors, write
a0 b
α
α
0 c 0 β = En β .
b 0a
γ
γ
(1)
aα + bγ = cα ⇒ (a − c)α + bγ = 0;
cβ = cβ
(redundant)
; ⇒ (a − c)2 − b2 α = 0.
bα + aγ = cγ ⇒ (a − c)γ + bα = 0.
So (excluding the degenerate case a − c = ±b) α = 0, and hence also γ = 0.
(2)
aα + bγ = (a + b)α ⇒
α − γ = 0;
cβ = (a + b)β ⇒
β = 0;
bα + aγ = (a + b)γ
(redundant).
So α = γ and β = 0.
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CHAPTER 3. FORMALISM
81
(3)
aα + bγ = (a − b)α ⇒
α + γ = 0;
cβ = (a − b)β ⇒
β = 0;
bα + aγ = (a − b)γ
(redundant).
So α = −γ and β = 0.
Conclusion: The (normalized) eigenvectors of H are
0
1
1
|s1 = 1 , |s2 = √ 0 ,
2 1
0
(a) Here |S(0) = |s1 , so
1
1
|s3 = √ 0 .
2 −1
0
|S(t) = e−iE1 t/ |s1 = e−ict/ 1 .
0
(b)
1
|S(0) = √ (|s2 + |s3 ) .
2
1
1
1 −iE2 t/
1
1
1
|S(t) = √ e
|s2 + e−iE3 t/ |s3 = √ e−i(a+b)t/ √ 0 + e−i(a−b)t/ √ 0
2
2
2 1
2 −1
e−ibt/ + eibt/
cos(bt/)
1
= e−iat/
.
0
0
= e−iat/
2
−ibt/
ibt/
−i
sin(bt/)
e
−e
Problem 3.38
(a) H:
E1 = ω, E2 = E3 = 2ω;
A:
1
0
0
|h1 = 0 , |h2 = 1 , |h3 = 0 .
0
0
1
−a λ
0
λ −a
= a2 (2λ − a) − (2λ − a)λ2 = 0 ⇒ a1 = 2λ, a2 = λ, a3 = −λ.
0
0 0 (2λ − a)
010
α
α
λβ = aα
λα = aβ
λ 1 0 0 β = a β ⇒
002
γ
γ
2λγ = aγ
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82
CHAPTER 3. FORMALISM
(1)
λβ = 2λα ⇒ β = 2α,
λα = 2λβ ⇒ α = 2β,
2λγ = 2λγ;
α = β = 0;
0
|a1 = 0 .
1
(2)
λβ = λα ⇒ β = α,
λα = λβ ⇒ α = β,
2λγ = λγ; ⇒ γ = 0.
1
1
1 .
|a2 = √
2 0
(3)
λβ = −λα ⇒ β = −α,
λα = −λβ ⇒ α = −β,
2λγ = −λγ; ⇒ γ = 0.
1
1
−1 .
|a3 = √
2
0
B:
(2µ − b) 0 0 0
−b µ = b2 (2µ − b) − (2µ − b)µ2 = 0 ⇒ b1 = 2µ, b2 = µ, b3 = −µ.
0
µ −b
200
α
α
2µα = bα
µγ = bβ
µ 0 0 1 β = b β ⇒
γ
µβ = bγ
010
γ
(1)
2µα = 2µα,
µγ = 2µβ ⇒ γ = 2β,
µβ = 2µγ ⇒ β = 2γ;
β = γ = 0;
1
|b1 = 0 .
0
(2)
2µα = µα ⇒ α = 0,
µγ = µβ ⇒ γ = β,
µβ = µγ; ⇒ β = γ.
0
1
|b2 = √ 1 .
2 1
(3)
2µα = −µα ⇒ α = 0,
µγ = −µβ ⇒ γ = −β,
µβ = −µγ; ⇒ β = −γ.
0
1
1 .
|b3 = √
2 −1
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CHAPTER 3. FORMALISM
83
(b)
1 0 0
c1
H = S(0)|H|S(0) = ω c∗1 c∗2 c∗3 0 2 0 c2 = ω |c1 |2 + 2|c2 |2 + 2|c3 |2 .
0 0 2
c3
0 1 0
c1
A = S(0)|A|S(0) = λ c∗1 c∗2 c∗3 1 0 0 c2 = λ c∗1 c2 + c∗2 c1 + 2|c3 |2 .
0 0 2
c3
2 0 0
c1
B = S(0)|B|S(0) = µ c∗1 c∗2 c∗3 0 0 1 c2 = µ 2|c1 |2 + c∗2 c3 + c∗3 c2 .
0 1 0
c3
(c)
|S(0) = c1 |h1 + c2 |h2 + c3 |h3 −iE1 t/
⇒
|h1 + c2 e
|h2 + c3 e−iE3 t/ |h3 = c1 e−iωt |h1 + c2 e−2iωt |h2 + c3 e−2iωt |h3
iωt
1
0
0
c1 e
= e−2iωt c1 eiωt 0 + c2 1 + c3 0 = e−2iωt c2 .
0
0
1
c3
|S(t) = c1 e
−iE2 t/
H: h1 = ω, probability |c1 |2 ;
A: a1 = 2λ,
a2 = λ,
h2 = h3 = 2ω, probability (|c2 |2 + |c3 |2 ).
iωt
c1 e
a1 |S(t) = e−2iωt 0 0 1 c2 = e−2iωt c3 ⇒ probability |c3 |2 .
c3
iωt
c1 e
1
1
a2 |S(t) = e−2iωt √ 1 1 0 c2 = √ e−2iωt c1 eiωt + c2 ⇒
2
2
c
3
a3 = −λ,
1 ∗ −iωt
c e
+ c∗2
2 1
1
c1 eiωt + c2 =
|c1 |2 + |c2 |2 + c∗1 c2 e−iωt + c∗2 c1 eiωt .
2
iωt
c1 e
1
1
a3 |S(t) = e−2iωt √ 1 −1 0 c2 = √ e−2iωt c1 eiωt − c2 ⇒
2
2
c
probability =
3
1 ∗ −iωt
1
− c∗2 c1 eiωt − c2 =
c1 e
|c1 |2 + |c2 |2 − c∗1 c2 e−iωt − c∗2 c1 eiωt .
2
2
Note that the sum of the probabilities is 1.
iωt
c1 e
B: b1 = 2µ, b1 |S(t) = e−2iωt 1 0 0 c2 = e−2iωt c1 ⇒ probability |c1 |2 .
c3
iωt
c1 e
1
1
b2 = µ, b2 |S(t) = e−2iωt √ 0 1 1 c2 = √ e−2iωt (c2 + c3 ) ⇒
2
2
c
probability =
3
probability =
1 ∗
1
(c + c∗2 ) (c1 + c2 ) =
|c1 |2 + |c2 |2 + c∗1 c2 + c∗2 c1 .
2 1
2
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84
CHAPTER 3. FORMALISM
b3 = −µ,
iωt
c1 e
1
1
b3 |S(t) = e−2iωt √ 0 1 −1 c2 = √ e−2iωt (c2 − c3 ) ⇒
2
2
c3
1 ∗
1
(c − c∗3 ) (c2 − c3 ) =
|c2 |2 + |c3 |2 − c∗2 c3 − c∗3 c2 .
2 2
2
Again, the sum of the probabilities is 1.
probability =
Problem 3.39
(a)
Expanding in a Taylor series: f (x + x0 ) =
n
∞
1 n d
x0
f (x).
n!
dx
n=0
n
∞
d
1 n ip
d
ip
But p =
f (x) = eipx0 / f (x).
, so
= . Therefore f (x + x0 ) =
x
i dx
dx
n! 0 n=0
(b)
n
∞
1 n ∂
Ψ(x, t + t0 ) =
Ψ(x, t);
t
n! 0 ∂t
n=0
i
∂Ψ
= HΨ.
∂t
∂
[Note: It is emphatically not the case that i ∂t
= H. These two operators have the same effect only when
(as here) they are acting on solutions to the (time-dependent) Schrödinger equation.] Also,
2
∂
∂Ψ
∂
i
= H 2 Ψ,
Ψ = i (HΨ) = H i
∂t
∂t
∂t
provided H is not explicitly dependent on t. And so on. So
n
∞
i
1 n
Ψ(x, t + t0 ) =
t0 − H Ψ = e−iHt0 / Ψ(x, t).
n!
n=0
(c)
Qt+t0 = Ψ(x, t + t0 )|Q(x, p, t + t0 )|Ψ(x, t + t0 ).
But Ψ(x, t + t0 ) = e−iHt0 / Ψ(x, t), so, using the hermiticity of H to write e−iHt0 /
†
= eiHt0 / :
Qt+t0 = Ψ(x, t)|eiHt0 / Q(x, p, t + t0 )e−iHt0 / |Ψ(x, t).
If t0 = dt is very small, expanding to first order, we have:
iH
∂Q
iH
dQ
Qt +
dt = Ψ(x, t)| 1 +
dt Q(x, p, t) +
dt 1 −
dt |Ψ(x, t)
dt
∂t
iH
iH
∂Q
i
∂Q
= Q(x, p, t) +
dt Q − Q
dt +
dt = Q + [H, Q]dt +
dt
∂t
∂t
i
∂Q
= Qt + [H, Q]dt + dt.
∂t
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CHAPTER 3. FORMALISM
∴
85
dQ
i
∂Q
= [H, Q] + . QED
dt
∂t
Problem 3.40
(a) For the free particle, V (x) = 0, so the time-dependent Schrödinger equation reads
∞
∂Ψ
1
2 ∂ 2 Ψ
i
.
Ψ(x, t) = √
eipx/ Φ(p, t) dp ⇒
=−
∂t
2m ∂x2
2π −∞
∞
∞ 2
∂Ψ
p
∂Φ
1
1
∂2Ψ
√
− 2 eipx/ Φ dp. So
eipx/
=
=√
dp,
2
∂t
∂t
∂x
2π −∞
2π −∞
∞
∞
2
1
∂Φ
p
1
√
dp = √
Φ dp.
eipx/ i
eipx/
∂t
2m
2π −∞
2π −∞
But two functions with the same Fourier transform are equal (as you can easily prove using Plancherel’s
theorem), so
i
p2
∂Φ
=
Φ.
∂t
2m
1
ip2
dΦ = −
dt
Φ
2m
⇒
Φ(p, t) = e−ip
2
t/2m
Φ(p, 0).
(b)
−ax2 ilx
Ψ(x, 0) = Ae
Φ(p, 0) = √
e
1
2π
Φ(p, t) =
,
2a
π
A=
1/4 ∞
2a
π
1/4
(Problem2.43(a)).
e−ipx/ e−ax eilx dx =
2
−∞
2
2
1
e−(l−p/) /4a e−ip t/2m ;
2
1/4
(2πa )
2
1
e−(l−p/) /4a
(2πa2 )1/4
|Φ(p, t)|2 = √
(Problem2.43(b)).
2
1
e−(l−p/) /2a .
2πa (c)
∞
2
1
pe−(l−p/) /2a dp
2πa −∞
−∞
[Let y ≡ (p/) − l, so p = (y + l) and dp = dy.]
∞
2
√
=
(y + l)e−y /2a dy [but the first term is odd]
2πa −∞
∞
2
2l
πa
2l
= √
= l [as in Problem 2.43(d)].
e−y /2a dy = √
2
2πa 0
2πa
p =
∞
p|Φ(p, t)|2 dp = √
∞
∞
2
1
2
2 −(l−p/)2 /2a
p =
p |Φ(p, t)| dp = √
p e
dp = √
(y 2 + 2yl + l2 )e−y /2a dy
2πa −∞
2πa −∞
−∞
∞
∞
2
2
22
= √
y 2 e−y /2a dy + l2
e−y /2a dy
2πa 0
0
3
√
22
a
πa
= √
2 π
= (a + l2 )2 [as in Problem 2.43(d)].
+ l2
2
2
2πa
2
∞
2
2
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86
CHAPTER 3. FORMALISM
p2
2 2
1
2 a
1 2
2 a
1 2
; H =
p =
(l +a) =
p2 +
. But H0 =
p 0 =
(Problem 2.22(d)).
2m
2m
2m
2m
2m
2m
2m
1
So H =
p2 + H0 . QED Comment: The energy of the traveling gaussian is the energy of the
2m
same gaussian at rest, plus the kinetic energy (p2 /2m) associated with the motion of the wave packet
as a whole.
(d) H =
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
87
Chapter 4
Quantum Mechanics in Three
Dimensions
Problem 4.1
(a)
[x, y] = xy − yx = 0, etc., so [ri , rj ] = 0.
∂
[px , py ]f =
i ∂x
∂f
i ∂y
∂
−
i ∂y
∂f
i ∂x
= −
2
∂2f
∂2f
−
∂x∂y ∂y∂x
=0
(by the equality of cross-derivatives), so [pi , pj ] = 0.
[x, px ]f =
i
∂f
∂
∂f
∂f
x
−
(xf ) =
x
−x
− f = if,
∂x ∂x
i
∂x
∂x
so [x, px ] = i (likewise [y, py ] = i and [z, pz ] = i).
[y, px ]f =
i
∂f
∂
∂f
∂f
∂y
y
−
(yf ) =
y
−y
= 0 (since
= 0). So [y, px ] = 0,
∂x ∂x
i
∂x
∂y
∂x
and same goes for the other “mixed” commutators. Thus [ri , pj ] = −[pj , ri ] = iδij .
(b) The derivation of Eq. 3.71 (page 115) is identical in three dimensions, so
i
dx
= [H, x];
dt
p2
1 2
1 2
+ V, x =
[px + p2y + p2z , x] =
[p , x]
2m
2m
2m x
1
1
=
(px [px , x] + [px , x]px ) =
[(−i)px + (−i)px ] = −i px .
2m
2m
m
[H, x] =
dx
i
∴
=
dt
dr
1
1
−i px = px . The same goes for y and z, so:
= p.
m
m
dt
m
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88
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
dpx p2
∂V
i
= [H, px ]; [H, px ] =
+ V, px = [V, px ] = i
(Eq. 3.65)
dt
2m
∂x
0
1 0
1
i
dp
∂V
∂V
= (i)
= −
. Same for y and z, so:
= −∇V .
∂x
∂x
dt
(c) From Eq. 3.62: σx σpx
1
1 ≥ [x, px ] = i = . Generally, σri σpj ≥ δij .
2i
2i
2
2
Problem 4.2
2 ∂ 2 ψ ∂ 2 ψ ∂ 2 ψ
= Eψ (inside the box). Separable solutions: ψ(x, y, z) =
+
+
2m ∂x2
∂y 2
∂z 2
X(x)Y (y)Z(z). Put this in, and divide by XY Z:
(a) Equation 4.8 ⇒ −
1 d2 X
1 d2 X
1 d2 Z
2m
+
+
= − 2 E.
2
2
X dx
Y dy
Z dz 2
The three terms on the left are functions of x, y, and z, respectively, so each must be a constant. Call the
separation constants kx2 , ky2 , and kz2 (as we’ll soon seen, they must be positive).
d2 X
= −kx2 X;
dx2
d2 Y
= −ky2 Y ;
dy 2
d2 Z
= −kz2 Z,
dz 2
with E =
2 2
(k + ky2 + kz2 ).
2m x
Solution:
X(x) = Ax sin kx x + Bx cos kx x;
Y (y) = Ay sin ky y + By cos ky y;
Z(z) = Az sin kz z + Bz cos kz z.
But X(0) = 0, so Bx = 0; Y (0) = 0, so By = 0; Z(0) = 0, so Bz = 0. And X(a) = 0 ⇒ sin(kx a) = 0 ⇒
kx = nx π/a (nx = 1, 2, 3, . . . ). [As before (page 31), nx = 0, and negative values are redundant.] Likewise
ky = ny π/a and kz = nz π/a. So
ψ(x, y, z) = Ax Ay Az sin
n π n π n π x
y
z
x sin
y sin
z ,
a
a
a
E=
We might as well normalize X, Y, and Z separately: Ax = Ay = Az =
3/2
n π n π n π 2
x
y
z
ψ(x, y, z) =
sin
x sin
y sin
z ;
a
a
a
a
E=
2 π 2 2
(n + n2y + n2z ).
2m a2 x
2/a. Conclusion:
π 2 2 2
(n + n2y + n2z );
2ma2 x
nx , ny , nz = 1, 2, 3, . . .
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
89
(b)
nx ny nz (n2x + n2y + n2z )
1 1 1
3
Energy
1 1 2
1 2 1
2 1 1
6
6
6
1 2 2
2 1 2
2 2 1
9
9
9
1 1 3
1 3 1
3 1 1
11
11
11
2 2 2
12
1
1
2
2
3
3
2
3
1
3
1
2
3
2
3
1
2
1
14
14
14
14
14
14
Degeneracy
E1 = 3
π 2 2
;
2ma2
d=1
E2 = 6
π 2 2
;
2ma2
d = 3.
E3 = 9
π 2 2
;
2ma2
d = 3.
E4 = 11
π 2 2
;
2ma2
d = 3.
E5 = 12
π 2 2
;
2ma2
d = 1.
E6 = 14
π 2 2
;
2ma2
d = 6.
(c) The next combinations are: E7 (322), E8 (411), E9 (331), E10 (421), E11 (332), E12 (422), E13 (431), and
E14 (333 and 511). The degeneracy of E14 is 4. Simple combinatorics accounts for degeneracies of 1
(nx = ny = nz ), 3 (two the same, one different), or 6 (all three different). But in the case of E14 there is
a numerical “accident”: 32 + 32 + 32 = 27, but 52 + 12 + 12 is also 27, so the degeneracy is greater than
combinatorial reasoning alone would suggest.
Problem 4.3
1
Eq. 4.32 ⇒ Y00 = √ P00 (cos θ); Eq. 4.27 ⇒ P00 (x) = P0 (x); Eq. 4.28 ⇒ P0 (x) = 1.
4π
Y21 = −
P2 (x) =
P21 (x) =
5 1 iφ 1
e P2 (cos θ);
4π 3 · 2
1
4·2
d
dx
1 − x2
2
x2 − 1
2
P21 (x) =
=
1 − x2
1
Y00 = √ .
4π
d
P2 (x);
dx
1 2
1
1 d 2
2(x − 1)2x =
x − 1 + x(2x) =
3x2 − 1 ;
8 dx
2
2
d 3 2 1
x −
= 1 − x2 3x; P21 (cos θ) = 3 cos θ sin θ. Y21 = −
dx 2
2
15 iφ
e sin θ cos θ.
8π
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90
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
|Y00 |2
Normalization:
|Y21 |2 sin θ dθ dφ =
1
sin θ dθ dφ =
4π
15
8π
π
2π
sin θ dθ
dφ =
0
0
π
2π
sin2 θ cos2 θ sin θ dθ
0
dφ =
0
15
4
1
(2)(2π) = 1. 4π
π
cos2 θ(1 − cos2 θ) sin θ dθ
0
π
15
15 2 2
cos3 θ cos5 θ 5 3
−
+
= 4 3−5 = 2−2 =1
4
3
5
0
=
Orthogonality:
15
8π
1
∗
Y00 Y21 sin θ dθ dφ = − √
4π
π
sin θ cos θ sin θ dθ
0
(sin3 θ)/3|π
0 =0
2π
0
eiφ dφ = 0. (eiφ )/i|2π
0 =0
Problem 4.4
A
1
1
A
dΘ
A
=
sec2 (θ/2) =
=
. Therefore
dθ
tan(θ/2) 2
2 sin(θ/2) cos(θ/2)
sin θ
d
With l = m = 0, Eq. 4.25 reads:
dθ
dΘ
sin θ
dθ
d
dθ
sin θ
dΘ
dθ
=
d
(A) = 0.
dθ
= 0. So A ln[tan(θ/2)] does satisfy Eq. 4.25. However,
π
Θ(0) = A ln(0) = A(−∞); Θ(π) = A ln tan
= A ln(∞) = A(∞). Θ blows up at θ = 0 and at θ = π.
2
Problem 4.5
Yll
(2l + 1) 1 ilφ l
e Pl (cos θ).
4π (2l)!
l
= (−1)
Pl (x) =
1
2l l!
d
dx
Pll (x) = (1 − x2 )l/2
l
(x2 − 1)l ,
so Pll (x) =
1
(1 − x2 )l/2
2l l!
d
dx
d
dx
l
Pl (x).
2l
(x2 − 1)l .
Now (x2 − 1)l = x2l + · · · , where all the other terms involve powers of x less than 2l, and hence give zero when
differentiated 2l times. So
2l
n
d
1
d
(2l)!
Pll (x) = l (1 − x2 )l/2
x2l . But
xn = n!, so Pll = l (1 − x2 )l/2 .
2 l!
dx
dx
2 l!
∴
Yll
l
= (−1)
(2l + 1) ilφ (2l)!
1
e
(sin θ)l =
l
4π(2l)!
2 l!
l!
(2l + 1)!
4π
1
− eiφ sin θ
2
l
.
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
7 1 2iφ 2
· e P3 (cos θ);
4π 5!
Y32 =
P32 (x) = (1 − x2 )
d
dx
91
2
P3 (x);
P3 (x) =
1
8 · 3!
d
dx
3
(x2 − 1)3 .
2
1 d 2
d
1
6x(x2 − 1)2 =
(x − 1)2 + 4x2 (x2 − 1)
8 · 3 · 2 dx
8 dx
1 3
1
1
=
4x(x2 − 1) + 8x(x2 − 1) + 4x2 · 2x =
x − x + 2x3 − 2x + 2x3 =
5x3 − 3x .
8
2
2
P3 =
P32 (x)
1
1 − x2
=
2
Y32 =
d
dx
2
5x3 − 3x =
7 1
1
15e2iφ cos θ sin2 θ =
4π 5!
4
Check that
Yll
∂Yll
= Aeilφ l(sin θ)l−1 cos θ;
∂θ
(2l + 1)!
4π
1
−
2
l
≡ A, so
Yll = A(eiφ sin θ)l .
∂Yll
= l cos θYll ;
∂θ
∂ 2 Yll
∂Y l
= −l2 Yll .
= l cos θ sin θ l − l sin2 θYll = l2 cos2 θ − l sin2 θ Yll .
∂θ
∂φ2
So the left side of Eq. 4.18 is l2 (1 − sin2 θ) − l sin2 θ − l2 Yll = −l(l + 1) sin2 θ Yll , which matches the right side.
1 105
Check that Y32 satisfies Eq. 4.18: Let B ≡
, so Y32 = Be2iφ sin2 θ cos θ.
4 2π
∂Y32
∂Y32
∂
∂
3
2iφ
2
= Be
sin θ
= Be2iφ sin θ
2 sin2 θ cos2 θ − sin4 θ
2 sin θ cos θ − sin θ ; sin θ
∂θ
∂θ
∂θ
∂θ
sin θ
∂
∂θ
sin θ
105 2iφ 2
e sin θ cos θ.
2π
1
l!
satisfies Eq. 4.18: Let
d
1
1
1 − x2
15x2 − 3 = (1 − x2 )30x = 15x(1 − x2 ).
2
dx
2
sin θ
∂Yll
∂θ
= Be2iφ sin θ 4 sin θ cos3 θ − 4 sin3 θ cos θ − 4 sin3 θ cos θ = 4Be2iφ sin2 θ cos θ cos2 θ − 2 sin2 θ
= 4(cos2 θ − 2 sin2 θ)Y32 .
∂ 2 Y32
= −4Y32 . So the left side of Eq. 4.18 is
∂φ2
4(cos2 θ − 2 sin2 θ − 1)Y32 = 4(−3 sin2 θ)Y32 = −l(l + 1) sin2 θ Y32 ,
where l = 3, so it fits the right side of Eq. 4.18.
Problem 4.6
1
1 1
Pl (x)Pl (x)dx = l l
2 l! 2 l !
−1
1
−1
d
dx
l
(x − 1)
2
l
d
dx
l l
(x − 1)
2
dx.
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
If l = l , we may as well let l be the larger of the two (l > l ). Integrate by parts, pulling successively each
derivative off the first term onto the second:
1
1
l−1
l
d
d
l
l
2
l
2
l 2 l!2 l !
Pl (x)Pl (x)dx =
(x − 1)
(x − 1) dx
dx
−1
−1
1 l−1
l +1
d
d
2
l
2
l
−
(x − 1)
(x − 1) dx
dx
dx
−1
l +l
1
d
= . . . (boundary terms) . . . + (−1)l
(x2 − 1)l
(x2 − 1)l dx.
dx
−1
l +l
But (d/dx)
(x2 − 1)l = 0, because (x2 − 1)l is a polynomial whose highest power is 2l , so more than 2l
derivatives will kill it, and l + l > 2l . Now, the boundary terms are of the form:
d
dx
l−n
(x − 1)
2
l
d
dx
l +n−1
+1
(x − 1) , n = 1, 2, 3, . . . , l.
l
2
−1
Look at the first term: (x2 − 1)l = (x2 − 1)(x2 − 1) . . . (x2 − 1); l factors. So 0, 1, 2, . . . , l − 1 derivatives will
still leave at least one overall factor of (x2 − 1). [Zero derivatives leaves l factors; one derivative leaves l − 1 :
d/dx(x2 −1)l = 2lx(x2 −1)l−1 ; two derivatives leaves l−2 : d2 /dx2 (x2 −1)l = 2l(x2 −1)l−1 +2l(l−1)2x2 (x2 −1)l−2 ,
1
and so on.] So the boundary terms are all zero, and hence −1 Pl (x)Pl (x)dx = 0.
This leaves only the case l = l . Again the boundary terms vanish, but this time the remaining integral does
not:
2l
1
1
d
(2l l!)2
[Pl (x)]2 dx = (−1)l
(x2 − 1)l
(x2 − 1)l dx
dx
−1
−1
(d/dx)2l (x2l )=(2l)!
= (−1)l (2l)!
−1
Let x ≡ cos θ, so dx = − sin θ dθ,
1
(x2 − 1)l dx = 2(2l)!
(sin θ)2l (− sin θ)dθ =
π/2
1
(1 − x2 )l dx.
0
(1 − x2 ) = sin2 θ,
0
(1 − x2 )l dx =
0
1
θ : π/2 → 0. Then
π/2
(sin θ)2l+1 dθ
0
(2)(4) · · · (2l)
(2l l!)2
(2l l!)2
=
=
=
.
(1)(3)(5) · · · (2l + 1)
1 · 2 · 3 · · · · (2l + 1)
(2l + 1)!
∴
1
−1
[Pl (x)]2 dx =
1
(2l l!)2
2
2(2l)!
=
. So
l
2
(2 l!)
(2l + 1)!
2l + 1
1
−1
Pl (x)Pl (x)dx =
2
δll . QED
2l + 1
Problem 4.7
(a)
cos x sin x
1 d cos x
n1 (x) = −(−x)
=− 2 −
.
x dx
x
x
x
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
93
2
1 d
1 d
cos x
1 d cos x
n2 (x) = −(−x)2
= −x2
x dx
x
x dx
x dx
x
d 1 −x sin x − cos x
d sin x cos x
= −x
=x
+ 3
·
dx x
x2
dx x2
x
2
x cos x − 2x sin x −x3 sin x − 3x2 cos x
=x
+
x4
x6
cos x
sin x sin x 3 cos x
3
3
1
=
−2 2 − 2 −
cos x − 2 sin x.
= − 3−
3
x
x
x
x
x
x
x
(b) Letting sin x ≈ x and cos x ≈ 1, and keeping only the lowest power of x:
n1 (x) ≈ −
1
1
1
+ x ≈ − 2.
2
x
x
x
As x → 0, this blows up.
3
3
1
3
n2 (x) ≈ − 3 −
− 2 x ≈ − 3 , which again blows up at the origin.
x
x
x
x
Problem 4.8
(a)
u = Arj1 (kr) = A
sin(kr) cos(kr)
A sin(kr)
−
=
− cos(kr) .
k2 r
k
k
(kr)
du
cos(kr) sin(kr)
A k 2 r cos(kr) − k sin(kr)
+ k sin(kr) = A
+ sin(kr) .
=
−
dr
k
(kr)2
kr
(kr)2
d2 u
−k 2 r sin(kr) − k cos(kr) k 3 r2 cos(kr) − 2k 2 r sin(kr)
=A
−
+ k cos(kr)
2
dr
(kr)2
(kr)4
sin(kr) cos(kr) cos(kr)
sin(kr)
= Ak −
−
+2
+ cos(kr)
−
(kr)
(kr)2
(kr)2
(kr)3
2
2
1
= Ak 1 −
cos(kr)
+
−
sin(kr) .
(kr)2
(kr)3
(kr)
With V = 0 and l = 1, Eq. 4.37 reads:
Ak 1 −
2
(kr)2
= Ak cos(kr) −
cos(kr) +
d2 u
2
2mE
− 2 u = − 2 u = −k 2 u. In this case the left side is
dr2
r
2
1
−
(kr)3
(kr)
sin(kr)
= −k 2 u.
kr
2
sin(kr) −
(kr)2
sin(kr)
− cos(kr)
(kr)
So this u does satisfy Eq. 4.37.
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94
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
z
cos z
(b) Equation 4.48 ⇒ j1 (z) = 0, where z = ka. Thus sin
= 0, or tan z = z. For high z (large n,
z2 − z
if n = 1, 2, 3, . . . counts the allowed energies in increasing order), the intersections occur slightly below
z = (n + 12 )π.
2 k 2
2 z 2
2 π 2
∴E=
=
=
2m
2ma2
2ma2
1
n+
2
2
. QED
z
tan z
π/2
Problem 4.9
For r ≤ a, u(r) = A sin(kr), with k ≡
state) E < 0 ⇒:
3π/2
5π/2
z
2m(E + V0 )/. For r ≥ a, Eq. 4.37 with l = 0, V = 0, and (for a bound
√
d2 u
2m
= − 2 Eu = κ2 u, with κ ≡ −2mE/ ⇒ u(r) = Ceκr + De−κr .
2
dr
κr
But the Ce term blows up as r → ∞, so u(r) = De−κr .
1
1
κ
Continuity of u at r = a : A sin(ka) = De−κa
divide: tan(ka) = − , or − cot ka = .
Continuity of u at r = a : Ak cos(ka) = −Dκe−κa
k
κ
k
√
2mV0 a2 /2 − z 2
κ
2mV0
Let ka ≡ z;
=
. Let z0 ≡
a. − cot z = (z0 /z)2 − 1. This is exactly the
k
z
same transcendental equation we encountered in Problem 2.29—see graph there. There is no solution if z0 < π/2,
which is to say, if 2mV0 a2 /2 < π 2 /4, or V0 a2 < π 2 2 /8m. Otherwise, the ground state energy occurs somewhere
between z = π/2 and z = π:
E + V0 =
2 π 2
2 k 2 a2
2 2
2 π 2
=
z , so
< (E0 + V0 ) <
2
2
2
2ma
2ma
8ma
2ma2
Problem 4.10
R30 (n = 3, l = 0) : Eq. 4.62 ⇒ v(ρ) =
Eq. 4.76 ⇒ c1 =
Eq. 4.73 ⇒ ρ =
/
2(1 − 3)
c0 = −2c0 ;
(1)(2)
r
;
3a
j=0 cj ρ
j
c2 =
Eq. 4.75 ⇒ R30 =
(precise value depends on V0 ).
.
2(2 − 3)
1
2
c1 = − c1 = c0 ;
(2)(3)
3
3
c3 =
2(3 − 3)
c2 = 0.
(3)(4)
r
2 r 2
1 −ρ
1 r −r/3a
c0 − 2c0
ρe v(ρ) =
e
+ c0
r
r 3a
3a 3
3a
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
R30 =
c 0
3a
2 r
2 r 2 −r/3a
e
+
.
3 a
27 a
1−
2(2 − 3)
1
c0 = − c0 ;
(1)(4)
2
R31 (n = 3, l = 1) : c1 =
R31
95
c2 =
2(3 − 3)
c1 = 0.
(2)(5)
c 1 r −r/3a
1 r 2 −r/3a
1 r
0
c0 − c0
r
1
−
=
e
.
=
e
r 3a
2 3a
9a2
6 a
2(3 − 3)
c0 = 0.
(1)(6)
R32 (n = 3, l = 2) : c1 =
R32 =
c 1 r 3 −r/3a
0
r2 e−r/3a .
e
(c0 ) =
r 3a
27a3
Problem 4.11
(a)
∞
Eq. 4.31 ⇒
c 0
|R|2 r2 dr = 1. Eq. 4.82 ⇒ R20 =
2a
0
1=
c 2
0
2a
∴ c0 =
∞
3
a
0
2
.
a
z 2 −z 2
c2 a
1−
e z dz = 0
2
4
∞
r −r/2a
.
e
2a
Let z ≡
r
.
a
1 4 −z
c20 a
24
a
z − z + z e dz =
2−6+
= c20 .
4
4
4
2
2
0
1−
3
1
Eq. 4.15 ⇒ ψ200 = R20 Y00 . Table 4.3 ⇒ Y00 = √ .
4π
1
∴ ψ200 = √
4π
2 1 r −r/2a
r −r/2a
1 1 1−
e
1−
e
⇒ ψ200 = √
.
a 2a
2a
2a
2πa 2a
(b)
R21
R21
c0
= 2 re−r/2a ;
4a
1 1
=√
re−r/2a ;
6a 2a2
ψ210
c 2 ∞
3
c20 a
0
5
4 −z
24 = ac20 , so c0 =
1=
a
z
e
dz
=
4a2
16
2
0
ψ21±1
1 1
=√
re−r/2a
6a 2a2
1 1
=√
re−r/2a
6a 2a2
'
'
3
sin θe±iφ
8π
∓
3
cos θ
4π
(
= √
(
2
.
3a
1 1
re−r/2a sin θe±iφ ;
= ∓√
πa 8a2
1
1
re−r/2a cos θ.
2πa 4a2
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96
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Problem 4.12
(a)
L0 = ex e−x = 1.
L2 = ex
d
dx
2
L1 = ex
d −x
e x = ex e−x − e−x x = 1 − x.
dx
e−x x2 = ex
d
2xe−x − e−x x2
dx
= ex 2e−x − 2xe−x + e−x x2 − 2xe−x = 2 − 4x + x2 .
L3 = ex
d
dx
3
e−x x3 = ex
d
dx
2
−e−x x3 + 3x2 e−x
d −x 3
e x − 3x2 e−x − 3x2 e−x + 6xe−x
dx
= ex −e−x x3 + 3x2 e−x + 6x2 e−x − 12xe−x − 6xe−x + 6e−x
= ex
= 6 − 18x + 9x2 − x3 .
(b)
v(ρ) = L52 (2ρ);
L52 (x) = L57−5 (x) = (−1)5
7
d
L7 (x) = e
dx
5
d
= ex
dx
4
d
= ex
dx
3
d
x
=e
dx
x
7 −x
x e
x
=e
d
dx
d
dx
6
5
L7 (x).
7x6 e−x − x7 e−x
42x5 e−x − 7x6 e−x − 7x6 e−x + x7 e−x
210x4 e−x − 42x5 e−x − 84x5 e−x + 14x6 e−x + 7x6 e−x − x7 e−x
840x3 e−x − (210 + 630)x4 e−x
+ (126 + 126)x5 e−x − (21 + 7)x6 e−x + x7 e−x
2
d
x
=e
2520x2 e−x − (840 + 3360)x3 e−x
dx
+(840 + 1260)x4 e−x − (252 + 168)x5 e−x + (28 + 7)x6 e−x − x7 e−x
d
= ex
5040xe−x − (2520 + 12600)x2 e−x + (4200 + 8400)x3 e−x
dx
− (2100 + 2100)x4 e−x + (420 + 210)x5 e−x − (35 + 7)x6 e−x + x7 e−x
= ex 5040e−x − (5040 + 30240)xe−x + (15120 + 37800)x2 e−x
− (12600 + 8400 + 8400)x3 e−x + (2100 + 2100 + 3150)x4 e−x
− (630 + 252)x5 e−x + (42 + 7)x6 e−x − x7 e−x
= 5040 − 35280x + 52920x2 − 29400x3 + 7350x4 − 882x5 + 49x6 − x7 .
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
L52 = −
d
dx
97
5
−882x5 + 49x6 − x7
= − −882(5 · 4 · 3 · 2) + 49(6 · 5 · 4 · 3 · 2)x − 7 · 6 · 5 · 4 · 3x2
= 60 (882 × 2) − (49 × 12)x + 42x2 = 2520(42 − 14x + x2 ).
v(ρ) = 2520(42 − 28ρ + 4ρ2 ) = 5040 21 − 14ρ + 2ρ2 .
(c)
Eq. 4.62 ⇒ v(ρ) =
∞
cj ρj . Eq. 4.76 ⇒ c1 =
j=0
c2 =
2(4 − 5)
1
2
c1 = − c1 =
c0 ;
(2)(7)
7
21
c3 =
2(3 − 5)
2
c0 = − c0 .
(1)(6)
3
2(5 − 5)
c2 = 0.
(3)(8)
c0
2
2
v(ρ) = c0 − c0 ρ + c0 ρ2 =
21 − 14ρ + 2ρ2 .
3
21
21
Problem 4.13
(a)
ψ=√
r =
1
e−r/a ,
πa3
4
a3
∞
so rn =
r3 e−2r/a dr =
0
1
πa3
rn e−2r/a r2 sin θ dr dθ dφ =
3
4 a 4
3!
= a;
3
a
2
2
r2 =
4
a3
∞
4π
πa3
∞
rn+2 e−2r/a dr.
0
r4 e−2r/a dr =
0
4 a 5
4!
= 3a2 .
a3
2
(b)
x = 0;
x2 =
1 2
r = a2 .
3
(c)
1 1
re−r/2a sin θeiφ
ψ211 = R21 Y11 = − √
πa 8a2
(Problem 4.11(b)).
1
1
2
2 −r/a
x =
r
e
sin
θ
r2 sin2 θ cos2 φ r2 sin θ dr dθ dφ
πa (8a2 )2
∞
π
2π
1
5
6 −r/a
=
r
e
dr
sin
θ
dθ
cos2 φ dφ
64πa5 0
0
0
1
2·4
1
7
=
6!a
2
· 2π = 12a2 .
64πa5
1·3·5
2
2
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98
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Problem 4.14
ψ=√
1
πa3
dp
4
= 3
dr
a
4
4
P = |ψ|2 4πr2 dr = 3 e−2r/a r2 dr = p(r) dr; p(r) = 3 r2 e−2r/a .
a
a
2 −2r/a
8r −2r/a r
−2r/a
2
2re
= 3e
1−
+r − e
= 0 ⇒ r = a.
a
a
a
e−r/a ;
Problem 4.15
1
1 (a) Ψ(r, t) = √ ψ2 1 1 e−iE2 t/ + ψ2 1 −1 e−iE2 t/ = √ (ψ2 1 1 + ψ2 1 −1 ) e−iE2 t/ ;
2
2
E2 =
E1
2
.
=−
4
8ma2
From Problem 4.11(b):
i
1 1
ψ2 1 1 + ψ2 1 −1 = − √
re−r/2a sin θ eiφ − e−iφ = − √
re−r/2a sin θ sin φ.
πa 8a2
πa 4a2
Ψ(r, t) = − √
i
re−r/2a sin θ sin φ e−iE2 t/ .
2πa 4a2
(b)
1
e2 1
1
e2
|Ψ|2 −
r2 e−r/a sin2 θ sin2 φ
d3 r =
−
r2 sin θ dr dθ dφ
4
4π90 r
(2πa)(16a )
4π90
r
∞
π
2π
1
2
2
4
3
2
3 −r/a
4
=
(π)
−
r e
dr
sin θ dθ
sin φ dφ = −
3!a
6
32πa5
ma2
32πma
3
0
0
0
V =
= −
2
1
1
= E1 = (−13.6eV) = −6.8eV
4ma2
2
2
(independent of t).
Problem 4.16
En (Z) = Z 2 En ;
E1 (Z) = Z 2 E1 ;
a(Z) = a/Z;
R(Z) = Z 2 R.
Lyman lines
range from
from ni = 2 to ni = ∞ (with nf = 1);the wavelengths
range
3
1
1
1
4
1
1
= R ⇒ λ2 =
down to
= R ⇒ λ1 = .
=R 1−
=R 1−
λ2
4
4
3R
λ1
∞
R
1
1
1
−8
For Z = 2 : λ1 =
=
= 2.28 × 10 m to λ2 =
= 3.04 × 10−8 m, ultraviolet.
4R
4(1.097 × 107 )
3R
1
4
For Z = 3 : λ1 =
= 1.01 × 10−8 m to λ2 =
= 1.35 × 10−8 m, also ultraviolet.
9R
27R
Problem 4.17
(a) V (r) = −G
Mm
.
r
So
e2
→ GM m translates hydrogen results to the gravitational analogs.
4π90
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
(b) Equation 4.72: a =
=
(6.6726 × 10−11
4π90
e2
2
,
m
so
ag =
99
2
GM m2
(1.0546 × 10−34 Js)2
= 2.34 × 10−138 m.
m3 /kg · s2 )(1.9892 × 1030 kg)(5.98 × 1024 kg)2
(c) Equation 4.70 ⇒ En = −
1
m
(GM m)2 2 .
2
2
n
1
Mm
Mm
mv 2
1
GM m
mv 2 − G
. But G 2 =
⇒ mv 2 =
, so
2
ro
ro
ro
2
2ro
1
m
GM m
GM m2
ro
Ec = −
=−
(GM m)2 2 ⇒ n2 =
ro =
⇒ n=
2
2ro
2
n
2
ag
Ec =
ro
.
ag
1.496 × 1011
= 2.53 × 1074 .
2.34 × 10−138
ro = earth-sun distance = 1.496 × 1011 m ⇒ n =
(d)
∆E = −
So
G 2 M 2 m3
22
1
1
− 2 .
2
(n + 1)
n
1
1
1
− 2 ≈ 2
2
(n + 1)
n
n
∆E =
1
1
1
= 2
≈ 2
2
2
(n + 1)
n (1 + 1/n)
n
1−
2
−1
n
=−
2
;
n3
∆E =
1−
2
n
.
G 2 M 2 m3
.
2 n3
(6.67 × 10−11 )2 (1.99 × 1030 )2 (5.98 × 1024 )3
= 2.09 × 10−41 J.
(1.055 × 10−34 )2 (2.53×74 )3
Ep = ∆E = hν =
hc
.
λ
λ = (3 × 108 )(6.63 × 10−34 )/(2.09 × 10−41 ) = 9.52 × 1015 m.
But 1 ly = 9.46 × 1015 m. Is it a coincidence that λ ≈ 1 ly? No: From part (c), n2 = GM m2 ro /2 , so
'
(
3/2
GM m2 ro
ro3
ch
2π3
2 n 3
λ=
= c 2π
= c2π 2 2 3 = c 2 2 3
.
∆E
G M m
G M m
2
GM
But (from (c)) v = GM/ro = 2πro /T , where T is the period of the orbit (in this case one year), so
T = 2π ro3 /GM , and hence λ = cT (one light year). [Incidentally, the same goes for hydrogen: The
wavelength of the photon emitted in a transition from a highly excited state to the next lower one is equal
to the distance light would travel in one orbital period.]
Problem 4.18
f |L± g = f |Lx g ± if |Ly g = Lx f |g ± iLy f |g = (Lx ∓ iLy )f |g = L∓ f |g, so (L± )† = L∓ .
Now, using Eq. 4.112, in the form L∓ L± = L2 − L2z ∓ Lz :
flm |L∓ L± flm = flm |(L2 − L2z ∓ Lz )flm = flm | 2 l(l + 1) − 2 m2 ∓ 2 m flm = 2 [l(l + 1) − m(m ± 1)] flm |flm = 2 [l(l + 1) − m(m ± 1)]
m±1
m±1
2 m±1 m±1
2
|Am
= |Am
|fl
= |Am
= L± flm |L± flm = Am
l fl
l fl
l | fl
l | .
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100
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Conclusion:
Am
l =
l(l + 1) − m(m ± 1).
Problem 4.19
(a)
[Lz , x] = [xpy − ypx , x] = [xpy , x] − [ypx , x] = 0 − y[px , x] = iy. [Lz , y] = [xpy − ypx , y] = [xpy , y] − [ypx , y] = x[py , y] − 0 = −ix. [Lz , z] = [xpy − ypx , z] = [xpy , z] − [ypx , z] = 0 − 0 = 0. [Lz , px ] = [xpy − ypx , px ] = [xpy , px ] − [ypx , px ] = py [x, px ] − 0 = ipy . [Lz , py ] = [xpy − ypx , py ] = [xpy , py ] − [ypx , py ] = 0 − px [y, py ] = −ipx . [Lz , pz ] = [xpy − ypx , pz ] = [xpy , pz ] − [ypx , pz ] = 0 − 0 = 0. (b)
[Lz , Lx ] = [Lz , ypz − zpy ] = [Lz , ypz ] − [Lz , zpy ] = [Lz , y]pz − [Lz , py ]z
= −ixpz + ipx z = i(zpx − xpz ) = iLy .
(So, by cyclic permutation of the indices, [Lx , Ly ] = iLz .)
(c)
[Lz , r2 ] = [Lz , x2 ] + [Lz , y 2 ] + [Lz , z 2 ] = [Lz , x]x + x[Lz , x] + [Lz , y]y + y[Lz , y] + 0
= iyx + xiy + (−ix)y + y(−ix) = 0.
[Lz , p2 ] = [Lz , p2x ] + [Lz , p2y ] + [Lz , p2z ] = [Lz , px ]px + px [Lz , px ] + [Lz , py ]py + py [Lz , py ] + 0
= ipy px + px ipy + (−ipx )py + py (−ipx ) = 0.
(d) It follows from (c) that all three components
of L commute with r2 and p2 , and hence with the whole
√
2
2
Hamiltonian, since H = p /2m + V ( r ). QED
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
101
Problem 4.20
(a)
dLx i
1 2
= [H, Lx ]. [H, Lx ] =
[p , Lx ] + [V, Lx ].
dt
2m
The first term is zero (Problem 4.19(c)); the second would be too if V were a function only of r = |r|, but
in general
Equation 3.71 ⇒
[H, Lx ] = [V, ypz − zpy ] = y[V, pz ] − z[V, py ]. Now (Problem 3.13(c)):
[V, pz ] = i
Thus
∂V
∂V
∂V
∂V
and [V, py ] = i
. So [H, Lx ] = yi
− zi
= i[r × (∇V )]x .
∂z
∂y
∂z
∂y
dLx = −[r × (∇V )]x , and the same goes for the other two components:
dt
dL
= [r × (−∇V )] = N. QED
dt
(b)
If V (r) = V (r), then ∇V =
∂V
dL
r̂, and r × r̂ = 0, so
= 0. QED
∂r
dt
Problem 4.21
(a)
∂
∂f
∂
∂f
+ i cot θ
e−iφ
− i cot θ
∂θ
∂φ
∂θ
∂φ
2
2
∂ f
2 iφ
−iφ ∂ f
2 ∂f
e
− e
+ cot θ
− i − csc θ
∂θ2
∂φ
∂θ ∂φ
2
∂f
∂ f
∂f
∂2f
−iφ
−iφ
+i cot θ −ie
− i cot θ
+e
− i cot θ 2
∂θ
∂φ
∂φ ∂θ
∂φ
2
2
2
∂ f
∂ f
∂f
∂2f
2
2 ∂f
2 ∂f
2 ∂ f
−
− i cot θ
+ cot θ
− i cot θ
+ i cot θ
+ cot θ 2
+ i csc θ
∂θ2
∂φ
∂θ ∂φ
∂θ
∂φ
∂φ ∂θ
∂φ
2
2
∂
∂
∂
∂
−2
+ cot θ
+ cot2 θ 2 + i(csc2 θ − cot2 θ)
f, so
∂θ2
∂θ
∂φ
∂φ
2
∂
∂
∂2
∂
2
−2
+
cot
θ
θ
+i
+
cot
. QED
2
2
∂θ
∂θ
∂φ
∂φ
L+ L− f = −2 eiφ
=
=
=
L+ L− =
∂
, Eq. 4.112 ⇒ L2 = L+ L− + L2z − Lz , so, using (a):
i ∂φ
2
2
∂
∂
∂2
∂
∂
2
2
2
2 ∂
L = −
+ cot θ 2 + i
−
+ cot θ
−
∂θ2
∂θ
∂φ
∂φ
∂φ2
i ∂φ
2
2
2
∂
∂
∂
∂
∂
∂
∂
1 ∂2
2
2
2
= −
+ cot θ
+ cot θ
+ (cot θ + 1) 2 + i
−i
= −
+
∂θ2
∂θ
∂φ
∂φ
∂φ
∂θ2
∂θ sin2 θ ∂φ2
1 ∂
∂
1 ∂2
= −2
. QED
sin θ
+
sin θ ∂θ
∂θ
sin2 θ ∂φ2
(b) Equation 4.129 ⇒ Lz =
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102
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Problem 4.22
(a) L+ Yll = 0 (top of the ladder).
(b)
Lz Yll = lYll ⇒
∂ l
Y = lYll ,
i ∂φ l
∂Yll
= ilYll , and hence Yll = f (θ)eilφ .
∂φ
so
[Note: f (θ) is the “constant” here—it’s constant with respect to φ . . . but still can depend on θ.]
∂
df ilφ
∂ l
iφ
L+ Yl = 0 ⇒ e
+ i cot θ
f (θ)eilφ = 0, or
e + if cot θ il eilφ = 0, so
∂θ
∂φ
dθ
df
df
= l cot θf ⇒
= l cot θdθ ⇒
dθ
f
ln f = ln(sinl θ) + K ⇒ ln
f
sinl θ
df
=l
f
=K⇒
cos θ
dθ ⇒ ln f = l ln(sin θ) + constant.
sin θ
f
= constant ⇒ f (θ) = A sinl θ.
sinl θ
Yll (θ, φ) = A(eiφ sin θ)l .
(c)
1=A
2l
2
sin θ sin θ dθ dφ = 2πA
π
sin(2l+1) θ dθ = 2πA2 2
2
0
= 4πA2
(2 · 4 · 6 · · · · · 2l)2
(2l l!)2
= 4πA2
,
1 · 2 · 3 · 4 · 5 · · · · · (2l + 1)
(2l + 1)!
(2 · 4 · 6 · · · · · (2l))
1 · 3 · 5 · · · · · (2l + 1)
A=
so
1
2l+1 l!
(2l + 1)!
,
π
the same as Problem 4.5, except for an overall factor of (−1)l , which is arbitrary anyway.
Problem 4.23
L+ Y21
= e
iφ
∂
∂
+ i cot θ
∂θ
∂θ
−
15
sin θ cos θeiφ
8π
=−
15 iφ iφ
cos θ
e (cos2 θ − sin2 θ) + i
e
sin θ cos θ ieiφ
8π
sin θ
=−
15 2iφ
e
cos2 θ − sin2 θ − cos2 θ =
8π
√
= 2 · 3 − 1 · 2 Y22 = 2Y22 .
∴ Y22 =
15
eiφ sin θ
8π
1
4
15 iφ
e sin θ
2π
2
2
.
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
103
Problem 4.24
(a)
H=2
1
mv 2
2
= mv 2 ;
a
|L| = 2 mv = amv,
2
so L2 = a2 m2 v 2 ,
and hence H =
L2
.
ma2
But we know the eigenvalues of L2 : 2 l(l + 1); or, since we usually label energies with n:
En =
2 n(n + 1)
(n = 0, 1, 2, . . . ).
ma2
(b) ψnm (θ, φ) = Ynm (θ, φ), the ordinary spherical harmonics. The degeneracy of the nth energy level is the
number of m-values for given n: 2n + 1.
Problem 4.25
(1.6 × 10−19 )2
= 2.81 × 10−15 m.
4π(8.85 × 10−12 )(9.11 × 10−31 )(3.0 × 108 )2
1
2 2 v 2
L = = Iω =
mr
= mrv so
2
5
r
5
rc =
v=
5
(5)(1.055 × 10−34 )
=
= 5.15 × 1010 m/s.
4mr
(4)(9.11 × 10−31 )(2.81 × 10−15 )
Since the speed of light is 3 × 108 m/s, a point on the equator would be going more than 100 times the speed
of light. Nope : This doesn’t look like a very realistic model for spin.
Problem 4.26
(a)
2
[Sx , Sy ] = Sx Sy − Sy Sx =
4
2
i 0
−i
=
−
0 −i
0
4
(b)
σx σ x =
σx σy =
σy σ x =
0 1
0 −i
0 −i
0 1
−
1 0
i 0
i 0
1 0
2
2i 0
1 0
0
=
= i
= iSz . i
4 0 −2i
2 0 −1
10
01
= 1 = σy σy = σz σz ,
i 0
= iσz ;
0 −i
σy σz =
−i 0
0 i
so σj σj = 1 for j = x, y, or z.
0 i
i 0
= −iσz ;
σz σy =
= iσx ;
σz σx =
0 −i
= −iσx ;
−i 0
0 1
−1 0
σx σz =
= iσy ;
0 −1
1 0
= −iσy .
Equation 4.153 packages all this in a single formula. c
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104
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Problem 4.27
(a)
χ† χ = |A|2 (9 + 16) = 25|A|2 = 1 ⇒ A = 1/5.
(b)
Sx = χ† Sx χ =
1 −3i 4
25 2
Sy = χ† Sy χ =
1 −3i 4
25 2
1 −3i 4
Sz = χ Sz χ =
25 2
†
0 1
1 0
3i
4
−3i 4
=
=
(12i + 12i) = 0.
4
3i
50
50
12
0 −i
3i
−4i
−3i 4
=
=
(−12 − 12) = − .
i 0
4
−3
50
50
25
7
3i
3i
−3i
4
=
=
(9 − 16) = − .
4
−4
50
50
50
1 0
0 −1
(c)
Sx2 = Sy2 = Sz2 =
2
2
(always, for spin 1/2), so σS2 x = Sx2 − Sx 2 =
− 0, σSx = .
4
4
2
σS2 y = Sy2 − Sy 2 =
−
4
σS2 z
=
Sz2 2
− Sz =
−
4
2
12
25
2
7
50
2 =
2
7
49 2
(625 − 576) =
, σSy =
.
2500
2500
50
2
2 =
2
12
576 2
(625 − 49) =
, σSz =
.
2500
2500
25
(d)
σSx σSy =
7 ? 7
· ≥ |Sz | = · 2 50
2
2 50
σSy σSz =
7
12 ? · ≥ |Sx | = 0
50
25
2
σSz σSx =
12
12
? · ≥ |Sy | = · 25
2
2
2 25
(right at the uncertainty limit). (trivial). (right at the uncertainty limit). c
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
105
Problem 4.28
∗ ∗
a b
Sx =
2
Sy =
=
Sz =
∗ ∗
a b
2
∗ ∗ b
a
a b
=
= (a∗ b + b∗ a) = Re(ab∗ ).
b
a
2
2
01
10
∗ ∗ −ib
0 −i
a
a b
=
i 0
b
ia
2
(−ia∗ b + iab∗ ) = i(ab∗ − a∗ b) = − Im(ab∗ ).
2
2
∗ ∗
a b
2
S2x
2
=
4
S2z
2
=
4
01
10
2 1 0
2
01
=
=
;
10
4 01
4
1 0
0 −1
∗ ∗
a
a
a b
=
= (a∗ a − b∗ b) = (|a|2 − |b|2 ).
b
−b
2
2
2
1 0
0 −1
1 0
0 −1
Sx2 + Sy2 + Sz2 =
=
2
;
4
S2y
2
=
4
2
0 −i
0 −i
=
;
i 0
i 0
4
so Sx2 = Sy2 = Sz2 =
2
.
4
3 2 ?
1 1
3
= s(s + 1)2 = ( + 1)2 = 2 = S 2 . 4
2 2
4
Problem 4.29
(a)
Sy =
2
2
0 −i
;
i 0
−λ −i/2
2
2
i/2 −λ = λ − 4 ⇒ λ = ± 2 (of course).
α
0 −i
α
=±
⇒ −iβ = ±α;
i 0
β
2 β
(y)
χ+
1
=√
2
1
;
i
(y)
χ−
1
=√
2
1
|α|2 + |β|2 = 1 ⇒ |α|2 + |α|2 = 1 ⇒ α = √ .
2
1
.
−i
(b)
†
1
1
a
(y)
c+ = χ+
χ = √ 1 −i
= √ (a − ib);
b
2
2
†
1
1
a
(y)
= √ (a + ib);
c− = χ−
χ= √ 1 i
b
2
2
1
+ , with probability |a − ib|2 .
2
2
1
− , with probability |a + ib|2 .
2
2
1 ∗
[(a + ib∗ )(a − ib) + (a∗ − ib∗ )(a + ib)]
2
1 2
=
|a| − ia∗ b + iab∗ + |b|2 + |a|2 + ia∗ b − iab∗ + |b|2 = |a|2 + |b|2 = 1. 2
P+ + P − =
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106
(c)
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
2
, with probability 1.
4
Problem 4.30
Sr = S · r̂ = Sx sin θ cos φ + Sy sin θ sin φ + Sz cos θ
0
sin θ cos φ
0
−i sin θ sin φ
cos θ
0
=
+
+
sin θ cos φ
0
i sin θ sin φ
0
0 − cos θ
2
cos θ
sin θ(cos φ − i sin φ)
cos θ e−iφ sin θ
=
.
=
− cos θ
2 sin θ(cos φ + i sin φ)
2 eiφ sin θ − cos θ
( cos θ − λ) e−iφ sin θ 2
2
2
2
2
2
θ
+
λ
−
cos
sin2 θ = 0 ⇒
=
−
eiφ sin θ (− cos θ − λ)
4
4
2
2
λ2 =
2
2
2
(sin2 θ + cos2 θ) =
⇒ λ = ± (of course).
4
4
2
cos θ e−iφ sin θ
iφ
e sin θ − cos θ
α
α
⇒ α cos θ + βe−iφ sin θ = ±α;
=±
β
2 β
β = eiφ
(±1 − cos θ)
α.
sin θ
sin(θ/2)
α. Normalizing:
cos(θ/2)
1
sin2 (θ/2) 2
θ
θ
cos(θ/2)
(r)
iφ
1 = |α|2 +|β|2 = |α2 |+ 2
sin
χ
=
.
|α| = |α|2
⇒
α
=
cos
,
β
=
e
,
+
eiφ sin(θ/2)
cos (θ/2)
cos2 (θ/2)
2
2
Upper sign: Use 1 − cos θ = 2 sin2 θ2 , sin θ = 2 sin θ2 cos θ2 . Then β = eiφ
θ
cos(θ/2)
1
cos2 (θ/2) 2
Lower sign: Use 1 + cos θ = 2 cos2 , β = −eiφ
α; 1 = |α|2 +
|α| = |α|2 2
.
2
sin(θ/2)
sin2 (θ/2)
sin (θ/2)
−iφ
sin(θ/2)
e
(r)
−iφ
Pick α = e
.
sin(θ/2); then β = − cos(θ/2), and χ− =
− cos(θ/2)
Problem 4.31
1
0
0
There are three states: χ+ = 0 , χ0 = 1 , χ− = 0 .
0
0
1
Sz χ+ = χ+ , Sz χ0 = 0, Sz χ− = −χ− ,
1 0 0
⇒ Sz = 0 0 0 .
0 0 −1
From Eq. 4.136:
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
107
√
√
0 1 0
0 0 0
√
√
S+ χ+ = 0,√
S+ χ0 = √2χ+ , S+ χ− = 2χ0
⇒ S+ = 2 0 0 1 , S− = 2 1 0 0 .
S− χ+ = 2χ0 , S− χ0 = 2χ− , S− χ− = 0
0 0 0
0 1 0
010
1
1 0 1 ,
Sx = (S+ + S− ) = √
2
2 010
0 −1 0
i
1
1 0 −1 .
Sy = (S+ − S− ) = √
2i
2 0 1 0
Problem 4.32
(a) Using Eqs. 4.151 and 4.163:
1
(x)
(x)†
c+ = χ+ χ = √ 1 1
2
cos α2 eiγB0 t/2
sin α2 e−iγB0 t/2
1 α
α
= √ cos eiγB0 t/2 + sin e−iγB0 t/2 .
2
2
2
α
α
1
α
α
(x)
(x)
P+ (t) = |c+ |2 =
cos e−iγB0 t/2 + sin eiγB0 t/2 cos eiγB0 t/2 + sin e−iγB0 t/2
2
2
2
2
2
1 2α
α
α iγB0 t
2 α
−iγB0 t
=
cos
+ sin
+ sin cos
e
+e
2
2
2
2
2
1
1
α
α
=
1 + 2 sin cos cos(γB0 t) =
[1 + sin α cos(γB0 t)] .
2
2
2
2
1 1
(y)
(b) From Problem 4.29(a): χ+ = √
.
2 i
1 1
α
α
cos α2 eiγB0 t/2
(y)
(y)†
c+ = χ+ χ = √ 1 −i
= √ cos eiγB0 t/2 − i sin e−iγB0 t/2 ;
α iγB0 t/2
sin 2 e
2
2
2
2
α
1
α
α
α
(y)
(y)
P+ (t) = |c+ |2 =
cos e−iγB0 t/2 + i sin eiγB0 t/2 cos eiγB0 t/2 − i sin e−iγB0 t/2
2
2
2
2
2
1 2α
α
α iγB0 t
2 α
−iγB0 t
=
−e
cos
+ sin
+ i sin cos
e
2
2
2
2
2
1
1
α
α
=
1 − 2 sin cos sin(γB0 t) =
[1 − sin α sin(γB0 t)] .
2
2
2
2
(c)
(z)
χ+ =
1
;
0
(z)
c+ = 1 0
cos α2 eiγB0 t/2
sin α2 e−iγB0 t/2
= cos
α iγB0 t/2
;
e
2
(z)
(z)
P+ (t) = |c+ |2 = cos2
α
.
2
Problem 4.33
(a)
γB0 1 0
H = −γB · S = −γB0 cos ωt Sz = −
cos ωt
.
0 −1
2
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108
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
(b)
χ(t) =
i
1
α(t)
, with α(0) = β(0) = √ .
β(t)
2
∂χ
γB0 γB0 α̇
1 0
α
α
= i
= Hχ = −
cos ωt
=−
cos ωt
.
0 −1
β
−β
β̇
∂t
2
2
α̇ = i
γB0
2
cos ωt α ⇒
α(t) = Aei(γB0 /2ω) sin ωt ;
β̇ = −i
γB0
2
dα
=i
α
γB0
2
cos ωt dt ⇒ ln α =
1
α(0) = A = √ ,
2
iγB0 sin ωt
+ constant.
2
ω
1
so α(t) = √ ei(γB0 /2ω) sin ωt .
2
1
cos ωt β ⇒ β(t) = √ e−i(γB0 /2ω) sin ωt .
2
1
χ(t) = √
2
ei(γB0 /2ω) sin ωt
.
e−i(γB0 /2ω) sin ωt
(c)
(x)
(x)†
c− = χ− χ =
= i sin
i(γB /2ω) sin ωt 0
1 i(γB0 /2ω) sin ωt
1
e
− e−i(γB0 /2ω) sin ωt
(1 − 1) −i(γB0 /2ω) sin ωt =
e
e
2
2
γB0
γB0
(x)
(x)
sin ωt . P− (t) = |c− |2 = sin2
sin ωt .
2ω
2ω
(d) The argument of sin2 must reach π/2 (so P = 1) ⇒
γB0
πω
π
= , or B0 =
.
2ω
2
γ
Problem 4.34
(a)
1
(1)
(2) 1
S− |1 0 = (S− + S− ) √ (↑↓ + ↓↑) = √ [(S− ↑) ↓ +(S− ↓) ↑ + ↑ (S− ↓)+ ↓ (S− ↑)] .
2
2
√
√
1
But S− ↑= ↓, S− ↓= 0 (Eq. 4.143), so S− |10 = √ [ ↓↓ +0 + 0 + ↓↓] = 2 ↓↓= 2|1 − 1.
2
(b)
1
(1)
(2) 1
S± |0 0 = (S± + S± ) √ (↑↓ − ↓↑) = √ [(S± ↑) ↓ −(S± ↓) ↑ + ↑ (S± ↓)− ↓ (S± ↑)] .
2
2
1
1
S+ |0 0 = √ (0 − ↑↑ + ↑↑ −0) = 0; S− |0 0 = √ ( ↓↓ −0 + 0 − ↓↓) = 0. 2
2
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
109
(c)
S 2 |1 1 = (S (1) )2 + (S (2) )2 + 2S(1) · S(2) ↑↑
= (S 2 ↑) ↑ + ↑ (S 2 ↑) + 2 [(Sx ↑)(Sx ↑) + (Sy ↑)(Sy ↑) + (Sz ↑)(Sz ↑)]
3 2
3
i i
↑↑ + 2 ↑↑ +2
↓ ↓+ ↓
↓+ ↑ ↑
4
4
2 2
2
2
2 2
2 3
= 2 ↑↑ +2
↑↑ = 22 ↑↑= 22 |1 1 = (1)(1 + 1)2 |1 1, as it should be.
2
4
=
S 2 |1 − 1 = (S (1) )2 + (S (2) )2 + 2S(1) · S(2) ↓↓
32
32
↓↓ +
↓↓ +2 [(Sx ↓)(Sx ↓) + (Sy ↓)(Sy ↓) + (Sz ↓)(Sz ↓)]
4
4
3 2
i
i
= ↓↓ +2
↑
↑ + − ↑
− ↑ + − ↓
− ↓
2
2
2
2
2
2
2
3 2
2
= ↓↓ +2
↓↓= 22 ↓↓= 22 |1 − 1. 2
4
=
Problem 4.35
(a) 1/2 and 1/2 gives 1 or zero; 1/2 and 1 gives 3/2 or 1/2; 1/2 and 0 gives 1/2 only. So baryons can have
spin 3/2 or spin 1/2 (and the latter can be acheived in two distinct ways). [Incidentally, the lightest
baryons do carry spin 1/2 (proton, neutron, etc.) or 3/2 (∆, Ω− , etc.); heavier baryons can have higher
total spin, but this is because the quarks have orbital angular momentum as well.]
(b) 1/2 and 1/2 gives spin 1 or spin 0. [Again, these are the observed spins for the lightest mesons: π’s and
K’s have spin 0, ρ’s and ω’s have spin 1.]
Problem 4.36
(a) From the 2 × 1 Clebsch-Gordan table we get
|3 1 =
1
|2 2|1 − 1 +
15
8
|2 1|1 0 +
15
6
|2 0|1 1,
15
so you might get 2 (probability 1/15), (probability 8/15), or (probability 6/15).
(b) From the 1× 12 table: |1 0| 12 − 12 =
2 3
3 |2
− 12 +
15/42 and 3/42 , respectively. Thus you get
1 1
3 |2
− 12 . So the total is 3/2 or 1/2, with l(l+1)2 =
15 2
3
(probability 2/3), or 2 (probability 1/3).
