A Modern Approach to Quantum Mechanics · Problem 13.8

Photons and Atoms — John S. Townsend

Physics — Quantum Mechanics · plane-wave-expansion-partial-waves

Problem

Our goal is to establish (13.60), which, with the aid of (13.59), can be put in the form

e^{ikr\cos\theta} = \sum_{l=0}^\infty i^l\sqrt{4\pi(2l+1)}\, j_l(kr)Y_{l,0}(\theta)

(a) Explain why the expansion of the plane wave must be of the form

e^{ikr\cos\theta} = \sum_{l=0}^\infty c_l j_l(kr) Y_{l,0}(\theta)

(b) Use the fact that

|l,0\rangle = \frac{1}{\sqrt{(2l)!}}\left(\frac{\hat{L}_-}{\hbar}\right)^l |l,l\rangle

(see Problem 9.18) to show

c_l j_l(kr) = \frac{1}{\sqrt{(2l)!}}\int d\Omega\, Y^*_{l,l}\left\{\left[\frac{1}{i}e^{i\phi}\left(i\frac{\partial}{\partial\theta} - \cot\theta\frac{\partial}{\partial\phi}\right)\right]^l e^{ikr\cos\theta}\right\} = \frac{1}{\sqrt{(2l)!}}\int d\Omega\, Y^*_{l,l}\left[(-1)^l e^{il\phi}\sin^l\theta\frac{d^l}{d(\cos\theta)^l}e^{ikr\cos\theta}\right]

where the last step follows from the explicit form (9.142) for the raising operator in position space. (c) Use the explicit form (9.146) for

Y_{l,l}(\theta,\phi)

to express this result in the form

c_l j_l(kr) = (ikr)^l\frac{2^l l!}{\sqrt{(2l)!}}\sqrt{\frac{4\pi}{(2l+1)!}}\int d\Omega\,|Y_{l,l}(\theta,\phi)|^2 e^{ikr\cos\theta}

(d) Finally, isolate

c_l

by evaluating this expression as

r\to 0

. Hint: For small

r

j_l(kr)

behaves as

(kr)^l/(2l+1)!!

, where

(2l+1)!! = (2l+1)(2l-1)\cdots 5\cdot 3\cdot 1