VÓRTICES EN SUPERCONDUCTORES ! " # $ % & " & ' ) ' ) ) & ) # (! * $+ ' ' , - . ' ' / -*- ( 0 *1 " ' * ) % 23 5 2 # % 42 3 ns e v ( r ) = j s ( r ) 1 Ecin = dV m v 2 n s 2 52 3 ' # 56 3 (o lenta variación de v(r) ) " 7 E = E0 + Ecin + Emag & E mag &' 48% ' * $+ ' h2 = dV 8π 4π js rot h = c EC de corrientes persistentes !! / -*- ( 0 *1 " & ' E = E0 + Ecin + Emag ( 1 2 2 2 E = E0 + dV h + λ L rot h 8π ) 2 mc y se define la longitud λ L como λL = 4π ns e 2 * & δ E=0 5 ' h + λ2L rot rot h = 0 rot h = * $+ 1/ 2 4π js c ne 2 rot j = h mc # / -*- ( 0 *1 9 > * : ?@6 # ! ; ; & '4 % * $+ 2< & 3' rot h = 6= 4π js , div h = 0 c 4' ∂h =0 ∂z A! 4 A! 4 % 2 2 % 4$3 B 46 46= % 6= 24 # # % 43 d h 4π js = c dz ne 2 rot j = h mc B 3 # / -*- ( 0 *1 * 9 : ne 2 rot j = h mc d2 h h = 2 2 dz λL d h 4π js = c dz 2 mc λ2L = 4π ns e 2 4$ h( z ) = h(0) exp(− z / λL ) ' 9 d js ne 2 h = d z mc $ λ D: & λ ?? < < & & 4 $ # % c h ( 0) exp(− z / λ L ) ' j y ( z) = − 4π λ L C $ # / -*- ( 0 *1 * h( z ) = h(0) exp(− z / λL ) ?? C j y ( z) = − c h(0) exp(− z / λ L ) 4π λ L $ ¿h(z), j(z)? d>>λ d<<λ H0 I z d z d H0 z 1( 0E / / > ( 07-" F ( @ / "1 ) 7 ( " 0 ; ; 0 < 5 G < 52 3 % 2 3 5 & H 5 ξ= B ∆ ' 52 3 → 5, → 7 , $ & ∆ B % G , H ,! ∆I J B I ,D ∆ δ ≈ ∆J 5, 5 δ$ δ ≈ 4 ξ0 = ' vF π∆ ' 4 1( 0E / / > ( 07-" F ( @ / "1 ) 7 ( " 0 / 0 & ξ0 = λL = vF π∆ 2 mc 4π ns e 2 ' 4 1/ 2 1( 0E / / > ( 07-" F ( @ / "1 ) 7 ( " 0 0 λ 4 5 % 5 & G λ 0 > ( < L 5, λ KK ξ= 5 0 2ξ= H' 3 > B 3 5, < L " & $ < & B 5B 0 2( 2 4 < " L 3 E > 7"1 ( / E "01 7 C-M1 "-*>1 *- ( N0 "1 > < ' ! ! ! & & & ! & # (! ( & ! 0 & ! & > B & 4 " Ψ = Ψ e iδ * µM h2 1 2 (− i ∇ − 2eA)Ψ + FS − FN = α Ψ + Ψ + * 2 2 2m 2 β Ψ ( x, y ) 4 Ψ =∆ )6= α ; β cerca de Tc : α = α ' (T − Tc ), β = cte. F min 2 Ψ =− F = − ev × B 1 7 ( 0O @ / ,-7-/-@ ∇×E = − ∇×E = − @ ∂B ∂t >1 0 ( " - "01 7 B ≡ ∇×A ∂B ∂ ∂A = − ∇× A ∴ E = − ∂t ∂t ∂t "E -" F ( / *1 * ( 01 ∂p = − eE ∴ ∂t ' ∂k ∂A =e ∂t ∂t k ( B) = k o + eA 0 & - ! & # > B ! & # B 5 ! ' µM h2 1 2 (− i ∇ − 2eA)Ψ + FS − FN = α Ψ + Ψ + * 2 2 2m 1 2 δF = dr δΨ * αΨ + β Ψ Ψ + * (− i ∇ − 2eA)2 Ψ + CC + 2m 2 β 4 + dr dA 4 [ 1 e rot h − Ψ * (− i ∇ − 2eA)Ψ + CC 4π m 25 ] Ψ = Ψ e iδ =0 B 3' 1 2 2 ( ) − ∇ − 2 Ψ + β Ψ Ψ = −α Ψ i e A * 2m 25 @ [ 5 e j = ∇ × h = * Ψ * (− i ∇ − 2eA)Ψ + Ψ (i ∇ − 2eA)Ψ * m 3' ] 7 0 & ! & ! A ' 2 3' 1 2 2 ( ) − ∇ − 2 Ψ + β Ψ Ψ = −α Ψ i e A 2m * ' d 2Ψ − + αΨ + β Ψ 3 = 0, elegimos Ψ real 2 2m dx 2 ' d 2Ψ − + αΨ + β Ψ 3 = 0, elegimos Ψ real 2 2m dx Ψ = 0, metal normal ξ (T ) = 2 m * α (T − Tc ) 1 2 5 0 Ψ = Ψ0 ; Ψ02 = −α / β > 0, superconductor si (α < 0 → T < TC ) Esto es teoría de Landau para transiciones de fase de 2º orden 2 , 0 & ! & 2 3' 4 A ' 4e 2 2 rot j = − Ψ0 h m *B j = ∇×h = [ e Ψ * (− i ∇ − 2eA)Ψ + Ψ (i ∇ − 2eA)Ψ * * m 2 3 ] ne 2 rot j = h mc * $+ h( z ) = h(0) exp(− z / λ L ) c m*β λ (T ) = e * 4πα T − Tc 5 0 1 2 0 & ! & ! 4 @ λ ξ 5 B 5 ξ (T ) = 2 m * α (T − Tc ) 1 2 1 c m*β λ (T ) = e * 4πα T − Tc - 5 0 , 24 2 # $ & α 3 0 & ! " ' 7 5 & > B 0 ! 0 ξ (T ) = ' ! 2 m * α (T − Tc ) 1 2 c m*β λ (T ) = e * 4πα T − Tc 0' , 1 λ(T) m*c β κ= = ξ(T) e* 2π B 5 # κ κ > 1 2 $ 5 - P 5 Abrikosov (1957) 2 0 & ! & > B 7 " ' " 1 2 2 ( ) − ∇ − 2 Ψ + β Ψ Ψ = −α Ψ i e A 2m * Ψ= ( B < L ' 1 2 ( ) − i ∇ − 2 e Ψ = −α Ψ = E Ψ A 2m * E = −α = 2 2m * ξ 2 (T ) = 2 2 m * ξ 2 ( 0) 1− > T Tc 0 # 5 B # [ j= 2 β Ψ Ψ→0 Ψ = Ψ e iδ e j = ∇ × h = * Ψ * (− i ∇ − 2eA)Ψ + Ψ (i ∇ − 2eA)Ψ * m ] 2e 2 2 ( ) Ψ ∇ δ − 2 e A ≡ 2 e Ψ vs * m ∇δ = m * v s + 2eA 0 & ! " ' B " # $ C 4! # " p' # p dl = 2π n 2Q 3 2 2 p2 ∇ =− ; p = −i ∇ 2m 2m E= p dl = 1 1 2 * 2 ( ) 2 − i ∇ − e A = ( m v ) s 2m * 2m * (2eA + m v )dl = L h * Lo hemos sacado antes s < L 2 B 5 n3 5 < 0 & ! " ' " # $ ∇δ = m * v s + 2eA ' 1 5 p dl = ' (2eA + m v )dl = L h * # , % ' # % ' ∇δ dl = L 2π s @ " # ' Φ' = Φ + Φ = A dl Φ0 = h 