The spontaneous emission rate for the 21-cm hyperfine line in hydrogen (Section 7.5) can be obtained from Equation 11.63, except that this is a magnetic dipole transition, not an electric one: φ → 1/c M = 1/c 〈1 | μe + μp) | 0 〉, where μe = - e/me Se, μp = 5.59e/2mp Sp. are the magnetic moments of

Question

The spontaneous emission rate for the 21-cm hyperfine line in hydrogen (Section 7.5) can be obtained from Equation 11.63, except that this is a magnetic dipole transition, not an electric one:

[latex] A=\frac{\omega_{0}^{3}| ho |^{2}}{3 \pi \epsilon_{0} \hbar c^{3}} [/latex] (11.63).

[latex] \wp ightarrow \frac{1}{c} M =\frac{1}{c}\left\langle 1\left|\left(\mu_{e}+\mu_{p} ight) ight| 0 ight angle [/latex],

where

[latex] \mu_{e}=-\frac{e}{m_{e}} S _{e}, \quad \mu _{p}=\frac{5.59 e}{2 m_{p}} S _{p} [/latex].

are the magnetic moments of the electron and proton (Equation 7.89), and [latex] |0 angle [/latex], [latex] |1 angle [/latex] are the singlet and triplet configurations (Equations 4.175 and 4.176). Because [latex] m_{p} \gg m_{e} [/latex], the proton contribution is negligible, so

[latex] \mu_{p}=\frac{g_{p} e}{2 m_{p}} S _{p}, \quad \mu _{e}=-\frac{e}{m_{e}} S _{e} [/latex] (7.89).

[latex] \left\{\begin{array}{l} |11 angle=|\uparrow \uparrow angle \ |10 angle=\frac{1}{\sqrt{2}}(|\uparrow \downarrow angle+|\downarrow \uparrow angle) \ |1-1 angle=|\downarrow \downarrow angle \end{array} ight\} \quad s=1 ext { (triplet) } [/latex] (4.175).

[latex] \left\{|00 angle=\frac{1}{\sqrt{2}}(|\uparrow \downarrow angle-|\downarrow \uparrow angle) ight\} \quad s=0 ext { (singlet) } [/latex] (4.176).

[latex] A=\frac{\omega_{0}^{3} e^{2}}{3 \pi \epsilon_{0} \hbar c^{5} m_{e}^{2}}\left|\left\langle 1\left| S _{e} ight| 0 ight angle ight|^{2} [/latex].

Work out [latex] \left|\left\langle 1\left| S _{e} ight| 0 ight angle ight|^{2} [/latex] (use whichever triplet state you like). Put in the actual numbers, to determine the transition rate and the lifetime of the triplet state.
Answer: [latex] 1.1 imes 10^{7} ext { years. } [/latex].

Accepted Answer

Choosing the triplet configuration [latex] |1 angle=\left|\uparrow_{e} \uparrow_{p} ight angle [/latex] we have

[latex] \left\langle 1\left| S _{e} ight| 0 ight angle=\frac{1}{\sqrt{2}}\left[\left\langle\uparrow_{e} \uparrow_{p}\left| S _{e} ight| \uparrow_{e} \downarrow_{p} ight angle-\left\langle\uparrow_{e} \uparrow_{p}| S | \downarrow_{e} \uparrow_{p} ight angle ight] [/latex].

Since [latex] S_e [/latex] doesn't effect the spin of the proton the first term vanishes and we have

[latex] \left\langle 1\left| S _{e} ight| 0 ight angle=-\frac{1}{\sqrt{2}}\left\langle\uparrow_{e}\left| S _{e} ight| \downarrow_{e} ight angle [/latex].

[latex] =-\frac{1}{\sqrt{2}}\left(\begin{array}{c} \frac{1}{2}\left\langle\uparrow_{e}\left|S_{e}^{+}+S_{e}^{-} ight| \downarrow_{e} ight angle \ \frac{1}{2 i}\left\langle\uparrow_{e}\left|S_{e}^{+}-S_{e}^{-} ight| \downarrow e ight angle \ \left\langle\uparrow_{e}\left|S_{e}^{z} ight| \downarrow_{e} ight angle \end{array} ight) [/latex].

[latex] =\frac{i \hbar}{2 \sqrt{2}}\left(\begin{array}{l} i \ 1 \ 0 \end{array} ight) [/latex].

We can then compute

[latex] \left|\left\langle 1\left| S _{e} ight| 0 ight angle ight|^{2}=\frac{\hbar^{2}}{8}(1+1)=\frac{\hbar^{2}}{4} [/latex].

Using the expression given in the problem we then have

[latex] A=\frac{\omega_{0}^{3} e^{2}}{3 \pi \epsilon_{0} \hbar c^{5} m_{e}^{2}} \frac{\hbar^{2}}{4} [/latex].

The frequency [latex] ω_0[/latex] is that of the 21cm line transition (see Equation 7.97),

[latex] \Delta E=\frac{4 g_{p} \hbar^{4}}{3 m_{p} m_{e}^{2} c^{2} a^{4}}=5.88 imes 10^{-6} eV [/latex] (7.97).

[latex] \omega_{0}=\frac{\Delta E}{\hbar}=\frac{4 g_{p} \hbar^{3}}{3 m_{p} m_{e}^{2} c^{2} a^{4}} [/latex],

giving

[latex] au=\frac{1}{A}=\frac{81}{64} \frac{1}{\alpha^{13}} \frac{1}{g_{p}^{3}}\left(\frac{m_{p}}{m_{e}} ight)^{3} \frac{\hbar}{m_{e} c^{2}}=1.1 imes 10^{7} yr [/latex].

Question 11.33: The spontaneous emission rate for the 21-cm hyperfine line i...

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