# Klein-Gordon linear potential solution

1. Jan 21, 2009

### pellman

I have an exact solution to the Klein-Gordon equation with linear potential. But I am only an amateur physics enthusiast with no incentive (or time) to do anything with it, nor familiarity enough with the physics to know if it is interesting and, if so, interesting to whom. It has been sitting on my desk for a couple of years now.

It is a quite simple solution but not given in terms of energy eigenstates. It is the relativistic analog to a very simple solution of the linear potential Schrodinger case--which itself is so simple you could write it on the palm of your hand, but which appears to be generally unknown since it also is not in terms of energy eigenstates and so (I presume) not useful.

If you are interested in having it to further develop it into a paper or incorporate it into your research, let me know.

If you are reading this any time after a week of posting this, send me a PM since I probably won't see replies to the discussion thread.

Todd

2. Jan 21, 2009

### Hans de Vries

The solution(s) should simply be wavefunctions of accelerating particles.
See for instance. "The Lorentz force from the Klein Gordon equation"

http://www.physics-quest.org/Book_Lorentz_force_from_Klein_Gordon.pdf

Which should become more evident if you take the charge-current density of your solution.

\begin{aligned} &j^o ~~=~~~~ &\frac{i\hbar e}{2m}\left(~\psi^*\frac{\partial \psi}{\partial x^o}-\frac{\partial \psi^*}{\partial x^o}\psi ~ \right) ~~-~~ &\frac{e}{c}~\Phi~\psi^*\psi \\ &j^i ~~=~~ - &\frac{i\hbar e}{2m}\left(~\psi^*\frac{\partial \psi}{\partial x^i}-\frac{\partial \psi^*}{\partial x^i}\psi ~ \right) ~~-~~ &e\,A^i~\psi^*\psi \end{aligned}

Regards, Hans

3. Jan 21, 2009

### pellman

A point of interest is that in both the Schrodinger and K-G cases the explicit solutions are NOT the same as a free particle viewed from a constantly accelerating frame. In other words, you can't make the solution for a linear potential -Fx look like the free particle solution by substituting x --> x - F/m .

In the Schrodinger case though it is true that any expectation value is the same as the free particle case viewed from an accelerated frame, so the equivalence principle would still hold experimentally.

I can't make the same claim for the Klein-Gordon case because .... I don't know what the probabilistic interpretation of a K-G wave is! (I think its still an open question.)