EM radiation fields from a single electron transition

Question

This question is about a description for the EM fields created when an electron decays into a lower energy state spontaneously in a single, isolated atom.

The treatment that I recall from grad school of absorption and emission starts from the model of a plane wave perturbing the atomic Hamiltonian. This model works conceptually for me to describe absorption and stimulated emission on a sort of statistical scale (many atoms), and obviously produces good real-world predictions for things like transition probabilities.

However, consider a single atom in a diffuse medium (say, a gas with low density) which, for some reason, has an electron in an excited state. Perhaps it absorbed some radiation or was jostled by other atoms. Let's consider the case that the excited state is one which decays quickly. The electron may move to a lower energy state, "emitting a photon", and some form of traveling EM fields will be radiated. This could happen even without interacting with neighbors or a passing EM plane wave. Right?

So what traveling EM fields are generated by such a transition? I imagine a sort of wave packet, perhaps the sum of multipole fields, but with a limited spatial/time extent, moving outward from the atom. Perhaps, the initial and final electron orbital states (themselves having multipole components) in the transition would determine the multipole components of the EM wave.

However, this does not seem to really work; one issue is, how would such a packet have a single frequency, as "the photon" does? Limited spatial extent would suggest a superposition of frequencies. A second issue is that, if "the photon" later interacts with some matter somewhere else (say, it is absorbed by a different atom), what happens to the wave front far away from the site of this interaction (say, in the opposite direction from the location of the original emission)?

Does anyone have an explanation, or link to a good treatment of this?

Thank you

Answer 1

Although you can explain the excitation of an electron with time dependent perturbation theory, by the introduction of a periodic perturbation with frequency related to the energy difference between the initial and excited state, the decay of an electron of an isolated atom into a lower energy state (with the emission of a photon) is a little more complicated.

For an isolated hydrogen-like atom, for example, all energy levels are by definition eigenstates of the Hamiltonian, so their time evolution is simply given by a phase, and no transition to other energy level is possible.

In order to explain the spontaneous emission of a photon by an excited electron in an isolated atom, one need to quantize the electromagnetic field, i.e., promote the electromagnetic field to operators satisfying certain commutation relations, that describe the creation and annihilation of photons in a Fock space, that will describe our system together with the Hilbert space of the electron states.

In this formalism, the electromagnetic field and and its interaction with the atom are described by a Hamiltonian $H=H_{\text{Atom}} + H_{\text{EM field}} + H_I$, and the transition amplitudes can no be calculated $\mathcal{A} \propto ~ \langle f,\text{one photon}| H_I|i,\text{no photon}\rangle$.

The book Quantum Mechanics, Volume 3: Fermions, Bosons, Photons, Correlations, and Entanglement by Claude Cohen-Tannoudji, Bernard Diu and Franck Laloë has a good treatment of those phenomena in chapter XX.

Answer 2

This is a very interesting question. As the author said and it is THE point: what is the shape of the EM field? QED provides a plane wave with a "creation operator". So physicaly speaking this is a photon with an infinite spatial extension by one single electron states transition... this does not make sense. A spherical wave for instance, or multipole extension wave should have more sense...

I would like to answer properly to this question, but it is difficult. I tried something using a current probability like to describe transition between states. I asked already on the forum but nobody answered since it is very specific. But using the maxwell equation for the potential vector (EM field) and a transition probablity current like as current source for inhomogeneous vector wave equation: that could provide the shape of the EM field produced by the single electron transition. I tried my best to answer, using my phone, see my topic to go further.

EDIT:

Photoelectric effect, spontaneous emission (with the single electron jump leading to a single photon creation) and Lamb shift have been theorized properly by Roger Boudet using a relativistic transition current if you are interested in.

Answer 3

This probably will not answer all of the sub questions you asked. This is aimed at what I understand to be the core question you are asking: "How can an isolated atom emit a photon in a wavepacket, when the photon should have a definite frequence related to the transition energy"?

Keep in mind that the Hydrogen atom has 2 variables, the separation of the electron and the nucleus (which is what we use when solving the schrodinger equation), but also the center of mass coordinate. We usually drop the latter since it behaves as a free particle subject to no potential and therefore does not add much to the discussion.

To be fully realistic though we should say that the whole atom sits in a wavepacket (so that we can give it a well defined position). Specifically the center of mass coordinate is smeared out over a wavepacket. Then both the electron and the nucleus are in this smeared out wavepacket, but their separation corresponds to the "electron orbitals" we think of.

So now if there is an electronic transition, the atom emits a photon with frequency given by the transition energy, at least in relation to their "separation" coordinate, including things like multipole moments as you mention. But keep in mind the whole system is smeared out over a wavepacket, so any photon emitted relative to the whole system would be smeared out over the wavepacket as well.

This is at least my understanding of the phenomena, and of your question. I will not attempt an in depth answer to your second question, but I think it ultimately boils down to: "photons and EM waves are not the same thing". A classical EM wave corresponds to a superposition of photon states without a well defined "number of photons". If you are interested in the dynamics of a single photon state, the "wavefront" in the opposite direction it is detected in does not look or act like an EM wave, but instead more like a "quantum wavefunction". However you need to be careful when considering photons and there are plenty of in depth answers on this site related to that. In particular the following:

"What exactly is a photon": What exactly is a photon?

"What is the relation between electromagnetic wave and photon?": What is the relation between electromagnetic wave and photon?