Question

An electron in the $\mathit{n}\mathbf{=}\mathbf{3}\mathbf{,}\mathit{l}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{m}\mathbf{=}\mathbf{0}$state of hydrogen decays by a sequence of (electric dipole) transitions to the ground state.

(a) What decay routes are open to it? Specify them in the following way:

$\mathbf{|}\mathbf{300}\mathbf{⟩}\mathbf{\to }\mathbf{|}\mathit{n}\mathcal{l}\mathit{m}\mathbf{⟩}\mathbf{\to }\mathbf{|}{\mathbf{n}}^{\mathbf{\text{'}}}{\mathbf{l}}^{\mathbf{\text{'}}}{\mathbf{m}}^{\mathbf{\text{'}}}\mathbf{⟩}\mathbf{\to }\mathbf{\cdots }\mathbf{\to }\mathbf{|}\mathbf{100}\mathbf{⟩}\mathbf{.}$

(b) If you had a bottle full of atoms in this state, what fraction of them would decay via each route?

(c) What is the lifetime of this state? Hint: Once it’s made the first transition, it’s no longer in the state |300\rangle∣300⟩, so only the first step in each sequence is relevant in computing the lifetime.

Short answer

(a)$\left(|300⟩\to |200⟩\text{and}|300⟩\to |1\begin{array}{ccc}1& 0& 0\end{array}⟩\right)$

(b)$|⟨21\pm 1|r|300⟩{|}^2=2|⟨21\pm 1\left|x\right|300⟩{|}^2=K^2/3$

(c)$\tau =\frac{1}{R}=1.58\times {10}^{-7}s$

Step-by-step solution

(a) Specifying the decay routes.

$\left(|300⟩\to |200⟩\text{and}|300⟩\to |1\begin{array}{ccc}1& 0& 0\end{array}⟩\text{violate}\Delta l=\pm 1\text{rule.}\right)$

(b) Fraction of decay via each route. 

From Eq. 11.76:

$\{\begin{array}{l}\text{if}{m}^{\text{'}}=m,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}then}⟨n^{\text{'}}{\mathcal{l}}^{\text{'}}{m}^{\text{'}}\left|x\right|n\mathcal{l}m⟩=⟨n^{\text{'}}{\mathcal{l}}^{\text{'}}{m}^{\text{'}}\left|y\right|n\mathcal{l}m⟩=0\\ \text{if}{m}^{\text{'}}=m\pm 1,\text{then}⟨n^{\text{'}}{\mathcal{l}}^{\text{'}}{m}^{\text{'}}\left|x\right|n\mathcal{l}m⟩=\pm i\left(n^{\text{'}}{\mathcal{l}}^{\text{'}}{m}^{\text{'}}\right|y|n\mathcal{l}m⟩\\ \text{and}⟨n^{\text{'}}{\mathcal{l}}^{\text{'}}{m}^{\text{'}}\left|z\right|n\mathcal{l}m⟩=0\end{array}$(11.76).

$\begin{array}{l}⟨210\left|r\right|300⟩=⟨210\left|z\right|300⟩\widehat{k}\\ ⟨21\pm 1\left|r\right|300⟩=⟨21\pm 1\left|x\right|300⟩\widehat{i}+⟨21\pm 1\left|y\right|300⟩\widehat{j}\end{array}$

$\begin{array}{l}\pm ⟨21\pm 1\left|x\right|300⟩=i⟨21\pm 1\left|y\right|.\\ Thus|⟨210|r|300⟩{|}^2=|⟨210\left|z\right|300⟩{|}^2\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}|⟨21\pm 1|r|300⟩{|}^2=2|⟨21\pm 1\left|x\right|300⟩{|}^2\end{array}$

So there are really just two matrix elements to calculate.

${\psi }_{21m}=R_{21}{Y}_1^m,{\psi }_{300}=R_{30}{Y}_0^0.$From Table 4.3:

$\begin{array}{c}\int Y_1^0{Y}_0^0\cos \theta \sin \theta d\theta d\varphi =\sqrt{\frac{3}{4\pi }}\sqrt{\frac{1}{4\pi }}{\int }_0^{\pi }{\cos }^2\theta \sin \theta d\theta {\int }_0^{2\pi }d\varphi \\ ={\frac{\sqrt{3}}{4\pi }(-\frac{{\cos }^3\theta }{3})|}_0^{\pi }\left(2\pi \right)\\ =\frac{\sqrt{3}}{2}\left(\frac{2}{3}\right)\\ =\frac{1}{\sqrt{3}}\end{array}$

