Worked Problem for Scattering on a Delta-Shell Potential

Consider an attractive delta-shell potential ($\lambda >0$) of the form:

$V({\textbf {r}})=-{\frac {\hbar ^{2}\lambda }{2m}}\delta (r-a)$

1) Derive the equation for the phase shift caused by this potential for arbitrary angular momentum.

2) Obtain the expression for the s-wave phase shift.

3) Obtain the scattering amplitude for the s-wave.

Solutions:

$\left[-{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}}{dr^{2}}}+{\frac {\hbar ^{2}l(l+1)}{2mr^{2}}}+V(r)\right]u_{l}(r)=Eu_{l}(r)$

where $\psi ({\textbf {r}})=f_{l}(r)Y_{m}^{l}(\theta ,\phi )$ and $f_{l}(r)={\frac {u_{l}(r)}{r}}$

In region one, r < a, $f_{1l}(r)=Cj_{l}(kr)\!$ where $k={\sqrt {\frac {2mE}{\hbar ^{2}}}}$

In region two, r > a, $f_{2l}(r)=Aj_{l}(kr)+Bn_{l}(kr)\!$

Invoking continuity of the wave function on either side of the boundary:

$Cj_{l}(ka)=Aj_{l}(ka)+Bn_{l}(ka)\!$

The first derivative is discontinuous due to the behavior of the delta function, so we must find the second condition needed a slightly different way. The delta function is most easily evaluated with an integral, so we consider the integral of the Schrodinger equation from a+$\epsilon$ to a $-\epsilon$:

$\int _{a-\epsilon }^{a+\epsilon }\left[{\frac {d^{2}}{dr^{2}}}+{\frac {l(l+1)}{r^{2}}}-\lambda \delta (r-a)-k^{2}\right]u_{l}(r)dr=0$

Now, we take the limit as $\epsilon \rightarrow 0$, and note that only the following two terms remain (the other integrals have the same value on either side of a):

$\int _{a-\epsilon }^{a+\epsilon }{\frac {d^{2}}{dr^{2}}}u_{l}(r)dr=\int _{a-\epsilon }^{a+\epsilon }\lambda \delta (r-a)dr$

Which now becomes

${\frac {d}{dr}}u_{l}(r)|_{a+\epsilon }-{\frac {d}{dr}}u_{l}(r)|_{a-\epsilon }=\lambda u_{l}(a)=f'_{1l}(a)-f'_{2l}(a)$

Which, combined with the above boundary condition for continuity gives that:

$Akj'_{l}(ka)+Bkn'_{l}(ka)-Ckj'_{l}(ka)=-\lambda Cj_{l}(ka)\!$

and

$Akj'_{l}(ka)+Bkn'_{l}(ka)-Ckj'_{l}(ka)=-\lambda Aj_{l}(ka)+Bn_{l}(ka)\!$

As usual, let $tan(\delta _{l})={\frac {-B}{A}}$

By solving the two equations obtained from the boundary conditions for the ratio ${\frac {-B}{A}}$, we find:

$tan(\delta _{l})={\frac {\lambda j_{l}^{2}(ka)}{kj_{l}(ka)n'_{l}(ka)-kn_{l}(ka)j'_{l}(ka)+\lambda n_{l}(ka)j_{l}(ka)}}$

For s-waves, set $l=0\!$

Therefore:

$tan(\delta _{0})={\frac {\lambda j_{0}^{2}(ka)}{kn_{0}(ka)j_{1}(ka)-kn_{1}(ka)j_{0}(ka)+\lambda n_{0}(ka)j_{0}(ka)}}$

which simplifies to:

$tan(\delta _{0})={\frac {\lambda sin^{2}(ka)}{k-\lambda sin(ka)cos(ka)}}$

From here, recall that the scattering amplitude $f_{k}(\theta )={\frac {1}{k}}\sum _{l=0}^{\infty }(2l+1)e^{i\delta _{l}(k)}sin(\delta _{l}(k))P_{l}(cos(\theta ))$

For $l=0\!$ and in conjunction with the derived result for $\delta _{l}\!$ above:

$f_{k}(\theta )={\frac {e^{i\delta _{0}}}{k}}sin(\delta _{0})$