# Phase shift in the spherical delta function shell

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 11.7.

We can apply 3-d partial wave analysis using phase shifts to the problem of the spherical delta function shell. Restricting our attention to the ${l=0}$ term, we found earlier that the wave function for points outside the sphere is

$\displaystyle \psi_{ext}=\frac{A}{kr}\left[\sin kr+ka_{0}e^{ikr}\right] \ \ \ \ \ (1)$

where

 $\displaystyle a_{0}$ $\displaystyle =$ $\displaystyle -\frac{\beta e^{-ika}\sin^{2}ka}{\left(\beta\sin ka+ka\cos ka-iak\sin ka\right)k}\ \ \ \ \ (2)$ $\displaystyle k$ $\displaystyle \equiv$ $\displaystyle \frac{\sqrt{2mE}}{\hbar}\ \ \ \ \ (3)$ $\displaystyle \beta$ $\displaystyle \equiv$ $\displaystyle \frac{2ma\alpha}{\hbar^{2}} \ \ \ \ \ (4)$

and ${a}$ is the radius of the sphere, and ${\alpha}$ is the strength of the delta function in the potential: ${V\left(r\right)=\alpha\delta\left(r-a\right)}$.

By comparing this form of the wave function with the phase shift form, we found that

$\displaystyle a_{l}=\frac{1}{k}e^{i\delta_{l}}\sin\delta_{l} \ \ \ \ \ (5)$

To find the phase shift from 2, we need to put ${ka_{0}}$ in modulus-argument form. We can grind through the calculations by multiplying 2 top and bottom by the complex conjugate of the denominator and then finding the real and imaginary parts. This is just rather tedious algebra, so I got Maple to do it for me, with the results:

 $\displaystyle \Re\left(ka_{0}\right)$ $\displaystyle =$ $\displaystyle -\frac{\beta\sin^{2}\left(ka\right)\left(ka+\beta\sin\left(ka\right)\cos\left(ka\right)\right)}{\left(ka\right)^{2}+\beta^{2}+2ka\beta\sin\left(ka\right)\cos\left(ka\right)-\beta^{2}\cos^{2}\left(ka\right)}\ \ \ \ \ (6)$ $\displaystyle \Im\left(ka_{0}\right)$ $\displaystyle =$ $\displaystyle \frac{\beta^{2}\sin^{4}\left(ka\right)}{\left(ka\right)^{2}+\beta^{2}+2ka\beta\sin\left(ka\right)\cos\left(ka\right)-\beta^{2}\cos^{2}\left(ka\right)} \ \ \ \ \ (7)$

From this, we get

 $\displaystyle \delta_{0}$ $\displaystyle =$ $\displaystyle \arctan\left(\frac{\Im ka_{0}}{\Re ka_{0}}\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \arctan\left(-\frac{\beta\sin^{2}\left(ka\right)}{ka+\beta\sin\left(ka\right)\cos\left(ka\right)}\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\arctan\left(\frac{\beta\sin^{2}\left(ka\right)}{ka+\beta\sin\left(ka\right)\cos\left(ka\right)}\right) \ \ \ \ \ (10)$

For some reason, Griffiths wants to express the answer using cotangents, so using ${\arctan x=\mbox{arccot}\frac{1}{x}}$, we have

 $\displaystyle \delta_{0}$ $\displaystyle =$ $\displaystyle -\mbox{arccot}\left(\frac{ka+\beta\sin\left(ka\right)\cos\left(ka\right)}{\beta\sin^{2}\left(ka\right)}\right)\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\mbox{arccot}\left(\cot\left(ka\right)+\frac{ka}{\beta\sin^{2}\left(ka\right)}\right) \ \ \ \ \ (12)$