Magnetic Flux Quantization from Gauge Symmetry

We've shown that we can compute the function $f(\vec{r})$ from the vector potential.

$$\bgroup\color{black}f(\vec{r})=\int\limits_{\vec{r}_0}^{\vec{r}}d\vec{r}\cdot\vec{A}\egroup$$

A superconductor excludes the magnetic field so we have our field free region. If we take a ring of superconductor, as shown, we get a condition on the magnetic flux through the center.

Consider two different paths from $\vec{r}_0$ to $\vec{r}$.

$$\bgroup\color{black}f_1(\vec{r})-f_2(\vec{r})=\oint d\vec{r}\... ...vec{\nabla}\times\vec{A} =\int d\vec{S}\cdot\vec{B}=\Phi\egroup$$

The difference between the two calculations of f is the flux.

Now $f$ is not a physical observable so the $f_1-f_2$ does not have to be zero, but, $\psi$ does have to be single valued.

$$\begin{eqnarray*} \psi_1 & = & \psi_2 \\ & \Rightarrow & e^{-i{e\over\hbar c}f... ...2)=2n\pi \\ & \Rightarrow & \Phi=f_1-f_2={2n\pi\hbar c\over e} \end{eqnarray*}$$

The flux is quantized.

Magnetic flux is observed to be quantized in a region enclosed by a superconductor. however, the fundamental charge seen is $2e$.