Superconducting Wavefunction Phase (Feynman Lectures) In Volume 3, Section 21-5 of the Feynman lectures (superconductivity), Feynman 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 Feynman 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, Feynman 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 Laplace 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 $A=B_0x\hat{y}$.  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 Laplace'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 Feynman say $\theta=0$?  This seems important as the next step results in the a London equation.  [EDIT: Jack Chi, 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 Feynman's.]
Thanks!
 A: There is no magnetic field in a superconductor: this is the Meissner effect. $\vec J$ may be nonzero on the boundary in order to make the magnetic field zero in the interior. A good way to understand this is Ginzburg-Landau theory. The free energy (from wikipedia) 
$$F = \alpha |\psi|^2 + \frac{\beta}{2} |\psi|^4  + \frac{1}{2m} \left| (-i \hbar \vec \nabla - 2 e \vec A) \psi \right|^2 + \frac{|\vec B|^2}{2 \mu_0}$$
includes a free energy cost to nonzero $\vec J$ and nozero $\vec B$. So, the ground state wavefunction "expels" the magnetic field.
In general, there is a "no node" theorem from Feynman that predicts that the ground state of a bosonic system has a positive definite wavefunction. A superconductor is a clear example of this, insofar that we consider Cooper pairs to be bosonic. Superfluids are the same: the wavefunction is real everywhere.
What you've started to derive is the wavefunction of a Landau level in the symmetric gauge, which describes a charged particle in a magnetic field.
A: After more reading from other resources, including the pages mentioned by emarti (thank you!), I cannot fill in the step Feynman left out.
However, if, instead of taking the divergence of $J=\frac{\hbar}{m}(\nabla \theta -\frac{q}{\hbar}A)\rho$ like Feynman, you take the curl, you quickly arrive at $\nabla \times\Lambda  J=-B$, where $\Lambda=\frac{m^*}{n^*{q^*}^2}$, the second London equation.  And this equation leads directly to the Meissner effect.  I still don't know why Feynman took the divergence.
For more context on this approach: http://web.mit.edu/6.763/www/FT03/Lectures/Lecture10.pdf
A: I had the similar issue when I read this part. My handwaving answer is that $\theta$ is continuous from inside the superconductor to the outside vacuum. In vacuum, $\theta=0$ and homogeneous.
   In your choice $\nabla\theta=\frac{qB_0x}{\hbar}\hat{y}$, you had missed the $\hat{x}$ component. (Do calculate $\theta$ and then take the gradient.) Therefore, $J\neq 0$ everywhere.
   But indeed, take the curl is a much clear way.
