Phy5646/An Example of spontaneous emission calculation

(Submitted by Team 1)

This example was taken from "Theory and Problems of Quantum Mechanics", Yoaf Peleg, et al, p. 298.

---

Problem: Find the transition rate of spontaneous emission for a hydrogen atom in the first excited state.

---

Solution: The transition rate for $2p\rightarrow 1s$ is given by

$\Gamma _{2p\rightarrow 1s}={\frac {4}{3}}{\frac {e^{2}\omega _{21}^{3}}{\hbar c^{3}}}\left|\langle 100|{\vec {r}}|21m'\rangle \right|^{2}$

where $|21m'\rangle$ is one of the three $2p\!$ states. In particular, for the hydrogen atom in the first excited state,

$\left|\langle 100|{\vec {r}}|21m'\rangle \right|^{2}={\frac {1}{3}}|\langle 100\left|{\vec {r}}|21m\rangle \right|^{2}={\frac {5}{9}}a_{0}^{2}$

where $a_{0}={\frac {\hbar ^{2}}{me^{2}}}$ is the Bohr radius. Therefore,

$\Gamma _{2p\rightarrow 1s}={\frac {20}{27}}{\frac {e^{2}\omega _{21}^{3}}{\hbar c^{3}}}a_{0}^{2}={\frac {20}{27}}\alpha {\frac {\omega _{21}^{3}}{c^{2}}}a_{0}^{2}={\frac {20}{27}}{\frac {\omega _{21}^{3}}{c^{4}}}{\frac {\hbar ^{2}}{m^{2}\alpha }}$

where $\alpha ={\frac {e^{2}}{\hbar c}}\cong {\frac {1}{137}}$ is the fine structure constant. However,

$\hbar \omega _{21}={\frac {3}{4}}{\frac {e^{2}}{2a_{0}}}={\frac {3}{4}}{\frac {\alpha ^{2}mc^{2}}{2}}$

Hence,

$\Gamma _{2p\rightarrow 1s}={\frac {20}{8^{3}}}\alpha ^{5}{\frac {mc^{2}}{\hbar }}={\frac {5}{48}}\alpha ^{3}\omega _{21}\cong 6.25\times 10^{-8}sec^{-1}$