4
4
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110
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Problem 4.37
(1)
2
2
(1)
(1)
(1)
Using Eq. 4.179: [S 2 , Sz ] = [S (1) , Sz ] + [S (2) , Sz ] + 2[S(1) · S(2) , Sz ]. But [S 2 , Sz ] = 0 (Eq. 4.102), and
anything with superscript (2) commutes with anything with superscript (1). So
+
,
[S 2 , Sz(1) ] = 2 Sx(2) [Sx(1) , Sz(1) ] + Sy(2) [Sy(1) , Sz(1) ] + Sz(2) [Sz(1) , Sz(1) ]
+
,
= 2 −iSy(1) Sx(2) + iSx(1) Sy(2) = 2i(S(1) × S(2) )z.
[S 2 , Sz(1) ] = 2i(Sx(1) Sy(2) − Sy(1) Sx(2) ), and [S 2 , S(1) ] = 2i(S(1) × S(2) ). Note that [S 2 , S(2) ] = 2i(S(2) × S(1) ) =
−2i(S(1) × S(2) ), so [S 2 , (S(1) + S(2) )] = 0.]
Problem 4.38
(a)
−
2
2m
∂2ψ ∂2ψ ∂2ψ
+
+
∂x2
∂y 2
∂z 2
1
+ mω 2 x2 + y 2 + z 2 ψ = Eψ.
2
Let ψ(x, y, z) = X(x)Y (y)Z(z); plug it in, divide by XY Z, and collect terms:
−
2 1 d 2 X
1
+ mω 2 x2
2m X dx2
2
2 1 d 2 Y
2 1 d2 Z
1
1
2 2
2 2
mω
mω
+ −
+
−
= E.
+
y
+
z
2m Y dy 2
2
2m Z dz 2
2
The first term is a function only of x, the second only of y, and the third only of z. So each is a constant
(call the constants Ex , Ey , Ez , with Ex + Ey + Ez = E). Thus:
−
2 d 2 X
1
2 d 2 Y
1
2 d 2 Z
1
2 2
2 2
+
x
X
=
E
X;
−
+
y
Y
=
E
Y
;
−
+ mω 2 z 2 Z = Ez Z.
mω
mω
x
y
2
2
2
2m dx
2
2m dy
2
2m dz
2
Each of these is simply the one-dimensional harmonic oscillator (Eq. 2.44). We know the allowed energies
(Eq. 2.61):
Ex = (nx + 12 )ω; Ey = (ny + 12 )ω; Ez = (nz + 12 )ω; where nx , ny , nz = 0, 1, 2, 3, . . . .
So E = (nx + ny + ny + 32 )ω = (n + 32 )ω,
with n ≡ nx + ny + nz .
(b) The question is: “How many ways can we add three non-negative integers to get sum n?”
If nx = n, then ny = nz = 0; one way.
If nx = n − 1, then ny = 0, nz = 1, or else ny = 1, nz = 0; two ways.
If nx = n − 2, then ny = 0, nz = 2, or ny = 1, nz = 1, or ny = 2, nz = 0; three ways.
And so on. Evidently d(n) = 1 + 2 + 3 + · · · + (n + 1) =
(n + 1)(n + 2)
.
2
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
111
Problem 4.39
Eq. 4.37: −
2 d 2 u
1
2 l(l + 1)
2 2
u = Eu.
mω
+
r
+
2m dr2
2
2m r2
Following Eq. 2.71, let ξ ≡
or
1
2 mω l(l + 1)
mω
2
2 mω d2 u
u = Eu,
+ mω 2
r. Then −
ξ +
2
2m dξ
2
mω
2m ξ2
d2 u
l(l + 1)
2E
= ξ2 +
− K u, where K ≡
(as in Eq. 2.73).
2
2
dξ
ξ
ω
At large ξ,
At small ξ,
2
d2 u
≈ ξ 2 u, and u ∼ ( )e−ξ /2 (see Eq. 2.77).
dξ 2
d2 u
l(l + 1)
≈
u, and u ∼ ( )ξ l+1 (see Eq. 4.59).
dξ 2
ξ2
So let u(ξ) ≡ ξ l+1 e−ξ
2
/2
v(ξ).
[This defines the new function v(ξ).]
2
2
2
du
= (l + 1)ξ l e−ξ /2 v − ξ l+2 e−ξ /2 v + ξ l+1 e−ξ /2 v .
dξ
2
2
2
2
d2 u
= l(l + 1)ξ l−1 e−ξ /2 v − (l + 1)ξ l+1 e−ξ /2 v + (l + 1)ξ l e−ξ /2 v − (l + 2)ξ l+1 e−ξ /2 v
2
dξ
+ ξ l+3 e−ξ /2 v − ξ l+2 e−ξ /2 v + (l + 1)ξ l e−ξ /2 v − ξ l+2 e−ξ /2 v + ξ l+1 e−ξ /2 v
✭2✭
2
2✘
2
✘
−ξ
/2
✭✭
✭l−1
✘e✘
=✭
l(l✭
+✭
1)ξ
e−ξ /2 v − (2l + 3)ξ l+1 e−ξ /2 v + ✘
ξ l+3
v + 2(l + 1)ξ l e−ξ /2 v
✭2✭
2
2
2✘
2
✘
−ξ
/2
✭✭
✭l−1
✘e✘
− 2ξ l+2 e−ξ /2 v + ξ l+1 e−ξ /2 v = ✘
ξ l+3
v +✭
l(l✭
+✭
1)ξ
e−ξ /2 v − Kξ l+1 e−ξ /2 v.
2
2
2
Cancelling the indicated terms, and dividing off ξ l+1 e−ξ
l+1
v + 2v
− ξ + (K − 2l − 3) v = 0.
ξ
Let v(ξ) ≡
∞
aj ξ j , so v =
j=0
∞
∞
jaj ξ j−1 ;
v =
j=0
j(j − 1)aj ξ j−2 + 2(l + 2)
j=2
∞
2
2
/2
∞
2
, we have:
j(j − 1)aj ξ j−2 . Then
j=2
jaj ξ j−2 − 2
j=1
∞
jaj ξ j + (K − 2l − 3)
j=1
∞
aj ξ j = 0.
j=0
In the first two sums, let j → j + 2 (rename the dummy index):
∞
j=0
j
(j + 2)(j + 1)aj+2 ξ + 2(l + 1)
∞
j=0
(j + 2)aj+2 ξ − 2
j
∞
j=0
jaj ξ + (K − 2l − 3)
j
∞
aj ξ j = 0.
j=0
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112
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Note: the second sum should start at j = −1; to eliminate this term (there is no compensating one in ξ −1 ) we
must take a1 = 0. Combining the terms:
∞
[(j + 2)(j + 2l + 3)aj+2 + (K − 2j − 2l − 3)aj ] = 0, so aj+2 =
j=0
(2j + 2l + 3 − K)
aj .
(j + 2)(j + 2l + 3)
Since a1 = 0, this gives us a single sequence: a0 , a2 , a4 , . . . . But the series must terminate (else we get the
wrong behavior as ξ → ∞), so there occurs some maximal
(even)
number jmax such that ajmax+2 = 0. Thus
1
3
K = 2jmax + 2l + 3. But E = ωK, so E = jmax + l +
ω. Or, letting jmax + l ≡ n,
2
2
En = (n + 32 )ω, and n can be any nonnegative integer.
[Incidentally, we can also determine the degeneracy of En . Suppose n is even; then (since jmax is even)
l = 0, 2, 4, . . . , n. For each l there are (2l + 1) values for m. So
n
d(n) =
(2l + 1). Let j = l/2;
l=0,2,4,...
( n2 )( n2 +
=4
then d(n) =
n/2
(4j + 1) = 4
j=0
1)
2
+ ( n2 + 1) = ( n2 + 1)(n + 1) =
n/2
j+
j=0
n/2
1
j=0
(n + 1)(n + 2)
, as before (Problem 4.38(b)).]
2
Problem 4.40
(a)
d
i
r · p = [H, r · p].
dt
[H, r · p] =
3
[H, ri pi ] =
i=1
[p2 , ri ] =
3
([H, ri ]pi + ri [H, pi ]) =
i=1
[pj pj , ri ] =
j=1
[V, pi ] = i
3
3
j=1
(pj [pj , ri ] + [pj , ri ]pj ) =
3 1 2
[p , ri ]pi + ri [V, pi ] .
2m
i=1
3
[pj (−iδij ) + (−iδij )pj ] = −2ipi .
j=1
3
1
∂V
∂V
(Problem 3.13(c)). [H, r · p] =
(−2i)pi pi + ri i
∂ri
2m
∂ri
i=1
2
p
= i − + r · ∇V .
m
For stationary states
d
p2
r · p = − r · ∇V = 2T − r · ∇V .
dt
m
d
r · p = 0, so 2T = r · ∇V . QED
dt
(b)
V (r) = −
e2 1
e2 1
e2 1
r̂
⇒
r
·
∇V
=
⇒ ∇V =
= −V.
4π90 r
4π90 r2
4π90 r
So 2T = −V .
But T = V = En , so T − 2T = En , or T = −En ; V = 2En . QED
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
113
(c)
V =
1
mω 2 r2 ⇒ ∇V = mω 2 r r̂ ⇒ r · ∇V = mω 2 r2 = 2V.
2
But T + V = En , so T = V =
So 2T = 2V , or T = V .
1
En . QED
2
Problem 4.41
i ∇Ψ · ∇Ψ∗ + Ψ(∇2 Ψ∗ ) − ∇Ψ∗ · ∇Ψ − Ψ∗ (∇2 Ψ) =
2m
∂Ψ
2 2
But the Schrödinger equation says i
=−
∇ Ψ + V Ψ,
∂t 2m
∂Ψ
2m
2m
∂Ψ∗
∇2 Ψ = 2 V Ψ − i
, ∇2 Ψ∗ = 2 V Ψ∗ + i
.
∂t
∂t
(a) ∇ · J =
i Ψ(∇2 Ψ∗ ) − Ψ∗ (∇2 Ψ) .
2m
so
Therefore
i 2m
∂Ψ
∂Ψ∗
∗
∗
−
Ψ
Ψ
V
Ψ
V
Ψ
−
i
+
i
2m 2
∂t
∂t
∗
i
∂Ψ
∂Ψ
∂
∂
= i Ψ
+ Ψ∗
= − (Ψ∗ Ψ) = − |Ψ|2 . ∂t
∂t
∂t
∂t
∇·J =
1 1
Ψ2 1 1 = − √
re−r/2a sin θeiφ e−iE2 t/ .
πa 8a2
∂Ψ
1 ∂Ψ
1 ∂Ψ
∇Ψ =
r̂ +
θ̂ +
φ̂, so
∂r
r ∂θ
r sin θ ∂φ
(b) From Problem 4.11(b),
In spherical coordinates,
r −r/2a
1 1 1
∇Ψ2 1 1 = − √
1−
sin θeiφ e−iE2 t/ r̂ + re−r/2a cos θeiφ e−iE2 t/ θ̂
e
2
2a
r
πa 8a
1
i
1
r
+
re−r/2a sin θ ieiφ e−iE2 t/ φ̂ = 1 −
r̂ + cot θ θ̂ +
φ̂ Ψ2 1 1 .
r sin θ
2a
sin θ
r
Therefore
J =
=
i r i
r i
1
1−
r̂ + cot θ θ̂ −
φ̂ − 1 −
r̂ − cot θ θ̂ −
φ̂ |Ψ2 1 1 |2
2m
2a
sin θ
2a
sin θ
r
i (−2i)
1 1 r2 e−r/a sin2 θ
re−r/a sin θ φ̂.
|Ψ2 1 1 |2 φ̂ =
φ̂ =
4
2m r sin θ
m πa 64a
r sin θ
64πma5
2 −r/a
r
e
sin
θ
r̂
×
φ̂
,
while
r̂
×
φ̂
= −θ̂ and ẑ · θ̂ = − sin θ, so
64πma5
r × Jz =
r2 e−r/a sin2 θ, and hence
64πma5
2
2 −r/a
r
Lz = m
e
sin
θ
r2 sin θ dr dθ dφ
64πma5
∞
π
2π
4
3
4 −r/a
5
=
4!a
(2π) = ,
r e
dr
sin θ dθ
dφ =
5
64πa5 0
64πa
3
0
0
(c) Now r × J =
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114
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
as it should be, since (Eq. 4.133) Lz = m, and m = 1 for this state.
Problem 4.42
(a)
ψ=√
1
−r/a
πa3
e
1
1
√
⇒ φ(p) =
3/2
(2π)
πa3
e−ip·r/ e−r/a r2 sin θ dr dθ dφ.
With axes as suggested, p · r = pr cos θ. Doing the (trivial) φ integral:
π
∞
2π
1
2 −r/a
√
φ(p) =
r
e
e−ipr cos θ/ sin θ dθ dr.
(2πa)3/2 π 0
0
π
2
pr −ipr cos θ/ π
ipr/
− e−ipr/ =
e
e
sin
.
=
ipr
ipr
pr
0
e−ipr cos θ/ sin θ dθ =
0
1
√
2
1
φ(p) =
3/2 p
(a)
π 2
∞
re−r/a sin
pr 0
re−r/a sin
pr 0
1
2i
∞
dr.
re−r/a eipr/ dr −
∞
re−r/a e−ipr/ dr
1
1
1
1
(2ip/a)2
=
−
2
2 = 2i 2i (1/a − ip/)2
2
2
(1/a + ip/)
(1/a) + (p/)
dr =
=
φ(p) =
∞
0
0
(2p/)a3
2.
[1 + (ap/)2 ]
2 1 1 2pa3
1
1
=
a3/2 πp [1 + (ap/)2 ]2
π
2a
3/2
1
2.
[1 + (ap/)2 ]
(b)
|φ|2 d3 p = 4π
∞
p2 |φ|2 dp = 4π
0
∞
From math tables:
0
0
∞
p2
4 dp =
[1 + (ap/)2 ]
1
π2
2a
3 0
∞
p2
4
[1 + (ap/)2 ]
x2
π −5/2
dx =
,
m
(m + x2 )4
32
dp.
so
−5
3 3
8
π π 32 a 3 π =
;
|φ|2 d3 p =
= 1. a
32 a
32 a
π 32 a
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
115
(c)
p2 =
0
∞
p2 |φ|2 d3 p =
1
π2
2a
3
∞
4π
0
p4
4 dp.
[1 + (ap/)2 ]
π
x4
4
m−3/2 . So p2 =
dx
=
2
4
[m + x ]
32
π
2a
From math tables:
3 8
−3
π 2
= 2.
a
32 a
a
(d)
T =
1 2
1 2
2 m2
=
p =
2
2m
2m a
2m 4
e2
4π90
2
=
m
22
e2
4π90
2
= −E1 ,
which is consistent with Eq. 4.191.
Problem 4.43
(a) From Tables 4.3 and 4.7,
ψ321 =
R32 Y21
1 r 2 −r/3a
−
=
e
81 30 a3/2 a
4
√
15
1
1
iφ
= −√
r2 e−r/3a sin θ cos θeiφ .
sin θ cos θe
8π
π 81a7/2
(b)
1
1
2
4 −2r/3a
2
r
|ψ| d r =
e
sin
θ
cos
θ
r2 sin θ dr dθ dφ
π (81)2 a7
∞
π
1
6 −2r/3a
=
2π
r e
dr
(1 − cos2 θ) cos2 θ sin θ dθ
π(81)2 a7
0
0
π
7
2
3a
cos3 θ cos5 θ =
6!
−
+
(81)2 a7
2
3
5
0
2 3
=
2
38 a7
6·5·4·3·2
37 a7 2 2
3·5 4
−
=
·
= 1. 27 3 5
4
15
(c)
2
∞
4
1 1
rs+6 e−2r/3a dr
81
30 a7 0
0
s+7
5
3
8
1
3a
3a
(s + 6)! 3a
=
(s + 6)!
= (s + 6)!
=
.
15(81)2 a7
2
2
720
6!
2
rs =
∞
rs |R32 |2 r2 dr =
Finite for s > −7 .
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116
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Problem 4.44
(a) From Tables 4.3 and 4.7,
ψ433 =
(b)
R43 Y33
1
1 r 3 −r/4a
√
=
e
768 35 a3/2 a
'
35
sin3 θ cos θe3iφ
64π
−
(
= −
1
√
r3 e−r/4a sin3 θe3iφ .
6144 πa9/2
1
6
6 −r/2a
r = r|ψ| d r =
r
r
e
sin
θ
r2 sin θ dr dθ dφ
(6144)2 πa9
∞
π
2π
1
7
9 −r/2a
=
r
e
dr
sin
θ
dθ
dφ
(6144)2 πa9 0
0
0
1
2·4·6
=
9!(2a)10 2
(2π) = 18a.
2
9
(6144) πa
3·5·7
2 3
(c) Using Eq. 4.133: L2x + L2y = L2 − L2z = 4(5)2 − (3)2 = 112 , with probability 1.
Problem 4.45
(a)
b
a 2 −2r/a a3 −2r/a
2r
4π
4
−2r/a 2
− −1
P = |ψ| d r =
e
r dr = 3 − r e
+ e
πa3 0
a
2
4
a
b
2
2
2r 2r
2b
b
=− 1+
+ 2 e−2r/a = 1 − 1 +
+ 2 2 e−2b/a .
a
a
a
a
0
2
3
b
0
(b)
1 2 −5
P =1− 1+9+ 9 e ≈1− 1+9+
2
92
93
93
92
≈1−1+9−
+
− 9 + 92 −
−
2
6
2
2
3
3
1 2b
4 b
=
=
.
6 a
3 a
92
1 2
93
1−9+
9
−
2
2
3!
1
93
1
1
3
+
=9
− +
2
6 2 2
(c)
|ψ(0)|2 =
1
1
4
4
⇒ P ≈ πb3 3 =
πa3
3
πa
3
3
b
. a
(d)
4
P =
3
10−15
0.5 × 10−10
3
=
4
2 × 10−5
3
3
=
4
32
· 8 × 10−15 =
× 10−15 = 1.07 × 10−14 .
3
3
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
117
Problem 4.46
(a) Equation 4.75 ⇒ Rn(n−1) =
So
1 n −ρ
ρ e v(ρ), where
r
v(ρ) = c0 , and hence
∞
1=
r
;
na
Eq. 4.76 ⇒ c1 =
Rn(n−1) = Nn rn−1 e−r/na , where
∞
|R| r dr = (Nn )
2 2
ρ≡
2
0
2n −2r/na
r e
0
2(n − n)
c0 = 0.
(1)(2n)
c0
.
(na)n
Nn ≡
2n+1
n 2
na
2
dr = (Nn ) (2n)!
; Nn =
.
2
na
na(2n)!
2
(b)
r =
l
∞
|R| r
2 l+2
0
r =
r =
∞
r2n+l e−2r/na dr.
0
2
na
2
dr =
Nn2
2n+1
2
na
2n+2
1
1
na
= n+
(2n + 1)!
na.
(2n)!
2
2
2n+1
2
2n+3
1
1
na
na
= (2n + 2)(2n + 1)
= n+
(2n + 2)!
(n + 1)(na)2 .
(2n)!
2
2
2
(c)
σr2
2
1
1
2
= r − r =
n+
(na)2
(n + 1)(na) − n +
2
2
1
1
r
1
=
σr = √
n+
(na)2 =
r2 ;
.
2
2
2(n + 1/2)
2n + 1
2
2
R10
a
Maxima occur at:
R 32
r
6a
R 26 25
r
650a
r
dRn,n−1
1 n−1 −r/na
e
= 0 ⇒ r = na(n − 1).
= 0 ⇒ (n − 1)rn−2 e−r/na −
r
dr
na
Problem 4.47
Here are a couple of examples: {32, 28} and {224,56}; {221, 119} and {119, 91}. For further discussion see
D. Wyss and W. Wyss, Foundations of Physics 23, 465 (1993).
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118
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Problem 4.48
(a) Using Eqs. 3.64 and 4.122: [A, B] = [x2 , Lz ] = x[x, Lz ] + [x, Lz ]x = x(−iy) + (−iy)x = −2ixy.
2 2
Equation 3.62 ⇒ σA
σB ≥
1
(−2i)xy
2i
(b) Equation 4.113 ⇒ B = Lz = m;
2
= 2 xy2 ⇒ σA σB ≥ |xy|.
B 2 = L2z = m2 2 ;
σB = m2 2 − m2 2 = 0.
so
(c) Since the left side of the uncertainty principle is zero, the right side must also be: xy = 0, for eigenstates
of Lz .
Problem 4.49
(a) 1 = |A|2 (1 + 4 + 4) = 9|A|2 ;
(b)
A = 1/3.
5 4
5
4
, with probability ; − , with probability . Sz =
+
2
9
2
9
92 9
−
2
=
.
18
(c) From Eq. 4.151,
(x)
c+
(x)
c−
13
1 1
9+4
1
3 − 2i
1 − 2i
(x)
=
.
=
χ= √ 1 1
= √ (1 − 2i + 2) = √ ; |c+ |2 =
2
3 2
9
·
2
18
3 2
3 2
†
1 1
1+4
1
1 + 2i
5
1 − 2i
(x)
(x)
= χ−
χ = √ 1 −1
= √ (1 − 2i − 2) = − √ ; |c− |2 =
=
.
2
3 2
9
·
2
18
3 2
3 2
(x)
χ+
†
13
5
, with probability
; − , with probability
.
2
18
2
18
13 5
Sx =
+
18 2 18
−
2
=
2
.
9
(d) From Problem 4.29(a),
†
17
1 1
1 + 16
1
1 − 4i
1 − 2i
(y)
(y)
(y)
c+ = χ+
=
.
χ = √ 1 −i
= √ (1 − 2i − 2i) = √ ; |c+ |2 =
2
3 2
9·2
18
3 2
3 2
†
1
1 1
1
1
1
1 − 2i
(y)
(y)
(y)
c− = χ−
=
.
χ= √ 1 i
= √ (1 − 2i + 2i) = √ ; |c− |2 =
2
3 2
9·2
18
3 2
3 2
17
1
, with probability
; − , with probability
.
2
18
2
18
1
17 +
Sy =
18 2 18
−
2
=
4
.
9
Problem 4.50
(1)
(1)
(2)
We may as well choose axes so that â lies along the z axis and b̂ is in the xz plane. Then Sa = Sz , and Sb
(2)
(2)
(1) (2)
cos θ Sz + sin θ Sx . 0 0|Sa Sb |0 0 is to be calculated.
=
1 (2)
Sa(1) Sb |0 0 = √ Sz(1) (cos θ Sz(2) + sin θ Sx(2) ) (↑↓ − ↓↑)
2
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
119
1
= √ [(Sz ↑)(cos θ Sz ↓ + sin θ Sx ↓) − (Sz ↓)(cos θ Sz ↑ + sin θ Sx ↑)]
2
1
=√
2
=
↑
2
cos θ − ↓ + sin θ
↑ − − ↓ cos θ
↑ + sin θ
↓
(using Eq. 4.145)
2
2
2
2
2
2
1
2
1
1
cos θ √ (− ↑↓ + ↓↑) + sin θ √ (↑↑ + ↓↓) =
− cos θ|0 0 + sin θ √ (|1 1 + |1 − 1) .
4
4
2
2
2
(2)
(2)
so Sa(1) Sb = 0 0|Sa(1) Sb |0 0 =
2
1
2
0 0| − cos θ|0 0 + sin θ √ (|1 1 + |1 − 1) = − cos θ0 0|0 0
4
4
2
(2)
(by orthogonality), and hence Sa(1) Sb = −
2
cos θ.
4
QED
Problem 4.51
(a) First note from Eqs. 4.136 and 4.144 that
1
[S+ |s m + S− |s m]
2
=
s(s + 1) − m(m + 1)|s m + 1 + s(s + 1) − m(m − 1)|s m − 1
2
Sx |s m =
1
[S+ |s m − S− |s m]
2i
=
s(s + 1) − m(m + 1)|s m + 1 − s(s + 1) − m(m − 1)|s m − 1
2i
Sy |s m =
Now, using Eqs. 4.179 and 4.147:
(1)
(2)
(1)
(2)
(1)
(2)
S 2 |s m = (S (1) )2 + (S (2) )2 + 2(Sx Sx + Sy Sy + Sz Sz )
= A S 2 | 12 12 |s2 m − 12 + | 12 21 S 2 |s2 m − 12 A| 12
1
2 |S2
m − 12 + B| 12 − 12 |s2 m + 12 Sx | 12 21 +2
Sx |s2 m −
1
2
+
Sy | 12 12 Sy |s2 m −
1
2
+
Sz | 12 12 Sz |s2 m −
1
2
+B
S 2 | 12 − 12 |s2 m + 12 + | 12 − 12 S 2 |s2 m + 12 +2 Sx | 12 − 12 Sx |s2 m + 12 + Sy | 12 − 12 Sy |s2 m + 12 + Sz | 12 − 12 = A 34 2 | 12 21 |s2 m − 12 + 2 s2 (s2 + 1)| 12 12 |s2 m − 12 1
1 +2 2 | 2 − 2 2
s2 (s2 + 1) − (m − 12 )(m + 12 )|s2 m + 12 Sz |s2 m + 12 c
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120
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
1
3
3
+ s2 (s2 + 1) − (m − 2 )(m − 2 )|s2 m − 2 i
1
1 + 2 | 2 − 2 2i
s2 (s2 + 1) − (m − 12 )(m + 12 )|s2 m + 12 − s2 (s2 + 1) − (m − 12 )(m − 32 )|s2 m − 32 + 2 | 12
1
2 (m
− 12 )|s2 m − 12 +B
3 2 1
4 |2
+2
1 1 2|2 22
+ −i
| 12
2
− 12 |s2 m + 12 + 2 s2 (s2 + 1)| 12 − 12 |s2 m + 12 1 2 2i
s2 (s2 + 1) − (m +
1
2 )(m
+
3
2 )|s2
3
4
+2 B
+ s2 (s2 + 1) + m −
3
4
+
s2 (s2 + 1) − (m +
1
2 )(m
−
1
2 )|s2
m−
1
2
s2 (s2 + 1) − (m + 12 )(m + 32 )|s2 m + 32 1
1
1
− s2 (s2 + 1) − (m + 2 )(m − 2 )|s2 m − 2 +
= 2 A
m+
3
2
1
2
+ s2 (s2 + 1) − m −
+B
1
2
= 2 s(s + 1)|s m = 2 s(s + 1) A| 12
−
2
| 12
−
1
2 (m
+
1
2 )|s2
m+
1
2
s2 (s2 + 1) − m2 +
+A
1
2 |s2
| 12
1
4
1
2 |s2
m − 12 s2 (s2 + 1) − m2 +
1
4
| 12 − 12 |s2 m + 12 m − 12 + B| 12 − 12 |s2 m + 12 .
A s2 (s2 + 1) + 1 + m +B s2 (s2 + 1) − m2 + 1 = s(s + 1)A,
4
4
B s2 (s2 + 1) + 1 − m +A s2 (s2 + 1) − m2 + 1 = s(s + 1)B,
4
4
or
A s2 (s2 + 1) − s(s + 1) + 1 + m +B s2 (s2 + 1) − m2 + 1 = 0,
A(a + m) +Bb = 0
4
4
or
,
B(a − m) +Ab = 0
B s2 (s2 + 1) − s(s + 1) + 1 − m +A s2 (s2 + 1) − m2 + 1 = 0,
4
4
where a ≡ s2 (s2 + 1) − s(s + 1) + 14 , b ≡ s2 (s2 + 1) − m2 + 14 . Multiply by (a − b) and b, then subtract:
A(a2 − m2 ) + Bb(a − m) = 0; Bb(a − m) + Ab2 = 0 ⇒ A(a2 − m2 − b2 ) = 0 ⇒ a2 − b2 = m2 , or:
2
s2 (s2 + 1) − s(s + 1) + 14 − s2 (s2 + 1) + m2 − 14 = m2 ,
2
2
s2 (s2 + 1) − s(s + 1) + 14 = s22 + s2 + 14 = s2 + 12 , so
s2 (s2 + 1) − s(s + 1) +
Add
1
4
= ± s2 +
1
2
;
s(s + 1) = s2 (s2 + 1) ∓ s2 +
to both sides:
s2 + s +
So
1
4
1
4
s+
1
2
s+
1
2
= s+
1 2
2
= s2 (s2 + 1) ∓ s2 +
1
2
+
1
2
2
s2 + s2 − s2 −
=
s22 + s2 + s2 +
s2 − 12
1
= ±s2
⇒ s = ±s2 − 2 =
−s2− 12
.
1
s2 + 2
= ±(s2 + 1) ⇒ s = ±(s2 + 1) − 12 =
3
−s2 − 2
1
2
+ 14 .
1
2
+
1
2
= s22
1
2
+
1
2
= (s2 + 1)2
.
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
But s ≥ 0, so the possibilities are s = s2 ± 1/2.
121
Then:
1
1
1
a = s22 + s2 − s2 ±
s2 ± + 1 +
2
2
4
1
1
1
1
1
1 1
2
2
= s2 + s2 − s2 ∓ s2 − s2 ∓ s2 − ∓ + = ∓s2 ∓ = ∓ s2 +
.
2
2
4 2 4
2
2
2
1
1
1
1
2
b=
s2 + s2 +
s2 +
s2 + + m
− m2 =
− m2 =
s2 + − m .
4
2
2
2
+ m = ∓A s2 + 12 ∓ m = −Bb = −B
s2 + 12 + m
⇒ A s2 + 12 ∓ m = ±B s2 + 12 ± m. But |A|2 + |B|2 = 1,
∴ A ∓ s2 +
1
2
|A| + |A|
2
2
s2 +
s2 +
1
2
1
2
∓m
±m
⇒ A=
=
s2 +
1
2
−m
so
|A|2
1
1
(2s2 + 1)
s2 + ± m + s2 + ∓ m =
|A|2 .
1
2
2
(s2 + 2 ± m)
(s2 + 12 ± m)
s2 ± m +
.
2s2 + 1
1
2
s2 +
B = ±A s2 +
1
2
∓m
1
2
±m
= ±
s2 ∓ m + 12
.
2s2 + 1
(b) Here are four examples:
(i) From the 1/2 × 1/2 table (s2 = 1/2), pick s = 1 (upper signs), m = 0. Then
1
1
1
1
2 +0+ 2
2 −0+ 2
√1 ;
A=
=
B
=
= √12 .
1+1
1+1
2
(ii) From the 1 × 1/2 table (s2 = 1), pick s = 3/2 (upper signs), m = 1/2. Then
1 1
1 1
1+ 2 + 2
1− 2 + 2
2
A=
=
;
B
=
= √13 .
2+1
3
2+1
(iii) From the 3/2 × 1/2 table (s2 = 3/2), pick s = 1 (lower signs), m = −1. Then
3
3
√
1
1
3
2 +1+ 2
2 −1+ 2
A=
=
;
B
=
−
= − 12 .
3+1
2
3+1
(iv) From the 2 × 1/2 table (s2 = 2), pick s = 3/2 (lower signs), m = 1/2. Then
1 1
1 1
2− 2 + 2
2+ 2 + 2
2
3
A=
=
;
B
=
−
=
−
4+1
5
4+1
5.
These all check with the values on Table 4.8, except that the signs (which are conventional) are reversed
in (iii) and (iv). Normalization does not determine the sign of A (nor, therefore, of B).
Problem 4.52
| 32
1
0
0
0
0
1
0
3 −1
3 −3
3
0
| 32 21 =
2 = 0 ;
0 ; | 2 2 = 1 ; | 2 2 = 0 .
0
0
0
1
√
3 1
S+ | 32 32 = 0, S+ | 32 21 = 3| 32 32 , S+ | 32 −1
2 = 2| 2 2 ,
S− | 32
3
2
=
√
3| 32
1
2 ,
S− | 32
1
2
= 2| 32
−1
2 ,
S− | 32
−1
2 =
√
Equation 4.136 ⇒
S+ | 32
3| 32
−3
2 −3
2 ,
=
√
3| 32
S− | 32
−1
2 ;
−3
2 = 0.
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122
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
√
0 3
0 0
So: S+ =
0 0
0 0
−λ
√
3
0
0
√
3
−λ
2
0
0
2
−λ
√
3
0
2
0
0
0
√0
3
0
√ ; S− =
0
3
0
0
0
0
2
0
0
0
√0
3
0
√0
3
1
0
; Sx = (S+ + S− ) =
0
2
2 0
0
0
√
3
0
2
0
0
2
√0
3
0
√0 .
3
0
√
0 −λ 2 0 √ √ 3 2 √0 0
√ = −λ 2 −λ 3 − 3 0 −λ 3
√
3
0 √3 −λ
0
3 −λ
−λ
√ √ √
= −λ −λ3 + 3λ + 4λ − 3 3λ2 − 3 3 = λ4 − 7λ2 − 3λ2 + 9 = 0,
or λ4 − 10λ2 + 9 = 0;
(λ2 − 9)(λ2 − 1) = 0;
λ = ±3, ±1. So the eigenvalues of Sx are
3
1
2 , 2 ,
− 12 , − 32 .
Problem 4.53
From Eq. 4.135, Sz |s m = m|s m. Since s is fixed, here, let’s just identify the states by the value of m (which
runs from −s to +s). The matrix elements of Sz are
Snm = n|Sz |m = mn|m = mδn m .
It’s a diagonal matrix, with elements m, ranging from m = s in
right corner:
s 0
0 ···
0 s − 1 0 · · ·
Sz = 0 0 s − 2 · · ·
.. ..
.. . .
. .
.
.
0
0
0
the upper left corner to m = −s in the lower
0
0
0
..
.
.
· · · −s
From Eq. 4.136,
S± |s m = s(s + 1) − m(m ± 1) |s (m ± 1) = (S+ )nm = n|S+ |m = (s ∓ m)(s ± m + 1) |s (m ± 1).
(s − m)(s + m + 1) n|m + 1 = bm+1 δn (m+1) = bn δn (m+1) .
All nonzero elements have row index (n) one greater than the column index (m), so they are on the diagonal
just above the main diagonal (note that the indices go down, here: s, s − 1, s − 2 . . . , −s):
0
0
0
S+ = .
..
0
0
Similarly
(S− )nm = n|S− |m = 0
bs 0
0 bs−1 0
0 0 bs−2
.. ..
..
. .
.
0 0
0
0 0
0
···
···
···
..
.
0
0
0
..
.
.
· · · b−s+1
···
0
(s + m)(s − m + 1) n|m − 1 = bm δn (m−1) .
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
123
This time the nonzero elements are on the diagonal just below
0 0 0 ···
b s 0 0 · · ·
S− = 0 bs−1 0 · · ·
.. .. .. . .
. . . .
0
0
the main diagonal:
0 0
0 0
0 0
.
..
..
.
.
0 · · · b−s+1 0
1
To construct Sx = 12 (S+ + S− ) and Sy = 2i
(S+ − S− ), simply add and subtract the matrices S+ and S− :
0 bs
0
bs
0
0 ···
0
0
0
0 ···
0
0
bs 0 bs−1 0 · · ·
−bs
0
0
0
bs−1
0 ···
0
0
0 bs−1 0 bs−2 · · ·
0 −bs−1
0
0
0 bs−2 · · ·
0
0
0
0
0
0
0
0 −bs−2 0 · · ·
Sx = 0 0 bs−2 0 · · ·
; Sy =
.
2 .. ..
2i ..
..
.. . .
..
..
..
..
.. . .
..
..
. .
.
.
.
.
.
.
.
.
.
.
.
.
0 0
0
0 ···
0 b−s+1
0
0
0
0 ···
0
b−s+1
0 0
0
0 · · · b−s+1 0
0
0
0
0 · · · −b−s+1 0
Problem 4.54
L+ Ylm = l(l + 1) − m(m + 1)Ylm±1 (Eqs. 4.120 and 121). Equation 4.130 ⇒
∂
∂
iφ
e
+ i cot θ
Blm eimφ Plm (cos θ) = l(l + 1) − m(m + 1)Blm+1 ei(m+1)φ Plm+1 (cos θ).
∂θ
∂φ
d
Blm
− m cot θ Plm (cos θ) = l(l + 1) − m(m + 1)Blm+1 Plm+1 (cos θ).
dθ
cos θ
d
d
x
dx d
d
Let x ≡ cos θ; cot θ =
;
=√
=
= − sin θ
= − 1 − x2 .
2
sin θ
dθ dx
dx
dx
1 − x dθ
d
1
dP m
x
Plm (x) = −Blm √
(1 − x2 ) l + mxPlm = −Blm Plm+1
Blm − 1 − x2
− m√
dx
dx
1 − x2
1 − x2
=
l(l + 1) − m(m + 1)Blm+1 Plm+1 (x).
⇒
Blm+1 = −1
l(l + 1) − m(m + 1)
Blm .
Now l(l + 1) − m(m + 1) = (l − m)(l + m + 1), so
Blm+1 = √
−1
−1
√
Blm ⇒ Bl1 = √ √
Bl0 ;
l−m l+1+m
l l+1
Bl3 = √
Bl2 = √
−1
1
√
Bl1 = Bl0 ;
l−1 l+2
l(l − 1) (l + 1)(l + 2)
−1
−1
√
Bl0 ,
Bl2 = l−2 l+3
(l + 3)(l + 2)(l + 1)l(l − 1)(l − 2)
etc.
Evidently there is an overall sign factor (−1)m , and inside the square root the quantity is [(l + m)!/(l − m)!].
(l − m)!
Thus: Blm = (−1)m
C(l) (where C(l) ≡ Bl0 ), for m ≥ 0. For m < 0, we have
(l + m)!
−Bl0
Bl−1 = (l + 1)l
; Bl−2 = −1
(l + 2)(l − 1)
Bl−1 = 1
(l + 2)(l + 1)l(l − 1)
Bl0 ,
etc.
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Thus Bl−m = Blm , so in general: Blm = (−1)m
(l−|m|)!
(l+|m|! C(l).
Now, Problem 4.22 says:
(2l + 1)! iφ
(e sin θ)l = Bll eilφ Pll (cos θ). But
π
l
l
2l
d
1
d
(1 − x2 )l/2 d
(2l)!
l
2 l/2
2
l
Pl (x) = (1 − x )
(x − 1) =
(x2l − . . . ) = l (1 − x2 )l/2 ,
l
l
dx 2 l! dx
2 l!
dx
2 l!
Yll =
1
2l l!
(2l)!
so
(2l)!
Pll (cos θ) = l (sin θ)l . Therefore
2 l!
1
2l l!
(2l + 1)! iφ
(2l)!
1
(e sin θ)l = Bll eilφ l (sin θ)l ⇒ Bll =
π
2 l!
(2l)!
(2l + 1)!
=
π
1
C(l),
(2l)!
But Bll = (−1)l
(2l + 1)
.
π(2l)!
so C(l) = (−1)l
2l + 1
,
π
and hence
Blm = (−1)l+m
(2l + 1) (l − |m|)!
.
π
(l + |m|)!
This agrees with Eq. 4.32 except for the overall sign, which of course is purely conventional.
Problem 4.55
(a) For both terms, l = 1, so
(b) 0, P =
2 (1)(2) = 22 , P = 1.
1
2
, or , P = .
3
3
(c)
3 2
, P = 1.
4
(d)
1
2
, P = , or − , P = .
2
3
2
3
(e) From the 1 ×
1
2
Clebsch-Gordan table (or Problem 4.51):
2 1 −1
2 3 1
2 √1 3 1
2 1
√1 | 1 1 |1 0 +
√1
√1 | 1 1 +
|
|1
1
=
|
−
|
+
3 2 2
3 2 2
3
3|2
3 2 2
3
3 2 2
3 2 2
√ 15 2
3
8
1
= 2 32 | 32 12 + 13 | 12 12 . So s = 32 or 12 .
, P = , or 2 , P = .
4
9
4
9
(f )
1
2
1
, P = 1.
2
(g)
|ψ| = |R21 |
2
2
√
1 02 †
2 12 †
2 0∗ 1 †
†
1∗ 0
|Y | (χ χ+ ) +
Y1 Y1 (χ+ χ− ) +Y1 Y1 (χ− χ+ ) + |Y1 | (χ− χ− )
3 1 + 3
3
1
0
0
1
3
1 1 1 r2 −r/a 3
= |R21 |2 |Y10 |2 + 2|Y11 |2 = ·
· 3 · 2e
cos2 θ + 2
sin2 θ
3
3 24 a a
4π
8π
=
1
[Tables 4.3, 4.7]
1
1
3
r2 e−r/a ·
r2 e−r/a .
(cos2 θ + sin2 θ) =
3 · 24 · a5
4π
96πa5
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
125
(h)
1
|R21 |2
3
|Y10 |2 sin2 θ dθ dφ =
1 2 −r/a
1
1
1 2 −r/a
r e
=
r e
.
|R21 |2 = ·
3
3 24a3
72a5
Problem 4.56
(a) Equation 4.129 says Lz =
∂
i ∂φ ,
so this problem is identical to Problem 3.39, with p̂ → Lz and x → φ.
(b) First note that if M is a matrix such that M2 = 1, then
1
1
1
1
eiMφ = 1 + iMφ + (iMφ)2 + (iMφ)3 + · · · = 1 + iMφ − φ2 − iM φ3 + · · ·
2
3!
2
3!
1
1
1
1
= (1 − φ2 + φ4 − · · · ) + iM(φ − φ3 + φ5 − · · · ) = cos φ + iM sin φ.
2
4!
3!
5!
0 1
π
π
iπσx /2
2
So R = e
= cos 2 + iσx sin 2 (because σx = 1 – see Problem 4.26) = iσx = i
.
1 0
01
1
0
Thus Rχ+ = i
=i
= iχ− ; it converts “spin up” into “spin down” (with a factor of i).
10
0
1
(c)
iπσy /4
R=e
1
Rχ+ = √
2
π
π
1
1
= cos + iσy sin = √ (1 + iσy ) = √
4
4
2
2
1
1 0
0 −i
1 1
+i
= √
.
0 1
i 0
2 −1 1
1
1
1
1
(x)
=√
= √ (χ+ − χ− ) = χ− (Eq. 4.151).
0
−1
2
2
1 1
−1 1
What had been spin up along z is now spin down along x (see figure).
z
y
y'
x
z'
x'
(d) R = eiπσz = cos π + iσz sin π = −1; rotation by 360◦ changes the sign of the spinor. But since the sign
of χ is arbitrary, it doesn’t matter.
(e)
(σ · n̂)2 = (σx nx + σy ny + σz nz )(σx nx + σy ny + σz nz )
= σx2 n2x + σy2 n2y + σz2 n2z + nx ny (σx σy + σy σx ) + nx nz (σx σz + σz σx ) + ny nz (σy σz − σz σy ).
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126
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
But σx2 = σy2 = σz2 = 1, and σx σy + σy σx = σx σz + σz σx = σy σz + σz σy = 0 (Problem 4.26), so
(σ · n̂)2 = n2x + n2y + n2z = 1. So ei(σ·n̂)φ/2 = cos
φ
φ
+ i(σ · n̂) sin . QED
2
2
Problem 4.57
(a)
1
x + a2 / py ,
2
1
[p1 , p2 ] =
px − /a2 y,
2
1
[q1 , p1 ] =
x + a2 / py ,
2
1
[q2 , p2 ] =
x − a2 / py ,
2
[q1 , q2 ] =
x − a2 / py = 0, because [x, py ] = [x, x] = [py , py ] = 0.
px + /a2 y = 0, because [y, px ] = [y, y] = [px , px ] = 0.
1
1
px − /a2 y = ([x, px ] − [py , y]) = [i − (−i)] = i.
2
2
1
2
px + /a y = ([x, px ] − [py , y]) = i.
2
[See Eq. 4.10 for the canonical commutators.]
(b)
1
q12 − q22 =
x2 +
2
1 2
2
2
p1 − p2 =
p −
2 x
So
a2
(xpy + py x) +
(px y + ypx ) +
a2
a2
a2
2
a2
p2y − x2 + (xpy + py x) −
2
y −
2
p2x
− 2 (px y + ypx ) −
a
a2
a2
2
p2y =
2
y
2
2a
xpy .
=−
2
ypx .
a2
2
a2 2
(q1 − q22 ) +
(p − p22 ) = xpy − ypx = Lz .
2
2a
2 1
(c)
H=
1 2 1
a2 2
p + mω 2 x2 =
p + 2 x2 = H(x, p).
2m
2
2
2a
Then H(q1 , p1 ) =
a2 2
2
p +
q ≡ H1 ,
2 1 2a2 1
H(q2 , p2 ) =
a2 2
2
p +
q ≡ H2 ;
2 2 2a2 2
Lz = H1 − H2 .
(d) The eigenvalues of H1 are (n1 + 12 ), and those of H2 are (n2 + 12 ), so the eigenvalues of Lz are
(n1 + 12 ) − (n2 + 12 ) = (n1 − n2 ) = m, and m is an integer, because n1 and n2 are.
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
127
Problem 4.58
a
From Problem 4.28 we know that in the generic state χ =
(with |a|2 + |b|2 = 1),
b
Sz =
|a|2 − |b|2 ,
2
Sx = Re(ab∗ ),
Sy = −Im(ab∗ );
Sx2 = Sy2 =
2
.
4
Writing a = |a|eiφa , b = |b|eiφb , we have ab∗ = |a||b|ei(φa −φb ) = |a||b|eiθ , where θ ≡ φa − φb is the phase
difference between a and b. Then
Sx = Re(|a||b|eiθ ) = |a||b| cos θ,
σS2 x = Sx2 − Sx 2 =
We want σS2 x σS2 y =
Sy = −Im(|a||b|eiθ ) = −|a||b| sin θ.
2
− 2 |a|2 |b|2 cos2 θ;
4
σS2 y = Sy2 − Sy 2 =
2
− 2 |a|2 |b|2 sin2 θ.
4
2
Sz 2 , or
4
2
2
2 2
1 − 4|a|2 |b|2 cos2 θ
1 − 4|a|2 |b|2 sin2 θ =
|a|2 − |b|2
4
4
4 4
2
.
1 − 4|a|2 |b|2 cos2 θ + sin2 θ + 16|a|4 |b|4 sin2 θ cos2 θ = |a|4 − 2|a|2 |b|2 + |b|4 .
1 + 16|a|4 |b|4 sin2 θ cos2 θ = |a|4 + 2|a|2 |b|2 + |b|4 = |a|2 + |b|2
2
=1
⇒ |a|2 |b|2 sin θ cos θ = 0.
So either θ = 0 or π, in which case a and b are relatively real, or else θ = ±π/2, in which case a and b are
relatively imaginary (these two options subsume trivially the solutions a = 0 and b = 0).
Problem 4.59
(a)
Start with Eq. 3.71:
H=
dr
i
= [H, r].
dt
1
1 2
(p − qA) · (p − qA) + qϕ =
p − q(p · A + A · p) + q 2 A2 + qϕ.
2m
2m
[H, x] =
1 2
q
[p , x] −
[(p · A + A · p), x].
2m
2m
[p2 , x] = [(p2x + p2y + p2z ), x] = [p2x , x] = px [px , x] + [px , x]px = px (−i) + (−i)px = −2ipx .
[p · A, x] = [(px Ax + py Ay + pz Az ), x] = [px Ax , x] = px [Ax , x] + [px , x]Ax = −iAx .
[A · p, x] = [(Ax px + Ay py + Az pz ), x] = [Ax px , x] = Ax [px , x] + [Ax , x]px = −iAx .
[H, x] =
1
q
i
(−2ipx ) −
(−2iAx ) = − (px − qAx );
2m
2m
m
[H, r] = −
i
(p − qA).
m
dr
1
= (p − qA). QED
dt
m
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128
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
(b)
We define the operator v ≡
H=
1
(p − qA);
m
dv
i
∂v
= [H, v] + ;
dt
∂t
1
m
mv 2 + qϕ ⇒ [H, v] = [v 2 , v] + q[ϕ, v];
2
2
[ϕ, px ] = i
∂ϕ
∂x
[ϕ, v] =
∂v
q ∂A
=−
.
∂t
m ∂t
1
[ϕ, p].
m
(Eq. 3.65), so [ϕ, p] = i∇ϕ, and [ϕ, v] =
i
∇ϕ.
m
[v 2 , vx ] = [(vx2 + vy2 + vz2 ), vx ] = [vy2 , vx ] + [vz2 , vx ] = vy [vy , vx ] + [vy , vx ]vy + vz [vz , vx ] + [vz , vx ]vz .
1
q
[(py − qAy ), (px − qAx )] = − 2 ([Ay , px ] + [py , Ax ])
2
m
m
q
iq
∂Ay
∂Ax
iq
= − 2 i
− i
= − 2 (∇ × A)z = − 2 Bz .
m
∂x
∂y
m
m
1
q
[vz , vx ] = 2 [(pz − qAz ), (px − qAx )] = − 2 ([Az , px ] + [pz , Ax ])
m
m
q
iq
∂Ax
iq
∂Az
= − 2 i
− i
= 2 (∇ × A)y = 2 By .
m
∂x
∂y
m
m
[vy , vx ] =
∴ [v 2 , vx ] =
[v 2 , v] =
iq
iq
(−vy Bz − Bz vy + vz By + By vz ) = 2 [−(v × B)x + (B × v)x ] .
m2
m
iq
[(B × v) − (v × B)] . Putting all this together:
m2
dv
i
=
dt
0
qi
m iq
(B × v − v × B) +
∇ϕ
2 m2
m
1
−
q ∂A
.
m ∂t
0
1
dv
q
∂A
q
[] m
= (v × B) − (B × v) + q −∇ϕ −
= (v × B − B × v) + qE. Or, since
dt
2
dt
2
v×B−B×v =
1
1
q
[(p − qA) × B − B × (p − qA)] =
[p × B − B × p] −
[A × B − B × A] .
m
m
m
[Note: p does not commute with B, so the order does matter in the first term. But A commutes with B,
so B × A = −A × B in the second.]
m
dv
q
q2
= qE +
p × B − B × p − A × B. QED
dt
2m
m
(c) Go back to Eq. , and use E = E, v × B = v × B; B × v = B × v = −v × B. Then
m
dv
= qv × B + qE. QED
dt
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
129
Problem 4.60
(a)
E = −∇ϕ = −2Kz k̂.
î
ĵ
k̂ B = ∇ × A = ∂/∂x ∂/∂y ∂/∂z = B0 k̂.
−B0 y/2 B0 x/2 0 (b) For time-independent potentials Eq. 4.205 separates in the usual way:
1
∇ − qA ·
∇ − qA ψ + qϕψ = Eψ,
2m i
i
−
2 2
q2 2
iq
∇ ψ+
[∇ · (Aψ) + A · (∇ψ)] +
A + qϕψ = Eψ.
2m
2m
2m
2 2
iq
−
∇ ψ+
[2A · (∇ψ) + ∇ · (Aψ)] +
2m
2m
or
But ∇ · (Aψ) = (∇ · A)ψ + A · (∇ψ),
so
q2 2
A + qϕ ψ = Eψ.
2m
This is the time-independent Schrödinger equation for electrodynamics. In the present case
B0
∂ψ
∂ψ
B2
∇ · A = 0, A · (∇ψ) =
x
−y
, A2 = 0 x2 + y 2 , ϕ = Kz 2 .
2
∂y
∂x
4
qB0
q 2 B02 2
∂
∂
2 2
But Lz =
x
−y
, so −
∇ ψ−
Lz ψ +
x + y 2 + qKz 2 ψ = Eψ.
i
∂y
∂x
2m
2m
8m
Since Lz commutes with H, we may as well pick simultaneous eigenfunctions of both:
Lz ψ = m̄ψ,
where m̄ = 0, ±1, ±2, . . . (with the overbar to distinguish the magnetic quantum number from the mass).
Then
2 2 (qB0 )2 2
qB0 −
∇ +
x + y 2 + qKz 2 ψ = E +
m̄ ψ.
2m
8m
2m
Now let ω1 ≡ qB0 /m, ω2 ≡ 2Kq/m, and use cylindrical coordinates (r, φ, z):
2 1 ∂
1
1
1
∂ψ
1 ∂2ψ ∂2ψ
2
2
2
2 2
−
+ mω1 x + y + mω2 z ψ = E + m̄ω1 ψ.
r
+ 2
+
2m r ∂r
∂r
r ∂φ2
∂z 2
8
2
2
∂2ψ
∂
1
1
, so
= − 2 L2z ψ = − 2 m̄2 2 ψ = −m̄2 ψ.
i ∂φ
∂φ2
R(r)Φ(φ)Z(z) :
2
d2 Z
1 d
dR
m̄2
1
−
ΦZ
r
− 2 RΦZ + RΦ 2 +
mω12 r2 +
2m
r dr
dr
r
dz
8
But Lz =
Use separation of variables: ψ(r, φ, z) =
1
1
2 2
mω2 z RΦZ = E + m̄ω1 RΦZ.
2
2
Divide by RΦZ and collect terms:
2
1
2 1 d 2 Z
1
1
1 d
dR
m̄2
2 2
−
=
E
+
+
z
r
− 2 + mω12 r2 + −
mω
m̄ω
1 .
2
2m rR dr
dr
r
8
2m Z dz 2
2
2
The first term depends only on r, the second only on z, so they’re both constants; call them Er and Ez :
2 1 d
dR
m̄2
1
2 d 2 Z 1
1
−
r
− 2 R + mω12 r2 R = Er R; −
+ mω22 z 2 Z = Ez Z; E = Er +Ez − m̄ω1 .
2m r dr
dr
r
8
2m dz 2 2
2
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
The z equation is a one-dimensional harmonic oscillator, and we can read off immediately that Ez =
(n2 + 1/2)ω2 , with
√n2 = 0, 1, 2, . . . . The r equation is actually a two-dimensional harmonic oscillator; to
get Er , let u(r) ≡ r R, and follow the method of Sections 4.1.3 and 4.2.1:
√
u
dR √
u
u
dR
u
d
u
dR
R= √ ,
= √ − 3/2 , r
= ru − √ ,
r
= r u + 3/2 ,
dr
dr
dr
dr
r
r 2r
2 r
4r
1 d
u
u
2
u
1 u 1
m̄2 u
1
dR
u
u
√ +
√ − 2 √
+ mω12 r2 √ = Er √
r
= √ + 5/2 ; −
2
r dr
dr
2m
r
8
r 4r
r 4r
r
r
r
r
2
2
u
1
1
d u
−
+ mω12 r2 u = Er u.
+
− m̄2
2
2
2m dr
4
r
8
This is identical to the equation we encountered in Problem 4.39 (the three-dimentional harmonic oscillator), only with ω → ω1 /2, E → Er , and l(l + 1) → m̄2 − 1/4, which is to say, l2 + l + 1/4 = m̄2 , or
(l + 1/2)2 = m̄2 , or l = |m̄| − 1/2. [Our present equation depends only on m̄2 , and hence is the same for
either sign, but the solution to Problem 4.39 assumed l + 1/2 ≥ 0 (else u is not normalizable), so we need
|m| here.] Quoting 4.39:
E = (jmax + l + 3/2)ω → Er = (jmax + |m̄| + 1)ω1 /2,
where
jmax = 0, 2, 4, . . . .
E = jmax + |m̄| + 1)ω1 /2 + (n2 + 1/2)ω2 − m̄ω1 /2 = (n1 + 12 )ω1 + (n2 + 12 )ω2 ,
where n1 = 0, 1, 2, . . . (if m̄ ≥ 0, then n1 = jmax /2; if m̄ < 0, then n1 = jmax /2 − m̄).
Problem 4.61
(a)
B = ∇ × A = ∇ × A + ∇ × (∇λ) = ∇ × A = B.
[∇ × ∇λ = 0, by equality of cross-derivatives: (∇ × ∇λ)x =
∂A
E = −∇ϕ −
= −∇ϕ + ∇
∂t
[Again: ∇
∂Λ
∂t
=
∂Λ
∂t
−
∂
∂y
∂λ
∂z
−
∂
∂z
∂λ
∂y
= 0, etc.]
∂A
∂
∂A
− (∇Λ) = −∇ϕ −
= E.
∂t
∂t
∂t
∂
(∇Λ) by the equality of cross-derivatives.]
∂t
(b)
∇ − qA − q(∇Λ) eiqΛ/ Ψ = q(∇Λ)eiqΛ/ Ψ + eiqΛ/ ∇Ψ − qAeiqΛ/ Ψ − q(∇Λ)eiqΛ/ Ψ
i
i
= eiqΛ/ ∇Ψ − qAeiqΛ/ Ψ.
i
∇ − qA − q(∇Λ)
i
2
iqΛ/
e
Ψ=
∇ − qA − q(∇Λ)
i
iqΛ/
e
∇Ψ − qAeiqΛ/ Ψ
i
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CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
131
iq
q
(∇Λ · ∇Ψ)eiqΛ/ + eiqΛ/ ∇2 Ψ − (∇ · A)eiqΛ/ Ψ − q 2 (A · ∇Λ)eiqΛ/ Ψ
i
q
q
− eiqΛ/ A · (∇Ψ) − eiqΛ/ (A · ∇Ψ) + q 2 A2 eiqΛ/ Ψ
i
i
q iqΛ/
− e
(∇Λ · ∇Ψ) + q 2 (A · ∇Λ)eiqΛ/ Ψ
i
-
= eiqΛ/ −2 ∇2 Ψ + iq(∇ · A)Ψ + 2iq(A · ∇Ψ) + q 2 A2 Ψ
.
−iq(∇Λ) · (∇Ψ) − q 2 (A · ∇Λ)Ψ + iq(∇Λ) · (∇Ψ) + q 2 (A · ∇Λ)Ψ
2 = eiqΛ/
∇ − qA Ψ .
i
= −2
So:
1
2m
∇ − qA
i
[using Eq. 4.205]
2
+ qϕ
iqΛ/
Ψ =e
1
2m
∇ − qA
i
2
∂Λ
+ qϕ − q
Ψ
∂t
∂Λ
∂ iqΛ/ ∂Ψ
∂Ψ
= eiqΛ/ i
−q
Ψ = i
e
. QED
Ψ = i
∂t
∂t
∂t
∂t
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132
CHAPTER 5. IDENTICAL PARTICLES
Chapter 5
Identical Particles
Problem 5.1
(a)
(m1 + m2 )R = m1 r1 + m2 r2 = m1 r1 + m2 (r1 − r) = (m1 + m2 )r1 − m2 r
r1 = R +
⇒
m2
µ
r=R+
r. m1 + m2
m1
(m1 + m2 )R = m1 (r2 + r) + m2 r2 = (m1 + m2 )r2 + m1 r ⇒ r2 = R −
m1
µ
r=R−
r. m1 + m2
m2
Let R = (X, Y, Z), r = (x, y, z).
∂
∂X ∂
∂x ∂
=
+
∂x1
∂x1 ∂X
∂x1 ∂x
m1
∂
µ
∂
=
+ (1)
=
(∇R )x + (∇r )x ,
m1 + m2 ∂X
∂x
m2
∂
∂X ∂
∂x ∂
(∇2 )x =
=
+
∂x
∂x2 ∂X
∂x2 ∂x
2
∂
m2
∂
µ
=
(∇R )x − (∇r )x ,
− (1)
=
m1 + m2 ∂X
∂x
m1
(∇1 )x =
so
∇1 =
µ
∇R + ∇ r . m2
so
∇2 =
µ
∇ R − ∇r . m1
(b)
µ
∇21 ψ = ∇1 · (∇1 ψ) = ∇1 ·
∇R ψ + ∇ r ψ
m2
µ
µ
µ
=
∇R ·
∇R ψ + ∇r ψ + ∇r ·
∇R ψ + ∇r ψ
m2
m2
m2
2
µ
µ
=
∇2R ψ + 2
(∇r · ∇R )ψ + ∇2r ψ.
m2
m2
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CHAPTER 5. IDENTICAL PARTICLES
Likewise,
∇22 ψ =
µ
m1
133
2
∇2R ψ − 2
µ
(∇r · ∇R ) + ∇2r ψ.
m1
2 2
2 2
∴ Hψ = −
∇1 ψ −
∇ ψ + V (r1 , r2 )ψ
2m1
2m2 2
2
2µ
1 2
µ2
2µ
1 2
µ2
2
2
=−
ψ
∇
+
∇
·
∇
+
∇
+
∇
−
∇
·
∇
+
∇
r
R
r
R
2 m1 m22 R m1 m2
m1 r m2 m21 R m2 m1
m2 r
2
1
1
1
1
µ2
2
+ V (r)ψ = −
∇R +
∇2r ψ + V (r)ψ = Eψ.
+
+
2 m1 m 2 m2
m1
m1
m2
But
−
1
1
+
m1
m2
=
m 1 + m2
µ2
1
= , so
m 1 m2
µ
m1 m 2
2
2 2
∇2R ψ −
∇ ψ + V (r)ψ = Eψ.
2(m1 + m2 )
2µ r
1
1
+
m2
m1
=
µ
1
m 1 m2
+
=
.
m1 m2
m1 m2 (m1 + m2 ) m1 + m2
(c) Put in ψ = ψr (r)ψR (R), and divide by ψr ψR :
−
2
2 1 2
1 2
∇R ψR + −
∇ ψr + V (r) = E.
2(m1 + m2 ) ψR
2µ ψr r
The first term depends only on R, the second only on r, so each must be a constant; call them ER and
Er , respectively. Then:
−
2
∇ 2 ψ R = ER ψ R ;
2(m1 + m2 )
−
2 2
∇ ψr + V (r)ψr = Er ψr ,
2µ
with
ER + Er = E.
Problem 5.2
∆E1
∆m
m−µ
m(m + M ) M
m
=
=
=
−
=
.
E1
µ
µ
mM
M
M
The fractional error is the ratio of the electron mass to the proton mass:
9.109 × 10−31 kg
= 5.44 × 10−4 . The percent error is 0.054% (pretty small).
1.673 × 10−27 kg
(a) From Eq. 4.77, E1 is proportional to mass, so
∆(1/λ)
∆R
∆µ
(1/λ2 )∆λ
∆λ
=
=
=−
=−
.
(1/λ)
R
µ
(1/λ)
λ
But µ = mM/(m + M ), where m = electron mass, and M =
(b) From Eq. 4.94, R is proportional to m, so
So (in magnitude) ∆λ/λ = ∆µ/µ.
nuclear mass.
∆µ =
=
m(2mp )
mmp
mmp
(2m + 2mp − m − 2mp )
−
=
m + 2mp
m + mp
(m + mp )(m + 2mp )
m2 mp
mµ
.
=
(m + mp )(m + 2mp )
m + 2mp
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134
CHAPTER 5. IDENTICAL PARTICLES
∆λ
m
m
∆µ
m
≈
, so ∆λ =
λh , where λh is the hydrogen wavelength.
=
=
λ
µ
m + 2mp
2mp
2mp
1
=R
λ
∴ ∆λ =
(c) µ =
1 1
−
4 9
5
36
36
R⇒λ=
=
m = 6.563 × 10−7 m.
36
5R
5(1.097 × 107 )
9.109 × 10−31
(6.563 × 10−7 )m = 1.79 × 10−10 m.
2(1.673 × 10−27 )
mm
m
= , so the energy is half what it would be for hydrogen: (13.6/2)eV = 6.8 eV.
m+m
2
mp m µ
m p mµ
mp + me
mµ (mp + me )
; R ∝ µ, so R is changed by a factor
·
=
, as compared
mp + m µ
mp + mµ
mp me
me (mp + mµ )
with hydrogen. For hydrogen, 1/λ = R(1−1/4) = 34 R ⇒ λ = 4/3R = 4/3(1.097×107 ) m = 1.215×10−7 m,
and λ ∝ 1/R, so for muonic hydrogen the Lyman-alpha line is at
(d) µ =
λ=
me (mp + mµ )
1 (1.673 × 10−27 + 206.77 × 9.109 × 10−31 )
(1.215 × 10−7 m) =
(1.215 × 10−7 m)
mµ (mp + me )
206.77
(1.673 × 10−27 + 9.109 × 10−31 )
= 6.54 × 10−10 m.
Problem 5.3
The energy of the emitted photon, in a transition from vibrational state ni to state nf , is
Ep = (ni + 12 )ω − (nf + 12 )ω = nω, (where n ≡ ni − nf ). The frequency of the photon is
Ep
k
nω
n
ν=
=
=
. The splitting of this line is given by
h
2π
2π µ
n√
1 n
k ∆µ
1 ∆µ
1
∆ν = = ν
.
k − 3/2 ∆µ =
2π
2 2π µ µ
2 µ
2µ
Now
µ=
mh mc
=
mh + mc
1
mc
1
+
1
mh
∆ν =
⇒ ∆µ = −1
1
mc
+
1
mh
2
−
1
∆mc
m2c
=
µ2
∆mc .
m2c
1 µ∆mc
1 (∆mc /mc )
.
= ν
ν
2
m2c
2
1 + mc
mh
Using the average value (36) for mc , we have ∆mc /mc = 2/36, and mc /mh = 36/1, so
∆ν =
1 (1/18)
1
ν=
ν = 7.51 × 10−4 ν.
2 (1 + 36)
(36)(37)
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CHAPTER 5. IDENTICAL PARTICLES
135
Problem 5.4
(a)
|ψ± |2 d3 r1 d3 r2
1=
∗
= |A|2
[ψa (r1 )ψb (r2 ) ± ψb (r1 )ψa (r2 )] [ψa (r1 )ψb (r2 ) ± ψb (r1 )ψa (r2 )] d3 r1 d3 r2
|ψa (r1 )|2 d3 r1
= |A|2
ψb (r1 )∗ ψa (r1 )d3 r1
±
|ψb (r2 )|2 d3 r2 ±
ψa (r1 )∗ ψb (r1 )d3 r1
ψa (r2 )∗ ψb (r2 )d3 r2 +
ψb (r2 )∗ ψa (r2 )d3 r2
|ψb (r1 )|2 d3 r1
|ψa (r2 )|2 d3 r2
√
= |A|2 (1 · 1 ± 0 · 0 ± 0 · 0 + 1 · 1) = 2|A|2 =⇒ A = 1/ 2.
(b)
1 = |A|
2
∗
[2ψa (r1 )ψa (r2 )] [2ψa (r1 )ψa (r2 )] d3 r1 d3 r2
= 4|A|2
|ψa (r1 )|2 d3 r1
|ψa (r2 )|2 d3 r2 = 4|A|2 .
A = 1/2.
Problem 5.5
(a)
−
2 ∂ 2 ψ
2 ∂ 2 ψ
−
= Eψ
2
2m ∂x1
2m ∂x22
√
πx 2
1
ψ=
sin
sin
a
a
2πx2
a
(for 0 ≤ x1 , x2 ≤ a, otherwise ψ = 0).
− sin
2πx1
a
sin
πx 2
a
√ 2
πx πx d2 ψ
2
2πx1
π 2
2πx2
2π
1
2
=
sin
sin
−
sin
+
sin
dx21
a
a
a
a
a
a
a
√ 2
πx πx d2 ψ
2π
2πx2
π 2
2
2πx1
1
2
−
sin
+
sin
=
sin
sin
dx22
a
a
a
a
a
a
a
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CHAPTER 5. IDENTICAL PARTICLES
−
d2 ψ d2 ψ
+ 2
dx21
dx2
2
2m
=−
d2 ψ d2 ψ
+ 2
dx21
dx2
π 2
a
=
+
2π
a
2 ψ = −5
5π 2 2
ψ = Eψ,
2ma2
π2
ψ,
a2
with E =
5π 2 2
= 5K.
