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I'm asking specifically for the schrodinger equation. Is there a unique correspondence between the energy eigenfunctions $\phi_i(x)$ and the potential term $U(x) = V(x)\phi(x)$? Furthermore, is this still true for a more general potential term that is a functional of $\phi(x)$ such as: $U(x) \equiv F(\phi(x))$? If not, I'm interested in what statements can be made.

anon01
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    See here. Even in finite dimensions, the eigenvectors don't uniquely determine the matrix. You need both the eigenvalues and the eigenvectors to uniquely define the matrix (and I don't remember whether or not this is enough in infinite dimensions). – march Nov 06 '15 at 01:25
  • @march thanks, I had that question in mind when posting this. I was actually approaching this from the other direction: if you compare two potentials that appear different (ie, gauge transformed) is it enough to show they have the same eigenvectors (and maybe energies, as you point out) to say they are indeed equivalent hamiltonians that share expectation values for all observables? – anon01 Nov 06 '15 at 02:20
  • I'm not sure what you mean. If you have two potentials that appear different, it's not enough to show that the Hamiltonians share a set of eigenvectors in order to say that they are identical, but if you have extra information (like that they're related by a gauge transform, although how would you know that to begin with?), then perhaps there's something you can do. I don't know the answer to that question however. – march Nov 06 '15 at 03:25

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