Question 11.32: Show that the spontaneous emission rate (Equation 11.63) for...

Show that the spontaneous emission rate (Equation 11.63) for a transition from n, \ell \text { to } n^{\prime}, \ell^{\prime} in hydrogen is

A=\frac{\omega_{0}^{3}| S |^{2}}{3 \pi \epsilon_{0} \hbar c^{3}}             (11.63).

\frac{e^{2} \omega^{3} I^{2}}{3 \pi \epsilon_{0} \hbar c^{3}} \times \begin{cases}\frac{\ell+1}{2 \ell+1}, & \ell^{\prime}=\ell+1 \\ \frac{\ell}{2 \ell+1} & \ell^{\prime}=\ell-1\end{cases}                 (11.132).

where

I \equiv \int_{0}^{\infty} r^{3} R_{n \ell}(r) R_{n^{\prime} \ell^{\prime}}(r) d r                 (11.133).

(The atom starts out with a specific value of m, and it goes to any of the states m^{\prime} \text { consistent with the selection rules: } m^{\prime}=m+1, m \text {, 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 \text { and }\left|n^{\prime} \ell^{\prime} m^{\prime}\right\rangle \text { for the case } \ell^{\prime}=\ell+1 \text {. }

From these, determine the quantity

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

Then do the same for \ell^{\prime}=\ell-1 . You may find useful the following recursion formulas (which hold for m ≥0):

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

and the orthogonality relation Equation 4.33.

\int_{0}^{\pi} \int_{0}^{2 \pi}\left[Y_{\ell}^{m}(\theta, \phi)\right]^{*}\left[Y_{\ell^{\prime}}^{m^{\prime}}(\theta, \phi)\right] \sin \theta d \theta d \phi=\delta_{\ell \ell^{\prime}} \delta_{m m^{\prime}} .                   (4.33).

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.
\ell^{\prime}=\ell+1

From Eq. 11.76, \left\langle n^{\prime} \ell^{\prime} m^{\prime}|z| n \ell m\right\rangle=0 \text { unless } m^{\prime}=m \text {, } so the only nonzero z term is

\left\{\begin{array}{l} \text { if } m^{\prime}=m, \quad \text { then }\left\langle n^{\prime} \ell^{\prime} m^{\prime}|x| n \ell m\right\rangle=\left\langle n^{\prime} \ell^{\prime} m^{\prime}|y| n \ell m\right\rangle=0 \\ \text { if } m^{\prime}=m \pm 1, \text { then }\left\langle n^{\prime} \ell^{\prime} m^{\prime}|x| n \ell m\right\rangle=\pm i\left(n^{\prime} \ell^{\prime} m^{\prime}|y| n \ell m\right\rangle \\ \text { and }\left\langle n^{\prime} \ell^{\prime} m^{\prime}|z| n \ell m\right\rangle=0 \end{array}\right.                    (11.76).

\left\langle n^{\prime} \ell^{\prime} m|z| n \ell m\right\rangle=\int R_{n^{\prime} \ell^{\prime}}\left(Y_{\ell^{\prime}}^{m}\right)^{*} r \cos \theta R_{n \ell} Y_{\ell}^{m} r^{2} d r \sin \theta d \theta d \phi .

=I \sqrt{\frac{\left(2 \ell^{\prime}+1\right)}{4 \pi} \frac{\left(\ell^{\prime}-|m|\right) !}{\left(\ell^{\prime}+|m|\right) !}} \sqrt{\frac{(2 \ell+1)}{4 \pi} \frac{(\ell-|m|) !}{(\ell+|m|) !}} 2 \pi \int_{0}^{\pi} P_{\ell^{\prime}}^{m} P_{\ell}^{m} \cos \theta \sin \theta d \theta .            [ \star 1] .

This is independent of the sign of m, so we might as well assume m \geq 0 . The integral (changing variables to x \equiv \cos \theta ) is (using Eq. 11.134)

\operatorname{Int}_{\theta}=\int_{-1}^{1} x P_{\ell+1}^{m}(x) P_{\ell}^{m}(x) d x=\frac{1}{(2 \ell+1)}\left[(\ell+m) \int_{-1}^{1} P_{\ell+1}^{m} P_{\ell-1}^{m} d x+(\ell-m+1) \int_{-1}^{1} P_{\ell+1}^{m} P_{\ell+1}^{m} d x\right] .