2ma2
(b) Distinguishable:
ψ22 = (2/a) sin (2πx1 /a) sin (2πx2 /a) , with E22 = 8K (nondegenerate).
ψ13 = (2/a) sin (πx1 /a) sin (3πx2 /a)
, with E13 = E31 = 10K (doubly degenerate).
ψ31 = (2/a) sin (3πx1 /a) sin (πx2 /a)
Identical Bosons:
ψ22 = (2/a) sin (2πx1 /a) sin (2πx2 /a), E22 = 8K (nondegenerate).
√
ψ13 = ( 2/a) [sin (πx1 /a) sin (3πx2 /a) + sin (3πx1 /a) sin (πx2 /a)], E13 = 10K
Identical Fermions:
√
ψ13 = ( 2/a) sin πxa 1 sin
√
ψ23 = ( 2/a) sin
2πx1
a
3πx2
a
− sin
− sin
3πx2
a
sin
3πx1
a
sin
3πx1
a
sin
πx2
a
, E13 = 10K
2πx2
a
(a) Use Eq. 5.19 and Problem 2.4, with xn = a/2 and x2 n = a2
(nondegenerate).
, E23 = 13K
Problem 5.6
1
3
−
(nondegenerate).
1
2(nπ)2
(nondegenerate).
.
1
1
1
1
+ 2
−
6 2π 2 n2
m
a
(m−n)π
nπ
1 a
(b) xmn = a2 0 x sin mπ
x − cos (m+n)π
x dx
a x sin a x dx = a 0 x cos
a
a
2
(m−n)π
(m−n)π
a
ax
= a1
sin
cos
x
+
x
(m−n)π
a
(m−n)π
a
2
a
(m+n)π
(m+n)π
a
ax
− (m+n)π cos
x − (m+n)π sin
x a
a
(x1 − x2 )2 = a2
=
1
a
a
(m−n)π
2
1
3
−
1
2(nπ)2
+ a2
1
3
(cos[(m − n)π] − 1) −
−
1
2(mπ)2
a
(m+n)π
2
−2·
a
2
·
a
2
= a2
.
0
(cos[(m + n)π] − 1) .
But cos[(m ± n)π] = (−1)m+n , so
xmn
a = 2 (−1)m+n − 1
π
1
1
−
2
(m − n)
(m + n)2
So Eq. 5.21 ⇒ (x1 − x2 ) = a
2
2
1
1
−
6 2π 2
=
1
1
+ 2
n2
m
a(−8mn)
π 2 (m2 −n2 )2 ,
0,
−
if m and n have opposite parity,
if m and n have same parity.
128a2 m2 n2
.
π 4 (m2 − n2 )4
(The last term is present only when m, n have opposite parity.)
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CHAPTER 5. IDENTICAL PARTICLES
(c) Here Eq. 5.21 ⇒ (x1 − x2 )2 = a2
137
1
1
−
6 2π 2
1
1
+ 2
n2
m
+
128a2 m2 n2
.
π 4 (m2 − n2 )4
(Again, the last term is present only when m, n have opposite parity.)
Problem 5.7
(a) ψ(x1 , x2 , x3 ) = ψa (x1 )ψb (x2 )ψc (x3 ).
(b) ψ(x1 , x2 , x3 ) =
(c) ψ(x1 , x2 , x3 ) =
√1 [ψa (x1 )ψb (x2 )ψc (x3 )
6
+ ψa (x1 )ψc (x2 )ψb (x3 ) + ψb (x1 )ψa (x2 )ψc (x3 )
.
+ψb (x1 )ψc (x2 )ψa (x3 ) + ψc (x1 )ψb (x2 )ψa (x3 ) + ψc (x1 )ψa (x2 )ψb (x3 )]
− ψa (x1 )ψc (x2 )ψb (x3 ) − ψb (x1 )ψa (x2 )ψc (x3 )
.
+ψb (x1 )ψc (x2 )ψa (x3 ) − ψc (x1 )ψb (x2 )ψa (x3 ) + ψc (x1 )ψa (x2 )ψb (x3 )]
√1 [ψa (x1 )ψb (x2 )ψc (x3 )
6
Problem 5.8
ψ = A ψ(r1 , r2 , r3 , . . . , rZ ) ± ψ(r2 , r1 , r3 , . . . , rZ ) + ψ(r2 , r3 , r1 , . . . , rZ ) + etc. ,
where “etc.” runs over all permutations of the arguments r1 , r2 , . . . , rZ , with a + sign for all even permutations
(even number of transpositions ri ↔ rj , starting from r1 , r2 , . . . , rZ ), and ± for all odd permutations (+
√ for
bosons, − for fermions). At the end of the process, normalize the result to determine A. (Typically A = 1/ Z!,
but this may not be right if the starting function is already symmetric under some interchanges.)
Problem 5.9
(a) The energy of each electron is E = Z 2 E1 /n2 = 4E1 /4 = E1 = −13.6eV, so the total initial energy is
2 × (−13.6) eV= −27.2 eV. One electron drops to the ground state Z 2 E1 /1 = 4E1 , so the other is left
with 2E1 − 4E1 = −2E1 = 27.2 eV.
(b) He+ has one electron; it’s a hydrogenic ion (Problem 4.16) with Z = 2, so the spectrum is
1/λ = 4R 1/n2f − 1/n2i , where R is the hydrogen Rydberg constant, and ni , nf are the initial and final
quantum numbers (1, 2, 3, . . . ).
Problem 5.10
(a) The ground state (Eq. 5.30) is spatially symmetric, so it goes with the symmetric (triplet) spin configuration. Thus the ground state is orthohelium, and it is triply degerate. The excited states (Eq. 5.32) come
in ortho (triplet) and para (singlet) form; since the former go with the symmetric spatial wave function,
the orthohelium states are higher in energy than the corresponding (nondegenerate) para states.
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138
CHAPTER 5. IDENTICAL PARTICLES
(b) The ground state (Eq. 5.30) and all excited states (Eq. 5.32) come in both ortho and para form. All are
quadruply degenerate (or at any rate we have no way a priori of knowing whether ortho or para are higher
in energy, since we don’t know which goes with the symmetric spatial configuration).
Problem 5.11
(a)
0
1
|r1 − r2 |
1
=
∞
= 2π
8
πa3
2 1
=
r1 r 2
r12
+
r22
= 4πe
1
r1
r1
0
r12 + r22 − 2r1 r2 cos θ2
∞
r1
= 4π
dθ2 r22 dr2
π
1
− 2r1 r2 cos θ2 =
r12 + r22 + 2r1 r2 − r12 + r22 − 2r1 r2
r1 r 2
0
r1
r22 e−4r2 /a dr2
2/r1 (r2 < r1 )
2/r2 (r2 > r1 )
r22 e−4r2 /a dr2
∞
+
0
0
=−
1
r1
sin θ2
1
[(r1 + r2 ) − |r1 − r2 |] =
r1 r2
−4r1 /a
d r2 d3 r1
π
e
0
=
3
r12 + r22 − 2r1 r2 cos θ2
−4(r1 +r2 )/a
e−4(r1 +r2 )/a
r2 e−4r2 /a dr2 .
r1
a 2 −4r2 /a a a 2 −4r2 /a
4r2
1
− r2 e
−
−1
=
+
e
r1
4
2 4
a
r1
0
a
ar1 −4r1 /a a2 −4r1 /a a2
e
.
r12 e−4r1 /a +
+ e
−
4r1
2
8
8
r2 e−4r2 /a dr2 =
a 2
4
e−4r2 /a
−
∞
ar1 −4r1 /a a2 −4r1 /a
4r2
+ e
.
− 1 =
e
a
4
16
r1
a3 −4r1 /a
a2
a3
a2 −8r1 /a
ar1
ar1
−
−
+
e
e
+ −
+
32r1
a
8
32r1
4
16
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CHAPTER 5. IDENTICAL PARTICLES
=
0
πa2
8
32
= 4
a
32
= 4
a
a
a −4r1 /a
e−8r1 /a .
e
− 2+
r1
r1
1
1
|r1 − r2 |
a
8
=
· 4π
πa4
∞
−4r1 /a
r1 e
a·
a −4r1 /a
a
e−8r1 /a r12 dr1
e
− 2+
r1
r1
∞
0
dr1 − 2
0
139
∞
r12 e−8r1 /a dr1
−a
0
a 2
4
−2·2
a 3
8
−a·
∞
−8r1 /a
r1 e
dr1
0
a 2 8
32
=
a
1
1
1
−
−
16 128 64
=
5
.
4a
(b)
Vee ≈
e2
4π90
0
1
|r1 − r2 |
1
=
5 e2 1
5m
=
4 4π90 a
4 2
e2
4π90
2
=
5
5
(−E1 ) = (13.6 eV) = 34 eV.
2
2
E0 + Vee = (−109 + 34)eV = −75 eV, which is pretty close to the experimental value (−79 eV).
Problem 5.12
(a) Hydrogen: (1s); helium: (1s)2 ; lithium: (1s)2 (2s); beryllium: (1s)2 (2s)2 ;
boron: (1s)2 (2s)2 (2p); carbon: (1s)2 (2s)2 (2p)2 ; nitrogen: (1s)2 (2s)2 (2p)3 ;
oxygen: (1s)2 (2s)2 (2p)4 ; fluorine: (1s)2 (2s)2 (2p)5 ; neon: (1s)2 (2s)2 (2p)6 .
These values agree with those in Table 5.1—no surprises so far.
(b) Hydrogen: 2 S1/2 ;
helium: 1 S0 ;
lithium: 2 S1/2 ;
beryllium 1 S0 . (These four are unambiguous,
because the orbital angular momentum is zero in all cases.) For boron, the spin (1/2) and orbital (1)
angular momenta could add to give 3/2 or 1/2, so the possibilities are 2 P3/2 or 2 P1/2 . For carbon, the
two p electrons could combine for orbital angular momentum 2, 1, or 0, and the spins could add to 1 or 0:
1
S0 , 3 S1 , 1 P1 , 3 P2 , 3 P1 , 3 P0 , 1 D2 , 3 D3 , 3 D2 , 3 D1 . For nitrogen, the 3 p electrons can add to orbital angular
momentum 3, 2, 1, or 0, and the spins to 3/2 or 1/2:
2
4
S1/2 , 4 S3/2 , 2 P1/2 , 2 P3/2 , 4 P1/2 , 4 P3/2 , 4 P5/2 , 2 D3/2 , 2 D5/2 ,
D1/2 , 4 D3/2 , 4 D5/2 , 4 D7/2 , 2 F5/2 , 2 F3/2 , 4 F3/2 , 4 F5/2 , 4 F7/2 , 4 F9/2 .
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CHAPTER 5. IDENTICAL PARTICLES
Problem 5.13
(a) Orthohelium should have lower energy than parahelium, for corresponding states (which is true).
(b) Hund’s first rule says S = 1 for the ground state of carbon. But this (the triplet) is symmetric, so the
orbital state will have to be antisymmetric. Hund’s second rule favors L = 2, but this is symmetric, as
you can see most easily by going to the “top of the ladder”: |2 2 = |1 11 ||1 12 . So the ground state of
carbon will be S = 1, L = 1. This leaves three possibilities: 3 P2 , 3 P1 , and 3 P0 .
(c) For boron there is only one electron in the 2p subshell (which can accommodate a total of 6), so Hund’s
third rule says the ground state will have J = |L − S|. We found in Problem 5.12(b) that L = 1 and
S = 1/2, so J = 1/2, and the configuration is
2
P1/2 .
(d) For carbon we know that S = 1 and L = 1, and there are only two electrons in the outer subshell, so
Hund’s third rule says J = 0, and the ground state configuration must be
3
P0 .
For nitrogen Hund’s first rule says S = 3/2, which is symmetric (the top of the ladder is | 32 32 =
Hund’s second rule favors L = 3, but this is also symmetric. In fact, the only
antisymmetric orbital configuration here is L = 0. [You can check this directly by working out the
Clebsch-Gordan coefficients, but it’s easier to reason as follows: Suppose the three outer electrons are in
the “top of the ladder” spin state, so each one has spin up (| 12 12 ); then (since the spin states are all the
same) the orbital states have to be different: |1 1, |1 0, and |1−1. In particular, the total z-component of
orbital angular momentum has to be zero. But the only configuration that restricts Lz to zero is L = 0.]
The outer subshell is exactly half filled (three electrons with n = 2, l = 1), so Hund’s third rule says
| 12 12 1 | 12 21 2 | 12 12 3 ).
J = |L − S| = |0 − 32 | = 3/2. Conclusion: The ground state of nitrogen is
4
S3/2 . (Table 5.1 confirms
this.)
Problem 5.14
S = 2; L = 6; J = 8. (1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (3d)10 (4s)2 (4p)6 (4d)10 (5s)2 (5p)6 (4f )10 (6s)2 .
definite (36 electrons)
likely (30 electrons)
Problem 5.15
Divide Eq. 5.45 by Eq. 5.43, using Eq. 5.42:
Etot /N q
3
2 (3π 2 N q)5/3 1
2m
=
= .
2
2/3
2
2
2/3
EF
N q (3π N q/V )
5
10π mV
Problem 5.16
N
atoms
moles
gm
NA
2
Nq
(3ρπ 2 )2/3 . ρ =
=
=
×
×
=
· d, where NA is Avogadro’s
2m
V
V
mole
gm
volume
M
23
number (6.02 × 10 ), M = atomic mass = 63.5 gm/mol, d = density = 8.96 gm/cm3 .
(6.02 × 1023 )(8.96 gm/cm3 )
ρ=
= 8.49 × 1022 /cm3 = 8.49 × 1028 /m3 .
(63.5 gm)
(1.055 × 10−34 J · s)(6.58 × 10−16 eV · s)
EF =
(3π 2 8.49 × 1028 /m3 )2/3 = 7.04 eV.
(2)(9.109 × 10−31 kg)
(a) EF =
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CHAPTER 5. IDENTICAL PARTICLES
141
(b)
7.04 eV =
v2
14.08
v
1
= 2.76 × 10−5 ⇒ = 5.25 × 10−3 ,
(0.511 × 106 eV/c2 )v 2 ⇒ 2 =
2
c
.511 × 106
c
so it’s nonrelativistic.
v = (5.25 × 10−3 ) × (3 × 108 ) = 1.57 ×106 m/s.
(c)
T =
7.04 eV
= 8.17 × 104 K.
8.62 × 10−5 eV/K
P =
(3π 2 )2/3 2 5/3
(3π 2 )2/3 (1.055 × 10−34 )2
2
ρ
(8.49 × 1028 )5/3 N/m = 3.84 × 1010 N/m2 .
=
5m
5(9.109 × 10−31 )
(d)
Problem 5.17
(3π 2 )2/3 2
P =
5m
Nq
V
5/3
= AV
−5/3
dP
= −V A
⇒ B = −V
dV
−5
3
V −5/3−1 =
5
5
AV −5/3 = P.
3
3
For copper, B = 53 (3.84 × 1010 N/m2 ) = 6.4 × 1010 N/m2 .
Problem 5.18
(a) Equations 5.59 and 5.63 ⇒ ψ = A sin kx + B cos kx;
A sin ka = eiKa − cos ka B. So
iKa
A sin ka
A
cos kx = iKa
e
sin kx − sin kx cos ka + cos kx sin ka
(eiKa − cos ka)
(e
− cos ka)
.
AeiKa
= C sin kx + e−iKa sin[k(a − x)] , where C ≡ iKa
.
e
− cos ka
ψ = A sin kx +
(b) If z = ka = jπ, then sin ka = 0, Eq. 5.64 ⇒ cos Ka = cos ka = (−1)j ⇒ sin Ka = 0, so eiKa =
cos Ka + i sin Ka = (−1)j , and the constant C involves division by zero. In this case we must go back to
Eq. 5.63, which is a tautology (0=0) yielding no constraint on A or B, Eq. 5.61 holds automatically, and
Eq. 5.62 gives
2mα
kA − (−1)j k A(−1)j − 0 = 2 B ⇒ B = 0. So ψ = A sin kx.
Here ψ is zero at each delta spike, so the wave function never “feels” the potential at all.
Problem 5.19
We’re looking for a solution to Eq. 5.66 with β = 10 and z π: f (z) = cos z + 10
Mathematica gives z = 2.62768.
So
E=
sin z
= 1.
z
2 k 2
2 z 2
(2.62768)2
z2 α
=
=
eV = 0.345 eV.
=
2
2m
2ma
2β a
20
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142
CHAPTER 5. IDENTICAL PARTICLES
Problem 5.20
Positive-energy solutions. These are the same as before, except that α (and hence also β) is now a negative
number.
Negative-energy solutions. On 0 < x < a we have
√
d2 ψ
−2mE
2
=
κ
ψ,
where
κ
≡
⇒ ψ(x) = A sinh kx + B cosh kx.
dx2
According to Bloch’s theorem the solution on −a < x < 0 is
ψ(x) = e−iKa [A sinh κ(x + a) + B cosh κ(x + a)] .
Continuity at x = 0 ⇒
or A sinh κa = B eiKa − cosh κa .
B = e−iKa [A sinh κa + B cosh κa] ,
(1)
The discontinuity in ψ (Eq. 2.125) ⇒
κA−e−iKa κ [A cosh κa + B sinh κa] =
2mα
2mα
B, or A 1 − e−iKa cosh κa = B
+ e−iKa sinh κa . (2)
2
2 κ
Plugging (1) into (2) and cancelling B:
eiKa − cosh κa
1 − e−iKa cosh κa =
2mα
sinh κa + e−iKa sinh2 κa.
2 κ
2mα
sinh κa.
2 κ
mα
cos Ka = cosh κa + 2 sinh κa.
κ
eiKa − 2 cosh κa + e−iKa cosh2 κa − e−iKa sinh2 κa =
eiKa + e−iKa = 2 cosh κa +
2ma
sinh κa,
2 κ
This is the analog to Eq. 5.64. As before, we let β ≡ mαa/2 (but remember it’s now a negative number), and
this time we define z ≡ −κa, extending Eq. 5.65 to negative z, where it represents negative-energy solutions.
In this region we define
sinh z
f (z) = cosh z + β
.
(3)
z
In the Figure I have plotted f (z) for β = −1.5, using Eq. 5.66 for postive z and (3) for negative z. As
before, allowed energies are restricted to the range −1 ≤ f (z) ≤ 1, and occur at intersections of f (z) with the
N horizontal lines cos Ka = cos(2πn/N a), with n = 0, 1, 2 . . . N − 1. Evidently the first band (partly negative,
and partly positive) contains N states, as do all the higher bands.
1
0
-1
0
π
2π
3π
4π
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CHAPTER 5. IDENTICAL PARTICLES
143
Problem 5.21
2πn
n
Equation 5.56 says K =
⇒ Ka = 2π ; at the bottom of page 227 we found that n = 0, 1, 2, . . . , N − 1.
Na
N
Each value of n corresponds to a distinct state. To find the allowed energies we draw N horizontal lines on
Figure 5.6, at heights cos Ka = cos(2πn/N ), and look for intersections with f (z). The point is that almost all
of these lines come in pairs—two different n’s yielding the same value of cos Ka:
N = 1 ⇒ n = 0 ⇒ cos Ka = 1. Nondegenerate.
N = 2 ⇒ n = 0, 1 ⇒ cos Ka = 1, −1. Nondegenerate.
N = 3 ⇒ n = 0, 1, 2 ⇒ cos Ka = 1, − 12 , − 12 . The first is nondegenerate, the other two are degenerate.
N = 4 ⇒ n = 0, 1, 2, 3 ⇒ cos Ka = 1, 0, −1, 0. Two are nondegenerate, the others are degenerate.
Evidently they are doubly degenerate (two different n’s give same cos Ka) except when cos Ka = ±1, i.e., at
the top or bottom of a band. The Bloch factors eiKa lie at equal angles in the complex plane, starting with
1 (see Figure, drawn for the case N = 8); by symmetry, there is always one with negative imaginary part
symmetrically opposite each one with positive imaginary part; these two have the same real part (cos Ka).
Only points which fall on the real axis have no twins.
sin(Ka)
n=2
n=3
n=1
n=0
n=4
n=5
cos(Ka)
n=7
n=6
Problem 5.22
(a)
'
(3
5πxA
7πxB
17πxC
5πxA
17πxB
7πxC
sin
sin
sin
− sin
sin
sin
a
a
a
a
a
a
7πxA
17πxB
5πxC
7πxA
5πxB
17πxC
+ sin
sin
sin
− sin
sin
sin
a
a
a
a
a
a
17πxA
5πxB
7πxC
17πxA
7πxB
5πxC
+ sin
sin
sin
− sin
sin
sin
.
a
a
a
a
a
a
1
ψ(xA , xB , xC ) = √
6
(b) (i)
'
ψ=
2
a
2
a
(3
sin
11πxA
a
sin
11πxB
a
sin
11πxC
a
.
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144
CHAPTER 5. IDENTICAL PARTICLES
(ii)
'
(3
19πxC
a
a
a
πx πx πx 19πxB
πxC
19πxA
A
B
C
+ sin
sin
sin
+ sin
sin
sin
.
a
a
a
a
a
a
1
ψ = √
3
2
a
sin
πx A
sin
πx B
sin
(iii)
'
(3
5πxA
7πxB
17πxC
5πxA
17πxB
7πxC
sin
sin
+ sin
sin
sin
a
a
a
a
a
a
7πxA
17πxB
5πxC
7πxA
5πxB
17πxC
+ sin
sin
sin
+ sin
sin
sin
a
a
a
a
a
a
17πxA
5πxB
7πxC
17πxA
7πxB
5πxC
+ sin
sin
sin
+ sin
sin
sin
.
a
a
a
a
a
a
1
ψ = √
6
2
a
sin
Problem 5.23
(a) En1 n2 n3 = (n1 + n2 + n3 + 32 )ω = 92 ω ⇒ n1 + n2 + n3 = 3. (n1 , n2 , n3 = 0, 1, 2, 3 . . . ).
n1
0
0
3
0
0
1
1
2
2
1
State
n2 n3
0
3
3
0
0
0
1
2
2
1
0
2
2
0
0
1
1
0
1
1
Configuration
(N0 , N1 , N2 . . . )
# of States
(2,0,0,1,0,0 . . . )
3
(1,1,1,0,0,0 . . . )
6
(0,3,0,0,0 . . . )
1
Most probable configuration:
Possible single-particle energies:
E0 = ω/2 :
E1 = 3ω/2 :
E2 = 5ω/2 :
E3 = 7ω/2 :
P0
P1
P2
P3
= 12/30 = 4/10.
= 9/30 = 3/10.
= 6/30 = 2/10.
= 3/30 = 1/10.
(1,1,1,0,0,0 . . . ).
Most probable single-particle energy:
E0 = 12 ω.
(b) For identical fermions the only configuration is (1,1,1,0,0,0 . . . ) (one state), so this is also the most
probable configuration. The possible one-particle energies are
E0 (P0 = 1/3),
E1 (P1 = 1/3),
E2 (P2 = 1/3),
and they are all equally likely, so it’s a 3-way tie for the most probable energy.
(c) For identical bosons all three configurations are possible, and there is one state for each. Possible oneparticle energies: E0 (P0 = 1/3), E1 (P1 = 4/9), E2 (P2 = 1/9), E3 (P3 = 1/9). Most probable energy: E1 .
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CHAPTER 5. IDENTICAL PARTICLES
145
Problem 5.24
∞
<
1
(distinguishable),
Eq.
5.74
⇒
Q
=
6
N
n!
n=1
∞
<
Here N = 3, and dn = 1 for all states, so:
1
Eq.
5.75
⇒
Q
=
(fermions),
N
!(1
− Nn )!
n
n=1
Eq. 5.77 ⇒ Q = 1
(bosons).
(In the products, most factors are 1/0! or1/1!, both of which are 1, so I won’t write them.)
1
= 1
(distinguishable),
Q=6×
3!
1
1
Configuration 1 (N11 = 3, others 0):
Q=
(fermions),
×
= 0
3!
(−2)!
Q= 1
(bosons).
1
1
(distinguishable),
Q=6× ×
= 3
1! 2!
1
1
Configuration 2 (N5 = 1, N13 = 2):
Q=
×
= 0
(fermions),
1!0! 2!(−1)!
Q= 1
(bosons).
1
1
Q = 6× ×
(distinguishable),
= 3
2! 1!
1
1
Configuration 3 (N1 = 2, N19 = 1):
Q=
×
= 0
(fermions),
2!(−1)! 1!0!
Q= 1
(bosons).
1
1
1
Q = 6 × 1! × 1! × 1! = 6 (distinguishable),
1
1
1
Configuration 4 (N5 = N7 = N17 = 1):
Q=
×
×
= 1
(fermions),
1!0!
1!0!
1!0!
Q= 1
(bosons).
All of these agree with what we got “by hand” at the top of page 231.
Problem 5.25
N = 1 : - can put the ball in any of d baskets, so d ways.
- could put both balls in any of the d baskets : d ways, or
N = 2 : - could put one in one basket (d ways), the other in another(d − 1) ways—but it
doesn’t matter which is which, so divide by 2.
Total: d + 12 d(d − 1) = 12 d(2 + d − 1) = 12 d(d + 1) ways.
- could put all three in one basket : d ways, or
N = 3 : - 2 in one basket, one in another : d(d − 1) ways, or
- 1 each in 3 baskets : d(d − 1)(d − 2)/3! ways.
Total: d + d(d − 1) + d(d − 1)(d − 2)/6 = 16 d(6 + 6d − 6 + d2 − 3d + 2) = 16 d(d2 + 3d + 2)
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146
CHAPTER 5. IDENTICAL PARTICLES
d(d + 1)(d + 2)
ways.
6
=
N =4:
-
all in one basket: d ways, or
3 in one basket, 1 in another: d(d − 1) ways, or
2 in one basket, 2 in another: d(d − 1)/2 ways, or
2 in one basket, one each in others: d(d − 1)(d − 2)/2, or
all in different baskets: d(d − 1)(d − 2)(d − 3)/4!
Total: d + d(d − 1) + d(d − 1)/2 + d(d − 1)(d − 2)/2 + d(d − 1)(d − 2)(d − 3)/24
1
= 24
(24 + 24d − 24 + 12d − 12 + 12d2 − 36d + 24 + d3 − 6d2 + 11d − 6)
=
1
3
24 d(d
+ 6d2 + 11d + 6) =
d(d + 1)(d + 2)(d + 3)
ways.
24
d(d + 1)(d + 2) · · · (d + N − 1)
d+N −1
(d + N − 1)!
The general formula seems to be f (N, d) =
.
=
=
N
N!
N !(d − 1)!
Proof: How many ways to put N identical balls in d baskets? Call it f (N, d).
- Could put all of them in the first basket: 1 way.
- Could put all but one in the first basket; there remains 1 ball for d − 1 baskets: f (1, d − 1) ways.
- Could put all but two in the first basket; there remain 2 for d − 1 baskets: f (2, d − 1) ways.
..
.
- Could put zero in the first basket, leaving N for d − 1 baskets: f (N, d − 1) ways.
/N
Thus: f (N, d) = f (0, d−1)+f (1, d−1)+f (2, d−1)+· · ·+f (N, d−1) = j=0 f (j, d−1) (where f (0, d) ≡ 1).
/N −1
It follows that f (N, d) = j=0 f (j, d − 1) + f (N, d − 1) = f (N − 1, d) + f (N, d − 1). Use this recursion relation
to confirm the conjectured formula by induction:
d+N −1
N
?
=
d+N −2
d+N −2
(d + N − 2)!
(d + N − 2)!
+
=
+
N −1
N
(N − 1)!(d − 1)!
N !(d − 2)!
(d + N − 2)!
(d + N − 1)!
=
(N + d − 1) =
=
N !(d − 1)!
N !(d − 1)!
It works for N = 0 :
d−1
0
= 1, and for d = 1 :
N
N
d+N −1
. d−1
= 1 (which is obviously correct for just one basket). QED
Problem 5.26
A(x, y) = (2x)(2y) = 4xy; maximize, subject to the constraint (x/a)2 + (y/b)2 = 1.
b
(x,y)
a
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CHAPTER 5. IDENTICAL PARTICLES
147
G(x, y, λ) ≡ 4xy + λ (x/a)2 + (y/b)2 − 1 .
∂G
2λ
2λy
= 4x + 2 = 0 ⇒ 4x = − 2
∂y
b
b
λx
− 2
2a
⇒ 4x =
∂G
λx
2λx
= 4y + 2 = 0 ⇒ y = − 2 .
∂x
a
2a
λ2
x ⇒ x = 0 (minimum), or else λ = ±2ab.
a2 b2
2abx
b
= ∓ x. We may as well pick x and y positive, (as in the figure); then y = (b/a)x (and
2a2
a √
∂G
x 2 y 2
x2 b2 x2
2
λ = −2ab).
=0⇒
+
= 1 (of course), so 2 + 2 2 = 1, or 2 x2 = 1, or x = a/ 2, and hence
∂λ
a
b
a
a b
a
√
√
a b
y = ba/(a 2) ⇒ y = b/ 2. A = 4 √ √ = 2ab.
2 2
So y = ∓
Problem 5.27
(a) ln(10!) = ln(3628800) = 15.1044; 10 ln(10) − 10 = 23.026 − 10 = 13.0259;
2.0785; 2.0785/15.1044 = 0.1376, or
14% .
(b) The percent error is:
ln(z!) − z ln(z) + z
× 100.
ln(z!)
z
20
100
50
90
85
89
15.1044 − 13.0259 =
%
5.7
0.89
1.9
0.996
1.06
1.009
Since my calculator cannot compute factorials greater than 69! I used Mathematica to construct the table.
Evidently, the smallest integer for which the error is < 1% is 90.
Problem 5.28
∞
V
Equation 5.108 ⇒ N =
k 2 n(9) dk, where n(9) is given (as T → 0) by Eq. 5.104.
2π 2 0
√
kmax
3
2
V
2mEF
V kmax
2 kmax
2
So N =
k dk =
, where kmax is given by
= µ(0) = EF ⇒ kmax =
.
2π 2 0
2π 2 3
2m
V
N=
(2mEF )3/2 . Compare Eq. 5.43, which says
6π 2 3
2/3
Nq
Nq
2
(2mEF )3/2
V
EF =
, or
= 3π 2
(2mEF )3/2 .
3π 2
, or N =
2m
V
3
V
3π 2 q3
Here q = 1, and Eq. 5.108 needs an extra factor of 2 on the right, to account for spin, so the two formulas agree.
kmax
5
V
V 2
V 2 kmax
Equation 5.109 ⇒ Etot =
⇒ Etot =
k 4 dk =
(2mEF )5/2 .
2
2
4π m 0
4π m 5
20π 2 m3
V 2 5
Compare Eq. 5.45, which says Etot =
. Again, Eq. 5.109 for electrons has an extra factor of 2, so
k
10π 2 m max
the two agree.
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148
CHAPTER 5. IDENTICAL PARTICLES
Problem 5.29
(a) Equation 5.103, n(9) > 0 ⇒
allowed energies 9.
1
e(5−µ)/kB T
−1
> 0 ⇒ e(5−µ)/kB T > 1 ⇒
(9 − µ)
> 0 ⇒ 9 > µ(T ), for all
kB T
2 2
k → 0 (as k → 0, in the continuum limit), so µ(T ) is always negative.
2m
2 π 2 1
1
1
(Technically, the lowest energy is
, but we take the dimensions lx ly lz to be very large
+
+
2m lx2
ly2
lz2
∞
1
k2
in the continuum limit.) Equation 5.108 ⇒ N/V =
dk. The integrand is
2
2
2π 2 0 e( k /2m−µ)/kB T − 1
always positive, and the only T dependence is in µ(T ) and kB T . So, as T decreases, (2 k 2 /2m) − µ(T )
must also decrease, and hence −µ(T ) decreases, or µ(T ) increases (always negative).
√
√
∞
N
k2
2mkB T 1/2
2mkB T 1 −1/2
2 k 2
1
(c)
;
dk
=
dx.
dk.
Let
x
≡
=
,
so
k
=
x
x
V
2π 2 0 e2 k2 /2mkB T − 1
2mkB T
2
3/2 ∞ 1/2
∞ 3/2−1
N
1
2mkB T
1
x
x
=
dx,
where
dx = Γ(3/2)ζ(3/2).
V
2π 2
2
2 0 ex − 1
ex − 1
0
2/3
3/2
√
mkB T
N
N
2π2
Now Γ(3/2) = π/2; ζ(3/2) = 2.61238, so
= 2.612
;
T
=
.
c
V
2π2
mkB 2.612V
(b) For a free particle gas, E =
(d)
3
N
mass/volume
0.15 × 103 kg/m
=
=
= 2.2 × 1028 /m3 .
V
mass/atom
4(1.67 × 10−27 kg)
Tc =
2π(1.05 × 10−34 J · s)2
4(1.67 × 10−27 kg)(1.38 × 10−23 J/K)
2.2 × 1028
2.61 m3
2/3
= 3.1 K.
Problem 5.30
(a)
ω = 2πν =
2πc
,
λ
ρ(ω)|dω| = 8π
so
dω = −
2πc
dλ,
λ2
and
ρ(ω) =
(2πc)3
.
π 2 c3 λ3 (e2πc/kB T λ − 1)
2πc 16π 2 c
−
= ρ(λ) dλ ⇒ ρ(λ) =
.
dλ
2
3
2πc/k
T
λ
5
2πc/k
B
B T λ − 1)
λ
λ (e
− 1)
λ (e
1
(For density, we want only the size of the interval, not its sign.)
(b) To maximize, set dρ/dλ = 0:
−5
e2πc/kB T λ (2πc/kB T )
0 = 16π c 6 2πc/k T λ
−
B
λ (e
− 1)
λ5 (e2πc/kB T λ − 1)2
2
1
− 2
λ
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CHAPTER 5. IDENTICAL PARTICLES
149
5
4
5-x
3
2
5e -x
1
1
2
5
4
3
⇒ 5(e
2πc/kB T λ
− 1) = e
2πc/kB T λ
2πc
kB T λ
.
Let x ≡ 2πc/kB T λ; then 5(ex − 1) = xex ; or 5(1 − e−x ) = x, or 5e−x = 5 − x. From the graph, the
solution occurs slightly below x = 5.
Mathematica says x = 4.966, so λmax =
2πc 1
(4.966)kB T
=
(6.626 × 10−34 J · s)(2.998 × 108 m/s) 1
=
(4.966)(1.3807 × 10−23 J/K)
T
2.897 × 10−3 m·K/T.
Problem 5.31
From Eq. 5.113:
∞
∞
E
ω3
ρ(ω) dω = 2 3
dω.
=
V
π c 0 (eω/kB T − 1)
0
E
= 2 3
V
π c
=
kB T
4 ∞
0
Let x ≡
ω
.
kB T
Then
(kB T )4
x3
(kB T )4
π4
dx = 2 3 3 Γ(4)ζ(4) = 2 3 3 · 6 ·
=
x
e −1
π c π c 90
4
π 2 kB
3
15c 3
T4
J
π 2 (1.3807 × 10−23 J/K)4
T 4 = 7.566 × 10−16 3 4 T 4 . QED
15(2.998 × 108 m/s)3 (1.0546 × 10−34 J · s)3
m K
Problem 5.32
From Problem 2.11(a),
x0 = x1 = 0;
x2 0 =
;
2mω
x2 1 =
3
.
2mω
From Eq. 3.98,
x01 =
∞
−∞
xψ0 (x)ψ1 (x) dx = 0|x|1 =
√
√
1 δ0 0 + 0 δ1 −1 =
2mω
(a) Equation 5.19 ⇒ (x1 − x2 )2 d =
3
2
+
−0=
.
2mω 2mω
mω
(b) Equation 5.21 ⇒ (x1 − x2 )2 + =
2
−2
=
.
mω
2mω
mω
.
2mω
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150
CHAPTER 5. IDENTICAL PARTICLES
(c) Equation 5.21 ⇒ (x1 − x2 )2 − =
3
2
+2
=
.
mω
2mω
mω
Problem 5.33
(a) Each particle has 3 possible states: 3 × 3 × 3 = 27.
(b) All in same state: aaa, bbb, ccc ⇒ 3.
2 in one state: aab, aac, bba, bbc, cca, ccb ⇒ 6 (each symmetrized).
3 different states: abc (symmetrized) ⇒ 1.
Total: 10.
(c) Only abc (antisymmetrized) =⇒ 1.
Problem 5.34
(
'
n2y
2 k 2
π 2 2 n2x
πnx πny
=
. Each state is represented by an
Equation 5.39 ⇒ Enx ny =
+
,
,
with
k
=
2m
lx2
ly2
2m
lx
ly
intersection on a grid in “k-space”—this time a plane—and each state occupies an area π 2 /lx ly = π 2 /A (where
A ≡ lx ly is the area of the well). Two electrons per state means
1 2
Nq
πkF =
4
2
π2
A
, or kF =
Nq
2π
A
1/2
= (2πσ)1/2 ,
where σ ≡ N q/A is the number of free electrons per unit area.
∴ EF =
π2 σ
2 kF2
2
=
2πσ =
.
2m
2m
m
ky
kx
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CHAPTER 5. IDENTICAL PARTICLES
151
Problem 5.35
(a)
V =
4 3
πR ,
3
so E =
2 (3π 2 N q)5/3
10π 2 m
4 3
πR
3
−2/3
22
15πmR2
=
9
πN q
4
5/3
.
(b) Imagine building up a sphere by layers. When it has reached mass m, and radius r, the work necessary
to bring in the next increment dm is: dW = −(Gm/r) dm. In terms of the mass density ρ, m = 43 πr3 ρ,
and dm = 4πr2 drρ, where dr is the resulting increase in radius. Thus:
4
dr
16π 2 2 4
dW = −G πr3 ρ 4πr2 ρ
=−
ρ Gr dr,
3
r
3
and the total energy of a sphere of radius R is therefore
R
NM
16π 2 2
16π 2 ρ2 R5
Egrav = −
ρ G
G. But ρ =
r4 dr = −
, so
3
15
4/3πR3
0
Egrav = −
16π 2 R5 9N 2 M 2
3 N 2M 2
=
−
G
G
.
15
16π 2 R6
5
R
(c)
Etot
A
B
= 2− ,
R
R
22
where A ≡
15πm
9
πN q
4
dEtot
2A
B
= − 3 + 2 = 0 ⇒ 2A = BR,
dR
R
R
R=
R=
4
9π
9π
4
9π
4
5/3 N 5/3
N2
2/3
(6.673 ×
10−11
so
5/3
R=
2
q 5/3 =
GmM 2
4
2A
=
B
15πm
9π
4
3
GN 2 M 2 .
5
B≡
and
2/3
9
πN q
4
5/3
5
.
3GN 2 M 2
2
q 5/3
.
GmM 2 N 1/3
(1.055 × 10−34 J · s)2 (1/2)5/3
N −1/3
Nm2 /kg2 )(9.109 × 10−31 kg)(1.674 × 10−27 kg)2
= (7.58 × 1025 m)N −1/3 .
(d) Mass of sun:
1.989 × 1030 kg,
so N =
1.989 × 1030
= 1.188 × 1057 ; N −1/3 = 9.44 × 10−20 .
1.674 × 10−27
R = (7.58 × 1025 )(9.44 × 10−20 ) m = 7.16 × 106 m (slightly larger than the earth).
(e)
2
From Eq. 5.43: EF =
2m
Nq
3π
4/3πR3
2
2/3
2
=
2mR2
9π
Nq
4
2/3
.
Numerically:
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152
CHAPTER 5. IDENTICAL PARTICLES
EF =
(1.055 × 10−34 J · s)2
1
9π
(1.188 × 1057 )
2(9.109 × 10−31 kg)(7.16 × 106 m)2 4
2
or, in electron volts: EF =
2/3
= 3.102 × 10−14 J,
3.102 × 10−14
eV = 1.94 × 105 eV.
1.602 × 10−19
Erest = mc2 = 5.11 × 105 eV, so the Fermi energy (which is the energy of the most energetic electrons) is
comparable to the rest energy, so they are getting fairly relativistic.
Problem 5.36
(a)
V 2
cV
k dk ⇒ Etot = 2
2
π
π
dE = (ck)
So Etot =
kF
k 3 dk =
0
cV 4
k ;
4π 2 F
kF =
3π 2 N q
V
4/3
1/3
.
c
(3π 2 N q)4/3 V −1/3 .
4π 2
(b)
4
c
V = πR3 ⇒ Edeg =
(3π 2 N q)4/3
3
4π 2 R
4π
3
−1/3
c
=
3πR
9
πN q
4
Adding in the gravitational energy, from Problem 5.35(b),
4/3
A B
3
c 9
Etot =
and B ≡ GN 2 M 2 .
− , where A ≡
πN q
R R
3π 4
5
.
dEtot
(A − B)
= 0 ⇒ A = B,
=−
dR
R2
but there is no special value of R for which Etot is minimal. Critical value: A = B(Etot = 0) ⇒
4/3
c 9
3
= GN 2 M 2 , or
πN q
3π 4
5
Nc =
15 √
5π
16
= 2.04 × 1057 .
c
G
3/2
q2
15 √
=
5π
3
M
16
1.055 × 10−34 J · s × 2.998 × 108 m/s
6.673 × 10−11 N · m2 /kg2
3/2
(1/2)2
(1.674 × 10−27 kg)3
(About twice the value for the sun—Problem 5.35(d).)
(c) Same as Problem 5.35(c), with m → M and q → 1, so multiply old answer by (2)5/3 m/M :
R = 25/3
(9.109 × 10−31 )
(7.58 × 1025 m)N −1/3 = (1.31 × 1023 m)N −1/3 . Using N = 1.188 × 1057 ,
(1.674 × 10−27 )
R = (1.31 × 1023 m)(9.44 × 10−20 ) = 12.4 km. To get EF , use Problem 5.35(e) with q = 1, the new R,
and the neutron mass in place of m:
2 7.16 × 106
9.11 × 10−31
EF = 22/3
(1.94 × 105 eV) = 5.60 × 107 eV = 56.0 MeV.
1.24 × 104
1.67 × 10−27
The rest energy of a neutron is 940 MeV, so a neutron star is reasonably nonrelativistic.
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CHAPTER 5. IDENTICAL PARTICLES
153
Problem 5.37
(a) From Problem 4.38: En = (n + 32 )ω, with n = 0, 1, 2, . . . ; dn = 12 (n + 1)(n + 2).
From Eq. 5.103, n(9) = e−(5−µ)/kB T , so Nn = 12 (n + 1)(n + 2)e(µ− 2 ω)/kB T e−nω/kB T .
∞
∞
3
1
N=
Nn = e(µ− 2 ω)/kB T
(n + 1)(n + 2)xn , where x ≡ e−ω/kB T . Now
2
n=0
n=0
3
∞
∞
∞
∞
1
x
d
1
x
n
n+1
x ⇒
x
⇒
(n + 1)xn ⇒
=
(n + 1)xn .
=
=
=
2
1 − x n=0
1 − x n=0
dx 1 − x
(1
−
x)
n=0
n=0
∞
∞
x2
d
2x
x2
n+2
=
=
(n
+
1)x
,
and
hence
(n + 1)(n + 2)xn+1 =
.
2
2
(1 − x)
dx (1 − x)
(1 − x)3
n=0
n=0
∞
(n + 1)(n + 2)xn =
n=0
2
.
(1 − x)3
So N = eµ/kB T e− 2 ω/kB T
eµ/kB T = N (1 − e−ω/kB T )3 e 2 ω/kB T ;
3
E=
∞
n=0
N n En =
1
3
(1 −
e−ω/kB T )3
.
µ = kB T ln N + 3 ln(1 − e−ω/kB T ) + 32 ω/kB T .
∞
3
1
(n + 3/2)(n + 1)(n + 2)xn .
ωe(µ− 2 ω)/kB T
2
n=0
From above,
∞
∞
2x3/2
d
2x3/2
n+3/2
=
=
(n
+
1)(n
+
2)x
⇒
(n + 3/2)(n + 1)(n + 2)xn+1/2 , or
3
(1 − x)3
dx
(1
−
x)
n=0
n=0
∞
d
(n + 3/2)(n + 1)(n + 2)xn = 1/2
dx
x
n=0
E=
E=
1
2x3/2
(1 − x)3
=
2
x1/2
3 1/2
2x
3x3/2
+
(1 − x)3
(1 − x)4
3
3
1
3(1 + e−ω/kB T )
ωe(µ− 2 ω)/kB T
. But e(µ− 2 ω)/kB T = N (1 − e−ω/kB T )3 ,
2
(1 − e−ω/kB T )4
=
3(1 + x)
.
(1 − x)4
so
1 + e−ω/kB T
3
N ω
.
2
1 − e−ω/kB T
(b) kB T << ω (low temperature) ⇒ e−ω/kB T ≈ 0, so E ≈ 32 N ω (µ ≈ 32 ω). In this limit, all particles
are in the ground state, E0 = 32 ω.
(c) kB T >> ω (high temperature) ⇒ e−ω/kB T ≈ 1 − (ω/kB T ), so E ≈ 3N kB T
(µ ≈ kB T [ln N + 3 ln (ω/kB T )]) . The equipartition theorem says E = N # 12 kB T , where # is the number
of degrees of freedom for each particle. In this case #/2 = 3, or # = 6 (3 kinetic, 3 potential, for each
particle—one of each for each direction in space).
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154
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
Chapter 6
Time-Independent Perturbation
Theory
Problem 6.1
(a)
ψn0 (x) =
En1 =
a
nπ nπ 2
2
a
sin2
sin
x , so En1 = ψn0 |H |ψn0 = α
x δ x−
dx.
a
a
a 0
a
2
nπ a 2α
nπ 2α
sin2
=
sin2
=
a
a 2
a
2
0,
if n is even,
2α/a, if n is odd.
For even n the wave function is zero at the location of the perturbation (x = a/2), so it never “feels” H .
(b) Here n = 1, so we need
0
ψm
|H |ψ10 =
2α
a
sin
mπ π mπ π 2α
mπ a
2α
x δ x−
sin
x dx =
sin
sin
=
sin
.
a
2
a
a
2
2
a
2
This is zero for even m, so the first three nonzero terms will be m = 3, m = 5, and m = 7. Meanwhile,
π 2 2
0
E10 − Em
=
(1 − m2 ), so
2ma2
(2α/a) sin(mπ/2) 0
2α 2ma2 −1 0
1
−1
ψ
=
ψ3 +
ψ50 +
ψ70 + . . .
m
0
0
2
2
E
−
E
a
π
1
−
9
1
−
25
1
−
49
m
1
m=3,5,7,...
4maα 2 1
3π
1
5π
1
7π
= 2 2
sin
x −
sin
x +
sin
x + ...
π a 8
a
24
a
48
a
3π
1
5π
1
7π
mα
a
= 2 2
sin
x − sin
x + sin
x + ... .
π 2
a
3
a
6
a
ψ11 =
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
155
Problem 6.2
√
(a) En = (n + 12 )ω , where ω ≡ k(1 + 9)/m = ω 1 + 9 = ω(1 + 12 9 − 18 92 +
√
En = (n + 12 )ω 1 + 9 = (n + 12 )ω(1 + 12 9 − 18 92 + · · · ).
1 3
16 9
· · · ),
so
(b) H = 12 k x2 − 12 kx2 = 12 kx2 (1 + 9 − 1) = 9( 12 kx2 ) = 9V , where V is the unperturbed potential energy. So
En1 = ψn0 |H |ψn0 = 9n|V |n, with n|V |n the expectation value of the (unperturbed) potential energy
in the nth unperturbed state. This is most easily obtained from the virial theorem (Problem 3.31), but it
can also be derived algebraically. In this case the virial theorem says T = V . But T + V = En . So
V = 12 En0 = 12 (n + 12 )ω;
En1 = 25 (n + 12 )ω, which is precisely the 91 term in the power series from
part (a).
Problem 6.3
(a) In terms of the one-particle states (Eq. 2.28) and energies (Eq. 2.27):
πx πx 2
π 2 2
1
2
Ground state: ψ10 (x1 , x2 ) = ψ1 (x1 )ψ1 (x2 ) =
sin
sin
; E10 = 2E1 =
.
a
a
a
ma2
First excited state: ψ20 (x1 , x2 ) =
√
πx 2
1
sin
sin
a
a
=
(b)
E11
=
=−
ψ10 |H
4V0
a
|ψ10 a
sin4
2πx2
a
[ψ1 (x1 )ψ2 (x2 ) + ψ2 (x1 )ψ1 (x2 )]
2πx1
a
+ sin
sin
πx 2
a
E20 = E1 + E2 =
;
5 π 2 2
.
2 ma2
2 a a
πx πx 2
1
2
= (−aV0 )
sin2
sin2
δ(x2 − x2 ) dx1 dx2
a
a
a
0
0
πx a
0
√1
2
dx = −
4V0 a
a π
π
sin4 y dy = −
0
4V0 3π
3
·
= − V0 .
π
8
2
E21 = ψ20 |H |ψ20 = (−aV0 )
2
a2
a
sin
πx 1
a
0
sin
2πx2
a
+ sin
2πx1
a
sin
πx 2
a
2
δ(x1 − x2 ) dx1 dx2
πx 2
2πx
2πx
sin
sin
+ sin
sin
dx
a
a
a
a
0
8V0 a 2 πx 2 2πx
8V0 a π 2
=−
sin
dx = −
·
sin
sin y sin2 (2y) dy
a 0
a
a
a π 0
π
8V0
32V0 π
=−
sin2 y sin2 y cos2 y dy = −
(sin4 y − sin6 y) dy
·4
π
π
0
0
32V0 3π 5π
=−
−
= −2V0 .
π
8
16
2V0
=−
a
a
πx c
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156
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
Problem 6.4
(a)
0
ψm
|H|ψn0 2
= α
a
a
sin
0
nπ mπ nπ mπ a
2α
x δ x−
sin
x dx =
sin
sin
,
a
2
a
a
2
2
which is zero unless both m and n are odd—in which case it is ±2α/a. So Eq. 6.15 says
En2 =
m=n, odd
En2
=
2m
2α
a
2α
π
2
(En0
2
1
.
0 )
− Em
0,
m=n, odd
if n is even;
1
, if n is odd.
(n2 − m2 )
1
1
1
1
=
−
.
To sum the series, note that 2
(n − m2 )
2n m + n m − n
1 1
1
for n = 1:
=
−
2 3,5,7,... m + 1 m − 1
1
=
2
1 1 1
1 1 1 1
+ + + ··· − − − − ···
4 6 8
2 4 6 8
for n = 3:
=
1
6
=
π 2 2 2
n , so
2ma2
But Eq. 2.27 says En0 =
1
=
2
1
−
2
Thus,
1
=− ;
4
1
1
1 −
6 1,5,7,... m + 3 m − 3
1 1
1
1 1 1 1 1
1
+ +
+ ··· + − − − − −
···
4 8 10
2 2 4 6 8 10
=
1
6
−
1
6
=−
1
.
36
In general, there is perfect cancellation except for the “missing” term 1/2n in the first sum, so the total
1
1
1
0,
if n is even;
2
is
−
=−
. Therefore: En =
2
−2m (α/πn) , if n is odd.
2n
2n
(2n)2
(b)
H =
1
9kx2 ;
2
m|x2 |n =
=
0
ψm
|H |ψn0 =
1
9km|x2 |n. Using Eqs. 2.66 and 2.69:
2
m|(a2+ + a+ a− + a− a+ + a2− )|n
2mω
(n + 1)(n + 2)m|n + 2 + nm|n + (n + 1)m|n + n(n − 1)m|n − 2 .
2mω
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
So, for m = n,
En2 =
=
0
ψm
|H
|ψn0 =
157
1
(n + 1)(n + 2) δm,n+2 + n(n − 1) δm,n−2 .
k9
2
2mω
2
2 (n + 1)(n + 2) δ
+
n(n
−
1)
δ
m,n+2
m,n−2
9ω
4
(n + 12 )ω − (m + 12 )ω
m=n
92 ω [(n + 1)(n + 2) δm,n+2 + n(n − 1) δm,n−2 ]
16
(n − m)
m=n
=
92 ω (n + 1)(n + 2)
n(n − 1)
92 ω
1
1
+
=
− (n + 1)(n + 2) + n(n − 1)
16
n − (n + 2)
n − (n − 2)
16
2
2
=
92 ω
1
1
92 ω
−n2 − 3n − 2 + n2 − n =
(−4n − 2) = −92 ω n +
32
32
8
2
(which agrees with the 92 term in the exact solution—Problem 6.2(a)).
Problem 6.5
(a)
En1 = ψn0 |H |ψn0 = −qEn|x|n = 0 (Problem 2.12).
From Eq. 6.15 and Problem 3.33: En2 = (qE)2
|m|x|n|2
(n − m)ω
m=n
=
√
√
(qE)2 [ n + 1 δm,n+1 + n, δm,n−1 ]2
(qE)2 [(n + 1) δm,n+1 + n δm,n−1 ]
=
ω 2mω
(n − m)
2mω 2
(n − m)
m=n
=
(b) −
(qE)2
n
(qE)2
(n + 1)
(qE)2
+
=
[−(n + 1) + n] = −
.
2
2
2mω n − (n + 1) n − (n − 1)
2mω
2mω 2
2 d 2 ψ
+
2m dx2
=
m=n
1
mω 2 x2 − qEx ψ = Eψ.
2
1
mω 2 x2 − qEx
2
=
1
mω 2 x +
2
With the suggested change of variables,
qE
mω 2
2
− qE x +
qE
mω 2
1
qE
(qE)2
1
(qE)2
1
1 (qE)2
2
2
+ mω 2 2 4 − qEx −
= mω 2 x −
.
mω 2 x + mω 2 x
2
2
2
mω
2
m ω
mω
2
2 mω 2
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158
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
So the Schrödinger equation says
−
2 d 2 ψ
1
1 (qE)2
2
ψ,
+ mω 2 x ψ = E +
2
2m dx
2
2 mω 2
which is the Schrödinger equation for a simple harmonic oscillator, in the variable x . The constant on
the right must therefore be (n + 12 )ω, and we conclude that
1
1 (qE)2
En = (n + )ω −
.
2
2 mω 2
The subtracted term is exactly what we got in part (a) using perturbation theory. Evidently all the higher
corrections (like the first-order correction) are zero, in this case.
Problem 6.6
(a)
0
0
ψ+
|ψ−
= (α+ ψa0 + β+ ψb0 )|(α− ψa0 + β− ψb0 )
∗
∗
∗
∗
= α+
α− ψa0 |ψa0 + α+
β− ψa0 |ψb0 + β+
α− ψb0 |ψa0 + β+
β− ψb0 |ψb0 ∗
∗
1
= α+
α− + β+
β− . But Eq. 6.22 ⇒ β± = α± (E±
− Waa )/Wab ,
0
0
∗
ψ+
|ψ−
= α+
α− 1 +
so
1
1
− Waa )(E−
− Waa )
(E+
α∗ α− 1
1
= + 2 |Wab |2 + (E+
− Waa )(E−
− Waa ) .
∗
Wab Wab
|Wab |
The term in square brackets is:
1 1
1
1
2
1
[ ] = E+
E− − Waa (E+
+ E−
) + |Wab |2 + Waa
. But Eq. 6.27 ⇒ E±
= 12 [(Waa + Wbb ) ±
1
1
shorthand for the square root term. So E+
+ E−
= Waa + Wbb , and
1 1
E+
E− =
Thus
√
], where
√
is
√ 1
1
(Waa + Wbb )2 − ( )2 =
(Waa + Wbb )2 − (Waa − Wbb )2 − 4|Wab |2 = Waa Wbb − |Wab |2 .
4
4
2
[ ] = Waa Wbb − |Wab |2 − Waa (Waa + Wbb ) + |Wab |2 + Waa
= 0,
so
0
0
ψ+
|ψ−
= 0. QED
(b)
0
0
∗
∗
∗
∗
ψ+
|H |ψ−
= α+
α− ψa0 |H |ψa0 + α+
β− ψa0 |H |ψb0 + β+
α− ψb0 |H |ψa0 + β+
β− ψb0 |H |ψb0 ∗
∗
∗
∗
= α+
α− Waa + α+
β− Wab + β+
α− Wba + β+
β− Wbb
∗
= α+
α− Waa + Wab
1
1
(E−
− Waa )
(E 1 − Waa )
(E 1 − Waa ) (E−
− Waa )
+ Wba + ∗
+ Wbb + ∗
Wab
Wab
Wab
Wab
∗
1
1
= α+
α− Waa + E−
− Waa + E+
− Waa + Wbb
But we know from (a) that
1
1
(E+
− Waa )(E−
− Waa )
.
|Wab |2
1
1
(E+
− Waa )(E−
− Waa )
= −1, so
|Wab |2
0
0
∗
1
1
ψ+
|H |ψ−
= α+
α− [E−
+ E+
− Waa − Wbb ] = 0. QED
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
159
(c)
0
0
∗
∗
∗
∗
ψ±
|H |ψ±
= α±
α± ψa0 |H |ψa0 + α±
β± ψa0 |H |ψb0 + β±
α± ψb0 |H |ψa0 + β±
β± ψb0 |H |ψb0 1
(E±
− Waa )
(E 1 − Wbb )
+ |β± |2 Wba ±
+ Wbb
Wab
Wba
= |α± |2 Waa + Wab
(this time I used Eq. 6.24 to express α in terms of β, in the third term).
0
0
1
1
1
1
∴ ψ±
|H |ψ±
= |α± |2 (E±
) + |β± |2 (E±
) = |α± |2 + |β± |2 E±
= E±
. QED
Problem 6.7
(a) See Problem 2.46.
(b) With a → n, b → −n, we have:
Waa = Wbb = −
Wab = −
V0
L
V0
L
L/2
L/2
e−x
2
/a2
−L/2
e−x
2
V0
L
dx ≈ −
/a2 −4πnix/L
e
−L/2
dx ≈ −
∞
e−x
2
/a2
−∞
V0
L
∞
e−(x
dx = −
2
V0 √
a π.
L
/a2 +4πnix/L)
−∞
dx = −
V0 √ −(2πna/L)2
.
a πe
L
(We did this integral in Problem 2.22.) In this case Waa = Wbb , and Wab is real, so Eq. 6.26 ⇒
√ V0 a 2
1
1 ∓ e−(2πna/L) .
or E±
= − π
L
1
E±
= Waa ± |Wab |,
√
2
1
− Waa )
(E−
± π(V0 a/L)e−(2πna/L)
√
= ∓α. Evidently, the “good” linear
(c) Equation 6.22 ⇒ β = α
=α
Wab
− π(V0 a/L)e−(2πna/L)2
combinations are:
1 1 ψ+ = αψn − αψ−n = √ √ ei2πnx/L − e−i2πnx/L = i
2 L
ψ− = αψn + αψ−n =
1
E+
2
cos
L
2
= ψ+ |H |ψ+ = (−V0 )
L
1
E−
= ψ− |H |ψ− =
2
(−V0 )
L
2πnx
.
L
L/2
−x2 /a2
e
L/2
−L/2
e−x
2
/a2
2πnx
L
and
Using Eq. 6.9, we have :
2
sin
−L/2
2
sin
L
2πnx
L
cos2
2πnx
L
dx,
dx.
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160
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
sin2 θ = (1 − cos 2θ)/2, and cos2 θ = (1 + cos 2θ)/2, so
∞
∞
2
2
2
2
V0 ∞ −x2 /a2
4πnx
V0
4πnx
1
E±
≈−
e
e−x /a dx ∓
e−x /a cos
dx = −
dx
1 ∓ cos
L −∞
L
L
L
−∞
−∞
But
=−
√ V0 a √
2
2
V0 √
π a ∓ a πe−(2πna/L) = − π
1 ∓ e−(2πna/L) ,
L
L
same as (b).
(d) Af (x) = f (−x) (the parity operator). The eigenstates are even functions (with eigenvalue +1) and odd
functions (with eigenvalue −1). The linear combinations we found in (c) are precisely the odd and even
linear combinations of ψn and ψ−n .
Problem 6.8
Ground state is nondegenerate; Eqs. 6.9 and 6.31 ⇒
3
2
3
a V0
E =
a
a
sin2
1
2
= 8V0 sin
π
4
2
sin
0
π
2
π π π a
a
3a
x sin2
y sin2
z δ(x − )δ(y − )δ(z − ) dx dy dz
a
a
a
4
2
4
2
sin
3π
4
1
1
= 8V0
(1)
= 2V0 .
2
2
First excited states (Eq. 6.34):
2π
a
a
3a
2 π
2 π
2
Waa = 8V0
sin
x sin
y sin
z δ(x − )δ(y − )δ(z − ) dx dy dz
a
a
a
4
2
4
1
= 8V0
(1)(1) = 4V0 .
2
π π 2π
a
a
3a
2
sin
= 8V0
x sin
y sin2
z δ(x − )δ(y − )δ(z − ) dx dy dz
a
a
a
4
2
4
1
1
= 8V0
(0)
= 0.
2
2
Wbb
2
2
Wcc = 8V0
π π 2π
a
a
3a
x sin2
y sin2
z δ(x − )δ(y − )δ(z − ) dx dy dz
a
a
a
4
2
4
sin
1
= 8V0 (1)(1)
= 4V0 .
2
2
Wab = 8V0 sin
Wac = 8V0 sin
π
4
π
4
sin
sin
π
2
π
2
sin(π) sin
2
sin
π
2
3π
2
sin
3π
2
sin
3π
4
sin
3π
4
= 0.
= 8V0
1
√
2
(1)(1)(−1)
1
√
2
= −4V0 .
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
Wbc = 8V0 sin
3π
sin
sin(π) sin
sin
= 0.
2
2
4
(1 − λ) 0
−1
−1 0 = 4V0 D; det(D − λ) = 0
−λ
0 = −λ(1 − λ)2 + λ = 0
1
−1
0 (1 − λ) π
4
1 0
W = 4V0 0 0
−1 0
161
π
λ = 0,
π
or (1 − λ)2 = 1 ⇒ 1 − λ = ±1 ⇒ λ = 0
⇒
or λ = 2.
So the first-order corrections to the energies are 0, 0, 8V0 .
Problem 6.9
1
(a) χ1 = 0 , eigenvalue V0 ;
0
0
0
χ2 = 1 , eigenvalue V0 ;
χ3 = 0 , eigenvalue 2V0 .
0
1
[V0 (1 − 9) − λ]
0
0
0
[V0 − λ]
9V0 = 0;
(b) Characteristic equation: det(H − λ) = 0
9V0 [2V0 − λ] [V0 (1 − 9) − λ][(V0 − λ)(2V0 − λ) − (9V0 )2 ] = 0 ⇒ λ1 = V0 (1 − 9).
(V0 − λ)(2V0 − λ) − (9V0 )2 = 0 ⇒ λ2 − 3V0 λ + (2V02 − 92 V02 ) = 0 ⇒
V 9V02 − 4(2V02 − 92 V02 )
V0 0
=
3 ± 1 + 492 ≈
3 ± (1 + 292 ) .
2
2
2
V0 V0 λ2 =
3 − 1 + 492 ≈ V0 (1 − 92 ); λ3 =
3 + 1 + 492 ≈ V0 (2 + 92 ).
2
2
λ=
(c)
3V0 ±
−1 0 0
H = 9V0 0 0 1 ;
0 10
= 9V0
E31 = χ3 |H |χ3 = 9V0
−1 0 0
0
0 0 1 0 0 10
0 1 0
1
0
0 0 1 1 = 0 (no first-order correction).
0
|χm |H |χ3 |2
E32 =
;
0
E30 − Em
m=1,2
χ2 |H |χ3 = 9V0
E30 − E20 = 2V0 − V0 = V0 .
χ1 |H |χ3 = 9V0
−1 0 0
0
0
1 0 0 0 0 1 0 = 9V0 1 0 0 1 = 0,
0 1 0
1
0
0
0 1 0 0 = 9V0 .
1
So E32 = (9V0 )2 /V0 = 92 V0 .
E3 = E30 + E31 + E32 = 2V0 + 0 + 92 V0 = V0 (2 + 92 )
Through second-order, then,
(same as we got for λ3 in (b)).
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162
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
(d)
Waa = χ1 |H |χ1 = 9V0
−1 0 0
1
−1
1 0 0 0 0 1 0 = 9V0 1 0 0 0 = −9V0 .
0 1 0
0
0
Wbb = χ2 |H |χ2 = 9V0
0
−1 0 0
0
0 1 0 0 0 1 1 = 9V0 0 1 0 0 = 0.
0 1 0
0
1
Wab = χ1 |H |χ2 = 9V0
−1 0 0
0
0
1 0 0 0 0 1 1 = 9V0 1 0 0 0 = 0.
0 1 0
0
1
Plug the expressions for Waa , Wbb , and Wab into Eq. 6.27:
1
E±
=
1
−9V0 + 0 ±
2
92 V02 + 0 =
To first-order, then, E1 = V0 − 9V0 ,
we got in (b).
1
(−9V0 ± 9V0 ) = {0, −9V0 }.
2
E2 = V0 , and these are consistent (to first order in 9) with what
Problem 6.10
Given a set of orthonornal states {ψj0 } that are degenerate eigenfunctions of the unperturbed Hamiltonian:
Hψj0 = E 0 ψj0 ,
construct the general linear combination,
ψ0 =
ψj0 |ψl0 = δjl ,
n
αj ψj0 .
j=1
It too is an eigenfunction of the unperturbed Hamiltonian, with the same eigenvalue:
H 0 ψ0 =
n
αj H 0 ψj0 = E 0
j=1
n
αj ψj0 = E 0 ψ 0 .
j=1
We want to solve the Schrödinger equation Hψ = Eψ for the perturbed Hamiltonian H = H 0 + λH .
Expand the eigenvalues and eigenfunctions as power series in λ:
E = E 0 + λE 1 + λ2 E 2 + . . . ,
ψ = ψ 0 + λψ 1 + λ2 ψ 2 + . . . .
Plug these into the Schrödinger equation and collect like powers:
(H 0 + λH )(ψ 0 + λψ 1 + λ2 ψ 2 + . . . ) = (E 0 + λE 1 + λ2 E 2 + . . . )(ψ 0 + λψ 1 + λ2 ψ 2 + . . . )
⇒
H 0 ψ 0 + λ(H 0 ψ 1 + H ψ 0 ) + . . . = E 0 ψ 0 + λ(E 0 ψ 1 + E 1 ψ 0 ) + . . .
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
163
The zeroth-order terms cancel; to first order
H 0 ψ1 + H ψ0 = E 0 ψ1 + E 1 ψ0 .
Take the inner product with ψj0 :
ψj0 |H 0 ψ 1 + ψj0 |H ψ 0 = E 0 ψj0 |ψ 1 + E 1 ψj0 |ψ 0 .
But
ψj0 |H 0 ψ 1 = H 0 ψj0 |ψ 1 = E 0 ψj0 |ψ 1 , so the first terms cancel, leaving
ψj0 |H ψ 0 = E 1 ψj0 |ψ 0 .