2e 1 m * v s dl = L Φ 0 2e B # % L' Q Φ 0 ≅ 2 ⋅10 −7 G ⋅ cm 2 = 20 G ⋅ µm 2 = 2 mT ⋅ µm 2 B # $ 0 & ! " Φ' = Φ + ' " F70 " # $ 1 m * v s dl = L Φ 0 2e 5 r' 2 µ H π r Φ h m*v s = L− = L− M 2π r 2π r Φ0 Φ0 h 5 & ) 5 B< LK= < r < 5 # 5 Q 4 $ ' 5 < 0 & ! " ' 7 5 & > B λ(T) m*c β κ= = ξ(T) e* 2π e 2 κ = 2 2 λ H cm c H cm 1 Φ0 = 2 2πλξ ! 0 ! ' 0' Porque… 0 α2 Fn − Fs 0 = β 2 H cm Fn − Fs 0 = 8π ξ (T ) = 2 m * α (T − Tc ) 1 2 c m*β λ (T ) = e * 4πα T − Tc T.Landau: , 1 2 & 4' 24 C' 2 ρ6# ( = 3 5 3 # 5 4 ( C ) H FN − FS = C 8π B = (1 − ρ ) hn + ρ ⋅ 0 = (1 − ρ ) hn 2 2 2 HC B2 + F = FN − ρ 8π 8π (1 − ρ ) 2 BH HC B2 BH = FN − ρ + − G ( B, ρ ) = F − 4π 8π 8π (1 − ρ ) 4π H h F = FN − ρ C + (1 − ρ ) n 8π 8π B 2 & < (! ' 3 B = H C (1 − ρ ) ρ 6 < & λ II ξ= # (! λ KK ξ= hn = H C # ξ λ 2 & λ II ξ= 2 9S λ G L 5 < H 2 2 H H H BH G ( B, ρ = 0) = F − = FN + C − C = FN − C 4π 8π 4π 8π H BH G ( B = 0, ρ = 1) = F − = FN − C 4π 8π 2ρ6 C6=3' 2 : L )" JRπ (! L ξ > 5 ξ " & ( '2G #5 B & 2 (1 (! HC B2 BH BH G ( B, ρ ) = F − = FN − ρ + − 4π 8π 8π (1 − ρ ) 4π 0 ( 2ρ6= )6C6) 3' # # H3 ( G=energía/u.de vol→ G×l = energía/u.de área H γ ≈ C ξ0 8π 2 & λ KK ξ= 2 0 & 2 (! 4 " 4& # B 3 3 4$ )" λ En N : h( z ) = H C En S : h( z ) = H C exp(− z / λ L ) ( 2 HC ( ) / h 2 hH dh dz + − + λ2 G = dvol FN − 8π 8π 4π 8π en S 2 ! # $%π π " 2 E ∝ vs , vs ∝ js d h 4π = js dz c & λ KK ξ= 2 % # γ= ∞ 0 2 2 2 < 3 γ : tensión superficial = energía/u.de área +γ S S : área de la pared (dh / dz ) h 2 hH C − + λ2 dz 8π 4π 8π 2 & (! (dh / dz ) HC h 2 hH + − + λ2 G = dr FN − 8π 8π 4π 8π en S 0 H G = dvol FN − C 8π # En S : 2 H =− C λ 8π h( z ) = H C exp(− z / λ L ) 5 ' & 5 ( & λ KK ξ= # 0 H λ KK ξ= & 0 5 ' & > 9 B ' 9" B " (! # % # % 25 5 3 : 5 : - 5 & E = E0 + ' 5 ' ε= * & 2 h + λ2 rot rot h = 0, r > ξ h + λ2 rot rot h = Φ 0 δ (r ), / ( 1 2 dr h 2 + λ2L rot h 8π 42 3 1 8π 5 ( dr h 2 + λ2 rot h r >ξ 2 & 5 ) " ξ ) > 