Worked Problem for Scattering on a Delta-Shell Potential

The effective one-dimensional Schrödinger equation is

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

In region one, given by $r<a,\!$

$u_{1l}(r)=Cj_{l}(kr),\!$

where $k={\sqrt {\frac {2mE}{\hbar ^{2}}}}.$

In region two, given by $r>a,\!$

$u_{2l}(r)=Aj_{l}(kr)+Bn_{l}(kr).\!$

Continuity of the wave function on either side of the boundary requires that

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

The first derivative is discontinuous due to the behavior of the delta function, so we must find the second condition in a manner similar to how we found it for the one-dimensional delta function potential. We first integrate the effective Schrödinger equation from $a+\epsilon \!$ to $a-\epsilon :\!$

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

We now 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 $r=a\!$):

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

This becomes

$\left.{\frac {du_{l}(r)}{dr}}\right|_{a+\epsilon }-\left.{\frac {du_{l}(r)}{dr}}\right|_{a-\epsilon }=u'_{2l}(a)-u'_{1l}(a)=\lambda u_{l}(a).$

This, combined with the above continuity condition, gives us

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

and

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

We now define the phase shift,

$\tan {\delta _{l}}=-{\frac {B}{A}}.$

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

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

For the $s\!$ wave case, $l=0.\!$ In this case,

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

or

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

From here, recall that the scattering amplitude is given by

$f_{k}(\theta )={\frac {1}{k}}\sum _{l=0}^{\infty }(2l+1)e^{i\delta _{l}(k)}\sin {\delta _{l}(k)}P_{l}(\cos {\theta }).$

Keeping only the $l=0\!$ term and in conjunction with the result for $\delta _{l}\!$ derived above, we find that

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

Back to Central Potential Scattering and Phase Shifts