Now, it follows from Eq. 4.33 that the associated Legendre functions satisfy the orthogonality relation

\int_{0}^{\pi} \int_{0}^{2 \pi}\left[Y_{\ell}^{m}(\theta, \phi)\right]^{*}\left[Y_{\ell^{\prime}}^{m^{\prime}}(\theta, \phi)\right] \sin \theta d \theta d \phi=\delta_{\ell \ell^{\prime}} \delta_{m m^{\prime}} .                   (4.33).

\int_{-1}^{1} P_{\ell^{\prime}}^{m}(x) P_{\ell}^{m}(x) d x=\frac{2}{(2 \ell+1)} \frac{(\ell+|m|) !}{(\ell-|m|) !} \delta_{\ell \ell^{\prime}} ,      [ \star 2] .

so

\operatorname{Int}_{\theta}=\frac{(\ell-m+1)}{(2 \ell+1)} \frac{2}{(2 \ell+3)} \frac{(\ell+1+m) !}{(\ell+1-m) !}=\frac{2}{(2 \ell+1)(2 \ell+3)} \frac{(\ell+1+m) !}{(\ell-m) !} .

Putting this into Eq. [ \star 1] :

\left\langle n^{\prime}(\ell+1) m|z| n \ell m\right\rangle=\frac{I}{2} \sqrt{(2 \ell+3) \frac{(\ell+1-m) !}{(\ell+1+m) !}} \sqrt{(2 \ell+1) \frac{(\ell-m) !}{(\ell+m) !}} \frac{2}{(2 \ell+1)(2 \ell+3)} \frac{(\ell+1+m) !}{(\ell-m) !} .

=I \sqrt{\frac{(\ell+1)^{2}-m^{2}}{(2 \ell+1)(2 \ell+3)}}           [ \star 3] .

From Eq. 11.76, \left\langle n^{\prime} \ell^{\prime} m^{\prime}|x| n \ell m\right\rangle=0 \text { unless } m^{\prime}=m \pm 1 \text {; let's start with } m+1 \text { : }

\left\{\begin{array}{l} \text { if } m^{\prime}=m, \quad \text { then }\left\langle n^{\prime} \ell^{\prime} m^{\prime}|x| n \ell m\right\rangle=\left\langle n^{\prime} \ell^{\prime} m^{\prime}|y| n \ell m\right\rangle=0 \\ \text { if } m^{\prime}=m \pm 1, \text { then }\left\langle n^{\prime} \ell^{\prime} m^{\prime}|x| n \ell m\right\rangle=\pm i\left(n^{\prime} \ell^{\prime} m^{\prime}|y| n \ell m\right\rangle \\ \text { and }\left\langle n^{\prime} \ell^{\prime} m^{\prime}|z| n \ell m\right\rangle=0 \end{array}\right.                    (11.76).

  \left\langle n^{\prime} \ell^{\prime}(m+1)|x| n \ell m\right\rangle=\int R_{n^{\prime} \ell^{\prime}}\left(Y_{\ell^{\prime}}^{m+1}\right)^{*} r \sin \theta \cos \phi R_{n \ell} Y_{\ell}^{m} r^{2} d r \sin \theta d \theta d \phi .

=I \sqrt{\frac{\left(2 \ell^{\prime}+1\right)}{4 \pi} \frac{\left(\ell^{\prime}-m-1\right) !}{\left(\ell^{\prime}+m+1\right) !}} \sqrt{\frac{(2 \ell+1)}{4 \pi} \frac{(\ell-m) !}{(\ell+m) !}} \int_{0}^{\pi} P_{\ell^{\prime}}^{m+1} P_{\ell}^{m} \sin ^{2} \theta d \theta \int_{0}^{2 \pi} \cos \phi e^{-i(m+1) \phi} e^{i m \phi} d \phi         [ \star 4] .

\operatorname{Int}_{\phi}=\frac{1}{2} \int_{0}^{2 \pi}\left(e^{i \phi}+e^{-i \phi}\right) e^{-i \phi} d \phi=\frac{1}{2} \int_{0}^{2 \pi}\left(1+e^{-2 i \phi}\right) d \phi=\pi         [ \star 5] .

