# Weak interaction of $(\nu^c)_R$ fields of the SM and $N_R$ fields in type-I seesaw extension of the SM

The antiparticle of $\nu_L$ is given by its charge conjugated field i.e., $(\nu_L)^c$ which is equal to $(\nu^c)_R$. Both $\nu_L$ and $(\nu^c)_R$ are part of the Standard Model (SM) of massless neutrinos. See the answer here by JeffDror.

$\bullet$ How can $(\nu^c)_R$ take part in weak interaction which always couple the left-chiral fields? I'm looking for the relevant piece of the Lagrangian.

$\bullet$ Why a Dirac mass cannot be written down using the left-handed $\nu_L$ field and the right-handed $(\nu^c)_R$ field?

$\bullet$ If $(\nu^c)_R$ fields can take part in weak interaction, does it mean $N_R$ fields (fields introduced in the type-I seesaw extension of the SM) can also take part in weak interaction?

I assume you are not asking for a tutorial of the Dirac & Majorana equations or C, amply covered in standard texts like Li & Cheng, Schwartz, or Ramond’s “Journeys Beyond the Standard Model” —the horse’s mouth! You might prefer Klauber. I’ll just answer your narrower questions with the standard mnemonic.

First, go back to pre superKamiokande days, so, then without a weak singlet $\nu_R$ sterile neutrino, so neutrino Dirac masses —or a seesaw speculated upon in 1979. So all neutrinos and antineutrinos have weak charge (doublet indices partnered with those of the charged leptons), conjugate to each other.

$\nu_L$ destroys a LH neutrino and creates a RH antineutrino.

Its hermitian conjugate, $\overline{\nu_L}$, creates a LH neutrino and destroys a RH antineutrino.

The SU(2)-symmetric charge current vertex (+h.c.) is then proportional to $$W^-_\mu \overline{e_L} \gamma^\mu \nu_L + W^+_\mu \overline{\nu_L } \gamma^\mu e_L ,$$ where $e_L$ destroys a LH electron, and creates a RH positron, as before, and $\overline{e_L}$ creates a LH electron and destroys a RH positron. Since $W_\mu^-$ destroys a $W^-$ and creates a $W^+$, the first term of this charge-conserving (and weak isospin conserving) interaction destroys a LH neutrino and creates a LH electron and a $W^+$, etc. The second term, likewise, destroys a RH antineutrino and creates a RH positron and a $W^-$, etc, for all line reversals of your choice.

• So, your sweeping judgment that the weak vertices “only couple to LH fields” is unwarranted when the chirality-reversing bars sheltering antiparticles are considered. Now, since $\nu_L^c=C \overline{\nu_L}$ does what $\overline{\nu_L}$ does ; and $\overline{\nu_L} ^c= -\nu_L C^{-1}$ does what $\nu_L$ does; and, mutatis mutandis for the electrons, the above weak vertex presents, perversely, as $$W^-_\mu e_L ^c\gamma^\mu \nu_L + W^+_\mu \nu_L ^c\gamma^\mu e_L ,$$ where the evident transposes are understood, so that a c superscript does what an overbar does. Again, the first term destroys a LH neutrino and creates a LH electron and a $W^+$, or else creates a RH antineutrino and destroys a RH positron and a $W^-$; etc for all line reversals of your choice.

• Because $\overline{\nu_L} (\nu_L)^c$ is simply the definition of a Majorana mass, not the Dirac mass term; it preserves Lorentz invariance, but fails to obey the Dirac equation, as all QFT texts detail. It violates lepton number, as it destroys a LH neutrino and a RH antineutrino, or else creates a RH neutrino and a RH antineutrino. It is a source or sink term for lepton umber 2. (A corresponding term for the electron would also violate lepton number by 2, but, crucially electric charge by 2, as well, and of course would fail to be an isosinglet.) Therefore, if you looked at the weak indices, suppressed, you’d see that this term cannot be a weak isosinglet, so it is forbidden. (Except in the unrenormalizable, dimension 5, “Weinberg term”, where each neutrino’s index is saturated to a singlet by the neutral component of a Higgs each, whose v.e.v. sinks into the vacuum taking away one total unit of weak isospin. I gather this is what you have in mind for your next question.)

