Radiative decay of 2p state of hydrogen

4.4 Radiative decay of $2 p$ state of hydrogen

Summary: This is the classic test-case of the theory we’ve developed so far. It passes with ﬂying colours!

We wish to calculate the total decay rate of the $2 p$ state of hydrogen. We start with the expression for spontaneous decay deduced from the preceding arguments:

R_{i \to f} = \frac{π e^{2}}{ε_{0} ℏ^{2}} ℏ ω D (ω \hat{k}) {|⟨ f | ε \cdot r | i ⟩|}^{2}

where $ω \equiv ω_{f i}$ is taken as read. The expression $D (ω \hat{k})$ is the density of states factor for photons with a particular direction of propagation $\hat{k}$ (which must be perpendicular to $ε$ ): $D (ω \hat{k}) = ω^{2} ∕ {(2 π c)}^{3}$ . It is just $D (k)$ written in terms of $ω$ so that $\int D (ω \hat{k}) d ω = \int D (k) d k$ . See section A.10 for more details.

Now we don’t care about the direction in which the ﬁnal photon is emitted, nor about its polarisation, and so we need to integrate and sum the decay rate to a particular photon mode, as written above, over these. ( $ε$ is the polarisation vector.) We can pick any of the $2 p$ states (with $m = 0, \pm 1$ ) since the decay rate can’t depend on the direction the angular momentum is pointing, so we will choose $m = 0$ . Then $⟨ 100 | ε \cdot r | 210 ⟩ = ⟨ 100 | ε_{z} z | 210 ⟩$ as we can see by writing $ε \cdot r$ in terms of the $l = 1$ spherical harmonics (see A.2). There are two polarisation states and both are perpendicular to $k$ ; if we use spherical polar coordinates in $k$ -space, $(k, θ_{k}, ϕ_{k})$ , we can take the two polarisation vectors to be $ε^{(1)} = θ_{k}$ and $ε^{(2)} = ϕ_{k}$ . Only the ﬁrst contributes since $ε_{z}^{(1)} = - sin θ_{k}$ but $ε_{z}^{(2)} = 0$ .

We will need to integrate over all directions of the emitted photon. Hence the decay rate is

\begin{array}{rcl} R_{i \to f} & = & \frac{π e^{2}}{ε_{0} ℏ^{2}} ℏ ω \frac{ω^{2}}{8 π^{3} c^{3}} \int {sin}^{2} θ_{k} {|⟨ 100 | z | 210 ⟩|}^{2} sin θ_{k} d θ_{k} d ϕ_{k} \\ = & \frac{e^{2} ω^{3}}{8 π^{2} c^{3} ℏ ε_{0}} \frac{8 π}{3} {|⟨ 100 | z | 210 ⟩|}^{2} \\ = & \frac{4 α ω^{3}}{3 c^{2}} \frac{2^{15} a_{0}^{2}}{3^{10}} where ℏ ω = (3 ∕ 4) E_{Ry} \\ = & {(\frac{2}{3})}^{8} \frac{α^{5} m c^{2}}{ℏ} = 6.265 \times 1 0^{8} s^{- 1} \end{array}

This matches the experimental value of the $2^{2} p_{1 ∕ 2}$ state to better than one part in $1 0^{- 5}$ , at which point the eﬀects of ﬁne structure enter. The fact that this rate is very much smaller than the frequency of the emitted photon ( $2.5 \times 1 0^{15}$ Hz) justiﬁes the use of the long-time approximation which led to Fermi’s golden rule.

Gasiorowicz ch 17.2,4
Shankar ch 18.5
Townsend ch 14.8

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4.4 Radiative decay of 2p state of hydrogen

4.4 Radiative decay of $2 p$ state of hydrogen