Carsten Timm: Theory of superconductivity

Thus the energy is
∫
E 2 = 2E core + d 2 r 1 2 E · E
= 2E core − 1 2 2γ α β
∫
= 2E core − γ α ∫∞
β 2
−∞
d 2 r
dx
∫ ∞
0
(
)
|r + R/2| 2 (r − R/2) − |r − R/2| 2 2
(r + R/2)
|r − R/2| 2 |r + R/2| 2
dy
× |r + R/2|4 |r − R/2| 2 + |r − R/2| 4 |r + R/2| 2 − 2 |r + R/2| 2 |r − R/2| 2 (r 2 − R 2 /4)
|r − R/2| 4 |r + R/2| 4 . (7.38)
We introduce elliptic coordinates σ, τ according to
x = R 2
στ, (7.39)
y = R 2
√
σ2 − 1 √ 1 − τ 2 , (7.40)
where σ ∈ [1, ∞[, τ ∈ [−1, 1]. Then
E 2 = 2E core − 2γ α β
( R
2
) 2 ∫
dσ dτ
σ 2 − τ 2
√
σ2 − 1 √ 1 − τ 2
×
( R
2
) 4 ( )
(σ + τ)
4 R 2
(σ − τ) 2 + ( )
R 4 ( )
(σ − τ)
4 R 2
(σ + τ)
2
− 2 ( R
2
2
( R
2
) 2
(σ + τ)
2 ( R
2
= 2E core − 2γ α ∫
dσ dτ
β
= 2E core − 2γ α ∫
dσ dτ
β
= 2E core − 8πγ α ∫
dσ
β
) 4 ( )
(σ − τ)
4 R 4
(σ + τ)
4
2
2
) 2
(σ − τ)
2 ( R
2
) 2
(σ 2 τ 2 + (σ 2 − 1)(1 − τ 2 ) − 1)
2
1
√
σ2 − 1 √ (σ + τ) 2 + (σ − τ) 2 − 2(σ 2 + τ 2 − 2)
1 − τ 2 σ 2 − τ 2
1
√
σ2 − 1 √ 4
1 − τ 2
σ 2 − τ 2
1
σ(σ 2 − 1) . (7.41)
We have to keep in mind that the integrals in real space have a lower cutoff r 0 . The minimum separation from
the vortex at R ˆx/2 is (σ − 1)R/2. For this separation to equal r 0 , the lower cutoff for σ must be
With this cutoff, we obtain
which for R ≫ r 0 becomes
E 2 = 2E core − 8πγ α β
We absorb an R-independent constant into E core and finally obtain
σ 0 = 1 + 2r 0
R . (7.42)
1
2 ln (R + 2r 0) 2
4r 0 (R + r 0 ) , (7.43)
E 2 = 2E core − 4πγ α β ln R
4r 0
. (7.44)
E 2 ≡ 2E core + V int (R) = 2E core − 4πγ α ln R . (7.45)
β r
} {{ } 0
> 0
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