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  1. The symmetry of neutrino mass matrix $M_\nu$ is often realized as $$G^TM_\nu G=M_\nu$$ where $G$ is an element of the corresponding symmetry group. Is this because the neutrinos are Majorana in nature?

  2. Since a symmetry is always a symmetry of the Lagrangian, isn't it also necessary to impose the same symmetry on the charged lepton mass matrix?

  3. If yes, will be be implemented as $G^\dagger M_lG=M_l$ owing to the Dirac nature of the charged leptons?

EDIT: In the first question, $M_\nu$ could be the effective (Majorana) mass matrix obtained after type-I seesaw, for example.

If $M_\nu$ corresponds to the neutrino mass in Standard model extended only with sterile right-handed neutrinos and nothing else. In that case, $M_\nu$ is Dirac type. In that case, how should a flavour symmetry be implemented to the Lagrangian?

In this reference, if I understand correctly, neutrinos are taken to be Majorana from start and no Dirac contribution is assumed.

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    $\begingroup$ You should state more clearly, what kind of symmetry you are talking about. From the formula you give in question 1, I assume you talk about flavor symmetries among right-handed neutrinos. Is this correct? $\endgroup$ – Neuneck Dec 15 '16 at 13:12
  • $\begingroup$ @Neuneck- I've added the EDIT section to clarify. $\endgroup$ – SRS Dec 15 '16 at 13:37
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The symmetries you consider are flavor symmetries. These mix the three fields that correspond to the different generations or families of Standard Model matter.

Before mass terms are added, there is a $U(3)^5$ flavor symmetry in the Standard Model without right-handed neutrinos. Adding right-handed neutrinos also adds another $U(3)$ symmetry. These are global redefinitions of the fields, taking linear combinations and adding phases in such a way that the kinetic terms are invariant. So each $U(3)$ factor is linked to one of the fundamental Standard Model matter fields, which are

  • the left-handed quark doublet,
  • right-handed up-quarks
  • right-handed down-quarks
  • the left-handed lepton doublet,
  • right-handed electrons

and possibly

  • right-handed neutrinos.

In general, since these are only global symmetries, they need not be respected by the mass terms. For general Yukawa matrices, the $U(3)$ symmetries can then be used to simplify these matrices. For example, in the quark sector we can use the $U(3)^3$ symmetry to diagonalize one matrix through a bi-unitary transformation. Trying to do the same with the other Yukawa matrix would require an overall $U(3)^4$ symmetry, which is not there. We are stuck with one unitary matrix mixing different mass eigenstates in interactions - the CKM matrix.

In the lepton sector, things are similar. We have a global $U(3)^3$ symmetry that we can use to simplify the mass matrices. As an example, we can use the freedom to re-define the right-handed electrons and left-handed lepton doublets in oder to diagonalize the charged lepton mass matrix right away. This leaves us with only the $U(3)$ symmetry of the right-handed neutrinos to simplify the neutrino masses overall.

If the neutrino masses arise from just a Yukawa matrix with tiny entries (which is a possibility), we can again produce a unitary matrix governing neutrino mixing, the PMNS matrix.

If there is a see-saw mechanism, there are actually two matrices that go into neutrino masses: The Yukawa matrix, which can be a general complex matrix and the Majorana mass matrix which has to be symmetric. $$ \mathcal L_m \supset v_1 \nu_i \nu^c_j Y_\nu^{ij} + \nu^c_i \nu^c_j M_\nu^{ij}$$ We can always diagonalize the Majorana mass matrix with a (complex) orthogonal transformation, but this already eats up a lot of our freedom to simplify the Yukwa matrix.

Threfore, one sometimes further restricts the shape of the mass matrix $M_\nu$ or the Yukawa matrix $Y_\nu$. If $$ G^T M_\nu G = M_\nu$$ for a group of transformations $G$, this means that we can diagonalize $M_\nu$ and still have all the transformations $G$ left to simplify the Yukawa matrix. If we further impose some symmetries on $Y_\nu$, we can even achieve that the mixing matrix in the neutrino sector is more constrained than just an arbitrary unitary matrix. As an example, one can constrain $M_\nu$ and $Y_\nu$ such that the mixing matrix has the tri-bi-maximal shape.

