Why Does an Electron Emit a Photon?

Question

I've seen similar questions asked, but I didn't think they were quite the same.

My understanding, and it is probably an incorrect one, is that an electron in an orbital shell has a stable balance of kinetic energy and potential energy. When a photon of the right frequency disturbs this, the electron is lifted to a new shell with a new stable balance between kinetic energy and potential energy.

I would think that this is a new stable situation. Why, then, would an electron ever lose energy and drop back to the old level?

This is, in fact, a unique question on this site. I am not asking about stimulated emission (using bouncing light off the electron to decelerate it). I would prefer an answer to why the vis-viva equations (Hamiltonians) don't apply, so as much as I appreciate answers sending me to databases of Einstein A coefficients, like HITRAN (thank you), this is only the answer to my question in the same way that "because I say so" is an answer to every question.

The closest to what might be an answer (for anyone following behind me) is this Atomic Natural Line Width

Which I haven't finished parsing yet, but my understanding so far is : that the vis-viva (Hamiltonian) equations are only for the atom. And considered this way alone, spontaneous emission would not occur. You need to consider the field effect of other things in the media. These combined have a net slowing effect, generally.

So, thank you for the help finding an answer in other questions in the comments, and thank you to the two people who tried to field an answer before this was question was cut-off barely 24 hours after it was asked by a bunch of people who provided wrong answers only (so far; it's a lot of links to partial answers, which all require more research on my part).

This useless nonsense is why I stopped posting on Stack Exchange and went somewhere helpful like Codidact. Thanks for the reminder!

Answer 1

I will try to answer this question from quantum electrodynamics's perspective, because the process of particle creation and annihilation (or you can also say particle absorption and emission) is better explained by quantum field theory.

The core idea is that, due to the interaction term in the Lagrangian (or you can also say the Hamiltonian),A process with initial state of one electron and a final state of one electron(with lower energy) have a nonzero amplitude, so you can calculate a nonzero probability for this event to happen. In the same way, you can also calculate a nonzero probability for the electron to fall back to its original state, which means that, both procedure you are asking is not prohibited by laws of physics.

Related mathematics is discussed in detail in,David Tong's lecture note, and I would try to provide an outline for the explanation.

When calculating an amplitude in quantum electrodynamics, we work in interaction picture, a detailed explanation can be found here in p51.

I will only state the result here: the formula for calculating an amplitude that involves particle interaction is $$\lim_{t_{\pm} \to \pm \infty}\langle f|U(t_+,t_-)|i\rangle$$where $\langle f|$ stands for initial state and $|i\rangle$ stands for final state, and $$\lim_{t_{\pm} \to \pm \infty}\langle f|U(t_+,t_-)|i\rangle =\langle f |T\exp\left( -i \int_{t_0}^{t} H_I(t')\,dt' \right) |i \rangle= \ 1 - i \int_{t_0}^{t} dt' \, H_I(t') + (-i)^2 \int_{t_0}^{t} dt' \int_{t_0}^{t'} dt'' \, H_I(t') H_I(t'') + \cdots \ $$

where the exponent is defined by Taylor expansion.

From Yukawa theory, the interaction term in the Hamiltonian is $H_{I}=\int d^3x\bar{\psi}\gamma^\mu\psi A_{\mu}$, and to construct initial states and final states with particles, you take creation operators and annihilation operators and act on the vacuum state $| 0 \rangle $. calculate the amplitude (you can use wick's theorem or Feynman's diagram to simplify the procedure) and you will find that both procedure you are asking have a nonzero amplitude, which means that both procedure you are asking are possible.

For detailed treatments on Yukawa theory, creation and annihilation operators and the mathematical formalism of quantum electrodynamics, please refer to David Tong's lecture note,For detailed treatments on wick's theorem and Feynman diagram, please refer to Peskin & Schroeder's book "An Introduction to quantum field theory"(chapter 4.4 and 4.6).

Reference:

1: David Tong's lecture note,https://www.damtp.cam.ac.uk/user/tong/qft.html

2: "An Introduction to quantum field theory" Michael E. Peskin, Daniel V. Schroeder

Answer 2

Systems prefer to be in the lowest possible energy state (i.e. ground state). They prefer to minimize the Helmholtz free energy essentially dictated by the second law of thermodynamics. This is why an excited electron in the atom emits a photon to go back to the ground state.

See this Wikipedia article: https://en.wikipedia.org/wiki/Principle_of_minimum_energy