Changing variables (x \equiv \cos \theta) , and using Eqs. 11.135 and [ \star 2] :

\operatorname{Int}_{\theta}=\int_{-1}^{1} \sqrt{1-x^{2}} P_{\ell+1}^{m+1}(x) P_{\ell}^{m}(x) d x=\frac{1}{(2 \ell+1)}\left[\int_{-1}^{1} P_{\ell+1}^{m+1} P_{\ell+1}^{m+1} d x-\int_{-1}^{1} P_{\ell+1}^{m+1} P_{\ell-1}^{m+1} d x\right] .

=\frac{2}{(2 \ell+1)(2 \ell+3)} \frac{(\ell+m+2) !}{(\ell-m) !} .

\text { Thus }[\star 4] \text { becomes }

\left\langle n^{\prime}(\ell+1)(m+1)|x| n \ell m\right\rangle=\frac{I}{2} \sqrt{(2 \ell+3) \frac{(\ell-m) !}{(\ell+m+2) !}} \sqrt{(2 \ell+1) \frac{(\ell-m) !}{(\ell+m) !}} \frac{1}{(2 \ell+1)(2 \ell+3)} \frac{(\ell+m+2) !}{(\ell-m) !} .

=\frac{I}{2} \sqrt{\frac{(\ell+m+2)(\ell+m+1)}{(2 \ell+1)(2 \ell+3)}}             [ \star 6] .

Now we do the same for m^{\prime}=m-1 :

\left\langle n^{\prime} \ell^{\prime}(m-1)|x| n \ell m\right\rangle=\int R_{n^{\prime} \ell^{\prime}}\left(Y_{\ell^{\prime}}^{m-1}\right)^{*} r \sin \theta \cos \phi R_{n \ell} Y_{\ell}^{m} r^{2} d r \sin \theta d \theta d \phi

=I \sqrt{\frac{\left(2 \ell^{\prime}+1\right)}{4 \pi} \frac{\left(\ell^{\prime}-m+1\right) !}{\left(\ell^{\prime}+m-1\right) !}} \sqrt{\frac{(2 \ell+1)}{4 \pi} \frac{(\ell-m) !}{(\ell+m) !}} \int_{0}^{\pi} P_{\ell^{\prime}}^{m-1} P_{\ell}^{m} \sin ^{2} \theta d \theta \int_{0}^{2 \pi} \cos \phi e^{-i(m-1) \phi} e^{i m \phi} d \phi \cdot[ \star 7] .

\operatorname{Int}_{\phi}=\frac{1}{2} \int_{0}^{2 \pi}\left(e^{i \phi}+e^{-i \phi}\right) e^{i \phi} d \phi=\frac{1}{2} \int_{0}^{2 \pi}\left(e^{2 i \phi}+1\right) d \phi=\pi .         [\star 8]   .

Changing variables (x \equiv \cos \theta) , and using Eqs. 11.135 and [\star 2] :

\operatorname{Int}_{\theta}=\int_{-1}^{1} \sqrt{1-x^{2}} P_{\ell+1}^{m-1}(x) P_{\ell}^{m}(x) d x=\frac{1}{(2 \ell+3)}\left[\int_{-1}^{1} P_{\ell+2}^{m} P_{\ell}^{m} d x-\int_{-1}^{1} P_{\ell}^{m} P_{\ell}^{m} d x\right] .

=-\frac{2}{(2 \ell+1)(2 \ell+3)} \frac{(\ell+m) !}{(\ell-m) !}.

\text { and }[\star 7] \text { becomes }

\left\langle n^{\prime}(\ell+1)(m-1)|x| n \ell m\right\rangle=-\frac{I}{4} \sqrt{(2 \ell+3) \frac{(\ell-m+2) !}{(\ell+m) !}} \sqrt{(2 \ell+1) \frac{(\ell-m) !}{(\ell+m) !}} \frac{2}{(2 \ell+1)(2 \ell+3)} \frac{(\ell+m) !}{(\ell-m) !} .

=-\frac{I}{2} \sqrt{\frac{(\ell-m+2)(\ell-m+1)}{(2 \ell+1)(2 \ell+3)}} .        [\star 9] .