Now use
ψ0 =
n
αl ψl0 ,
and exploit the orthonormality of {ψl0 }:
l=1
n
αl ψj0 |H
|ψl0 =E
l=1
1
n
αl ψj0 |ψl0 = E 1 αj ,
l=1
or, defining
Wjl ≡ ψj0 |H |ψl0 ,
n
Wjl αl = E 1 αl .
l=1
This (the generalization of Eq. 6.22 for the case of n-fold degeneracy) is the eigenvalue equation for the matrix
W (whose jlth element, in the {ψj0 } basis, is Wjl ); E 1 is the eigenvalue, and the eigenvector (in the {ψj0 } basis)
is χj = αj . Conclusion: The first-order corrections to the energy are the eigenvalues of W. QED
Problem 6.11
m
(a) From Eq. 4.70: En = −
22
e2
4π90
2 1
1
= − mc2
n2
2
1 e2
c 4π90
2
1
α2 mc2
=
−
.
n2
2n2
(b) I have found a wonderful solution—unfortunately, there isn’t enough room on this page for the proof.
Problem 6.12
Equation 4.191 ⇒ V = 2En ,
e2
−
4π90
2 2 e
1
e2 1
m
for hydrogen. V = −
.
; En = −
2
4π90 r
2
4π90
n2
0 1
2 2 1
1
e
m
= −2
2
r
2
4π90
n2
⇒
0 1 1
1
1
me2
=
=
2
2
r
4π90 n
an2
So
(Eq. 4.72). QED
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164
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
Problem 6.13
In Problem 4.43 we found (for n = 3, l = 2, m = 1) that
s = 0 : 1 =
s = −1 :
0
0
1
r2
1
r3
1
=
1
4!
6!
3!
=
6!
(s + 6)!
6!
3a
2
s
.
(of course). 0 1
−1
1
1
1 2
5! 3a
= ·
=
=
r
6! 2
6 3a
9a
s = −2 :
s = −3 :
6!
(1) = 1
6!
rs =
3a
2
3a
2
−2
=
−3
Eq. 6.55 says
2
1
4
=
·
6 · 5 9a2
135a2
1
1
=
32 a
9a
Eq. 6.56 says
1
1
8
=
=
·
3
6 · 5 · 4 27a
405a3
. 1
2
=
(5/2) · 27 · a2
135a2
. 1
1
Eq. 6.64 says
=
3
2(5/2)3 · 27 · a
405a3
.
For s = −7 (or smaller) the integral does not converge: 1/r7 = ∞ in this state; this is reflected in the fact
that (−1)! = ∞.
Problem 6.14
Equation 6.53 ⇒ Er1 = −
Er1
1
=−
2mc2
1
n+
2
But Problem 2.12
1
Er1 = −
2mc2
From Eq. 2.69:
x4 =
2
1
ω −2 n+
2
2
2
x4 =
2
1
1
E = (n + )ω, V = mω 2 x2
2
2
⇒
1 2 4 4
1
2 2
ω mω x + m ω x .
2
4
1 x2 = (n + )
, so
2 mω
⇒
1
n+
2
1 2
E − 2EV + V 2 . Here
2mc2
1
2 ω 2 − n +
2
2
1
mω 4
2 ω 2 + m2 ω 4 x4 = − 2 x4 .
4
8c
2
a2 + a+ a− + a− a+ + a2−
4m2 ω 2 +
a2+ + a+ a− + a− a+ + a2− ,
2
n| a2+ a2− + a+ a− a+ a− + a+ a− a− a+ + a− a+ a+ a− + a− a+ a− a+ + a2− a2+ |n.
4m2 ω 2
(Note that only terms with equal numbers of raising and lowering operators will survive). Using Eq. 2.66,
x4 =
2
n| a2+
n(n − 1) |n − 2 + a+ a− (n |n) + a+ a− (n + 1) |n
2
2
4m ω
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
+a− a+ (n |n) + a− a+ (n + 1) |n + a2−
=
(n + 1)(n + 2) |n + 2
2
n|
n(n
−
1)
n(n
−
1)
|n
+ n (n |n) + (n + 1) (n |n)
4m2 ω 2
+n (n + 1) |n + (n + 1) (n + 1) |n +
=
165
(n + 1)(n + 2)
(n + 1)(n + 2) |n
2 n(n − 1) + n2 + (n + 1)n + n(n + 1) + (n + 1)2 + (n + 1)(n + 2)
2
2
4m ω
=
Er1
2mω
2
(n − n + n + n + n + n + n + n + 2n + 1 + n + 3n + 2) =
2
2
2
2
2
mω 4
2
3
=− 2 ·
· 3(3n2 + 2n + 1) = −
2
2
8c
4m ω
32
2
2 ω 2
mc2
2mω
2
(6n2 + 6n + 3).
(2n2 + 2n + 1).
Problem 6.15
Quoting the Laplacian in spherical coordinates (Eq. 4.13), we have, for states with no dependence on θ or φ:
d
2 d
p2 = −2 ∇2 = − 2
r2
.
r dr
dr
Question: Is it Hermitian?
Using integration by parts (twice), and test functions f (r) and g(r):
∞
∞
1 d
d
2
2
2 dg
2
2
2 dg
f |p g = −
f 2
f
r
4πr dr = −4π
r
dr
r dr
dr
dr
dr
0
0
∞
df dg
dg ∞
= −4π2 r2 f −
r2
dr
dr 0
dr
dr
0
∞
∞
∞
d
2
2 dg 2 df 2 df
= −4π r f − r g +
r
g dr
dr 0
dr 0
dr
dr
0
∞
dg
df = −4π2 r2 f
+ p2 f |g.
− r2 g
dr
dr 0
The boundary term at infinity vanishes for functions f (r) and g(r) that go to zero exponentially; the boundary
term at zero is killed by the factor r2 , as long as the functions (and their derivatives) are finite. So
f |p2 g = p2 f |g,
and hence p2 is Hermitian.
Now we apply the same argument to
4 d
p = 2
r dr
4
d 1 d
r
dr r2 dr
2
d
r
dr
2
,
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166
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
integrating by parts four times:
∞
d
1 d
4
4
2 d
2 dg
f |p g = 4π
f
r
r
dr
dr
dr r2 dr
dr
0
∞ ∞
1 d
1 d
4
2 d
2 dg
2 df d
2 dg
= 4π r f
r
r
dr
r
−
dr r2 dr
dr
dr dr r2 dr
dr
0
0
∞ ∞
1 d
1 d
df d
d
4
2 d
2 dg
2 dg
2 df
2 dg
= 4π
r f
r
−
r
r
r
dr
+
dr r2 dr
dr
dr dr
dr
r2 dr
dr dr
dr
0
0
∞
d 1 d
df d
d
dg 2 dg
2 dg
2 df
= 4π4 r2 f
r
−
r
+
r
dr r2 dr
dr
dr dr
dr
dr
dr dr 0
∞
dg
d 1 d
df
r2
dr
−
r2
2 dr
dr
r
dr
dr
0
d 1 d
df d
d
dg
1 d
2 dg
2 dg
2 df
2 d
2 df
= 4π4 r2 f
g
r
−
r
+
r
−
r
r
dr r2 dr
dr
dr dr
dr
dr
dr dr
dr r2 dr
dr
∞
d
d 1 d
df
+
r2
r2
g dr
2 dr
dr
dr
r
dr
0
∞
1 d
1 d
4
2 d
2 dg
2 d
2 df
= 4π r f
r
−r g
r
dr r2 dr
dr
dr r2 dr
dr
0
∞
df
dg
df
d
d
dg
+ p4 f |g
− 4π4
r2
−
r2
dr dr
dr
dr
dr dr 0
∞
This time there are four boundary terms to worry about. Infinity is no problem; the trouble comes at r = 0.
If the functions f and g went to zero at the origin (as they do for states with l > 0) we’d be OK, but states
with l = 0 go like exp(−r/na). So let’s test the boundary terms using
f (r) = e−r/na ,
g(r) = e−r/ma .
In this case
dg
1 2 −r/ma
=−
r e
dr
ma
dg
1
r2 − 2mar e−r/ma
r2
=
dr
(ma)2
1
1
2 dg
r2 − 2mar e−r/ma .
r
= − e−r/na
dr
na
(ma)2
r2
d
dr
df d
dr dr
This goes to zero as r → 0, so the second pair of boundary terms vanishes—but not the first pair:
1 d
2ma −r/ma
1
2 dg
1
−
r
=
e
r2 dr
dr
(ma)2
r
d 1 d
1
2 dg
2(ma)2 + 2mar − r2 e−r/ma
r
=
2
3
2
dr r dr
dr
(ma) r
dg
d 1 d
1 r2 f
2(ma)2 + 2mar − r2 e−r/ma e−r/na
r2
=
2
3
dr r dr
dr
(ma)
This does not vanish as r → 0; rather, it goes to 2/ma. For these particular states, then,
1
8π4 1
4
f |p g =
−
+ p4 f |g,
a
m n
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
167
or, tacking on the normalization factor,
ψn00 = √
1
e−r/na ,
π (na)3/2
ψn00 |p4 ψm00 =
84 (n − m)
+ p4 ψn00 |ψm00 ,
a4 (nm)5/2
and hence p4 is not Hermitian, for such states.
Problem 6.16
(a)
[L · S, Lx ] = [Lx Sx + Ly Sy + Lz Sz , Lx ] = Sx [Lx , Lx ] + Sy [Ly , Lx ] + Sz [Lz , Lx ]
= Sx (0) + Sy (−iLz ) + Sz (iLy ) = i(Ly Sz − Lz Sy ) = i(L × S)x .
Same goes for the other two components, so [L · S, L] = i(L × S).
(b) [L · S, S] is identical, only with L ↔ S: [L · S, S] = i(S × L).
(c) [L · S, J] = [L · S, L] + [L · S, S] = i(L × S + S × L) = 0.
(d) L2 commutes with all components of L (and S) , so
L · S, L2 = 0.
L · S, S 2 = 0.
(e) Likewise,
(f ) L · S, J 2 = L · S, L2 + L · S, S 2 + 2 [L · S, L · S] = 0 + 0 + 0 =⇒ L · S, J 2 = 0.
Problem 6.17
With the plus sign, j = l + 1/2 (l = j − 1/2) :
1
Equation 6.65 ⇒ Eso
=
=
1
Efs
=
=
Er1
+
(En )2
2mc2
1
Eso
(En )2
=
2mc2
3+
Eq. 6.57 ⇒
Er1
(En )2
=−
2mc2
4n
−3 .
j
(En )2 n j(j + 1) − (j − 12 )(j + 12 ) − 34
mc2
(j − 12 )j(j + 12 )
n
(En )2 n(j 2 + j − j 2 + 14 − 34 )
(En )2
=
.
1
1
2
2
mc
mc j(j + 12 )
(j − 2 )j(j + 2 )
4n
2n
−
+3+
j
j(j + 12 )
1
(En )2
2n
4n
1
−
2
j
+
=
3
−
.
2
2mc2
j(j + 12 )
j + 12
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168
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
With the minus sign, j = l − 1/2 (l = j + 1/2) :
Equation 6.65 ⇒
=
1
Efs
=
1
Eso
Eq. 6.57 ⇒
Er1
(En )2
=−
2mc2
4n
−3 .
j+1
(En )2 n j(j + 1) − (j + 12 )(j + 32 ) − 34
=
mc2
(j + 12 )(j + 1)(j + 32 )
(En )2 n(j 2 + j − j 2 − 2j − 34 − 34 )
−n
(En )2
=
.
1
3
2
mc
mc2 (j + 1)(j + 12 )
(j + 2 )(j + 1)(j + 2 )
2n
4n
(En )2
(En )2
−3+
−
=
1
2
2mc
j+1
2mc2
(j + 1)(j + 2 )
(En )2
=
2mc2
4n
3−
j + 12
Problem 6.18
.
For both signs, then,
1
Efs
3−
2n
1
1
+
2
j
+
2
(j + 1)(j + 12 )
(En )2
=
2mc2
4n
3−
j + 12
. QED
1 1
36 2πc
5
; E1 = −13.6 eV;
−
= − E1 ⇒ λ = −
9 4
36
5 E1
36 (2π)(1.97 × 10−11 × 106 eV · cm)
c = 1.97 × 10−11 MeV·cm; λ =
= 6.55 × 10−5 cm= 655 nm.
5
(13.6 eV)
4n
3.00 × 108 m/s
c
(En )2
1
14
3−
:
= 4.58 × 10 Hz.
ν= =
Equation 6.66 ⇒ Ef s =
λ
6.55 × 10−7 m
2mc2
j + 12
E30 − E20 = hν =
2πc
= E1
λ
For n = 2: l = 0 or l = 1, so j = 1/2 or 3/2. Thus n = 2 splits into two levels :
j = 1/2 :
E21
(E2 )2
=
2mc2
j = 3/2 : E21 =
(E2 )2
2mc2
8
3−
1
3−
8
2
5 (E2 )2
5
=−
=−
2 mc2
2
=−
2
(E1 )2
1
(13.6 eV)2
5
= −5.66 × 10−5 eV.
=
−
4
mc2
32 (.511 × 106 eV)
1 (E2 )2
1
= − (3.62 × 10−4 eV) = −1.13 × 10−5 eV.
2 mc2
32
For n = 3: l = 0, 1 or 2, so j = 1/2, 3/2 or 5/2. Thus n = 3 splits into three levels :
j = 1/2 :
E31
(E3 )2
=
2mc2
j = 3/2 :
E31
(E3 )2
=
2mc2
j = 5/2 : E31 =
(E3 )2
2mc2
12
3−
1
12
3−
2
3−
12
3
9
(E3 )2
=−
= −9
mc2
2
1
92
(E1 )2
1
= − (3.62 × 10−4 eV) = −2.01 × 10−5 eV.
mc2
18
=−
1
3 (E3 )2
= − (3.62 × 10−4 eV) = −0.67 × 10−5 eV.
2 mc2
54
=−
1 (E3 )2
1
(3.62 × 10−4 eV) = −0.22 × 10−5 eV.
=−
2 mc2
162
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
169
0
E3
j= 5/2
j= 3/2
j=1/2
0
E2
1 2 3
4
5
6
j= 3/2
j= 1/2
There are six transitions here; their energies are (E30 + E31 ) − (E20 + E21 ) = (E30 − E20 ) + ∆E, where
∆E ≡ E31 − E21 . Let β ≡ (E1 )2 /mc2 = 3.62 × 10−4 eV. Then:
1
3
1
7
1
( → ) : ∆E =
− −
β=−
β = −8.80 × 10−6 eV.
−
2
2
18
32
288
(
3
3
→ ):
2
2
5
3
( → ):
2
2
(
1
1
→ ):
2
2
∆E =
1
1
11
−
− −
β=
β = 4.61 × 10−6 eV.
54
32
864
1
1
65
∆E =
−
+
β=
β = 9.08 × 10−6 eV.
162
32
2592
∆E =
5
32
−
1
18
β=
29
β = 36.45 × 10−6 eV.
288
3
1
( → ):
2
2
1
5
119
∆E =
−
+
β=
β = 49.86 × 10−6 eV.
54
32
864
5
1
( → ):
2
2
1
5
389
∆E =
−
+
β=
β = 54.33 × 10−6 eV.
162
32
2592
Conclusion: There are six lines; one of them ( 12 → 32 ) has a frequency less than the unperturbed line, the
other five have (slightly) higher frequencies. In order they are: 32 → 32 ; 52 → 32 ; 12 → 12 ; 32 → 12 ; 52 → 12 . The
frequency spacings are:
ν2 − ν1
ν3 − ν3
ν4 − ν3
ν5 − ν4
ν6 − ν5
=
=
=
=
=
(∆E2 − ∆E1 )/2π
(∆E3 − ∆E2 )/2π
(∆E4 − ∆E3 )/2π
(∆E5 − ∆E4 )/2π
(∆E6 − ∆E5 )/2π
=
=
=
=
=
3.23 × 109
1.08 × 109
6.60 × 109
3.23 × 109
1.08 × 109
Hz
Hz
Hz
Hz
Hz
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170
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
Problem 6.19
1
j+
2
n − (j +
=
2
−
α2
=
α
1
2)
+
j+
α
n 1−
α2
2n(j+ 12 )
1
j+
2
1 2
2
1−
≈
− α2
≈
n − (j + 12 ) +
≈1−
1 α2
2 n2
Enj ≈ mc
2
1+
2
≈
1
2
n− j+
1
j+
2
α
+ j+
1
2
1
2
−
1
1−
2
α2
2(j+ 12 )
=
α
j+
2 1
2
1
α2
= (j + ) −
.
2
2(j + 12 )
α
α2
2(j+ 12 )
n−
2 −1/2
α2
α2
≈ 1+ 2 1+
2
n
n(j + 12 )
− α2
α
j+
1
2
α2
n(j + 12 )
α2
α4
1− 2 + 4
2n
2n
13.6 eV
α2
=−
1
+
n2
n2
α
j+
α
α2
.
1+
n
2n(j + 12 )
1 +
+
3 α4
α2
α4
=
1
−
+
8 n4
2n2
2n4
−n
3
1 + 4
j+2
n
j+
1
2
3
−
4
−n
3
1 + 4
j+2
α2
α2 mc2
1
+
−1 =−
2n2
n2
−1/2
.
n
j+
1
2
3
−
4
,
confirming Eq. 6.67.
Problem 6.20
Equation 6.59 ⇒ B =
B=
=
e
1
L.
4π90 mc2 r3
Say L = , r = a; then
1
e
4π90 mc2 a3
(1.60 × 10−19 C)(1.05 × 10−34 J · s)
2
4π 8.9 × 10−12 C /N · m2 (9.1 × 10−31 kg) (3 × 108 m/s) (0.53 × 10−10 m)
2
3
= 12 T.
So a “strong” Zeeman field is Bext 10 T, and a “weak” one is Bext 10 T. Incidentally, the earth’s field
(10−4 T) is definitely weak.
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
171
Problem 6.21
For n = 2, l = 0 (j = 1/2) or l = 1 (j = 1/2 or 3/2). The eight states are:
|1 = |2 0
|2 = |2 0
1
2
|3 = |2 1
1 1
2 2
|4 = |2 1
1 1
2 2
1
2
− 12
gJ = 1 +
(1/2)(3/2) + (3/4)
3/2
=1+
= 2.
2(1/2)(3/2)
3/2
gJ = 1 +
−1/2
(1/2)(3/2) − (1)(2) + (3/4)
=1+
= 2/3.
2(1/2)(3/2)
3/2
− 12
In these four cases, Enj = −
|5 = |2 1
3 3
2 2
|6 = |2 1
3 1
2 2
|7 = |2 1
3
2
|8 = |2 1
3
2
13.6 eV
α2
1+
4
4
− 12
3
− 2
In these four cases, Enj
gJ = 1 +
2 3
−
1 4
5
= −3.4 eV 1 + α2 .
16
(3/2)(5/2) − (1)(2) + (3/4)
5/2
=1+
= 4/3.
2(3/2)(5/2)
15/2
α2
= −3.4 eV 1 +
4
2 3
−
2 4
1 2
= −3.4 eV 1 + α .
16
E1 = −3.4 eV 1 +
5 2
16 α
+ µB Bext .
E2 = −3.4 eV 1 +
5 2
16 α
− µB Bext .
E3 = −3.4 eV 1 +
5 2
16 α
+ 13 µB Bext .
E4 = −3.4 eV 1 +
5 2
16 α
− 13 µB Bext .
E5 = −3.4 eV 1 +
1 2
16 α
+ 2µB Bext .
E6 = −3.4 eV 1 +
1 2
16 α
+ 23 µB Bext .
E7 = −3.4 eV 1 +
1 2
16 α
− 23 µB Bext .
E8 = −3.4 eV 1 +
1 2
16 α
− 2µB Bext .
The energies are:
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172
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
E
µBBext
5 (slope 2)
6 (slope 2/3)
-3.4 (1+α2/16) eV
7 (slope -2/3)
8 (slope -2)
1 (slope 1)
3 (slope 1/3)
-3.4 (1+5α2/16) eV
4 (slope -1/3)
2 (slope -1)
Problem 6.22
1
Efs
= nlml ms |(Hr + Hso )|nlml ms = −
4n
2 ml ms
En2
e2
.
−3 +
2
2
2
2mc l + 1/2
8π90 m c l(l + 1/2)(l + 1)n3 a3
2En2
E1
α2
2E1
−
=
=
−
(13.6 eV). (Problem 6.11.)
mc2
mc2
n4
n4 2 2 2
Now
2 2
2 2
2 3
e
e2
e
(me
)
e
m
= α2 (13.6 eV).
8π9 m2 c2 a3 = 2 · 4π9 m2 c2 (4π9 2 )3 = 22 4π9
4π90 c
0
0
0
0
1
Efs
1
13.6 eV 2
3
m l ms
13.6 eV 2 3
l(l + 1) − ml ms
=
α −
α
+
+
=
−
. QED
n3
(l + 1/2) 4n l(l + 1/2)(l + 1)
n3
4n l(l + 1/2)(l + 1)
Problem 6.23
The Bohr energy is the same for all of them: E2 = −13.6 eV/22 = −3.4 eV. The Zeeman contribution is the
1
second term in Eq. 6.79: µB Bext (ml +2ms ). The fine structure is given by Eq. 6.82: Efs
= (13.6 eV/8)α2 {· · · } =
2
(1.7 eV)α {· · · }. In the table below I record the 8 states, the value of (ml + 2ms ), the value of {· · · } ≡
l(l + 1) − ml ms
3
−
, and (in the last column) the total energy, −3.4 eV [1−(α2 /2){· · · }]+(ml +2ms )µB Bext .
8
l(l + 1/2)(l + 1)
State = |nlml ms |1 = |2 0 0 12 |2 = |2 0 0 − 12 |3 = |2 1 1 12 |4 = |2 1 − 1 − 12 |5 = |2 1 0 12 |6 = |2 1 0 − 12 |7 = |2 1 1 − 12 |8 = |2 1 − 1 12 (ml + 2ms )
1
−1
2
−2
1
−1
0
0
{· · · }
−5/8
−5/8
−1/8
−1/8
−7/24
−7/24
−11/24
−11/24
Total Energy
-3.4 eV [1 + (5/16)α2 ] + µB Bext
-3.4 eV [1 + (5/16)α2 ] − µB Bext
-3.4 eV [1 + (1/16)α2 ] + 2µB Bext
-3.4 eV [1 + (1/16)α2 ] − 2µB Bext
-3.4 eV [1 + (7/48)α2 ] + µB Bext
-3.4 eV [1 + (7/48)α2 ] − µB Bext
-3.4 eV [1 + (11/48)α2 ]
-3.4 eV [1 + (11/48)α2 ]
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
173
Ignoring fine structure there are five distinct levels—corresponding to the possible values of (ml + 2ms ):
2
(d = 1); 1
(d = 2); 0
(d = 2); −1
(d = 2); −2
(d = 1).
Problem 6.24
e
e
Bext · L + 2S =
Bext 2ms = 2ms µB Bext (same as the Zeeman term in Eq. 6.79,
2m
2m
α2
3
13.6 eV
with ml = 0). Equation 6.67 ⇒ Enj = −
(since j = 1/2). So the total energy is
1
+
n
−
n2
n2
4
Equation 6.72 ⇒ Ez1 =
α2
13.6 eV
1
+
E=−
n2
n2
3
n−
4
+ 2ms µB Bext .
13.6 eV 2 3
3
13.6 eV 2
=
− 1 , which is the same as
Fine structure is the α term:
= −
α n−
α
n4
4
n3
4n
Eq. 6.82, with the term in square brackets set equal to 1. QED
1
Efs
2
Problem 6.25
1
Equation 6.66 ⇒ Efs
=
1
Efs
E1
=−
32
α2
2
E22
2mc2
3−
8
3−
j + 1/2
8
j + 1/2
=
E12
32mc2
3−
8
j + 1/2
;
E1
α2
(Problem 6.11), so
=−
2
mc
2
8
13.6 eV 2
8
=
α 3−
=γ 3−
.
64
j + 1/2
j + 1/2
8
For j = 1/2 (ψ1 , ψ2 , ψ6 , ψ8 ), Hfs1 = γ(3 − 8) = −5γ. For j = 3/2 (ψ3 , ψ4 , ψ5 , ψ7 ), Hfs1 = γ(3 − ) = −γ.
2
This confirms all the γ terms in −W (p. 281). Meanwhile, Hz = (e/2m)Bext (Lz + 2Sz ) (Eq. 6.71); ψ1 , ψ2 , ψ3 , ψ4
are eigenstates of Lz and Sz ; for these there are only diagonal elements:
Hz =
e
Bext (ml + 2ms ) = (ml + 2ms )β;
2m
Hz 11 = β;
This confirms the upper left corner of −W. Finally:
Hz 55
(Lz + 2Sz )|ψ5 = + 23 |1 0| 12 12
Hz 66
(Lz + 2Sz )|ψ6 = − 13 |1 0| 12 12
Hz 77
so
Hz 88
2
1
1
(Lz + 2Sz )|ψ7 = − 3 |1 0| 2 − 2
Hz 56
(Lz + 2Sz )|ψ8 = − 13 |1 0| 12 − 12
Hz 78
=
=
=
=
=
=
Hz 22 = −β;
Hz 33 = 2β;
Hz 44 = −2β.
(2/3)β,
(1/3)β,
−(2/3)β,
−(1/3)β, √
Hz 65 = −(√2/3)β,
Hz 87 = −( 2/3)β,
which confirms the remaining elements.
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174
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
Problem 6.26
There are eighteen n = 3 states (in general, 2n2 ).
WEAK FIELD
Equation 6.67 ⇒ E3j
α2
13.6 eV
1+
=−
9
9
3
3
−
j + 1/2 4
α2
= −1.51 eV 1 +
3
1
1
−
j + 1/2 4
.
Equation 6.76 ⇒ Ez1 = gJ mj µB Bext .
State |3 l j mj gJ (Eq. 6.75)
l = 0, j = 1/2
|3 0
1 1
2 2
l = 0, j = 1/2
|3 0
1
2
l = 1, j = 1/2
|3 1
1 1
2 2
l = 1, j = 1/2
|3 1
1
2
l = 1, j = 3/2
|3 1
3 3
2 2
l = 1, j = 3/2
|3 1
3 1
2 2
l = 1, j = 3/2
|3 1
3
2
−
1
2
l = 1, j = 3/2
|3 1
3
2
− 32 l = 2, j = 3/2
|3 2
3 3
2 2
l = 2, j = 3/2
|3 2
3 1
2 2
l = 2, j = 3/2
|3 2
3
2
−
1
2
l = 2, j = 3/2
|3 2
3
2
− 32 l = 2, j = 5/2
|3 2
5 5
2 2
l = 2, j = 5/2
|3 2
l = 2, j = 5/2
1
3
1
j+1/2
−
2
1/4
2
1/4
2/3
1/4
2/3
1/4
4/3
1/12
4/3
1/12
4/3
1/12
4/3
1/12
4/5
1/12
4/5
1/12
4/5
1/12
4/5
1/12
6/5
1/36
5 3
2 2
6/5
1/36
|3 2
5 1
2 2
6/5
1/36
l = 2, j = 5/2
|3 2
5
2
− 12 6/5
1/36
l = 2, j = 5/2
|3 2
5
2
−
3
2
6/5
1/36
l = 2, j = 5/2
|3 2
5
2
− 52 6/5
1/36
− 12 − 12 1
4
Total Energy
2
−1.51 eV 1 + α4 + µB Bext
2
−1.51 eV 1 + α4 − µB Bext
2
−1.51 eV 1 + α4 + 13 µB Bext
2
−1.51 eV 1 + α4 − 13 µB Bext
2
−1.51 eV 1 + α12 + 2µB Bext
2
−1.51 eV 1 + α12 + 23 µB Bext
2
−1.51 eV 1 + α12 − 23 µB Bext
2
−1.51 eV 1 + α12 − 2µB Bext
2
−1.51 eV 1 + α12 + 65 µB Bext
2
−1.51 eV 1 + α12 + 25 µB Bext
2
−1.51 eV 1 + α12 − 25 µB Bext
2
−1.51 eV 1 + α12 − 65 µB Bext
2
−1.51 eV 1 + α36 + 3µB Bext
2
−1.51 eV 1 + α36 + 95 µB Bext
2
−1.51 eV 1 + α36 + 35 µB Bext
2
−1.51 eV 1 + α36 − 35 µB Bext
2
−1.51 eV 1 + α36 − 95 µB Bext
2
−1.51 eV 1 + α36 − 3µB Bext
STRONG FIELD
Equation 6.79 ⇒ −1.51 eV + (ml + 2ms )µB Bext ;
13.6 eV 2
Equation 6.82 ⇒
α
27
1
l(l + 1) − ml ms
−
4
l(l + 1/2)(l + 1)
α2
= −1.51 eV
3
l(l + 1) − ml ms
1
−
l(l + 1/2)(l + 1) 4
.
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publisher.
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
Etot
1
= −1.51 eV(1 + α A) + (ml + 2ms )µB Bext , where A ≡
3
175
l(l + 1) − ml ms
1
−
l(l + 1/2)(l + 1) 4
2
.
These terms are given in the table below:
State |n l ml ms l=0
l=0
l=1
l=1
|3 0 0
(ml + 2ms )
A
1
1/4
1
2
|3 0 0 − 12 −1
|3 1 1 12 |3 1 − 1 − 12 1/4
2
1/12
−2
l=1
|3 1 0
l=1
|3 1 0 − 12 l=1
|3 1 − 1
1
2
0
7/36
l=1
|3 1 1 − 12 0
7/36
l=2
|3 2 2 12 3
1/36
l=2
1
2
1/12
|3 2 − 2 − 12 l=2
|3 2 1
l=2
|3 2 − 1 − 12 l=2
l=2
|3 2 0
1
5/36
−1
5/36
−3
1
2
2
7/180
−2
7/180
1
2
|3 2 0 − 12 1/36
1
1/20
−1
1/20
l=2
|3 2 − 1 12 0
11/180
l=2
|3 2 1 − 12 0
11/180
l=2
|3 2 − 2 12 −1
13/180
l=2
1
2
|3 2 2 −
1
13/180
Total Energy
−1.51 eV 1 +
−1.51 eV 1 +
−1.51 eV 1 +
−1.51 eV 1 +
−1.51 eV 1 +
−1.51 eV 1 +
−1.51 eV 1 +
−1.51 eV 1 +
−1.51 eV 1 +
−1.51 eV 1 +
−1.51 eV 1 +
−1.51 eV 1 +
−1.51 eV 1 +
−1.51 eV 1 +
−1.51 eV 1 +
−1.51 eV 1 +
−1.51 eV 1 +
−1.51 eV 1 +
α2
4 2
α
4
2
α
12 2
+ µB Bext
− µB Bext
+ 2µB Bext
− 2µB Bext
+ µB Bext
α
12
5α2
36 5α2
36 − µB Bext
7α2
36 7α2
36
α2
36 2
α
36
+ 3µB Bext
− 3µB Bext
+ 2µB Bext
7α2
180 7α2
180
α2
20 2
α
20
2
11α
180
− 2µB Bext
+ µB Bext
− µB Bext
11α2
180 13α2
180 13α2
180
− µB Bext
+ µB Bext
INTERMEDIATE FIELD
As in the book, I’ll use the basis |n l j mj (same as for weak field); then the fine structure matrix elements
are diagonal: Eq. 6.66 ⇒
1
Efs
γ≡
E32
=
2mc2
12
3−
j + 1/2
E12
=
54mc2
4
1−
j + 1/2
E1 α 2
=−
108
4
1−
j + 1/2
4
= 3γ 1 −
j + 1/2
,
13.6 eV 2
1
1
1
α . For j = 1/2, Efs
= −9γ; for j = 3/2, Efs
= −3γ; for j = 5/2, Efs
= −γ.
324
The Zeeman Hamiltonian is Eq. 6.71: Hz = 1 (Lz + 2Sz )µB Bext . The first eight states (l = 0 and l = 1) are
the same as before (p. 281), so the β terms in W are unchanged; recording just the non-zero blocks of −W:
'
(9γ − β), (9γ + β), (3γ − 2β), (3γ + 2β),
√
2
(3γ√− 23 β)
3 β
2
(9γ − 13 β)
3 β
( '
,
√
2
(3γ√+ 23 β)
3 β
2
(9γ + 13 β)
3 β
(
.
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176
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
The other 10 states (l = 2) must first be decomposed into eigenstates of Lz and Sz :
| 52
5
2
= |2 2| 12
1
2
=⇒ (γ − 3β)
| 52 − 52 = |2 − 2| 12 − 12 =⇒ (γ + 3β)
| 52
3
2
=
| 32
3
2
=
| 52
1
2
=
| 32
1
2
=
1
|2
5
2| 12 − 12 +
4
5 |2
2| 12 − 12 −
2
|2
5
1| 12 − 12 +
| 52 − 12 =
| 32 − 12 =
| 52 − 32 =
| 32 − 32 =
3
5 |2
1| 12 − 12 −
3
|2
5
2
5 |2
4
|2
5
1
5 |2
1| 12
1
2
1
5 |2
1| 12
1
2
3
|2
5
0| 12
1
2
0| 12
1
2
4
|2
5
0| 12 − 12 +
0| 12 − 12 −
2
5 |2
2
|2
5
− 1| 12 − 12 +
− 1| 12 − 12 −
3
5 |2
'
=⇒
1
2
− 1| 12
1
2
1
|2
5
4
5 |2
√
6
(γ √
− 35 β)
5 β
6
(3γ − 25 β)
5 β
− 1| 12
2
(γ − 95 β)
5β
2
β
(3γ
− 65 β)
5
=⇒
'
=⇒
− 2| 12
1
2
− 2| 12
1
2
√
(
6
(γ √
+ 35 β)
5 β
6
(3γ + 25 β)
5 β
=⇒
(
2
(γ + 95 β)
5β
2
β
(3γ
+ 65 β)
5
[Sample Calculation: For the last two, letting Q ≡ 1 (Lz + 2Sz ), we have
Q| 52 − 32 = −2 45 |2 − 1| 12 − 12 − 15 |2 − 2| 12 12 ;
Q| 32 − 32 = −2 15 |2 − 1| 12 − 12 + 45 |2 − 2| 12 12 .
52 − 32 |Q| 52 − 32 = (−2) 45 − 15 = − 95 ; 32 − 32 |Q| 32 − 32 = (−2) 15 − 45 = − 65 ;
52 − 32 |Q| 32 − 32 = −2 45 15 + 15 45 = − 45 + 25 = − 25 = 32 − 32 |Q| 52 − 32 .]
So the 18 × 18 matrix −W splits into six 1 × 1 blocks and six 2 × 2 blocks. We need the eigenvalues of the
2 × 2 blocks. This means solving 3 characteristic equations (the other 3 are obtained trivially by changing the
sign of β):
2
3γ − β − λ
3
9
γ− β−λ
5
3
γ− β−λ
5
1
2
9γ − β − λ − β 2 = 0 =⇒ λ2 + λ(β − 12γ) + γ(27γ − 7β) = 0.
3
9
6
4 2
33
2
2
2
= 0.
3γ − β − λ − β = 0 =⇒ λ + λ(3β − 4γ) + γ 3γ − γβ + 2β
5
25
5
11
2
6 2
2
3γ − β − λ − β = 0 =⇒ λ + λ(β − 4γ) + γ 3γ − β = 0.
5
25
5
The solutions are:
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
λ = −β/2 + 6γ ± (β/2)2 + βγ + 9γ 2
λ = −3β/2 + 2γ ± (β/2)2 + 35 βγ + γ 2
λ = −β/2 + 2γ ± (β/2)2 + 15 βγ + γ 2
⇒
91
92
93
94
95
=
=
=
=
=
177
E3 − 9γ + β
E3 − 3γ + 2β
E3 − γ + 3β
E3 − 6γ + β/2 + 9γ 2 + βγ + β 2 /4
E3 − 6γ + β/2 − 9γ 2 + βγ + β 2 /4
96 = E3 − 2γ + 3β/2 +
γ 2 + 35 βγ + β 2 /4
97 = E3 − 2γ + 3β/2 − γ 2 + 35 βγ + β 2 /4
98 = E3 − 2γ + β/2 + γ 2 + 15 βγ + β 2 /4
99 = E3 − 2γ + β/2 − γ 2 + 15 βγ + β 2 /4
(The other 9 9’s are the same, but with β → −β.) Here γ =
13.6 eV 2
324 α ,
and β = µB Bext .
In the weak-field limit (β γ):
2
94 ≈ E3 − 6γ + β/2 + 3γ 1 + β/9γ ≈ E3 − 6γ + β/2 + 3γ(1 + β/18γ) = E3 − 3γ + β.
3
1
95 ≈ E3 − 6γ + β/2 − 3γ(1 + β/18γ) = E3 − 9γ + β.
3
9
96 ≈ E3 − 2γ + 3β/2 + γ(1 + 3β/10γ) = E3 − γ + β.
5
6
97 ≈ E3 − 2γ + 3β/2 − γ(1 + 3β/10γ) = E3 − 3γ + β.
5
3
98 ≈ E3 − 2γ + β/2 + γ(1 + β/10γ) = E3 − γ + β.
5
2
99 ≈ E3 − 2γ + β/2 − γ(1 + β/10γ) = E3 − 3γ + β.
5
1.51 eV 2
2
Noting that γ = −(E3 /36)α = 36 α , we see that the weak field energies are recovered as in the first table.
In the strong-field limit (β γ):
94 ≈ E3 − 6γ + β/2 + β/2 1 + 4γ/β ≈ E3 − 6γ + β/2 + β/2(1 + 2γ/β) = E3 − 5γ + β.
95 ≈ E3 − 6γ + β/2 − β/2(1 + 2γ/β) = E3 − 7γ.
7
96 ≈ E3 − 2γ + 3β/2 + β/2(1 + 6γ/5β) = E3 − γ + 2β.
5
97 ≈ E3 − 2γ + 3β/2 − β/2(1 + 6γ/5β) = E3 −
13
γ + β.
5
9
98 ≈ E3 − 2γ + β/2 + β/2(1 + 2γ/5β) = E3 − γ + β.
5
11
γ.
5
Again, these reproduce the strong-field results in the second table.
In the figure below each line is labeled by the level number and (in parentheses) the starting and ending
slope; for each line there is a corresponding one starting from the same point but sloping down.
99 ≈ E3 − 2γ + β/2 − β/2(1 + 2γ/5β) = E3 −
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178
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
3(3)
E
6(9/5−> 2)
8(3/5−>1)
2(2)
Ε3
7(6/5−>1)
4(2/5−>1)
Ε 3 −γ
9(2/5−>0)
Ε 3 −3γ
1(1)
5(1/3−>0)
Ε 3 −9γ
Problem 6.27
I ≡ (a · r̂)(b · r̂) sin θ dθ dφ
= (ax sin θ cos φ + ay sin θ sin φ + az cos θ)(bx sin θ cos φ + by sin θ sin φ + bz cos θ) sin θ dθ dφ.
2π
But
0
2π
sin φ dφ =
2π
cos φ dφ =
0
sin φ cos φ dφ = 0, so only three terms survive :
0
(ax bx sin2 θ cos2 φ + ay by sin2 θ sin2 φ + az bz cos2 θ) sin θ dθ dφ.
I=
2
But
I=
0
cos φ dφ = π,
0
π
2π
2
sin φ dφ =
0
so
2π
2π
dφ = 2π, so
0
π(ax bx + ay by ) sin2 θ + 2πaz bz cos2 θ sin θ dθ.
π
sin3 θ dθ =
But
0
4
,
3
π
cos2 θ sin θ dθ =
0
2
,
3
2
4
4π
4π
I = π(ax bx + ay by ) + 2πaz bz =
(ax bx + ay by + az bz ) =
(a · b). QED
3
3
3
3
[Alternatively, noting that I has to be a scalar bilinear in a and b, we know immediately that I = A(a·b), where
A is some constant (same for all a and b). To determine A, pick a = b = k̂; then I = A = cos2 θ sin θ dθ dφ =
4π/3.]
√
For states with l = 0, the wave function is independent of θ and φ (Y00 = 1/ 4π), so
0
1 ∞
3(Sp · r̂)(Se · r̂) − Sp · Se
1
2 2
=
|ψ(r)| r dr
[3(Sp · r̂)(Se · r̂)] sin θ dθ dφ.
r3
r3
0
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
179
The first angular integral is 3(4π/3)(Sp · Se ) = 4π(Sp · Se ), while the second is −(Sp · Se ) sin θ dθ dφ =
−4π(Sp · Se ), so the two cancel, and the result is zero. QED [Actually, there is a little sleight-of-hand here,
since for l = 0, ψ → constant as r → 0, and hence the radial integral diverges logarithmically at the origin.
Technically, the first term in Eq. 6.86 is the field outside an infinitesimal sphere; the delta-function gives the
field inside. For this reason it is correct to do the angular integral first (getting zero) and not worry about the
radial integral.]
Problem 6.28
g
; we want reduced mass in a, but not in mp me (which come from
From Eq. 6.89 we see that ∆E ∝
mp me a3
Eq. 6.85); the notation in Eq. 6.93 obscures this point.
(a) g and mp are unchanged; me → mµ = 207me , and a → aµ .
From Eq. 4.72, a ∝ 1/m, so
a
207
mµ (reduced)
m µ mp
1
207
207
=
= 186.
=
=
·
=
=
(9.11×10−31 )
aµ
me
mµ + mp me
1 + 207(me /mp )
1.11
1 + 207 1.67×10−27 )
∆E = (5.88 × 10−6 eV) (1/207) (186)3 = 0.183 eV.
(b) g : 5.59 → 2;
a
mp (reduced)
m2e
1
1
=
=
·
= .
ap
me
me + me me
2
3
1
2
1.67 × 10−27
eV)
= 4.82 ×10−4 eV.
5.59
9.11 × 10−31
2
m p → me ;
∆E = (5.88 × 10−6
a
mm (reduced)
m e mµ
1
207
.
=
=
·
=
am
me
m e + mµ m e
208
3
2
1.67 × 10−27
207
−6
∆E = (5.88 × 10 )
= 1.84×10−5 eV.
5.59
(207)(9.11 × 10−31 )
208
(c) g : 5.59 → 2;
m p → mµ ;
Problem 6.29
Use perturbation theory:
1 1
e2
H =−
−
,
4π90 b
r
for
0 < r < b. ∆E = ψ|H |ψ,
with ψ ≡ √
1
πa3
e−r/a .
b
b
b
e2 1
1 1 −2r/a 2
1
e2
2 −2r/a
−2r/a
4π
r
dr
=
−
r
e
dr
−
re
dr
−
e
4π90 πa3
b
r
π90 a3 b 0
0
0
b
2
2
e2
1
2r
2r
a
a 2 −2r/a
a
−2r/a
−2r/a
=−
−
−
+a
e
e
− r e
−1 −
−1
π90 a3 b
2
2
a
2
a
0
a2 −2b/a
a3
a2
a 2 −2b/a a3 −2b/a
2b
2b
e2
−1 − e
−1 +
−
− b e
−
−
=−
+ e
π90 a3
2b
4b
a
4
a
4b
4
2
2
3
2
2
e
ab a
a
ab a
a a
=−
e−2b/a −
−
−
+
+
+
−1
π90 a3
2
2
4b
2
4
4 b
e2
a
a2 a
e2
a
a −2b/a
a2
−2b/a
=−
+
1
+
−
1
=
1
−
+
1
+
e
e
−
.
π90 a3
4
b
4 b
4π90 a
b
b
∆E = −
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180
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
Let +2b/a = 9 (very small). Then the term in square brackets is:
2
2
92
93
1−
+ 1+
1−9+
−
+ ···
9
9
2
6
2
3
2✄
92
2✄
9
9
92
=1
+ ✁9 −
−
+ ( )93 + · · · =
+ ( )93 + ( )94 · · ·
✁− ✄ +1
✁ + ✄ − ✁9 − 2
✁+
2
6
3
6
✄9
✄9
e2 1 4b2
To leading order, then, ∆E =
.
4π90 a 6a2
E = E1 = −
∆E
e2
=
E
4π90
me4
;
2(4π90 )2 2
−
a=
4π90 2
;
me2
so
Ea = −
e2
.
2(4π90 )
2
2(4π90 ) 2b2
4 b
=
−
.
e2
3a2
3 a
Putting in a = 5 × 10−11 m:
∆E
16
4
10−15
= − × 10−10 ≈ −5 × 10−10 .
=−
−11
E
3 5 × 10
3
By contrast,
fine structure:
∆E/E ≈ α2 = (1/137)2 = 5 × 10−5 ,
hyperfine structure: ∆E/E ≈ (me /mp )α2 = (1/1800)(1/137)2 = 3 × 10−8 .
So the correction for the finite size of the nucleus is much smaller (about 1% of hyperfine).
Problem 6.30
(a) In terms of the one-dimensional harmonic oscillator states {ψn (x)}, the unperturbed ground state is
|0 = ψ0 (x)ψ0 (y)ψ0 (z).
E01 = 0|H |0 = ψ0 (x)ψ0 (y)ψ0 (z)|λx2 yz|ψ0 (x)ψ0 (y)ψ0 (z) = λx2 0 y0 z0 .
But
y0 = z0 = 0.
So there is no change, in first order.
(b) The (triply degenerate) first excited states are
|1 = ψ0 (x)ψ0 (y)ψ1 (z)
|2 = ψ0 (x)ψ1 (y)ψ0 (z)
|3 = ψ1 (x)ψ0 (y)ψ0 (z)
In this basis the perturbation matrix is
Wij = i|H |j,
i = 1, 2, 3.
1|H |1 = ψ0 (x)ψ0 (y)ψ1 (z)|λx yz|ψ0 (x)ψ0 (y)ψ1 (z) = λx2 0 y0 z1 = 0,
2
2|H |2 = ψ0 (x)ψ1 (y)ψ0 (z)|λx2 yz|ψ0 (x)ψ1 (y)ψ0 (z) = λx2 0 y1 z0 = 0,
3|H |3 = ψ1 (x)ψ0 (y)ψ0 (z)|λx2 yz|ψ1 (x)ψ0 (y)ψ0 (z) = λx2 1 y0 z0 = 0,
1|H |2 = ψ0 (x)ψ0 (y)ψ1 (z)|λx2 yz|ψ0 (x)ψ1 (y)ψ0 (z) = λx2 0 0|y|11|z|0
2
=λ
[using Problems 2.11 and 3.33].
|0|x|1|2 = λ
2mω
2mω
1|H |3 = ψ0 (x)ψ0 (y)ψ1 (z)|λx2 yz|ψ1 (x)ψ0 (y)ψ0 (z) = λ0|x2 |1y0 1|z|0 = 0,
2|H |3 = ψ0 (x)ψ1 (y)ψ0 (z)|λx2 yz|ψ1 (x)ψ0 (y)ψ0 (z) = λ0|x2 |11|y|00 z0 = 0.
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
181
2
0 a0
W = a 0 0 , where a ≡ λ
.
2mω
0 0 0
−E a 0 2
a −E 0 = −E 3 + Ea2 = 0 ⇒ E = {0, ±a} = 0, ±λ
.
2mω
0 0 −E Eigenvalues of W :
Problem 6.31
(a) The first term is the nucleus/nucleus interaction, the second is the interaction between the nucleus of
atom 2 and the electron in atom 1, the third is between nucleus 1 and electron 2, and the last term is the
interaction between the electrons.
x x 2
1
1 x −1
1
=
1−
1+
+
=
+ ... ,
R−x
R
R
R
R
R
so
1 e2
H ∼
=
4π90 R
≈
1 e2
4π90 R
1− 1+
−
2x1 x2
R2
x 1
R
=−
+
x 2
1
R
− 1−
x 2
R
+
x 2
2
R
+ 1+
x1 − x2
R
+
x1 − x2
R
2 e2 x1 x2
. 2π90 R3
(b) Expanding Eq. 6.99:
1
1
e2
x2 − x2−
p2+ + p2− + k x2+ + x2− −
2m
2
4π90 R3 +
1
1
e2
=
(2x1 x2 ) = H 0 + H
p21 + p22 + k x21 + x22 −
2m
2
4π90 R3
H =
(c)
ω± =
k
m
e2
1∓
2π90 R3 k
1/2
(Eqs. 6.96 and 6.98).
2
1
1
e2
e2
∼
+ ... .
−
= ω0 1 ∓
2 2π90 R3 mω02
8 2π90 R3 mω02
2
1
e2
e2
−
+
2π90 R3 mω02
8 2π90 R3 mω02
2
1
e2
e2
1
1+
−
− ω0
2 2π90 R3 mω02
8 2π90 R3 mω02
2
2 2
1
e
1
1
1 e2
−
=
=
−
. 2ω0
4
2π90 R3 mω02
8 m2 ω03 2π90
R6
1
1
∆V ∼
= ω0 1 −
2
2
(d) In first order:
E01 = 0|H |0 = −
e2
e2
ψ
(x
)ψ
(x
)|x
x
|ψ
(x
)ψ
(x
)
=
−
x0 x0 = 0.
0
1
0
2
1
2
0
1
0
2
2π90 R3
2π90 R3
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182
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
In second order:
∞
|ψn |H |ψ0 |2
. Here |ψ0 = |0|0, |ψn = |n1 |n2 , so
E0 − En
n=1
2 ∞ ∞
e2
|n1 |x1 |0|2 |n2 |x2 |0|2
=
[use Problem 3.33]
3
2π90 R
E0,0 − En1 ,n2
n1 =1 n2 =1
2
e2
|1|x|0|2 |1|x|0|2
=
[zero unless n1 = n2 = 1]
2π90 R3
( 12 ω0 + 12 ω0 ) − ( 32 ω0 + 32 ω0 )
2 2
2 2
e2
1
e
1
=
−
=− 2 3
. 3
2π90 R
2ω0
2mω0
8m ω0 2π90
R6
E02 =
Problem 6.32
(a) Let the unperturbed Hamiltonian be H(λ0 ), for some fixed value λ0 . Now tweak λ to λ0 + dλ. The
perturbing Hamiltonian is H = H(λ0 + dλ) − H(λ0 ) = (∂H/∂λ) dλ (derivative evaluated at λ0 ).
The change in energy is given by Eq. 6.9:
dEn = En1 = ψn0 |H |ψn0 = ψn |
∂H
|ψn dλ (all evaluated at λ0 );
∂λ
so
∂En
∂H
= ψn |
|ψn .
∂λ
∂λ
[Note: Even though we used perturbation theory, the result is exact, since all we needed (to calculate the
derivative) was the infinitesimal change in En .]
(b) En = (n + 12 )ω;
H=−
2 d 2
1
+ mω 2 x2 .
2
2m dx
2
(i)
∂En
1
= (n + );
∂ω
2
V =
1
mω 2 x2 ,
2
so
∂H
= mωx2 ;
∂ω
1
so F-H ⇒ (n + ) = n|mωx2 |n. But
2
1
1
1
V = n| mω 2 x2 |n = ω(n + );
2
2
2
(ii)
∂En
1
= (n + )ω;
∂
2
2
∂H
d2
=
=−
∂
m dx2
1
2
so F-H ⇒ (n + )ω = n|T |n,
2
or
−
2 d 2
2m dx2
∂H
1
1
2 d 2
+ ω 2 x2 = −
=
∂m
2m2 dx2
2
m
1
1
So F-H ⇒ 0 = − T + V , or
m
m
Problems 2.12 and 3.31.
=
2
T;
T = 12 (n + 12 )ω.
(iii)
∂En
= 0;
∂m
V = 12 (n + 12 )ω.
T = V .
2 d 2
−
2m dx2
1
+
m
1
mω 2 x2
2
=−
1
1
T + V.
m
m
These results are consistent with what we found in
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
183
Problem 6.33
(a)
∂En
4
4me3
= En ;
=−
2
2
2
∂e
32π 90 (jmax + l + 1)2
e
4
e
En = −
e
2π90
0 1
0 1
2 2 m
1
1
8π90
1
e
8π90 E1
8π90
e2 m 1
, or
= − 2 En = − 2 2 = − 2 − 2
=
.
2
r
r
e
e n
e
2
4π90
n
4π90 2 n2
4π90 2
= a (by Eq. 4.72), so
me2
But
∂H
2e 1
=−
. So the F-H theorem says:
∂e
4π90 r
0 1
1
1
= 2 .
r
n a
(Agrees with Eq. 6.55.)
(b)
∂H
2
(2l + 1);
=
∂l
2mr2
∂En
2En
2me4
=−
=
;
2
2
2
∂l
32π 90 (jmax + l + 1)3
n
0
2En
2 (2l + 1)
−
=
n
2m
But −
4mE1
2
= 2,
2
a
1
r2
1
0
,
or
0
so
1
r2
1
r2
1
=−
1
=
n3 (l
so F-H says
4mE1
4mEn
=− 3
.
2
n(2l + 1)
n (2l + 1)2
1
.
+ 12 )a2
(Agrees with Eq. 6.56.)
Problem 6.34
Equation 4.53 ⇒ u =
But
l(l + 1) 2mEn
2m
−
− 2
2
2
r
me2
1
= (Eq. 4.72),
2
4π90 a
u =
and
−
e2
4π90
1
u.
r
2m m
2mEn
= 2
2
22
e2
4π90
2
1
1
= 2 2 . So
2
n
a n
l(l + 1)
2
1
u.
−
+
r2
ar n2 a2
∴
(ur u ) dr =
=−
s
urs
l(l + 1)
2
2
1
1
−
+ 2 2 u dr = l(l + 1)rs−2 − rs−1 + 2 2 rs 2
r
ar n a
a
n a
d
(urs )u dr = −
dr
(u rs u ) dr − s
(urs−1 u ) dr.
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184
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
(urs u ) dr = −
Lemma 1:
d
(urs )u dr = −
dr
(ur u ) dr = −sr
s
2
s−1
,
Lemma 2:
or
(u r
s+1
u ) dr = −
2
(u rs u) dr − s
s
(urs u ) dr = − rs−1 .
2
d s+1
u
(r u ) dr = −(s + 1)
dr
(u rs+1 u ) dr = −(s + 1)
⇒
urs−1 u dr
(u r u ) dr −
s
(u rs u ) dr,
or:
(u rs u ) dr = −
2
s+1
(u rs+1 u ) dr.
(u rs+1 u ) dr.
Lemma 3: Use in Lemma 2, and exploit Lemma 1:
l(l + 1)
1
2
2
+
(urs+1 u ) dr
−
s+1
r2
ar n2 a2
2
2
1
=−
l(l + 1) (urs−1 u ) dr −
(urs u ) dr + 2 2 (urs+1 u ) dr
s+1
a
n a
s+1 s
2
2
1
s − 1 s−2
s
=−
l(l + 1) −
r −
− rs−1 + 2 2 −
r s+1
2
a
2
n a
2
s−1
2
1
s
= l(l + 1)
rs−2 −
rs−1 + 2 2 rs .
s+1
a s+1
n a
(u rs u ) dr = −
Plug Lemmas 1 and 3 into :
2
1
l(l + 1)rs−2 − rs−1 + 2 2 rs a n
a
s−1
2
1
s(s − 1) s−2
s
s−2
= −l(l + 1)
r +
rs−1 − 2 2 rs +
r .
s+1
a s+1
n a
2
2
s
2
s−1
s(s − 1)
s
s−1
1+
rs−2 = 0.
r −
r + l(l + 1) 1 +
−
n2 a2
a
s+1
s+1
2
2s
s+1
2s+1
s+1
2(s + 1) s
2
(s2 − 1)
r − (2s + 1)rs−1 + 2s l2 + l −
rs−2 = 0, or, finally,
2
2
n a
a
4
(s + 1) s
sa2 2
r − a(2s + 1)rs−1 +
(4l + 4l + 1 −s2 )rs−2 = 0. QED
2
n
4 (2l+1)2
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
185
Problem 6.35
(a)
1
1 − a
n2
0 1
0 1
1
1
1
+0=0⇒
= 2 .
r
r
n a
2
1
r − 3a1 +
(2l + 1)2 − 1 a2
2
n
4
r =
0 1
1
1
2
a = 0 ⇒ 2 r = 3a − l(l + 1)a2 2 = 2 3n2 − l(l + 1) .
r
n
n a
n
a 2
3n − l(l + 1) .
2
2
a2 3 2
1
3 2
a 2
2
r
−
5ar
+
−
4
a
=
0
⇒
r
=
5a
−
l(l
+
1)
−
(2l
+
1)
3n
(2l + 1)2 − 4
2
2
n
2
n
2
2
a2 3 2
a2 r =
15n2 − 5l(l + 1) − 4l(l + 1) − 1 + 4 =
15n2 − 9l(l + 1) + 3
2
n
2
2
2 2 2 3a
n a
=
r2 =
5n2 − 3l(l + 1) + 1 ;
5n2 − 3l(l + 1) + 1 .
2
2
4 3
3
(2l + 1)2 − 9 a2 r = 0 =⇒
r − 7ar2 +
2
n
4
3
4 3
a
n2 a2 2
r = 7a
5n − 3l(l + 1) + 1 − [4l(l + 1) − 8] a2 3n2 − l(l + 1)
2
n
2
4
2
2
.
a3 4
2
2
=
35n − 21l(l + 1)n + 7n − [3l(l + 1) − 6] 3n − l(l + 1)
23
a =
35n4 − 21l(l + 1)n2 + 7n2 − 9l(l + 1)n2 + 3l2 (l + 1)2 + 18n2 − 6l(l + 1)
2
a3 =
35n4 + 25n2 − 30l(l + 1)n2 + 3l2 (l + 1)2 − 6l(l + 1) .
2
r3 =
n2 a3 35n4 + 25n2 − 30l(l + 1)n2 + 3l2 (l + 1)2 − 6l(l + 1) .
8
(b)
0
0+a
1
r2
1
1
−
(2l + 1)2 − 1 a2
4
0
1
r3
1
0
=0⇒
1
r2
1
0
= al(l + 1)
1
1
.
r3
(c)
0
al(l + 1)
1
r3
1
1
=
⇒
(l + 1/2)n3 a2
0
1
r3
1
=
1
.
l(l + 1/2)(l + 1)n3 a3
Agrees with Eq. 6.64.
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186
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
Problem 6.36
(a)
|1 0 0 = √
1
πa3
−r/a
e
(Eq. 4.80),
But the θ integral is zero:
0
π
Es1
1
= 1 0 0|H |1 0 0 = eEext 3
πa
e−2r/a (r cos θ)r2 sin θ dr dθ dφ.
π
sin2 θ cos θ sin θ dθ =
= 0. So Es1 = 0. QED
2 0
1 1 r −r/2a
√
|1
=
ψ
=
1
−
e
2
0
0
2a
2πa 2a
1 1
re−r/2a sin θeiφ
|2 = ψ2 1 1 = − √
2
8a
πa
(b) From Problem 4.11:
1
1
|3 = ψ2 1 0 = √
re−r/2a cos θ
2
4a
2πa
1 1
re−r/2a sin θe−iφ
|4 = ψ2 1−1 = √
πa 8a2
π
1|Hs |1 = {. . . }
cos θ sin θ dθ = 0
0
π
2
2|Hs |2 = {. . . }
sin θ cos θ sin θ dθ = 0
0 π
2
3|Hs |3 = {. . . }
cos θ cos θ sin θ dθ = 0
0 π
2
4|Hs |4 = {. . . }
sin θ cos θ sin θ dθ = 0
All matrix elements of Hs are zero
0 2π
iφ
except 1|Hs |3 and 3|Hs |1
1|Hs |2 = {. . . }
e dφ = 0
(which are complex conjugates,
0 2π
so only one needs to be evaluated).
−iφ
1|Hs |4 = {. . . }
e
dφ = 0
0
2π
2|Hs |3 = {. . . }
e−iφ dφ = 0
0
2π
−2iφ
2|Hs |4 = {. . . }
e
dφ = 0
0
2π
−iφ
3|Hs |4 = {. . . }
e
dφ = 0
0
r −r/2a −r/2a
e
1−
re
cos θ(r cos θ)r2 sin θ dr dθ dφ
2a
∞
eEext
r −r/a 4
2
=
1
−
(2π)
cos
θ
sin
θ
dθ
r dr
e
2πa8a3 2a 0
0
∞
∞
eEext 2
1
eEext
1
4 −r/a
5 −r/a
5
6
=
4!a
5!a
r
e
dr
−
r
e
dr
=
−
8a4 3
2a 0
12a4
2a
0
eEext
5
5
=
= eaEext (−3) = −3aeEext .
24a 1 −
12a4
2
1|Hs |3 = eEext √
1 1 1
1
√
2πa 2a π2πa 4a2
0
0
W = −3aeEext
1
0
0
0
0
0
1
0
0
0
0
0
.
0
0
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
187
We need the eigenvalues of this matrix. The characteristic equation is:
−λ 0 1 0 −λ 0 0 0 −λ 0 0 −λ 0 0 3
2
2 2
1 0 −λ 0 = −λ 0 −λ 0 + 1 0 0 = −λ(−λ) + (−λ ) = λ (λ − 1) = 0.
0 0 −λ 0 0 −λ 0 0 0 −λ
The eigenvalues are 0, 0, 1, and −1, so the perturbed energies are
E2 , E2 , E2 + 3aeEext , E2 − 3aeEext .
Three levels.
0
0
1
0
(c) The eigenvectors with eigenvalue 0 are |2 =
0 and |4 = 0 ; the eigenvectors with eigenvalues ±1
0
1
1
1
0
. So the “good” states are ψ2 1 1 , ψ2 1−1 , √1 (ψ2 0 0 + ψ2 1 0 ), √1 (ψ2 0 0 − ψ2 1 0 ).
are |± ≡ √
±1
2
2
2
0
pe 4 = −e
1 1
πa 64a4
2π
But
cos φ dφ =
0
r2 e−r/a sin2 θ r sin θ cos φî + r sin θ sin φĵ + r cos θk̂ r2 sin θ dr dθ dφ.
2π
sin φ dφ = 0,
0
0
π
4 π
sin θ = 0,
sin3 θ cos θ dθ = 4 pe 4 = 0.
so
0
Likewise
pe 2 = 0.
1
pe ± = − e (ψ1 ± ψ3 )2 (r)r2 sin θ dr dθ dφ
2
2
1 1 1
r r
=− e
1−
±
cos θ e−r/a r(sin θ cos φ î + sin θ sin φ ĵ + cos θ k̂)r2 sin θ dr dθ dφ
2
2 2πa 4a 2a
2a
2
e k̂ 1
r r
=−
2π
1
−
r3 e−r/a cos θ sin θ dr dθ.
±
cos
θ
2 2πa 4a2
2a
2a
π
π
But 0 cos θ sin θ dθ = 0 cos3 θ sin θ dθ = 0, so only the cross-term survives:
pe ±
1
r e
1−
r cos θ r3 e−r/a cos θ sin θ dr dθ
= − 3 k̂ ±
8a
a
2a
∞
π
e e 2
r 4 −r/a
1
2
1
−
=∓
cos
θ
sin
θ
dθ
e
dr
=
∓
k̂
k̂
r
4!a5 − 5!a6
4
4
8a
2a
8a
3
2a
0
0
1
5
5
= ∓ek̂
24a
1
−
±3ae
k̂.
=
12a4
2
Problem 6.37
(a) The nine states are:
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
l=0:
l = 1 :
l=2:
|3 0 0
|3 1 1
|3 1 0
|3 1 − 1
|3 2 2
|3 2 1
|3 2 0
|3 2 − 1
|3 2 − 2
= R30 Y00
= R31 Y11
= R31 Y10
= R31 Y1−1
= R32 Y22
= R32 Y21
= R32 Y20
= R32 Y2−1
= R32 Y2−2
Hs contains no φ dependence, so the φ integral will be:
2π
n l m|Hs |n l m = {· · · }
e−imφ eim φ dφ,
which is zero unless
m = m.
0
π
2
For diagonal elements: n l m|Hs |n l m = {· · · } 0 [Plm (cos θ)] cos θ sin θ dθ. But (p. 137 in the text)
m
Pl is a polynomial (even or odd) in cos θ, multiplied (if m is odd) by sin θ. Since sin2 θ = 1 − cos2 θ,
2
[Plm (cos θ)] is a polynomial in even powers of cos θ. So the θ integral is of the form
π
2j+1
(cos θ)
0
π
(cos θ)2j+2 sin θ dθ = −
= 0. All diagonal elements are zero.
(2j + 2) 0
There remain just 4 elements to calculate:
m = m = 0 : 3 0 0|Hs |3 1 0, 3 0 0|Hs |3 2 0, 3 1 0|Hs |3 2 0; m = m = ±1 : 3 1 ± 1|Hs |3 2 ± 1.
3 0 0|Hs |3 1 0 = eEext
R30 R31 r3 dr
8
2
1
1 1
√ 3/2
R30 R31 r dr = √
3/2
27 a 27 6 a a
3
Let
Y00 Y10 cos θ sin θ dθ dφ.
2r
2r2
1−
+
3a 27a2
From Table 4.7 :
r −r/3a 3
e−r/3a 1 −
r dr.
re
6a
x ≡ 2r/3a:
5 ∞ 3a
x2 24
x 4 −x
√
1−x+
1−
x e dx
6
4
35 2a4 2
0
∞
a
5
1
a
5
5
5
1
= √
1 − x + x2 − x3 x4 e−x dx = √
4! − 5! + 6! − 7!
4
12
24
4
12
24
2 2 0
2 2
√
= −9 2a.
R30 R31 r3 dr =
Y00 Y10
√
√
π
3
32
3
3
cos θ cos θ sin θ dθ dφ =
2π
=
.
cos θ sin θ dθ =
4π
2
3
3
0
'√ (
√
√
3
3 0 0|Hs |3 1 0 = eEext (−9 2a)
= −3 6aeEext .
3
3 0 0|Hs |3 2 0 = eEext R30 R31 r3 dr Y00 Y20 cos θ sin θ dθ dφ.
1
cos θ sin θ dθ dφ = √
4π
3
4π
√
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
1
Y00 Y20 cos θ sin θ dθ dφ = √
4π
5
16π
(3 cos2 θ − 1) cos θ sin θ dθ dφ = 0.
3 1 0|Hs |3 2 0 = eEext
189
3 0 0 |Hs |3 2 0 = 0.
R31 R32 r3 dr
Y10 Y20 cos θ sin θ dθ dφ.
r −r/3a 2 −r/3a 3
8
1 1 4
1 1
√ 3/2
√
R31 R32 r dr =
1
−
r e
r dr
re
6a
27 6 a a 81 30 a3/2 a2
√
7 ∞ 24
3a
x 6 −x
a
9 5
1
= √
1−
x e dx = √
a.
6! − 7! = −
4
4
2
38 5a6 2
24 5
0
3
3
5
cos θ(3 cos2 θ − 1) cos θ sin θ dθ dφ
4π 16π
√
√
π
15
15
3
1
4
2
=
2π
− cos5 θ + cos3 θ
(3 cos θ − cos θ) sin θ dθ =
8π
4
5
3
0
Y10 Y20 sin θ cos θ dθ dφ =
π
= √2 .
