quantum mechanics - Why does a force field leave the momentum operator unchanged in the Schrödinger equation?

The reasoning leading to the Schrödinger equation goes as follows:

A plane wave in empty space has the following form:

$$\psi = e^{i(kx-\omega t)}$$

Einstein had previously explained the photoelectric effect, i.e. the emission of electrons from a metal surface through light, by suggesting that light is made from photons containing the momentum:

$$p_\text{Photon}=\hbar k$$

and energy:

$$E_\text{Photon}=\hbar\omega$$

This was proven to be correct by experiment.

De Broglie then suggested that the same relations hold for electrons, so that, given the momentum and energy of an electron, one could find the wavelength and frequency of an electron's plane wave:

$$\begin{gather} k_\text{Electron}=\frac{p_\text{Electron}}{\hbar}\\ \omega_\text{Electron}=\frac{E_\text{Electron}}{\hbar} \end{gather}$$

Given the plane-wave function of an electron, one can then use certain operators to extract momentum and energy from it:

$$\begin{gather} \hat{p}\psi = -i\hbar\partial_x\psi = \hbar k \psi\\ \hat{E}\psi = i\hbar\partial_t\psi = \hbar\omega\psi \end{gather}$$

But for a free electron outside of a force field, the relationship between energy and momentum is given by:

$$E = \frac{p^2}{2m}$$

The Schrödinger Equation for a free electron can be derived from this equation by replacing energy and momentum by the extraction operators:

$$\hat{E}\psi = i\hbar\partial_t\psi = \frac{\hat{p}^2}{2m}\psi=-\frac{ \hbar^2\partial^2_x}{2m}\psi$$

Schrödinger now argued that placing the free electron in a Potential would modify the equation simply through the addition of the Potential Energy:

$$i\hbar\partial_t\psi =\biggl(-\frac{\hbar^2\partial^2_x}{2m}+V(x)\biggr)\psi$$

Now my question is: The introduction of a potential is going to mess up the plane waves very seriously, transforming them into something quite different, so the extraction operators for Energy and Momentum - which only work for plane waves, for which they were designed - are no longer going to work! How can the Schrödinger equation still hold up?

Answer

I found a proof that the momentum operator does not change when the wave it operates upon is not a plane wave:

Let $\phi(k) = \frac{1}{2\pi}\int\limits_{-\infty}^{\infty}e^{-ikx}\psi(x)dx$

be the Fourier transform of an (arbitrary) wave function in position space.

The expectation value of k - using the wave function $\psi(x)$ in position space - is:

$\overline{k}=\int\limits_{-\infty}^{\infty}\psi^*(x)k\psi(x)dx$

Rewriting the above expression with the Fourier Transform of $\psi(x)$ we get:

$\overline{k}=\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\psi^*(x)ke^{ikx}\phi(k) dxdk = \frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\psi^*(x)ke^{ikx}e^{-ikx'}\psi(x')dxdkdx'$

We can now write $ke^{-ikx'}$ in the above expression as $i\frac{\partial}{\partial x'} e^{ikx'}$:

$\overline{k}=\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\psi^*(x)e^{ikx}[i\frac{\partial}{\partial x'} e^{ikx'}]\psi(x')dxdkdx'$

Let us integrate the two last factors containing x' by parts, using the fact that the wave equation $\psi(x')$ will vanish at infinity:

$\overline{k}= \frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\psi^*(x)\frac{1}{i}\frac{\partial\psi(x')}{\partial x'}e^{ik(x-x')}dx dkdx'$

The Integral over k of the last exponential together with the factor $\frac{1}{2\pi}$is the Dirac Delta Function, so when we integrate over x, we end up with:

$\overline{k}=\int\limits_{-\infty}^{\infty}\psi^*(x')\frac{1}{i}\frac{\partial\psi(x')}{\partial x'}dx'$

Using the de Broglie equation p = $\hbar k$ and using x instead of x' for clarity we get:

$\overline{p}=\int\limits_{-\infty}^{\infty}\psi^*(x)\frac{\hbar}{i}\frac{\partial\psi(x)}{\partial x}dx$

where we see that the momentum operator $\hat{p}=-i\hbar\partial_x$

Exactly the same demonstration can be made with $\omega$ and $t$ instead of $k$ and $x$, and the result is that the Energy operator is $\hat{E}= i\hbar\partial_t$.

However, I'm still not completely convinced by this argument, because I don't really know what physical interpretation to give to the waves $e^{ikx}$ and $e^{-i\omega t}$ in which we have Fourier-analysed the wave function $\psi (x,t)$ under consideration. We have presumed that they are plane waves with Momentum $\hbar k$ and Energy $\hbar \omega$, respectively, but $e^{ikx}$ is time independent and $e^{-i\omega t}$ is space independent - Something like this is completely unphysical, much more so than a plane wave $e^{i(kx-\omega t)}$ which might not exist for practical purposes, but could at least theoretically exist and has definite momentum AND Energy.

But I believe this is is as good as it gets.