How is a photon created when an electron jumps to a lower orbit?

Question

Everyone is taught that when an electron jumps from a high to a lower orbit, the difference in energy is emitted as a photon.

However, how does the photon get created? See more exact phrasing of the question, below.

What is a photon is answered here: What exactly is a photon?

How a photon is made in general, is answered here: How are photons made?

For the latter question, the answers by ACuriousMind and John Rennie appear to be relevant. An exact answer to my question would be nice though, as it seems to be missing on Physics Stack Exchange.

My own reasoning, directly based on reading John Rennie's answer:

A photon has an associated quantum field. Energy gets added into this field (the photon field) when the electron moves to a lower orbit. The energy comes from the lost energy in the electron field -- the two fields transfer energy. The energy added equals that of a photon, hence a photon is created and emitted.

As John Rennie points out "If we transfer energy into the photon field then it appears as a photon".

However this seemingly provides no explanation at all as to how the photon is created, beyond introducing the notion of quantum fields. How is it that energy just "appears as a photon"? What is the best model to explain this?

Answer 1

A model which you may find helpful at the level of ordinary quantum mechanics (without field theory) is as follows:

Suppose the electron starts in a state $\psi_1(\vec r)$ with energy $E_1$, and finishes in a state $\psi_0(\vec r)$ with energy $E_0$, and $E_1 > E_0$ if we're talking about emission.

In between start and finish, during the emission process, it is a state $a(t)\psi_1 + b(t) \psi_0$, where $a^2+b^2=1$ and the exact form doesn't concern us. This is not an eigenstate of the time-independent Schroedinger equation and the full description needs to include the different time dependences of the 2 states

$\Psi(\vec r,t)=a(t) e^{-i E_1 t/ \hbar} \psi_1(\vec r) +b(t)e^{-i E_0 t/\hbar} \psi_0(\vec r) $

Now consider the mean value of, say, the electron's $x$ co-ordinate (or any other if you prefer). It is $\int \Psi^* x \Psi dV$ which is

$a^2 \int x |\psi_1|^2 dV + b^2 \int x |\psi_2|^2 dV +ab^*e^{i(E_0-E_1)t/\hbar}\int x \psi_1 \psi_0^* dV +a^*be^{-i(E_0-E_1)t/\hbar}\int x \psi_1^* \psi_0 dV $

The first two terms are probably zero, by symmetry, unless the potential is asymmetric, and are anyway slowly-varying with time and can be ignored. The third and fourth terms combine to give a term proportional to $\cos((E_1-E_0)t/\hbar)$.

So the mean position of the electron during this intermediate period is oscillating with frequency $\omega = {E_1-E_0 \over \hbar}$. This electron has a charge, so it is behaving like an oscillating dipole. Switching hats from QM to E&M, we remember that oscillating dipoles produce electromagnetic radiation. The frequency of that radiation is exactly right to balance energy conservation.

So your answer: photon is created as a chunk of dipole radiation from the charged electron as its mean position oscillates in the intermediate state between the initial and final energy levels.

(There are other ways of thinking about photons, of course. But this is a very nice example of the correspondence principle.)

Answer 2

TL;DR: there is a coupling between the electron and the electromagnetic field, which allows the transfer of energy and momentum from one to the other.

Atom, field and interaction
The net Hamiltonian of the atom and the electromagnetic field can be written as $$H=H_{atom}+H_{em} + H_{int},$$ where interaction term is actually the coupling of the electron to the part of the scalar and vector potentials, which are due to the EM field. One typically starts with solving separately the Hamiltonians for the atom and the EM field. The former gives us the energy levels (while the latter is usually already diagonal, as it describes free field, but there may be variations). The coupling is then included as a perturbation, and the first order contribution mixes the degenerate (and nearly degenerate) states of the atim and the field. E.g., the states with no photons and the atim excited has energy close to the state with the atom in its ground state and a photon present: $E_e=E_g+\hbar\omega_k$. Higher order processes (such as Raman scattering) appear in higher orders of perturbative expansion.

Jaynes-Cummings model
The simplest model of this type is Jaynes-Cummings model, with Hamiltonian: $$H_{atom}=E_e|e\rangle\langle e| + E_g|g\rangle\langle g|,\\ H_{em}=\hbar\omega_0a^\dagger a,\\ H_{int}=dE (a^\dagger + a)(|e\rangle\langle g|+|g\rangle\langle e|)$$ The interaction term is actually the product of the dipole moment of the atom $d(|e\rangle\langle g|+|g\rangle\langle e|)$ and the electric field due to photons, $E (a^\dagger + a)$. (There are some caveats if one uses thsi model for describing photon emission, I put it for illustartive purposes.)

Not only photons
The approach is by no means unique to photons - many interactions are discussed in this perturbative way.