# Could a proton be produced using the dynamical Casimir effect?

I think the answer to your question is 'no', but as I will point out below it is possible that there is some clever way to make proton production happen.

The first thing to note is that photons are always easy to produce as "soft" excitations due to their masslessness. Massive excitations are significantly more expensive to produce.

This doesn't yet address your specific question about protons. In the usual formulation of the Casimir effect, it is discussed within quantum electrodynamics (QED). The fundamental excitations of QED are the photon and electron/positron. In this context the dynamical Casimir effect would be expected to produce photons and electron/positron pairs, but there is no mechanism for generating other particles. And, of course, photon generation will have qualitative differences from the pair production processes due to the difference in particle masses.

If you want to think about protons, it is better to think about the dynamical Casimir effect in the context of quantum chromodynamics (QCD), which is a context where I have not personally seen it considered. Presumably, virtual particle production proceeds there as in QED, but again, the proton is not a fundamental particle of the theory. The new particles which could be produced directly are quarks and gluons. Protons are bound states of such particles, so proton production would require a subsequent binding of the particles produced by the dynamical Casimir effect. Perhaps there is a a clever way to arrange for this to happen, but it seems to me that protons cannot be produced directly.

Short answer: 'Yes' as told by @knzhou.

Long answer: How it actually happens? As reviewed here DCE is a variant of "bogoliubov transformation coefficient" calculation problem: you have annihilation, creation operator at one time, and then they mix with each other over time you measure the number of particle $$\langle 0|b^\dagger b|0\rangle$$ at later time to find particles created.

So how does proton buisness actually take place? Answer to this one will address both the issue of "acceleration" and "QCD" mentioned above complementary to the answer saying 'yes' and 'no' and how they're intertwined? As in QED, we have interaction of electrons (Dirac spinor $$\psi$$) with photons ($$A_\mu$$) when we produce photon, electron we have to accelerate the plates of setup which is crudely speaking a measure of energy scale at which we're probing. To produce proton we have to be a bit careful since if we jump directly to a very high energy scale we would be probing the inner structure of hadron which leads to the business of quarks and gluons (analogous to $$e^\pm$$ and photon) but we want protons so we have step down the energy scale so that we can work with Chiral Lagrangian it leads to business of nucleons and pions$$^{[1]}$$ (analogous to $$e^\pm$$ and photon).

Similiar kind of arguments are there in Hawking's paper on BH radiation where he worked out for spin$$-0$$ scalar field and concluded gravitons, photons will analogously be produced.

As for the actual calculation, I'm not quite sure how easy it is get the results because for Schwinger effect one has to take help of effective lagrangian (we're already using it in chiral lagrangian) and answer is obtained using an indirect approach. Let see if anyone here who had a working knowledge of effective field theory actually produce some closed form results.

$$[1]$$ Check section $$22.3$$ of Schwartz QFT