3' # % B & 5 & 5 & D * $+ 5 4 6= " λ2 2π r rot h = Φ 0 , rot h = −dh / dr , (ξ < r << λ ) h= Φ0 Φ0 λ λ + cte , si r << λ K → ln 0 r r 2πλ2 2πλ2 @ % 2 KKλ3' h= πλ Φ0 2πλ2 @ 2r & λ2 ε= 8π exp [− r / λ ] 5 ' λ2 dr (h × rot h ) = 2π ξ h(ξ ) rot h(ξ ) 8π núcleo Φ0 ε= 4πλ 2 ln λ ξ ξ > & Φ0 ε= 4πλ $ > S > 2 λ ln ξ " ξ B % 5 5 # % # % 5 # % Φ= Φ Φ2 2 & Tε3 Pero ε>0, ¿Cuándo entran los vórtices? El vórtice lleva: B = Φ 0 Entrarán vórtices si G<0: Hc1 > & ε3 G = F − BH 0 / 4π G = ε − Φ 0 H 0 / 4π H c1 = 4πε / Φ 0 5 H centro vórtice ≈ 2 H c1 & 5 5 / ' h + λ rot rot h = Φ 0 [δ (r − r1 ) + δ (r − r2 )] 2 F = 2 ε + 2h12 Φ 0 / 8π = 2 ε + U ( x = r1 − r2 ) Φ dh dU f =− = − 0 12 dx 4π dx f =− ; ; ; Φ 0 dh12 1 = − j12 Φ 0 4π dx c & ( (' ' h= Φ0 λ K 0 2πλ2 r Φ 0 h12 U12 = 4π ∝ (1 / r12 ) exp [− r12 / λ ] ∝ ln [λ / r12 ] & 5 5 / ' h + λ rot rot h = Φ 0 [δ (r − r1 ) + δ (r − r2 )] 2 Φ h U12 = 0 12 4π ; ; ; fuerza > < & & ( (' 2 " 0" * + " - ' Φ0 λ K 0 r 2πλ2 ∝ (1 / r12 ) exp [− r12 / λ ] ∝ ln [λ / r12 ] 3' G = nLε + U ij − ij ) Φ0 λ K 0 2πλ2 r h12 = h1 (r ) + h2 (r ); hi (r ) = ' & h= , G≅ . / + BH , B = nL Φ 0 4π 1 Φ λ B H C1 − H + m 0 2 K 0 4π 2 2πλ r 2Φ 0 B = nL Φ 0 = 3d2 " & Si el campo es grande, los vórtices casi se tocan. La zona S es “como capas muy finas”. (el problema de antes) H c , paralelo ≈ H cm H c 2 ≈ H cm λ d , ahora d ≈ ξ Φ0 λ = H cm κ , Exacto : H c 2 = 2 H cm κ = ξ 2πξ 2 E > 7"1 ( / E "01 7 C-M1 "-* >1 * - ( N0 "1 Diagrama de fase H - T Penetración del campo magnético: balance energético - fronteras N-S - fronteras S-exterior Parámetro de Ginzburg-Landau: H HC Longitud de penetración: λ Longitud de coherencia: ξ ψ S κ >> 1 : tipo II λ Η Aluminio λ (0) = 16 nm ξ (0) = 1600 nm ψ Tipo II H HC2 HC ξ NbSe2 λ (0) = 240 nm ξ (0) = 8 nm N 0 Η ξ HC = 100 - 1000 G κ(Τ)=λ(Τ)/ ξ(Τ) κ = 1/ 2 κ << 1 : tipo I λ Tipo I TC T HC1 < 100 G HC2 = 104 - 105 G HC1 S 0 N TC T