One of the remarkable features of a superconductor is the Meissner effect, which states that a current can flow through the sample without any dissipation. This phenomenon gives rise to what we call supercurrents, which occur due to a coherent effect. In contrast to the normal phase, or the Fermi liquid, where the current is more like a local property resulting from electron scattering with disorder or impurities, superconductors exhibit a different behavior. Interestingly, in the case of a simple s-wave superconductor, the normal phase and superconducting phase share the same current operator.

We consider a spinful Fermi gas with density-density interaction,

\[\begin{align} H & =\sum_{k\sigma}\xi(k)c_{k\sigma}^{\dagger}c_{k\sigma}-\int drUc_{\uparrow}^{\dagger}(r)c_{\downarrow}^{\dagger}(r)c_{\downarrow}(r)c_{\uparrow}(r)\\ & =\sum_{k\sigma}\xi(\mathbf{k})c_{\mathbf{k}\sigma}^{\dagger}c_{\mathbf{k}\sigma}-\frac{U}{V}\sum_{kk^{\prime}q}c_{\mathbf{k}\uparrow}^{\dagger}c_{-\mathbf{k}+\mathbf{q}\downarrow}^{\dagger}c_{-\mathbf{k}^{\prime}+\mathbf{q}\downarrow}c_{\mathbf{k}\prime\uparrow} \end{align}\]

The interaction is a uniform density-density interaction, and we do not expect a current from it. Therefore, regardless of the real phase, the current operator for electrons always have the form as

\[\mathbf{j}(\mathbf{k})=\sum_{\sigma}\nabla\xi(\mathbf{k})c_{\mathbf{k}\sigma}^{\dagger}c_{\mathbf{k}\sigma}\]

One should not confuse the current of an electron with the current of a quasiparticle. From the Landau’s Fermi liquid theory, we can learn that the current of quasiparticles can get renormalized by interaction or orbital hybridization.

We can apply the mean-field theory to obtain an ansatz for a ground state. We can make a mean-field ansatz by assuming

\[\Delta=-\frac{1}{V}\int dr\langle c_{\downarrow}(r)c_{\uparrow}(r)\rangle=-\frac{1}{V}\sum_k\langle c_{-\mathbf{k}\downarrow}c_{\mathbf{k}\uparrow}\rangle\]

Then we obtain a mean-field Hamiltonian \(H_{MF}\)

\[H_{MF}(\mathbf{k})=\sum_{\sigma}\xi(\mathbf{k})c_{\mathbf{k}\sigma}^{\dagger}c_{\mathbf{k}\sigma}+\Delta\left[c_{\mathbf{k}\uparrow}^{\dagger}c_{-\mathbf{k}\downarrow}^{\dagger}+c_{-\mathbf{k}\downarrow}c_{\mathbf{k}\uparrow}\right]\]

where we make a proper gauge choice such that the order parameter \(\Delta\) be a real number. We can diagonalize the mean-field Hamiltonian

\[H_{MF}(\mathbf{k})=E(\mathbf{k})\left[\gamma_{\mathbf{k}\uparrow}^{\dagger}\gamma_{\mathbf{k}\uparrow}+\gamma_{\mathbf{k}\downarrow}^{\dagger}\gamma_{\mathbf{k}\downarrow}\right]\]

by the Bogoliubov transformation

\[\begin{aligned} & \left[\begin{array}{c} \gamma_{\boldsymbol{k},\uparrow}\\ \gamma_{-\boldsymbol{k},\downarrow}^{\dagger} \end{array}\right]=\left[\begin{array}{cc} u_{\boldsymbol{k}} & -v_{\boldsymbol{k}}\\ v_{\boldsymbol{k}} & u_{\boldsymbol{k}} \end{array}\right]\left[\begin{array}{c} c_{\boldsymbol{k},\uparrow}\\ c_{-\boldsymbol{k},\downarrow}^{\dagger} \end{array}\right],\\ & \left[\begin{array}{c} c_{\boldsymbol{k},\uparrow}\\ c_{-\boldsymbol{k},\downarrow}^{\dagger} \end{array}\right]=\left[\begin{array}{cc} u_{\boldsymbol{k}} & v_{\boldsymbol{k}}\\ -v_{\boldsymbol{k}} & u_{\boldsymbol{k}} \end{array}\right]\left[\begin{array}{c} \gamma_{\boldsymbol{k},\uparrow}\\ \gamma_{-\boldsymbol{k},\downarrow}^{\dagger} \end{array}\right]. \end{aligned}\]

