- #1
FunkyDwarf
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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
[tex] -\psi''(r) +A(r) \psi(r) = 0 [/tex]
then one can invoke (in suitable circumstances) the semi-classical (WKB) approximation such that the solution is of the form
[tex] \psi(r) \approx \frac{1}{\sqrt{p(r)}} \sin\left(\int p(r) dr\right) [/tex]
where
[tex] p(r) = \int \sqrt{-A(r)}dr [/tex] (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:
[tex] \Phi(r) = \int p(r) dr = \epsilon \int f(r) dr[/tex]
If so, can I also extend this to the case where m !=0, ie write
[tex] \Phi(r) = \int p(r) dr = \int \sqrt{\epsilon^2 g(r)-m^2 h(r)} dr[/tex]
(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 [tex] \epsilon = \sqrt{p^2 + m^2} [/tex], but is the above the correct natural extension?
Hope that made sense, and thanks in advance for replies (if they're helpful =P)!
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
[tex] -\psi''(r) +A(r) \psi(r) = 0 [/tex]
then one can invoke (in suitable circumstances) the semi-classical (WKB) approximation such that the solution is of the form
[tex] \psi(r) \approx \frac{1}{\sqrt{p(r)}} \sin\left(\int p(r) dr\right) [/tex]
where
[tex] p(r) = \int \sqrt{-A(r)}dr [/tex] (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:
[tex] \Phi(r) = \int p(r) dr = \epsilon \int f(r) dr[/tex]
If so, can I also extend this to the case where m !=0, ie write
[tex] \Phi(r) = \int p(r) dr = \int \sqrt{\epsilon^2 g(r)-m^2 h(r)} dr[/tex]
(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 [tex] \epsilon = \sqrt{p^2 + m^2} [/tex], but is the above the correct natural extension?
Hope that made sense, and thanks in advance for replies (if they're helpful =P)!
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