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Carsten Timm: Theory of superconductivity

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This is called the BCS gap equation for reasons to be discussed below. When we have solved it, it is easy to<br />

obtain the original variational parameters in terms <strong>of</strong> ∆ k ,<br />

(<br />

)<br />

u 2 k = 1 ξ k<br />

1 + √ , (9.29)<br />

2 ξ<br />

2<br />

k<br />

+ ∆ 2 k<br />

(<br />

)<br />

vk 2 = 1 ξ k<br />

1 − √ , (9.30)<br />

2 ξ<br />

2<br />

k<br />

+ ∆ 2 k<br />

∆ k<br />

u k v k =<br />

2 √ ξk 2 + . (9.31)<br />

∆2 k<br />

The relative sign <strong>of</strong> u k and v k is thus the sign <strong>of</strong> ∆ k . The absolute sign <strong>of</strong>, say, u k is irrelevant because <strong>of</strong> the<br />

invariance <strong>of</strong> E BCS under simultaneous phase rotations <strong>of</strong> u k , v k (consider a phase factor <strong>of</strong> e iπ = −1).<br />

For our special interaction<br />

{<br />

−V 0 for |ξ k | , |ξ k ′| < ω D ,<br />

V kk ′ =<br />

(9.32)<br />

0 otherwise,<br />

the BCS gap equation becomes<br />

⎧<br />

⎪⎨ − 1 ∑<br />

∆ k ′<br />

(−V 0 )<br />

∆ k = N<br />

2 √ for |ξ<br />

ξ 2 k<br />

⎪⎩<br />

′ , |ξ k ′ | < ω D k<br />

+ ∆ 2 k | < ω D ,<br />

′ k ′<br />

0 otherwise,<br />

(9.33)<br />

which can be solved with the ansatz<br />

∆ k =<br />

{<br />

∆ 0 > 0 for |ξ k | < ω D ,<br />

0 otherwise.<br />

(9.34)<br />

We obtain<br />

∆ 0 = V 0<br />

N<br />

⇒ 1 = V 0<br />

N<br />

∑<br />

∆ 0<br />

2 √ ξ<br />

k ′<br />

k 2 + ∆ 2 ′ 0<br />

|ξ k ′ | < ω D<br />

∑ 1<br />

2 √ ξ<br />

k ′<br />

k 2 + ∆ 2 ′ 0<br />

|ξ k ′ | < ω D<br />

∫ω D<br />

= V 0<br />

(9.35)<br />

1<br />

dξ D(µ + ξ)<br />

2 √ , (9.36)<br />

ξ 2 + ∆ 2 0<br />

−ω D<br />

where D(ϵ) is the density <strong>of</strong> states per spin direction and per unit cell. If the density <strong>of</strong> states is approximately<br />

constant within ±ω D <strong>of</strong> the Fermi energy, we obtain<br />

⇒<br />

1 = V 0 D(E F ) 1 2<br />

sinh<br />

∫ω D<br />

−ω D<br />

dξ<br />

√ = V<br />

ξ2 + ∆ 2 0 D(E F ) Arsinh ω D<br />

(9.37)<br />

∆<br />

0<br />

0<br />

1<br />

V 0 D(E F ) = ω D<br />

∆ 0<br />

(9.38)<br />

1<br />

⇒ ∆ 0 = ω D 1<br />

. (9.39)<br />

sinh<br />

V 0 D(E F )<br />

In the so-called weak-coupling limit <strong>of</strong> small V 0 D(E F ), this result simplifies to<br />

(<br />

)<br />

∆ 0<br />

∼<br />

1<br />

= 2ωD exp −<br />

. (9.40)<br />

V 0 D(E F )<br />

87

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