Density of final states in photon absorption/emission by a hydrogen atom

Question

Consider a hydrogen atom in an electromagnetic field. The Hamiltonian is of the form

$$\hat{H}=\underbrace{\frac{\hat{p}^2}{2m}+V(r)}_{\text{atom}}+\underbrace{\sum_{\vec{k},\sigma}\hbar cka^{\dagger}_{\vec{k},\sigma}a_{\vec{k},\sigma}}_\text{radiation}+\underbrace{\hat{H}_1+\hat{H}_2}_{\text{interaction}}.\tag{1}\label{1}$$ In first order perturbation theory, only the $\hat{H}_1$ piece in the interaction part contributes to transition amplitudes. Let me consider the case of emission first.

Notation

• $\lvert\nu_\text{Atom}\rangle\rightarrow$ hydrogen state with energy eigenvalue $\epsilon_\nu$;
• $\lvert1_{\vec{k},\sigma}\rangle$ and $\lvert\boldsymbol{0}\rangle\rightarrow$ one $(\vec{k},\sigma)$ photon and the electromagnetic vacuum state respectively;
• $E_i$ and $E_f\rightarrow$total initial and final energy.

Note: I'm considering the atomic states in the transition to be assigned and the photon frequency to be determined by the arising conditions.$^1$

Emission

$$\lvert i\rangle=\lvert\boldsymbol{0}\rangle\otimes\lvert\alpha_\text{Atom}\rangle\longrightarrow\lvert f\rangle=\lvert1_{\vec{k},\sigma}\rangle\otimes\lvert{\beta_{\text{atom}}\rangle}\label{2}\tag{emission of a photon}$$ Although the atom has a discrete spectrum and I'm only considering the two levels involved in the atomic transition, there is a photon in the final state, so the photonic density of final states (DoS) is $$\rho_f(\hbar\omega_k)=\frac{V}{2\pi^2\hbar c}\omega_k^2d\Omega\label{3}\tag{DoS #1}$$ This only considers the photonic energy, so the condition $E_f=E_i$ in Fermi Golden rule easily becomes $\omega_k=\frac{E_\beta-E_\alpha}{\hbar}$ and the transition rate per unit solid angle is $$R_{i\to f}=\frac{2\pi}{\hbar}\lvert\langle f\lvert\hat{H}_1\rvert i\rangle\rvert^2\rho_f(\hbar\omega_k)\bigg\rvert_{\omega_k=\frac{E_\beta-E_\alpha}{\hbar}}\label{4}\tag{transition rate #1}.$$ This works fine.

Absorption

Nevertheless, in the case of absorption I have a problem with the DoS.

$$\lvert i\rangle=\lvert 1_{\vec{k},\sigma}\rangle\otimes\lvert\beta _\text{Atom}\rangle\longrightarrow\lvert f\rangle=\lvert\boldsymbol{0}\rangle\otimes\lvert{\alpha_{\text{atom}}\rangle}\label{5}\tag{absorption of a photon}.$$ In this case there is no photon in the final state and remember that my final atomic level is fixed, so what should I do with the density of final states to write the transition rate? Although I think that the presence of the photon in the initial state should lead to some continuum and thus to a DoS, I'm not allowed to write a photonic density of final state like \eqref{3} because in my final state I have no photons in this case. Also, the difference between emission and absorption should only happen due to the photonic piece of the transition amplitude, not in the DoS. So, how do I deal with the DoS in this case?

Update

Checking Landau&Lifshitz QED (volume 4, section 4: emission and absorption), I noticed that they use Fermi Golden rule in equations $(44.1)$ and $(44.2)$. They say that in the case of emission, the final states lie on a continuum as I've also said in my post. After that, they consider absorption and state only the amplitude is to be replaced, without mentioning the DoS. As its evident from $(44.6)$, they are assuming it is the same. This made the situation even more ambiguous.

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$^1$ _{In other words, we know which atomic states are involved in the transition but not the modes and polarization of the photon.}

Answer 1

In this case there is no photon in the final state and remember that my final atomic level is fixed...

No, the final level is not fixed. (At least not in the Fermi's Golden Rule expression, which is the right way to think about things--and how things might get "fixed".)

You must sum over final states in Fermi's Golden Rule, which in this case include a sum over all the electronic states (and no photon states, since there is no photon).

Your delta function in Fermi's Golden Run will take care of energy conservation, but you still must do the sum over all final states (which in this case is all the electronic states).

so what should I do with the density of final states to write the transition rate?

Use the electronic density of states.

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If you want an answer in terms of equations, Consider Fermi's Golden Rule in the following form: $$ R = 2\pi\sum_f |\langle\psi_f|\langle 0|H_1|\psi_0\rangle|\vec k,\epsilon\rangle|^2\delta(E_0+\hbar\omega - E_f)\;, $$

This leads to an expression for the absorption cross section $\sigma(\omega)$ as follows: $$ \sigma(\omega) = \frac{4\pi^2e^2\omega}{c}\sum_f|\langle \psi_f|\epsilon^*\cdot \vec r |\psi_0\rangle|^2\delta(E_0 + \hbar\omega - E_f)\;,\tag{1} $$ where $\vec r$ is the position operator of the sample electron and $-e$ is the electronic charge.

That sum over $f$ you see in Eq. (1) is a sum over the electronic states of the sample.

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Update:

Below is an example. Caveat: In the example I ignore electron spin degeneracy.

Suppose that the incident light polarization $\vec \epsilon$ can be chosen such that $\vec \epsilon\cdot\vec r = z$. Suppose further that the initial states is the 1s orbital. Then the matrix element selects final states with $\ell=1$ and $m=0$.

Eq. (1) becomes: $$ \sigma(\omega) = \frac{4\pi^2e^2\omega}{c}\sum_{n=2}^{\infty}|\langle \psi_{n,1,0}|z|\psi_{1,0,0}\rangle|^2\delta(E_{n=1} + \hbar\omega - E_n)\;, $$ where $E_n = \frac{-13.6\;\text{eV}}{n^2}$.

For an isolated hydrogen atom, this spectrum look like isolated peaks (the delta functions) at energies where the photon energy $\hbar\omega$ can cause transitions "up" from the 1s state to the np states (2p, 3p, 4p, etc).

For a solid, the isolated delta-function peaks are modified into "absorption edges," which pop up roughly at the atomic transition frequencies, but are not delta functions. The absorption edges have structure due to interatomic factors and broadening due to quasi-particle lifetime, core-hole, and other multi-particle effects that are often swept under the rug in a single quasi-particle picture.