Superconducting Wavefunction Phase (Feynman Lectures)
In Volume 3, Section 21-5 of the <PERSON> lectures (superconductivity), <PERSON> makes a step that I can't quite follow. To start, he writes the wavefunction of the ground state in the following form (21.17):
$\psi(r)=\rho(r)e^{i\theta(r)}$
If the density $\rho^2$ is approximated to be constant throughout a superconducting block, then <PERSON> says (21.18) that the (probability) current density can be written $J=\frac{\hbar}{m}\left( \nabla\theta-\frac{q}{\hbar}A \right)\rho$
By insisting that the divergence of the current be zero, <PERSON> shows that the Laplacian of the phase is zero. (Assuming $A$ is chosen to have zero divergence).
$\nabla^2 \theta =0$
I follow everything up to here.
Then he states that, for a single lump of superconducting material (by which I assume he means finite and simply-connected) this implies $\theta=0$.
I don't understand that step...I recognize that the <PERSON> equation has $\theta=0$ as its unique solution if the boundary conditions are $\theta=0$. But the implied boundary condition for the superconducting chunk I would assume is only $J=0$ normal to the boundary (so no current flows in/out), which is not equivalent to $\theta=0$.
Now, for concreteness, let me choose a B-field $B=B_0\hat{z}$. Then one choice of $A$ is $A=B_0x\hat{y}$. 474
This choice makes $\nabla\cdot A=0$. In fact, if we use this B-field, then we can set $\nabla\theta=\frac{qB_0x}{\hbar}\hat{y}$, so that $J=0$ everywhere. The divergence of $\nabla \theta$ is zero, so <PERSON>'s equation is satisfied, and we can integrate this up to get a wavefunction
$\psi(r)=\rho e^{i qB_0xy/\hbar}$
So what have I done wrong? Why does <PERSON> say $\theta=0$? This seems important as the next step results in the a London equation. [EDIT: <PERSON>, below, pointed out that this example was not valid. The reason is that my choice of $\nabla\theta$ has curl, and thus is not a possible gradient. Furthermore, my answer below lists an alternate route of derivation from <PERSON>'s.]
Thanks! 474