15
0
'
√ (
√
9 5
2
√
3 1 0|Hs |3 2 0 = eEext −
= −3 3aeEext .
a
2
15
∗
3 1 ± 1|Hs |3 2 ± 1 = eEext R31 R32 r3 dr
Y1±1 Y2±1 cos θ sin θ dθ dφ.
('
(
3
15
∓
sin θe∓iφ sin θ cos θe±iφ cos θ sin θ dθ dφ
Y2±1 cos θ sin θ dθ dφ = ∓
8π
8π
√
π
π
3 5
3√
cos3 θ cos5 θ 2
2
=
2π
+
cos θ(1 − cos θ) sin θ dθ =
5 −
8π
4
3
5
0
0
1
= √ .
5
' √ (
5
9
1
√
3 1 ± 1|Hs |3 2 ± 1 = eEext −9
= − aeEext .
a
2
2
5
'
∗
Y1±1
Thus the matrix representing Hs is (all empty boxes are zero; all numbers multiplied by −aeEext ):
(b) The perturbing matrix (below) breaks into a 3×3 block, two 2 × 2 blocks, and two 1 × 1 blocks, so we can
work out the eigenvalues in each block separately.
√
√
−λ 2 0 20
√
√
√0
3 × 3 : 3 3 2 0 1 ; 2 −λ 1 = −λ3 + λ + 2λ = −λ(λ2 − 3) = 0;
0 1 −λ
0 1 0
√
λ = 0, ± 3
2×2:
9
2
01
10
⇒
;
E11 = 0, E21 = 9aeEext , E31 = −9aeEext .
−λ 1 2
1 −λ = λ − 1 = 0 ⇒ λ = ±1.
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190
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
300 310 320 311 321 31-1 32-1 322 32-2
300
3√6
310 3√6
320
3√3
3√3
9/2
311
321
9/2
9/2
31-1
32-1
9/2
322
32-2
9
9
aeEext , E51 = − aeEext . From the other 2×2 we get E61 = E41 , E71 = E51 , and from the 1×1’s we
2
2
0
(degeneracy 3)
(9/2)aeEext
(degeneracy 2)
−(9/2)aeEext (degeneracy 2)
get E81 = E91 = 0. Thus the perturbations to the energy (E3 ) are:
9aeEext
(degeneracy 1)
(degeneracy 1)
−9aeEext
E41 =
Problem 6.38
1
Equation 6.89 ⇒ Ehf
=
µ0 gd e2
Sd · Se ;
3πmd me a3
Eq. 6.91 ⇒ Sd · Se =
1 2
(S − Se2 − Sd2 ).
2
Electron has spin 12 , so Se2 = 12 32 2 = 34 2 ; deuteron has spin 1, so Sd2 = 1(2)2 = 22 .
1
3 2
2
2
Total spin could be 32 [in which case S 2 = 32 52 2 = 15
4 ] or 2 [in which case S = 4 ]. Thus
1 15 2 3 2
2 4 − 4 − 22 = 12 2
3
µ0 gd e2 2
Sd · Se =
; the difference is 2 , so ∆E =
.
1 3 2 3 2
2
2πmd me a3
2
2
−
−
2
=
−
2 4
4
1
1
2gd e2 2
2gd 4
3 g d mp
⇒
µ
=
,
so
∆E
=
=
=
∆Ehydrogen (Eq. 6.98).
0
2
2
2
3
2
2
4
c
90 c
4π90 md me c a
md me c a 2 gpmd
c
2 g p md
4 5.59
ch
Now, λ = =
λh , and since md = 2mp , λd =
, so λd =
(21 cm) = 92 cm.
ν
∆E
3 gd mp
3 1.71
But
µ0 90 =
Problem 6.39
(a) The potential energy of the electron (charge −e) at (x, y, z) due to q’s at x = ±d alone is:
eq
1
1
V =−
. Expanding (with d x, y, z) :
+
4π90
(x + d)2 + y 2 + z 2
(x − d)2 + y 2 + z 2
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
≈
1
(x ± d)2 + y 2 + z 2
1
d
1∓
V =−
= (x2 ± 2dx + d2 + y 2 + z 2 )−1/2 = (d2 ± 2dx + r2 )−1/2 =
3 4x2
x
r2
− 2+
d 2d
8 d2
=
191
1
d
1±
2x r2
+ 2
d
d
−1/2
x
1
1
1 ∓ + 2 (3x2 − r2 ) .
d
d 2d
eq
x
2eq
x
1
1
eq
(3x2 − r2 )
1 − + 2 (3x2 − r2 ) + 1 + + 2 (3x2 − r2 ) = −
−
4π90 d
d 2d
d 2d
4π90 d 4π90 d3
= 2βd2 + 3βx2 − βr2 ,
where
e q
.
4π90 d3
β≡−
Thus with all six charges in place
H = 2(β1 d21 + β2 d22 + β3 d23 ) + 3(β1 x2 + β2 y 2 + β3 z 2 ) − r2 (β1 + β2 + β3 ). QED
1
(b) 1 0 0|H |1 0 0 =
e−2r/a H r2 sin θ dr dθ dφ
πa3
3
(β1 + β2 + β3 )
= V0 + 3
e−2r/a (β1 x2 + β2 y 2 + β3 z 2 )r2 sin θdr dθ dφ −
r2 e−2r/a r2 sin θ dr dθ dφ.
πa
πa3
I1 ≡
r e
I2 ≡
r sin θ dr dθ dφ = 4π
0
−2r/a
e
2 −2r/a 2
2
2
2
∞
a
r4 e−2r/a dr = 4π4!( )5 = 3πa5 .
2
2
(β1 x + β2 y + β3 z )r sin θ dr dθ dφ
r4 e−2r/a (β1 sin2 θ cos2 φ + β2 sin2 θ sin2 φ + β3 cos2 θ) sin θ dr dθ dφ.
2π
2π
2π
But
cos2 φ dφ =
sin2 φ dφ = π,
dφ = 2π. So
0
0
0
∞
π
=
r4 e−2r/a dr
π(β1 + β2 ) sin2 θ + 2πβ3 cos2 θ sin θ dθ.
0
0
π
π
4
2
3
But
sin θ dθ = ,
cos2 θ sin θ dθ = . So
3
3
0
0
a 5 4π
4π
= 4!
(β1 + β2 ) +
β3 = πa5 (β1 + β2 + β3 ).
2
3
3
=
3
(β1 + β2 + β3 )
πa5 (β1 + β2 + β3 ) −
3πa5 = V0 .
πa3
πa3
|2 0 0 = R20 Y00
|2 1 1 = R21 Y11
(functional forms in Problem 4.11).
(c) The four states are
−1
|2 1 − 1 = R21 Y1
|2 1 0 = R21 Y10
1 0 0|H |1 0 0 = V0 +
Diagonal elements: n l m|H |n l m = V0 + 3 β1 x2 + β2 y 2 + β3 z 2 − (β1 + β2 + β3 )r2 .
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192
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
For |2 0 0, x2 = y 2 = z 2 = 13 r2 (Y00 does not depend on φ, θ; this state has spherical symmetry),
so 2 0 0|H |2 0 0 = V0 .
(I could have used the same argument in (b).)
n2 a2 2
From Problem 6.35(a), r2 =
5n − 3l(l + 1) + 1 , so for n = 2, l = 1 : r2 = 30a2 . Moreover,
2
2π
2π
since x2 = {. . . }
cos2 φ dφ = {. . . }
sin2 φ dφ = y 2 , and x2 + y 2 + z 2 = r2 , it follows
0
0
1 2
1
(r − z 2 ) = 15a2 − z 2 . So all we need to calculate is z 2 .
2
2
1
1
r2 e−r/a cos2 θ(r2 cos2 θ)r2 sin θ dr dθ dφ
2 1 0|z 2 |2 1 0 =
2πa 16a4
that x2 = y 2 =
=
1
16a5
∞
r6 e−r/a dr
0
π
cos4 θ sin θ dθ =
0
2
1
6!a7 = 18a2 ;
5
16a
5
x2 = y 2 = 15a2 − 9a2 = 6a2 .
2 1 0|H |2 1 0 = V0 + 3(6a2 β1 + 6a2 β2 + 18a2 β3 ) − 30a2 (β1 + β2 + β3 )
= V0 − 12a2 (β1 + β2 + β3 ) + 36a2 β3 .
1 1
2 1 ± 1|z |2 1 ± 1 =
πa 64a4
2
1
=
32a5
∞
6 −r/a
r e
π
1
(1 − cos θ) cos θ sin θ dθ =
6!a7
32a5
2
dr
0
r2 e−r/a sin2 θ(r2 cos2 θ)r2 sin θ dr dθ dφ
0
2
2 2
−
3 5
= 6a2 ;
x2 = y 2 = 15a2 − 3a2 = 12a2 .
2 1 ± 1|H |2 1 ± 1 = V0 + 3(12a2 β1 + 12a2 β2 + 6a2 β3 ) − 30a2 (β1 + β2 + β3 )
= V0 + 6a2 (β1 + β2 + β3 ) − 18a2 β3 .
Off-diagonal elements:
We need 2 0 0|H |2 1 0, 2 0 0|H |2 1 ± 1, 2 1 0|H |2 1 ± 1, and 2 1 − 1|H |2 1 1.
Now n l m|V0 |n l m = 0, by orthogonality, and n l m|r2 |n l m = 0, by orthogonality of Ylm , so
all we need are the matrix elements of x2 and y 2 (|z 2 | = −|x2 | − |y 2 ). For 2 0 0|x2 |2 1 ± 1 and
2π
2π
2π
2 1 0|x2 |2 1 ± 1 the φ integral is 0 cos2 φe±iφ dφ = 0 cos3 φ dφ ± i 0 cos2 φ sin φ dφ = 0, and the
same goes for y 2 . So 2 0 0|H |2 1 ± 1 = 2 1 0|H |2 1 ± 1 = 0.
π
π
For 2 0 0|x2 |2 1 0 and 2 0 0|y 2 |2 1 0 the θ integral is 0 cos θ(sin2 θ) sin θ dθ = sin4 θ/40 = 0, so
2 0 0|H |2 1 0 = 0.
Finally:
1 1
2 1 − 1|x |2 1 1 = −
r2 e−r/a sin2 θe2iφ (r2 sin2 θ cos2 φ)r2 sin θ dr dθ dφ
πa 64a4
∞
π
2π
1
5
6 −r/a
=−
r e
dr
sin θdθ
e2iφ cos2 φ dφ
64πa5 0
0
0
2
6!a7
=−
16/15
π/2
16 π
1
6!a7
= −6a2 .
64πa5
15 2
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
For y 2 , the φ integral is
2π
0
193
e2iφ sin2 φ dφ = −π/2, so 2 1 − 1|y 2 |2 1 1 = 6a2 , and 2 1 − 1|z 2 |2 1 1 = 0.
2 1 − 1|H |2 1 1 = 3 β1 (−6a2 ) + β2 (6a2 ) = −18a2 (β1 − β2 ).
The perturbation matrix is:
2 1 0
2 1 1
2 1 -1
0
0
0
V0 − 12a2 (β1 + β2 ) + 24a2 β3
0
0
0
V0 + 6a2 (β1 + β2 ) − 12a2 β3
−18a2 (β1 − β2 )
0
−18a2 (β1 − β2 )
V0 + 6a2 (β1 + β2 ) − 12a2 β3
A B
The 2 × 2 block has the form
; its characteristic equation is (A − λ)2 − B 2 = 0, so A − λ = ±B,
B A
or
2
2
2
2
2 0 0
V0
0
0
0
0 0
1 0
1 1
1 -1
λ = A ∓ B = V0 + 6a2 (β1 + β2 ) − 12a2 β3 ± 18a2 (β1 − β2 ) =
91
9
The first-order corrections to the energy (E2 ) are therefore: 2
93
94
V0 + 24a2 β1 − 12a2 β2 − 12a2 β3 ,
V0 − 12a2 β1 + 24a2 β2 − 12a2 β3 .
= V0
= V0 − 12a2 (β1 + β2 − 2β3 )
= V0 − 12a2 (−2β1 + β2 + β3 )
= V0 − 12a2 (β1 − 2β2 + β3 )
(i) If β1 = β2 = β3 , then 91 = 92 = 93 = 94 = V0 : one level (still 4-fold degenerate).
(ii) If β1 = β2 = β3 , then 91 = V0 , 92 = V0 − 24a2 (β1 − β3 ), 93 = 94 = V0 + 12a2 (β1 − β3 ): three levels
(one remains doubly degenerate).
(iii) If all three β’s are different, there are four levels (no remaining degeneracy).
Problem 6.40
(a) (i) Equation 6.10:
0
H =
E00 =
H =
ψ00 =
(H 0 − E00 )ψ01 = −(H − E01 )ψ00 .
2
2 2
e2 1
2
2
∇ −
=−
∇ +
, since
−
2m
4π90 r
2m
ar
2
−
.
2ma2
eEext r cos θ; E01 = 0 (Problem6.36(a)).
1
√
e−r/a ; ψ01 = f (r)e−r/a cos θ.
πa3
a=
4π90 2
.
me2
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194
CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
Equation 4.13 ⇒
f e−r/a d
cos θ d
d
2 d
−r/a
+ 2
r
f
e
sin θ (cos θ)
2
r dr
dr
r sin θ dθ
dθ
cos θ d
f e−r/a d 1
2
=
f
−
r
f
e−r/a + 2
− sin2 θ
2
r dr
a
r sin θ dθ
cos θ
2 cos θ −r/a
1
2
1
−r/a
2
=
2r
f
f
−
+
r
−
+
f
e−r/a −
fe
f
e
f
r2
a
a
a2
r2
2
1
1
1
1
= cos θe−r/a f − f + 2 f + 2 f − f
− 2f 2 .
a
a
a
r
r
∇2 ψ01 =
Plug this into Eq. 6.10:
2
1
1
1
2
1
1
11
1
−r/a
−
f − f + 2f + 2 f − f
cos θe
− 2f 2 + 2f
− f 2 = −eEext r cos θ √
e−r/a ,
2m
a
a
a
r
r
ar
a
πa3
2
f − f
a
1
1
+ 2f − 2f 2 =
r
r
f (r) = A + Br + Cr2 ,
Now let
so
2meEext
√
2 πa3
f = B + 2Cr
r=
4γ
r,
a
and f = 2C.
where
γ≡
meEext
√ .
22 πa
Then
2
2
2
4γ
2C − (B + 2Cr) + (B + 2Cr) − 2 (A + Br + Cr2 ) =
r.
a
r
r
a
Collecting like powers of r:
r−2 :
r−1 :
r0 :
r1 :
A = 0.
2B − 2B = 0 (automatic).
2C − 2B/a + 4C − 2C = 0 ⇒ B = 2aC.
−4C/a = 4γ/a ⇒ C = −γ.
Evidently the function suggested does satisfy Eq. 6.10, with the coefficients A = 0, B = −2aγ, C = −γ;
the second-order correction to the wave function is
ψ01 = −γr(r + 2a)e−r/a cos θ.
(ii) Equation 6.11 says, in this case:
E02
meEext
√ eEext e−r/a (r cos θ)r(r + 2a)e−r/a cos θ r2 sin θ dr dθ dφ
=
= −√
πa3 22 πa
∞
π
m(eEext )2
4
−2r/a
=−
2π
r (r + 2a)e
dr
cos2 θ sin θ dθ
2πa2 2
0
0
2
a 5 cos3 θ π
a 6
eEext
= −m
+ 2a 4!
5!
−
a
2
2
3
ψ00 |H
= −m
1
|ψ01 eEext
a
2 27 6
a
8
2
= −m
3
2
3eEext a
2
0
2
.
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CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY
195
(b) (i) This is the same as (a) [note that E01 = 0, as before, since ψ00 is spherically symmetric, so cos θ = 0]
except for the r-dependence of H . So Eq. ⇒
1 1
1
2mep
2β
mep
1
√
√
f + 2f
= − 2 , where β ≡
.
−
− 2f 2 = −
r
a
r
r
4π90 2 πa3 r2
4π90 2 πa3
The solution this time it obvious: f (r) = β (constant). [For the general solution we would add the general
solution to the homogeneous equation (right side set equal to zero), but this would simply reproduce the
unperturbed ground state, ψ00 , which we exclude—see p. 253.] So
ψ01 = βe−r/a cos θ.
(ii) The electric dipole moment of the electron is
pe = −er cos θ = −eψ00 +ψ01 |r cos θ|ψ00 +ψ01 = −e ψ00 |r cos θ|ψ00 + 2ψ00 |r cos θ|ψ01 + ψ01 |r cos θ|ψ01 .
But the first term is zero, and the third is higher order, so
1
pe = −2e √
β e−r/a (r cos θ)e−r/a cos θ r2 sin θ dr dθ dφ
πa3
∞
π
a 4 2 mep
me2 p
3 −2r/a
2
= −2e
2π
3!
r e
dr
cos θ sin θ dθ = −
4π90 2 πa3
90 2 πa3
2
3
0
0
me2 p
3a4
2
me2 pa
=−
= −p.
=−
90 2 πa3
8
3
4π90 2
Evidently the dipole moment associated with the perturbation of the electron cloud cancels the dipole
moment of the nucleus, and the total dipole moment of the atom is zero.
(iii) The first-order correction is zero (as noted in (i)). The second-order correction is
E02
mep
cos θ
−r/a
√
e
e−r/a cos θ r2 sin θ dr dθ dφ
=
=√
r2
πa3
4π90 2 πa3
∞
π
a 2
(ep)2
(ep)2
−2r/a
2
= −m
2π
e
dr
cos θ sin θ dθ = −2m
(4π90 )2 2 πa3
(4π90 )2 2 a3 2
3
0
0
4
2
2
p
4
4 p
me
=
=
E1 .
−
3
2(4π90 )2 2 e2 a2
3 ea
ψ00 |H
|ψ01 1
ep
−
4π90
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196
CHAPTER 7. THE VARIATIONAL PRINCIPLE
Chapter 7
The Variational Principle
Problem 7.1
(a)
∞
V = 2αA
2
−2bx2
xe
2
dx = 2αA
0
2 b
α
H =
+√
.
2m
2πb
Hmin
2
=
2m
∞
1 −2bx2 αA2
α
− e
=
=
4b
2b
2b
0
2b
α
.
=√
π
2bπ
2
1 α
α m
∂H
=
− √ b−3/2 = 0 =⇒ b3/2 = √
; b=
∂b
2m 2 2π
2π 2
mα
√
2π2
2/3
α
+√
2π
'√
2π2
mα
(1/3
α2/3 2/3
= 1/3
m (2π)1/3
mα
√
2π2
2/3
.
1/3
1
3 α 2 2
.
+1 =
2
2 2πm
(b)
∞
V = 2αA
2
x4 e−2bx dx = 2αA2
2
0
2 b
3α
H =
+
.
2m 16b2
Hmin
2
=
2m
Problem 7.2
2
Normalize: 1 = 2|A|
0
∞
3αm
42
3
8(2b)2
π
3α
=
2b
16b2
π
2b
2b
3α
.
=
π
16b2
2
3α
∂H
3αm
=
−
= 0 =⇒ b3 =
; b=
∂b
2m 8b3
42
1/3
3α
+
16
42
3αm
2/3
α1/3 4/3 1/3 −1/3
=
3 4
m2/3
π
1
π
dx = 2|A|2 3 = 3 |A|2 . A =
(x2 + b2 )2
4b
2b
3αm
42
1/3
1 1
+
2 4
.
3
=
4
3α4
4m2
1/3
.
2b3
.
π
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CHAPTER 7. THE VARIATIONAL PRINCIPLE
197
∞
1
1
2
d2
2
Kinetic Energy: T = −
|A|
dx.
2 + b2 ) dx2
2m
(x2 + b2 )
−∞ (x
d2
1
−2
4x
2(3x2 − b2 )
d
−2x
But
= 2
+ 2x 2
=
, so
=
2
2
2
2
2
2
2
2
2
3
dx
(x + b )
dx (x + b )
(x + b )
(x + b )
(x2 + b2 )3
∞
∞
2 2b3 ∞ (3x2 − b2 )
1
1
42 b3
2
T = −
dx = −
dx − 4b
dx
3
2 + b2 )3
2 + b2 )4
2m π 0 (x2 + b2 )4
πm
(x
(x
0
0
42 b3
2
3π
2 5π
=−
=
3
−
4b
.
πm
16b5
32b7
4mb2
∞
3
1
x2
1
2
2
2 2b π
Potential Energy: V = mω |A| 2
dx
=
mω
= mω 2 b2 .
2 + b2 )2
2
(x
π
4b
2
0
H =
2
1
+ mω 2 b2 .
4mb2
2
Hmin
2
=
4m
√
Problem 7.3
∂H
2
1 2
2
4
+
mω
b
=
0
=⇒
b
=
=⇒ b2 = √
=−
.
∂b
2mb3
2m2 ω 2
2 mω
2mω 1
1 + mω 2 √
= ω
2
2 mω
'√
2
1
+ √
4
2 2
(
√
=
2
1
ω = 0.707 ω > ω.
2
2
A(x + a/2), (−a/2 < x < 0),
ψ(x) = A(a/2 − x), (0 < x < a/2),
0,
(otherwise).
2
3 a/2
a 3
12
2
a3 2
21 a
1 = |A| 2
− x dx = −2|A|
− x |A| ; A =
= |A|2
=
(as before).
2
3 2
3
3
12
a3
0
0
A,
(−a/2 < x < 0),
dψ
a
d2 ψ
a
=
Aδ
x
+
= −A, (0 < x < a/2),
−
2Aδ(x)
+
Aδ
x
−
.
dx
dx2
2
2
0,
(otherwise).
a/2
2
a
2
a
a 2
2 2 a
ψ Aδ x +
− 2Aδ(x) + Aδ x −
dx =
2Aψ(0) =
A
2m
2
2
2m
m
2
2
2
a 12
=6
(as before).
=
2m a3
ma2
a 2
α
α
2
V = −α |ψ|2 δ(x) dx = −α|ψ(0)|2 = −αA2
= −3 . H = T + V = 6
−3 .
2
a
ma2
a
T = −
Hmin
∂
α
2
2
+3 2 =0⇒a=4
H = −12
.
3
∂a
ma
a
mα
mα mα2 3 3 2 mα 2
3mα2
=
=6
−
3α
−
−
=
m 42
42
2
8 4
82
> −
mα2
. 22
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198
CHAPTER 7. THE VARIATIONAL PRINCIPLE
Problem 7.4
(a) Follow the proof in §7.1: ψ =
∞
∞
cn ψn , where ψ1 is the ground state. Since ψ1 |ψ = 0, we have:
n=1
cn ψ1 |ψ = c1 = 0; the coefficient of the ground state is zero. So
n=1
∞
H =
En |cn |2 ≥ Efe
n=2
(b)
1 = |A|2
∞
|cn |2 = Efe , since En ≥ Efe for all n except 1.
n=2
∞
−∞
x2 e−2bx dx = |A|2 2
2
T = −
2
|A|2
2m
2
∞
xe−bx
−∞
2
1
8b
π
=⇒ |A|2 = 4b
2b
2b
.
π
d2 −bx2 xe
dx
dx2
2
2
2
2
d −bx2
− 2bx2 e−bx = −2bxe−bx − 4bxe−bx + 4b2 x3 e−bx
e
dx
∞
2
2b
π
3
2
22 b 2b
1
−6bx2 + 4b2 x4 e−2bx dx = −
T = −
4b
2
2 −6b
+ 4b2
2m
π
m
π
8b 2b
32b2
0
42 b
3 3
32 b
=−
− +
=
.
m
4 8
2m
d2
xe−bx
dx2
=
∞
V =
1
mω 2 |A|2
2
H =
32 b 3mω
+
;
2m
8b
Hmin
x2 e−2bx x2 dx =
2
−∞
1
mω 2 4b
2
2b
3
2
π 32b2
π
2b
π
3mω 2
=
.
2b
8b
∂H
32
3mω 2
m2 ω 2
mω
2
=
−
.
=
0
=⇒
b
=
=⇒ b =
2
2
∂b
2m
8b
4
2
3
32 mω 3mω 2 2
3 3
=
+
= ω
+
= ω.
2m 2
8 mω
4 4
2
This is exact, since the trial wave function is in the form of the true first excited state.
Problem 7.5
0
(a) Use the unperturbed ground state (ψgs
) as the trial wave function. The variational principle says
0
0
0
0
0
0
0
0
0
0
0
ψgs |H|ψgs ≥ Egs . But H = H 0 + H , so ψgs
|H|ψgs
= ψgs
|H 0 |ψgs
+ ψgs
|H |ψgs
. But ψgs
|H 0 |ψgs
=
0
0
0
Egs (the unperturbed ground state energy), and ψgs |H |ψgs is precisely the first order correction to the
0
1
ground state energy (Eq. 6.9), so Egs
+ Egs
≥ Egs . QED
2
2
(b) The second order correction (Egs
) is Egs
=
|ψ 0 |H |ψgs |2
m
. But the numerator is clearly positive,
0 − E0
Egs
m
m=gs
0
0
2
and the denominator is always negative (since Egs
< Em
for all m), so Egs
is negative.
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CHAPTER 7. THE VARIATIONAL PRINCIPLE
199
Problem 7.6
He+ is a hydrogenic ion (see Problem 4.16); its ground state energy is (2)2 (−13.6 eV), or −54.4 eV. It takes
79.0 − 54.4 = 24.6 eV to remove one electron.
Problem 7.7
I’ll do the general case of a nucleus with Z0 protons. Ignoring electron-electron repulsion altogether gives
ψ0 =
Z03 −Z0 (r1 +r2 )/a
e
,
πa3
(generalizing Eq. 7.17)
and the energy is 2Z02 E1 . Vee goes like 1/a (Eqs. 7.20 and 7.25), so the generalization of Eq. 7.25 is Vee =
− 54 Z0 E1 , and the generalization of Eq. 7.26 is H = (2Z02 − 54 Z0 )E1 .
If we include shielding, the only change is that (Z − 2) in Eqs. 7.28, 7.29, and 7.32 is replaced by (Z − Z0 ).
Thus Eq. 7.32 generalizes to
5
5
H = 2Z 2 − 4Z(Z − Z0 ) − Z E1 = −2Z 2 + 4ZZ0 − Z E1 .
4
4
∂H
5
5
= −4Z + 4Z0 −
E1 = 0 =⇒ Z = Z0 − .
∂Z
4
16
Hmin
2
5
5
5
5
= −2 Z0 −
+ 4 Z0 −
Z0 −
Z0 −
E1
16
16
4
16
5
25
5
5
25
2
2
= −2Z0 + Z0 −
+ 4Z0 − Z0 − Z0 +
E1
4
128
4
4
64
5
25
(16Z0 − 5)2
E1 =
E1 ,
= 2Z02 − Z0 +
4
128
128
generalizing Eq. 7.34. The first term is the naive estimate ignoring electron-electron repulsion altogether; the
second term is Vee in the unscreened state, and the third term is the effect of screening.
5
11
=
= 0.688. The effective nuclear charge is less than 1, as expected.
16
16
112
121
=
E1 =
E1 = −12.9 eV.
128
128
Z0 = 1 (H− ): Z = 1 −
Hmin
27
5
272
729
=
= 1.69 (as before); Hmin =
E1 =
E1 = −77.5 eV.
16
16
128
128
43
1849
5
432
Z0 = 3 (Li+ ): Z = 3 −
=
= 2.69 (somewhat less than 3); Hmin =
E1 =
E1 = −196 eV.
16
16
128
128
Z0 = 2 (He): Z = 2 −
Problem 7.8
1
1
1
1
e−2r2 /a d3 r
D = aψ0 (r1 ) ψ0 (r1 ) = aψ0 (r2 ) ψ0 (r2 ) = a 3
r2
r1
πa
r1
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200
CHAPTER 7. THE VARIATIONAL PRINCIPLE
1
=
πa3
2
−a
√
e
1
[. . . ] =
rR
r 2 +R2 −2rR cos θ
r+R
e−2y/a y dy = −
|r−R|
1 2
2π
r sin θ dr dθ dφ =
r
πa3
∞
π
r
0
e− a
2
√
r 2 +R2 −2rR cos θ
sin θ dθ dr.
0
a
a a −2(r+R)/a r+R+
e
− e−2|r−R|/a |r − R| +
2rR
2
2
2 a −2R/a ∞ −2r/a a
D= 2 −
r+R+
e
e
dr
a
2R
2
0
−2R/a
R
−e
2R/a
e
0
=−
1
aR
+e−2R/a
a 2
e−2R/a
a 2
2
2
a
R−r+
dr − e2R/a
2
∞
−2r/a
e
R
a
r−R+
dr
2
a a 2r/a R
a a
+ R+
− e−2R/a R +
e
2
2
2
2
0
e2r/a
R
∞ 2
a a −2r/a ∞
2r
2r
2R/a a
−2r/a
−
−
e
e
− 1 − e2R/a −R +
− e
−
1
a
2
2
2
a
R
0
R
a2
aR a2
aR a2
a2
aR a2
a2 2R a2
aR a2
a2 2R a2
+
+
+
+
+
+ −
−
+
−
+
−
−
−
4
2
4
2
4
4
2
4
4 a
4
2
4
4 a
4
=−
1
aR
=−
1 −2R/a 2
a a −2R/a
a + aR + −a2 =⇒ D =
e
− 1+
e
aR
R
R
e−2R/a
(confirms Eq. 7.47).
1
1
1
X = aψ0 (r1 ) ψ0 (r2 ) = a 3
e−r1 /a e−r2 /a d3 r
r1
πa
r1
=
1
πa2
√
e−r/a e−
[. . . ] = −
X=
r 2 +R2 −2rR cos θ/a
2π
1 2
r sin θ dr dθ dφ =
r
πa2
∞
re−r/a
0
π
√
e−
r 2 +R2 −2rR cos θ/a
sin θ dθ dr.
0
a −(r+R)/a
(r + R + a) − e−|r−R|/a (|r − R| + a)
e
rR
2 a −R/a ∞ −2r/a
−
e
e
(r + R + a)dr
a2
R
0
−R/a
R
−e
(R − r + a)dr − e
R/a
0
2
=−
aR
e−R/a
a 2
2
∞
−2r/a
e
(r − R + a)dr
R
+ (R + a)
a
2
− e−R/a (R + a)R −
R2
2
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CHAPTER 7. THE VARIATIONAL PRINCIPLE
−e
R/a
=−
2
e−R/a
aR
201
∞ ∞
2
a
2r
−2r/a R/a a
−2r/a
− − 1 (−R + a) − e
e
−e
2
2
a
R
R
a2
R2
aR a2
aR a2
a2 2R a2
+
+
− R2 − aR +
+
−
−
−
4
2
2
2
2
2
4 a
4
2 −R/a
aR R2
R
−R/a
=−
e
−
=⇒ X = e
−
1+
aR
2
2
a
(confirms Eq. 7.48).
Problem 7.9
There are two changes: (1) the 2 in Eq. 7.38 changes sign . . . which amounts to changing the sign of I in
Eq. 7.43; (2) the last term in Eq. 7.44 changes sign . . . which amounts to reversing the sign of X. Thus Eq. 7.49
becomes
H = 1 + 2
D−X
E1 ,
1−I
and hence Eq. 7.51 becomes
Etot
D−X
2
1/x − (1 + 1/x) e−2x − (1 + x)e−x
2a
−1−2
= −1 + − 2
=
−E1
R
1−I
x
1 − (1 + x + x2 /3)e−x
2 1 − (1 + x + x2 /3)e−x − 1 + (x + 1)e−2x + (x + x2 )e−x
= −1 +
x
1 − (1 + x + x2 /3)e−x
2 (1 + x)e−2x + 23 x2 − 1 e−x
= −1 +
.
x
1 − (1 + x + x2 /3)e−x
F (x) =
The graph (with plus sign for comparison) has no minimum, and remains above −1, indicating that the energy
is greater than for the proton and atom dissociated. Hence, no evidence of bonding here.
F(x)
-0.5
-0.6
(+)
(−)
-0.7
-0.8
-0.9
2
4
6
8
x
-1.1
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202
CHAPTER 7. THE VARIATIONAL PRINCIPLE
Problem 7.10
According to Mathematica, the minimum occurs at x = 2.493, and at this point F = 0.1257.
mω 2 = V
=−
E1
F ,
a2
so
ω=
1
a
−(0.1257)E1
.
m
mp m p
1
= mp .
mp + m p
2
3 × 108 m/s
(0.1257)(13.6 eV)
ω=
= 3.42 × 1014 /s.
(0.529 × 10−10 m) (938 × 106 eV)/2
Here m is the reduced mass of the proton: m =
1
1
ω = (6.58 × 10−16 eV · s)(3.42 × 1014 /s) = 0.113 eV (ground state vibrational energy).
2
2
Mathematica says that at the minimum F = −1.1297, so the binding energy is (0.1297)(13.6 eV) = 1.76 eV.
Since this is substantially greater than the vibrational energy, it stays bound. The highest vibrational energy is
1
1.76
given by (n + 12 )ω = 1.76 eV, so n =
− = 7.29. I estimate eight bound vibrational states (including
0.226 2
n = 0).
Problem 7.11
(a)
1=
T =
V =
=
H =
|ψ|2 dx = |A|2
a/2
cos2
πx a
2
⇒
A=
2
.
a
a
2
d ψ
π
π 2 2
−
ψ 2 dx =
ψ 2 dx =
.
2m
dx
2m a
2ma2
a/2
1
1
mω 2 a 3 π/2 2
2
2 2
22
2
2 πx
x ψ dx = mω
x cos
y cos2 y dy
mω
dx =
2
2
a −a/2
a
a
π
−π/2
π/2
2
mω 2 a2 y 3
y
1
y cos 2y mω 2 a2 π 2
+
−
sin 2y +
−1 .
=
π3
6
4
8
4
4π 2
6
−π/2
π 2 2
mω 2 a2 π 2
mω 2 a π 2
∂H
π 2 2
+
+
−
1
;
=
−
−
1
=0 ⇒
2ma2
4π 2
6
∂a
ma3
2π 2
6
2
2
−a/2
2
a=π
Hmin
dx = |A|2
π 2 2 mω
=
2mπ 2 =
1
ω
2
π 2 /6 − 1 mω 2
+
2
4π 2
mω
2
π 2 /6 − 1
1/4
.
π2
2
2 −1 π
6
mω π 2 /6 − 1
π2
1
1
− 2 = ω(1.136) > ω.
3
2
2
[We do not need to worry about the kink at ±a/2. It is true that d2 ψ/dx2 has delta functions there, but
since ψ(±a/2) = 0 no “extra” contribution to T comes from these points.]
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CHAPTER 7. THE VARIATIONAL PRINCIPLE
203
(b) Because this trial function is odd, it is orthogonal to the ground state, so by Problem 7.4 H will give
an upper bound to the first excited state.
a
πx 1
1=
|ψ|2 dx = |B|2
sin2
dx = |B|2 a ⇒ B = √ .
a
a
−a
2 2
2 2 2 2
d ψ
π
π ψ 2 dx =
T = −
ψ 2 dx =
.
2m
dx
2m a
2ma2
a
πx 1
1
1
mω 2 a 3 π 2 2
V = mω 2 x2 ψ 2 dx = mω 2
x2 sin2
y sin y dy
dx =
2
2
a −a
a
2a π
−π
π
2
mω 2 a2 y 3
y
1
y cos 2y mω 2 a2 2π 2
=
−
−
sin 2y −
−1 .
= 4π 2
2π 3
6
4
8
4
3
−π
π 2 2
mω 2 a2 2π 2
mω 2 a 2π 2
∂H
π 2 2
H =
+
+
−1 ;
=−
−1 =0 ⇒
2ma2
4π 2
3
∂a
ma3
2π 2
3
mω
a=π
Hmin
π 2 2 mω
=
2mπ 2 =
1
ω
2
2π 2 /3 − 1 mω 2
+
2
4π 2
2
2π 2 /3 − 1
1/4
.
2π 2
2
2 −1 π
3
mω 2π 2 /3 − 1
4π 2
1
3
− 2 = ω(3.341) > ω.
3
2
2
Problem 7.12
We will need the following integral repeatedly:
∞
Γ
xk
1
dx = 2l−k−1
l
2
2
2b
(x + b )
0
k+1
2
Γ 2l−k−1
2
.
Γ(l)
(a)
∞
1=
−∞
|ψ| dx = 2|A|
2
2
0
∞
1
(x2 +
2n
b2 )
|A|2 Γ
dx = 4n−1
b
1
2
Γ 4n−1
2
Γ(2n)
⇒A=
b4n−1 Γ(2n)
.
Γ 12 Γ 4n−1
2
1
d
d2 ψ
2 2 ∞
−2nx
dx
ψ 2 dx = −
A
2
2 n
2
2 n+1
dx
2m
−∞
−∞ (x + b ) dx (x + b )
n2 2 ∞
1
1
2(n + 1)x2
=
dx
A
n+1 −
n+2
2
2 n
m
(x2 + b2 )
(x2 + b2 )
−∞ (x + b )
∞
∞
2n2 2
1
x2
=
A
2n+1 dx − 2(n + 1)
2n+2 dx
m
(x2 + b2 )
(x2 + b2 )
0
0
2n2 b4n−1 Γ(2n)
2(n + 1) Γ 32 Γ 4n+1
2 n(4n − 1)
1 Γ 12 Γ 4n−1
2
2
=
−
=
.
m Γ 12 Γ 4n−1
2b4n−1 Γ(2n + 1)
2b4n−1
Γ(2n + 2)
4mb2 (2n + 1)
2
2
T = −
2m
∞
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204
CHAPTER 7. THE VARIATIONAL PRINCIPLE
1
V = mω 2
2
= mω 2
H =
∞
1
ψ x dx = mω 2 2A2
2
−∞
2 2
b4n−1 Γ(2n)
1 Γ
1
4n−1 2b4n−3
Γ 2 Γ
2
3
2
2 n(4n − 1)
mω 2 b2
+
;
4mb2 (2n + 1)
(4n − 3)
0
∞
x2
2n
(x2 + b2 )
dx
Γ 4n−3
mω 2 b2
2
=
.
Γ(2n)
2(4n − 3)
∂H
2 n(4n − 1)
mω 2 b
=−
+
=0 ⇒
∂b
2mb3 (2n + 1)
(4n − 3)
1/4
n(4n − 1)(4n − 3)
.
mω
2(2n + 1)
mω 2
2(2n + 1)
n(4n − 1)(4n − 3)
2 n(4n − 1) mω
+
=
4m (2n + 1) n(4n − 1)(4n − 3) 2(4n − 3) mω
2(2n + 1)
1
2n(4n − 1)
8n2 − 2n
1
1
= ω
= ω
> ω. 2
2
(2n + 1)(4n − 3)
2
8n − 2n − 3
2
b =
Hmin
(b)
b4n−3 Γ(2n)
|B|2 Γ 32 Γ 4n−3
2
2|B|
.
⇒
B
=
dx
=
2n
b4n−3
Γ(2n)
Γ 32 Γ 4n−3
(x2 + b2 )
0
2
2 2 ∞
x
d
2nx2
1
−
dx
−
B
n+1
2
2 n
2
2 n
2m
(x2 + b2 )
−∞ (x + b ) dx (x + b )
2 B 2 ∞
4nx
4n(n + 1)x3
−2nx
x
−
−
dx
n+1 +
n+2
2m −∞ (x2 + b2 )n (x2 + b2 )n+1
(x2 + b2 )
(x2 + b2 )
∞
∞
4n2 B 2
x2
x4
3
2n+1 dx − 2(n + 1)
2n+2 dx
2m
(x2 + b2 )
(x2 + b2 )
0
0
5
4n−1
Γ
Γ
2n2 b4n−3 Γ(2n)
3 Γ 32 Γ 4n−1
2(n
+
1)
32 n(4n − 3)
2
2
2
−
=
.
m Γ 32 Γ 4n−3
2b4n−1 Γ(2n + 1)
2b4n−1
Γ(2n + 2)
4mb2 (2n + 1)
2
1=
T =
=
=
=
2
∞
1
V = mω 2 2B 2
2
x2
∞
x4
2n
b2 )
dx =
(x2 +
3 n(4n − 3) 3 mω 2 b2
H =
+
;
4mb2 (2n + 1)
2 (4n − 5)
0
2
b4n−3 Γ(2n)
1
2 Γ
mω 2
4n−5
2
2b
Γ 32 Γ 4n−3
2
5
2
Γ 4n−5
3 mω 2 b2
2
=
.
Γ(2n)
2 (4n − 5)
∂H
32 n(4n − 3)
3mω 2 b
=−
+
=0 ⇒
∂b
2mb3 (2n + 1)
(4n − 5)
1/4
n(4n − 3)(4n − 5)
.
mω
2(2n + 1)
3 mω 2
32 n(4n − 3) mω
2(2n + 1)
n(4n − 3)(4n − 5)
+
=
4m (2n + 1) n(4n − 3)(4n − 5) 2 (4n − 5) mω
2(2n + 1)
3
2n(4n − 3)
8n2 − 6n
3
3
= ω
= ω
> ω. 2
2
(2n + 1)(4n − 5)
2
8n − 6n − 5
2
b =
Hmin
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CHAPTER 7. THE VARIATIONAL PRINCIPLE
205
(c) As n → ∞, ψ becomes more and more “gaussian”. In the figures I have plotted the trial wave functions
for n = 2, n = 3, and n = 4, as well as the exact states (heavy line). Even for n = 2 the fit is pretty good,
so it is hard to see the improvement, but the successive curves do move perceptably toward the correct
result.
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
1
2
Analytically, for large n, b ≈
x2 + b2
n
mω
1
4
3
n · 4n · 4n
2 · 2n
1/4
=
2
4
3
2n
, so
mω
n
n
x2
mωx2
= b2n 1 + 2
≈ b2n 1 +
b
2n
→
2
b2n emωx
/2
.
Meanwhile, using Stirling’s approximation (Eq. 5.84), in the form Γ(z + 1) ≈ z z e−z :
2n−1
b4n−1 Γ(2n)
b4n−1 1
b4n−1 (2n − 1)2n−1 e−(2n−1)
2n − 1
√
A =
≈ √ √
2n − 3/2.
3
1
1 ≈
2n−3/2
3
π 2n −
π
e 2n − 2
Γ 2 Γ 2n − 2
e−(2n−3/2)
2
1 1 − 2n
1
3
3
1
1
But
≈
1
−
1
+
≈1+
−
=1+
;
3
2n
4n
4n 2n
4n
1 − 4n
n 2
2n−1
2 √
2n − 1
1
1
1/4
so
→
e
≈
1
+
= e.
4n
1 + 1/4n
2n − 32
1/4
b4n−1 √ √
2n 4n−1
2n
= √
e 2n =
⇒ A≈
b2n−1/2 . So
b
π
π
πe
1/4
1/4 mω 1/4 −mωx2 /2 mω 1/4 −mωx2 /2
2n
2n
2n−1/2 1 −mωx2 /2
ψ ≈
b
e
=
e
=
e
,
π
b2n
π
2n
π
2
which is precisely the ground state of the harmonic oscillator (Eq. 2.59). So it’s no accident that we get
the exact energies, in the limit n → ∞.
Problem 7.13
1 = |A|
2
−2br 2 2
e
2
r sin θ dr dθ dφ = 4π|A|
∞
2 −2br 2
r e
0
dr = |A|
2
π 3/2
2b
⇒ A=
2b
π
3/4
.
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publisher.
206
CHAPTER 7. THE VARIATIONAL PRINCIPLE
V = −
e2
|A|2 4π
4π90
2
T = −
|A|2
2m
∞
e−2br
0
2
1 2
e2
r dr = −
r
4π90
2b
π
3/2
4π
1
e2
2
=−
4b
4π90
2b
.
π
e−br (∇2 e−br ) r2 sin θ dr dθ dφ
−2b
2
2
2
2
1 d d
1 d
But (∇2 e−br ) = 2
r2 e−br = 2
−2br3 e−br = 2 3r2 − 2br4 e−br .
r dr
dr
r dr
r
3/2
3/2
∞
2
2
2
1
−
π
π
2b
2b
3
3
=
πb4
− 2b
(4π)(−2b)
(3r2 − 2br4 )e−2br dr =
2
2m π
m
π
8b 2b
32b
2b
0
2
2b
3
3
32 b
=
4πb
−
=
.
m
π
8b 16b
2m
H =
Hmin
32 b
e2
−
2
2m
4π90
2
2b
;
π
2
∂H
32
e2
=
−
∂b
2m
4π90
2 1
√ =0
π b
2
2 2 4m2
2
e2
e
e2
−
2
4π90
π 94
4π90
π 4π90
2 2
m
e
8
8
=− 2
=
E1 = −11.5 eV.
2
4π90
3π
3π
32
=
2m
2 2m
=
π 32
⇒
√
e2
4π90
b=
2
e2
4π90
m
2
2 2m
.
π 32
4
8
−
3π 3π
Problem 7.14
Let ψ = √
1
πb3
e−r/b (same as hydrogen, but with a → b adjustable). From Eq. 4.191, we have T = −E1 =
2
2
for hydrogen, so in this case T =
.
2
2ma
2mb2
∞
−µr
e2 4π
1
1
e2 4 ∞ −(µ+2/b)r
e2 4
e2
−2r/b e
2
V = −
e
dr
=
−
e
r
dr
=
−
=
−
.
r
3
3
3
2
2
4π90 πb 0
r
4π90 b 0
4π90 b (µ + 2/b)
4π90 b(1 + µb
2 )
H =
2
1
e2
−
.
2
2
2mb
4π90 b(1 + µb
2 )
∂H
2
1
2
e2
µ
e2 (1 + 3µb/2)
=− 3 +
=
−
+
+
=0
∂b
mb
4π90 b2 (1 + µb/2)2
b(1 + µb/2)3
mb3
4π90 b2 (1 + µb/2)3
2
m
4π90
e2
=b
(1 + 3µb/2)
,
(1 + µb/2)3
or
b
⇒
(1 + 3µb/2)
= a.
(1 + µb/2)3
This determines b, but unfortunately it’s a cubic equation. So we use the fact that µ is small to obtain a suitable
approximate solution. If µ = 0, then b = a (of course), so µa 1 =⇒ µb 1 too. We’ll expand in powers of
µb:
2 3µb
9
3µb
µb
3
6
a≈b 1+
≈ b 1 − (µb)2 + (µb)2 = b 1 − (µb)2 .
1−
+6
2
2
2
4
4
4
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CHAPTER 7. THE VARIATIONAL PRINCIPLE
207
Since the 34 (µb)2 term is already a second-order correction, we can replace b by a:
a
3
≈ a 1 + (µa)2 .
b≈ 4
1 − 34 (µb)2
Hmin =
≈
1
2
e2
−
2
3
3
2
2
2
4π9
0 a 1 + (µa)
2ma 1 + 4 (µa)
1+
4
3
e2 1
2
3
2
1
−
2
−
(µa)
1 − (µa)2
2ma2
4
4π90 a
4
1
2
2
(µa)
1−2
µa 2
µa
+3
2
2
3
3
3
3
= −E1 1 − (µa)2 + 2E1 1 − µa + (µa)2 − (µa)2 = E1 1 − 2(µa) + (µa)2 .
2
4
4
2
Problem 7.15
(a)
H=
λ=
1
2
Ea h
h Eb
Ea + Eb ±
; det(H − λ) = (Ea − λ)(Eb − λ) − h2 = 0 =⇒ λ2 − λ(Ea + Eb ) + Ea Eb − h2 = 0.
Ea2 + 2Ea Eb + Eb2 − 4Ea Eb + 4h2
⇒ E± =
1
2
Ea + Eb ±
(Ea − Eb )2 + 4h2 .
(b) Zeroth order: Ea0 = Ea , Eb0 = Eb . First order: Ea1 = ψa |H |ψa = 0, Eb1 = ψb |H |ψb = 0. Second
order:
|ψb |H |ψa |2
h2
|ψa |H |ψb |2
h2
Ea2 =
=−
; Eb2 =
=
;
Ea − Eb
Eb − Ea
Eb − Ea
Eb − Ea
E − ≈ Ea −
h2
;
(Eb − Ea )
E+ ≈ Eb +
h2
.
(Eb − Ea )
(c)
H = cos φ ψa + sin φ ψb |(H 0 + H )| cos φ ψa + sin φ ψb = cos2 φ ψa |H 0 |ψa + sin2 φ ψb |H 0 |ψb + sin φ cos φ ψb |H |ψa + sin φ cos φ ψa |H |ψb = Ea cos2 φ + Eb sin2 φ + 2h sin φ cos φ.
∂H
= −Ea 2 cos φ sin φ + Eb 2 sin φ cos φ + 2h(cos2 φ − sin2 φ) = (Eb − Ea ) sin 2φ + 2h cos 2φ = 0.
∂φ
tan 2φ = −
or
2h
= −9 where
Eb − Ea
sin2 2φ(1 + 92 ) = 92 ;
cos 2φ = √
∓1
1 + 92
9≡
sin 2φ = √
2h
.
Eb − Ea
±9
;
1 + 92
(sign dictated by tan 2φ =
sin 2φ
1 − sin2 2φ
= −9;
sin2 2φ = 92 (1 − sin2 2φ);
cos2 2φ = 1 − sin2 2φ = 1 −
92
1
=
;
1 + 92
1 + 92
sin 2φ
= −9).
cos 2φ
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208
CHAPTER 7. THE VARIATIONAL PRINCIPLE
1
1
1
1
1
1
2
√
√
; sin φ = (1 − cos 2φ) =
.
cos φ = (1 + cos 2φ) =
1∓
1±
2
2
2
2
1 + 92
1 + 92
1
1
1
9
1
1
(Eb − Ea + 2h9)
√
Hmin = Ea 1 ∓ √
+ Eb 1 ± √
± h√
=
E a + Eb ±
2
2
2
1 + 92
1 + 92
1 + 92
1 + 92
2
(Eb − Ea ) + 2h (Eb2h
(Eb − Ea + 2h9)
(E − Ea )2 + 4h2
−Ea )
√
b
=
=
= (Eb − Ea )2 + 4h2 ,
2
2
2
2
1+9
(Eb − Ea ) + 4h
1 + (Eb4h
−Ea )2
But
So
1
Ea + Eb ± (Eb − Ea )2 + 4h2
we want the minus sign (+ is maximum)
2
= 12 Ea + Eb − (Eb − Ea )2 + 4h2 .
Hmin =
(d) If h is small, the exact result (a) can be expanded: E± =
1
=⇒ E± ≈
2
so
1
2
2h2
Ea + Eb ± (Eb − Ea ) 1 +
(Eb − Ea )2
E+ ≈ E b +
h2
,
(Eb − Ea )
E− ≈ E a −
(Ea + Eb ) ± (Eb − Ea ) 1 +
=
4h2
(Eb −Ea )2
.
1
2h2
Ea + Eb ± (Eb − Ea ) ±
,
2
(Eb − Ea )
h2
,
(Eb − Ea )
confirming the perturbation theory results in (b). The variational principle (c) gets the ground state (E− )
exactly right—not too surprising since the trial wave function Eq. 7.56 is almost the most general state
(there could be a relative phase factor eiθ ).
Problem 7.16
For the electron,
γ= −e/m, so E± =
±eBz /2m (Eq. 4.161). For consistency with Problem 7.15, Eb > Ea ,
eBz eBz 1
0
so χb = χ+ =
, Ea = E − = −
.
, χa = χ− =
, Eb = E + =
0
1
2m
2m
(a)
eBx 01
χa |H |χa =
m 2
eBx 10
χb |H |χb =
2m
eBx 01
χa |H |χb =
2m
01
10
eBx 0
1
0 1
=
= 0;
1
0
2m
01
10
1
= 0;
0
01
10
eBx eBx 1
0
0 1
=
=
. So
0
1
2m
2m
eBx 1 0
χb |H |χa =
2m
0 1
1 0
eBx 0
=
;
1
2m
h=
eBx ,
2m
and the conditions of Problem 7.15 are met.
c
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publisher.
CHAPTER 7. THE VARIATIONAL PRINCIPLE
209
(b) From Problem 7.15(b),
Egs
h2
e
eBz (eBx /2m)2
≈ Ea −
=−
−
= −
(Eb − Ea )
2m
(eBz /m)
2m
(c) From Problem 7.15(c), Egs =
Egs
1
=−
2
eBz m
1
2
Ea + Eb −
2
+4
eBx 2m
Bx2
.
Bz +
2Bz
(Eb − Ea )2 + 4h2 (it’s actually the exact ground state).
2
= −
e 2
Bz + Bx2
2m
(which was obvious from the start, since the square root is simply the magnitude of the total field).
Problem 7.17
(a)
1
1
1
r1 = √ (u + v); r2 = √ (u − v); r12 + r22 = (u2 + 2u · v + v 2 + u2 − 2u · v + v 2 ) = u2 + v 2 .
2
2
2
(∇21 + ∇22 )f (r1 , r2 ) =
∂2f
∂2f
∂2f
∂2f
∂2f
∂2f
+ 2 + 2 +
+ 2 + 2
2
2
∂x1
∂y1
∂z1
∂x2
∂y2
∂z2
∂f
∂f ∂ux
∂f ∂vx
1
=
+
=√
∂x1
∂ux ∂x1
∂vx ∂x1
2
∂2f
1 ∂
=√
2
∂x1
2 ∂x1
=
1
2
So
∂f
∂f
+
∂ux
∂vx
∂f
∂f
−
∂ux
∂vx
=
∂2f
∂2f
+ 2
2
∂ux
∂vx
;
.
∂f
∂f ∂ux
∂f ∂vx
1
=
+
=√
∂x2
∂ux ∂x2
∂vx ∂x2
2
∂f
∂f
−
∂ux
∂vx
∂ 2 f ∂vx
∂ 2 f ∂ux
∂ 2 f ∂vx
∂ 2 f ∂ux
+
+
+
∂u2x ∂x1
∂ux ∂vx ∂x1
∂vx ∂ux ∂x1
∂vx2 ∂x1
;
1
=√
2
∂2f
∂2f
∂2f
−
2
+
∂u2x
∂ux ∂vx
∂vx2
∂2f
∂2f
+
2
∂x1
∂x22
∂f
∂f
+
∂ux
∂vx
1
=√
2
∂2f
∂2f
∂2f
+2
+ 2
2
∂ux
∂ux ∂vx
∂vx
∂2f
1 ∂
=√
∂x22
2 ∂x2
1
=
2
∂ 2 f ∂vx
∂ 2 f ∂ux
∂ 2 f ∂vx
∂ 2 f ∂ux
+
−
−
∂u2x ∂x2
∂ux ∂vx ∂x2
∂vx ∂ux ∂x2
∂vx2 ∂x2
.
, and likewise for y and z: ∇21 + ∇22 = ∇2u + ∇2v .
2
1
λ
(∇2u + ∇2v ) + mω 2 (u2 + v 2 ) − mω 2 2v 2
2m
2
4
2 2 1
2 2 1
1
2 2
∇ + mω u + −
∇ + mω 2 v 2 − λmω 2 v 2 . QED
= −
2m u 2
2m v 2
2
H =−
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210
CHAPTER 7. THE VARIATIONAL PRINCIPLE
√
√
(b) The energy is 32 ω (for the u part) and 32 ω 1 − λ (for the v part): Egs = 32 ω 1 + 1 − λ .
(c) The ground state for a one-dimensional oscillator is
ψ0 (x) =
mω 1/4
π
e−mωx
2
/2
(Eq. 2.59).
So, for a 3-D oscillator, the ground state is ψ0 (r) =
ψ(r1 , r2 ) =
H =
mω 3/2
π
mω 3/4 −mωr 2 /2
e
,
π
and for two particles
e− 2 (r1 +r2 ) . (This is the analog to Eq. 7.17.)
mω
2
2
3
3
ω + ω + Vee = 3ω + Vee (the analog to Eq. 7.19).
2
2
mω 3
λ
Vee = − mω 2
4
π
e−
2
2
mω
(r1 +r2 )
(r1 − r2 )2 d3 r1 d3 r2
(the analog to Eq. 7.20).
r12 −2r1 ·r2 +r22
The r1 · r2 term integrates to zero, by symmetry, and the r22 term is the same as the r12 term, so
mω 3 2
2
mω
λ
Vee = − mω 2
2 e− (r1 +r2 ) r12 d3 r1 d3 r2
4
π
∞
∞
mω 3
2
λ
2
2
−mωr22 / 2
= − mω
(4π)
e
r2 dr2
e−mωr1 / r14 dr1
2
π
0
0
2
4 5
π
π
3
3
8m ω 1 = − λω.
= −λ
π3
4 mω mω
8 mω
mω
4
3
λ
H = 3ω − λω = 3ω 1 −
.
4
4
The variational principle says this must exceed the exact ground-state energy (b); let’s check it:
√
√
λ
λ
λ √
λ2
3 3ω 1 −
> ω 1 + 1 − λ ⇔ 2 − > 1 + 1 − λ ⇔ 1 − > 1 − λ ⇔ 1 − λ +
> 1 − λ.
4
2
2
2
4
It checks. In fact, expanding the exact answer in powers of λ, Egs ≈ 32 ω(1 + 1 − 12 λ) = 3ω 1 −
we recover the variational result.
λ
4
Problem 7.18
1==
|ψ| d r1 d r2 = |A|
2 3
3
2
ψ12
3
d r1
ψ22
3
d r2 + 2
3
ψ1 ψ2 d r1
3
ψ1 ψ2 d r2 +
ψ22
3
d r1
ψ12 d3 r2
= |A|2 (1 + 2S 2 + 1),
where
S≡
3
ψ1 (r)ψ2 (r) d r =
(Z1 Z2 )3
πa3
e−(Z1 +Z2 )r/a 4πr2 dr =
y 3
2a3
4 y 3
=
.
a3 2
(Z1 + Z2 )3
x
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,
CHAPTER 7. THE VARIATIONAL PRINCIPLE
A2 =
1
6
211
.
2 1 + (y/x)
2
e2
H=−
(∇21 + ∇22 ) −
2m
4π90
Hψ = A
1
1
+
r1
r2
2
e2
−
(∇21 + ∇22 ) −
2m
4π90
+ −
+A
where Vee ≡
2
e2
(∇21 + ∇22 ) −
2m
4π90
e2
4π90
+
e2
1
,
4π90 |r1 − r2 |
Z1
Z2
+
r1
r2
Z1
Z2
+
r1
r2
ψ1 (r1 )ψ2 (r2 )
ψ2 (r1 )ψ1 (r2 )
Z1 − 1 Z2 − 1
Z2 − 1 Z1 − 1
ψ1 (r1 )ψ2 (r2 ) +
ψ2 (r1 )ψ1 (r2 ) + Vee ψ,
+
+
r1
r2
r1
r2
1
e2
.
4π90 |r1 − r2 |
The term in first curly brackets is (Z12 + Z22 )E1 ψ1 (r1 )ψ2 (r2 ) + (Z22 + Z12 )ψ2 (r1 )ψ1 (r2 ), so
Hψ = (Z12 + Z22 )E1 ψ
Z1 − 1 Z2 − 1
e2
Z2 − 1 Z1 − 1
ψ1 (r1 )ψ2 (r2 ) +
ψ2 (r1 )ψ1 (r2 ) + Vee ψ
+A
+
+
4π90
r1
r2
r1
r2
H =
(Z12
+
Z22 )E1
+ Vee + A
2
e2
4π90
Z1 − 1 Z2 − 1
× ψ1 (r1 )ψ2 (r2 ) + ψ2 (r1 )ψ1 (r2 ) +
r1
r2
ψ1 (r1 )ψ2 (r2 ) + Z2 − 1 + Z1 − 1
r1
r2
ψ2 (r1 )ψ1 (r2 )
.
1
1
= (Z1 − 1)ψ1 (r1 ) ψ1 (r1 ) + (Z2 − 1)ψ2 (r2 ) ψ2 (r2 )
r1
r2
1
+ (Z2 − 1)ψ1 (r1 ) ψ2 (r1 )ψ2 (r2 )|ψ1 (r2 )
r1
1
1
+ (Z1 − 1)ψ1 (r1 )|ψ2 (r1 )ψ2 (r2 ) ψ1 (r2 ) + (Z1 − 1)ψ2 (r1 ) ψ1 (r1 )ψ1 (r2 )|ψ2 (r2 )
r2
r1
1
1
+ (Z2 − 1)ψ2 (r1 )|ψ1 (r1 )ψ1 (r2 ) ψ2 (r2 ) + (Z2 − 1)ψ2 (r1 ) ψ2 (r1 )
r2
r1
1
+ (Z1 − 1)ψ1 (r2 ) ψ1 (r2 )
r2
0 1
0 1
1
1
1
1
= 2(Z1 − 1)
+ 2(Z1 − 1)
+ 2(Z1 − 1)ψ1 |ψ2 ψ1 ψ2 + 2(Z2 − 1)ψ1 |ψ2 ψ1 ψ2 .
r 1
r 2
r
r
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212
CHAPTER 7. THE VARIATIONAL PRINCIPLE
0 1
1
1
Z1
= ψ1 (r) ψ1 (r) =
;
r 1
r
a
But
+A
2
e2
4π90
0 1
Z2
1
=
,
r 2
a
so
H = (Z12 + Z22 )E1
1
1
1
2 (Z1 − 1)Z1 + (Z2 − 1)Z2 + (Z1 + Z2 − 2)ψ1 |ψ2 ψ1 ψ2 + Vee .
a
a
r
3
ψ1 |ψ2 = S = (y/x) , so
And
1
(Z1 Z2 )3
a
y3
−(Z1 +Z2 )r/a
ψ1 ψ2 =
4π
e
r
dr
=
r
πa3
2a3 Z1 + Z2
2
=
y3
.
2ax2
2 y 3 y 3 e
2 2
1
H = (x2 − y 2 )E1 + A2
+ Vee Z1 + Z22 − (Z1 + Z2 ) + (x − 2)
2
4π90 a
x 2x2
y6
1
1
1
= (x2 − y 2 )E1 + 4E1 A2 x2 − y 2 − x + (x − 2) 5 + Vee .
2
2
2
x
e2
1
ψ
ψ 4π90
|r1 − r2 | 1
e
ψ1 (r1 )ψ2 (r2 ) + ψ2 (r1 )ψ1 (r2 )
A2 ψ1 (r1 )ψ2 (r2 ) + ψ2 (r1 ) + ψ1 (r2 ) =
4π90
|r1 − r2 | e
1
1
ψ1 (r1 )ψ2 (r2 ) + 2ψ1 (r1 )ψ2 (r2 ) ψ2 (r1 )ψ1 (r2 )
=
A2 2ψ1 (r1 )ψ2 (r2 ) 4π90
|r1 − r2 |
|r1 − r2 | e
A2 (B + C), where
=2
4π90
Vee =
B ≡ ψ1 (r1 )ψ2 (r2 ) B=
Z13 Z23
(πa3 )2
1
ψ1 (r1 )ψ2 (r2 );
|r1 − r2 | e−2Z1 r1 /a e−2Z2 r2 /a
e−2Z2 r2 /a +
− 2r1 r2 cos θ2
r22
d3 r2
Z2 r1
πa3
1− 1+
= 3
e−2Z2 r1 /a
Z2 r1
a
Z 3 Z 3 (πa3 )
B = 1 3 22
4π
(πa ) Z23
4Z 3
= 31
a
0
∞
∞
−2Z1 r1 /a
e
0
1
ψ2 (r1 )ψ1 (r2 ).
|r1 − r2 | 1
d3 r1 d3 r2 . As on pp 300-301, the r2 integral is
|r1 − r2 |
1
r12
C ≡ ψ1 (r1 )ψ2 (r2 ) (Eq. 7.24, but with a →
2
a).
Z2
1
Z2 r1
1− 1+
e−2Z2 r1 /a r12 dr1
r1
a
r1 e−2Z1 r1 /a − r1 e−2(Z1 +Z2 )r1 /a −
Z2 2 −2(Z1 +Z2 )r1 /a
dr1
r e
a 1
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CHAPTER 7. THE VARIATIONAL PRINCIPLE
4Z 3
= 31
a
=
C=
=
a
2Z1
2
−
a
2(Z1 + Z2 )
2
Z2
2
−
a
y2
Z1 Z2
Z1 Z2
=
1+
2
a(Z1 + Z2 )
(Z1 + Z2 )
4ax
Z13 Z23
(πa3 )2
(Z1 Z2 )3
(πa3 )2
213
1+
a
2(Z1 + Z2 )
y2
4x2
e−Z1 r1 /a e−Z2 r2 /a e−Z2 r1 /a e−Z1 r2 /a
e−(Z1 +Z2 )(r1 +r2 )/a
3 =
Z13
a
1
1
Z2
−
−
Z12
(Z1 + Z2 )2
(Z1 + Z2 )3
.
1
d3 r1 d3 r2
|r1 − r2 |
1
d3 r1 d3 r2 .
|r1 − r2 |
4
The integral is the same as in Eq. 7.20, only with a → Z1 +Z
a. Comparing Eqs. 7.20 and 7.25, we see that the
2
integral itself was
2
5 πa3
5 2 5
(Z1 Z2 )3 5π 2
20 (Z1 Z2 )3
5 y6
45 a5
=
=
=
.
π a . So C =
4a
8
256
(πa3 )2 256 (Z1 + Z2 )5
a (Z1 + Z2 )5
16a x5
Vee = 2
e
4π90
A
y2
4ax
2
y2
1+ 2
4x
5 y6
y2
2
+
=
2A
(−2E
)
1
16a x5
4x
5y 4
y2
1+ 2 + 4
4x
4x
H = E1
y6
y2
2
1
2
1 2
1
2
x
y
(x
−
2)
x − y2 −
−
−
−
x
+
2
[1 + (y/x)6 ]
2
2
x5
[1 + (y/x)6 ] 4x
2
E1
= 6
(x + y 6 )
E1
= 6
(x + y 6 )
.
y2
5y 4
1+ 2 + 4
4x
4x
y4
1
1
1 y6
y6
y2
5y 6
(x − y 2 )(x6 + y 6 ) − 2x6 x2 − y 2 − x +
+
−
+
+
2
2
2 x4
x5
4x 16x3
16x5
2
E1
=
(x6 + y 6 )
1
1
1
1
5
x + x y − x6 y 2 − y 8 − 2x8 + x6 y 2 + 2x7 − x2 y 6 + 2xy 6 − x5 y 2 − x3 y 4 − xy 6
2
2
2
8
8
8
2 6
1 6 2 1 5 2 1 3 4 11 6 1 8
−x + 2x + x y − x y − x y + xy − y .
2
2
8
8
2
8
7
Mathematica finds the minimum of H at x = 1.32245, y = 1.08505, corresponding to Z1 = 1.0392, Z2 =
0.2832. At this point, Hmin = 1.0266E1 = −13.962 eV, which is less than −13.6 eV—but not by much!
Problem 7.19
The calculation is the same as before, but with me → mµ (reduced), where
mµ (reduced) =
mµ
1+
=1+
2mp
m µ md
mµ 2mp
mµ
=
=
. From Problem 6.28, mµ = 207me , so
mµ + md
mµ + 2mp
1 + mµ /2mp
207
2
(9.11 × 10−31 )
207 me
= 1.056; mµ (reduced) =
= 196 me .
