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If we have a neutrino $\nu_e$ and an antineutrino $\bar{\nu_e}$, one would expect that a possible interaction would be that they would annihilate each other, producing possibly two photons. I do not believe that this has been observed, however it would certainly be expected based on other particle/antiparticle pairs which have been observed.

If we consider the time reversal of that interaction, we would have two photons interacting, producing a neutrino/antineutrino pair.

One would naively expect it to be possible fairly easily; the rest mass of a neutrino is unclear; it is speculated to be around $1$ eV (perhaps within an order of magnitude); this level of energy can be achieved with visible or near ultraviolet photons.

However, it obviously doesn't happen (or, at best, happens extremely rarely); if it did happen, one would observe photons 'disappearing' (as the generated neutrinos are effectively invisible), and we don't see that. And, if it did happen, observing when it did happen would allow us to get a much better estimate of the neutrino rest mass.

So, my question is: do we know of a reason why this doesn't happen?

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    $\begingroup$ Are you wildly imagining an intermediary Higgs? Surely you know there are no SM tree diagrams underlying your amp and why. $\endgroup$
    – Cosmas Zachos
    Jun 3, 2021 at 13:31
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    $\begingroup$ Neutrino doesn't have electric charge and cannot emit/absorb a photon. $\endgroup$
    – Mitchell Porter
    Jun 3, 2021 at 13:34
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    $\begingroup$ @CosmasZachos: no, I don't believe I am wildly imaging anything, and certainly not an intermediate Higgs. As for an SM tree diagram, well, if neutrino/antineutrino annihilation is possible (and hence must have a consistant tree), it would appear to me that the time reversal would as well $\endgroup$
    – poncho
    Jun 3, 2021 at 13:36
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    $\begingroup$ Related: Neutrino annihilation and bosons. The short answer is that the annihilation cross-section for neutrinos is ultra-tiny unless they have enough energy to produce a $Z$ (at tree level.) $\endgroup$
    – Michael Seifert
    Jun 3, 2021 at 13:38
  • $\begingroup$ @MitchellPorter: hmmmm, is neutrino/antineutrino annihilation possible? If it is, what is the product? I didn't think that there was any possible way to expend that tiny amount of energy except photons... $\endgroup$
    – poncho
    Jun 3, 2021 at 13:38

3 Answers 3

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I thought about such a process at one loop level, please correct me if this won't work for some reason.

Feynman diagram

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  • $\begingroup$ Exactly! That should be possible.! So charged fields can interact with neutrino fields. $\endgroup$
    – user301229
    Jun 3, 2021 at 14:44
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    $\begingroup$ @Barbierium Anything is possible at one loop level, including the diagram with three W sides and one charged lepton side you asked in your comments about. But one must compare propagator suppressions at that level. The OP is quite focussed on tree diagrams. $\endgroup$
    – Cosmas Zachos
    Jun 3, 2021 at 14:51
  • $\begingroup$ Does the W+ stand for the W- also? $\endgroup$
    – user301229
    Jun 3, 2021 at 16:37
  • $\begingroup$ Yes, that looks possible - on the other hand, the rest mass of W is so huge I can't imagine that it has much of a probability of happening at all. $\endgroup$
    – poncho
    Jun 3, 2021 at 16:39
  • $\begingroup$ @CosmasZachos: BTW: I'm not at all focused on tree diagrams (unless, of course, that's the way to address my "why doesn't this happen" question) $\endgroup$
    – poncho
    Jun 3, 2021 at 16:40
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You know $Z\to \bar \nu \nu $ at tree level, routinely. However, by construction of the SM there is no $Z\gamma\gamma$ tree coupling: There are several answers on this site explaining why. (No trilinear couplings for the $W_3$, nor for the B, so not for their mixtures.)

The reason I mentioned an intermediary Higgs is since there is such a freakishly small $h\to \bar \nu \nu $ (presumably you recognize why: proportional to neutrino mass!), and an induced fermion loop coupling for $h\to \gamma\gamma $ exploited in its discovery. Presumably your can eyeball how truly, doubly freakishly, unlikely this is.

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  • $\begingroup$ Technically $h\to \bar\nu \nu$ is also at loop level in the Standard Model. To get it at tree-level you are assuming that a right handed neutrino exists. $\endgroup$
    – FrodCube
    Jun 3, 2021 at 14:56
  • $\begingroup$ Yes; at this stage of development (in contrast to the last millennium!), the Occam's razor default is that right-handed neutrinos exist--it takes an assumption to exclude them! Not to be really known until the Majorana option is settled. $\endgroup$
    – Cosmas Zachos
    Jun 3, 2021 at 15:02
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By the simple fact that photon fields don't interact with neutrino fields, they can't excite a neutrino pair. At tree level. At loop level, they could excite particle fields that do interact with it (W's, quarks, or electrons), but these kinds of interactions are highly suppressed (see comment by @CosmasZachos).

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  • $\begingroup$ This answer is not accurate. The process of the question can happen at one loop without any problem. It's just forbidden at tree-level. $\endgroup$
    – FrodCube
    Jun 3, 2021 at 14:52
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    $\begingroup$ @FrodCube Yeah, I realized after posting it... Next time I should wait a bit before answering. $\endgroup$
    – user301229
    Jun 3, 2021 at 14:53
  • $\begingroup$ Can we say that some conservation law is violated in this process? Like strangeness or something? $\endgroup$
    – Physiker
    Jun 3, 2021 at 15:02
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    $\begingroup$ @IndischerPhysiker No strict conservation law is violated. Actually strangeness is only conserved in the strong interactions, and the name of the game here is to beat trees by loops. $\endgroup$
    – Cosmas Zachos
    Jun 3, 2021 at 15:06

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