Relativistic E/p relations in the WKB Approximation

[FunkyDwarf]

EDIT: fixed minus sign issue =)

Hey,

I have what is probably a rather trivial question but I just want to ensure that I'm on the right track :)

If I have a wave equation of the form
-\psi''(r) +A(r) \psi(r) = 0

then one can invoke (in suitable circumstances) the semi-classical (WKB) approximation such that the solution is of the form

\psi(r) \approx \frac{1}{\sqrt{p(r)}} \sin\left(\int p(r) dr\right)
where
p(r) = \int \sqrt{-A(r)}dr (assuming phase at r = 0 is 0 and no absorption is occurring).

My question ultimately has to do with the relativistic E/p relation, specifically in the case where I have a particle in some central potential. For arguments sake let's say I cannot separate A(r) into a nice potential form, ie an energy part and a potential part.

Is it still fair to say that when m = 0 the momentum p(r) must be linear in energy ? ie can I write the phase as the following:
\Phi(r) = \int p(r) dr = \epsilon \int f(r) dr

If so, can I also extend this to the case where m !=0, ie write
\Phi(r) = \int p(r) dr = \int \sqrt{\epsilon^2 g(r)-m^2 h(r)} dr
(keeping in mind A(r) could be a total mess, i.e. very different radial dependences on energy and mass, say)

Obviously I can't use the normal \epsilon = \sqrt{p^2 + m^2}, but is the above the correct natural extension?

Hope that made sense, and thanks in advance for replies (if they're helpful =P)!

 

[azntoon]

Yes, it is correct to extend the semi-classical approximation to the case when m != 0. The expression you have written for the phase is correct, and you can use it to calculate the momentum as a function of energy and mass. Also, note that the expression you have written for the phase is also valid in the case when m = 0, so you don't need to make any special cases for m = 0.