Carsten Timm: Theory of superconductivity

We obtain
e −ik yy
(− 2
2m ∗ ∇2 − iω c x ∂ ∂y + 1 )
2 m∗ ωc 2 x 2 e ikyy f(x)
= − 2 d 2 f
2m ∗ dx 2 + 2 ky
2
2m ∗ f − iω c x ik y f + 1 2 m∗ ωc 2 x 2 f
= − 2
2m ∗ d 2 f
dx 2 + 1 2 m∗ ω 2 c
(
x 2 + 2 k y
m ∗ ω c
+
2 k 2 y
(m ∗ ) 2 ω 2 c
)
f = −αf. (6.117)
Defining x 0 := −k y /m ∗ ω c , we get
− 2
2m ∗ d 2 f
dx 2 + 1 2 m∗ ω 2 c (x − x 0 ) 2 f = −αf. (6.118)
This is the Schrödinger equation for a one-dimensional harmonic oscillator with shifted minimum. Thus we obtain
solutions f n (x) as shifted harmonic-oscillator eigenfunctions for
(
−α = ω c n + 1 )
, n = 0, 1, 2, . . . (6.119)
2
Solution can only exist if
−α = −α ′ (T − T c ) = α ′ (T c − T ) ≥ ω c
2 = eH
m ∗ c . (6.120)
Keeping T ≤ T c fixed, a solution can thus only exist for H ≤ H c2 with the upper critical field
Using ξ 2 = − 2 /2m ∗ α, we obtain
H c2 (T ) =
H c2 = − m∗ c
e
α. (6.121)
c
2e ξ 2 (T ) = Φ 0
2π ξ 2 (T ) . (6.122)
Note that since ξ ∼ 1/ √ T c − T close to T c , H c2 (T ) sets in linearly, as shown in the above sketch. H c2 should be
compared to the thermodynamic critical field H c ,
Hc 2 = − mc2 1
2e 2 λ 2 α = 2 c 2 1
8e 2 λ 2 ξ 2 = 1 ( ) 2 hc 1
8π 2 2e λ 2 ξ 2
Φ 0
⇒ H c =
2π √ 2 λξ = H c2 ξ
√
2 λ = H c2
√
2 κ
⇒ H c2 = √ 2 κ H c . (6.123)
For κ = 1/ √ 2, H c2 equals H c . This is the transition between a type-II and a type-I superconductor.
So far, our considerations have not told us what the state for H H c2 actually looks like. To find out,
one in principle has to solve the full, not the linearized, Ginzburg-Landau equation. (We have seen that the
linearized equation is equivalent to the Schrödinger equation for a particle in a uniform magnetic field. The
solutions are known: The eigenfunctions have uniform amplitude and the eigenenergies form discrete Landau
levels, very different from what is observed in the Shubnikov phase.) Abrikosov did this approximately using a
variational approach. He used linear combinations of solutions of the linearized equation as the variational ansatz
and assumed that the solution is periodic in two dimensions, up to a plane wave.
We call the periods in the x- and y-directions a x and a y , respectively. The function
ψ n (x, y) = e ikyy f n (x) (6.124)
has the period a y in y if
k y = 2π
a y
q, q ∈ Z. (6.125)
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