Can spontaneous emission be considered a thermodynamic process?

[Karl Coryat]

TL;DR

Wondering if entropy has anything to do with the smallest systems seeking low-energy states.

I realize that nothing causes an excited atom to emit a photon, and that it's a random process. But someone was asking me about why energized systems in general tend to lose their energy to the environment and move toward equilibrium. I mentioned that an inflated balloon, given a hole, will tend to deflate, and I gave the thermodynamic explanation which ultimately invokes statistical mechanics. Plenty of other macroscopic processes, even gravity, have thermodynamic, emergent explanations, at least in some theories.

What about on the atomic level? Is a system of an atom plus an emitted photon considered higher-entropy than the atom when excited? One could say that there are many more configurations of atom + photon than there are of an excited atom alone.

Basically, I want to learn how small a system can be and still have a "macrostate" for the purpose of considering a statistical, entropic explanation. Surely an ensemble of excited atoms, decaying exponentially, meets the description...but just one?

To be clear, again, I'm not looking to explain the particular emission, just why an excited atom tends in that direction.

(Note: A similar question was asked here a long time ago, but I find the answer unsatisfying. https://www.physicsforums.com/threads/a-thermodynamic-approach-to-spontaneous-emission.75515/ )

Thank you!

 

[hilbert2]

An ensemble of non-interacting excited atoms becoming photons and ground state atoms is quite equivalent to a single excited atom being likely to end up that way.

 

[kith]

Here's Feynman:

Incidentally, why does it go one way instead of the other way? Why does an atom radiate light? The answer has to do with entropy. When the energy is in the electromagnetic field, there are so many different ways it can be—so many different places where it can wander—that if we look for the equilibrium condition, we find that in the most probable situation the field is excited with a photon, and the atom is de-excited. It takes a very long time for the photon to come back and find that it can knock the atom back up again.

 

[atyy]

By itself, spontaneous emission can be obtained without thermodynamics.
https://en.wikipedia.org/wiki/Fermi's_golden_rule
http://www2.optics.rochester.edu/~stroud/cqi/stanford/15_single_mode_spont_emiss.pdf
http://www.tcm.phy.cam.ac.uk/~bds10/aqp/lec19_compressed.pdf

However, Einstein gave a famous argument relating spontaneous emission and stimulated emission.
http://hyperphysics.phy-astr.gsu.edu/hbase/optmod/eincoef.html
https://sites.google.com/site/pueng...lation-between-einstein-s-a-and-b-coefficient
L13.3 Einstein's B and A coefficients determined. Lifetimes and selection rules

 

[vanhees71]

Spontaneous emission is occurring also at $T=0$. It's not something specifically "thermal". It's the most simple phenomenon that really needs the quantization of the electromagnetic field (i.e., photons).

Let's take a hydrogen atom to have a specific example. The usual QM1 treatment is of course in the semi-classical approximation, i.e., you treat the electromagnetic field as classical and you take into account only the static coulomb potential due to the proton and calculate the energy eigenstates of the electron moving in this Coulomb field (let's for simplicity use the approximation where the proton is taken as infinitely heavy and at rest). In this approximation the lonely H atom will stay in any of the energy eigenstates once prepared in one. After all, that's what energy eigentstates are: the stationary states of the system.

Of course, now this is an approximation in view that the electromagnetic field in fact has to be quantized too. Now the total dynamical system consists of a proton, an electron and the electromagnetic field. The Heisenberg uncertainty relation for the em. field operators tells you that you cannot have a determined electromagnetic field, implying that even in the vacuum the electromagnetic field is fluctuating. Now having the electron and proton prepared in one of its excited states due to the fluctuating em. field there is some probability that a transition occurs such that the atom goes to a lower-energy state and a photon is emitted, and that can happen when initially the electromagnetic field is prepared in its ground state, i.e., no photons present. That's spontaneous emission (in a perturbative description).