Magnetic Flux Quantization from Gauge Symmetry

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


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 \bgroup\color{black}$\vec{r}_0$\egroup to \bgroup\color{black}$\vec{r}$\egroup.

\begin{displaymath}\bgroup\color{black}f_1(\vec{r})-f_2(\vec{r})=\oint d\vec{r}\...
=\int d\vec{S}\cdot\vec{B}=\Phi\egroup\end{displaymath}

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

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

\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}

The flux is quantized.

Magnetic flux is observed to be quantized in a region enclosed by a superconductor. however, the fundamental charge seen is \bgroup\color{black}$2e$\egroup.

Jim Branson 2013-04-22