# Why are neutrinos LH chiral

• I
Hello! I am a bit confused about neutrinos in the standard model. The vertex of the weak interaction charged current, implies that any neutrino interacting through the charged current must be left handed. However the neutral current allows coupling to the right handed particles, too (and we see that in the interactions with charged leptons). Why is it not possible for a Z boson to decay to a pair for neutrino-antineutrino, in which the neutrino is right handed? I.e. why don't we have any right handed neutrinos in the standard model? Thank you!

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mfb
Mentor
I.e. why don't we have any right handed neutrinos in the standard model?
Because we don't see any right handed neutrinos in nature and our models are made to reflect nature. Why don't we see any right handed neutrinos? If you find an answer to that question it's an almost certain Nobel Prize.

Astronuc and vanhees71
Because we don't see any right handed neutrinos in nature and our models are made to reflect nature. Why don't we see any right handed neutrinos? If you find an answer to that question it's an almost certain Nobel Prize.
I am not sure if this answers my question. For example one open question related to neutrinos is wether they are Dirac or Majorana. If they are Dirac, and they get mass by Yukawa coupling, that implies the existence of right handed neutrinos. So based on your answer, if we are sure that there are not right handed neutrinos, why can't we say for sure that the neutrinos are not Dirac particle? Or, if there are right handed neutrinos (and the Dirac option is still possible) why don't they couple to the Z boson (the same way right handed electrons do)?

jtbell
Mentor
if we are sure that there are not right handed neutrinos,
As far as I know, we're not sure that this is the case. All we know is that we have not detected right-handed neutrinos (yet). In general, it's difficult to prove a negative statement experimentally, e.g. that right-handed neutrinos do not exist.

As far as I know, we're not sure that this is the case. All we know is that we have not detected right-handed neutrinos (yet). In general, it's difficult to prove a negative statement experimentally, e.g. that right-handed neutrinos do not exist.
But my confusion is related to the Z boson. We know from the form of the vertex, that the Z boson couples to the right handed particles, too (at a different rate than to the left handed ones, tho). Yet, we didn't see any right handed neutrino experimentally. Why is that not a good indication that right handed neutrinos don't exist? Shouldn't they interact with the Z boson if they existed?

vanhees71
Gold Member
2019 Award
It's just a (most likely approximate) observational fact of nature, how the weak interaction (or rather the electroweak interaction) couples the fundamental particles (quarks, leptons, and the Higgs boson) through the appropriate gauge fields.

Historically the description of the weak interaction started with Fermi's QFT model for ##\beta## decay. Later on there was a lot of experimental confusion about the details of the interaction, and thus Feynman and Gell-Mann wrote down the most general model for couplings of the leptons and after some great experimental effort, including the discovery of (maximal) parity violation through the em. interaction, the entire story ended up with the discovery that the weak interaction is mediated by the current pattern ##V-A## (vector minus axial vector currents).

The trouble with these Fermi-like four-fermion point coupling models however was that they were not renormalizable. After the discovery of the non-Abelian gauge theories the idea came up that there should be a gauge theory for the electroweak interaction. At the same time the weak interaction is very short-ranged and the physicists thought there must be massive gauge bosons for describing such short-ranged interactions if mediated by gauge bosons. Unfortunately a naive mass term kills gauge invariance and spoils the consistency of the entire gauge model (violating unitarity besides all kinds of nasty artifacts).

Then there was quite some theoretical confusion, because what was clear through the work of Nambu, Weinberg et al. concerning spontaneous symmetry breaking, one can have chiral models with massive particles when the chiral symmetry is broken. Now the misunderstanding for a while was that this was thought to imply the existence of massless scalar or pseudoscalar bosons (Nambu-Goldstone modes), and no such thing was wanted in connection with the weak interactions.

However, and this is important, and as Anderson first figured out, the Nambu-Goldstone modes only occur in the physical spectrum where the spontaneously broken symmetry is global. As soon as you gauge the symmetry, i.e., make it a local symmetry, by introduction of the corresponding gauge bosons, the would-be-Nambu-Goldstone modes can be absorbed into the gauge fields by a clever choice of gauge (the socalled unitary gauge). Then the gauge bosons "eat" the Nambu Goldstone modes giving them a third polarization degree of freedom and make them massive without breaking the gauge symmetry at all (in fact, you cannot spontaneously break a local gauge symmetry according to Elitzur's theorem).

Then Higgs figured out that nevertheless, if you want a (superficially) renormalizable theory, there'll be always at least one massive scalar boson in the theory, the famous Higgs boson.

What was also unclear for a while was whether non-Abelian "Higgsed" gauge models are really renormalizable. The trouble is that in the unitary gauge it's not manifestly renormalizable. The trick was to introduce another gauge and a gauge-invariant renormalization scheme by 't Hooft and Veltman, i.e., the socalled ##R_{\xi}## gauges and dimensional regularization. In the ##R_{\xi}## gauge you apparently reintroduce the would-be-Nambu-Goldstone modes, but as it turns out together with the Faddeev Popov ghosts you need anyway to quantize the gauge theory properly, they just compensate unphysical degrees of freedom in the description of the S-matrix, i.e., at the end you get a unitary S-matrix with only physical degrees of freedom (in the usual perturbative sense of course).

The very exciting history of how the Standard Model was discovered in a very illustrative example for how model building works as a process involving both theory and experiment, see

F. Close, The Infinity Puzzle

Astronuc
But my confusion is related to the Z boson. We know from the form of the vertex, that the Z boson couples to the right handed particles, too
And if you look at that vertex more carefully, you will see that the coupling to the right handed particles is proportional to the electric charge.

Now, what is the electric charge of a neutrino?

vanhees71