It can be a bit confusing to disentangle all the transformations involved. It helps to first understand that there is a $U(3)$ symmetry for each fundamental fermion (that is massless before electro-weak symmetry breaking). These symmetries can then be used to simplify the mass matrices without introducing any family mixing. They are not enough to completely diagonalize all matrices that appear, therefore we are left with a unitary CKM matrix.

If we do not allow for the most general mass matrices, one can get a more specific, a more simple mixing matrix. This is a model assumption, though it is often motivated by some high-energy physics principle.

To re-state it in another way: Actually, flavor symmetries are the unitary transformations one can perform without altering the kinetic terms of the fermions. They can be used to simplify the mass part of the lagrangean. However, the term "flavor symmetry" is sometimes abused to atually refer to a symmetry imposed on the mass terms directly, before performing rotations in field space.

EDIT for clarification: The "flavor symmetry" $$G^T M_\nu G = M_\nu$$ is imposed by hand and is a priori independent of the freedom to rotate the fields in flavor space. The relation comes into play once you want to actually construct the mass eigenstates. It is only in that step that $G$ becomes a residual freedom to transform the mass eigenstates among each other, without disturbing the diagonal shape of the Majorana mass matrix.

You could just as well impose some symmetry on the Yukawa matrix. The proper generalization of the equation above would be $$ G^{-1} Y G = Y. $$ The restriction to the transpose $G^T$ in place of $G^{-1}$ comes from the symmetry of the Majorana mass matrix, making it diagonalizable with orthogonal matrices, for which $G^T = G^{-1}$.

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    $\begingroup$ @Neuneck- You said that without right-handed neutrinos, there is one $U(3)$ symmetry and probably you meant $l_{iL}\rightarrow U_{ij}l_{jL}$ where $l_{iL}$ is the lepton doublet of $i^{th}$ generation. But don't you think that there is another $U(3)$ symmetry given by $e_{iR}\rightarrow U^\prime_{ij}e_{jR}$ without introducing right-handed neutrinos? For SM is augmented with RH neitrinos shouldn't there be a $U(3)\times U^\prime(3)\times U^{\prime\prime}(3)$ global symmetry in the massless limit? $\endgroup$ – SRS Dec 15 '16 at 15:14
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    $\begingroup$ Indeed, there is a $U(3)^3$ symmetry in the massless case. What I am saying is that there is one $U(3)$ remaining after you use two to make the charged lepton masses diagonal. $\endgroup$ – Neuneck Dec 15 '16 at 15:33
  • $\begingroup$ @Neuneck- If I understand it correct, then the basis in which the charged lepton mass matrix is made diagonal, the Dirac neutrino mass matrix cannot be diagonal. However, one can diagonalize the RH Majorana mass matrix by rotating RH neutrinos by the residual $U(3)$. So simultaneously the charged lepton mass matrix and the RH Majorana mass matrix can be made diagonal. Is that correct? $\endgroup$ – SRS Dec 15 '16 at 15:51
  • $\begingroup$ @SRS Yes, that statement should hold for general Yukawa matrices and RH Majorana mass matrices. $\endgroup$ – Neuneck Dec 15 '16 at 15:52
  • $\begingroup$ Since, we have used up all freedom/symmetries in diagonalizing the RH Majorana mass matrix and charged lepton mass matrix, where does the symmetry $G^TM_\nu G=M_\nu$ come from? Moreover, if $M_\nu$ is Dirac-type, why should any symmetry be implemented as $G^TM_\nu G=M_\nu$? How can it help because a Dirac mass matrix is diagonalized by a bi-unitary transformation. I think, $G^TM_\nu G=M_\nu$ is useful when $M_\nu$ is the (effective) Majorana mass matrix after seesaw. $\endgroup$ – SRS Dec 15 '16 at 16:06

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