$\begin{array}{c}\int {\left(Y_1^{\pm 1}\right)}^{*}{Y}_0^0{\sin }^2\theta \cos \varphi d\theta d\varphi =\mp \sqrt{\frac{3}{8\pi }}\sqrt{\frac{1}{4\pi }}{\int }_0^{\pi }{\sin }^3\theta d\theta {\int }_0^{2\pi }\cos \varphi e^{\mp i\varphi }d\varphi \\ =\mp \frac{1}{4\pi }\sqrt{\frac{3}{2}}\left(\frac{4}{3}\right)[{\int }_0^{2\pi }{\cos }^2\varphi d\varphi \mp i{\int }_0^{2\pi }\cos \varphi \sin \varphi d\varphi ]\\ =\mp \frac{1}{\pi \sqrt{6}}(\pi \mp 0)\\ =\mp \frac{1}{\sqrt{6}}\end{array}$

From Table 4.7:

$\begin{array}{c}K\equiv {\int }_0^{\infty }{R}_{21}{R}_{30}{r}^{3}dr=\frac{1}{\sqrt{24}{a}^{3/2}}\frac{2}{\sqrt{27}{a}^{3/2}}{\int }_0^{\infty }\frac{r}{a}{e}^{-r/2a}[1-\frac{2}{3}\frac{r}{a}+\frac{2}{27}{\left(\frac{r}{a}\right)}^2]e^{-r/3a}{r}^{3}dr\\ =\frac{1}{9\sqrt{2}{a}^3}{a}^4{\int }_0^{\infty }(1-\frac{2}{3}u+\frac{2}{27}{u}^2)u^4{e}^{-5u/6}du\\ =\frac{a}{9\sqrt{2}}[4!{\left(\frac{6}{5}\right)}^5-\frac{2}{3}5!{\left(\frac{6}{5}\right)}^6+\frac{2}{27}6!{\left(\frac{6}{5}\right)}^7]\end{array}$

$\begin{array}{c}=\frac{a}{9\sqrt{2}}\frac{4!6^5}{5^6}(5-\frac{2}{3}6\cdot 5+\frac{2}{27}{6}^3)\\ =\frac{a}{9\sqrt{2}}\frac{4!6^5}{5^6}\\ =\frac{2^7{3}^4}{5^6}\sqrt{2}a\end{array}$

So,

$\begin{array}{c}⟨21\pm 1\left|x\right|300⟩=\int R_{21}{\left(Y_1^{\pm 1}\right)}^{*}\left(r\sin \theta \cos \varphi \right)R_{30}{Y}_0^0{r}^2\sin \theta drd\theta d\varphi \\ =K(\mp \frac{1}{\sqrt{6}})\end{array}$

$\begin{array}{c}⟨210\left|z\right|300⟩=\int R_{21}{Y}_1^0\left(r\cos \theta \right)R_{30}{Y}_0^0{r}^2\sin \theta drd\theta d\varphi \\ =K\left(\frac{1}{\sqrt{3}}\right)\end{array}$

$\begin{array}{c}|⟨210|r|300⟩{|}^2=|⟨210\left|z\right|300⟩{|}^2\\ =K^2/3;\\ |⟨21\pm 1|r|300⟩{|}^2=2|⟨21\pm 1\left|x\right|300⟩{|}^2\\ =K^2/3\end{array}$

Evidently the three transition rates are equal, and hence 1/3 go by each route.

(c) Lifetime of the state

For each mode,

$\begin{array}{c}A=\frac{{\omega }^3{e}^2|⟨r⟩{|}^2}{3\pi {ϵ}_0\hslash c^3}\\ \text{here,}\\ \omega =\frac{E_3-E_2}{\hslash }\\ =\frac{1}{\hslash }(\frac{E_1}{9}-\frac{E_1}{4})\\ =-\frac{5}{36}\frac{E_1}{\hslash }\end{array}$,

so the total decay rate is

$\begin{array}{c}R=3{(-\frac{5}{36}\frac{E_1}{\hslash })}^3\frac{e^2}{3\pi {ϵ}_0\hslash c^3}\frac{1}{3}{\left(\frac{2^7{3}^4}{5^6}\sqrt{2}a\right)}^2\\ =6{\left(\frac{2}{5}\right)}^9{\left(\frac{E_1}{m{c}^2}\right)}^2\left(\frac{c}{a}\right)\end{array}$

$\begin{array}{c}=6{\left(\frac{2}{5}\right)}^9{\left(\frac{13.6}{0.511\times {10}^6}\right)}^2\left(\frac{3\times {10}^8}{0.529\times {10}^{-10}}\right)/s\\ =6.32\times {10}^6/s\\ \tau =\frac{1}{R}\\ =1.58\times {10}^{-7}s\end{array}$