What happens when an anti-electron collides with a neutrino?

What happens when an anti-electron collides with a neutrino? If something does happen, is a photon released after the collision?

The only way the anti-electron $$e^+$$ and a neutrino $$\nu$$ can interact is by the weak interaction, i.e. mediated by a $$W$$ or $$Z$$ boson.

There are several reactions possible, here shown as Feynman diagrams. (I am using the convention of time running from left to right, and backward arrows for anti-particles.)

1. Scattering by exchanging a $$Z$$ boson:

2. Forming a $$W^+$$ which immediately decays again (as @AgniusVasiliauskas already showed in his answer):

3. Forming a $$W^+$$ which immediately decays into a quark/anti-quark pair, for example a $$u$$ quark (with charge $$+\frac{2}{3}$$) and a $$\bar{d}$$ anti-quark (with charge $$+\frac{1}{3}$$), which then immediately fuses into a $$\pi^+$$ meson. (This reaction is only possible if the incoming $$e^+$$ and $$\nu$$ together have sufficiently high energy to produce the rest mass of the $$\pi^+$$ meson):

• What would be an order of magnitude, or even coarser, idea for relative cross-sections of these processes? Dec 7, 2023 at 17:50
• @VladimirFГероямслава Very small. The CERN article Neutrino Electron scsttering (on page 13) reports experimental values for $\sigma(\nu e)/E_\nu$ around some $10^{-45}\text{cm}^2\text{MeV}^{-1}$. Dec 7, 2023 at 18:11

No it's not photon released after the collision, since the electro charge would not be conserved in this scenario. It's $$W^{\pm}$$ boson released after such electroweak process.

Note that even if you ask a seemingly easy question: "Do electron and positron (anti-electron) annihilate to release photons?", the answer may still surprise you: "it's only partially correct". Given enough energy, $$Z$$ boson or Higgs boson could also be directly released after the electron-positron annihilation: $$e^- +e^+ \to Z$$ $$e^- +e^+ \to Higgs$$

Of course, $$Z$$/$$W^{\pm}$$ boson and Higgs boson all have mass, as opposed to the massless photon. Therefore, it's a bit hard to directly observe these bosons after annihilation since these massive bosons can't travel far before bumping/turning into something else.

$$W^+$$ boson will be produced, but that reaction is very unstable : $$\nu_e+e^+ \to W^+ \to \nu_e + e^+ \tag 1$$ because $$W^\pm$$ boson has a half-life on the order $$10^{-25}~\text{s}$$, hence after forming it rapidly will decay back into electron neutrino and positron. Or you can check the Feynman diagram of such process if you like :

If something does happen, is a photon released after the collision?

is exactly known, and indeed determined by deep physics of asymptotic symmetries of gauge theories. As a rough summary, if some nontrivial interaction does take place between your positron and your neutrino, a photon is released after the collision about $$1/137$$th of the time.

An answer of 'no' to your second question either underestimates how cool quantum physics is or overestimates what symmetries there could exist in the Standard Model that could forbid a process with an outgoing photon.

Indeed, there is such a higher-order process you may visualize by attaching an outgoing photon line to one of the charged particle lines in any of the Feynman diagrams drawn in the other answers. A collider physicist would refer to such a photon as 'initial (final) state radiation' depending which line you chose.

In fact there is quite a nice symmetry understanding in quantum field theory of how likely such a process is---in the limit that we ask about a 'soft' (much lower-energy than that of the external scattering states) photon, there is a famous theorem due to Weinberg 1965. (The modern reader may want to refer to, e.g., Schwartz' QFT & the SM which was the book closest to my current position that gave a satisfactory treatment.)

E.g. Taking the inelastic, weak process in Thomas Fritsch' answer with an additional photon coming off any external line and sum over all such possible insertions of a photon with momentum $$q$$ and polarization $$\epsilon$$. The amplitude for this to occur $$\mathcal{M}$$, taking $$\mathcal{M}_0$$ as the amplitude for the original process to occur, is $$\mathcal{M} \approx e \mathcal{M}_0 \left[ \sum_i (\pm Q_i) \frac{p_i \cdot \epsilon}{p_i \cdot q} \right]$$ where $$e$$ is the electromagnetic charge with which the photon couples to charged particles, in the normal 'natural' units $$e^2/(4\pi) \equiv \alpha \simeq 1/137$$; we sum over all external lines $$i$$ with electromagnetic charge $$Q_i$$ and momentum $$p_i$$ and use the $$+$$ for incoming states and $$-$$ for outgoing states.

In quantum mechanics, probabilities are like amplitudes squared, and indeed a particle physicist computes a 'cross-section' for this process that integrates over low-energy-energy outgoing photons and is suppressed by $$\alpha$$ with respect to the process with no outgoing photon.

The most likely thing that would happen is nothing. Particles don't collide like billiard balls. They're vibrations of fields, and the fields are coupled together, but not strongly enough in this case for a significant interaction to be likely. It's something like crossing two laser beams: in theory there is a small interaction between them but it would be difficult to detect it even in a carefully conducted experiment.