Radiative TransitionsInitial questions: How can we compute the strengths of atomic lines? What does thepresence of forbidden lines tell us about the physical state?From static atoms we now move to their response when perturbed by an electromagneticwave. We are going to use a semiclassical approximation, in which the atom is treatedquantum mechanically but the radiation is treated classically. Recall that the classicalapproximation is one in which the discrete nature of photons does not matter. Ask class:given that, which of the Einstein coefficients should we be able to get in this way? The Bcoefficients (absorption and stimulated emission), which depend on the external radiationfield. The A coefficient, for spontaneous emission, requires a full quantum treatment andthe inclusion of virtual quanta.As our initial background, suppose that the unperturbed atom has a HamiltonianH0= p2/2m + eφatom. The eigenvectors of this Hamiltonian are |φni, with eigenvaluesEn: H0|φni = En|φni. Initially, before the wave comes by, let’s say that the system was inthe stationary state |φii. Thus in our semiclassical approximation the system will stay thatway for all time if it is not perturbed (in reality, virtual quanta will provide perturbationsand thus induce spontaneous emission if |φii is not the ground state).If a perturbation is now added then the Hamiltonian changes toH(t) = H0+ λˆW (t) (1)where we include the possibility of time dependence and we use the convention that themagnitude ofˆW is comparable to the magnitude of H0, but that λ ≪ 1. In formal analyses,this allows us to keep track of the perturbative order. The time-dependent Schr¨odingerequation can thus be rewritten as an evolution equation:i¯h∂ψ(t)∂t= H(t)ψ(t) , (2)with the initial condition ψ(0) = |φii. Because during the period of the perturbationH(t) 6= H0, |φii will not in general be an eigenfunction of H(t), thus the wavefunctionwill evolve. Note that because the full set of energy eigenfunctions (including ones withdegenerate energies) forms a complete orthonormal basis, ψ(t) can always be represented asa linear sum of the |φni functions. This means that after the wave has passed, there is someprobability that the system has undergone a transition to a different state. The probabilityof a transition to the state |φfi (note that after the perturbation we’re back to the originaleigenfunctions) isPif(t) = |hφf|ψ(t)i|2. (3)Now let’s specify that the perturbation is electromagnetic. In an electromagneticpotential φ = φatom+ φext, where φatomis the scalar potential for the atom, and φextis theexternal scalar potential, the nonrelativistic Hamiltonian isH = P2/2m + eφ . (4)Note that here P is the mechanical momentum, not the conjugate momentum p thatbecomes −i¯h∇ in the coordinate representation. The relation between the two isP = p − eA/c, where A is the external vector potential. As always with the potentials,we have a choice of gauge; in this case, a convenient one is the “Coulomb gauge” in which∇ · A = φext= 0, because then A commutes with p in their product. In this gauge,H = p2/2m − (e/mc)A · p + e2A2/(2mc2) + eφatom. (5)This can be thought of as the sum of the original (non-perturbed) Hamiltonian plusperturbations: H = H0+ H1+ H2, where H0= p2/2m + eφatom, H1= (e/mc)A · p, andH2= e2A2/(2mc2). Do we have to include both H1and H2? We can take their ratio to findout.H1/H2= (epA/mc)/(e2A2/2mc2) = (2ev/c)/(α2a0A) , (6)where α = e2/¯hc ≈ 1/137 is the fine structure constant and a0= ¯h2/me2= 5.3 × 10−9cmis the Bohr radius. What is the order of magnitude of the velocity? For a typical energyE = −meZ2e4/2¯h2, setting E = −12mev2gives v/c = Ze2/¯hc = Zα. That givesη ≡ H1/H2∼ 2Ze/(αa0A) . (7)What is the magnitude of A? We know that B = ∇ × A, so with E ∼ B we can estimateA ∼ λE, where λ is the wavelength of the electromagnetic wave. Note that in doing thiswe’re doing the “cancelling the d’s” approach to derivatives. This will be justified after thefact when we discover that H1≫ H2, meaning that real precision isn’t necessary most of thetime. For the energies of interest, λ ∼ a0/α. A few algebraic manipulations then produceη2∼4¯hω2παa20λE2. (8)What do we do with the E2? We know that the electromagnetic energy density is somethinglike E2or B2(with factors of 8π, but at this level of inaccuracy we don’t care). For atypical photon energy ¯hω, that means that the photon number density is nph∼ E2/¯hω.Putting this together, the ratio of H1to H2isη ∼ (npha30)−1/2. (9)Ask class: can we argue that in most circumstances this ratio is much larger than unity,so that H2may be ignored? The idea is that if there are typically far fewer than onephoton in a cubic Bohr radius, then η ≫ 1. One photon per cubic Bohr radius would beroughly 1025cm−3. Ask class: for a blackbody, how do we estimate the photon numberdensity? We could take the radiation energy density and divide by kT , for example. Or, ifwe happened to remember that there are about 20T3photons per cubic centimeter for atemperature of T Kelvin, that works as well. Either way, if T ≪ 108K our assumption thatH1≫ H2is okay.Let’s look at this from another perspective. It turns out that the A · p term representssingle-photon transitions, whereas the A2term represents two-photon transitions. Therefore,the H2term is only important if there is a reasonable probability that “near” the atom(within a volume a30) there is more than one photon. From a statistical standpoint, this onlyis likely if the average number of photons in that volume is greater than or comparable tounity. As we’ll see, however, there are times when a single-photon transition is forbidden,meaning that the comparatively rare two-photon transitions are needed to go from onestate to another.So let’s consider only H1. What does this do? If the Hamiltonian of a system isexplicitly time-independent (e.g., the Hamiltonian of a static atom in the nonrelativisticlimit), then once the system is in an eigenstate of that Hamiltonian it stays that way untilthere is some perturbation. The wave function is then ψ(t) = φi(r) exp(−iEit/¯h), where Eiis the energy of the ith eigenstate φi. Since the global phase factor exp(iEit/¯h) does notchange measurable quantities, the system is stationary. Ask class: what does that implyabout spontaneous emission in this picture? It wouldn’t happen; that’s