On changing frequency fields as a solution to in the Schrödinger equation

Question

In the initial variant, the SE had the following form: $$\Delta{\psi}=-\frac{2m}{\hbar}\left(W+\frac{e^{2}}{4\pi \epsilon r}\right)\tag{1}$$ Schrödinger and Weyl cut off the divergent power series of the radial function R(r) at a κ-th term. This enabled them to get the radial solutions.

Now, consider the time-independent SE: $$∇^2\psi=-\frac{2m}{\hbar}\left(E-V\right)\psi$$ This has the solution:$∇^2\psi=-k^{2}\psi\tag{2}$ where $$k=\pm\sqrt{\frac{2m}{\hbar}\left(E-V\right)}$$ Comparing $(1)$ and $(2)$, we get: $$k=\pm\sqrt{\frac{2m}{\hbar}\left(W+\frac{e^{2}}{4\pi \epsilon r}\right)}$$

This implies that the wave number k in Schrödinger’s radial equation varies continuously in the radial direction. How can there be a field, where the wave number, and hence the frequency, change from one point to another in the space of the field? Such wave objects do not exist in Nature, right?

Answer 1

To take your final questions one by one:

This implies that the wave number $k$ in Schrödinger’s radial equation varies continuously in the radial direction.

Yep, that's about right. Your calculations are correct, as is this much of your interpretation.

How can there be a field, where the wave number, and hence the frequency,

That bit, on the other hand, isn't. The wavenumber changes, but the frequency is the temporal rate of change. In this case it is given by $E/h$ and it is constant. In general, you can have objects without a well-defined frequency, but if your source has a single well-defined temporal frequency of oscillations then the signal will have that frequency throughout.

change from one point to another in the space of the field? Such wave objects do not exist in Nature, right?

Yes, they do, this is perfectly standard. As the simplest example, take light travelling through glass, water, air, or indeed any material that isn't vacuum. The frequency of the light stays the same, but the wavelength is given by $\lambda=c/n\nu$, and if the index of refraction changes then so does $\lambda$ and therefore so does $k=2\pi/\lambda$.

In fact, you can make a direct analogy between the Schrödinger equation as you've posed it and the propagation of light down a waveguide in the paraxial approximation, and you can use a spatial dependence of $n$ to make quantum "wells" with "eigenstates" (a.k.a. waveguide modes) and "tunnelling" (a.k.a. evanescent coupling), among other "quantum" (wave) phenomena.

If you want something closer to home, surface waves on water have a dispersion relation which depends on the water depth; this means in turn that if the bottom drops or rises then equal temporal frequencies will give different wavelengths and wavevectors.

There's nothing wrong, puzzling, or unphysical about this. (Moreover, the fact that the paper you cited makes such a ridiculous claim should be an indicator as to the quality of the rest of that paper.)