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 last 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 $1/\lambda$.

    4. Flux quantization

    Take a ring-shaped metal and establish a magnetic field through it. Cool the metal below its critical temperature and remove the field, then the magnetic flux through the ring becomes quantized.

    Explanation
    1. Inside the body of the ring, we can approximate $\bfB$ as zero due to the Meissner effect, and therefore also approximate $\bfj \propto \bfB$ as zero. Then we have $$\hbar \nabla \theta = q \bfA.$$