Advanced Solid State Physics

In the last step (integration by parts) we used thatfis some test function, i.e. it decays sufficiently
fast for|r|→∞. From the last line we deduce the desired result

δBa^2 (r)

δA(r′)

= 2∇×∇×A(r′)δ(r−r′). (294)

Using∇×∇×A=∇×Ba=μ 0 jSandδFδAS= 0we obtain the secondGinzburg - Landauequation
as

jS = −

q

2 m

[φ(−i~∇+qA)φ∗−φ∗(−i~∇−qA)φ]

=

q

m

Re[φ∗(−i~∇−qA)φ]. (295)

wherejSis called the supercurrent.

We now regard several interesting outcomes of this model: LetA= 0 andαβ|φ|^2. Then we obtain
in the one dimensional case from the firstGinzburg - Landauequation (291)

−

~^2

2 m

d^2

dx^2

φ=αφ (296)

with the solution φ(x) = exp (ix/ξ)whereξ = ~/

√

2 mα. ξmay be interpreted as the intrinsic
coherence length which will later turn out to be the average distance between the two electrons of a
Cooperpair.

Another interesting case is obtained by regarding the situation ofφ= 0atx= 0andφ=φ 0 as
x→∞. This situation resembles a boundary between normal and superconducting state, i.e. there is
a magnetic fieldHcin the normal region. We regard an extreme type I superconductor, in particular
the penetration depth satisfiesλLξ. We have to solve

−

~^2

2 m

d^2

dx^2

φ−αφ+β|φ|^2 φ= 0. (297)

A proper solution satisfying the boundary conditions is

φ(x) =

√α

β

tanh

(

x

√

2 ξ

)

. (298)

Inside the superconductor we obtainφ 0 =

√α

βas the solution of the minimization of−α|φ|

(^2) + 1
2 β|φ|
(^4).
Since inside the superconductor we haveFS=FN−α
2
2 β=FN−
1
2 μ 0 B
2
acwe obtain
Bac=
√
μ 0 α^2
β

. (299)

For a small penetration length (φ≈φ 0 ) we obtain from the secondGinzburg - Landauequation

jS=−

q^2

m

|φ 0 |^2 A. (300)