Introduction to Quantum Mechanics · Problem 11.32

Quantum Dynamics — David J. Griffiths

Physics — Quantum Mechanics · spontaneous-emission-rate-hydrogen

Problem

Show that the spontaneous emission rate (Equation 11.63) for a transition from

n, \ell

n', \ell'

in hydrogen is

\frac{e^2\omega^3 I^2}{3\pi\epsilon_0\hbar c^3} \times \begin{cases} \dfrac{\ell+1}{2\ell+1}, & \ell' = \ell+1, \\[6pt] \dfrac{\ell}{2\ell+1} & \ell' = \ell-1, \end{cases} \tag{11.132}

where

I \equiv \int_0^\infty r^3 R_{n\ell}(r) R_{n'\ell'}(r)\,dr. \tag{11.133}

(The atom starts out with a specific value of

m

, and it goes to any of the states

m'

consistent with the selection rules:

m' = m+1, m

, or

m-1

. Notice that the answer is independent of

m

.) *Hint:* First calculate all the nonzero matrix elements of

x

y

, and

z

between

|n\ell m\rangle

and

|n'\ell'm'\rangle

for the case

\ell' = \ell+1

. From these, determine the quantity

\left|\langle n', \ell+1, m+1|\,\mathbf{r}\,|n\ell m\rangle\right|^2 + \left|\langle n', \ell+1, m\,|\,\mathbf{r}\,|n\ell m\rangle\right|^2 + \left|\langle n', \ell+1, m-1|\,\mathbf{r}\,|n\ell m\rangle\right|^2.

Then do the same for

\ell' = \ell-1

. You may find useful the following recursion formulas (which hold for

m \geq 0

(2\ell+1)\,x\,P_\ell^m(x) = (\ell+m)P_{\ell-1}^m(x) + (\ell-m+1)P_{\ell+1}^m(x), \tag{11.134}

(2\ell+1)\sqrt{1-x^2}\,P_\ell^m(x) = P_{\ell+1}^{m+1}(x) - P_{\ell-1}^{m+1}(x), \tag{11.135}

and the orthogonality relation Equation 4.33.