Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

given a radial potential in 3 dimension and its Schroedinguer equation

$ -D^{2}U(r) + \frac{l(l+1)}{r^{2}}+V(r) $ here D means derivative with respect to 'r'

then if we apply quantum scattering how can we calculate the PHASE SHIFT ?? $\delta $ , for a general potential ?? .. for example with the condition $ V(r) \to \infty$ as $ r\to \infty$

share|cite|improve this question
Related: – Qmechanic Nov 17 '11 at 15:19
Do you mean V(r) goes to zero at large r? If V(r) goes to infinity, there are no scattering states, the motion is bound. Also V(r) needs to go to zero fast enough for the phase shift to approach a limit, generally this means V(r) goes to zero faster than 1/r. Electrostatic potential scattering with a 1/r potential has a logarithmically diverging phase shift, making the S-matrix undefined without a small photon mass. This is a form of infrared divergence. – Ron Maimon Nov 17 '11 at 15:20
thanks RON i forgot :) , yes the potential is $ V(r) \to 0 $ as $ r\to \infty $, whenever you say SCATTERING STATES you are referring to states with CONTINUUM energy in the form $ k^{2}+a^{2} $ here 'k' is the wave number and 'a' is a constant :) – Jose Javier Garcia Nov 17 '11 at 15:50
up vote 1 down vote accepted

For a general potential $V(r)$ you can use the variable-phase method.

I think you will find the following book very useful: Calogero, F. (1967), The Variable Phase Approach to Potential Scattering. Academic Press, New York.

Interesting to note is that Calogero emphasizes in his 1967 book that the numerical calculation of phase shifts using the variable-phase equation is well within the power of a simple desk calculator!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.