with the Bogolibov coefficients

\[\begin{align} u(\mathbf{k}) & =\sqrt{\frac{1}{2}\left(1+\frac{\xi(\mathbf{k})}{E(\mathbf{k})}\right)}\\ v(\mathbf{k}) & =\sqrt{\frac{1}{2}\left(1-\frac{\xi(\mathbf{k})}{E(\mathbf{k})}\right)} \end{align}\]

and the dispersion of quasiparticle \(E(\mathbf{k})=\sqrt{\xi^{2}(\mathbf{k})+\Delta^{2}}\). Accordingly the ground state is

\[\vert\mathrm{BCS}\rangle=\prod_{\mathbf{k}}(u(\mathbf{k})+v(\mathbf{k})c_{\mathbf{k}\uparrow}^{\dagger}c_{-\mathbf{k}\downarrow}^{\dagger})\vert0\rangle\]

It is easy to check \(\gamma_{\mathbf{k}\sigma}\vert\mathrm{BCS}\rangle=0\). The BCS ground state represents a condensation of Cooper pairs. The net momentum of a Cooper pair vanishes and thus we expect the current vanishes

\[\begin{align} \langle\mathrm{BCS}\vert J\vert\mathrm{BCS}\rangle & =\sum_{\mathbf{k}}\langle\mathrm{BCS}\vert\mathbf{j}(\mathbf{k})\vert\mathrm{BCS}\rangle\\ & =\sum_{\mathbf{k}}v^{2}(\mathbf{k})(\nabla\xi(\mathbf{k})-\nabla\xi(\mathbf{k}))=0 \end{align}\]

We need an excited state to show the supercurrents. There are two types of excitations. The first one is the Fermionic Bogolibov quasiparticle \(\gamma_{\mathbf{k}\sigma}^{\dagger}\vert\mathrm{BCS}\rangle\) with an energy \(E=E_{GS}+E(\mathbf{k})\). It has a finite energy gap. In fact, we have the second type of excitation, that is the gapless Bosonic Goldstone mode. By contrast it will cost at least energy \(2\Delta\) to create a Cooper pair state \(\gamma_{\mathbf{k}\uparrow}^{\dagger}\gamma_{\mathbf{-k}\downarrow}^{\dagger}\vert\mathrm{BCS}\rangle\), which can be taken as local excitations (in momentum space). Instead, to pursue the Goldstone mode, we can consider a new mean-field ansatz

\[\Delta_{\mathbf{q}}=\frac{1}{V}\sum_k\langle c_{-\mathbf{k}\downarrow}c_{\mathbf{k}+\mathbf{q}\uparrow}\rangle\]

or in the real space

\[\begin{align} \Delta_{\mathbf{q}} & =\int dre^{i\mathbf{q}\cdot\mathbf{r}}\langle c_{\downarrow}(\mathbf{r})c_{\uparrow}(\mathbf{r})\rangle \end{align}\]

Then we have the mean-field Hamiltonian

\[H_{\mathbf{q}}(\mathbf{k})=\xi(\mathbf{k}+\mathbf{q})c_{\mathbf{k}+\mathbf{q}\uparrow}^{\dagger}c_{\mathbf{k}+\mathbf{q}\uparrow}+\xi(\mathbf{k})c_{-\mathbf{k}\uparrow}^{\dagger}c_{-\mathbf{k}\uparrow}+\Delta_{\mathbf{q}}^{*}c_{\mathbf{k}+\mathbf{q}\uparrow}^{\dagger}c_{-\mathbf{k}\downarrow}^{\dagger}+\Delta_{\mathbf{q}}c_{-\mathbf{k}\downarrow}c_{\mathbf{k}+\mathbf{q}\uparrow}\]

Also we can act the Bogoliubov transformation

\[\begin{align} u_{\mathbf{q}}(\mathbf{k}) & =\sqrt{\frac{1}{2}\left(1+\frac{\xi(\mathbf{k}+\mathbf{q})+\xi(\mathbf{k})}{2\epsilon_{\mathbf{q}}(\mathbf{k})}\right)}\\ v_{\mathbf{q}}(\mathbf{k}) & =\sqrt{\frac{1}{2}\left(1-\frac{\xi(\mathbf{k}+\mathbf{q})+\xi(\mathbf{k})}{2\epsilon_{\mathbf{q}}(\mathbf{k})}\right)} \end{align}\]

with a quasiparticle dispersion \(E_{\mathbf{q\pm}}(\mathbf{k})\)

\[\begin{align} E_{\mathbf{q\pm}}(\mathbf{k}) & =\sqrt{\left(\frac{\xi(\mathbf{k}+\mathbf{q})+\xi(\mathbf{k})}{2}\right)^{2}+\vert\Delta_{\mathbf{q}}\vert^{2}}\pm\frac{\xi(\mathbf{k}+\mathbf{q})-\xi(\mathbf{k})}{2}\\ & \equiv\epsilon_{\mathbf{q}}(\mathbf{k})\pm\frac{\xi(\mathbf{k}+\mathbf{q})-\xi(\mathbf{k})}{2} \end{align}\]