−27
(1.67 × 10 )
1.056
This shrinks the whole molecule down by a factor of almost 200, bringing the deuterons much closer together,
as desired. The equilibrium separation for the electron case was 2.493 a (Problem 7.10), so for muons, R =
2.493
(0.529 × 10−10 m) = 6.73 × 10−13 m.
196
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214
CHAPTER 7. THE VARIATIONAL PRINCIPLE
Problem 7.20
(a)
−
2 ∂ 2 ψ ∂ 2 ψ
= Eψ.
+
2m ∂x2
∂y 2
Y
d2 X
d2 Y
2mE
+
X
= − 2 XY ;
dx2
dy 2
d2 X
= −kx2 X;
dx2
Let ψ(x, y) = X(x)Y (y).
d2 Y
= −ky2 Y,
dy 2
1 d2 X
1 d2 Y
2mE
+
=− 2 .
X dx2
Y dy 2
with kx2 + ky2 =
2mE
. The general solution to the y equation is
2
Y (y) = A cos ky y + B sin ky y; the boundary conditions Y (±a) = 0 yield ky =
nπ
π
with minimum
.
2a
2a
[Note that ky2 has to be positive, or you cannot meet the boundary conditions at all.] So
2
π2
2
E ≥
kx + 2 . For a traveling wave kx2 has to be positive. Conclusion: Any solution with E <
2m
4a
π 2 2
will be a bound state.
8ma2
(b)
y
a
II
I
a
x
Integrate over regions I and II (in the figure), and multiply by 8.
III = A2
∞
1−
x=a
= A2 a2
1
a
∞
y=0
1
y
a
2
e−2αx/a dx dy.
(1 − v)2 e−2αu du dv = A2 a2
0
y
x
, v ≡ , dx = a du, dy = a dv.
a
a
1
∞
e−2αu (1 − v)3 ×
3
2α 1
0
Let u ≡
2 2
=
A a
A2 a2 −2α
.
(−1) − e−2α =
e
6α
6α
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CHAPTER 7. THE VARIATIONAL PRINCIPLE
II =
1 2
A
2
a
a
1−
x=0
1
= A2 a2
2
y=0
1
1
xy
a2
2
e−2α dx dy
2 −2α
(1 − uv) e
0
0
215
1
du dv = A2 a2 e−2α
2
Normalizing:
H = −
1
(1 − uv)3 dv
−3v 0
1
(v 2 − 3v + 3)dv,
0
e2α
A2 a2 −2α 11 2 2 −2α
9α
= 1 ⇒ A2 = 2
+ A a e
e
.
6α
36
2a (6 + 11α)
8
2
∂2
2
∂2
ψ| 2 + 2 |ψ = −8
(JI + JII ). [Ignore roof-lines for the moment.]
2m
∂x
∂y
2m
0
2
2
✼
y
∂
y −αx/a
∂
1−
1
−
dx dy
e−αx/a
e
= A2
+
a
∂x2 ∂y 2
a
x=a y=0
2
2✄ 2 2
2 2
∞ a y
α
1
α
α A a −2α
2
−2αx/a
=A
1−
e
dx dy =
III =
= A2 αe−2α .
e
a
a
a
6
α
6
a
✚
x=a y=0
✁
JII
3
0
1
1
(1 − v) − 1
1
= − A2 a2 e−2α
dv = A2 a2 e−2α
2
3 0
v
6
3
1
2
1
v
11 2 2 −2α
v
= A2 a2 e−2α
.
− 3 + 3v =
A a e
6
3
2
36
0
1
1
∞
a
xy −α ∂ 2
xy −α
∂2
e
1
−
e dx dy = 0.
+
a2
∂x2
∂y 2
a2
0
0
xy
∂ y
∂2
1
−
=
− 2 = 0, and likewise for ∂ 2 /∂y 2 .]
[Note that
∂x2
a2
∂x
a
JI =
1 2
A
2
Hso
far
a
a
1−
2 2 α −2α
e
= − A2
.
3
m
Now the roof-lines; label them as follows:
I. Right arm: at y = 0 : KI .
II. Central square: at x = 0 and at y = 0 : KII .
III. Boundaries: at x = ±a and at y = ±a : KIII .
2
∞ a |y| −αx/a ∂
|y| −αx/a
2
∂2
2
KI = 4 −
1−
1−
+ 2
dx dy.
A
e
e
2m
a
∂x2
∂y
a
x=a y=−a
|y| = y θ(y) − θ(−y) ,
∂
|y|
1
✟✟ + ✟
✟✟ ,
1−
= − θ(y) − θ(−y) + ✟
yδ(y)
yδ(y)
∂y
a
a
∂2
1
2
|y|
= − δ(y) − δ(−y)(−1) = − δ(y).
1
−
2
∂y
a
a
a
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216
CHAPTER 7. THE VARIATIONAL PRINCIPLE
22 2
KI = −
A
m
∞
−2αx/a
e
x=a ♣
a
dx
y=−a
|y|
2
1−
− δ(y) dy
a
a
♠
∞
a −2α
e−2αx/a e−2α
2
=
e
♣=
=−
; ♠=− ,
(−2α/a) a
(−2α/a)
2α
a
22 A2 a
22 −2α 2
✁ e−2α − 2 ;
=−
KI =
A .
e
a
m ✁
mα
2α
✁
KII
2
x|y| −α ∂
x|y| −α
∂2
1− 2 e
1 − 2 e dx dy
= 4A
+ 2
a
∂x2
∂y
a
x=0 y=−a
a a 2
2 2 −2α
x|y|
2x
=−
1− 2
− 2 δ(y) dx dy
A e
m
a
a
x=0 y=−a
a
22 A2 −2α
2
22 −2α 2
=−
e
e
− ✁2✁
xdx;
KII =
A .
m
m
✁a
0
2
2
−
2m
a
a
a2
2
KIII
2
a a+5
2
∂
∂2
=8 −
ψ
+ 2 ψ dx dy.
2m
∂x2
∂y
y=0 x=a−5
In this region (x, y both positive)
ψ=A
2
−α
1
−
xy/a
e
(x
<
a)
ψ=A ,
−αx/a
1
−
y/a
e
(x
>
a)
y
x
1−
θ(x − a) + θ(a − x)
a
a
or
−α[θ(a−x)+ x
θ(x−a)]
a
.
e
✟ −α θ(a−x)+ x θ(x−a)
∂ψ
y
✘ + 1 θ(a − x) − x δ(a✟−✟x)
]
✘a)
a
✘−
=A − ✘
δ(x
e [
✟
∂x
a
a
a
✟
y
α
x
αx
✘
✘
✘
−α[θ(a−x)+ x
θ(x−a)]
✘
✘
a
✘ − a)
✘− x) − θ(x − a) − ✘✘δ(x
+ 1−
α✘
δ(a
θ(x − a) + θ(a − x) e
a
a
a
a
[Note: f (x) = xδ(x) should be zero—but perhaps we should check that this is still safe when we’re
planning to take it’s derivative: df /dx = δ(x) + x dδ/dx :
df
dδ
dδ
g dx = g δ(x) + x
dx = g(0) + gx dx
dx
dx
dx
✿0
✘
d
dg
✘
✘
= g(0) + ✘
gxδ(x)|
(gx)δ(x)dx = g(0) −
g+x
δ(x)dx
✘✘ x=0 −
dx
dx
= g(0) − g(0) − (xg )|x=0 = 0.
This confirms that f (x) can be taken to be zero even when differentiated.]
So δ(x − a) −
x
1
δ(a − x) = (a − x)δ(a − x) = 0. Hence the cancellations above, leaving
a
a
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CHAPTER 7. THE VARIATIONAL PRINCIPLE
217
x
∂ψ
y
= A − 2 θ(a − x)e−α[θ(a−x)+ a θ(x−a)]
∂x
a
α
y
x
−α[θ(a−x)+ x
θ(x−a)]
a
− θ(x − a) 1 −
θ(x − a) + θ(a − x) e
a
a
a
x
α
y
y
x
= Ae−α[θ(a−x)+ a θ(x−a)] − 2 θ(a − x) − θ(x − a) 1 −
θ(x − a) + θ(a − x)
a
a
a
a
x
A
y
y
= − e−α[θ(a−x)+ a θ(x−a)] θ(a − x) + αθ(x − a) 1 −
.
a
a
a
∂2ψ
y
y
A −α[θ(a−x)+ xa θ(x−a)]
− δ(a − x) + αδ(x − a) 1 −
=− e
∂x2
a
a
a
✟
✘ + 1 θ(x − a) + x δ(x✟−✟a)
✘−✘x)
− α −✘
δ(a
✟
a
a
✟
integral 0
A
αy y
= − e−α δ(x − a) α −
−
.
a
a
a
A
αy y
−
dx dy
ψ(x, y) − e−α δ(x − a) α −
a
a
a
y=0 x=a−5
42 A −α a
αy y
αy y αy 2
42 A2 −2α a
y2
=
α−
α−2
e
−
dy =
e
− + 2 + 2 dy
ψ(a, y)
ma
a
a
ma
a
a
a
a
y=0
0
KIII = −
42
m
a
a+5
A(1−y/a)e−α
42 A2 −2α
α a2✄
α a3
1 a3
1 a2
=
αa − ✁
2
e
+ 2
+ 2
−
ma
a 2
a 3
a 3
a
2
✁ ✁
a
42 A2 −2α
αa
a
✁
✁
✁
✟−✟
✟− +
αa
αa
=
e
+
✟
m
2
3
3
α
1
1
3 − 6 = 6 (2α−1)
=
42 A2
(2α − 1)e−2α ;
6m
KIII =
22 A2
(2α − 1)e−2α .
3m
2 2 α −2α 22 −2α 2 22 −2α 2 22 2
e
e
e
A (2α − 1)e−2α
H = − A2
+
A +
A +
3
m
mα
m
3m
2
2
2
2A2 e−2α 2
3
A2 e−2α 2
− α + + 2 + (2α − 1) =
− α + + 3 + 2α − 1
=
m
3
α
3
3m
α
✘
✘
2 −2α 2
2 −2α 2
−2α
✟
✘
2A e
2 2
3
2A e
e2α
α
2e
9✚
✟
=
α+2+
=
α2 + 2α + 3 = ✁
α + 2α + 3
2
3m
α
3mα
3 m✚
α
✁2 a (6 + 11α)
=
32 (α2 + 2α + 3)
.
ma2 (6 + 11α)
32 (6 + 11α)(2α + 2) − (α2 + 2α + 3)(11)
dH
=0
=
dα
ma2
(6 + 11α)2
⇒ (6 + 11α)(2α + 2) = 11(α2 + 2α + 3).
12α + 12 + 22α2 + 22α = 11α2 + 22α + 33 ⇒ 11α2 + 12α − 21 = 0.
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218
CHAPTER 7. THE VARIATIONAL PRINCIPLE
√
(12)2 + 4 · 11 · 21
−6 ± 36 + 231
α=
=
22
11
−6 ± 16.34
10.34
=
=
[α has to be positive] = 0.940012239.
11
11
−12 ± 2
Hmin
2 6 2
32 2(α + 1)
.
=
(α + 1) = 1.058
=
2
2
ma
11
11 ma
ma2
But Ethreshold =
π 2 2
2
=
1.2337
,
8 ma2
ma2
so E0 is definitely less than Ethreshold .
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CHAPTER 8. THE WKB APPROXIMATION
219
Chapter 8
The WKB Approximation
Problem 8.1
a
p(x) dx = nπ,
with n = 1, 2, 3, . . . and
p(x) =
2m[E − V (x)] (Eq. 8.16).
0
Here
a
p(x) dx =
√
0
2mE
a
2
+
2m(E − V0 )
4
⇒ E + E − V0 + 2 E(E − V0 ) =
2m
Square again:
nπ
a
a
2
=
√
2
= 4En0 ;
2m
a √
2
E+
E − V0 = nπ
2 E(E − V0 ) = (4En0 − 2E + V0 ).
2
4E(E − V0 ) = 4E 2 − 4EV0 = 16En0 + 4E 2 + V02 − 16EEn0 + 8En0 V0 − 4EV0
2
⇒ 16EEn0 = 16En0 + 8En0 V0 + V02 ⇒ En = En0 +
V2
V0
+ 0 0.
2
16En
V0
Perturbation theory gave En = En0 + ; the extra term goes to zero for very small V0 (or, since En0 ∼ n2 ), for
2
large n.
Problem 8.2
(a)
dψ
i
= f eif / ;
dx
d2 ψ
i
=
2
dx
if /
f e
i
+ (f )2 eif /
=
i
1
f − 2 (f )2 eif / .
d2 ψ
p2
i
1
p2
f − 2 (f )2 eif / = − 2 eif / =⇒ if − (f )2 + p2 = 0. QED
= − 2 ψ =⇒
2
dx
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220
CHAPTER 8. THE WKB APPROXIMATION
(b) f = f0 + f1 + 2 f2 + · · · =⇒ (f )2 = (f0 + f1 + 2 f2 + · · · )2 = (f0 )2 + 2f0 f1 + 2 [2f0 f2 + (f1 )2 ] + · · ·
f = f0 + f1 + 2 f2 + · · · .
i(f0 + f1 + 2 f2 ) − (f0 )2 − 2f0 f1 − 2 [2f0 f2 + (f1 )2 ] + p2 + · · · = 0.
0 : (f0 )2 = p2 ;
1 : if0 = 2f0 f1 ; 2 : if1 = 2f0 f2 + (f1 )2 ; . . .
df0
df1
i ±p
i
i f0
i d
(c)
=
= ±p =⇒ f0 = ± p(x)dx + constant ;
=
=
ln p =⇒ f1 = ln p + const.
dx
dx
2 f0
2 ±p
2 dx
2
if
i
i
i
ψ = exp
= exp
± p(x) dx + ln p + K
= exp ±
p dx p−1/2 eiK/
2
C
i
= √ exp ±
p dx . QED
p
Problem 8.3
1
γ=
1
|p(x)| dx =
2a
2m(V0 − E) dx =
0
2a 2m(V0 − E).
√
T ≈ e−4a 2m(V0 −E)/ .
From Problem 2.33, the exact answer is
1
T =
1+
V02
4E(V0 −E)
sinh2 γ
.
Now, the WKB approximation assumes the tunneling probability is small (p. 322)—which is to say that γ is
large. In this case, sinh γ = 12 (eγ − e−γ ) ≈ 12 eγ , and sinh2 γ ≈ 14 e2γ , and the exact result reduces to
16E(V0 − E)
1
e−2γ .
≈
T ≈
2
V02
V
2γ
0
1 + 16E(V0 −E) e
The coefficient in { } is of order 1; the dominant dependence on E is in the exponential factor. In this sense
T ≈ e−2γ (the WKB result).
Problem 8.4
I take the masses from Thornton and Rex, Modern Physics, Appendix 8. They are all atomic masses, but the
electron masses subtract out in the calculation of E. All masses are in atomic units (u): 1 u = 931 MeV/c2 .
The mass of He4 is 4.002602 u, and that of the α-particle is 3727 MeV/c2 .
U238 : Z = 92, A = 238, m = 238.050784 u → Th234 : m = 234.043593 u.
r1 = (1.07 × 10−15 m)(238)1/3 = 6.63 × 10−15 m.
E = (238.050784 − 234.043593 − 4.002602)(931) MeV = 4.27 MeV.
V =
2E
=
m
(2)(4.27)
× 3 × 108 m/s = 1.44 × 107 m/s.
3727
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CHAPTER 8. THE WKB APPROXIMATION
221
90
γ = 1.980 √
− 1.485 90(6.63) = 86.19 − 36.28 = 49.9.
4.27
τ=
(2)(6.63 × 10−15 ) 98.8
7.46 × 1021
21
e
s
=
7.46
×
10
s
=
yr = 2.4 × 1014 yrs.
1.44 × 107
3.15 × 107
Po212 : Z = 84, A = 212, m = 211.988842 u → Pb208 : m = 207.976627 u.
r1 = (1.07 × 10−15 m)(212)1/3 = 6.38 × 10−15 m.
E = (211.988842 − 207.976627 − 4.002602)(931) MeV = 8.95 MeV.
V =
2E
=
m
(2)(8.95)
× 3 × 108 m/s = 2.08 × 107 m/s.
3727
82
γ = 1.980 √
− 1.485 82(6.38) = 54.37 − 33.97 = 20.4.
8.95
τ=
(2)(6.38 × 10−15 ) 40.8
e
s = 3.2 × 10−4 s.
2.08 × 107
These results are way off—but note the extraordinary sensitivity to nuclear masses: a tiny change in E produces
enormous changes in τ .
√
Much more impressive results are obtained when you plot the logarithm of lifetimes against 1/ E, as in
Figure 8.6. Thanks to David Rubin for pointing this out. Some experimental values are listed below (all energies
in MeV):
A
E
τ
238 4.198 4.468 × 109 yr
A
E
τ
236 4.494 2.342 × 107 yr
224 7.488 0.79 s
5
234 4.775 2.455 × 10 yr
Uranium (Z = 92):
Protactinium (Z = 91): 222 8.540 2.9 ms
232 5.320
68.9 yr
220 9.650 0.78 µs
230 5.888
20.8 day
218 9.614 0.12 ms
228 6.680
9.1 min
226 7.570
0.35 s
A
232
Thorium (Z = 90): 230
228
226
E
τ
4.012 1.405 × 1010 yr
4.687 7.538 × 104 yr
5.423
1.912 yr
6.337
30.57 min
A
226
224
Radium (Z = 88):
222
220
218
E
4.784
5.685
6.559
7.455
8.389
τ
1600 yr
3.66 day
38 s
18 ms
25.6 µs
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222
CHAPTER 8. THE WKB APPROXIMATION
Problem 8.5
(a) V (x) = mgx.
(b)
2 d 2 ψ
d2 ψ
2m2 g
−
+
mgxψ
=
Eψ
=⇒
=
2m dx2
dx2
2
E
x−
mg
E
. Let y ≡ x −
, and α ≡
mg
2m2 g
2
1/3
.
d2 ψ
E
d2 ψ
3
=
α
yψ.
Let
z
≡
αy
=
α(x
−
= zψ. This is the Airy equation (Eq. 8.36), and
),
so
dy 2
mg
dz 2
the general solution is ψ = aAi(z) + bBi(z). However, Bi(z) blows up for large z, so b = 0 (to make ψ
E
normalizable). Hence ψ(x) = aAi α(x − mg
) .
Then
(c) Since V (x) = ∞ for x < 0, we require ψ(0) = 0; hence Ai [α(−E/mg)] = 0. Now, the zeros of Ai are
an (n = 1, 2, 3, . . . ). Abramowitz and Stegun list a1 = −2.338, a2 = −4.088, a3 = −5.521, a4 = −6.787,
2 1/3
αEn
mg
etc. Here −
an , or En = −( 12 mg 2 2 )1/3 an . In this case
= an , or En = −
an = −mg
mg
α
2m2 g
1
1
2 2
2 2
−34
J·s)2 = 5.34 × 10−68 J3 ; ( 12 mg 2 2 )1/3 = 3.77 × 10−23 J.
2 mg = 2 (0.1 kg)(9.8 m/s ) (1.055 × 10
E1 = 8.81 × 10−23 J,
E2 = 1.54 × 10−22 J,
E3 = 2.08 × 10−22 J,
E4 = 2.56 × 10−22 J.
(d)
2T = x
dV
(Eq. 3.97);
dX
here
But T + V = H = En ,
For the electron,
1
mg 2 2
2
dV
= mg,
dx
3
V = En ,
2
so
1/3
=
so
x
dV
= mgx = V ,
dx
or V =
x =
T =
2
En . But V = mgx,
3
1
(9.11 × 10−31 )(9.8)2 (1.055 × 10−34 )2
2
E1 = 1.84 × 10−32 J = 1.15 × 10−13 eV.
so
1/3
1
V .
2
so x =
2En
.
3mg
= 7.87 × 10−33 J.
2(1.84 × 10−32 )
= 1.37 × 10−3 = 1.37 mm.
3(9.11 × 10−31 )(9.8)
Problem 8.6
(a)
x2
Eq. 8.47 =⇒
0
x2
p(x) dx =
0
√
1
p(x) dx = (n − )π, where p(x) = 2m(E − mgx) and E = mgx2 =⇒ x2 = E/mg.
4
x2
√
2
3/2 2m
E − mgx dx = 2m −
(E − mgx)
3mg
0
0
2 2 1
2 2 1
=−
(E − mgx2 )3/2 − E 3/2 =
E 3/2 .
3 mg
3 mg
x2
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CHAPTER 8. THE WKB APPROXIMATION
223
V(x)
E
mgx
x2
1
1
√
(2E)3/2 = (n − )π,
4
3 mg
or
En =
9
8π
2
x
mg 2 2 (n − 14 )2
1/3
.
(b)
9 2 2 2
π mg 8
1/3
=
9 2
π (0.1)(9.8)2 (1.055 × 10−34 )2
8
1/3
= 1.0588 × 10−22 J.
2/3
3
= 8.74 × 10−23 J,
4
2/3
7
−22
E2 = (1.0588 × 10 )
= 1.54 × 10−22 J,
4
2/3
11
−22
E3 = (1.0588 × 10 )
= 2.08 × 10−22 J,
4
2/3
15
E4 = (1.0588 × 10−22 )
= 2.56 × 10−22 J.
4
E1 = (1.0588 × 10−22 )
These are in very close agreement with the exact results (Problem 8.5(c)). In fact, they agree precisely
(to 3 significant digits), except for E1 (for which the exact result was 8.81 × 10−23 J).
(c) From Problem 8.5(d),
x =
n=
2En
,
3mg
so
1=
2 (1.0588 × 10−22 )
3
(0.1)(9.8)
n−
1
4
2/3
, or
n−
1
4
2/3
= 1.388 × 1022 .
1
+ (1.388 × 1022 )3/2 = 1.64 × 1033 .
4
Problem 8.7
x2
p(x) dx =
x1
1
n−
2
1
n−
2
x2
π = mω
−x2
π;
p(x) =
1
2m E − mω 2 x2 ;
2
2E
− x2 dx = 2mω
mω 2
0
x2
x2 = −x1 =
1
ω
2E
.
m
x22 − x2 dx = mω x x22 − x2 + x22 sin−1 (x/x2 )
x2
0
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224
CHAPTER 8. THE WKB APPROXIMATION
π
π
πE
2E
=
mωx22 = mω
.
2
2
mω 2
ω
= mωx22 sin−1 (1) =
En = n −
1
2
ω (n = 1, 2, 3, . . . )
Since the WKB numbering starts with n = 1, whereas for oscillator states we traditionally start with n = 0,
letting n → n + 1 converts this to the usual formula En = (n + 12 )ω. In this case the WKB approximation
yields the exact results.
Problem 8.8
(a)
1
mω 2 x22 = En =
2
n+
1
2
ω
(counting n = 0, 1, 2, . . . );
x2 =
(2n + 1)
.
mω
(b)
Vlin (x) =
1
1
mω 2 x22 + (mω 2 x2 )(x − x2 ) =⇒ Vlin (x2 + d) = mω 2 x22 + mω 2 x2 d.
2
2
V (x2 + d) − Vlin (x2 + d)
=
V (x2 )
1
2
2 mω (x2
x2 + 2x2 d + d2 − x22 − 2x2 d
= 2
=
x22
+ d)2 − 12 mω 2 x22 − mω 2 x2 d
1
2 2
2 mω x2
d
x2
2
= 0.01.
d = 0.1 x2 .
(c)
α=
2m
mω 2 x2
2
1/3
(Eq. 8.34),
2m2 ω 2 (2n + 1)2 2
≥ (50)3 ;
2
m2 ω 2
nmin = 125.
so
0.1 x2
or (2n + 1)2 ≥
2m2 ω 2
x2
2
1/3
≥ 5 =⇒
(50)3
= 62500;
2
2m2 ω 2 4
x2
2
2n + 1 ≥ 250;
1/3
n≥
≥ 50.
249
= 124.5.
2
However, as we saw in Problems 8.6 and 8.7, WKB may be valid at much smaller n.
Problem 8.9
Shift origin to the turning point.
0
1
1
(x < 0)
De− x |p(x )| dx
|p(x)|
ix ψWKB =
x
i
1
Be 0 p(x ) dx + Ce− 0 p(x ) dx (x > 0)
|p(x)|
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CHAPTER 8. THE WKB APPROXIMATION
225
overlap 1
overlap 2
ψP
ψ
ψ
WKB
WKB
E
Nonclassical
Classical
x
0
patching region
Linearized potential in the patching region:
d2 ψp
2mV (0)
=
xψp = −α3 xψp , whereα ≡
dx2
2
V (x) ≈ E + V (0)x. Note : V (0) is negative.
2m|V (0)|
2
1/3
.
ψp (x) = aAi(−αx) + bBi(−αx). (Note change of sign, as compared with Eq. 8.37).
p(x) =
2m[E − E − V (0)x] =
−2mV (0)x =
2m|V (0)|x =
√
√
α3 2 x = α3/2 x.
Overlap region 1 (x < 0):
0
|p(x )| dx = α
0
3/2
x
√
−x dx = α
3/2
x
ψWKB ≈
0
2
2 3/2
2
3/2 3/2
− (−x )
= (−αx)3/2 .
= 3 α (−x)
3
3
x
3/2
2
1
De− 3 (−αx) . For large positive argument (−αx 1) :
1/2 α3/2 (−x)1/4
3/2
3/2
2
2
1
1
ψp ≈ a √
e− 3 (−αx) + b √
e 3 (−αx) . Comparing ⇒ a = 2D
2 π(−αx)1/4
π(−αx)1/4
π
;
α
b = 0.
Overlap region 2 (x > 0):
x
x
|p(x )| dx = α3/2
0
x
2
2
(x )3/2 = (αx)3/2 .
3
3
0
x dx = α3/2
0
ψWKB ≈
ψp (x) ≈ a √
√
1
1/2 α3/4 x1/4
2
Bei 3 (αx)
3/2
+ Ce−i 3 (αx)
2
3/2
. For large negative argument (−αx −1) :
1 iπ/4 i 2 (αx)3/2
1
2
π
a
3/2
−iπ/4 −i 23 (αx)3/2
3
√
(remember : b = 0).
sin
+
e
−
e
e
(αx)
=
e
3
4
π(αx)1/4
π(αx)1/4 2i
Comparing the two: B =
a
2i
a
α iπ/4
, C=−
e
π
2i
α −iπ/4
.
e
π
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226
CHAPTER 8. THE WKB APPROXIMATION
Inserting the expression for a from overlap region 1 : B = −ieiπ/4 D; C = ie−iπ/4 D. For x > 0, then,
x
i
π
2D
−iD i x p(x ) dx +i π
1
4 − e− 0 p(x ) dx −i 4
ψWKB = =
e 0
sin
p(x)
p(x)
Finally, switching the origin back to x1 :
x1
1
D
−
|p(x )| dx
x
e
,
(x
<
x
)
;
1
|p(x)|
ψWKB (x) =
x
2D
1
π
sin
p(x )dx +
, (x > x1 ).
x1
4
p(x)
x
p(x ) dx +
0
π
.
4
QED
Problem 8.10
At x1 , we have an upward-sloping turning point. Follow the method in the book. Shifting origin to x1 :
1 i 0 p(x ) dx
− i x0 p(x ) dx
x
(x < 0)
+
B
Ae
p(x)
ψWKB (x) =
x
1
1 1 x |p(x )| dx
Ce 0
+ D− 0 |p(x )| dx (x > 0)
p(x)
In overlap region 2, Eq. 8.39 becomes ψWKB ≈
1
1/2 α3/4 x1/4
whereas Eq. 8.40 is unchanged. Comparing them =⇒ a = 2D
In overlap region 1, Eq. 8.43 becomes ψWKB ≈
2
Ce 3 (αx)
π
,
α
3/2
+ De− 3 (αx)
b=C
2
3/2
,
π
.
α
1
i 23 (−αx)3/2
−i 23 (−αx)3/2
Ae
,
+
Be
1/2 α3/4 (−x)1/4
and Eq. 8.44 (with b = 0) generalizes to
ψp (x) ≈ √
a
2
π
2
π
b
sin (−αx)3/2 +
cos (−αx)3/2 +
+√
1/4
1/4
3
4
3
4
π(−αx)
π(−αx)
1
i 23 (−αx)3/2 iπ/4
−i 23 (−αx)3/2 −iπ/4
= √
. Comparing them =⇒
e
+
(ia
+
b)e
e
(−ia
+
b)e
2 π(−αx)1/4 )
α
π
A=
A=
−ia + b
2
C
− iD eiπ/4 ;
2
iπ/4
e
;
B=
B=
α
π
ia + b
2
e−iπ/4 . Putting in the expressions above for a and b :
C
+ iD e−iπ/4 .
2
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CHAPTER 8. THE WKB APPROXIMATION
227
These are the connection formulas relating A, B, C, and D, at x1 .
At x2 , we have a downward-sloping turning point, and follow the method of Problem 8.9. First rewrite the
middle expression in Eq. 8.52:
1 x2
x
x
1
1
1
|p(x )| dx + |p(x )| dx
− 1 x2 |p(x )| dx − |p(x )| dx
x2
x2
ψWKB = .
+ De x1
Ce x1
|p(x)|
x
Let γ ≡ x12 |p(x)| dx, as before (Eq. 8.22), and let C ≡ De−γ , D ≡ Ceγ . Then (shifting the origin to x2 ):
0
0
1
1
1
x |p(x )| dx + D e− x |p(x )| dx
, (x < 0);
C
e
|p(x)|
ψWKB =
x
i
1
(x > 0).
F e 0 p(x ) dx ,
p(x)
1/3
√
2m|V (0)|
In the patching region ψp (x) = aAi(−αx) + bBi(−αx), where α ≡
; p(x) = α3/2 x.
2
0
2
In overlap region 1 (x < 0):
|p(x )| dx = (−αx)3/2 , so
3
x
3/2
2
1
π
− 23 (−αx)3/2
3 (−αx)
e
+
D
e
C
a = 2
D
1/2 α3/4 (−x)1/4
α
Comparing
=⇒
3/2
3/2
2
2
a
b
π
ψp ≈ √
e− 3 (−αx) + √
e 3 (−αx)
b =
C
2 π(−αx)1/4
π(−αx)1/4
α
ψWKB ≈
x
In overlap region 2 (x > 0):
p(x ) dx =
0
ψp ≈ √
3/2
2
2
1
(αx)3/2 =⇒ ψWKB ≈ 1/2 3/4 1/4 F ei 3 (αx) .
3
α x
a
2
π
2
π
b
sin (αx)3/2 +
cos (αx)3/2 +
+√
1/4
1/4
3
4
3
4
π(αx)
π(αx)
3/2
3/2
2
2
1
iπ
−i π
4 ei 3 (αx)
4 e−i 3 (αx)
(−ia
+
b)e
. Comparing =⇒ (ia + b) = 0;
= √
+
(ia
+
b)e
2 π(αx)1/4
F =
α
π
C =
α
b = e−iπ/4 F,
π
−ia + b
2
α iπ/4
.
e
π
eiπ/4 = b
D =
1
2
b=
π −iπ/4
F;
e
α
a=i
α
i
a = e−iπ/4 F. D = eγ e−iπ/4 F ;
π
2
π −iπ/4
F.
e
α
C=
i −γ −iπ/4
e e
F.
2
These are the connection formulas at x2 . Putting them into the equation for A:
−γ
C
i −γ −iπ/4
e
A=
F − ieγ e−iπ/4 F eiπ/4 = i
− iD eiπ/4 =
e e
− eγ F.
2
4
4
2
F e−2γ
1
T = =
=
2 .
−γ
A
(eγ − e 4 )2
1 − (e−γ/2 )2
If γ 1, the denominator is essentially 1, and we recover T = e−2γ
(Eq. 8.22).
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228
CHAPTER 8. THE WKB APPROXIMATION
Problem 8.11
x2 x2
√
1
ν
π = 2
Equation 8.51 ⇒ n −
2m(E − αx ) dx = 2 2mE
2
0
0
1/ν
1/ν
1 1 −1
α ν
zE
E
z ≡ x , so x =
; dx =
z ν dz. Then
E
α
α
ν
n−
1
2
√
π = 2 2mE
√
= 2 2mE
E
1
1
ν+2
E
α
E
α
1/ν
1/ν
1
ν
1
√
√
1
z ν −1 1 − z dz = 2 2mE
0
1−
E
α
1/ν
α ν
x dx;
E
E = αxν2 . Let
1 Γ(1/ν)Γ(3/2)
ν Γ( ν1 + 32 )
√
1/ν
+ 1) 12 π √
Γ( ν1 + 1)
E
=
.
2πmE
1
3
α
Γ( ν + 2 )
Γ( ν1 + 32 )
Γ( ν1
(n − 1 )π 1/ν Γ( ν1 + 32 )
= √ 2
;
α
Γ( ν1 + 1)
2πm
1
En =
n−
2
π Γ( ν1 + 32 )
2mα Γ( ν1 + 1)
2ν
)
( ν+2
α.
π Γ(2)
2α
1
1
n−
α = (n − )
.
2
2mα Γ(3/2)
2
m
For a harmonic oscillator, with α = 12 mω 2 , En = n − 12 ω (n = 1, 2, 3, . . . ). For ν = 2: En =
Problem 8.12
2 a2
V (x) = −
sech2 (ax). Eq. 8.51 =⇒
m
√
= 2 2a
x2
sech2 (ax) +
0
E=−
2 a2
sech2 (ax2 )
m
1
dx =
a
−1
√ √
z 1−z
Limits :
1
z
1
n−
2
2m E +
π = 2
0
2 a2
sech2 (ax) dx
m
mE
dx.
2 a2
defines x2 . Let b ≡ −
mE
,
2 a2
1
1
1
√ dz = −
√
dz.
2a z 1 − z
2 z
x = 0 =⇒ z = sech2 (0) = 1
mE
x = x2 =⇒ z = sech (ax2 ) = − 2 2 = b
a
2
x2
z ≡ sech2 (ax),
Then
1
n−
2
.
n−
1
2
so that x =
√
1
sech−1 z, and hence
a
z2 √
√
z−b
1
√
π = 2 2a −
dz.
2a
z1 z 1 − z
π=
√ 2
b
1
1
z
z−b
dz.
1−z
1
(z − b)
1
b
z−b
= =
− .
1−z
z (1 − z)(z − b)
(1 − z)(z − b) z (1 − z)(z − b)
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CHAPTER 8. THE WKB APPROXIMATION
1
n−
2
π =
√
1
229
1
dz − b
(1 − z)(z − b)
2
1
1
dz
−b + (1 + b)z − z 2
1
√
1−z √
−1 (1 + b)z − 2b
−1
= 2 −2 tan
− b sin
z−b
z(1 − b)
b
√ √
√
√ π √ π √ π
−1
−1
−1
= 2 −2 tan (0) + 2 tan (∞) − b sin (1) + b sin−1 (−1) = 2 0 + 2 − b − b
2
2
2
√
√
√
√
(n − 12 )
1
1
√
= 2π(1 − b);
b=1− √
= 1 − b;
n−
.
2
2
2
b
b
z
√
√
Since the left side is positive, the right side must also be: (n − 12 ) < 2, n < 12 + 2 = 0.5 + 1.414 = 1.914.
So the only possible n is 1; there is only one bound state (which is correct—see Problem 2.51).
For n = 1,
√
9
1
1
1
b=1− √ + = − √ ;
8
2 8
2
1
b=1− √ ;
2 2
2 a2
E1 = −
m
9
1
−√
8
2
= −0.418
2 a2
.
m
2 a2
. Not bad.
m
The exact answer (Problem 2.51(c)) is −0.5
Problem 8.13
n−
1
4
r0
2m [E − V0 ln(r/a)] dr;
E = V0 ln(r0 /a) defines r0 .
r0 r0 √
= 2m
V0 ln(r0 /a) − V0 ln(r/a) dr = 2mV0
ln(r0 /r) dr.
π =
0
0
Let x ≡ ln(r0 /r),
1
n−
4
1
n−
4
r0 =
π =
0
or r = r0 e−x =⇒ dr = −r0 e−x dx.
so ex = r0 /r,
x2
2mV0 (−r0 )
√
−e
xe
dx. Limits :
x1
π =
∞
2mV0 r0
√
−x
xe
dx =
r = 0 =⇒ x1 = ∞
r = r0 =⇒ x2 = 0
2mV0 r0 Γ(3/2) =
0
2π
1
n−
mV0
4
⇒
En = V0 ln
a
3
En+1 − En = V0 ln n +
4
1
− V0 ln n −
4
2π
mV0
n−
= V0 ln
1
4
n + 3/4
n − 1/4
.
√
2mV0 r0
π
.
2
1
+ V0 ln
= V0 ln n −
4
a
2π
mV0
.
, which is indeed independent of m (and a).
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230
CHAPTER 8. THE WKB APPROXIMATION
Problem 8.14
1
n−
2
1
n−
2
π =
r1
where A ≡ −
r2
e2 1
4π90 E
π =
√
r2
√
2 l(l + 1)
e2 1
dr
=
−
2m E +
−2mE
4π90 r
2m r2
r1
B≡−
and
r2
−2mE
√
r1
2 l(l + 1)
2m E
1
n−
2
π =
√
r2
−2mE
r1
B
A
− 2 dr,
r
r
are positive constants, since E is negative.
−r2 + Ar − B
dr.
r
Let r1 and r2 be the roots of the polynomial in the numerator:
−1 +
− r2 + Ar − B = (r − r1 )(r2 − r).
√
(r − r1 )(r2 − r)
√ 2
π √
dr = −2mE ( r2 − r1 ) .
r
2
√
√
1
2 n−
= −2mE (r2 + r1 − 2 r1 r2 ) . But − r2 + Ar − B = −r2 + (r1 + r2 )r − r1 r2
2
=⇒ r1 + r2 = A; r1 r2 = B. Therefore
1
2 n−
2
=
=
e2
4π90
−
√
√ √
−2mE A − 2 B = −2mE
e2
4π90
−
'
e2 1
−
−2
4π90 E
2 l(l + 1)
−
2m E
(
2m
− 2 l(l + 1).
E
2m
1 = 2 n − + l(l + 1) ;
E
2
−
E
=
2m
42
(e2 /4π90 )2
2 .
n − 12 + l(l + 1)
−(m/22 )(e2 /4π90 )2
−13.6 eV
E=
2 = 2 .
1
1
n − 2 + l(l + 1)
n − 2 + l(l + 1)
Problem 8.15
(a)
D
(i) ψWKB (x) = |p(x)|
(ii) ψWKB (x) = 1
p(x)
1
−
e
i
Be x
x2
|p(x )| dx
x2
x
p(x ) dx
(x > x2 );
+ Ce− i
x2
x
p(x ) dx
(x1 < x < x2 );
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CHAPTER 8. THE WKB APPROXIMATION
(iii) ψWKB (x) = 1
|p(x)|
Equation 8.46 =⇒ (ii) ψWKB
1
Fe
x1
x
231
|p(x )| dx
1
2D
sin
=
p(x)
+ Ge− 1
x1
x
|p(x )| dx
x2
p(x )dx +
x
π
4
(0 < x < x1 ).
(x1 < x < x2 ).
To effect the join at x1 , first rewrite (ii):
2D
1 x
π
π
1 x2
1 x
2D
(ii) ψWKB = p(x ) dx −
p(x ) dx +
p(x ) dx − θ −
sin
sin
= −
,
4
4
p(x)
p(x)
x1
x1
x1
where θ is defined in Eq. 8.58. Now shift the origin to x1 :
10
0
1
1
x |p(x )| dx + Ge− x |p(x )| dx
(x
<
0)
F
e
|p(x)|
ψWKB =
.
x
2D
π
1
p(x ) dx − θ −
(x > 0)
sin
−
0
4
p(x)
Following Problem 8.9: ψp (x) = aAi(−αx) + bBi(−αx), with α ≡
0
|p(x )| dx =
Overlap region 1 (x < 0):
x
2m|V (0)|
2
2
3/2
1
− 23 (−αx)3/2
3 (−αx)
F
e
+
Ge
1/2 α3/4 (−x)1/4
=⇒ a = 2G
3/2
2
a
b
− 23 (−αx)3/2
3 (−αx)
√
≈ √
e
+
e
2 π(−αx)1/4
π(−αx)1/4
x
Overlap region 2 (x > 0):
p(x ) dx =
0
=⇒ ψWKB ≈ −
ψp ≈ √
2D
1/2 α3/4 x1/4
sin
1
πα1/4
−2D
=⇒
π
;
α
b=F
π
.
α
2
(αx)3/2 .
3
2
π
(αx)3/2 − θ −
,
3
4
a
2
π
2
π
b
sin (αx)3/2 +
cos (αx)3/2 +
+√
.
1/4
1/4
3
4
3
4
π(αx)
π(αx)
Equating the two expressions:
=√
√
; p(x) = α3/2 x.
2
(−αx)3/2 .
3
ψWKB ≈
ψp
1/3
2D
−2D 1 i 2 (αx)3/2 −iθ −iπ/4
−i 23 (αx)3/2 iθ iπ/4
3
e
e
−
e
e
e
e
1/2 α3/4 2i
b 2
3/2
3/2
3/2
2
2
a i 2 (αx)3/2 iπ/4
e
− e−i 3 (αx) e−iπ/4 +
e 3
ei 3 (αx) eiπ/4 + e−i 3 (αx) e−iπ/4
2i
2
π −iθ −iπ/4
e e
= (a + ib)eiπ/4 ,
α
π iθ iπ/4
= (−a + ib)e−iπ/4 ,
e e
α
or
or
π −iθ
ie
(a + ib) = 2D
α
π iθ
(a − ib) = −2D
ie
α
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232
CHAPTER 8. THE WKB APPROXIMATION
2a = 2D
=⇒
2ib = 2D
π
i(e−iθ − eiθ ) ⇒ a = 2D
α
π
i(e−iθ + eiθ ) ⇒ b = 2D
α
π
sin θ,
α
π
cos θ.
α
Combining these with the results from overlap region 1 =⇒
π
= 2D
α
2G
π
sin θ,
α
or
G = D sin θ;
D
Putting these into (iii) : ψWKB (x) = |p(x)|
F
1
2 cos θe π
= 2D
α
π
cos θ,
α
|p(x )| dx
+ sin θe− x1
x
1
or
x1
x
F = 2D cos θ.
|p(x )| dx
(0 < x < x1 ).
(b)
1
Odd(−) case: (iii) =⇒ ψ(0) = 0 ⇒ 2 cos θe 1
x1
|p(x )| dx =
0
x
1
0
|p(x )| dx
+ sin θe− 1
x
1
0
|p(x )| dx
= 0.
1
φ, with φ defined by Eq. 8.60. So sin θe−φ/2 = −2 cos θeφ/2 , or tan θ = −2eφ .
2
d|p(x)| 1
D
φ/2
−φ/2
Even(+) case: (iii) =⇒ ψ (0) = 0 ⇒ −
2
cos
θe
+
sin
θe
2 (|p(x)|)3/2 dx 0
+
Now
D
|p(x)|
2 cos θe
1
x
1
0
|p(x )| dx
x
1
1
1
1 |p(x )| dx
−
0
− |p(0)| + sin θe
|p(0)|
= 0.
√ 1
d|p(x)|
d 1
dV dV
d|p(x)| =
, and
2m[V (x) − E] = 2m √
= 0, so
= 0.
dx
dx
2 V − E dx
dx 0
dx 0
2 cos θeφ/2 = sin θe−φ/2 , or tan θ = 2eφ . Combining the two results: tan θ = ±2eφ . QED
(c)
tan θ = tan
sin n + 12 π + 9
1
(−1)n cos 9
cos 9
1
=
n+
=−
≈− .
π+9 =
1
2
(−1)n+1 sin 9
sin 9
9
cos n + 2 π + 9
1
1 −φ
1
1
φ
So − ≈ ±2e , or 9 ≈ ∓ e , or θ − n +
π ≈ ∓ e−φ ,
9
2
2
2
so θ ≈
1
n+
2
1
π ∓ e−φ . QED
2
[Note: Since θ (Eq. 8.58) is positive, n must be a non-negative integer: n = 0, 1, 2, . . . . This is like
harmonic oscillator (conventional) numbering, since it starts with n = 0.]
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publisher.
CHAPTER 8. THE WKB APPROXIMATION
233
V(x)
-a
x1
a
x2
x
(d)
1
θ =
2
=
x2
x1
z2
1
2m E − mω 2 (x − a)2 dx. Let z = x − a (shifts the origin to a).
2
1
1
2m E − mω 2 z 2 dz, where E = mω 2 z22 .
2
2
0
z2 2
mω
= mω
z z22 − z 2 + z22 sin−1 (z/z2 )
z22 − z 2 dz =
0
=
z2
= mω z22 sin−1 (1) = π mω z22 ,
2 0
π mω 2E
πE
=
.
2 mω 2
ω
πE
Putting this into Eq. 8.61 yields
≈
ω
1
n+
2
1
π ∓ e−φ ,
2
or
En±
≈
1
n+
2
ω ∓
ω −φ
e . QED
2π
(e)
+
−
1 Ψ(x, t) = √ ψn+ e−iEn t/ + ψn− e−iEn t/ =⇒
2
|Ψ(x, t)|2 =
−
+
−
+
1 +2
|ψn | + |ψn− |2 + ψn+ ψn− ei(En −En )t/ + e−i(En −En )t/ .
2
En− − En+
ω
1 ω
≈ 2 e−φ = e−φ , so
2π
π
1 + 2
ω −φ
|Ψ(x, t)|2 =
ψn (x) + ψn− (x)2 + ψn+ (x)ψn− (x) cos
e t .
2
π
(Note that the wave functions (i), (ii), (iii) are real ). But
It oscillates back and forth, with period
τ=
2π
2π 2 φ
e . QED
=
(ω/π) e−φ
ω
(f )
1
φ=2
x1
2m
0
x1
1
2√
2mE
mω 2 (x − a)2 − E dx =
2
0
mω 2
(x − a)2 − 1 dx.
2E
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234
CHAPTER 8. THE WKB APPROXIMATION
Let z ≡
m
ω(a − x),
2E
dx = −
so
m
ωa ≡ z0
x = 0 =⇒ z =
Limits:
.
2E
x = x1 =⇒ radicand = 0 =⇒ z = 1
2E 1
dz.
m ω
z0
2√
2E 1 z0 2
4E z0 2
4E 1 2
2mE
z − 1 dz =
z − 1 dz =
z z − 1 − ln(z + z 2 − 1) m ω 1
ω 1
ω 2
1
2E
z0 z02 − 1 − ln z0 + z02 − 1 ,
=
ω
φ=
where z0 = aω
m
.
2E
V (0) = 12 mω 2 a2 , so
V (0) E ⇒
m 2 2
ω a E ⇒ aω
2
m
1,
2E
or z0 1.
In that case
φ≈
2E 2
2E 2
2E 2 2 m
mωa2
z0 − ln(2z0 ) ≈
z0 =
a ω
=
.
ω
ω
ω
2E
This, together with Eq. 8.64, gives us the period of oscillation in a double well.
Problem 8.16
(a) En ≈
n2 π 2 2
.
2m(2a)2
With n = 1,
E1 =
π 2 2
.
8ma2
(b)
V(x)
-a
a
x
V0
E1
V(x)
x0
a
-a
E
x
tunneling
E1
(c)
1 x0
V 0 − E1
.
|p(x)| dx. αx0 = V0 − E1 ⇒ x0 =
a
α
p(x) = 2m [E − V (x)]; V (x) = −αx, E = E1 − V0 .
√
√
√
√
= 2m(E1 − V0 + αx) = 2mα x − x0 ; |p(x)| = 2mα x0 − x.
√
√
x0
x0
√
1√
2
2mα
2 2mα
3/2 3/2
γ =
− (x0 − x)
2mα
x0 − x dx =
= 3 (x0 − a) .
3
a
a
γ =
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CHAPTER 8. THE WKB APPROXIMATION
235
Now x0 − a = (V0 − E1 − aα)/α, and αa 2 /ma2 ≈ E1 V0 , so we can drop E1 and αa. Then
√
3/2
8mV03
2 2mα V0
γ≈
.
=
3 α
3α
Equation 8.28 ⇒ τ =
4a 2γ
e ,
v
where
τ=
1
π 2 2
π 2 2
⇒ v2 =
,
mv 2 ≈
2
2
8ma
4m2 a2
or v =
π
.
2ma
So
4a
8ma2 2γ
2ma e2γ =
e .
π
π
(d)
2
(8) 9.1 × 10−31 10−10
e2γ = 2 × 10−19 e2γ ;
π (1.05 × 10−34 )
3
(8) (9.1 × 10−31 ) (20 × 1.6 × 10−19 )
γ =
= 4.4 × 104 ;
(3) (1.6 × 10−19 ) (7 × 106 ) (1.05 × 10−34 )
τ =
4
e2γ = e8.8×10 = 10log e
8.8×104
= 1038,000 .
τ = 2 × 10−19 × 1038,000 s = 1038,000 yr.
Seconds, years . . . it hardly matters; nor is the factor out front significant. This is a huge number—the
age of the universe is about 1010 years. In any event, this is clearly not something to worry about.
Problem 8.17
Equation 8.22 ⇒ the tunneling probability: T = e−2γ , where
1 x0 γ =
2m(V − E) dx. Here V (x) = mgx, E = 0, x0 = R2 + (h/2)2 − h/2 (half the diagonal).
0
√
x
x0
2m √
m 2 3/2 0
2m 3/2
=
mg
x1/2 dx =
2g x =
2g x0 .
3
3
0
0
√
I estimate: h = 10 cm, R = 3 cm, m = 300 gm; let g = 9.8 m/s2 . Then x0 = 9 + 25 − 5 = 0.83 cm, and
(2)(0.3)
γ=
(2)(9.8) (0.0083)3/2 = 6.4 × 1030 .
−34
(3)(1.05 × 10 )
Frequency of “attempts”: say f = v/2R. We want the product of the number of attempts (f t) and the
probability of toppling at each attempt (T ), to be 1:
v −2γ
2R 2γ
=1 ⇒ t=
e
e .
2R
v
Estimating the thermal velocity: 12 mv 2 = 12 kB T (I’m done with the tunneling probability; from now on T
is the temperature, 300 K) ⇒ v = kB T /m.
30
30
m 2γ
0.3
13×1030
= (5 × 108 ) × 105.6×10 s
t = 2R
e = 2(0.03)
e12.8×10 = 5 × 108 10log e
−23
kB T
(1.4 × 10 )(300)
t
= 16 × 105.6×10
30
yr.
Don’t hold your breath.
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236
CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
Chapter 9
Time-Dependent Perturbation Theory
Problem 9.1
ψnlm = Rnl Ylm . From Tables 4.3 and 4.7:
ψ100 = √
ψ210 = √
1
πa3
e−r/a ;
1
32πa3
ψ200 = √
r −r/2a
cos θ;
e
a
1
8πa3
1−
r −r/2a
;
e
2a
ψ21±1 = ∓ √
1
64πa3
r r/2a
sin θ e±iφ .
e
a
But r cos θ = z and r sin θe±iφ = r sin θ(cos φ ± i sin φ) = r sin θ cos φ ± ir sin θ sin φ = x ± iy. So |ψ|2 is an
even function of z in all cases, and hence z|ψ|2 dx dy dz = 0, so Hii = 0. Moreover, ψ100 is even in z, and
so are ψ200 , ψ211 , and ψ21−1 , so Hij = 0 for all except
H100,210 = −eE √
eE
=− √
4 2πa4
or
∞
1
πa3
√
1
1
3
32πa a
r4 e−3r/2a dr
0
eE
e−r/a e−r/2a z 2 d3 r = − √
4 2πa4
π
cos2 θ sin θ dθ
0
0
2π
e−3r/2a r2 cos2 θ r2 sin θ dr dθ dφ
eE
dφ = − √
4!
4 2πa4
2a
3
5
2
2π = −
3
28
√
35 2
eEa,
−0.7449 eEa.
Problem 9.2
i
ċa = − Hab e−iω0 t cb ;
i
ċb = − Hba eiω0 t ca . Differentiating with respect to t :
i
i
i
i
c̈b = − Hba iω0 eiω0 t ca + eiω0 t ċa = iω0 − Hba eiω0 t ca − Hba eiωo t − Hab e−iω0 t cb , or
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CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
c̈b = iω0 ċb −
237
1
1
|H |2 cb . Let α2 ≡ 2 |Hab |2 . Then c̈b − iω0 ċb + α2 cb = 0.
2 ab
This is a linear differential equation with constant coefficients, so it can be solved by a function of the form
cb = eλt :
i
1
λ2 − iω0 λ + α2 = 0 =⇒ λ =
iω0 ± −ω02 − 4α2 = (ω0 ± ω) , where ω ≡ ω02 + 4α2 .
2
2
The general solution is therefore
cb (t) = Aei(ω0 +ω)/2 + Bei(ω0 −ω)/2 = eiω0 t/2 Aeiωt/2 + Be−iωt/2 , or
cb (t) = eiω0 t/2 [C cos (ωt/2) + D sin (ωt/2)] . But cb (0) = 0,
so C = 0,
and hence
cb (t) = Deiω0 t/2 sin (ωt/2) . Then
ċb = D
ca =
ω0
i
iω0 iω0 t/2
ω
ω
e
sin (ωt/2) + eiω0 t/2 cos (ωt/2) = Deiω0 t/2 cos (ωt/2) + i sin (ωt/2) = − Hba eiω0 t ca .
2
2
2
ω
i ω −iω0 t/2 ω0
D cos (ωt/2) + i sin (ωt/2) . But ca (0) = 1,
e
Hba 2
ω
ω0
ca (t) = e−iω0 t/2 cos (ωt/2) + i sin (ωt/2) ,
ω
2Hba iω0 t/2
cb (t) =
e
sin (ωt/2) ,
iω
where ω ≡
ω02 + 4
so
i ω
D = 1. Conclusion:
Hba 2
|2
|Hab
2 .
ω02
4|Hab |2
2
sin
(ωt/2)
+
sin2 (ωt/2)
2 ω2
ω2 1
|H |2
= cos2 (ωt/2) + 2 ω02 + 4 ab
sin2 (ωt/2) = cos2 (ωt/2) + sin2 (ωt/2) = 1. ω
2
|ca |2 + |cb |2 = cos2 (ωt/2) +
Problem 9.3
This is a tricky problem, and I thank Prof. Onuttom Narayan for showing me the correct solution. The safest
approach is to represent the delta function as a sequence of rectangles:
(1/29), −9 < t < 9,
δ5 (t) =
0,
otherwise.
Then Eq. 9.13 ⇒
t < −9 :
t > 9 :
ca (t) = 1, cb (t) = 0,
ca (t) = a, cb (t) = b,
iα −iω0 t
e
cb ,
ċa = − 25
−9
<
t
<
9
:
∗
iω0 t
ca .
ċb = − iα
25 e
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238
CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
In the interval −9 < t < 9,
d2 cb
iα∗
iω0 eiω0 t ca + eiω0 t
=
−
dt2
29
−iα −iω0 t
e
cb
29
=−
iα
|α|2
iα∗
i29 dcb
dcb
iω0 ∗
−
cb = iω0
−
cb .
29
α dt
29
dt
(29)2
Thus cb satisfies a homogeneous linear differential equation with constant coefficients:
d2 cb
|α|2
dcb
+
−
iω
cb = 0.
0
dt2
dt
(29)2
Try a solution of the form cb (t) = eλt :
iω0 ±
|α|2
λ − iω0 λ +
=0⇒λ=
2
(29)
2
or
λ=
The general solution is
iω0
iω
± , where ω ≡
2
2
−ω02 − |α|2 /(9)2
,
2
ω02 + |α|2 /(9)2 .
cb (t) = eiω0 t/2 Aeiωt/2 + Be−iωt/2 .
But
cb (−9) = 0 ⇒ Ae−iω5/2 + Beiω5/2 = 0 ⇒ B = −Ae−iω5 ,
so
cb (t) = Aeiω0 t/2 eiωt/2 − e−iω(5+t/2) .
Meanwhile
iω 2i9 −iω0 t
2i9
iω0 iωt/2
e
ċb = ∗ e−iω0 t/2 A
− e−iω(5+t/2) +
e
eiωt/2 + e−iω(5+t/2)
∗
α
α
2
2
9 −iω0 t/2 = − ∗e
A (ω + ω0 )eiωt/2 + (ω − ω0 )e−iω(5+t/2) .
α
ca (t) =
But
Thus
9 i(ω0 −ω)5/2
29ω i(ω0 −ω)5/2
α∗ i(ω−ω0 )5/2
e
A
[(ω
+
ω
)
+
(ω
−
ω
)]
=
−
e
A,
so
A
=
−
.
e
0
0
α∗
α∗
29ω
1 −iω0 (t+5)/2 ca (t) =
(ω + ω0 )eiω(t+5)/2 + (ω − ω0 )e−iω(t+5)/2
e
2ω
ω(t + 9)
ω0
ω(t + 9)
−iω0 (t+5)/2
=e
cos
+ i sin
;
2
ω
2
iα∗ iω0 (t−5)/2 iω(t+5)/2
ω(t + 9)
iα∗ iω0 (t−5)/2
− e−iω(t+5)/2 = −
sin
cb (t) = −
e
e
e
.
29ω
9ω
2
ca (−9) = 1 = −
ω0
a = ca (9) = e−iω0 5 cos(ω9) + i sin(ω9) ,
ω
b = cb (9) = −
iα∗
sin(ω9).
9ω
This is for the rectangular pulse; it remains to take the limit 9 → 0: ω → |α|/9,
a → cos
|α|
ω0 9
+i
sin
|α|
|α|
→ cos
|α|
,
so
iα∗
b→−
sin
|α|
|α|
,
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CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
239
and we conclude that for the delta function
ca (t) =
1,
t < 0,
cos(|α|/), t > 0;
0,
cb (t) =
−i
t < 0,
α∗
sin(|α|/), t > 0.
α
Obviously, |ca (t)|2 + |cb (t)|2 = 1 in both time periods. Finally,
Pa→b = |b|2 = sin2 (|α|/).
Problem 9.4
(a)
i
−iω0 t
Eq. 9.10 =⇒ ċa = − ca Haa + cb Hab e
(these are exact, and replace Eq. 9.13).
i
Eq. 9.11 =⇒ ċb = − cb Hbb + ca Hba eiω0 t
Initial conditions:
Zeroth order:
ca (0) = 1,
ca (t) = 1,
cb (0) = 0.
cb (t) = 0.
i t
c
(t)
=
1
−
H (t ) dt
ċa = − i Haa
a
=⇒
0 aa
First order:
t
i
iω0 t
=⇒ cb (t) = − i
Hba (t )eiω0 t dt
ċb = − Hba e
0
2
i t
1 t
i t
1+
Haa (t ) dt
Haa (t ) dt = 1 +
Haa (t ) dt
= 1 (to first order in H ).
0
0
0
i t
i t
|cb |2 = −
Hba (t )eiω0 t dt
H (t )e−iω0 t dt = 0 (to first order in H ).
0
0 ab
|ca |2 = 1 −
So |ca |2 + |cb |2 = 1 (to first order).
(b)
i
d˙a = e t
0
Haa
(t ) dt
t
i
i
i
Haa ca + e 0 Haa (t ) dt ċa . But ċa = − ca Haa + cb Hab e−iω0 t
Two terms cancel, leaving
t
i
i i t d˙a = − e 0 Haa (t ) dt cb Hab e−iω0 t . But cb = e− 0 Hbb (t ) dt db .
i it i
= − e 0 [Haa (t )−Hbb (t )]dt Hab e−iω0 t db , or d˙a = − eiφ Hab e−iω0 t db .
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240
CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
Similarly,
i
d˙b = e t
0
Hbb
(t ) dt
t
i
i
i
Hbb cb + e 0 Hbb (t ) dt ċb . But ċb = − cb Hbb + ca Hba eiω0 t .
t
i
i it = − e 0 Hbb (t ) dt ca Hba eiω0 t . But ca = e− 0 Haa (t ) dt da .
i
i it = − e 0 [Hbb (t )−Haa (t )]dt Hba eiω0 t da = − e−iφ Hba eiω0 t da . QED
(c)
Initial conditions:
Zeroth order:
ca (0) = 1 =⇒ da (0) = 1;
da (t) = 1,
cb (0) = 0 =⇒ db (0) = 0.
db (t) = 0.
i
d˙a = 0 =⇒ da (t) = 1 =⇒ ca (t) = e− First order:
i
i
d˙b = − e−iφ Hba eiω0 t =⇒ db = −
i it cb (t) = − e− 0 Hbb (t )dt
t
t
t
0
Haa
(t ) dt
.
e−iφ(t ) Hba (t )eiω0 t dt =⇒
0
e−iφ(t ) Hba (t )eiω0 t dt .
0
These don’t look much like the results in (a), but remember, we’re only working to first order in H ,
t
so ca (t) ≈ 1 − i 0 Haa (t ) dt (to this order), while for cb , the factor Hba in the integral means it is
already first order and hence both the exponential factor in front and e−iφ should be replaced by 1. Then
t
cb (t) ≈ − i 0 Hba (t )eiω0 t dt , and we recover the results in (a).
Problem 9.5
First order:
cb (t) = b.
i
ib t
−iω0 t
(1)
b =⇒ ca (t) = a −
Hab (t )e−iω0 t dt .
ċa = − Hab e
0
i
ia t
(1)
iω0 t
ċb = − Hba e
a =⇒ cb (t) = b −
Hba (t )eiω0 t dt .
0
Second order:
c(2)
a (t)
(0)
c(0)
a (t) = a,
Zeroth order:
ib
=a−
i
ia
ċa = − Hab e−iω0 t b −
t
−iω0 t
Hab (t )e
0
a
dt − 2
t
Hba (t )eiω0 t dt
=⇒
0
t
−iω0 t
Hab (t )e
0
t
iω0 t
Hba (t )e
dt
dt .
0
To get cb , just switch a ↔ b (which entails also changing the sign of ω0 ):
t
t
ia t
b
(2)
iω0 t
iω0 t
−iω0 t
cb (t) = b −
H (t )e
dt − 2
H (t )e
Hab (t )e
dt dt .
0 ba
0 ba
0
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CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
241
Problem 9.6
i
(2)
(1)
For H independent of t, Eq. 9.17 =⇒ cb (t) = cb (t) = − Hba
t
eiω0 t dt =⇒
0
t
H
= − ba eiω0 t − 1 . Meanwhile Eq. 9.18 =⇒
ω0
0
t
t
t
1
1
(2)
2
−iω0 t
iω0 t
2 1
ca (t) = 1 − 2 |Hab |
e
e
dt dt = 1 − 2 |Hab |
1 − e−iω0 t dt
iω0 0
0
0
'
(
t
1
i
e−iω0 t i
2
t +
= 1+
|H |
|H |2 t +
e−iω0 t − 1 .
= 1+
ω0 2 ab
iω0
ω0 2 ab
iω0
(2)
cb (t)
eiω0 t
i
= − Hba
iω0
0
For comparison with the exact answers (Problem 9.2), note first that cb (t) is already first order (because of
the Hba in front), whereas ω differs from ω0 only in second order, so it suffices to replace ω → ω0 in the exact
formula to get the second-order result:
cb (t) ≈
2Hba iω0 t/2
2Hba iω0 t/2 1 iω0 t/2
H
e
sin (ω0 t/2) =
e
− e−iω0 t/2 = − ba eiω0 t − 1 ,
e
iω0
iω0
2i
ω0
in agreement with the result above. Checking ca is more difficult. Note that
4|Hab |2
|Hab |2
|Hab |2
ω0
|Hab |2
ω = ω0 1 +
1
+
2
=
ω
≈
ω
+
2
;
.
≈
1
−
2
0
0
2
2
ω 0 2
ω 0 2
ω 0 2
ω
ω02 2
Taylor expansion:
ω0 t |Hab |2 t
|Hab |2 t
+
≈
cos
(ω
t/2)
−
sin (ω0 t/2)
cos(x + 9) = cos x − 9 sin x =⇒ cos (ωt/2) = cos
0
ω 0 2 ω 0 2
2
2
2
ω0 t |Hab | t
|H | t
≈ sin (ω0 t/2) + ab 2 cos (ω0 t/2)
+
sin(x + 9) = sin x + 9 cos x =⇒ sin (ωt/2) = sin
2
ω 0 2
ω0 ω0 t
ω0 t
ω0 t
ω0 t
|H |2 t
|H |2
|H |2 t
ca (t) ≈ e−iω0 t/2 cos
sin
− ab 2 sin
+ i 1 − 2 2ab 2
+ ab 2 cos
2
ω0 2
ω0 2
ω0 2
2
ω0 t
ω0 t
ω0 t
ω0 t
|H |
ω0 t
2i
= e−iω0 t/2 cos
sin
+ i sin
− ab2 t sin
− i cos
+
2
2
ω0 2
2
ω0
2
2
|H |
2i 1
= e−iω0 t/2 eiω0 t/2 − ab2 −iteiω0 t/2 +
eiω0 t/2 − e−iω0 t/2
ω0 ω 2i
|H |2
1
1
i
= 1 − ab2 −it +
|H |2 t +
1 − e−iω0 t = 1 +
e−iω0 t − 1 , as above. ω0 ω0
ω0 2 ab
iω0
Problem 9.7
(a)
ċa = −
i
Vab eiωt e−iω0 t cb ;
2
ċb = −
i
Vba e−iωt eiω0 t ca .
2
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242
CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
Differentiate the latter, and substitute in the former:
Vba i(ω0 − ω)ei(ω0 −ω)t ca + ei(ω0 −ω)t ċa
2
Vba i(ω0 −ω)t
Vab −i(ω0 −ω)t
Vba i(ω0 −ω)t
|Vab |2
= i(ω0 − ω) −i
e
e
e
−i
ca − i
cb = i(ω0 − ω)ċb −
cb .
2
2
2
(2)2
c̈b = −i
d2 cb
dcb
|Vab |2
|Vab |2
+ i(ω − ω0 )
cb = 0. Solution is of the form cb = eλt : λ2 + i(ω − ω0 )λ +
= 0.
+
2
2
dt
dt
4
42
1
(ω − ω0 )
|Vab |2
2
λ=
=i −
−i(ω − ω0 ) ± −(ω − ω0 ) −
± ωr , with ωr defined in Eq. 9.30.
2
2
2
i −
General solution: cb (t) = Ae
(ω−ω0 )
+ωr
2
t
i −
+ Be
(ω−ω0 )
+ωr
2
t
= e−i(ω−ω0 )t/2 Aeiωr t + Be−iωr t ,
or, more conveniently: cb (t) = e−i(ω−ω0 )t/2 [C cos(ωr t) + D sin(ωr t)] . But cb (0) = 0, so C = 0 :
ω0 − ω
cb (t) = De
ei(ω0 −ω)t/2 sin(ωr t) + ωr ei(ω0 −ω)t/2 cos(ωr t) ;
sin(ωr t). ċb = D i
2
2 i(ω−ω0 )t
2 i(ω−ω0 )t/2
ω0 − ω
ca (t) = i
e
ċb = i
e
D i
sin(ωr t) + ωr cos(ωr t) . But ca (0) = 1 :
Vba
Vba
2
2
−iVba
1=i
Dωr , or D =
.
