VÓRTICES EN SUPERCONDUCTORES

VÓRTICES EN SUPERCONDUCTORES
!
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&
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)
#
(!
* $+
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,
-
. '
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/ -*- ( 0 *1
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)
% 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