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  1. Why is it said that the Dirac mass term conserves the fermion number but the Majorana mass term does not? Can someone explain this mathematically?

  2. Which breakdown of symmetry is responsible for Majorana mass term? Is this also related to breakdown of $SU(2)\times U(1)$ symmetry like Dirac mass?

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Non-conservation of charge in Majorana terms

The Dirac mass term is $m\bar\psi \psi$ where one field-factor $\bar\psi$ is complex conjugated (aside from other transpositions included in the Dirac conjugation) and the other is not.

So one may assign a fermion number $1$ to $\psi$ which means that $\bar\psi$ automatically carries $-1$ and in the product, the lepton numbers add up to $(+1)+(-1)=0$, so it is conserved.

On the other hand, the Majorana mass term is of the form $m\chi\chi$ without a complex conjugation, so if $\chi$ carried a charge $Q$ such as a fermion number, $m\chi\chi$ would carry $2Q$ which would be nonzero and the charge conservation would be violated by this term.

In other words, the Dirac mass term destroys a particle and creates a new one, or destroys/creates a particle-antiparticle pair, or destroys an anti-particle and creates a new one. This conserves the charge. The Majorana mass term may destroy (or create) two particles that are identical (Majorana fermions are identical to their antiparticles) which means that they can't carry any conserved charges except for a binary "charged" conserved modulo 2, in a ${\mathbb Z}_2$ group

Breaking to get Majorana mass terms

The Dirac mass terms in the Standard Model are obtained from the Yukawa couplings $h\bar\psi\psi$ when the Higgs field gets a vev, i.e. after a spontaneous symmetry breaking.

On the other hand, the Majorana mass terms are generated by loops or, more likely, the integration out of heavier degrees of freedom. Most likely, there exist right-handed neutrinos and there is a Dirac mass term $m \nu_L\nu_R$ (a 2-component way to write Dirac mass terms) with a coefficient comparable to the weak scale. This Dirac mass term has the same symmetry-breaking origin as in the case of the electron Dirac field.

However, there's also a symmetry-unrelated Majorana term $M\nu_R \nu_R$ with $M$ comparable to the huge GUT scale. This mass term doesn't break any symmetry – the right-handed neutrino is an uncharged Standard Model singlet – so it must be expected to be a part of the Lagrangian with a generic coefficient. When the heavy degree of freedom $\nu_R$, the right-handed neutrino, is integrated out, we are only left with the well-known left-handed neutrino with its own Majorana term with a very small Majorana mass comparable to $m_\nu\sim m_{H}^2 / m_{GUT}$. I used $m_H$ for the electroweak scale (the Higgs mass). This small mass is smaller than the Higgs mass by the same factor by which the Higgs mass is smaller than the GUT scale, therefore the "seesaw" (going to the opposite direction from the electroweak scale than the GUT scale on the log scale).

There may be other ways to generate the Majorana mass terms for the left-handed neutrinos, too, including effects of sterile neutrinos and other fields. In some cases, one may be forced to go to loops. However, in all cases, it's important to realize that the Majorana mass term ultimately doesn't violate any symmetry or principle. At low energies, only the electromagnetic $U(1)$ is conserved, along with the QCD $SU(3)$, and the neutrino mass term doesn't violate either because the left-handed neutrino is electrically and color-neutral.

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  • $\begingroup$ Dear @Lubos I have a question regarding U(1) charges of Majorana fields and is related to your answer. You said that if the Majorana field $\chi$ carried a charge $Q$, then the Majorana mass would violate that charge by 2Q units. I agree. However, if the field $\psi$ is assigned a charge $Q$ under some U(1), $\psi^c$ would have a charge $-Q$. Now for Majorana fields, $\psi=\psi^c$ which implies $Q=-Q\Rightarrow Q=0$. Then there is no question of violation. Am I wrong? $\endgroup$ – SRS May 17 '17 at 15:21
  • $\begingroup$ Yup, that's a completely equivalent way of showing that the Majorana field has $Q=0$. Note that even your equation actually showed $2Q=0$. So a Majorana field can carry and often does carry a nontrivial charge under some $Z_2$ discrete group (or another discrete group with that subgroup). $\endgroup$ – Luboš Motl May 19 '17 at 4:15
  • $\begingroup$ The OP has asked a separate follow-up question here. $\endgroup$ – tparker Oct 7 '18 at 18:17

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