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Required fields*

Energy Eigenfunction Completeness

It's my understanding that eigenfunctions are complete (span the space). I don't know what the solution to the (Time-Dependent) Schrödinger Equation is, but whatever it is, any solution (no matter the potential $V$) can be expanded in terms of say position eigenfunctions or momentum eigenfunctions. I'd like to emphasize the phrase - no matter the potential $V$ - with doubt because this ties into my question. Energy eigenfunctions can also be used to represent a general solution. However, this is where my question begins:

Consider a set of energy eigenfunctions $\psi_n$ which satisfy by definition $\hat{H}\psi_n = E_n\psi_n$. It seems to me that the sum $\Psi = \sum c_n\psi_ne^{-iE_nt/\hbar}$ is a general solution to the Schrödinger equation only when the potential $V$ of the Schrödinger equation matches the potential $V$ in the Time-Independent Schrödinger Equation used to find the $\psi_n$'s. Is this correct? If it is, could one say that energy eigenfunctions are complete only with respect to the specific potential $V$ from which they are derived while (say) momentum eigenfunctions are complete with respect to any potential. If this is not true, if energy eigenfunctions of the Time-Independent Schrödinger equation $\hat{H}\psi_n = E_n\psi_n$ are complete with respect to any potential $V$ used in the (Time-Dependent) Schrödinger equation, why can't we use the $\psi_n$'s of say the infinite square well to construct general solutions $\Psi$ of the delta-function well, finite potential well, free particle, etc. Why are we always solving the time-independent Schrödinger equation when we can just use the energy eigenfunctions of the infinite square well?

Answer*

You have to be a bit more careful. When you say that "momentum eigenfunctions are complete with respect to any potential" you're saying that any function can be decomposed as
$$f(x) = \int \frac{dp}{2\pi} \, e^{ipx} \hat{f}(p)\,,
\quad
\hat{f}(p) = \int dx\, e^{-ipx} f(x)\,.$$
But this is formal nonsense: for a general function $f(x)$, the integral defining $\hat{f}(p)$ doesn't exist (it diverges). You need to impose some conditions on $f$, for instance that 
$$
\int dx \, |f(x)|^2 < \infty\,.
$$
By imposing such constraints (there are various choices you can make), you're defining a class of functions which is precisely a Hilbert space. This leads you down the mathematician's way to define the Fourier transform.

The same holds for the type of problem you're interested in, namely a complete basis of wavefunctions $\{\psi_n\}$ with respect to some potential $V(x)$. There you have a decomposition
$$
f(x) = \sum_n f_n \psi_n(x)\,,
\quad
f_n = \int dx\, w(x)\, \psi_n^*(x)\, f(x)
$$
where this is possibly some weight function $w(x)$. Again the integral defining $f_n$ diverges if you plug in a general function $f(x)$. Once more, you have to work within some Hilbert space $\mathcal{H}$ to make sense of these manipulations. If you do the exercise, you will see that the boundary conditions at infinity (or at some boundary points, if you're working on a finite interval) are essentially encoded by the potential $V(x)$.

This whole business is called Sturm-Liouville theory, and if you want to understand QM at any serious level, you should spend some time studying it (it's not difficult).

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