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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):


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.]


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Neumann boundary conditions imply that $\theta$ must be constant, see – jjcale Sep 12 '13 at 19:14
Neumann boundary conditions on what quantity? If on $\theta$, I agree with you, but, as mentioned, part of the difficulty is trying to find any physical justification for those boundary conditions. – Sam Bader Sep 12 '13 at 23:13

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.

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Sorry, yes, J may be non-zero at the boundary; it is only the boundary-normal component which must be zero. I edited above to fix that typo. – Sam Bader Feb 3 '13 at 23:37
Though I still miss how we get from "theta solves the Laplace equation" to "theta=0 everywhere". This argument from the Feynman lectures arrives at the second London equation from the section that I excerpted above. Then it derives the Meissner effect from the second London equation. So I'd rather avoid being circular by using the Meissner effect in completing this part if possible. I apologize for my confusion, and I hope we can figure this out. Thanks! – Sam Bader Feb 3 '13 at 23:49

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.

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up vote 0 down vote accepted

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:

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