•••The selection rule (11.46) that radiative transitions are forbidden unless Δl = ±1 is r...

•••The selection rule (11.46) that radiative transitions are forbidden unless Δl = ±1 is related to the fact that photons have spin 1, as we mentioned in Section 11.8. The details of the argument are surprisingly complicated and require a greater knowledge of quantum mechanics than we have developed. Nevertheless, you can reach the right conclusion arguing semiclassically, as follows: To be definite, consider a hydrogen atom whose electron has initial orbital angular momentum L and spin S and which is excited by a photon to a state with L′ and S′. Conservation of angular momentum requires that

where Lγ and Sγ are the orbital and spin angular momenta of the incoming photon. Notice first that the electric field has no effect on the electron’s spin,* so that S′ = S and these two terms cancel from (11.53).

(a) Next you can show that Lγ = 0, as follows: If it is to interact with the atom, the incoming photon must come within a distance of order aB of the nucleus; thus, to interact with the atom, its orbital angular momentum must be of order aBp or less, where p is the linear momentum of the photon. Given that the photon’s energy has to be of order ER, show that Lγ must be of order αℏ or less, where α is the so-called fine-structure constant α = ke2/ℏc ≈ 1/137. Since the only possible values of L are with l equal to an integer, this shows that l = 0 and Lγ = 0. Thus, (11.53) reduces to

 

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(b) The maximum magnitude of L′ occurs when L and Sγ are parallel. By taking the z component of (11.54) and remembering that the maximum value of Lz is lℏ (and similarly for all angular momenta), show that the maximum possible value of l′ is l + 1. Show similarly that the minimum possible value is l − 1. This leaves only three possibilities, l′ = l + 1 or l or l − 1, and Problem 1 has shown that l′ = l is not allowed, so you’re home.

* Recall that the spin centers the energy through the term −μ • B, which involves the magnetic field, and that the effects of B are usually negligible compared to those of ℰ.

Problem 1

•• The general proof of the selection rule (11.46) is beyond our scope, but you can prove it in a few special cases, (a) It is a fact that the wave function ψnlm(r) = Rnl(r)Өlm(θ)eimϕ satisfies

[This property is often described by saying that the wave function has parity (−1)l.] Use the angular functions listed in Table 8.1 to prove (11.51) for all wave functions with l = 0,1, or 2. (b) The probability of a radiative transition (n, l, m → n′,l′, m′) is given by (11.45), which now takes the form

(This is for radiation polarized with ℰ in the x direction. For isotropic unpolarized radiation, we must average over this and the two corresponding expressions with x replaced by y and by z.) Use the property (11.51) to prove that the transition probability (11.52) is zero if l′ − l (whether the integrand is x, y, or z). This proves that transitions with Δl = 0 are forbidden — a particular case of (11.46).