Meanwhile, Eq. 11.76 says \left|\left\langle n^{\prime} \ell^{\prime} m^{\prime}|y| n \ell m\right\rangle\right|^{2}=\left|\left\langle n^{\prime} \ell^{\prime} m^{\prime}|y| n \ell m\right\rangle\right|^{2} , so

\left\{\begin{array}{l} \text { if } m^{\prime}=m, \quad \text { then }\left\langle n^{\prime} \ell^{\prime} m^{\prime}|x| n \ell m\right\rangle=\left\langle n^{\prime} \ell^{\prime} m^{\prime}|y| n \ell m\right\rangle=0 \\ \text { if } m^{\prime}=m \pm 1, \text { then }\left\langle n^{\prime} \ell^{\prime} m^{\prime}|x| n \ell m\right\rangle=\pm i\left(n^{\prime} \ell^{\prime} m^{\prime}|y| n \ell m\right\rangle \\ \text { and }\left\langle n^{\prime} \ell^{\prime} m^{\prime}|z| n \ell m\right\rangle=0 \end{array}\right.                    (11.76).

\left|\left\langle n^{\prime}(\ell+1)(m+1)| r | n \ell m\right\rangle\right|^{2}+\left|\left\langle n^{\prime}(\ell+1) m| r | n \ell m\right\rangle\right|^{2}+\left|\left\langle n^{\prime}(\ell+1)(m-1)| r | n \ell m\right\rangle\right|^{2} .

=2\left[\frac{I}{2} \sqrt{\frac{(\ell+m+2)(\ell+m+1)}{(2 \ell+1)(2 \ell+3)}}\right]^{2}+\left[I \sqrt{\frac{(\ell+1)^{2}-m^{2}}{(2 \ell+1)(2 \ell+3)}}\right]^{2}+2\left[-\frac{I}{2} \sqrt{\frac{(\ell-m+2)(\ell-m+1)}{(2 \ell+1)(2 \ell+3)}}\right]^{2} .

=\frac{I^{2}}{2}\left\{\frac{(\ell+m+2)(\ell+m+1)+2\left[(\ell+1)^{2}-m^{2}\right]+(\ell-m+2)(\ell-m+1)}{(2 \ell+1)(2 \ell+3)}\right\} .

=I^{2} \frac{\left(2 \ell^{2}+5 \ell+3\right)}{(2 \ell+1)(2 \ell+3)}=I^{2} \frac{(\ell+1)}{(2 \ell+1)}     [\star 10] .

Therefore, | \wp |^{2}   (summed over the three allowed transitions) is e^{2} I^{2}(\ell+1) /(2 \ell+1) , and the spontaneous emission rate (Eq. 11.63) is

A=\frac{\omega_{0}^{3}| S |^{2}}{3 \pi \epsilon_{0} \hbar c^{3}}             (11.63).

A_{\ell \rightarrow \ell+1}=\frac{e^{2} \omega^{3} I^{2}}{3 \pi \epsilon_{0} \hbar c^{3}} \frac{(\ell+1)}{(2 \ell+1)}     [\star 11] .

\ell^{\prime}=\ell-1 .

Return to Eq [\star 1] . This time the integral is

\operatorname{Int}_{\theta}=\int_{-1}^{1} x P_{\ell-1}^{m}(x) P_{\ell}^{m}(x) d x=\frac{1}{(2 \ell+1)}\left[(\ell+m) \int_{-1}^{1} P_{\ell-1}^{m} P_{\ell-1}^{m} d x+(\ell-m+1) \int_{-1}^{1} P_{\ell-1}^{m} P_{\ell+1}^{m} d x\right] .

=\frac{(\ell+m)}{(2 \ell+1)} \frac{2}{(2 \ell-1)} \frac{(\ell-1+m) !}{(\ell-1-m) !}=\frac{2}{(2 \ell-1)(2 \ell+1)} \frac{(\ell+m) !}{(\ell-m-1) !} .

Therefore

\left\langle n^{\prime}(\ell-1) m|z| n \ell m\right\rangle=\frac{I}{2} \sqrt{(2 \ell-1) \frac{(\ell-1-m) !}{(\ell-1+m) !}} \sqrt{(2 \ell+1) \frac{(\ell-m) !}{(\ell+m) !}} \frac{2}{(2 \ell-1)(2 \ell+1)} \frac{(\ell+m) !}{(\ell-m-1) !} .

=I \sqrt{\frac{\ell^{2}-m^{2}}{(2 \ell-1)(2 \ell+1)}} .        [\star 12] .

