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Carsten Timm: Theory of superconductivity

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we obtain<br />

(<br />

)<br />

2ξ k |u k | |v k | e i(β k−α k ) + |∆ k | |v k | 2 e i(2β k−ϕ k ) − |u k | 2 e i(ϕ k−2α k )<br />

= 0. (10.16)<br />

A special solution <strong>of</strong> this equation (we do not require the general solution) is given by<br />

From the last equation we obtain<br />

α k = 0, (10.17)<br />

β k = ϕ k , (10.18)<br />

(<br />

2ξ k |u k | |v k | + |∆ k | |v k | 2 − |u k | 2) = 0. (10.19)<br />

(<br />

4ξk 2 |u k | 2 |v k | 2 = |∆ k | 2 |v k | 4 − 2 |v k | 2 |u k | 2 + |u k | 4) (10.20)<br />

⇒<br />

(<br />

4 ξk 2 + |∆ k | 2) ( |u k | 2 |v k | 2 = |∆ k | 2 |v k | 4 + 2 |v k | 2 |u k | 2 + |u k | 4) ( = |∆ k | 2 |v k | 2 + |u k | 2) 2<br />

= |∆k | 2 (10.21)<br />

|∆ k |<br />

⇒ |u k | |v k | =<br />

(10.22)<br />

2<br />

√ξk 2 + |∆ k| 2<br />

so that<br />

|u k | 2 − |v k | 2 = 2ξ k |u k | |v k |<br />

|∆ k |<br />

=<br />

ξ k<br />

√ξ 2 k + |∆ k| 2 . (10.23)<br />

Together with |u k | 2 + |v k | 2 = 1 we thus find<br />

⎛<br />

⎞<br />

|u k | 2 = 1 ξ k<br />

⎝1 + √ ⎠ , (10.24)<br />

2<br />

ξk 2 + |∆ k| 2<br />

⎛<br />

⎞<br />

|v k | 2 = 1 ξ k<br />

⎝1 − √ ⎠ . (10.25)<br />

2<br />

ξk 2 + |∆ k| 2<br />

Restoring the phases in Eq. (10.22), we also conclude that<br />

∆ k<br />

u k v k =<br />

2<br />

√ξ . (10.26)<br />

k 2 + |∆ k| 2<br />

The BCS Hamiltonian now reads, ignoring a constant,<br />

⎛<br />

⎞<br />

H BCS = ∑ ⎝<br />

ξk<br />

2 √ + |∆ k | 2<br />

√ ⎠ ( )<br />

γ †<br />

k ξk 2 + |∆ k| 2 ξk 2 + |∆ k| 2 k↑ γ k↑ + γ † −k,↓ γ −k,↓<br />

= ∑ √ ( )<br />

ξk 2 + |∆ k| 2 γ † k↑ γ k↑ + γ † −k,↓ γ −k,↓ . (10.27)<br />

k<br />

Using ξ −k = ξ k and the plausible assumption |∆ −k | = |∆ k |, we obtain the simple form<br />

H BCS = ∑ kσ<br />

E k γ † kσ γ kσ (10.28)<br />

with the dispersion<br />

E k :=<br />

√<br />

ξ 2 k + |∆ k| 2 . (10.29)<br />

91

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