Vba
2ωr
i(ω0 −ω)t/2
cb (t) = −
i
Vba ei(ω0 −ω)t/2 sin(ωr t),
2ωr
ca (t) = ei(ω−ω0 )t/2 cos(ωr t) + i
ω0 − ω
2ωr
sin(ωr t) .
(b)
Pa→b (t) = |cb (t)| =
2
and the denominator,
|Vab |
2ωr
2
sin2 (ωr t).
The largest this gets (when sin2 = 1) is
|Vab |2 /2
,
4ωr2
4ωr2 = (ω − ω0 )2 + |Vab |2 /2 , exceeds the numerator, so P ≤ 1 (and 1 only if ω = ω0 ).
2
2
ω0 − ω
|Vab |
sin2 (ωr t) +
sin2 (ωr t)
2ωr
2ωr
(ω − ω0 )2 + (|Vab |/)2
= cos2 (ωr t) +
sin2 (ωr t) = cos2 (ωr t) + sin2 (ωr t) = 1. 4ωr2
|ca |2 + |cb |2 = cos2 (ωr t) +
(c) If
|Vab |2 2 (ω − ω0 )2 ,
Eq. 9.28.
then ωr ≈
1
|ω − ω0 |,
2
and Pa→b ≈
0
|Vab |2 sin2 ω−ω
2 t
,
2
(ω − ω0 )2
confirming
(d) ωr t = π =⇒ t = π/ωr .
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CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
243
Problem 9.8
Spontaneous emission rate (Eq. 9.56): A =
R=
π
|℘|2 ρ(ω),
390 2
with
ρ(ω) =
ω 3 |℘|2
. Thermally stimulated emission rate (Eq. 9.47):
3π90 c3
ω3
π 2 c3 (eω/kB T − 1)
(Eq. 9.52).
So the ratio is
A
ω 3 |℘|2 390 2 π 2 c3 eω/kB T − 1
=
·
·
= eω/kB T − 1.
R
3π90 c3 π|℘|2
ω 3
The ratio is a monotonically increasing function of ω, and is 1 when
eω/kb t = 2,
or
ω
= ln 2,
kB T
ν=
ω=
kB T
ln 2,
or
ν=
ω
kB T
=
ln 2. For T = 300 K,
2π
h
(1.38 × 10−23 J/K)(300 K)
ln 2 = 4.35 × 1012 Hz.
(6.63 × 10−34 J · s)
For higher frequencies, (including light, at 1014 Hz), spontaneous emission dominates.
Problem 9.9
(a) Simply remove the factor eω/kB T − 1 in the denominator of Eq. 5.113:
ρ0 (ω) =
ω 3
.
π 2 c3
(b) Plug this into Eq. 9.47:
Rb→a =
ω 3
ω 3 |℘|2
π
|℘|2 2 3 =
,
2
390 π c
3π90 c3
reproducing Eq. 9.56.
Problem 9.10
N (t) = e−t/τ N (0) (Eqs. 9.58 and 9.59).
so t/τ = ln 2, or t1/2 = τ ln 2.
After one half-life, N (t) = 12 N (0), so
1
2
= e−t/τ , or
2 = et/τ ,
Problem 9.11
28
√ a. As
35 2
for x and y, we noted that |1 0 0, |2 0 0, and |2 1 0 are even (in x, y), whereas |2 1 ± 1 is odd. So the only
In Problem 9.1 we calculated the matrix elements of z; all of them are zero except 1 0 0|z|2 1 0 =
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244
CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
non-zero matrix elements are 1 0 0|x|2 1 ± 1 and 1 0 0|y|2 1 ± 1. Using the wave functions in Problem 9.1:
1
∓1
1
√
1 0 0|x|2 1 ± 1 = √
e−r/a re−r/2a sin θ e±iφ (r sin θ cos φ)r2 sin θ dr dθ dφ
3
3
a
πa
8 πa
∞
π
2π
1
3
4 −3r/2a
=∓
r
e
dr
sin
θ
dθ
(cos φ ± i sin φ) cos φ dφ
8πa4 0
0
0
5
∓1
2a
4
27
=
4!
a.
(π)
=
∓
8πa4
3
3
35
5 2π
∓1
2a
4
1 0 0|y|2 1 ± 1 =
4!
(cos φ ± i sin φ) sin φ dφ
4
8πa
3
3
0
5
∓1
2a
4
27
=
4!
(±iπ) = −i 5 a.
4
8πa
3
3
3
1 0 0|r|2 0 0 = 0;
√
27 2
1 0 0|r|2 1 0 =
a k̂;
35
℘2 = 0 (for |2 0 0 → |1 0 0),
Meanwhile,
E2 − E 1
1
ω=
=
and |℘|2 = (qa)2
E1
− E1
4
=−
1 0 0|r|2 1 ± 1 =
27 a
∓
î
−
i
ĵ
, and hence
35
215
(for |2 1 0 → 1 0 0 and |2 1 ± 1 → |1 0 0).
310
3E1
,
4
so for the three l = 1 states:
2
33 E13 (ea)2 215
1
c
29 E13 e2 a2
210 E1
=
−
=
6
3
10
3
8
4
3
8
2
2 3
3π90 c
3 π 90 c
3
mc
a
2
10
8
13.6
(3.00 × 10 m/s)
1
2
= 8
= 6.27 × 108 /s; τ =
= 1.60 × 10−9 s
3
0.511 × 106
(0.529 × 10−10 m)
A
A=−
for the three l = 1 states (all have the same lifetime); τ = ∞
for the l = 0 state.
Problem 9.12
[L2 , z] = [L2x , z] + [L2y , z] + [L2z , z] = Lx [Lx , z] + [Lx , z]Lx + Ly [Ly , z] + [Ly , z]Ly + Lz [Lz , z] + [Lz , z]Lz
[Lx , z] = [ypz − zpy , z] = [ypz , z] − [zpy , z] = y[pz , z] = −iy,
[Ly , z] = [zpx − xpz , z] = [zpx , z] − [xpz , z] = −x[pz , z] = ix,
But
[Lz , z] = [xpy − ypx , z] = [xpy , z] − [ypx , z] = 0.
So:
[L2 , z] = Lx (−iy) + (−iy)Lx + Ly (ix) + (ix)Ly = i(−Lx y − yLx + Ly x + xLy ).
But
Lx y = Lx y − yLx + yLx = [Lx , y] + yLx = iz + yLx ,
Ly x = Ly x − xLy + xLy = [Ly , x] + xLy = −iz + xLy .
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CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
245
So: [L2 , z] = i(2xLy − iz − 2yLx − iz) =⇒ [L2 , z] = 2i(xLy − yLx − iz).
2 2 .
L , [L , z] = 2i [L2 , xLy ] − [L2 , yLx ] − i[L2 , z]
.
= 2i [L2 , x]Ly + x[L2 , Ly ] − [L2 , y]Lx − y[L2 , Lx ] − i(L2 z − zL2 ) .
But
[L2 , Ly ] = [L2 , Lx ] = 0 (Eq. 4.102), so
.
2 2 L , [L , z] = 2i (yLz − zLy − ix) Ly − 2i (zLx − xLz − iy) Lx − i L2 z − zL2 , or
2 2 2
L , [L , z] = −2 2yLz Ly
−2zL2y
−
−2ixLy + 2xLz Lx + 2iyLx − L z + zL
2zL2x
2
2
−2z(L2x +L2y +L2z )+2zL2z
= −2
2
2yLz Ly − 2ixLy + 2xLz Lx + 2iyLx + 2zL2z − 2zL2 − L2 z + zL2
= −22 zL2 + L2 z − 42 (yLz − ix) Ly + (xLz + iy) Lx + zLz Lz
Lz y
Lz x
= 22 zL2 + L2 z − 42 (Lz yLy + Lz xLx + Lz zLz ) = 22 (zL2 + L2 z). QED
Lz (r·L)=0
Problem 9.13
|n 0 0 =
Rn0 (r)Y00 (θ, φ)
1
= √ Rn0 (r),
4π
so
1
n 0 0|r|n 0 0 =
4π
Rn 0 (r)Rn0 (r)(x î + y ĵ + z k̂) dx dy dz.
But the integrand is odd in x, y, or z, so the integral is zero.
Problem 9.14
(a)
|2 1 0
|2 1 1
|3 0 0 →
→ |1 0 0.
|2 1−1
(|3 0 0 → |2 0 0 and |3 0 0 → |1 0 0 violate ∆l = ±1 rule.)
(b)
From Eq. 9.72:
2 1 0|r|3 0 0 = 2 1 0|z|3 0 0 k̂.
From Eq. 9.69:
2 1 ± 1|r|3 0 0 = 2 1 ± 1|x|3 0 0 î + 2 1 ± 1|y|3 0 0 ĵ.
From Eq. 9.70:
± 2 1 ± 1|x|3 0 0 = i2 1 ± 1|y|3 0 0.
Thus
|2 1 0|r|3 0 0|2 = |2 1 0|z|3 0 0|2
and |2 1 ± 1|r|3 0 0|2 = 2|2 1 ± 1|x|3 0 0|2 ,
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246
CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
so there are really just two matrix elements to calculate.
ψ21m = R21 Y1m ,
ψ300 = R30 Y00 . From Table 4.3:
Y10 Y00
cos θ sin θ dθ dφ =
Y1±1
∗
3
4π
1
4π
π
2
cos θ sin θ dθ
0
Y00 sin2 θ cos φ dθ dφ = ∓
1
= ∓
4π
3
2
0
3
8π
1
4π
2π
√ √ π
3
3 2
cos3 θ 1
dφ =
(2π) =
−
=√ .
4π
3
2
3
3
0
π
2π
sin3 θ dθ
0
cos φe∓iφ dφ
0
2π
2π
4
1
1
2
cos φ dφ ∓ i
cos φ sin φ dφ = ∓ √ (π ∓ 0) = ∓ √ .
3
π 6
6
0
0
From Table 4.7:
∞
∞
2r
2
r −r/2a
1
2 r 2 −r/3a 3
3
√
K ≡
1−
e
R21 R30 r dr = √
r dr
e
+
3 a 27 a
24a3/2 27a3/2 0 a
0
6
7 5
∞
1
2
6
6
2
a
2
2
6
= √
1 − u + u2 u4 e−5u/6 du = √ 4!
a4
− 5!
+ 6!
3
3
27
5
3
5
27
5
9 2a
9 2
0
a 4! 65
2
a 4! 65
27 3 4 √
2
= √
5 − 6 · 5 + 63 = √
= 6 2 a.
6
6
3
27
5
9 2 5
9 2 5
So:
1
R21 (Y1±1 )∗ (r sin θ cos φ)R30 Y00 r2 sin θ dr dθ dφ = K ∓ √
.
6
1
2 1 0|z|3 0 0 =
R21 Y10 (r cos θ)R30 Y00 r2 sin θ dr dθ dφ = K √
.
3
2 1 ± 1|x|3 0 0 =
|2 1 0|r|3 0 0|2 = |2 1 0|z|3 0 0|2 = K 2 /3;
|2 1 ± 1|r|3 0 0 |2 = 2|2 1 ± 1|x|3 0 0 |2 = K 2 /3.
Evidently the three transition rates are equal, and hence 1/3 go by each route.
(c) For each mode,
decay rate is
ω 3 e2 |r|2
A=
;
3π90 c3
here
E3 − E2
1
ω =
=
E1
E1
−
9
4
=−
5 E1
,
36 so the total
2
9 3
2 5 E1
c
2
E1
e2
1 2 7 34 √
R=3 −
2a = 6
36 3π90 c3 3
56
5
mc2
a
9 2 2
13.6
3 × 108
1
=6
/s = 6.32 × 106 /s. τ =
= 1.58 × 10−7 s.
6
−10
5
0.511 × 10
0.529 × 10
R
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CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
247
Problem 9.15
(a)
Ψ(t) =
cn (t)e−iEn t/ ψn .
cn e−iEn t/ En ψn +
HΨ = i
∂Ψ
;
∂t
H = H0 + H (t);
cn e−iEn t/ H ψn = i
H0 ψn = En ψn . So
i ċn e−iEn t/ ψn + i −
cn En e−iEn t/ ψn .
The first and last terms cancel, so
cn e−iEn t/ H ψn = i
ċn e−iEn t/ ψn . Take the inner product with ψm :
cn e−iEn t/ ψm |H |ψn = i
ċn e−iEn t/ ψm |ψn .
ψm |ψn = δmn , and define
Assume orthonormality of the unperturbed states,
cn e−iEn t/ Hmn = i ċm e−iEm t/ ,
(b) Zeroth order:
i
ċN = − HN N ,
ċm = −
or
Hmn ≡ ψm |H |ψn .
i
cn Hmn ei(Em −En )t/ .
n
cN (t) = 1,
or
cm (t) = 0 for m = N . Then in first order:
i t
cN (t) = 1 −
H (t ) dt , whereas for m = N :
0 NN
i
ċm = − HmN ei(Em −EN )t/ , or
cm (t) = −
i
t
HmN (t )ei(Em −EN )t / dt .
0
(c)
t
i(EM −EN )t / e
ei(EM −EN )t/ − 1
i
ei(EM −EN )t / dt = − HM N
= −HM N
i(EM − EN )/ EM − EN
0
0
HM N
EM − E N
=−
ei(EM −EN )t/2 2i sin
t .
(EM − EN )
2
i
cM (t) = − HM N
PN →M = |cM |2 =
t
4|HM N |2
sin2
(EM − EN )2
EM − EN
t .
2
(d)
1 t iωt
i
cM (t) = − VM N
e
+ e−iωt ei(EM −EN )t / dt
2 0
t
ei(ω+EM −EN )t /
ei(−ω+EM −EN )t / iVM N
+
=−
.
2
i(ω + EM − EN )/ i(−ω + EM − EN )/ 0
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248
CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
If EM > EN , the second term dominates, and transitions occur only for ω ≈ (EM − EN )/:
iVM N
EM − EN − ω
1
cM (t) ≈ −
ei(EM −EN −ω)t/2 2i sin
t , so
2 (i/)(EM − EN − ω)
2
PN →M = |cM |2 =
(EM
|VM N |2
sin2
− EN − ω)2
EM − EN − ω
t .
2
If EM < EN the first term dominates, and transitions occur only for ω ≈ (EN − EM )/:
iVM N
EM − EN + ω
1
i(EM −EN +ω)t/2
cM (t) ≈ −
2i sin
e
t , and hence
2 (i/)(EM − EN + ω)
2
PN →M =
(EM
|VM N |2
sin2
− EN + ω)2
EM − EN + ω
t .
2
Combining the two results, we conclude that transitions occur to states with energy EM ≈ EN ± ω, and
PN →M =
EM − EN ± ω
t .
2
Vba = −℘E0 (Eq. 9.34). The rest is as before (Section 9.2.3), leading to Eq. 9.47:
(e) For light,
RN →M =
(EM
|VM N |2
sin2
− EN ± ω)2
π
|℘|2 ρ(ω), with ω = ±(EM − EN )/
390 2
(+ sign ⇒ absorption, − sign ⇒ stimulated emission).
Problem 9.16
For example (c):
cN (t) = 1 −
|cN |2 = 1 +
i
H t;
NN
cm (t) = −2i
1
|H |2 t2 ,
2 N N
|cm |2 = 1 +
m
HmN
ei(Em −EN )t/2 sin
(Em − EN )
|cm |2 = 4
|HmN |2
sin2
(Em − EN )2
Em − EN
t
2
(m = N ).
Em − EN
t , so
2
Em − EN
|HmN |2
t2
2
2
|H
|
+
4
sin
t
.
2 N N
(Em − EN )2
2
m=N
This is plainly greater than 1! But remember: The c’s are accurate only to first order in H ; to this order the
|H |2 terms do not belong. Only if terms of first order appeared in the sum would there be a genuine problem
with normalization.
For example (d):
cN
i
= 1 − VN N
0
t
t
sin(ωt ) i
i
cos(ωt ) dt = 1 − VN N
=⇒ cN (t) = 1 −
VN N sin(ωt).
ω 0
ω
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CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
cm (t) = −
|cN |2 = 1 +
VmN
2
249
ei(Em −EN +ω)t/ − 1 ei(Em −EN −ω)t/ − 1
+
(Em − EN + ω)
(Em − EN − ω)
|VN N |2
sin2 (ωt);
(ω)2
(m = N ).
So
and in the rotating wave approximation
Em − EN ± ω
|VmN |2
2
|cm | =
t
(m = N ).
sin
(Em − EN ± ω)2
2
/
Again, ostensibly
|cm |2 > 1, but the “extra” terms are of second order in H , and hence do not belong (to
first order).
/
You would do better to use 1 − m=N |cm |2 . Schematically: cm = a1 H + a2 H 2 + · · · , so |cm |2 =
2
a21 H 2 + 2a1 a2 H 3 + · · · , whereas cN = 1 + b1 H + b2 H 2 + · · · , so |cN |2 = 1 + 2b1 H + (2b2 + b21 )H 2 + · · · .
Thus knowing cm to first order (i.e., knowing a1 ) gets you |cm |2 to second order, but knowing cN to first order
(i.e., b1 ) does not get you |cN |2 to second order (you’d also need
/ b22). It is precisely this b2 term that would
cancel the “extra” (second-order) terms in the calculations of
|cm | above.
Problem 9.17
(a)
Equation 9.82 ⇒ ċm = −
i
ċm = − cm V0 (t);
− i
cm (t) = cm (0)e
i
cn Hmn ei(Em −En )t/ . Here Hmn = ψm |V0 (t)|ψn = δmn V0 (t).
n
dcm
i
i
= − V0 (t) dt ⇒ ln cm = −
cm
t
0
V0 (t ) dt
1
. Let Φ(t) ≡ −
V0 (t ) dt + constant.
t
V0 (t ) dt ;
cm (t) = eiΦ cm (0). Hence
0
|cm (t)|2 = |cm (0)|2 , and there are no transitions.
Φ(T ) = −
1
T
V0 (t) dt.
0
(b)
Eq. 9.84 ⇒ cN (t) ≈ 1 −
i
Eq. 9.85 ⇒ cm (t) = −
i
t
V0 (t ) dt = 1 + iΦ.
t
cN (t) = 1 + iΦ(t),
cm (t) = 0 (m = N ).
δmN V0 (t )ei(Em −EN )t / dt = 0 (m = N ).
0
0
The exact answer is cN (t) = eiΦ(t) , cm (t) = 0, and they are consistent, since eiΦ ≈ 1 + iΦ, to first order.
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250
CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
Problem 9.18
Use result of Problem 9.15(c).
H12
Here
En =
n2 π 2 2
,
2ma2
E2 − E1 =
so
3π 2 2
.
2ma2
π 2π
sin
x V0 sin
x dx
a
a
0
a/2
π 1
sin 3π
2V0 sin πa x
3π
4V0
V0
a x
=
−
sin
− sin
=
.
=
a
2(π/a)
2(3π/a) π
2
3
2
3π
2
=
a
a/2
0
Eq. 9.86 =⇒ P1→2 = 4
4V0
3π
2ma2
3π 2 2
2
2
sin
[Actually, in this case H11 and H22 are nonzero:
a/2
π V0
2
H11 = ψ1 |H |ψ1 = V0
x dx =
,
sin2
a
a
2
0
3π 2 t
4ma2
H22
=
16ma2 V0
sin
9π 3 2
2
= ψ2 |H |ψ2 = V0
a
3π 2 T
4ma2
2
.
a/2
2
sin
0
V0
2π
x dx =
.
a
2
However, this does not affect the answer, for according to Problem 9.4, c1 (t) picks up an innocuous phase factor,
while c2 (t) is not affected at all, in first order (formally, this is because Hbb is multiplied by cb , in Eq. 9.11, and
in zeroth order cb (t) = 0).]
Problem 9.19
Spontaneous absorption would involve taking energy (a photon) from the ground state of the electromagnetic
field. But you can’t do that, because the gound state already has the lowest allowed energy.
Problem 9.20
(a)
H = −γB · S = −γ (Bx Sx + By Sy + Bz Sz ) ;
γ
0 1
0 −i
1 0
H = −γ (Bx σx + By σy + Bz σz ) = −
Bx
+ By
+ Bz
1 0
i 0
0 −1
2
2
γ
γ
Bx − iBy
Brf (cos ωt + i sin ωt)
Bz
B0
=−
=−
−Bz
−B0
2 Bx + iBy
2 Brf (cos ωt − i sin ωt)
γ
Brf eiωt
B0
= −
.
−iωt
−B0
2 Brf e
(b) iχ̇ = Hχ ⇒
γ
γ
ȧ
a
Brf eiωt
Brf eiωt b
B0
B0 a
i
=−
=−
b
ḃ
2 Brf e−iωt −B0
2 Brf e−iωt a −B0 b
ȧ = i γ B0 a + Brf eiωt b = i Ωeiωt b + ω0 a ,
2
2
γ
i
ḃ = −i B0 b − Brf e−iωt a =
Ωe−iωt a − ω0 b .
2
2
⇒
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CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
251
(c) You can decouple the equations by differentiating with respect to t, but it is simpler just to check the quoted
results. First of all, they clearly satisfy the initial conditions: a(0) = a0 and b(0) = b0 . Differentiating a:
iω
ω
i
ω
ȧ =
[a0 (ω0 − ω) + b0 Ω]
a + −a0 sin(ω t/2) +
cos(ω t/2) eiωt/2
2
2
ω
2
i iωt/2
ω
ωa0 cos(ω t/2) + i [a0 (ω0 − ω) + b0 Ω] sin(ω t/2)
= e
2
ω
+ iω a0 sin(ω t/2) + [a0 (ω0 − ω) + b0 Ω] cos(ω t/2)
Equation 9.90 says this should be equal to
i
i iωt/2
Ω
iωt
Ωe b + ω0 a = e
Ωb0 cos(ω t/2) + i [b0 (ω − ω0 ) + a0 Ω] sin(ω t/2)
2
2
ω
ω0
+ ω0 a0 cos(ω t/2) + i [a0 (ω0 − ω) + b0 Ω] sin(ω t/2) .
ω
By inspection the cos(ω t/2) terms in the two expressions are equal; it remains to check that
ω
Ω
ω0
[a0 (ω0 − ω) + b0 Ω] + iω a0 = i [b0 (ω − ω0 ) + a0 Ω] + i [a0 (ω0 − ω) + b0 Ω] ,
ω
ω
ω
which is to say
i
a0 ω(ω0 − ω) + b0 ωΩ + a0 (ω )2 = b0 Ω(ω − ω0 ) + a0 Ω2 + a0 ω0 (ω0 − ω) + b0 ω0 Ω,
or
a0 ωω0 − ω 2 + (ω )2 − Ω2 − ω02 + ω0 ω = b0 [Ωω − ω0 Ω + ω0 Ω − ωΩ] = 0.
Substituting Eq. 9.91 for ω , the coefficient of a0 on the left becomes
2ωω0 − ω 2 + (ω − ω0 )2 + Ω2 − Ω2 − ω02 = 0. The check of b(t) is identical, with a ↔ b, ω0 → −ω0 , and ω → −ω.
(d)
Ω
b(t) = i sin(ω t/2)e−iωt/2 ;
ω
P (t) = |b(t)| =
2
Ω
ω
2
sin2 (ω t/2).
(e)
P(ω)
1
1/2
∆ω
ω0
ω
The maximum (Pmax = 1) occurs (obviously) at ω = ω0 .
P =
1
2
⇒ (ω − ω0 )2 = Ω2 ⇒ ω = ω0 ± Ω,
so
∆ω = ω+ − ω− = 2Ω.
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252
CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
(f ) B0 = 10, 000 gauss = 1 T; Brf = 0.01 gauss = 1 × 10−6 T. ω0 = γB0 . Comparing Eqs. 4.156 and
gp e
, where gp = 5.59. So
6.85, γ =
2mp
ω0
(5.59)(1.6 × 10−19 )
gp e
B0 =
=
(1) = 4.26 × 107 Hz.
2π
4πmp
4π(1.67 × 10−27 )
2Brf
∆ω
Ω
γ
∆ν =
= (4.26 × 107 )(2 × 10−6 ) = 85.2 Hz.
=
=
2Brf = νres
2π
π
2π
B0
νres =
Problem 9.21
(a)
H = −qE · r = −q(E0 · r)(k · r) sin(ωt). Write E0 = E0 n̂, k =
H = −q
ω
k̂. Then
c
E0 ω
qE0 ω
(n̂ · r)(k̂ · r) sin(ωt). Hba = −
b|(n̂ · r)(k̂ · r)|a sin(ωt).
c
c
This is the analog to Eq. 9.33: Hba = −qE0 b|n̂ · r|a cos ωt. The rest of the analysis is identical to the
dipole case (except that it is sin(ωt) instead of cos(ωt), but this amounts to resetting the clock, and clearly
has no effect on the transition rate). We can skip therefore to Eq. 9.56, except for the factor of 1/3, which
came from the averaging in Eq. 9.46:
A=
ω3 q2 ω2
q2 ω5
2
|b|(n̂
·
r)(
k̂
·
r)|a|
=
|b|(n̂ · r)(k̂ · r)|a|2 .
π90 c3 c2
π90 c5
(b) Let the oscillator lie along the x direction, so (n̂ · r) = n̂x x and k̂ · r = k̂x x. For a transition from n to n ,
we have
2
q2 ω5 A=
k̂
n |(a2+ +a+ a− +a− a+ +a2− )|n,
n̂
|n |x2 |n|2 . From Example 2.5, n |x2 |n =
x
x
π90 c5
2mω̄
where ω̄ is the frequency of the oscillator, not to be confused with ω, the frequency of the electromagnetic
wave. Now, for spontaneous emission the final state must be lower in energy, so n < n, and hence the
only surviving term is a2− . Using Eq. 2.66:
n |x2 |n =
n | n(n − 1)|n − 2 =
n(n − 1) δn ,n−2 .
2mω̄
2mω̄
Evidently transitions only go from |n to |n − 2, and hence
ω=
En − En−2
1
=
(n + 12 )ω̄ − (n − 2 + 12 )ω̄ = 2ω̄.
n |x2 |n =
n(n − 1) δn ,n−2 ;
mω
Rn→n−2 =
2
q2 ω5
2 (
k̂
n̂
)
n(n − 1).
x
x
π90 c5
m2 ω 2
It remains to calculate the average of (k̂x n̂x )2 . It’s easiest to reorient the oscillator along a direction r̂,
making angle θ with the z axis, and let the radiation be incident from the z direction (so k̂x → k̂r = cos θ).
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CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY
Averaging over the two polarizations (î and ĵ): n̂2r =
1
2
1
2
î2r + ĵr2
253
=
1
2
sin2 θ cos2 φ + sin2 θ sin2 φ =
sin2 θ. Now average overall directions:
π
1
1
1
k̂r2 n̂2r =
(1 − cos2 θ) cos2 θ sin θ dθ
sin2 θ cos2 θ sin θ dθ dφ =
2π
4π
2
8π
0
π
1
1 2 2
cos3 θ cos5 θ 1
=
=
−
+
−
=
.
4
3
5
4 3 5
15
0
R=
1 q 2 ω 3
n(n − 1).
15 π90 m2 c5
Comparing Eq. 9.63:
2
ω
R(forbidden)
= (n − 1) 2 .
R(allowed)
5
mc
For a nonrelativistic system, ω mc2 ; hence the term “forbidden”.
(c) If √
both the initial state and the final state have l = 0, the wave function is independent of angle (Y00 =
1/ 4π), and the angular part of the integral is:
4π
a|(n̂ · r)(k̂ · r)|b = · · · (n̂ · r)(k̂ · r) sin θ dθ dφ = · · ·
(n̂ · k̂) (Eq. 6.95).
3
But n̂ · k̂ = 0, since electromagnetic waves are transverse.
for forbidden transitions.
So R = 0 in this case, both for allowed and
Problem 9.22
[This is done in Fermi’s Notes on Quantum Mechanics (Chicago, 1995), Section 24, but I am looking for a more
accessible treatment.]
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254
CHAPTER 10. THE ADIABATIC APPROXIMATION
Chapter 10
The Adiabatic Approximation
Problem 10.1
(a)
Let
(mvx2 − 2Eni at)/2w = φ(x, t). Φn =
nπ √
∂Φn
1 1
v
sin
= 2 −
x eiφ +
∂t
2 w3/2
w
= −
i
nπ v
nπxv
∂φ
cot
−
x +i
Φn .
2w
w2
w
∂t
nπ 2
sin
x eiφ ,
w
w
so
nπ 2 nπx
− 2 v cos
x eiφ +
w
w
w
nπ ∂φ 2
sin
x i
eiφ
w
w
∂t
∂φ
1
2E i a
v
=
− n − 2 mvx2 − 2Eni at
∂t
2
w
w
=−
Eni a
v
− φ.
w
w
nπ ∂Φn
v
nπxv
Ei a
v
cot
= −i
+
x + i n + i φ Φn .
2
∂t
2w
w
w
w
w
2 ∂ 2 Φn
HΦn = −
.
2m ∂x2
∂φ
mvx
=
.
∂x
w
∂Φn
=
∂x
nπ 2 nπ
cos
x eiφ +
w w
w
nπ ∂φ
2
iφ
i
sin
x e
.
w
w
∂x
nπ
nπ ∂Φn
mvx =
cot
x +i
Φn .
∂x
w
w
w
nπ nπ imb
nπ 2
nπ
∂ 2 Φn
mvx 2
x +
Φn +
cot
x +i
= −
csc2
Φn .
2
∂x
w
w
w
w
w
w
So the Schrödinger equation (i∂Φn /∂t = HΦn ) is satisfied ⇔
−i
nπ v
nπxv
Eni a
v
cot
+
x
+
i
+i φ
2
2w
w
w
w
w
2
=−
2m
imv nπ
nπ nπ 2
mvx 2
2 nπ
−
x +
+
cot
x +i
csc
w
w
w
w
w
w
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CHAPTER 10. THE ADIABATIC APPROXIMATION
Cotangent terms: − i
nπxv w2
?
=−
255
2 nπ mvx nπvx
2 i
= −i 2 . 2m
w w
w
Remaining trig terms on right:
−
nπ 2
w
csc2
nπ nπ 2
nπ nπ 2 1 − cos2 (nπx/w)
nπ 2
cot2
.
x +
x =−
=
−
w
w
w
w
w
sin2 (nπx/w)
This leaves:
i
v
Ei a
v
+i n +i
2w
w
w
mvx2 − 2Eni at
2w
?
=
nπ 2 imv m2 v 2 x2
−
− 2 2
+
2m
w
w
w
2
2
iv✁✁ Eni a mv 2✚
x✚
x✚
vEni at ? n2 π 2
iv mv 2✚
−
− ✚ +
+ ✁✁ − ✚
=−
w
2mw
✁2
✁2 ✚2w
✚2w
−
Eni a
E i a2 ? n2 π 2
n2 π 2 2 a2
n2 π 2
(w − vt) = − n = −
⇔−
=−
= r.h.s. 2
w
w
2mw
2ma w
2mw
So Φn does satisfy the Schrödinger equation, and since Φn (x, t) = (· · · ) sin (nπx/w), it fits the boundary
conditions: Φn (0, t) = Φn (w, t) = 0.
(b)
Equation 10.4 =⇒ Ψ(x, 0) =
Multiply by
2
a
2
sin
a
a
Ψ(x, 0) sin
0
cn Φn (x, 0) =
2
nπ
x e−imvx /2a
a
cn
nπ 2
2
sin
x eimvx /2a .
a
a
and integrate:
nπ 2
2 π
nπ
nπ
cn
sin
x e−imvx /2a dx =
x sin
x dx = cn .
a
a
a
a
0
δnn
So, in general:
cn =
2
a
2
cn =
π
a
cn =
e−imvx
0
π
2
/2a
2
a
sin
a
e−imvx
/2a
0
nπ a
sin
e−iαz sin(nz) sin(z) dz.
2
2
sin
nπ x Ψ(x, 0)dx. In this particular case,
a
π x dx. Let
a
π
x ≡ z;
a
dx =
a
dz;
π
mva 2
mvx2
mvz 2 a2
=
=
z .
2
2a
2a π
2π 2 QED
0
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256
CHAPTER 10. THE ADIABATIC APPROXIMATION
(c)
e−iE1 t/ ⇒ ω =
w(Te ) = 2a ⇒ a + vTe = 2a ⇒ vTe = a ⇒ Te a/v;
Ti =
8π
2π
4 ma2
2ma2 =
.
2
2
π π mav 2π 2 = 8πα 1,
so
Ti =
4ma2
.
π
Adiabatic ⇒ Te Ti ⇒
α 1. Then cn =
2
π
E1
2π
⇒ Ti =
= 2π , or
ω
E1
a
4ma2
4 mav
⇒
1, or
v
π
π π
sin(nz) sin(z)dz = δn1 .
Therefore
0
πx 2
i
2
sin
ei(mvx −2E1 at)/2w ,
w
w
Ψ(x, t) =
which (apart from a phase factor) is the ground state of the instantaneous well, of width w, as required
2
mva
by the adiabatic theorem. (Actually, the first term in the exponent, which is at most mva
2a = 2 1
and could be dropped, in the adiabatic regime.)
(d)
t
t
1
π2 π 2 2
1
1
dt = −
−
2
2m
2m
v a + vt
0 (a + vt )
0
π2 1
1
π 2 vt
π 2 t
=−
−
=−
=−
.
2mv a w
2mv aw
2maw
1
θ(t) = −
So (dropping the
(since −
mvx2
term, as explained in (c))
2w
E1i at
π 2 2 at
π 2 t
=−
=−
= θ):
2
w
2ma w
2maw
πx i
2
sin
e−iE1 at/w
w
w
2
πx iθ
sin
e .
w
w
Ψ(x, t) =
Ψ(x, t) =
can be written
This is exactly what one would naively expect: For a fixed well (of width a) we’d have Ψ(x, t) =
Ψ1 (x)e−iE1 t/ ; for the (adiabatically) expanding well, simply replace a by the (time-dependent) width
w, and integrate to get the accumulated phase factor, noting that E1 is now a function of t.
Problem 10.2
To show:
i
∂χ
= Hχ,
∂t
where χ is given by Eq. 10.31 and H is given by Eq. 10.25.
∂χ
=
∂t
(ω1 −ω)
λ
λt
λt
−
sin
cos
−
i
cos α2 e−iωt/2 −
2
λ
2
2
(ω1 +ω)
λt
λ
sin α2 eiωt/2 +
cos λt
2 − sin 2 − i
λ
2
iω
2
iω
2
cos
cos
λt
2
λt
2
−
−
i(ω1 −ω)
λ
i(ω1 +ω)
λ
sin
sin
λt
2
λt
2
cos
α
2
e−iωt/2
sin
α
2
eiωt/2
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CHAPTER 10. THE ADIABATIC APPROXIMATION
257
Hχ =
(ω1 −ω)
i(ω1 +ω)
λt
α −iωt/2
λt
λt
α iωt/2
−iωt
)
−
i
sin(
)
cos
e
+
e
sin
α
cos(
)
−
sin(
)
sin
e
cos α cos( λt
2
λ
2
2
2
λ
2
2
ω1
2
(ω
−ω)
i(ω
+ω)
λt
α −iωt/2
λt
λt
α iωt/2
1
1
eiωt cos α cos( λt
)
−
i
sin(
)
cos
e
−
cos
α
cos(
)
−
sin(
)
sin
e
2
λ
2
2
2
λ
2
2
(1) Upper elements:
λ
(ω1 − ω)
λt
α iω
i(ω1 − ω)
λt
α
λt
λt
✚
✚
i
−
i
cos
cos
−
−
sin
cos
−
sin
cos
✚
✚
✁
2
λ
2
2
λ
2
2
2✁
✚ 2
✚ 2
✁
λt
α
(ω1 − ω)
λt
α
λt
(ω1 + ω)
λt
? ✁ ω1
cos
=
sin
−i
sin
cos α cos✚✚ + cos
−i
sin
sin
α
,
2
λ
2
2
λ
2
2
2
✚ 2
✁
where = 2 sin
The sine terms:
sin
α
α
cos✚✚
2✚ 2
λt
2
−iλ −
α ?
iω(ω1 − ω) ω1 (ω1 − ω)
iω1 (ω1 + ω)
= 0.
+
cos α +
2 sin2
λ
λ
λ
2
i
λt
2
ω✚
+ (ω12 − ωω1 ) cos α + (ω12 + ωω1 )(1 − cos α)
sin
−ω✟2 − ω12 + 2ωω1 cos α − ωω1 + ✚
✟
λ
2
i
λt
✘✘
✘✘
= − sin
ωω✘
ω12✘
cos
α − ωω1 cos α + ω12 + ✘
ωω✘
ω12✘
cos
α − ωω1 cos α = 0. −ω✟12 + 2ωω1 cos α − ✘
1 +✘
1 −✘
✟
λ
2
The cosine terms:
λt
α
λt
cos
ω) + ✚
ω − ω1 cos α − ω1 2 sin2
(ω1 − ✚
= −ω1 cos
[−1 + cos α + (1 − cos α)] = 0. 2
2
2
(2) Lower elements:
α✟
iω
α✟
λ
(ω1 + ω)
λt
i(ω1 + ω)
λt
λt
λt
✟
✟
i
−i
cos
sin
+
−
sin
sin
− sin
cos
✁
2
λ
2
✟✟ 2
2
λ
2
✟✟ 2
2
2✁
✁
α
(ω1 − ω)
λt
α
λt
i(ω1 + ω)
λt
α
λt
? ✁ ω1
✚
✚
=
−i
sin
2 sin✚
cos2 − cos
−
sin
cos α sin✚
.
cos
2
λ
2
2
2
λ
2
2
✚ 2
✚ 2
✁
The sine terms:
α iω (ω + ω)
λt
iω(ω1 + ω) iω1 (ω1 − ω)
?
1
1
sin
−iλ +
+
2 cos2
−
cos α = 0.
2
λ
λ
2
λ
i
λt
2
2
− ω12 + 2ωω1 cos α + ωω1 + ✚
ω✚
+ (ω12 − ωω1 )(1 + cos α) − (ω12 + ωω1 ) cos α
sin
−✚
ω✚
λ
2
i
λt
2
✘✘
✘✘
=
ωω✘
ωω✘
ω12✘
cos
α − ωω1 cos α − ✘
ω12✘
cos
α − ωω1 cos α = 0. sin
−ω12 + 2ωω1 cos α + ✘
1 + ω1 − ✘
1 +✘
λ
2
The cosine terms:
λt
λt
2 α
cos
(ω1 + ✚
+ ω1 cos α = cos
[ω1 − ω1 (1 + cos α) + ω1 cos α] = 0. ω) − ✚
ω − ω1 2 cos
2
2
2
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258
CHAPTER 10. THE ADIABATIC APPROXIMATION
As for Eq. 10.33:
cos
λt
2
−i
=
α=
α
,
β
cos
λt
2
= cos
λt
2
β=
(ω1 − ω cos α)
sin
λ
−
cos
λt
2
= cos
λt
2
−
λt
2
e−iωt/2
cos α2
iωt
e sin α2
+i
ω
sin α sin
λ
λt
2
e−iωt/2
sin α2
iωt
−e cos α2
with
iω1
−
sin
λ
λt
2
i(ω1 − ω)
sin
λ
iω1
−
sin
λ
λt
2
i(ω1 + ω)
sin
λ
λt
2
α iω
α
α
cos +
cos α cos + sin α sin
sin
2
λ 2 2
cos(α− α
2 )=cos
cos
α −iωt/2
e
2
λt
2
α iωt/2
e
2
e−iωt/2
(confirming the top entry).
sin( α
2 −α)=− sin
sin
λt
2
α
2
α iω
α
α
sin
sin +
cos α sin − sin α cos
2
λ 2 2
λt
2
eiωt/2
α
2
(confirming the bottom entry).
λt
λt
λt
(ω1 − ω cos α)2
ω2
2
2
2
sin
sin
α
sin
+
+
2
λ2
2
λ2
2
λt
λt
1
= cos2
+ 2 ω12 − 2ωω1 cos α + ω 2 cos2 α + ω 2 sin2 α sin2
2
λ 2
|c+ |2 + |c− |2 = cos2
= cos2
λt
2
+ sin2
λt
2
ω 2 +ω12 −2ωω1 cos α=λ2
= 1. Problem 10.3
(a)
ψn (x) =
nπ 2
sin
x . In this case R = w.
w
w
nπ √
∂ψn
1 1
sin
= 2 −
x +
∂R
2 w3/2
w
nπ 2 nπ − 2 x cos
x ;
w
w
w
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CHAPTER 10. THE ADIABATIC APPROXIMATION
259
1 w
∂ψn
∂ψn
ψn ψn
=
dx
∂R
∂R
0
w
nπ nπ nπ 1
2nπ w
=− 2
sin2
x sin
x dx − 3
x cos
x dx
w 0
w
w
w w 0
0
1
2nπ
2 sin( w x)
1 w nπ w
2nπ
=− 2
− 3
x dx
x sin
w
2
w 0
w
1
2nπ
nπ w 2
wx
2nπ
=−
sin
−
x −
cos
x
2w w3
2nπ
w
2nπ
w
w
0
w2
1
nπ
1
1
=−
− 3 −
cos(2nπ) = −
+
= 0.
2w w
2nπ
2w 2w
So Eq. 10.42 =⇒ γn (t) = 0.
(If the eigenfunctions are real, the geometric phase vanishes.)
(b)
Equation 10.39 =⇒ θn (t) =
θn = −
n2 π 2 2mv
w2
w1
1
t
0
n2 π 2 2
n2 π 2 dt = −
2
2mw
2m
1
n2 π 2 dw
=
w2
2mv
1 dt
dw;
w2 dw
w2
n2 π 2 1
1
1 .
=
−
w w1
2mv
w2
w1
(c) Zero.
Problem 10.4
√
ψ=
mα −mα|x|/2
e
. Here R = α,
so
√
√ ∂ψ
m 1 1
mα
m|x| −mα|x|/2
−mα|x|/2
√
e
+
.
e
=
− 2
∂R
2 α
∂ψ
ψ
=
∂R
√
mα 1
2
√
2
m m mα
m
m2 α
−2mα|x|/2
−
|x| e
=
− 4 |x| e−2mα|x|/ .
3
2
α
2
0 2 2 2
1
∞
∂ψ
m2 α ∞ −2mαx/2
m
m
2m2 α
−2mαx/2
ψ
e
dx − 4
xe
dx = 2
=2
−
4
∂R
22 0
2mα
2mα
0
1
1
−
= 0. So Eq. 10.42 =⇒ γ(t) = 0.
=
2α 2α
mα2
E=− 2 ,
2
1
so θ(t) = −
0
T
mα2
− 2
2
dt =
m
23
α2
α1
α2
dt
m
dα = 3
dα
2 c
α2
α1
α2 dα =
m
α3 − α13 .
62 c 2
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260
CHAPTER 10. THE ADIABATIC APPROXIMATION
Problem 10.5
According to Eq. 10.44 the geometric phase is
Rf
γn (t) = i
ψn |∇R ψn · dR.
Ri
Now
ψn |ψn = 1,
so
∇R ψn |ψn = ∇R ψn |ψn + ψn |∇R ψn = ψn |∇R ψn ∗ + ψn |∇R ψn = 0,
and hence ψn |∇R ψn is pure imaginary. If ψn is real, then, ψn |∇R ψn must in fact be zero.
Suppose we introduce a phase factor to make the (originally real) wave function complex:
ψn = eiφn (R) ψn ,
where ψn is real.
Then ∇R ψn = eiφn ∇R ψn + i(∇R φn )eiφn ψn . So
ψn |∇R ψn = e−iφn eiφn ψn |∇R ψn + ie−iφn (∇R φn )eiφn ψn |ψn . But ψn |ψn = 1,
ψn |∇R ψn = 0
Rf
γn (t) = i
(as we just found), so ψn |∇R ψn = i∇R φn ,
and
and Eq. 10.44 =⇒
i∇R (φn ) · dR = − [φn (Rf ) − φn (Ri )] , so Eq. 10.38 gives:
Ri
Ψn (x, t) = ψn (x, t)e− i
t
0
En (t )dt −i[φn (Rf )−φn (Ri )]
e
.
The wave function picks up a (trivial) phase factor, whose only function is precisely to kill the phase factor we
put in “by hand”:
t
i
Ψn (x, t) = ψn (x, t)e− 0 En (t )dt eiφn (Ri ) = Ψn (x, t)eiφn (Ri ) .
In particular, for a closed loop φn (Rf ) = φn (Ri ), so γn (T ) = 0.
Problem 10.6
H=
e
B · S. Here
m
B = B0 sin θ cos φ î + sin θ sin φ ĵ + cos θ k̂ ; take spin matrices from Problem 4.31.
√
2 0 0
010
0 −i 0
eB0
√ sin θ cos φ 1 0 1 + sin θ sin φ i 0 −i + cos θ 0 0 √
0
H=
m 2
010
0 i 0
0 0 − 2
√
−iφ
2 cos θ e
sin θ
0
eB0 iφ
e sin θ
0
e−iφ
=√
√ sin θ .
2m
iφ
0
e sin θ − 2 cos θ
We need the “spin up” eigenvector:
Hχ+ =
eB0
χ+ .
m
√
√
2 cos θa + e−iφ sin θb = √2a.
2 cos θ e−iφ sin θ
0
a
a
(i)
√
eiφ sin θ
b = 2 b =⇒ (ii) eiφ sin θa + e−iφ sin θc = 2b.
0
e−iφ
√ sin θ
√
√
iφ
c
c
0
e sin θ − 2 cos θ
(iii) eiφ sin θb − 2 cos θc = 2c.
√
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CHAPTER 10. THE ADIABATIC APPROXIMATION
(i) ⇒ b =
√
iφ
2e
1 − cos θ
sin θ
Thus c = e2iφ tan2 (θ/2) a;
2
|a|2 1 + tan2 (θ/2) = |a|2
a = e−iφ cos2 (θ/2);
Pick
a=
√
iφ
2e tan (θ/2) a;
(ii) is redundant.
1
cos(θ/2)
then
−iφ
cos2 (θ/2)
√ e
χ+ = 2 sin (θ/2) cos (θ/2) .
eiφ sin2 (θ/2)
261
(iii) ⇒ b =
Normalize:
√
−iφ
2e
1 + cos θ
sin θ
c=
√
2e−iφ cot (θ/2) c.
|a|2 + 2 tan2 (θ/2)|a|2 + tan4 (θ/2)|a|2 = 1 ⇒
4
b=
= 1 ⇒ |a|2 = cos4 (θ/2) .
√
2 sin(θ/2) cos(θ/2)
and
c = eiφ sin2 (θ/2),
and
This is the spin-1 analog to Eq. 10.57.
1 ∂χ+
1 ∂χ+
∂χ+
r̂ +
θ̂ +
φ̂
∂r
r ∂θ
r sin θ ∂φ
−e−iφ cos (θ/2) sin (θ/2)
−ie−iφ cos2 (θ/2)
1 √ 2
1
φ̂.
0
=
2 cos (θ/2) − sin2 (θ/2) /2 θ̂ +
r
r sin θ
ieiφ sin2 (θ/2)
eiφ sin (θ/2) cos (θ/2)
∇χ+ =
1− cos2 (θ/2) [cos (θ/2) sin (θ/2)] + sin (θ/2) cos (θ/2) cos2 (θ/2) − sin2 (θ/2)
r
.
+ sin2 (θ/2) [sin (θ/2) cos (θ/2)] θ̂
.
1 - 2
+
cos (θ/2) −i cos2 (θ/2) + sin2 (θ/2) i sin2 (θ/2) φ̂
r sin θ
i 4
=
sin (θ/2) − cos4 (θ/2) φ̂
r sin θ
i 2
=
sin (θ/2) + cos2 (θ/2) sin2 (θ/2) − cos2 (θ/2) φ̂
r sin θ
i
i
=
(1)(− cos θ) φ̂ = − cot θ φ̂.
r sin θ
r
1
∂
i
−i ∂
i sin θ
i
∇ × χ+ |∇χ+ =
sin θ − cot θ r̂ = 2
(cos θ) r̂ = 2
r̂ = 2 r̂.
r sin θ ∂θ
r
r sin θ ∂θ
r sin θ
r
χ+ |∇χ+ =
Equation 10.51 =⇒ γ+ (T ) = i
i 2
r dΩ = −Ω.
r2
Problem 10.7
(a) Giving H a test function f to act upon:
1
Hf =
∇ − qA ·
∇f − qAf + qϕf
2m i
i
1
q
q
=
− 2 ∇ · (∇f ) −
∇ · (Af ) − A · (∇f ) + q 2 A · Af + qϕf.
2m
i
i
(∇·A)f +A·(∇f )
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262
CHAPTER 10. THE ADIABATIC APPROXIMATION
But ∇ · A = 0 and ϕ = 0 (see comments after Eq. 10.66), so
Hf =
1 2 2
− ∇ f + 2iqA · ∇f + q 2 A2 f ,
2m
i∇
(b) Apply
But
− qA ·
∇ − qA
i
or
H=
1 2 2
− ∇ + q 2 A2 + 2iqA · ∇ . QED
2m
to both sides of Eq. 10.78:
2
Ψ=
q
ig
∇ − qA ·
e ∇Ψ = −2 ∇ · (eig ∇Ψ ) − eig A · ∇Ψ .
i
i
i
∇ · (eig ∇Ψ ) = ieig (∇g) · (∇Ψ ) + eig ∇ · (∇Ψ )
∇g =
and
q
A,
so the right side is
q
−i2 eig A · ∇Ψ − 2 eig ∇2 Ψ + iqeig A · ∇Ψ = −2 eig ∇2 Ψ . QED
Problem 10.8
(a) Schrödinger equation:
2 d 2 ψ
−
= Eψ,
2m dx2
or
d2 ψ
= −k 2 ψ
dx2
(k ≡
√
2mE/)
0 < x < 12 a + 9,
1
2 a + 9 < x < a.
Boundary conditions: ψ(0) = ψ( 12 a + 9) = ψ(a) = 0.
Solution:
(1) 0 < x < 12 a + 9 :
ψ(x) = A sin kx + B cos kx. But ψ(0) = 0 ⇒ B = 0,
k( 12 a + 9) = nπ
ψ( 12 a + 9) = 0 ⇒
or else A = 0.
(2) 12 a + 9 < x < a :
ψ( 12 a
and
(n = 1, 2, 3, . . . ) ⇒ En = n2 π 2 2 /2m(a/2 + 9)2 ,
ψ(x) = F sin k(a − x) + G cos k(a − x). But ψ(a) = 0 ⇒ G = 0,
k( 12 a − 9) = n π
+ 9) = 0 ⇒
or else F = 0.
and
(n = 1, 2, 3, . . . ) ⇒ En = (n )2 π 2 2 /2m(a/2 − 9)2 ,
either E1 =
The ground state energy is
or else
π 2 2
(n = 1), with F = 0,
2m( 12 a + 9)2
π 2 2
E1 =
(n = 1), with A = 0.
2m( 12 a − 9)2
Both are allowed energies, but E1 is (slightly) lower (assuming 9 is positive), so the ground state is
ψ(x) =
0,
2
1
2 a+5
sin
πx
1
2 a+5
, 0 ≤ x ≤ 12 a + 9;
1
2a
+ 9 ≤ x ≤ a.
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CHAPTER 10. THE ADIABATIC APPROXIMATION
263
ψ(x)
a x
a_ a_ + ε
2 2
(b)
2 d 2 ψ
−
+f (t)δ(x− 12 a−9)ψ = Eψ
2m dx2
⇒
ψ(x) =
A sin kx,
0 ≤ x < 12 a + 9,
F sin k(a − x), 12 a + 9 < x ≤ a,
√
where
k≡
2mE
.
Continuity in ψ at x = 12 a + 9 :
A sin k
1
2a
+ 9 = F sin k a − 12 a − 9 = F sin k
1
2a
−9
⇒F =A
sin k
sin k
1
2a
1
2a
+9
.
−9
Discontinuity in ψ at x = 12 a + 9 (Eq. 2.125):
−F k cos k(a−x)−Ak cos kx =
2mf
A sin kx ⇒ F cos k
2
1
2a
− 9 +A cos k
1
2a
+9 =−
2mf
2 k
A sin k
1
2a
2T
A sin k 12 a + 9 .
z
2T
1
1
1
1
sin k 2 a + 9 cos k 2 a − 9 + cos k 2 a + 9 sin k 2 a − 9 = −
sin k 12 a + 9 sin k 12 a − 9 .
z
2T 1 1
1
sin k 2 a + 9 + 2 a − 9 = −
cos k 12 a + 9 − 12 a + 9 − cos k 12 a + 9 + 12 a − 9 .
z
2
A
sin k
sin k
1
2a
1
2a
+9
cos k
−9
sin ka = −
1
2a
T
(cos 2k9 − cos ka)
z
(c)
sin z =
− 9 + A cos k
T
(cos z − 1)
z
⇒
⇒
1
2a
+9 =−
z sin z = T [cos z − cos(zδ)].
z
cos z − 1
=
= − tan(z/2)
T
sin z
⇒
z
tan(z/2) = − .
T
Plot tan(z/2) and −z/T on the same graph, and look for intersections:
tan(z/2)
π
2π
3π
z
-z/T
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+9 .
264
CHAPTER 10. THE ADIABATIC APPROXIMATION
As t : 0 → ∞, T : 0 → ∞, and the straight line rotates counterclockwise from 6 o’clock to 3 o’clock,
2 π 2
so the smallest z goes from π to 2π, and the ground state energy goes from ka = π ⇒ E(0) =
2ma2
2 π 2
(appropriate to a well of width a) to ka = 2π ⇒ E(∞) =
(appropriate for a well of width a/2.
2m(a/2)2
T
z
(d) Mathematica yields the following table:
(e) Pr =
Ir
1
,
=
Ir + Il
1 + (Il /Ir )
0
3.14159
1
3.67303
5
4.76031
20
5.72036
100
6.13523
1000
6.21452
where
a/2+5
1
1
Il =
x−
sin(2kx) A sin kx dx = A
2
4k
0
0
a
a
a
1
1
29
1
29
2
2
+9 −
sin 2k
+9
= A 1+
−
sin ka + ka
=A
2 2
4k
2
4
a
ka
a
a
1
= A2 1 + δ − sin(z + zδ) .
4
z
a/2+5
2
2
2
a
F 2 sin2 k(a − x) dx. Let
Ir =
a/2+5
0
= −F 2
u ≡ a − x, du = −dx.
a/2−5
sin2 ku du = F 2
sin2 ku du =
0
a/2−5
1
a 2
F 1 − δ − sin(z − zδ) .
4
z
Il
A2 [1 + δ − (1/z) sin(z + zδ)]
. But (from (b))
= 2
Ir
F [1 − δ − (1/z) sin(z − zδ)]
=
I+
,
I−
where
I± ≡ 1 ± δ −
A2
sin2 k(a/2 − 9)
sin2 [z(1 − δ)/2]
=
=
.
F2
sin2 k(a/2 + 9)
sin2 [z(1 + δ)/2]
1
sin z(1 ± δ) sin2 [z(1 ∓ δ)/2].
z
Pr =
1
.
1 + (I+ /I− )
Using δ = 0.01 and the z’s from (d), Mathematica gives
T
Pr
0
0.490001
1
0.486822
5
0.471116
20
0.401313
100
0.146529
1000
0.00248443
As t : 0 → ∞ (so T : 0 → ∞), the probability of being in the right half drops from almost 1/2 to zero—the
particle gets sucked out of the slightly smaller side, as it heads for the ground state in (a).
(f )
T=0
T=1
T=5
T=20
T=100
T=1000
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CHAPTER 10. THE ADIABATIC APPROXIMATION
Problem 10.9
(a) Check the answer given:
t
xc = ω
f (t ) sin [ω(t − t )] dt =⇒ xc (0) = 0. 0
265
t
ẋc = ωf (t) sin [ω(t − t)] + ω 2
f (t ) cos [ω(t − t )] dt = ω 2
0
ẍc = ω f (t) cos [ω(t − t)] − ω
f (t ) cos [ω(t − t )] dt ⇒ ẋc (0) = 0. 0
2
t
t
f (t ) sin [ω(t − t )] dt = ω 2 f (t) − ω 2 xc .
3
0
Now the classical equation of motion is m(d2 x/dt2 ) = −mω 2 x + mω 2 f . For the proposed solution,
m(d2 xc /dt2 ) = mω 2 f − mω 2 xc , so it does satisfy the equation of motion, with the appropriate boundary
conditions.
(b) Let
z ≡ x − xc
(so ψn (x − xc ) = ψn (z),
and z depends on t as well as x).
∂Ψ
i
dψn
1
m
xc
mω 2
=
(−ẋc )ei{} + ψn ei{}
− (n + )ω + mẍc (x − ) − ẋ2c +
f xc
∂t
dz
2
2
2
2
1
ẋ2
mω 2
[ ] = −(n + )ω +
2x(f − xc ) + x2c − c2 .
2
2
ω
∂Ψ
dψn i{}
1
ẋ2c
mω 2
2
.
= −ẋc
e + iΨ −(n + )ω +
2x(f − xc ) + xc − 2
∂t
dz
2
2
ω
∂Ψ
i
dψn i{}
=
e + ψn ei{} (mẋc );
∂x
dz
∂2Ψ
d2 ψn i{}
dψn i{} i
=
e +2
e
(mẋc ) −
2
2
∂x
dz
dz
mẋc
2
ψn ei{} .
2 ∂ 2 Ψ 1
+ mω 2 x2 Ψ − mω 2 f xΨ
2m ∂x2
2
2
2 d2 ψn i{}
2 dψn i{} imẋc
1
2 mẋc
=−
e
−
Ψ + mω 2 x2 Ψ − mω 2 f xΨ.
2
e
+
2
2m dz
2m dz
2m
2
HΨ = −
But −
2 d 2 ψ n
1
1
+ mω 2 z 2 ψn = (n + )ωψn ,
2m dz 2
2
2
so
✟
1 ✟✟
1
dΨ✟
m
1
n i{}
✟
HΨ = (n✟
+✟ )ωΨ − mω 2 z 2 Ψ − iẋ
e + ẋ2c Ψ + mω 2 x2 Ψ − mω 2 f xΨ
c
✟
2
2
2
✟ dz
✟ 2
✟
dψ
∂Ψ
1 ✟ mω 2
1
✟
✟
?
n i{}
✟
= i
+ )ω +
= −i
ẋ✟
e − Ψ −(n
(2xf − 2xxc + x2c − 2 ẋ2c )
c
✟
✟
∂t
dz
2
2
ω
✟
✟
2
1
m
1
✘ ? mω
✘2✘
− mω 2 z 2 + ẋ2c + mω 2 x2 − ✘
mω
fx = −
2
2
2
2
?
z 2 − x2 = −2xxc + x2c ;
'
✟ − 2xxc +
2xf
✟
x2c
1
− 2 ẋ2c
ω
(
?
z 2 = (x2 − 2xxc + x2c ) = (x − xc )2 . c
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266
CHAPTER 10. THE ADIABATIC APPROXIMATION
(c)
Eq. 10.90 ⇒ H = −
H=
−
2 ∂ 2
1
1
+ mω 2 x2 − 2xf + f 2 − mω 2 f 2 . Shift origin:
2m ∂x2
2
2
u ≡ x − f.
2 ∂ 2
1
1
+ mω 2 u2 − mω 2 f 2 .
2
2m ∂u
2
2
The first term is a simple harmonic oscillator in the variable u; the second is a constant (with respect
to position). So the eigenfunctions are ψn (u), and the eigenvalues are harmonic oscillator ones, (n +
1
less the constant: En = (n + 12 )ω − 12 mω 2 f 2 .
2 )ω,
t
1 d
d
(d) Note that sin [ω(t − t )] =
cos [ω(t − t )], so xc (t) =
f (t )
cos [ω(t − t )] dt , or
ω dt
dt
0
t t t df df
xc (t) = f (t ) cos [ω(t − t )] −
cos [ω(t − t )] dt = f (t) −
cos [ω(t − t )] dt
dt
dt
0
0
0
(since f (0) = 0). Now, for an adiabatic process we want df /dt very small; specifically:
df
ωf (t)
dt
(0 < t ≤ t). Then the integral is negligible compared to f (t), and we have xc (t) ≈ f (t). (Physically,
this says that if you pull on the spring very gently, no fancy oscillations will occur; the mass just moves
along as though attached to a string of fixed length.)
(e) Put xc ≈ f
into Eq. 10.92, using Eq. 10.93:
i
Ψ(x, t) = ψn (x, t)e 2
−(n+ 12 )ωt+mf˙(x−f /2)+ mω
2
t
0
f 2 (t )dt
.
The dynamic phase (Eq. 10.39) is
1
θn (t) = −
0
t
1
mω 2
En (t ) dt = −(n + )ωt +
2
2
t
f 2 (t ) dt ,
Ψ(x, t) = ψn (x, t)eiθn (t) eiγn (t) ,
so
0
˙
confirming Eq. 10.94, with the geometric phase given (ostensibly) by γn (t) = m
f (x − f /2). But the
eigenfunctions here are real, and hence(Problem 10.5) the geometric phase should be zero. The point is that
˙
(in the adiabatic approximation) f˙ is extremely small (see above), and hence in this limit m
f (x−f /2) ≈ 0
(at least, in the only region of x where ψn (x, t) is nonzero).
Problem 10.10
(a)
ċm = −
j
∂ψn iγn i(θn −θm )
δjn eiγn ψm |ψ˙j ei(θj −θm ) = −ψm |
e e
∂t
cm (t) = cm (0) −
t
ψm |
0
⇒
∂ψn iγn i(θn −θm )
e e
dt .
∂t
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CHAPTER 10. THE ADIABATIC APPROXIMATION
267
(b) From Problem 10.9:
ψn (x, t) = ψn (x − f ) = ψn (u),
where
u ≡ x − f,
and ψn (u) is the nth state of the ordinary harmonic oscillator;
But
p̂ =
∂
,
i ∂u
m|p|n = i
ψm |
so
ψm |
i
∂ψn
= − f˙m|p|n,
∂t
∂ψn ∂u
∂ψn
∂ψn
=
= −f˙
.
∂t
∂u ∂t
∂u
where (from Problem 3.33):
√
mω √
mδn,m−1 − nδm,n−1 . Thus:
2
√
mω √
mδn,m−1 − nδm,n−1 .
2
∂ψn
= f˙
∂t
Evidently transitions occur only to the immediately adjacent states, n ± 1, and
(1) m = n + 1 :
t
cn+1 = −
f˙
0
mω √
n + 1 eiγn ei(θn −θn+1 ) dt .
2
But γn = 0, because the eigenfunctions are real (Problem 10.5), and (Eq. 10.39)
1
1
1
1
θn = − (n + )ωt =⇒ θn − θn+1 = −(n + ) + (n + 1 + ) ωt = ωt.
2
2
2
So
cn+1 = −
mω √
n+1
2
t
f˙eiωt dt .
0
(2) m = n − 1:
t
cn−1 = −
0
−f˙
mω √
n eiγn ei(θn −θn−1 ) dt ;
2
1
1
θn − θn−1 = −(n + ) + (n − 1 + ) ωt = −ωt.
2
2
cn−1 =
mω √
n
2
t
f˙e−iωt dt .
0
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268
CHAPTER 11. SCATTERING
Chapter 11
Scattering
Problem 11.1
(a)
q1
b
θ
φ
r
q2
Conservation of energy:
E=
1
m(ṙ + r2 φ̇2 ) + V (r),
2
Conservation of angular momentum: J = mr2 φ̇. So
ṙ2 +
du
dφ
φ̇ =
V (r) =
2
=
1
− 2
u
2m
(E − V ) − u2 ;
J2
du J 2
J du
u =−
. Then:
dφ m
m dφ
du
=
dφ
q1 q2 1
.
4π9 r
J
.
mr2
J2
2
= (E − V ). We want r as a function of φ (not t).
2
2
m r
m
dr
dr du dφ
ṙ =
=
=
dt
du dφ dt
where
2m
(E − V ) − u2 ;
J2
Also, let u ≡ 1/r.
J du
−
m dφ
dφ = 2
+
Then
J2 2
2
u = (E − V ), or
m2
m
du
du
=
,
2m
I(u)
2
J 2 (E − V ) − u
where
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CHAPTER 11. SCATTERING
I(u) ≡
269
2m
(E − V ) − u2 . Now, the particle q1 starts out at r = ∞ (u = 0), φ = 0, and the point
J2
of closest approach is rmin (umax ), Φ :
umax
Φ=
0
du
√ . It now swings through an equal angle Φ
I
on the way out, so
Φ + Φ + θ = π,
or
θ = π − 2Φ.
umax
θ =π−2
0
du
.
I(u)
So far this is general ; now we put in the specific potential:
2mE
2m q1 q2
− 2
u − u2 = (u2 − u)(u − u1 ), where u1 and u2 are the two roots.
J2
J 4π90
(Since du/dφ = I(u), umax is one of the roots; setting u2 > u1 , umax = u2 .)
I(u) =
u2
θ =π−2
du
= π + 2 sin−1
u
−2u + u1 + u2 2
u2 − u1
0
(u2 − u)(u − u1 )
u1 + u2
−1
−1
= π + 2 sin (−1) − sin
u2 − u1
π
u
u1 + u2
+
u
1
2
−1
−1
= π + 2 − − sin
= −2 sin
.
2
u2 − u1
u2 − u1
0
Now J = mvb, E = 12 mv 2 , where v is the incoming velocity, so J 2 = m2 b2 (2E/m) = 2mb2 E, and hence
2m/J 2 = 1/b2 E. So
1
1 1 q1 q2
q1 q2
A
1
I(u) = 2 − 2
, so
− I(u) = u2 + 2 u − 2 .
u − u2 . Let A ≡
b
b
E 4π90
4π90 E
b
b
To get the roots:
Thus
A
1
1
A
2
u + 2 u − 2 = 0 =⇒ u =
− 2±
b
b
2
b
2
A
2b
u2 = 2 −1 + 1 +
,
2b
A
−1
θ = 2 sin
2
,
or
1 + (2b/A)
2b
A
2
=
cos2 (θ/2)
1 − sin2 (θ/2)
=
;
sin2 (θ/2)
sin2 (θ/2)
1
2
A
2b
.
= 2 −1 ± 1 +
2b
A
2
A
2b
u1 = 2 −1 − 1 +
;
2b
A
1
4
A2
+ 2
4
b
b
2
1 + (2b/A)
θ
= sin
;
2
2b
= cot(θ/2),
A
or
1+
b=
2b
A
u1 + u2
=
u2 − u1
2
=
−1
2
1 + (2b/A)
1
;
sin (θ/2)
2
q1 q2
cot(θ/2).