\text { From }\left\langle n^{\prime} \ell^{\prime} m^{\prime}|x| n \ell m\right\rangle \text { with } m^{\prime}=m+1 \text {, Eqs. }[\star 4] \text { and }[ \star 5]   are unchanged; this time

\operatorname{Int}_{\theta}=\int_{-1}^{1} \sqrt{1-x^{2}} P_{\ell-1}^{m+1}(x) P_{\ell}^{m}(x) d x=\frac{1}{(2 \ell+1)}\left[\int_{-1}^{1} P_{\ell-1}^{m+1} P_{\ell+1}^{m+1} d x-\int_{-1}^{1} P_{\ell-1}^{m+1} P_{\ell-1}^{m+1} d x\right] .

=-\frac{2}{(2 \ell-1)(2 \ell+1)} \frac{(\ell+m) !}{(\ell-m-2) !} .

\text { and }[\star 4] \text { becomes} 

\left\langle n^{\prime}(\ell-1)(m+1)|x| n \ell m\right\rangle=-\frac{I}{2} \sqrt{(2 \ell-1) \frac{(\ell-m-2) !}{(\ell+m) !}} \sqrt{(2 \ell+1) \frac{(\ell-m) !}{(\ell+m) !}} \frac{1}{(2 \ell-1)(2 \ell+1)} \frac{(\ell+m) !}{(\ell-m-2) !} .

=-\frac{I}{2} \sqrt{\frac{(\ell-m)(\ell-m-1)}{(2 \ell-1)(2 \ell+1)}}       [\star 13] .

Now we do the same for m^{\prime}=m-1 . \text { Eqs. }[\star 7] \text { and }[\star 8]   are unchanged, the θ integral is

\operatorname{Int}_{\theta}=\int_{-1}^{1} \sqrt{1-x^{2}} P_{\ell-1}^{m-1}(x) P_{\ell}^{m}(x) d x=\frac{1}{(2 \ell-1)}\left[\int_{-1}^{1} P_{\ell}^{m} P_{\ell}^{m} d x-\int_{-1}^{1} P_{\ell-2}^{m} P_{\ell}^{m} d x\right] .

=\frac{2}{(2 \ell-1)(2 \ell+1)} \frac{(\ell+m) !}{(\ell-m) !} .

\text { and }[\star 7] \text { becomes }

\left\langle n^{\prime}(\ell-1)(m-1)|x| n \ell m\right\rangle=\frac{I}{2} \sqrt{(2 \ell-1) \frac{(\ell-m) !}{(\ell+m-2) !}} \sqrt{(2 \ell+1) \frac{(\ell-m) !}{(\ell+m) !}} \frac{1}{(2 \ell-1)(2 \ell+1)} \frac{(\ell+m) !}{(\ell-m) !} .

=\frac{I}{2} \sqrt{\frac{(\ell+m)(\ell+m-1)}{(2 \ell-1)(2 \ell+1)}}           [\star 14] .

Thus

\left|\left\langle n^{\prime}(\ell-1)(m+1)| r | n \ell m\right\rangle\right|^{2}+\left|\left\langle n^{\prime}(\ell-1) m| r | n \ell m\right\rangle\right|^{2}+\left|\left\langle n^{\prime}(\ell-1)(m-1)| r | n \ell m\right\rangle\right|^{2} .

=2\left[-\frac{I}{2} \sqrt{\frac{(\ell-m)(\ell-m-1)}{(2 \ell-1)(2 \ell+1)}}\right]^{2}+\left[I \sqrt{\frac{\ell^{2}-m^{2}}{(2 \ell-1)(2 \ell+1)}}\right]^{2}+2\left[\frac{I}{2} \sqrt{\frac{(\ell+m)(\ell+m-1)}{(2 \ell-1)(2 \ell+1)}}\right]^{2} .

=\frac{I^{2}}{2}\left\{\frac{(\ell-m)(\ell-m-1)+2\left(\ell^{2}-m^{2}\right)+(\ell+m)(\ell+m-1)}{(2 \ell-1)(2 \ell+1)}\right\} .

=I^{2} \frac{\left(2 \ell^{2}-\ell\right)}{(2 \ell-1)(2 \ell+1)}=I^{2} \frac{\ell}{(2 \ell+1)}         [\star 15] .

and the emission rate is

A_{\ell \rightarrow \ell-1}=\frac{e^{2} \omega^{3} I^{2}}{3 \pi \epsilon_{0} \hbar c^{3}} \frac{\ell}{(2 \ell+1)}             [\star 16] .

(Of course, I is different for the two cases \ell \rightarrow \ell \pm 1 ).

Related Answered Questions