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$2P\rightarrow 1S$ 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 (1149) and (1150), 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 Sect. 9.4, 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),
\end{displaymath} (1151)

where the radial functions $R_{n,l}$ are given in Sect. 9.4, and the spherical harmonics $Y_{l,m}$ are given in Sect. 8.7. Some straight-forward, but tedious, integration reveals that
$\displaystyle \langle 1,0,0\vert x\vert 2,1,\pm 1\rangle$ $\textstyle =$ $\displaystyle \pm \frac{2^7}{3^5} a_0,$ (1152)
$\displaystyle \langle 1,0,0\vert y\vert 2,1,\pm 1\rangle$ $\textstyle =$ $\displaystyle {\rm i} \frac{2^7}{3^5} a_0,$ (1153)
$\displaystyle \langle 1,0,0\vert z\vert 2,1,0\rangle$ $\textstyle =$ $\displaystyle \sqrt{2}  \frac{2^7}{3^5} a_0,$ (1154)

where $a_0$ is the Bohr radius specified in Eq. (679). All of the other possible $2P\rightarrow 1S$ matrix elements are zero because of the selection rules. If follows from Eq. (1144) that the modulus squared of the dipole moment for the $2P\rightarrow 1S$ transition takes the same value
d^2 = \frac{2^{15}}{3^{10}} (e a_0)^2
\end{displaymath} (1155)

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 Eq. (678). Hence, the energy of the photon emitted during a $2P\rightarrow 1S$ transition is

\hbar \omega = E_0/4 - E_0 = -\frac{3}{4} E_0 = 10.2 {\rm eV}.
\end{displaymath} (1156)

This corresponds to a wavelength of $1.215\times 10^{-7}$ m.

Finally, according to Eq. (1131), the $2P\rightarrow 1S$ transition rate is written

w_{2P\rightarrow 1S} = \frac{\omega^{ 3} d^2}{3\pi \epsilon_0 \hbar c^3},
\end{displaymath} (1157)

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}
\end{displaymath} (1158)

with the aid of Eqs. (1155) and (1156). 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}.
\end{displaymath} (1159)

Incidentally, since 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
\Delta E_{2P} \sim \frac{\hbar}{\tau_{2P}}\sim 4\times 10^{-7} {\rm eV}.
\end{displaymath} (1160)

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_{2P}}{\hbar \omega}\sim 4 \times 10^{-8}.
\end{displaymath} (1161)

next up previous
Next: Intensity Rules Up: Time-Dependent Perturbation Theory Previous: Selection Rules
Richard Fitzpatrick 2010-07-20