8π90 E
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270
CHAPTER 11. SCATTERING
(b)
b
D(θ) =
sin θ
=
(c)
db 1
. Here db = q1 q2
−
.
dθ dθ
8π90 E
2 sin2 (θ/2)
1
q1 q2 cos(θ/2) q1 q2
1
=
2 sin(θ/2) cos(θ/2) 8π90 E sin(θ/2) 8π90 E 2 sin2 (θ/2)
σ=
D(θ) sin θ dθ dφ = 2π
q1 q2
8π90 E
2 π
0
q1 q2
16π90 E sin2 (θ/2)
2
.
sin θ
dθ.
sin4 (θ/2)
This integral does not converge,
for near θ =
5 0 (and again near π) we have sin θ ≈ θ, sin(θ/2) ≈ θ/2, so
5
the integral goes like 16 0 θ−3 dθ = − 8θ−2 0 → ∞.
Problem 11.2
r
θ
x
Two dimensions:
eikr
ψ(r, θ) ≈ A eikx + f (θ) √ .
r
One dimension:
ψ(x) ≈ A eikx + f (x/|x|)e−ikx .
Problem 11.3
Multiply Eq. 11.32 by Pl (cos θ) sin θ dθ and integrate from 0 to π, exploiting the orthogonality of the Legendre polynomials (Eq. 4.34)—which, with the change of variables x ≡ cos θ, says
π
2
Pl (cos θ)Pl (cos θ) sin θ dθ =
δll .
2l + 1
0
The delta function collapses the sum, and we get
(1)
2il jl (ka) + ikal hl (ka) = 0,
and hence (dropping the primes)
al = −
jl (ka)
(1)
ikhl (ka)
. QED
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CHAPTER 11. SCATTERING
271
Problem 11.4
Keeping only the l = 0 terms, Eq. 11.29 says that in the exterior region:
eikr
eikr
sin(kr)
sin(kr)
(1)
ψ ≈ A j0 (kr) + ika0 h0 (kr) P0 (cos θ) = A
+ ika0 −i
=A
+ a0
kr
kr
kr
r
(r > a).
In the internal region Eq. 11.18 (with nl eliminated because it blows up at the origin) yields
ψ(r) ≈ bj0 (kr) = b
sin(kr)
kr
(r < a).
The boundary conditions hold independently for each l, as you can check by keeping the summation over l and
exploiting the orthogonality of the Legendre polynomials:
(1) ψ continuous at r = a:
A
sinka
eika
sin ka
+ a0
=b
.
ka
a
ka
(2) ψ discontinuous at r = a: Integrating the radial equation across the delta function gives
2
2
2
2 l(l + 1)
d u
−
u dr ⇒ −
dr
+
αδ(r − a) +
∆u + αu(a) = 0, or
2
2
2m
dr
2m r
2m
Now
u = rR,
so u = R+rR ;
∆u = ∆R+a∆R = a∆R =
∆u =
2mα
u(a).
2
2mα
2mα
β
aR(a), or ∆ψ = 2 ψ(a) = ψ(a).
2
a
✭✭
A b ✟✟✟ β sin(ka)
A b
✭ika
✭a✭
sin ka = b
k cos(ka) + a0 ik 2 eika − 2 ✭sin(ka)
k cos(ka) + ✟
.
−
✭✭+
0 ke
✭
ka
ka
ka
ka2
a
ka
✟
β
A cos(ka) + ia0 keika = b cos(ka) +
sin(ka) .
ka
The indicated terms cancel (by (1)), leaving
Using (1) to eliminate b:
β A cos(ka) + ia0 keika = cot(ka) +
sin(ka) + a0 keika A.
ka
✘ + ia keika = cos(ka)
✘ + β sin(ka) + a k cot(ka)eika + β a0 eika .
✘
✘✘
cos(ka)
0
0
✘
✘✘
ka
a
ia0 keika 1 + i cot(ka) + i
ia0 k(1 + ika) 1 +
a0 = −
aβ
.
1+β
β
β
cos(ka)
1
=
sin(ka). But ka 1, so sin(ka) ≈ ka, and cot(ka) =
≈
.
ka
ka
sin(ka)
ka
i
(1 + β) = β;
ka
ia0 k 1 +
Equation 11.25 ⇒ f (θ) ≈ a0 = −
Equation 11.27
i
i
(1 + β) + ika − 1 − β ≈ ia0 k
(1 + β) = β.
ka
ka
⇒ σ = 4πD = 4π
aβ
1+β
aβ
.
1+β
Equation 11.14 ⇒ D = |f |2 =
aβ
1+β
2
2
.
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272
CHAPTER 11. SCATTERING
Problem 11.5
(a) In the region to the left
ψ(x) = Aeikx + B −ikx
(x ≤ −a).
In the region −a < x < 0, the Schrödinger equation gives
−
where k =
2m(E + V0 )/.
h2 d 2 ψ
− V0 ψ = Eψ
2m dx2
⇒
d2 ψ
= −k ψ
dx2
The general solution is
ψ = C sin(k x) + D cos(k x)
But ψ(0) = 0 implies D = 0, so
ψ(x) = C sin(k x)
(−a ≤ x ≤ 0).
The continuity of ψ(x) and ψ (x) at x = −a says
Ae−ika + Beika = −C sin(k a),
Divide and solve for B:
ikAe−ika − ikB ika = k C cos(k a).
ikAe−ika − ikBeika
= −k cot(k a),
Ae−ika + Beika
ikAe−ika − ikBeika = −Ae−ika k cot(k a) − Beika k cot(k a),
Beika [−ik + k cot(k a)] = Ae−ika [−ik − k cot(k a)] .
B = Ae−2ika
(b)
|B|2 = |A|2
k − ik cot(k a)
.
k + ik cot(k a)
k − ik cot(k a)
k + ik cot(k a)
·
= |A|2 . k + ik cot(k a)
k − ik cot(k a)
(c) From part (a) the wave function for x < −a is
ψ(x) = Aeikx + Ae−2ika
k − ik cot(k a) −ikx
e
.
k + ik cot(k a)
But by definition of the phase shift (Eq. 11.40)
ψ(x) = A eikx − ei(2δ−kx) .
so
k − ik cot(k a)
= −e2iδ .
k + ik cot(k a)
√
This is exact. For a very deep well, E V0 , k = 2mE/ 2m(E + V0 )/ = k , so
e−2ika
e−2ika
−ik cot(k a)
= −e2iδ ;
ik cot(k a)
e−2ika = e2iδ ;
δ = −ka.
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CHAPTER 11. SCATTERING
273
Problem 11.6
1 iδl
e sin δl ,
k
From Eq. 11.46,
al =
and Eq. 11.33,
But (Eq. 11.19)
hl (x) = jl (x) + inl (x),
(1)
eiδl sin δl = i
al = i
jl (ka)
(1)
khl (ka)
, it follows that eiδl sin δl = i
jl (ka)
(1)
hl (ka)
.
so
1
1 − i(n/j)
jl (ka)
(n/j) + i
=i
=i
=
,
jl (x) + inl (x)
1 + i(n/j)
1 + (n/j)2
1 + (n/j)2
(writing (n/j) as shorthand for nl (ka)/jl (ka)). Equating the real and imaginary parts:
cos δl sin δl =
(n/j)
;
1 + (n/j)2
sin2 δl =
1
.
1 + (n/j)2
Dividing the second by the first, I conclude that
tan δl =
1
,
(n/j)
or
δl = tan−1
jl (ka)
.
nl (ka)
Problem 11.7
r > a : u(r) = A sin(kr + δ);
r < a : u(r) = B sin kr + D cos kr = B sin kr,
because u(0) = 0 =⇒ D = 0.
Continuity at r = a =⇒ B sin(ka) = A sin(ka + δ) =⇒ B = A
sin(ka + δ)
sin(ka + δ)
. So u(r) = A
sin kr.
sin(ka)
sin(ka)
From Problem 11.4,
du sin(ka + δ)
β
β
∆
k cos(ka) = A sin(ka + δ).
= u(a) ⇒ Ak cos(ka + δ) − A
dr r=a
a
sin(ka)
a
cos(ka + δ) −
sin(ka + δ)
β
cos(ka) =
sin(ka + δ),
sin(ka)
ka
sin(ka) cos(ka + δ) − sin(ka + δ) cos(ka) =
sin(ka − ka − δ) =
− sin δ = β
β
sin(ka) [sin(ka) cos δ + cos(ka) sin δ] ,
ka
sin2 (ka)
[cos δ + cot(ka) sin δ] ;
ka
cot δ = − cot(ka) −
β
sin(ka + δ) sin(ka),
ka
ka
;
β sin2 (ka)
−1 = β
sin2 (ka)
[cot δ + cot(ka)] .
ka
cot δ = − cot(ka) +
ka
.
β sin2 (ka)
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274
CHAPTER 11. SCATTERING
Problem 11.8
eikr
1
=⇒ ∇G = −
4πr
4π
G=−
1
1 ikr
∇e + eikr ∇
r
r
=⇒
1
1
1
1 2 ikr
ikr
ikr 2
∇ G = ∇ · (∇G) = −
2 ∇
· (∇e ) + ∇ (e ) + e ∇
.
4π
r
r
r
2
But ∇
1
1
= − 2 r̂;
r
r
∇(eikr ) = ikeikr r̂;
(see reference in footnote 12) =⇒ ∇ e
2 ikr
1
∇
= −4πδ 3 (r). So
r
2
But
∇2 eikr = ik∇ · (eikr r̂) = ik
1 d 2 ikr
(r e )
r2 dr
ik
= 2 (2reikr + ikr2 eikr ) = ikeikr
r
2
+ ik ;
r
1
1
1 ikr 2
ikr
2 − 2 r̂ · ike r̂ + ike
+ ik − 4πeikr δ 3 (r) .
∇ G=−
4π
r
r
r
2
eikr δ 3 (r) = δ 3 (r), so
∇2 G = δ 3 (r) −
2ik 2ik k 2
eikr
1 ikr
− 2 + 2 −
e
= δ 3 (r) + k 2
= δ 3 (r) − k 2 G.
4π
r
r
r
4πr
(∇2 + k 2 )G = δ 3 (r). QED
Therefore
Problem 11.9
ψ=√
1
πa3
−r/a
e
;
e2
2 1
V =−
=−
4π90 r
ma r
√
(Eq. 4.72);
k=i
−2mE
i
= .
a
In this case there is no “incoming” wave, and ψ0 (r) = 0. Our problem is to show that
ik|r−r0 |
m
e
−
V (r0 )ψ(r0 ) d3 r0 = ψ(r).
2
2π
|r − r0 |
We proceed to evaluate the left side (call it I):
−|r−r0 |/a
m e
2
1
1 −r0 /a 3
√
I= −
e
d r0
−
3
2π2
ma
|r
−
r
|
r
0
0
πa
√
− r2 +r2 −2rr0 cos θ/a −r0 /a
0
e
e
1
1
√
=
r02 sin θ dr0 dθ dφ.
2
2
2πa πa3
r + r0 − 2rr0 cos θ r0
(I have set the z0 axis along the—fixed—direction r, for convenience.) Doing the φ integral (2π):
√
∞
π − r 2 +r02 −2rr0 cos θ/a
1
e
−r0 /a
I= √
sin θ dθ dr0 . The θ integral is
r0 e
r2 + r02 − 2rr0 cos θ
a πa3 0
0
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CHAPTER 11. SCATTERING
π
0
275
√2 2
e− r +r0 −2rr0 cos θ/a
a −(r+r0 )/a
a −√r2 +r02 −2rr0 cos θ/a π
−|r−r0 |/a
e
.
e
=
−
−
e
sin
θ
dθ
=
−
rr0
rr0
0
r2 + r02 − 2rr0 cos θ
1
I=− √
r πa3
1
=− √
r πa3
1
=− √
r πa3
1
=− √
r πa3
e−r0 /a e−(r0 +r)/a − e−|r0 −r|/a dr0
0
∞
r
e−r/a
e−2r0 /a dr0 − e−r/a
dr − er/a
∞
0
0
e−2r0 /a dr0
r
a
∞ e
− e−2r0 /a (r) − e
−e
2
2
r
a
1
a r/a −2r/a −r/a
−r/a
−r/a
=√
− re
− e e
e
= ψ(r). QED
e
2
2
πa3
−r/a
a
∞
−r/a
r/a
Problem 11.10
For the potential in Eq. 11.81, Eq. 11.88 =⇒
2m
f (θ) = − 2 V0
κ
0
a
2mV0
r sin(κr) dr = − 2
κ
For low-energy scattering (ka 1):
where (Eq. 11.89) κ = 2k sin(θ/2).
sin(κa) ≈ κa −
f (θ) ≈ −
1
(κa)3 ;
3!
a
1
r
2mV0
sin(κr) − cos(κr) = − 2 3 [sin(κa) − κa cos(κa)] ,
κ2
κ
κ
0
1
cos(κa) = 1 − (κa)2 ;
2
so
2mV0
1
1
2 mV0 a3
3
3
κa
−
=
−
κa
+
−
,
(κa)
(κa)
2 κ3
6
2
3 2
in agreement with Eq. 11.82.
Problem 11.11
sin(κr) =
=
So
1
2i
1 iκr
e − e−iκr ,
2i
so
∞
e−µr sin(κr) dr =
0
1
2i
∞
e−(µ−iκ)r − e−(µ+iκ)r dr
0
∞
κ
1
e−(µ−iκ)r
e−(µ+iκ)r 1
1
1 µ + iκ − µ + iκ
= 2
=
.
−
−
=
−(µ − iκ) −(µ + iκ) 0
2i µ − iκ µ + iκ
2i
µ2 + κ2
µ + κ2
f (θ) = −
2mβ
2mβ
κ
=− 2 2
. QED
2 κ µ2 + κ2
(µ + κ2 )
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276
CHAPTER 11. SCATTERING
Problem 11.12
Equation 11.91 =⇒ D(θ) = |f (θ)|2 =
σ=
D(θ) sin θ dθ dφ = 2π
2k
sin(θ/2) ≡ x,
µ
Let
σ = 2π
2
2mβ
2
so
2mβ
2
2
2mβ
2
1
µ4
2 sin(θ/2) =
2
1
,
(µ2 + κ2 )2
π
0
where Eq. 11.89 ⇒ κ = 2k sin(θ/2).
1
2 2 sin(θ/2) cos(θ/2) dθ.
2
1 + (2k/µ) sin2 (θ/2)
µ
x,
k
and
cos(θ/2) dθ =
µ
dx. Then
k
1 µ 2 x1
x
θ = 0 =⇒ x = x0 = 0,
dx.
The
limits
are
2 2
θ = π =⇒ x = x1 = 2k/µ.
µ4 k
x0 (1 + x )
So
2k/µ
2
2
2mβ
1
1
1
1
1
2mβ
σ = 2π
−
1−
=π
2
2
2
2
2
(µk)
2 (1 + x ) 0
(µk)
1 + (2k/µ)2
2
2
2mβ
1
1
1
4(k/µ)2
4mβ
2mE
=π
=
π
. But k 2 =
, so
2
(µk)2 1 + 4k 2 /µ2
2
µ2 µ2 + 4k 2
2
σ=π
4mβ
µ
2
(µk)2
1
.
+ 8mE
Problem 11.13
(a)
V (r) = αδ(r − a). Eq. 11.80 =⇒ f = −
2mα
f = − 2 a2 ;
D = |f | =
2
2mα 2
a
2
m
2π2
V (r) d3 r = −
2
m
α4π
2π2
;
σ = 4πD = π
4mα 2
a
2
∞
δ(r − a)r2 dr.
0
2
.
(b)
Eq. 11.88 =⇒ f = −
2m
α
2 κ
0
∞
rδ(r − a) sin(κr) dr = −
2mα
a sin(κa)
2 κ
(κ = 2k sin(θ/2)).
(c) Note first that (b) reduces to (a) in the low-energy regime (ka 1 =⇒ κa 1). Since Problem 11.4
was also for low energy, what we must confirm is that Problem 11.4 reproduces (a) in the regime for
which the Born approximation holds. Inspection shows that the answer to Problem 11.4 does reduce to
f = −2mαa2 /2 when β 1, which is to say when f /a 1. This is the appropriate condition, since
(Eq. 11.12) f /a is a measure of the relative size of the scattered wave, in the interaction region.
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CHAPTER 11. SCATTERING
277
Problem 11.14
F=
1 q1 q2
r̂;
4π90 r2
F⊥ =
1 q1 q2
cos φ;
4π90 r2
b
cos φ = ,
r
so
1 q1 q2 b
;
4π90 r3
F⊥ =
dt =
dx
.
v
F
φ
q1
r
∞
−∞
F⊥ dt =
1 q1 q2 b
4π90 v
dx
=2
2
(x + b2 )3/2
tan θ =
∞
0
∞
−∞
dx
. But
(x2 + b2 )3/2
∞
dx
2x
= 2,
= √
2
2
3/2
2
2
2
b2
(x + b )
b x + b 0
I⊥
1
q1 q2
q1 q2 1
=
.
=
1
2
mv
4π90 b( 2 mv )
4π90 bE
1
q1 q2
b=
=
4π90 E tan θ
b
x
q2
I⊥ =
q1 q2
8π90 E
r
θ = tan−1
so
I⊥ =
1 2q1 q2
.
4π90 bv
q1 q2
.
4π90 bE
(2 cot θ).
The exact answer is the same, only with cot(θ/2) in place of 2 cot θ. So I must show that cot(θ/2) ≈ 2 cot θ,
for small θ (that’s the regime in which the impulse approximation should work). Well:
cot(θ/2) =
cos(θ/2)
1
2
≈
= ,
sin(θ/2)
θ/2
θ
for small θ,
while 2 cot θ = 2
cos θ
1
≈ 2 . So it works.
sin θ
θ
Problem 11.15
First let’s set up the general formalism. From Eq. 11.101:
ψ(r) = ψ0 (r) + g(r − r0 )V (r0 )ψ0 (r0 ) d3 r0 + g(r − r0 )V (r0 )
g(r0 − r1 )V (r1 )ψ0 (r1 ) d3 r1 d3 r0 + · · ·
Put in ψ0 (r) = Aeikz ,
mA
2π2
m 2 +
A
2π2
ψ(r) = Aeikz −
g(r) = −
m eikr
:
2π2 r
eik|r−r0 |
V (r0 )eikz0 d3 r0
|r − r0 |
ik|r0 −r1 |
eik|r−r0 |
e
V (r0 )
V (r1 )eikz1 d3 r1 d3 r0 .
|r − r0 |
|r0 − r1 |
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278
CHAPTER 11. SCATTERING
In the scattering region r r0 , Eq. 11.73 =⇒
eikr −ik·r0
eik|r−r0 |
≈
e
,
|r − r0 |
r
with
k ≡ kr̂,
so
m eikr
ψ(r) = A eikz −
e−ik·r0 V (r0 )eikz0 d3 r0
2π2 r
ik|r0 −r1 |
m 2 eikr e
−ik·r0
ikz1 3
3
e
V (r1 )e
V (r0 )
d r1 d r0
2π2
r
|r0 − r1 |
f (θ, φ) = −
m
2π2
ei(k −k)·r V (r) d3 r +
ik|r−r0 |
m 2 e
−ik·r
e
V
(r)
V (r0 )eikz0 d3 r0 d3 r.
2
2π
|r − r0 |
I simplified the subscripts, since there is no longer any possible ambiguity. For low-energy scattering we drop
the exponentials (see p. 414):
m
f (θ, φ) ≈ −
2π2
m 2 1
V (r) d r +
V (r)
V (r0 ) d3 r0 d3 r.
2π2
|r − r0 |
3
Now apply this to the potential in Eq. 11.81:
a
1
1
V (r0 ) d3 r0 = V0
r02 sin θ0 dr0 dθ0 dφ0 .
|r − r0 |
|r
−
r
|
0
0
Orient the z0 axis along r, so |r − r0 | = r2 + r02 − 2rr0 cos θ0 .
π
a
1
1
3
2
V (r0 ) d r0 = V0 2π
r0
sin θ0 dθ0 dr0 . But
2
2
|r − r0 |
r + r0 − 2rr0 cos θ0
0
0
π
1
1
sin θ0 dθ0 =
2
2
rr0
r + r0 − 2rr0 cos θ0
0
r2
+
r02
π
1
2/r, r0 < r;
− 2rr0 cos θ0 =
[(r0 + r) − |r0 − r|] =
2/r0 , r0 > r.
rr
0
0
Here r < a (from the “outer” integral), so
1
1
V (r0 ) d3 r0 = 4πV0
|r − r0 |
r
V (r)
r
r02
0
a
r0 dr0 = 4πV0
dr0 +
r
1
V (r0 ) d3 r0 d3 r = V0 (2πV0 )4π
|r − r0 |
f (θ) = −
a
1 r3
1 2
1 2
2
2
+ (a − r ) = 2πV0 a − r .
r 3
2
3
a3
1 2 2
1 a5
32 2 2 5
a − r r dr = 8π 2 V02 a2 −
π V0 a .
=
3
3
3 5
15
2
0
m
4 3 m 2 32 2 2 5
V
πa +
π V0 a = −
0
2π2 3
2π2
15
2mV0 a3
32
1−
4
5
mV0 a2
2
.
Problem 11.16
d2
2
+ k G(x) = δ(x)
dx2
1
(analog to Eq. 11.52). G(x) = √
2π
eisx g(s) ds
(analog to Eq. 11.54).
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CHAPTER 11. SCATTERING
279
d2
1
1
1
2
2
2
isx
+k G= √
(−s + k )g(s)e ds = δ(x) =
.
eisx ds =⇒ g(s) = √
dx2
2π
2π
2π(k 2 − s2 )
G(x) =
1
2π
G(x) = −
1
2π
1
G(x) = +
2π
∞
−∞
eisx
ds. Skirt the poles as in Fig. 11.10.
− s2
k2
? ? eisx
s+k
eisx
s−k
1
1
ds = − 2πi
s−k
2π
1
1
ds =
2πi
s+k
2π
In either case, then, G(x) = −
ψ(x) = G(x − x0 )
i ik|x|
.
e
2k
For x > 0, close above:
eikx
eisx =
−i
. For x < 0, close below:
s + k s=k
2k
eisx e−ikx
.
=
−i
s − k s=−k
2k
(Analog to Eq. 11.65.)
2m
i 2m
V (x0 )ψ(x0 ) dx0 = −
2
2k 2
eik|x−x0 | V (x0 )ψ(x0 ) dx0 ,
plus any solution ψ0 (x) to the homogeneous Schrödinger equation:
2
d
2
ψ0 (x) = 0. So:
+
k
dx2
im
ψ(x) = ψ0 (x) − 2
k
∞
−∞
eik|x−x0 | V (x0 )ψ(x0 ) dx0 .
Problem 11.17
For the Born approximation let ψ0 (x) = Aeikx , and ψ(x) ≈ Aeikx .
im ∞ ik|x−x0 |
ψ(x) ≈ A eikx − 2
e
V (x0 )eikx0 dx0
k −∞
im x ik(x−x0 )
im ∞ ik(x0 −x)
= A eikx − 2
e
V (x0 )eikx0 dx0 − 2
e
V (x0 )eikx0 dx0 .
k −∞
k x
ψ(x) = A eikx −
im ikx
e
2 k
x
−∞
V (x0 ) dx0 −
im −ikx
e
2 k
∞
e2ikx0 V (x0 ) dx0 .
x
Now assume V (x) is localized; for large positive x, the third term is zero, and
im ∞
ikx
ψ(x) = Ae
V (x0 ) dx0 . This is the transmitted wave.
1− 2
k −∞
For large negative x the middle term is zero:
im −ikx ∞ 2ikx0
ikx
ψ(x) = A e − 2 e
e
V (x0 )dx0 .
k
−∞
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280
CHAPTER 11. SCATTERING
Evidently the first term is the incident wave and the second the reflected wave:
2
m 2 ∞
2ikx
.
R=
e
V
(x)
dx
2
k
−∞
If you try in the same spirit to calculate the transmission coefficient, you get
2
2
m 2 ∞
im ∞
T = 1 − 2
V (x)dx = 1 + 2
V (x)dx ,
k −∞
k
−∞
which is nonsense (greater than 1). The first Born approximation gets R right, but all you can say to this order
is T ≈ 1 (you would do better using T = 1 − R).
Problem 11.18
Delta function:
V (x) = −αδ(x).
∞
−∞
e2ikx V (x) dx = −α,
so
R=
mα 2
2 k
,
or, in terms of energy (k 2 = 2mE/2 ):
R=
m2 α2
mα2
=
;
2mE2
22 E
mα2
.
22 E
T =1−R= 1−
The exact answer (Eq. 2.141) is
1
mα2
22 E
≈1−
mα2
,
22 E
so they agree provided E mα
.
22
1+
−V0 (−a < x < a)
Finite square well: V (x) =
.
0
(otherwise)
a
∞
a
e2ikx V0
V0 e2ika − e−2ika
2ikx
2ikx
= − sin(2ka).
e
V (x) dx = −V0
e
dx = −V0
=−k
2ik
2i
k
−∞
−a
−a
m 2 V
0
So R = 2
sin(2ka)
k
k
2
.
V0
T =1−
sin
2E
2a √
2mE
2
.
If E V0 , the exact answer (Eq. 2.169) becomes
2
V0
V0
2a √
2a √
−1
T ≈1+
2mE
=⇒ T ≈ 1 −
2mE
sin
sin
2E
2E
2
,
so they agree provided E V0 .
Problem 11.19
The Legendre polynomials satisfy Pl (1) = 1 (see footnote 30, p. 124), so Eq. 11.47 ⇒
f (0) =
∞
1
(2l + 1)eiδl sin δl . Therefore
k
Im[f (0)] =
l=0
∞
1
(2l + 1) sin2 δl ,
k
l=0
and hence (Eq. 11.48):
σ=
4π
Im[f (0)]. QED
k
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CHAPTER 11. SCATTERING
281
Problem 11.20
Using Eq. 11.88 and integration by parts:
2
2
2m ∞
2mA ∞ d
1
f (θ) = − 2
rAe−µr sin(κr) dr = − 2
− e−µr sin(κr) dr
κ 0
κ 0 dr
2µ
∞ ∞
2mA
−µr 2
−µr 2 d
−
=
sin(κr)
e
e
[sin(κr)]
dr
2µ2 κ
dr
0
0
√
∞
mA
π −κ2 /4µ
mA
−µr 2
=
e
cos(κr) dr = − 2
0−κ
√ e
µ2 κ
µ
2 µ
0
√
2
mA π
= − 2 3/2 e−κ /4µ , where κ = 2k sin(θ/2) (Eq. 11.89).
2 µ
From Eq. 11.14, then,
dσ
πm2 A2 −κ2 /2µ
e
,
=
dΩ
44 µ3
and hence
σ =
=
=
=
2
2
dσ
πm2 A2
e−4k sin (θ/2)/2µ sin θ dθ dφ
dΩ =
dΩ
44 µ3
π 2 m2 A2 π −2k2 sin2 (θ/2)/µ
e
sin θ dθ; write sin θ = 2 sin(θ/2) cos(θ/2)
24 µ3 0
π 2 m2 A2 1 −2k2 x2 /µ
2π 2 m2 A2 1 −2k2 x2 /µ
e
2x 2 dx =
xe
dx
24 µ3 0
4 µ3
0
1
2π 2 m2 A2 µ −2k2 x2 /µ π 2 m2 A2 −2k2 /µ
−
e
e
=
−
−
1
4 µ3
4k 2
24 µ2 k 2
and let x ≡ sin(θ/2)
0
π 2 m2 A 2 −2k2 /µ
=
1
−
e
.
24 µ2 k 2
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282
CHAPTER 12. AFTERWORD
Chapter 12
Afterword
Problem 12.1
Suppose, on the contrary, that
α|φa (1)|φb (2) + β|φb (1)|φa (2) = |ψr (1)|ψs (2),
for some one-particle states |ψr and |ψs . Because |φa and |φb constitute a complete set of one-particle states
(this is a two-level system), any other one-particle state can be expressed as a linear combination of them. In
particular,
|ψr = A|φa + B|φb , and |ψs = C|φa + D|φb ,
for some complex numbers A, B, C, and D. Thus
α|φa (1)|φb (2) + β|φb (1)|φa (2) = A|φa (1) + B|φb (1) C|φa (2) + D|φb (2)
= AC|φa (1)|φa (2) + AD|φa (1)|φb (2) + BC|φb (1)|φa (2) + BD|φb (1)|φb (2).
(i)
(ii)
(iii)
(iv)
Take
Take
Take
Take
the
the
the
the
inner
inner
inner
inner
product
product
product
product
with
with
with
with
φa (1)|φb (2)|:
φa (1)|φa (2)|:
φb (1)|φa (2)|:
φb (1)|φb (2)|:
α = AD.
0 = AC.
β = BC.
0 = BD.
(ii) ⇒ either A = 0 or C = 0. But if A = 0, then (i) ⇒ α = 0, which is excluded by assumption, whereas if
C = 0, then (iii) ⇒ β = 0, which is likewise excluded. Conclusion: It is impossible to express this state as a
product of one-particle states. QED
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APPENDIX. LINEAR ALGEBRA
283
Appendix A
Linear Algebra
Problem A.1
(a) Yes; two-dimensional.
(b) No; the sum of two such vectors has az = 2, and is not in the subset. Also, the null vector (0,0,0) is not
in the subset.
(c) Yes; one-dimensional.
Problem A.2
(a) Yes;
1, x, x2 , . . . , xN −1 is a convenient basis. Dimension: N.
(b) Yes; 1, x2 , x4 , . . . .
(c) No.
Dimension N/2 (if N is even) or (N + 1)/2 (if N is odd).
The sum of two such “vectors” is not in the space.
(d) Yes; (x − 1), (x − 1)2 , (x − 1)3 , . . . , (x − 1)N −1 .
(e) No.
Dimension: N − 1.
The sum of two such “vectors” would have value 2 at x = 0.
Problem A.3
Suppose |α = a1 |e1 + a2 |e2 + · · · an |en and |α = b1 |e1 + b2 |e2 + · · · + bn |en . Subtract: 0 = (a1 − b1 )|e1 +
(a2 − b2 )|e2 + · · · + (an − bn )|en . Suppose aj = bj for some j; then we can divide by (aj − bj ) to get:
|ej = −
(a1 − b1 )
(a2 − b2 )
(an − bn )
|e1 −
|e2 − · · · − 0|ej − · · · −
|en ,
(aj − bj )
(aj − bj )
(aj − bj )
so |ej is linearly dependent on the others, and hence {|ej } is not a basis. If {|ej } is a basis, therefore, the
components must all be equal (a1 = b1 , a2 = b2 , . . . , an = bn ). QED
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284
APPENDIX. LINEAR ALGEBRA
Problem A.4
(i)
e1 |e1 = |1 + i|2 + 1 + |i|2 = (1 + i)(1 − i) + 1 + (i)(−i) = 1 + 1 + 1 + 1 = 4. "e1 " = 2.
|e1 =
1
1
i
(1 + i) î + ĵ + k̂.
2
2
2
(ii)
1
1
e1 |e2 = (1 − i)(i) + (3) +
2
2
−i
2
1=
1
(i + 1 + 3 − i) = 2.
2
|e2 ≡ |e2 − e1 |e2 |e1 = (i − 1 − i)î + (3 − 1)ĵ + (1 − i)k̂ = (−1)î + (2)ĵ + (1 − i)k̂.
e2 |e2 = 1 + 4 + 2 = 7.
1
|e2 = √ [−î + 2ĵ + (1 − i)k̂].
7
(iii)
e1 |e3 =
1
28 = 14;
2
√
2
e2 |e3 = √ 28 = 8 7.
7
|e3 = |e3 − e1 |e3 |e1 − e2 |e3 |e2 = |e3 − 7|e1 − 8|e2 = (0 − 7 − 7i + 8)î + (28 − 7 − 16)ĵ + (0 − 7i − 8 + 8i)k̂ = (1 − 7i)î + 5ĵ + (−8 + i)k̂.
"e3 "2 = 1 + 49 + 25 + 64 + 1 = 140.
1
|e3 = √ [(1 − 7i)î + 5ĵ + (−8 + i)k̂].
2 35
Problem A.5
From Eq. A.21:
α|β
α|β
γ|γ = γ| |β −
|α = γ|β −
γ|α. From Eq. A.19:
α|α
α|α
α|β
α|β
|α|β|2
γ|β = β|γ = β| |β −
|α = β|β −
β|α = β|β −
, which is real.
α|α
α|α
α|α
∗
α|β
α|β
γ|α∗ = α|γ = α| |β −
|α = α|β −
α|α = 0. γ|α = 0. So (Eq. A.20) :
α|α
α|α
γ|γ = β|β −
|α|β|2
≥ 0, and hence |α|β|2 ≤ α|αβ|β. QED
α|α
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APPENDIX. LINEAR ALGEBRA
285
Problem A.6
α|β = (1 − i)(4 − i) + (1)(0) + (−i)(2 − 2i) = 4 − 5i − 1 − 2i − 2 = 1 − 7i;
α|α = 1 + 1 + 1 + 1 = 4;
β|β = 16 + 1 + 4 + 4 = 25;
cos θ =
β|α = 1 + 7i;
1 + 49
1
=√ ;
4 · 25
2
θ = 45◦ .
Problem A.7
Let
|γ ≡ |α + |β;
γ|γ = γ|α + γ|β.
γ|α∗ = α|γ = α|α + α|β =⇒ γ|α = α|α + β|α.
γ|β∗ = β|γ = β|α + β|β =⇒ γ|β = α|β + β|β.
"(|α + |β)"2 = γ|γ = α|α + β|β + α|β + β|α.
But α|β + β|α = 2Re(α|β) ≤ 2|α|β| ≤ 2
α|αβ|β (by Schwarz inequality), so
"(|α + |β)" ≤ "α" + "β" + 2"α""β" = ("α" + "β") , and hence "(|α + |β)" ≤ "α" + "β". QED
2
2
2
2
Problem A.8
(a)
1
1
0
2
1
3 .
3i (3 − 2i) 4
−3 (1 + 3i)
3i
(−2 + 0 − 1) (0 + 1 + 3i) (i + 0 + 2i)
(4 + 0 + 3i) (0 + 0 + 9) (−2i + 0 + 6) = (4 + 3i)
9
(6 − 2i) .
6i
(6 − 2i)
6
(4i + 0 + 2i) (0 − 2i + 6) (2 + 0 + 4)
(b)
(c)
(−2 + 0 + 2) (2 + 0 − 2) (2i + 0 − 2i)
0
0 0
0 3 .
BA = (0 + 2 + 0) (0 + 0 + 0) (0 + 3 + 0) = 2
(−i + 6 + 4i) (i + 0 − 4i) (−1 + 9 + 4)
(6 + 3i) −3i 12
−3
(1 + 3i)
3i
9
(3 − 2i) .
[A, B] = AB − BA = (2 + 3i)
(−6 + 3i) (6 + i)
−6
(d)
−1 2 2i
1 0 −2i .
i 3 2
(e)
−1 1 −i
2 0 3 .
−2i 2i 2
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286
APPENDIX. LINEAR ALGEBRA
(f )
−1 2 −2i
1 0 2i .
−i 3 2
(g) 4 + 0 + 0 − 1 − 0 − 0 = 3.
(h)
B−1
1
= C̃;
3
|13 02 |
C = − 03 −i
2
0 −i 1 0
BB−1
0 0|
−
| 0i 13 |
2 0 −i
i 2
2| −i
20
i 2 2−i − | i 3 | = −3i 3 −6 .
2
0
i 0 2
− 0 0
|0 1|
(4 + 0 − 1) (−6i + 0 + 6i) (2i + 0 − 2i)
3 0
1
1
(0 + 0 + 0) (0 + 3 + 0)
(0 + 0 + 0) = 0 3
=
3
3
(2i + 0 − 2i) (3 + 9 − 12) (−1 + 0 + 4)
0 0
det A = 0 + 6i + 4 − 0 − 6i − 4 = 0.
No;
B−1
2 −3i i
1
0 3 0 .
=
3
−i −6 2
0
1 0 0
0 = 0 1 0 . 3
0 0 1
A does not have an inverse.
Problem A.9
(a)
−i + 2i + 2i
3i
2i + 0 + 6 = 6 + 2i .
−2 + 4 + 4
6
(b)
2
−i −2i 2 1 − i = −2i − 2i(1 − i) + 0 = −2 − 4i.
0
(c)
2 0 −i
2
4
i 2i 2 0 1 0 1 − i = i 2i 2 1 − i = 4i + 2i(1 − i) + 2(3 − i) = 8 + 4i.
i 3 2
0
3−i
(d)
i
2i (−1 + i)
2i 2 (1 + i) 0 = 4i (−2 + 2i)
2
4 (2 + 2i)
0
0 .
0
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APPENDIX. LINEAR ALGEBRA
287
Problem A.10
(a) S =
1
(T + T̃);
2
A=
1
(T − T̃).
2
(b) R =
1
(T + T∗ );
2
M=
1
(T − T∗ ).
2
(c) H =
1
(T + T† );
2
K=
1
(T − T† ).
2
Problem A.11
@ ki = (ST)ik =
(ST)
n
Sij Tjk =
j=1
n
@ = T̃S̃. QED
T̃kj S̃ji = (T̃S̃)ki ⇒ ST
j=1
@ ∗ = (T̃S̃)∗ = T̃∗ S̃∗ = T† S† . QED
(ST)† = (ST)
(T−1 S−1 )(ST) = T−1 (S−1 S)T = T−1 T = I ⇒ (ST)−1 = T−1 S−1 . QED
U† = U−1 , W† = W−1 ⇒ (WU)† = U† W† = U−1 W−1 = (WU)−1 ⇒ WU is unitary.
H = H† , J = J† ⇒ (HJ)† = J† H† = JH;
the product is hermitian ⇔ this is HJ, i.e. ⇔ [H, J] = 0 (they commute).
(U + W)† = U† + W† = U−1 + W−1 = (U + W)−1 .
?
No; the sum of two unitary matrices is not unitary.
(H + J)† = H† + J† = H + J. Yes; the sum of two hermitian matrices is hermitian.
Problem A.12
U† U = I
(U† U)ik = δik =⇒
=⇒
n
†
Uij
Ujk =
j=1
(j)
Construct the set of n vectors a
i
≡ Uij (a
(j)
a(i)† a(k) =
n
∗
Uji
Ujk = δik .
j=1
is the j-th column of U; its i-th component is Uij ). Then
n
∗
a(i) j a(k) j =
j=1
n
∗
Uji
Ujk = δik ,
j=1
so these vectors are orthonormal. Similarly,
UU† = I
=⇒
(UU† )ik = δik
=⇒
n
†
Uij Ujk
=
j=1
n
∗
Ukj
Uij = δki .
j=1
This time let the vectors b(j) be the rows of U: b(j) i ≡ Uji . Then
(k)† (i)
b
b
=
n
j=1
∗
b(k) j b(i) j
=
n
∗
Ukj
Uij = δki ,
j=1
so the rows are also orthonormal.
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288
APPENDIX. LINEAR ALGEBRA
Problem A.13
H† = H (hermitian) ⇒ det H = det(H† ) = det(H̃∗ ) = (det H̃)∗ = (det H)∗ ⇒ det H is real. U† = U−1 (unitary) ⇒ det(UU† ) = (det U)(det U† ) = (det U)(det Ũ)∗ = | det U|2 = det I = 1, so det U = 1. S̃ = S−1 (orthogonal) ⇒ det(SS̃) = (det S)(det S̃) = (det S)2 = 1, so det S = ±1. Problem A.14
(a)
î = cos θ î + sin θ ĵ;
ĵ = − sin θ î + cos θ ĵ;
cos θ − sin θ 0
k̂ = k̂. Ta = sin θ cos θ 0 .
0
0
1
y
y'
θ
x'
θ
x
z, z'
(b)
î = ĵ;
ĵ = k̂;
k̂ = î.
0 0 1
Tb = 1 0 0 .
0 1 0
y, x'
x, z'
z, y'
(c)
î = î;
ĵ = ĵ;
k̂ = −k̂.
1 0 0
Tc = 0 1 0 .
0 0 −1
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APPENDIX. LINEAR ALGEBRA
(d)
289
cos θ
T̃a Ta = − sin θ
0
010
001
1
T̃b Tb = 0 0 1 1 0 0 = 0
100
010
0
sin θ 0
cos θ − sin θ 0
1 0
cos θ 0 sin θ cos θ 0 = 0 1
0 1
0
0
1
0 0
0 0
1 0 0
1
1 0 . T̃c Tc = 0 1 0 0
0 1
0 0 −1
0
det Ta = cos2 θ + sin2 θ = 1.
det Tb = 1.
0
0 . 1
0 0
1 0 0
1 0 = 0 1 0 . 0 −1
0 0 1
det Tc = -1.
Problem A.15
y
y'
θ
x, x'
θ
z
î = î;
ĵ = cos θ ĵ + sin θ k̂;
z'
1 0
0
Tx (θ) = 0 cos θ − sin θ .
0 sin θ cos θ
k̂ = cos θ k̂ − sin θ ĵ.
y, y'
x'
θ
x
z
θ
z'
î = cos θ î − sin θ k̂;
ĵ = ĵ;
k̂ = cos θ k̂ + sin θ î.
î = ĵ;
ĵ = −î;
k̂ = k̂.
0 −1 0
S = 1 0 0 .
0 0 1
cos θ 0 sin θ
1 0 .
Ty (θ) = 0
− sin θ 0 cos θ
S−1
0 1 0
= −1 0 0 .
0 0 1
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290
APPENDIX. LINEAR ALGEBRA
STx S−1
STy S−1
−1 0
1 0
0 0 0 cos θ
0 1
0 sin θ
−1 0
0
0 0 − cos θ
0 1
− sin θ
−1 0
cos θ 0 sin θ
0 1 0
0 0 0
1 0 −1 0 0
0 1
− sin θ 0 cos θ
0 0 1
−1 0
0 cos θ sin θ
1
0
0
0 0 −1
0
0 = 0 cos θ sin θ = Tx (−θ).
0 1
0 − sin θ cos θ
0 − sin θ cos θ
0
= 1
0
0
= 1
0
0
= 1
0
0
= 1
0
0
0 1 0
− sin θ −1 0 0
cos θ
0 0 1
1
0
cos θ 0 sin θ
0 − sin θ = 0
1 0 = Ty (θ).
0 cos θ
− sin θ 0 cos θ
Is this what we would expect? Yes, for rotation about the x axis now means rotation about the y axis, and
rotation about the y axis has become rotation about the −x axis—which is to say, rotation in the opposite
direction about the +x axis.
Problem A.16
From Eq. A.64 we have
Af Bf = SAe S−1 SBe S−1 = S(Ae Be )S−1 = SCe S−1 = Cf . Suppose S† = S−1 and He = He† (S unitary, He hermitian). Then
Hf † = (SHe S−1 )† = (S−1 )† He† S† = SHe S−1 = Hf , so Hf is hermitian. In an orthonormal basis, α|β = a† b (Eq. A.50). So if {|fi } is orthonormal, α|β = af † bf . But bf = Sbe
(Eq. A.63), and also af † = ae† S† . So α|β = ae† S† Sbe . This is equal to ae† be (and hence {|ei } is also
orthonormal), for all vectors |α and |β ⇔ S† S = I, i.e. S is unitary.
Problem A.17
Tr(T1 T2 ) =
n
(T1 T2 )ii =
i=1
Is
T1 T2 T3 =
T2 T1 T3 =
(T1 )ij (T2 )ji =
i=1 j=1
Tr(T1 T2 T3 ) = Tr(T2 T1 T3 )?
n n
00
10
10
00
00
10
01
00
10
00
=
=
(T2 )ji (T1 )ij =
j=1 i=1
No.
0
T1 =
0
01
00
n n
Counterexample:
1
0 0
, T2 =
,
0
1 0
0 1
0 0
0 0
1 0
0 0
1 0
0 0
0 0
=
=
n
(T2 T1 )jj = Tr(T2 T1 ).
j=1
T3 =
1 0
.
0 0
1 0
0 0
=⇒ Tr(T1 T2 T3 ) = 1.
0 0
0 0
=⇒ Tr(T2 T1 T3 ) = 0.
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APPENDIX. LINEAR ALGEBRA
291
Problem A.18
Eigenvalues:
(cos θ − λ) − sin θ = (cos θ − λ)2 + sin2 θ = cos2 θ − 2λ cos θ + λ2 + sin2 θ = 0, or λ2 − 2λ cos θ + 1 = 0.
sin θ
(cos θ − λ)
λ=
2 cos θ ±
√
4 cos2 θ − 4
= cos θ ± − sin2 θ = cos θ ± i sin θ = e±iθ .
2
So there are two eigenvalues, both of them complex. Only if sin θ = 0 does this matrix possess real eigenvalues,
i.e., only if θ = 0 or π.
Eigenvectors:
cos θ − sin θ
α
±iθ α
=e
=⇒ cos θ α − sin θ β = (cos θ ± i sin θ)α ⇒ β = ∓iα. Normalizing:
sin θ cos θ
β
β
1
=√
2
(1)
a
1
;
−i
(2)
a
1
=√
2
1
.
i
Diagonalization:
1
(1)
(S−1 )11 = a1 = √ ;
2
−1
S
1
=√
2
1 1
;
−i i
1 1
2 1
1 1
=
2 1
STS−1 =
1
(2)
(S−1 )12 = a1 = √ ;
2
1
inverting: S = √
2
−i
(1)
(S−1 )21 = a2 = √ ;
2
i
(2)
(S−1 )22 = a2 = √ .
2
1 i
.
1 −i
1 1 i
i
cos θ − sin θ
1 1
(cos θ + i sin θ) (cos θ − i sin θ)
=
−i
sin θ cos θ
−i i
(sin θ − i cos θ) (sin θ + i cos θ)
2 1 −i
iθ
iθ
iθ
−iθ
1
i
e
2e
e
e
0
0
=
=
. −i
0 2e−iθ
0 e−iθ
−ieiθ ie−iθ
2
Problem A.19
(1 − λ)
1 = (1 − λ)2 = 0 =⇒ λ = 1
0
(1 − λ)
α
α
=
=⇒ α + β = α =⇒ β = 0;
β
β
11
01
(only one eigenvalue).
a=
1
0
(only one eigenvector—up to an arbitrary constant factor). Since the eigenvectors do not span the
space,
this
1 0
matrix cannot be diagonalized. [If it could be diagonalized, the diagonal form would have to be
, since
0 1
the only eigenvalue is 1. But in that case I = SMS−1 . Multiplying from the left by S−1 and on the right by
S : S−1 IS = S−1 SMS−1 S = M. But S−1 IS = S−1 S = I. So M = I, which is false.]
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292
APPENDIX. LINEAR ALGEBRA
Problem A.20
Expand the determinant (Eq. A.72) by minors, using the first column:
(T22 − λ) . . .
...
n
..
.
.
+
.
.
det(T − λ1) = (T11 − λ) Tj1 cofactor(Tj1 ).
j=2
.
.
.
(Tnn − λ)
But the cofactor of Tj1 (for j > 1) is missing two of the original diagonal elements: (T11 − λ) (from the first
column), and (Tjj − λ) (from the j-th row). So its highest power of λ will be (n − 2). Thus terms in λn and
λn−1 come exclusively from the first term above. Indeed, the same argument applied now to the cofactor of
(T11 − λ) – and repeated as we expand that determinant – shows that only the product of the diagonal elements
contributes to λn and λn−1 :
(T11 − λ)(T22 − λ) · · · (Tnn − λ) = (−λ)n + (−λ)n−1 (T11 + T22 + · · · + Tnn ) + · · ·
Evidently then, Cn = (−1)n , and Cn−1 = (−1)n−1 Tr(T). To get C0 – the term with no factors of λ – we simply
set λ = 0. Thus C0 = det(T). For a 3 × 3 matrix:
(T11 − λ)
T12
T13 T21
(T22 − λ)
T23 T31
T32
(T33 − λ)
= (T11 − λ)(T22 − λ)(T33 − λ) + T12 T23 T31 + T13 T21 T32
− T31 T13 (T22 − λ) − T32 T23 (T11 − λ) − T12 T21 (T33 − λ)
= −λ3 + λ2 (T11 + T22 + T33 ) − λ(T11 T22 + T11 T33 + T22 T33 ) + λ(T13 T31 + T23 T32 + T12 T21 )
+ T11 T22 T33 + T12 T23 T31 + T13 T21 T32 − T31 T13 T22 − T32 T23 T11 − T12 T21 T33
= −λ3 + λ2 Tr(T) + λC1 + det(T),
with
C1 = (T13 T31 + T23 T32 + T12 T21 ) − (T11 T22 + T11 T33 + T22 T33 ).
Problem A.21
The characteristic equation is an n-th order polynomial, which can be factored in terms of its n (complex) roots:
(λ1 − λ)(λ2 − λ) · · · (λn − λ) = (−λ)n + (−λ)n−1 (λ1 + λ2 + · · · + λn ) + · · · + (λ1 λ2 · · · λn ) = 0.
Comparing Eq. A.84, it follows that Tr(T) = λ1 + λ2 + · · · λn and det(T) = λ1 λ2 · · · λn .
QED
Problem A.22
(a)
[Tf1 , Tf2 ] = Tf1 Tf2 −Tf2 Tf1 = STe1 S−1 STe2 S−1 −STe2 S−1 STe1 S−1 = STe1 Te2 S−1 −STe2 Te1 S−1 = S[Te1 , Te2 ]S−1 = 0. c
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APPENDIX. LINEAR ALGEBRA
293
(b) Suppose SAS−1 = D and SBS−1 = E, where D and E are diagonal:
d1 0 · · · 0
e1 0 · · ·
0 d2 · · · 0
0 e2 · · ·
D=.
, E=.
. . ..
..
..
..
. .
.
0 0 · · · dn
0 0 ···
0
0
.. .
.
en
Then
[A, B] = AB − BA = (S−1 DS)(S−1 ES) − (S−1 ES)(S−1 DS) = S−1 DES − S−1 EDS = S−1 [D, E]S.
But diagonal matrices always commute:
d1 e1 0 · · · 0
0 d2 e2 · · · 0
DE = .
. = ED,
..
..
. ..
0
0 · · · dn en
so
[A, B] = 0. QED
Problem A.23
(a)
M† =
1 1
;
1 −i
MM† =
2
(1 − i)
,
(1 + i)
2
M† M =
2
(1 + i)
;
(1 − i)
2
[M, M† ] =
0 −2i
= 0.
2i 0
No.
(b) Find the eigenvalues:
(1 − λ)
1 = (1 − λ)(i − λ) − 1 = i − λ(1 + i) + λ2 − 1 = 0;
1
(i − λ)
√
(1 + i) ± (1 + i)2 − 4(i − 1)
(1 + i) ± 4 − 2i
λ=
=
.
2
2
Since there are two distinct eigenvalues, there must be two linearly independent eigenvectors, and that’s
enough to span the space. So this matrix is diagonalizable, even though it is not normal.
Problem A.24
Let |γ = |α + c|β, for some complex number c. Then
γ|T̂ γ = α|T̂ α + cα|T̂ β + c∗ β|T̂ α + |c|2 β|T̂ β, and
T̂ γ|γ = T̂ α|α + c∗ T̂ β|α + cT̂ α|β + |c|2 T̂ β|β.
Suppose T̂ γ|γ = γ|T̂ γ for all vectors. For instance, T̂ α|α = α|T̂ α and T̂ β|β = β|T̂ β), so
cα|T̂ β + c∗ β|T̂ α = cT̂ α|β + c∗ T̂ β|α, and this holds for any complex number c.
In particular, for c = 1: α|T̂ β+β|T̂ α = T̂ α|β+T̂ β|α, while for c = i: α|T̂ β−β|T̂ α = T̂ α|β−T̂ β|α.
(I canceled the i’s). Adding: α|T̂ β = T̂ α|β. QED
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294
APPENDIX. LINEAR ALGEBRA
Problem A.25
(a)
T† = T̃∗ =
(b)
(c)
1 1−i
= T. 1+i 0
√
(1 − λ) (1 − i) = −(1 − λ)λ − 1 − 1 = 0; λ2 − λ − 2 = 0; λ = 1 ± 1 + 8 = 1 ± 3 .
(1 + i) (0 − λ)
2
2
1
(1 − i)
α
α
=2
=⇒ α + (1 − i)β = 2α =⇒ α = (1 − i)β.
(1 + i)
0
β
β
1
|α|2 + |β|2 = 1 =⇒ 2|β|2 + |β|2 = 1 =⇒ β = √ .
3
λ1 = 2, λ2 = −1.
1
a(1) = √
3
1
(1 − i)
α
α
=−
=⇒ α + (1 − i)β = −α;
(1 + i)
0
β
β
1
3
2|β|2 + |β|2 = 1 =⇒ |β|2 = 1; β =
4
2
(1)† (2)
a
a
1
= √ (1 + i) 1
3 2
2
.
3
1
a(2) = √
6
1−i
.
1
1
α = − (1 − i)β.
2
i−1
.
2
1
(i − 1)
= √ (i − 1 − 1 − i + 2) = 0. 2
3 2
(d)
1
(1)
Eq. A.81 =⇒ (S−1 )11 = a1 = √ (1 − i);
3
1
(1)
(S−1 )21 = a2 = √ ;
3
1
=√
3
−1
S
−1
STS
√ (1 − i) (i −√
1)/ 2
;
1
2
1
(2)
(S−1 )12 = a1 = √ (i − 1);
6
2
(2)
(S−1 )22 = a2 = √ .
6
1
) =√
3
−1 †
S = (S
(1 + i)√ √1
.
(−i − 1)/ 2 2
√ 1
(1 − i)
(1 − i) (i −√
1)/ 2
(1 + i)
0
1
2
√ 1
1 6 0
(1 + i)√ √1
2(1 − i) (1 − √
i)/ 2
2 0
=
=
. =
0 −1
2
− 2
3 −(1 + i)/ 2 2
3 0 −3
1
=
3
(1 + i)√ √1
−(1 + i)/ 2 2
(e)
Tr(T) = 1;
det(T) = 0 − (1 + i)(1 − i) = −2.
Tr(STS−1 ) = 2 − 1 = 1. det(STS−1 ) = −2. c
2005
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
APPENDIX. LINEAR ALGEBRA
295
Problem A.26
(a)
det(T) = 8 − 1 − 1 − 2 − 2 − 2 = 0.
(b)
Tr(T) = 2 + 2 + 2 = 6.
(2 − λ)
i
1 −i (2 − λ)
i = (2 − λ)3 − 1 − 1 − (2 − λ) − (2 − λ) − (2 − λ) = 8 − 12λ + 6λ2 − λ3 − 8 + 3λ = 0.
1
−i (2 − λ)
−λ3 + 6λ2 − 9λ = −λ(λ2 − 6λ + 9) = −λ(λ − 3)2 = 0.
λ1 = 0, λ2 = λ3 = 3.
λ1 + λ2 + λ3 = 6 = Tr(T). λ1 λ2 λ3 = 0 = det(T).
0 0 0
Diagonal form: 0 3 0 .
0 0 3
(c)
2 i 1
α
−i 2 i β = 0 =⇒ 2α + iβ + γ = 0
.
−iα + 2β + iγ = 0 =⇒ α + 2iβ − γ = 0
1 −i 2
γ
Add the two equations: 3α + 3iβ = 0 =⇒ β = iα;
2α − α + γ = 0 =⇒ γ = −α.
(1)
a
α
1
= iα . Normalizing: |α|2 + |α|2 + |α|2 = 1 =⇒ α = √ .
3
−α
(1)
a
1
1
i
.
=√
3 −1
2 i 1
α
α
=⇒ −α + iβ + γ = 0,
2α + iβ + γ = 3α
−i 2 i β = 3 β =⇒ −iα + 2β + iγ = 3β =⇒ α − iβ − γ = 0,
1 −i 2
γ
γ
α − iβ + 2γ = 3γ
=⇒ α − iβ − γ = 0.
The three equations are redundant – there is only one condition here: α − iβ − γ = 0. We could pick
γ = 0, β = −iα, or β = 0, γ = α. Then
α
α
(2)
(3)
a0 = −iα ; a0 = 0 .
0
α
(2)
But these are not orthogonal, so we use the Gram-Schmidt procedure (Problem A.4); first normalize a0 :
a(2)
1
1
−i .
=√
2 0
† (3)
a(2) a0
1
1/2
1
1
† (3)
α
α
α
(3)
= √ 1 i 0 0 = √ . So a0 − (a(2) a0 ) a(2) = α 0 − −i = α i/2 .
2
2
2
0
1
1
1
c
2005
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
296
APPENDIX. LINEAR ALGEBRA
|α|2
Normalize:
1 1
3
+ + 1 = |α|2 = 1
4 4
2
=⇒
2
.
3
α=
a(3)
1
1
i .
=√
6 2
Check orthogonality:
†
a(1) a(2)
(1) † (3)
a
a
1
1
1
= √ 1 −i −1 −i = √ (1 − 1 + 0) = 0. 6
6
0
1
1
1
= √ 1 −i −1 i = √ (1 + 1 − 2) = 0. 3 2
3 2
2
(d) S−1 is the matrix whose columns are the eigenvectors of T (Eq. A.81):
S−1
√
√
2
√3
1 √
=√
2
i
−
3i
6 −√2 0
STS−1
√
√
√
2 −√ 2 i − 2
√
1
S = (S−1 )† = √ 3
3i
0 .
6
1
−i
2
1
i ;
2
√
√
√
√
√
2 i 1
2 − 2i − 2
√2
√3
1 √ √
=
2i − 3i
3
3i
0 −i 2 i √
6
1 −i 2
1
−i
2
− 2 0
√
0 3 √3 3
0 −3
3 i 3i
0
0
6
1
0 0 0
0 0 0
1
i = 0 18 0 = 0 3 0 . 6
0 0 18
0 0 3
2
Problem A.27
(a) Û α|Û β = Û † Û α|β = α|β. (b) Û |α = λ|α =⇒ Û α|Û α = |λ|2 α|α. But from (a) this is also α|α. So |λ| = 1. (c) Û |α = λ|α, Û |β = µ|β =⇒ |β = µÛ −1 |β, so Û † |β =
1
|β = µ∗ |β (from (b)).
µ
β|Û α = λβ|α = Û † β|α = µβ|α, or (λ − µ)β|α = 0. So if λ = µ, then β|α = 0. QED
Problem A.28
(a) (i)
0 0 4
M2 = 0 0 0 ;
0 0 0
0 0 0
M3 = 0 0 0 ,
0 0 0
so
c
2005
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
APPENDIX. LINEAR ALGEBRA
297
1 1 5
100
013
0 0 4
1
eM = 0 1 0 + 0 0 4 + 0 0 0 = 0 1 4 .
2
0 0 1
001
000
0 0 0
(ii)
2
M =
−θ2 0
0 −θ2
= −θ2 I;
M3 = −θ3 M;
M4 = θ4 I;
etc.
1 2
θ3 0 1
θ4
0 1
e =I+θ
− θ I−
+ I + ···
−1 0
2
3! −1 0
4!
2
4
3
5
θ
θ
θ
θ
0 1
= 1−
+
− ··· I + θ −
+
− ···
−1 0
2
4!
3!
5!
10
0 1
cos θ sin θ
= cos θ
+ sin θ
=
.
01
−1 0
− sin θ cos θ
M
(b)
SMS−1 = D =
..
0
M −1
Se S
0
d1
for some S.
.
dn
1 2
1 3
= S I + M + M + M + · · · S−1 . Insert SS−1 = I :
2
3!
1
1
SeM S−1 = I + SMS−1 + SMS−1 SMS−1 + SMS−1 SMS−1 SMS−1 + · · ·
2
3!
1 2
1 3
D
= I + D + D + D + · · · = e . Evidently
2
3!
det(eD ) = det(SeM S−1 ) = det(S) det(eM ) det(S−1 ) = det(eM ). But
d21
D2 =
0
..
0
.
d2n
,
d31
D3 =
0
..
.
d3n
,
Dk =
dk1
0
..
.
,
so
dkn
0
0
d
e 1
d21
d31
0
0
0
0
d1
1 ..
1 ..
..
eD = I + . . .
+
+
+ ··· =
.
.
.
.
2
3!
2
3
dn
0
dn
0
dn
0
dn
0
e
det(eD ) = ed1 ed2 · · · edn = e(d1 +d2 +···dn ) = eTr D = eTr M (Eq. A.68), so det(eM ) = eTr M . QED
(c) Matrices that commute obey the same algebraic rules as ordinary numbers, so the standard proofs of
ex+y = ex ey will do the job. Here are two:
c
2005
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
298
APPENDIX. LINEAR ALGEBRA
m
n
(i) Combinatorial Method: Use the binomial theorem (valid if multiplication is commutative):
M+N
e
∞
∞
n ∞ n
1
1 n
1
n
Mm Nn−m =
=
(M + N) =
Mm Nn−m .
m
n!
n!
m!(n
−
m)!
n=0
n=0
m=0
n=0 m=0
Instead of summing vertically first, for fixed n (m : 0 → n), sum horizontally first, for fixed m (n :
m → ∞, or k ≡ n − m : 0 → ∞)—see diagram (each dot represents a term in the double sum).
eM+N =
∞
∞
1 m 1 k
M
N = eM eN . QED
m!
k!
m=0
k=0
(ii) Analytic Method: Let
S(λ) ≡ eλM eλN ;
dS
= MeλM eλN + eλM NeλN = (M + N)eλM eλN = (M + N)S.
dλ
(The second equality, in which we pull N through eλM , would not hold if M and N did not commute.)
Solving the differential equation: S(λ) = Ae(M+N)λ , for some constant A. But S(0) = I, so A = 1,
and hence eλM eλN = eλ(M+N) , and (setting λ = 1) we conclude that eM eN = e(M+N) . [This method
generalizes most easily when M and N do not commute—leading to the famous Baker-CampbellHausdorf lemma.]
0 1
0 0
As a counterexample when [M, N] = 0, let M =
, N=
. Then M2 = N2 = 0, so
0 0
−1 0
11
1 0
1 1
1 0
0 1
M
N
M N
e =I+M=
, e =I+N=
; e e =
=
.
01
−1 1
0 1
−1 1
−1 1
But (M + N) =
0 1
cos(1) sin(1)
, so (from a(ii)): eM+N =
.
−1 0
− sin(1) cos(1)
The two are clearly not equal.
(d)
eiH =
∞
∞
∞
1 n n
1
1
i H =⇒ (eiH )† =
(−i)n (H† )n =
(−i)n Hn = e−iH (for H hermitian).
n!
n!
n!
n=0
n=0
n=0
(eiH )† (eiH ) = e−iH eiH = ei(H−H) = I, using (c).
So eiH is unitary. c
2005
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
299
2nd Edition – 1st Edition Problem Correlation Grid
N = New
M = 1/e problem number (modified for 2/e)
X = 2/e problem number (unchanged from 1/e)
Chapter 1
2/e
1/e
Chapter 2
2/e
1/e
1
2N
3
4
5
6
7
8
9
10
11
12
13
14
15
16N
17N
18N
1
2
3
4
5
6
7N
8N
9N
10
11
12
13
14N
15
16
17
18
19N
20
21N
22
23
24
25N
26
27
28
29
30
31
32
33
34
35
36
37
38
39N
40N
41N
42
43
44
1
6
7
8
11
12
13
14
2
3
4
5
9M
10
1
2
3
5
6M
7
13M
14
37
17M
15
16
18
19M
20
22
23
24
25
26
27
28
29
30
31
32
33
41M
4M
36
3.48
38
40
39
Chapter 2 (cont.)
2/e
1/e
45
46
47
48N
49
50
51
52
53
54N
55N
56N
42
43
44
45
47
48M
34M, 35M
49
300
2nd Edition – 1st Edition Problem Correlation Grid
N = New
M = 1/e problem number (modified for 2/e)
(M) = 1/e problem number (distant model for 2/e)
X = 2/e problem number (unchanged from 1/e)
Chapter 3
2/e
1/e
Chapter 4
2/e
1/e
1N
2N
3N
4N
5N
6N
7N
8N
9N
10N
11
12
13
14
15N
16
17
18
19
20
21
22N
23N
24
25
26N
27N
28
29N
30N
31
32
33
34
35N
36N
37N
38N
39
40N
1
2
3
4
5
6
7
8
9
10
11
12
13
14N
15N
16
17
18
19
20
21N
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41N
42
(33M)
(21M)
(12M)
38
51
41M
39
42
43
44
45
46
57M
57M
25M
52M
53
56
50
49M
55
1
2
3
4
5
6
7M
8
9M
10
11
12
13
17
16
19
20
21
22
23
25
26
27
28
29
30
31M
32
33
34
35
36
37
38
39
40
41
42
Chapter 4 (cont.)
2/e
1/e
43
44N
45
46
47N
48N
49N
50
51
52
53N
54
55
56
57
58N
59
60
61
43
14
15
44
45M
46
47
48
49
50
51
52M
53
301
2nd Edition – 1st Edition Problem Correlation Grid
N = New
M = 1/e problem number (modified for 2/e)
X = 2/e problem number (unchanged from 1/e)
Chapter 5
2/e
1/e
Chapter 6
2/e
1/e
Chapter 7
2/e
1/e
1
2
3N
4
5
6
7
8
9
10
11
12
13
14
15N
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32N
33
34
35
36
37
1
2
3
4
5
6
7
8
9
10N
11
12
13
14
15N
16
17
18
19
20
21
22
23
24
25
26
27
28
29N
30N
31N
32
33
34
35
36
37
38
39
40N
1
2
3
4
5
6
7
8
9
10
11N
12N
13
14
15
16
17
18
19
20N
1
2
3
4
5
6
7
8
9
10
11M
11M
12
13
14
15M
16M
17M
18
19M
20
21M
22
23
24
25
26
27M
28
29
30
31
32
33
1M
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
1
2M
3M
4
5
6
7
8
9
10
11
12
13
14
15
16
17
302
2nd Edition – 1st Edition Problem Correlation Grid
N = New
M = 1/e problem number (modified for 2/e)
X = 2/e problem number (unchanged from 1/e)
Chapter 8
2/e
1/e
Chapter 9
2/e
1/e
Chapter 10
2/e
1/e
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16N
17N
1
2
3
4
5
6
7
8
9N
10
11
12
13
14
15
16
17
18
19
20
21
22N
1
2
3
4
5
6
7
8N
9
10
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
1
2
3M
4
5
6
7
8
9
10
11
12
13
14
15
16
17
21
19M
20
1
3M
4
5
6
8
9
10
11M
303
2nd Edition – 1st Edition Problem Correlation Grid
N = New
M = 1/e problem number (modified for 2/e)
X = 2/e problem number (unchanged from 1/e)
Chapter 11
2/e
1/e
Chapter 12
2/e
1/e
Appendix
2/e
1/e
1
2
3
4
5N
6N
7N
8
9
10
11
12
13
14
15
16
17
18
19N
20N
1N
1
2
3
4
5
6
7
8
9
10
11
12
13N
14N
15
16
17
18
19
20
21
22
23N
24
25
26
27
28
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.9
3.10.
3.11
3.12
3.16
3.13
3.14
3.15
3.17
3.18
3.19
3.20.
3.40M
3.21M
3.22
3.23
3.24
3.47