Coulomb excitation [closed]

Question

I have read that a very large number of partial waves around 200 and a large matching radius of 300 fm is required to obtain the cross section in Coulomb excitation. This is certainly far greater than normally being used. Why are these values so high. 1. How does one arrive at the maximum number of partial waves (or maximum L the angular momentum of incoming wave) that has to be incorporated.2. The matching distance is chosen at a very large value? Why is it neccessary to have matching at such a large distance. I hope this adresses the isssue.

Answer 1

Those numbers actually depend on the energy you are looking at (compared with the masses of the scattered particles). And if you mean differential cross section then also on the angle where you want the look.

And your partial waves could be the ordinary (spherical Bessel functions times $Y_{lm}$'s) or Coulomb wave functions times $Y_{lm}$'s. In the latter case you only need a small number of partial waves to describe the scattering by forces additional to the Coulomb force since that force is then implicit in the formalism. So I assume you mean the ordinary partial waves.

But with ordinary partial waves you can never describe the solution with the Coulomb potential, you'd have to "screen" the potential with some cut-off at large distance to make things converge for all angles and energies. If you want that to work correctly for low energies the cutoff has to be sufficiently long-range and then you need a lot of partial waves. At high energies you would need less. (At very high energies you can often completely ignore the Coulomb interaction since other forces will dominate.)

So in conclusion, the problem arises because of the long range of the Coulomb force. In fact we say it has infinite range, which means that even infinitely many partial waves cannot describe the total wavefunction, which is why special Coulomb wave functions were invented to serve as basis functions in the expansion.