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BCS ground state and variational equation
This section shows the proper BCS model, following De Gennes 4-3 and Tinkham 3.4. The sketch provided in the Cooper pair section is qualitatively the same, but the final equation comes out different, so you can skip this, but it is instructive to have a look at the result below.
The solution is obtained by a variational principle with a trial wave function in the grand canonical system. The reason for this is that at very large numnerb {$N$} of particles, few particles (e.g. 2-3) are very much less than {$\sqrt N$}, which in turns is very much less than {$N$}. Therefore observables do not depend on the fluctuation of few particles and the canonical and grand canonical solution give essentially the same result. Allowing for variable number of particles the trial wave function may be written as the following combination
{$$\begin{equation}|\Psi\rangle = \prod_{\mathbf k}(\cos\theta_{\mathbf k} + \sin\theta_{\mathbf k}c^\dagger_{\mathbf k \uparrow}c^\dagger_{-\mathbf k \downarrow})|0\rangle\end{equation}$$}
Inspection of this wave function shows that it contains a mixture of occupied and unoccupied states for each pair. The sine term represents the amplitude of the occupied pair state and the cosine term the amplitude of the unoccupied pair state and normalization is granted by the trigonometric identity. Actually, calculating the average on {$|\Psi\rangle$} of the particle number {${\cal N}=\sum_{\mathbf k \sigma}c^\dagger_{\mathbf k \sigma}c_{\mathbf k \sigma}$} notice that for each value of {$\mathbf k$} only two terms do not vanish. They are the {$\sin^2 \theta_{\mathbf k}=\sin^2\theta_{-\mathbf k}$} terms, for {$\sigma=\uparrow$} and {$\sigma=\downarrow$} respectively, since for all other terms inevitably a factor {$c_{\mathbf k \sigma}|0\rangle =0$} or {$\langle 0|c^\dagger_{\mathbf k \sigma}=0$} appears. Therefore
{$$\langle\Psi|{\cal N}|\Psi\rangle = 2\sum_{\mathbf k}\sin^2\theta_{\mathbf k}$$}
In the grand canonical system the energy eigenvalue must be minimized together with the particle number, with a Lagrange multiplier equal to the chemical potential, to all effects the Fermi energy {$\epsilon_F$} in a superconductor, i.e. by using an equivalent Hamiltonian
{$$\begin{equation}{\cal H}-\epsilon_F{\cal N}=\sum_{\mathbf k\sigma} (\epsilon_{\mathbf k}-\epsilon_F)c^\dagger_{\mathbf k \sigma}c_{\mathbf k \sigma} + {\delta \cal V}\end{equation}$$}
Notice that
- we define a quasiparticle energy {$\varepsilon_{\mathbf k}=\epsilon_{\mathbf k}-\epsilon_F$}, that reproduces the condition under which Cooper pairs are bound however small the actractive potential;
- we assume that {$\epsilon_{\mathbf k}$} already includes mean field (or better) contributions from the Coulomb and exchange potential, so that {$\delta{\cal V}$} is only the residual attractive potential discussed above;
- {$\delta{\cal V}$} gives rise to transitions from pair state {$\mathbf k\uparrow,-\mathbf k \downarrow$} to pair state {$\mathbf q\uparrow,-\mathbf q \downarrow$} with matrix element {$V_{\mathbf k\mathbf q}$}
- for the transition probability to be non zero the initial state must be occupied and the final state unoccupied.
