Superconductivity

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

Meissner effect

Normally, it is easy to establish a magnetic field through a metal. However, when the metal is cooled below its critical temperature, the field is expelled.

Explanation

1. Quantum mechanics. We start with the Schrödinger equation for the state $\psi$ of a particle with charge $q$ and mass $m$ moving in an electromagnetic field $(\bfA,\phi)$. Writing $\phat \coloneqq \frac{\hbar}{i}\nabla$, the equation is $$ - \frac{\hbar}{i} \frac{\partial\psi}{\partial t} = \frac{1}{2m}\bigl(\phat - q \bfA\bigr) \cdot \bigl(\phat - q \bfA\bigr) \psi + q\phi \psi.$$ From this, we find that the probability density $P \coloneqq \psi^* \psi$ is locally conserved, with associated current $$\bfJ \coloneqq \frac{1}{2}\biggl( \psi^* \biggl(\frac{\phat- q \bfA}{m} \biggr) \psi + \psi \biggl(\frac{\phat-q \bfA}{m}\biggr)^*\psi^*\biggr).$$ Writing $\psi = \sqrt{P} \cdot e^{i\theta}$, a little algebra gives $$\bfJ = \frac{\hbar}{m} \Bigl( \nabla \theta - \frac{q}{\hbar} \bfA \Bigr) P.$$
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. Field expulsion. In the final steady state, physics dictates that $\rho$ and $\bfE$ are constant in space. Taking the curl of the last equation gives $$\nabla \times \bfj = -\frac{q\rho}{m}\bfB.$$ Taking the curl of Ampere's equation and combining with the preceding equation gives $$\nabla^2 \bfB = \lambda^2 \bfB,$$ where $\lambda \coloneqq \sqrt{\mu_0 q \rho/m}$. The solution decays exponentially with the distance $d$ into the material as $\exp(-\lambda d)$; so the characteristic penetration depth is $1/\lambda$.

Flux quantization

Take a ring-shaped metal and establish a magnetic field through it. When the metal is cooled below its critical temperature, the magnetic flux through the ring is quantized.