Daily Archives: Sat, 13 June 2015

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)