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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.


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

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    $\begingroup$ I would think that since there is only one final state here (because there is only one vacuum), you don't need a density of states. You only need a density of states when there is a continuum of final-state energies and there is a continuum of states "at" the final-state energy, but I'm not sure. $\endgroup$
    – march
    Feb 22, 2023 at 16:24
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    $\begingroup$ Density of final states is very singular here (one state), so the derivation of the golden rule need not be valid. The golden rule is a sort of simplification that sometimes makes sense, sometimes it need not (and then we have to go back to more detailed ways of describing the interaction). However, there is still the density of the initial states, so maybe you can derive a variant of the golden rule that gives rate of transition from any of the set of initial states to the single one final state. $\endgroup$ Feb 22, 2023 at 18:39
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    $\begingroup$ Atoms jumping from ground state to excited state suggested by the various uses of the golden rule is not really an exact mathematical description, it is more of a simplification of verbal description of the evolution of the psi functions, that for some purposes is enough, but it can't explain all details of light-matter interaction, like excitation by a non-resonant EM wave, Rabi oscillations, etc. $\endgroup$ Feb 22, 2023 at 18:51
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    $\begingroup$ Recall how the golden rule is derived - we sum, against the naive desire to apply rules of quantum theory - eigenstate probabilities, not amplitudes, for a class of final states (maybe we could do so for initial states, but this is not usual). This summation of probabilities over many states is what causes the oscillatory evolution of expansion coefficients to produce an increase in the sum probability that is linear in time. If the sum reduces to a single term due to a single final state, there is no suppresion of oscillation and no linear increase in time, there will be oscillations instead. $\endgroup$ Feb 22, 2023 at 19:42
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    $\begingroup$ When deriving the Einstein coefficients for a pair of states, both for emission and absorption, there has to be summation over a class of different states, or at least different characteristics of the radiation. In non-relativistic semi-classical QT, this is provided by the equilibrium radiation - there is infinity of possible directions and phases of the electric field in equilibrium radiation, and this means to get probability of excitation, or emission, we have to sum over all possibilities. This makes the golden rule viable, I think. $\endgroup$ Feb 22, 2023 at 19:49

1 Answer 1

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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.


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.


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.

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  • $\begingroup$ I said the final level is fixed because I imagined a transition between two specific atomic levels by means of a photon absorption (e.g. $\lvert 1 0 0\rangle\to\lvert 2 1 0\rangle for hydrogen); in such case the only part to be determined would be the photon energy, but maybe the problem at hand is that I'm reasoning the other way around. The photon energy should be known from the beginning as it is in the initial state, so the problem should be fixing the atomic level in the case of absorption, right? $\endgroup$ Mar 5, 2023 at 1:17
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    $\begingroup$ "...but maybe the problem at hand is that I'm reasoning the other way around." Yes. The photon energy and the initial state energy restrict the energy of the final state via the delta function that is already explicitly in Fermi's golden rule. This does not necessarily limit you to a single final state, since more than one state can have the same energy (even in hydrogen). Yes, you are effectively limited in some sense to only final states that are accessible, but that all comes out naturally from Fermi's Golden Rule. $\endgroup$
    – hft
    Mar 5, 2023 at 1:56
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    $\begingroup$ The final states in emission are clearly different (for example you need to account for the final photon states). The initial state in emission is also potentially different. For example, you can not emit a photon if you atom is already in the 1s state, but you can absorb a photon. $\endgroup$
    – hft
    Mar 5, 2023 at 1:57
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    $\begingroup$ There is a small typo in equation (1). The subscript of the last energy variable in the delta function should be $E_f$ instead of $E_m$. $\endgroup$
    – Hans Wurst
    Mar 5, 2023 at 9:02
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    $\begingroup$ Here we are assuming the initial state is fixed. If you would like to average over initial states, you can. For example, it is fairly typical to average over the polarizations. But in this example we assume a fixed photon initial state and fixed electron initial state. $\endgroup$
    – hft
    Mar 7, 2023 at 3:50

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