In Ginzburg-Landau theory, superconductivity is due to "supercarriers" of mass $m_{s}$ and charge $q_{s}$, which are described by a macroscopic wavefunction $\psi$ with "Mexican hat" potential

$$V=\alpha(T)|\psi|^{2}+\frac{1}{2} \beta|\psi|^{4}$$

Here, $\beta>0$ is constant and $\alpha(T)$ is a function of temperature $T$ such that $\alpha(T)>0$ for $T>T_{c}$ but $\alpha(T)<0$ for $T<T_{c}$, where $T_{c}$ is a critical temperature. In the presence of a magnetic field $\mathbf{B}=\boldsymbol{\nabla} \times \mathbf{A}$, the total energy of the superconducting system is

$$E\left[\psi, \psi^{*}, \mathbf{A}\right]=\int d^{3} x\left[\frac{1}{2 \mu_{0}} A_{k, j}\left(A_{k, j}-A_{j, k}\right)+\frac{\hbar^{2}}{2 m_{s}}\left|\psi_{, k}+i \frac{q_{s}}{\hbar} A_{k} \psi\right|^{2}+V\right]$$

Use this to derive the equations

$$-\frac{\hbar^{2}}{2 m_{s}}\left(\nabla-i \frac{q_{s}}{\hbar} \mathbf{A}\right)^{2} \psi+\left(\alpha+\beta|\psi|^{2}\right) \psi=0$$

and

$$\boldsymbol{\nabla} \times \mathbf{B} \equiv \boldsymbol{\nabla}(\boldsymbol{\nabla} \cdot \mathbf{A})-\nabla^{2} \mathbf{A}=\mu_{0} \mathbf{j}$$

where

$$\begin{aligned} \mathbf{j} &=-\frac{i q_{s} \hbar}{2 m_{s}}\left(\psi^{*} \boldsymbol{\nabla} \psi-\psi \boldsymbol{\nabla} \psi^{*}\right)-\frac{q_{s}^{2}}{m_{s}}|\psi|^{2} \mathbf{A} \\ &=\frac{q_{s}}{2 m_{s}}\left[\psi^{*}\left(-i \hbar \boldsymbol{\nabla}-q_{s} \mathbf{A}\right) \psi+\psi\left(i \hbar \boldsymbol{\nabla}-q_{s} \mathbf{A}\right) \psi^{*}\right] \end{aligned}$$

Suppose that we write the wavefunction as

$$\psi=\sqrt{n_{s}} e^{i \theta}$$

where $n_{s}$ is the (real positive) supercarrier density and $\theta$ is a real phase function. Given that

$$\left(\boldsymbol{\nabla}-\frac{i q_{s}}{\hbar} \mathbf{A}\right) \psi=0$$

show that $n_{s}$ is constant and that $\hbar \boldsymbol{\nabla} \theta=q_{s} \mathbf{A}$. Given also that $T<T_{c}$, deduce that (*) allows such solutions for a certain choice of $n_{s}$, which should be determined. Verify that your results imply $\mathbf{j}=\mathbf{0}$. Show also that $\mathbf{B}=\mathbf{0}$ and hence that ( $\dagger$ ) is solved.

Let $\mathcal{S}$ be a surface within the superconductor with closed boundary $\mathcal{C}$. Show that the magnetic flux through $\mathcal{S}$ is

$$\Phi \equiv \int_{\mathcal{S}} \mathbf{B} \cdot \mathbf{d} \mathbf{S}=\frac{\hbar}{q_{s}}[\theta]_{\mathcal{C}}$$

Discuss, briefly, flux quantization.