$$\require{cancel}$$

12.11: 2P-1S Transitions in Hydrogen

Let us calculate the rate of spontaneous emission between the first excited state (i.e., $$n=2$$) and the ground-state (i.e., $$n'=1$$) of a hydrogen atom. Now, the ground-state is characterized by $$l'=m'=0$$. Hence, in order to satisfy the selection rules ([e13.133]) and ([e13.134]), the excited state must have the quantum numbers $$l=1$$ and $$m=0,\,\pm 1$$. Thus, we are dealing with a spontaneous transition from a $$2P$$ to a $$1S$$ state. Note, incidentally, that a spontaneous transition from a $$2S$$ to a $$1S$$ state is forbidden by our selection rules.

According to Section [s10.4], the wavefunction of a hydrogen atom takes the form $\psi_{n,l,m}(r,\theta,\phi) = R_{n,l}(r)\,Y_{l,m}(\theta,\phi),$ where the radial functions $$R_{n,l}$$ are given in Section [s10.4], and the spherical harmonics $$Y_{l,m}$$ are given in Section [sharm]. Some straightforward, but tedious, integration reveals that \begin{aligned} \langle 1,0,0|x|2,1,\pm 1\rangle &=\pm \frac{2^{\,7}}{3^{\,5}}\,a_0,\\[0.5ex] \langle 1,0,0|y|2,1,\pm 1\rangle &= {\rm i}\,\frac{2^{\,7}}{3^{\,5}}\,a_0,\\[0.5ex] \langle 1,0,0|z|2,1,0\rangle &=\sqrt{2}\, \frac{2^{\,7}}{3^{\,5}}\,a_0,\end{aligned} where $$a_0$$ is the Bohr radius specified in Equation ([e9.57]). All of the other possible $$2P\rightarrow 1S$$ matrix elements are zero because of the selection rules. It follows from Equation ([e13.128]) that the modulus squared of the dipole moment for the $$2P\rightarrow 1S$$ transition takes the same value $\label{e13.139} d^{\,2} = \frac{2^{\,15}}{3^{\,10}}\,(e\,a_0)^{\,2}$ for $$m=0$$, $$1$$, or $$-1$$. Clearly, the transition rate is independent of the quantum number $$m$$. It turns out that this is a general result.

Now, the energy of the eigenstate of the hydrogen atom characterized by the quantum numbers $$n$$, $$l$$, $$m$$ is $$E = E_0/n^{\,2}$$, where the ground-state energy $$E_0$$ is specified in Equation ([e9.56]). Hence, the energy of the photon emitted during a $$2P\rightarrow 1S$$ transition is $\label{e13.140} \hbar\,\omega = E_0/4 - E_0 = -\frac{3}{4}\,E_0 = 10.2\,{\rm eV}.$ This corresponds to a wavelength of $$1.215\times 10^{-7}$$ m.

Finally, according to Equation ([e3.115]), the $$2P\rightarrow 1S$$ transition rate is written $w_{2P\rightarrow 1S} = \frac{\omega^{\,3}\,d^{\,2}}{3\pi\,\epsilon_0\,\hbar\,c^{\,3}},$ which reduces to $w_{2P\rightarrow 1S} = \left(\frac{2}{3}\right)^8\,\alpha^{\,5}\,\frac{m_e\,c^{\,2}}{\hbar} = 6.27\times 10^8\,{\rm s}^{-1}$ with the aid of Equations ([e13.139]) and ([e13.140]). Here, $$\alpha=1/137$$ is the fine-structure constant. Hence, the mean life-time of a hydrogen $$2P$$ state is $\tau_{2P} = (w_{2P\rightarrow 1S})^{-1} = 1.6\,{\rm ns}.$ Incidentally, because the $$2P$$ state only has a finite life-time, it follows from the energy-time uncertainty relation that the energy of this state is uncertain by an amount ${\mit\Delta} E_{2P} \sim \frac{\hbar}{\tau_{2P}}\sim 4\times 10^{-7}\,{\rm eV}.$ This uncertainty gives rise to a finite width of the spectral line associated with the $$2P\rightarrow 1S$$ transition. This natural line-width is of order

$$\frac{\Delta \lambda}{\lambda} \sim \frac{\Delta E_{2 P}}{\hbar \omega} \sim 4 \times 10^{-8}$$
