Phase shifts for a localized Coulomb and harmonic potential

[phywithAK]

Homework Statement

Finding the phase shifts for a localized coulomb and harmonic potential.

Relevant Equations

V(r)={
-μ/r ; r<a (where μ is a constant)

0 ; r>a
}
and
V(r)={
(1/2)k*r^2 ; r<a

0 ; r>a
}

I am struggling over a problem and i could really use some help in this.
So it's about finding phase shifts in a localized sphere of coulomb and harmonic potential. I tried solving the radial Schrödinger equation for both of them by using power series method, but still i am having problem finding the phase shifts or at least i am not confident enough if i am going the right way or is there a better approach. I would really appreciate if anyone help me by throwing some light on this or at least on the solution of the equation for a finite range.
Thank you

 

[Juan Carlos]

Hello,

First of all, it would be nice to see explicitly your attempt of the solution. However, I don't think that you should use the power series method to solve the Radial Schroedinger equation at $r<a$. In this approach, the problem arises once you impose boundary conditions.

NOTE: I will assume that $\mu>0$.This is one way, maybe more adequate and straightforward, to get phase shifts:

As always, for spherical systems, the radial Schroedinger equation is the starting point:
$$ -\frac{\hbar^2}{2m}\left(\frac{\partial^2}{\partial r^2}\ +\ \frac{2}{r}\frac{\partial}{\partial r}\ -\ \frac{\ell(\ell+1)}{r^2}\right)\psi(r)\ +\ V(r)\psi(r)\ =\ E\,\psi(r)\ ,\qquad\ell=0,1,2,...$$

For both cases (a localized sphere of coulomb and harmonic potential) the potential $V(r)$ vanishes at $r>a$, and the previous equation becomes exactly solvable. In this domain (II), the solutions are well-known:
$$\psi_\ell^{II}(r)\ =\ C_{\ell}(k)\left[j_\ell(kr)\ -\ \tan\delta_{\ell}(k)\,n_\ell(kr)\right]\ ,\qquad k\ =\ \sqrt{2mE/\hbar^2}\qquad (r>a)$$
where $C_\ell(k)$ is a constant (r-idenpendent), and $\delta_{\ell}(k)$ the phase shifts.

Now, the situation inside the region $r<a$ looks complicated, but it is not. For those potentials you are working with, solutions are given in terms of the confluent hypergeometric function or something related. You just have to look for the regular solutions at the origin $(r=0)$.For example, if $V(r)=-\dfrac{\mu}{r}$ then, the general solution for the wave function looks like

$$\psi^{I}_\ell(r)\ =\ e^{-i k r} r^l \left(A_\ell(k)\,U\left(\kappa,\nu,2 i k r\right)+B_\ell(k)\,L_{-\kappa}^{\nu}(2 i k r)\right),\qquad (r<a)$$
where
$$ \kappa = \ell+1+\frac{i k \mu }{2 E }\ ,\qquad \nu\ =\ 2\ell+1\ ,$$
while $U$ and $L$ are the confluent hypergeometric function and generalized Laguerre polynomial, respectively. (I asked Mathematica for the solution).

When $r \rightarrow 0$ the wave function $\psi^{I}(r)$ must remain finite, then we have to set $A_\ell(k)=0$: $U$ diverges at $r=0$.

So, you have the solution for $(r<a)$.

The next step is to use the appropriate boundary conditions: continuous wave function as well as its derivative at the matching point. The straightforward way to guarantee the fulfillment of both conditions is by matching the logarithmic derivatives at $r=a$:

$$
\frac{\partial_r\psi^{I}(a)}{\psi^{I}(a)}\ =\ \frac{\partial_r\psi^{II}(a)}{\psi^{II}(a)}
\ .$$
Finally, you solve for $\delta_\ell$.

A similar approach should work for the harmonic potential.