Meissner effect

Based on The Feynman Lectures on Physics, Volume III, Lecture 21

The effect “If you have a piece of metal at a high temperature (so that it is a normal conductor) and establish a magnetic field through it, and then you lower the temperature below the critical temperature (where the metal becomes a superconductor), the field is expelled.”

Assume the piece of metal is in a lump, i.e., simply connected.

The explanation

1. We start with the Schrödinger equation for the state $\psi$ of a particle with charge $q$ and mass $m$ moving in an electromagnetic field $(\bfA,\phi)$: $$ - \frac{\hbar}{i} \frac{\partial\psi}{\partial t} = \frac{1}{2m}\bigl(\phat - q \textbf{A}\bigr) \cdot \bigl(\phat - q \textbf{A}\bigr) \psi + q\phi \psi,$$ where $\phat \coloneqq \frac{\hbar}{i}\nabla$.
   Using the above, we find that the probability density $P \coloneqq \psi^* \psi$ is locally conserved, with associated current $$\bfJ \coloneqq \frac{1}{2}\biggl( \psi^* \biggl(\frac{\phat- q \bfA}{m} \biggr) \psi + \psi \biggl(\frac{\phat-q \bfA}{m}\biggr)^*\psi^*\biggr).$$ This means that $$ \frac{\partial P}{\partial t} = - \nabla \cdot \bfJ.$$ Writing $\psi = \sqrt{P} \cdot e^{i\theta}$, we can do a little algebra to express $\bfJ$ as $$\bfJ = \frac{\hbar}{m} \Bigl( \nabla \theta - \frac{q}{\hbar} \bfA \Bigr) P.$$
2. Macroscopic wavefunction. Below the critical temperature, electrons form pairs that all occupy the same lowest-energy state $\psi$, which is possible because each pair is a boson. In this scenario, the charge density $\rho$ is proportional to the probability density $P$, and the current density $\bfj$ is proportional to $\bfJ$. The constant of proportionality in both cases is the total charge in the metal. So the last equation implies $$\bfj = \frac{\hbar}{m} \Bigl( \nabla \theta - \frac{q}{\hbar} \bfA \Bigr) \rho.$$
3. In the final steady state, physics dictates that $\rho$ is constant in space and time. Let's work in the gauge such that $\nabla \cdot \bfA = 0$ inside the metal and $\bfnhat \cdot \bfA = 0$ on its boundary. Then the last displayed equation gives $\nabla^2 \theta = 0$ inside the metal and $\bfnhat \cdot \nabla \theta = 0$ on its boundary. We know that $\psi$ must be smooth inside the metal. But because the metal is in a lump, we may also choose $\theta$ such that it is also smooth inside the metal. Then, we can apply vector calculus to deduce $$\int_V \nabla \cdot(\theta \nabla \theta) \, \mathrm{d}V = \int_V |\nabla\theta|^2 \mathrm{d}V + \int_V \theta \nabla^2 \theta \, \mathrm{d}V = \int_V |\nabla\theta|^2 \mathrm{d}V,$$ and $$\int_V \nabla \cdot(\theta \nabla \theta) \, \mathrm{d}V = \int_S \theta (\bfnhat \cdot \nabla \theta) \, \mathrm{d}S = 0.$$ So $\nabla \theta = 0$ must be zero everywhere inside the metal. Therefore, $$\bfj = -\frac{\rho q}{m} \bfA.$$
4. Now classical electromagnetism (Maxwell's equations) imposes $$\nabla^2 \bfA = - \frac{1}{\epsilon_0 c^2} \bfj.$$ Combining the last two equations gives $$\nabla^2 \bfA = \lambda^2 \bfA,$$ where $\lambda \coloneqq (\rho \, \frac{q}{\epsilon_0 mc^2})^{1/2}$.
   The solution to the last equation decays with the distance $d$ into the material as $\exp(-\lambda d)$. The characteristic length scale of penetration is therefore $1/\lambda$.