Similarly, we have the ground state

\[\vert\mathrm{BCS}(\mathbf{q})\rangle=\prod_{\mathbf{k}}(u_{\mathbf{q}}(\mathbf{k})+v_{\mathbf{q}}(\mathbf{k})c_{\mathbf{k}+\mathbf{q}\uparrow}^{\dagger}c_{-\mathbf{k}\downarrow}^{\dagger})\vert0\rangle\]

When \(\mathbf{q}\) approaches \(0\), the state \(\vert\mathrm{BCS}(\mathbf{q})\rangle\) reduces to \(\vert\mathrm{BCS}\rangle\).Now let us evaluate the current operator

\[\begin{align} \langle\mathrm{BCS(\mathbf{q})}\vert\mathbf{J}\vert\mathrm{BCS(\mathbf{q})}\rangle & =\sum_{\mathbf{k}}\langle\mathrm{BCS(\mathbf{q})}\vert\mathbf{j}(\mathbf{k})\vert\mathrm{BCS(\mathbf{q})}\rangle\\ & =\sum_{\mathbf{k}}\langle\mathrm{BCS(\mathbf{q})}\vert\sum_{\sigma}\nabla\xi(\mathbf{k})c_{\mathbf{k}\sigma}^{\dagger}c_{\mathbf{k}\sigma}\vert\mathrm{BCS(\mathbf{q})}\rangle\\ & =\sum_{\mathbf{k}}\left(\nabla\xi(\mathbf{k}+\mathbf{q})-\nabla\xi(\mathbf{k})\right)v_{\mathbf{q}}^{2}(\mathbf{k}) \end{align}\]

If the \(\mathbf{q}\) is vanishingly small, we can assume \(\Delta_{\mathbf{q}}=\Delta\) or

\[\Delta_{\mathbf{q}}(\mathbf{r})=\Delta e^{i\mathbf{q}\cdot\mathbf{r}}\]

By noting

\[\begin{align} \epsilon_{\mathbf{q}}(\mathbf{k}) & =\sqrt{\left(\frac{\xi(\mathbf{k}+\mathbf{q})+\xi(\mathbf{k})}{2}\right)^{2}+\vert\Delta_{\mathbf{q}}\vert^{2}}\\ v_{\mathbf{q}}^{2}(\mathbf{k}) & =v^{2}(\mathbf{k})+\mathcal{O}(q) \end{align}\]

we can approximate

\[\langle\mathrm{BCS(\mathbf{q})}\vert J_{a}\vert\mathrm{BCS(\mathbf{q})}\rangle=\sum_{\mathbf{k}}\sum_{b}q_{b}\partial_{b}\partial_{a}\xi(\mathbf{k})v^{2}(\mathbf{k})\] \[\begin{align} \langle\mathrm{BCS(\mathbf{q})}\vert j_{a}\vert\mathrm{BCS(\mathbf{q})}\rangle & =\sum_{\mathbf{k}}\left(\nabla\xi(\mathbf{k}+\mathbf{q})-\nabla\xi(\mathbf{k})\right)v_{\mathbf{q}}^{2}(\mathbf{k})\\ & =\frac{q_{a}}{m}\sum_{\mathbf{k}}v^{2}(\mathbf{k})=\frac{q_{a}}{m}n_{s} \end{align}\]

where \(n_{s}\) is the superfluid stiffness,

\[n_{s}=\sum_{\mathbf{k}}v^{2}(\mathbf{k})\]

and the superfluid weight is

\[D_{s}=\frac{n_{s}}{m}\]

The supercurrent in a superconductor is considered global because it involves the integral over all momentum states. In other words, the state represented by \(\vert \mathrm{BCS}(\mathbf q)\rangle\), which consists of Bogoliubov quasiparticles, undergoes a change in its wave function throughout the entire system, indicating a global excitation.

In contrast, in a Fermi liquid theory, when an electron is excited from the Fermi surface, it carries charge currents that involve only a single momentum point on the Fermi surface. Therefore, we can consider this type of current as local, as it is confined to a specific region or point in the system.