• Enter sterile EW singlets (indexless) $\nu_R$ and the seesaw. Now the superheavy N will be mostly $\nu_R$ with a tiny contamination $O(m_D/M)$ component of (indexed!) $\nu_L^{~c}$, which, ipso facto, as you are surmising, does couple to the Ws. Nevertheless, this is a minuscule component of order 100GeV/$10^{15}$GeV ~ $10^{-13}$, so, negligible. The major handle on this mixing, instead is through lepton number violation involving the other, the light eigenvector of the seesaw mass matrix, not N; it will enter into the neutrinoless double β decay.

• Can you kindly explain how does $\nu_L$ create a RH antineutrino? All I know is that a Dirac neutrino field $\nu$ will destroy $\nu$-particle and create a $\bar{\nu}$-particle. If you prefer, I can ask it as a separate question. @CosmasZachos – SRS Jun 2 '18 at 16:19
• Yes, this is absolutely a broader, more basic question on the interpretation of Weyl spinors, so a different question. Check your thinking imagining there are no RH neutrinos and LH antineutrinos, and thinking of how the kinetic term of the LH neutrino could possibly survive, and the line reversals involved. – Cosmas Zachos Jun 2 '18 at 16:39

Charge conjugation $(\nu_L)^c$ of left-handed neutrino $\nu_L$ is nominally right-handed, ie. $$P_R (\nu_L)^c = (\nu_L)^c, P_L (\nu_L)^c = 0.$$ However, $(\nu_L)^c$ still walks and quacks in a semi-left-handed way. It carries, mutatis mutandis, left-handed charges. For example, $(\nu_L)^c$ has non-zero hypercharge. On the other hand, the hypercharge of $(\nu_R)^c$ remains zero, even though $(\nu_R)^c$ is nominally left-handed.

• I wouldn't call $(\nu_L)^c$ a "semi-left handed" field or merely "nominally right-handed" at all. The handedness of a field just refers to its chirality, and $(\nu_L)^c$ is right-chiral and hence right-handed, end of story. – knzhou Jun 8 '18 at 16:20
• Unless you're defining "left-handed" to mean "has nonzero hypercharge", which seems to me a rather strange and artificial definition. – knzhou Jun 8 '18 at 16:21
• Folks usually allege that "left-handed" neutrinos participate in electroweak processes, while "right-handed" neutrinos won't. "semi-left-handed" or "semi-right-handed" is used in that sense. – MadMax Jun 8 '18 at 16:31
• People say that, and it's completely correct. A left-helicity neutrino particle participates in an electroweak process, while a right-helicity neutrino particle doesn't. This doesn't say much about the chirality of the fields involved. – knzhou Jun 8 '18 at 16:58

The issue in that the question mixes up particles and fields. For example, the question speaks of "the antiparticle of a field", which doesn't make sense. In essence:

• We observe that only left-helicity neutrinos and right-helicity antineutrinos seem to exist, where a neutrino/antineutrino is defined to have positive/negative lepton number, respectively
• In general, a left-chiral massless $X$ field annihilates left-helicity $X$ particles and creates right-helicity anti-$X$ particles. The same is true swapping the words "left" and "right".
• Hence the particles observed can be accounted for with a left-chiral neutrino field $\nu_L$, which annihilates left-helicity neutrinos and creates right-helicity antineutrinos.
• The charge conjugate field $\nu_L^c$ is a right-chiral antineutrino field, because it annihilates right-helicity antineutrinos and creates left-helicity neutrinos.
• Thus, the exact same set of particles can be described by either a left-chiral neutrino field, or a right-chiral antineutrino field. So there's really no paradox as described in the question.

A lot of confusion in particle physics comes from mixing up particles and fields. In particular, there are two completely different definitions of charge conjugation for particles and for fields. We have used the notion of field charge conjugation, which never yields new particles, but particle charge conjugation can. For a lot more discussion about these subtleties, see my answer here.