Therefore the trial function the gives non vanishing contributions for
{$$\langle \Psi|\delta{\cal V}|\Psi\rangle = \sum_{\mathbf k\mathbf q}V_{\mathbf k\mathbf q}\sin\theta_{\mathbf k}\cos\theta_{\mathbf q}\sin\theta_{\mathbf q}\cos\theta_{\mathbf k} $$}
The grand canonical energy, Eq. (2), to be minimized with respect to the variational parameter {$\theta_{\mathbf k}$} defined in Eq. (1), is then
{$$E = \langle\Psi|{\cal H} - \epsilon_F{\cal N}|\Psi\rangle = 2\sum_{\mathbf k} \varepsilon_{\mathbf k}\sin^2\theta_{\mathbf k} + \frac 1 4 \sum_{\mathbf k\mathbf q}V_{\mathbf k\mathbf q}\sin2\theta_{\mathbf k}\sin2\theta_{\mathbf q}$$}
yielding
{$$\begin{equation}\frac {\partial E}{\partial \theta_{\mathbf k^\prime}} = 0 = 4\varepsilon_{\mathbf k^\prime}\sin\theta_{\mathbf k^\prime} \cos\theta_{\mathbf k^\prime}+\frac 1 4 2\sum_{\mathbf q}2\cos2\theta_{\mathbf k^\prime}\sin2\theta_{\mathbf q}V_{\mathbf k\mathbf q}\end{equation}$$}
where the second term is counted twice because state {${\mathbf k^\prime}$} occurs in both sums over {$\mathbf k$} and {$\mathbf q$}. The minimizing {$\theta_{\mathbf k}$} obeys the BCS equation
{$$2\varepsilon_{\mathbf k} \tan 2\theta_{\mathbf k} = - \sum_{\mathbf q} V_{\mathbf k\mathbf q} \sin2\theta_{\mathbf q}$$}
If we define a gap energy (half the second member of the BCS equation) and a total energy respectively as
{$$\Delta_{\mathbf k} = -\frac 1 2 \sum_{\mathbf q} V_{\mathbf k\mathbf q} \sin2\theta_{\mathbf q}\qquad E_{\mathbf k} = \sqrt{\Delta_{\mathbf k}^2 + \varepsilon_{\mathbf k}^2}$$}
we get {$\tan 2\theta_{\mathbf k} = \frac {\Delta_{\mathbf k} }{\varepsilon_{\mathbf k}} $}, hence {$\sin 2\theta_{\mathbf k} = \frac {\Delta_{\mathbf k}} {E_{\mathbf k}}$} and {$\cos2\theta_{\mathbf k} = \frac {\varepsilon_{\mathbf k}} {E_{\mathbf k}}$} .
Index
BCS gap equation
We can thus substitute {$\sin 2\theta_{\mathbf k}$} in the BCS equation to get the ground state gap
{$$\Delta_{\mathbf k} = - \frac 1 2 \sum_{\mathbf q} V_{\mathbf k\mathbf q} \frac {\Delta_{\mathbf q}} {E_{\mathbf q}} = - \frac 1 2 \sum_{\mathbf q} V_{\mathbf k\mathbf q} \frac {\Delta_{\mathbf q}} {\sqrt{\Delta_{\mathbf q}^2 + \epsilon_{\mathbf q}^2}}$$}
This equation has a trivial solution, {$\Delta_{\mathbf k}=0$} corresponding to {$\sin\theta_{\mathbf k}=1,0$} for {$\varepsilon_{\mathbf k}\lessgtr 0$}, corresponding to the normal metal. The superconducting solution can be obtained in a simplified model, where {$V_{\mathbf k\mathbf q} =-V$} for {$|\varepsilon_{\mathbf k}|,|\varepsilon_{\mathbf q}| \lt\hbar\omega_D$} and zero otherwise. In this case {$\Delta_{\mathbf k} = \Delta,0$} for {$|\varepsilon_{\mathbf k}| \lessgtr \hbar\omega_D$} and the BCS equation becomes
{$$\frac 1 V = \frac 1 2 \sum_{\mathbf k} \frac 1 {\sqrt{\Delta^2 + \epsilon_{\mathbf k}^2}} = \rho(0)\int_{-\hbar\omega_D}^{\hbar\omega_D} \frac {d\epsilon}{2\sqrt{\Delta^2 + \epsilon^2}}$$}
Index
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