2
$\begingroup$

Are photoionization from ground and recombination to ground consider a reversible processes in the thermodynamic sense?


Ignoring any quantum broadening effects, consider a closed system of one Hydrogen and one ionizing photon. The photon could ionize the atom, which would later recombine emitting the same energy photon that was absorbed. This seems to be a cycle that ends in the same state, therefore I'm guessing each ionization and recombination must be reversible processes.

Now recombinations do not always end up in the ground state. Then the recombination to a different level other than ground would lead to a cascade that would result in at a minimum of two photons, with at most only one of ionizing energy. Given enough time and repetition of recombining to a state other than ground there will eventually be no ionizing photon.

Since photons have zero chemical potential, this doesn't necessarily mean the system is in a different state does it? Just interestingly the system can no longer perform this process.


If this single atom case is reversible, are the same processes in a system of multiple atoms and photons reversible? This seems different since now the atoms can thermalize amongst each other. Imagine the gas is initially at a lower temperature than the photon gas, then given enough time the two gases will equilibrate. As before eventually there will be no more ionizing photons and you could not reverse the gas back to its original state of being colder than the photon gas. It seems the gas is then in a state which cannot be taken back to it's initial state.

If the system is both the photon and gas is this necessarily a different state? Did I pull a slight of hand, and the problem with this scenario is that the collisions were the irreversible process, not the ionization and recombinations themselves?

$\endgroup$
  • $\begingroup$ The reverse process can occur exactly how it happened in the forward direction: if your atom emitted one photon, then another, it may absorb the second and then the first photon and end up in the original state. $\endgroup$ – Ruslan Aug 15 '17 at 6:09
  • $\begingroup$ Following your argumentation I would call it an irreversible process. Interaction with the EM field leads to increase of entropy because of the vast amount of degrees of freedom (continuous energy spectrum) in the EM field. In fact my understanding is that the pointlike interaction with the EM field causes any fermion field-configuration (e.g. plane wave) to decohere into pointlike particles unless you have only a single particle + EM field which would be protected from decoherence by momentum conservation. $\endgroup$ – Lukas Berns Aug 15 '17 at 6:21
1
$\begingroup$

I think your question is somewhat ill defined, and you need to understand that the word process has different meanings in different contexts. The thermodynamic process implies some interaction between two systems, while in unitary, Schrödinger equation like dynamics, it usually mean transitions between states as considered when drawing Feynman diagrams.

If you're considering a closed system as you describe at the beginning of your question you must consider to photon bath as part of your closed system which is a bit unnatural in my eyes.

Closed System Approach

None the less, in such a case, indeed after enough time (which is usually termed the recurrence time if I'm not mistaken), a photon will be emitted and the system will return to its original state. However I wouldn't call it a thermodynamic process. This is just unitary evolution according to Schrödinger equation. Since these dynamics are Hamiltonian they're inherently energy conserving which may look to you like reversibility. However the notion of reversibility is meaningless as there are no baths which interact in any way, and there are no thermodynamics.

Open System Approach

If on the other hand you consider an open system, i.e. you think of your atom as a system which is coupled to a thermal (photon) bath, then you may indeed think of the resulting thermodynamics.

First, note that dynamics with open systems which have infinite degrees of freedom such as a photon bath induce dissipation. This is relevant though it doesn't answer your question plainly.

Second, in such a case processes of emission and absorption induce energy (and momentum) transfer between the two systems. This thermodynamic process however isn't reversible. Perhaps the easiest way to see this is by looking at the detail-balance relation which implies that the ratio of emission and absorption rates must be: $$ \frac{\Gamma_{abs}}{\Gamma_{em}}=e^{-\beta\epsilon} $$ where $\beta$ is the inverse temperature of the thermal bath, and $\epsilon$ is the ionization energy.

$\endgroup$
  • $\begingroup$ Yes sorry, completely my mistake I should have said an open system. Since the process convincingly doesn't seem to be reversible is there any hope in using $\textrm{d} h = \delta q + \frac{\textrm{d}p}{\rho}$ in a fluid that is experiencing external heating from photoionization. The idea is that $\delta q$ contains the heating and cooling introduced by the ionization and recombination. Could we still imagine that the pdV work of the gas is reversible, so then $dh = \delta q + \delta w + pdV +Vdp =\delta q + Vdp$? Or would this heating make the any otherwise quasistatic process not quasistatic? $\endgroup$ – Novice C Aug 15 '17 at 7:29
  • $\begingroup$ You're using notation which usually reserved to big bulk systems which operate out of equilibrium but in steady state. These quantities don't always translate to the language microscopic systems (atoms) so naturally, and things must be done carefully. I'm no expert on the matter so I don't want to make wrong claims. However the following can be said: the absorption and emission processes create entropy and are therefore not reversible. The steady state regime however allows for some relations such as you proposed at least on a macroscopic scale, regarding the mice-scopic scale I don't know. $\endgroup$ – Yair M Aug 15 '17 at 7:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.