Carsten Timm: Theory of superconductivity

link. We employ the first Ginzburg-Landau equation for A ≡ 0 assuming ψ(r) to depend only on the coordinate
x along the wire,
ξ 2 f ′′ (x) + f(x) − f 3 (x) = 0 (11.6)
with
f(x) =
ψ(x) √
|ψ(±∞)| = − α β
ψ(x). (11.7)
0 L x
We assume the two bulk superconductors to be uniform and to have a relative phase of ∆ϕ. This allows us to
write
{
1 for x ≤ 0,
f(x) =
e i∆ϕ (11.8)
for x ≥ L.
For the wire we have to solve Eq. (11.6) with the boundary conditions
f(0) = 1, f(L) = e i∆ϕ . (11.9)
Since L ≪ ξ, the first term in Eq. (11.6) is larger than the other two by a factor of order ξ 2 /L 2 , unless ∆ϕ = 0,
in which case the solution is trivially f ≡ 1. It is thus sufficient to solve f ′′ (x) = 0, which has the solution
f(x) = L − x
L + x L ei∆ϕ . (11.10)
Inserting f(x) into the second Ginzburg-Landau equation (with A ≡ 0), we obtain
j s = i q
2m ∗ [(ψ′ ) ∗ ψ − ψ ∗ ψ ′ ] = −i e (
− α )
[(f ′ ) ∗ f − f ∗ f ′ ]
2m β
= −i e [(
2m 2n s − 1 L + 1 ) ( L − x
L e−i∆ϕ L + x ) ( L − x
L ei∆ϕ −
L + x ) (
L e−i∆ϕ
= −i en s
m
= −2 en s
mL
[
− x L 2 (
e i∆ϕ − e −i∆ϕ) − L − x
L 2
(
e i∆ϕ − e −i∆ϕ)] = 2 en s
m
− 1 L + 1 L ei∆ϕ )]
[
− x L − x
sin ∆ϕ −
L2 L 2
]
sin ∆ϕ
sin ∆ϕ. (11.11)
The current is obviously obtained by integrating over the cross-sectional area,
I s = − 2en s
m
A
L
sin ∆ϕ, (11.12)
so that we get
I c = − 2en s A
m L . (11.13)
The negative sign is due to the negative charge −2e of the Cooper pairs. The amplitude of the current-phase
relation is clearly |I c |.
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