Sometimes the general seesaw matrix:
$$\begin{pmatrix} M_L & M_1\\ M_2 & M_R \end{pmatrix}$$
and its just said that in order to get renormalizable interactions one must impose the condition $M_L=0$. Why is this so?
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$\begingroup$ @CosmasZachos Yes, indeed it is something I have studied myself in the past during my Master Thesis degree, but I found myself thinking on the 'why' is it so after all recently studying seesaws and the efect on neutrino oscillations of ultra-light dark fields. $\endgroup$– riemanniumCommented Feb 4, 2023 at 20:00
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Your question is virtually answered in this one.
The SU(2)×U(1) gauge-invariant fermion bilinears corresponding to each term of the matrix have their origins in the terms of dimension [d], $$ M_L: ~~~~~ \frac{1}{v}\left( L ^T i \sigma _2 \cdot \tilde\phi^* ~~\tilde \phi \cdot L^c \right) \qquad [5],\\ M_1: \qquad\qquad \qquad y_1 \overline{L}\cdot \tilde{\phi}~ R \qquad [4],\\ M_2: \qquad\qquad \qquad y_2 \overline{R}~ \tilde{\phi}^* \cdot L \qquad [4],\\ M_R: \qquad \qquad M R ^T i \sigma _2 C R \qquad [3]. $$
All fermion bilinears contribute dimension [3], and each Higgs, providing the v.e.v., contributes [1]. The non-diagonal ones are SM-like Dirac masses. The $M_R$ is a huge non-SM lepton number (doubly)violating Majorana one, and
- the $M_L$ is the famous dimension [5], so nonrenormalizable, Weinberg operator. It too is a Majorana mass violating lepton number by 2.
answered Feb 4, 2023 at 20:02
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$\begingroup$ So, it is due to be a non-renormalizable GUT-like term outside the SM? Can't it be non-zero in theories BSM? $\endgroup$ Commented Feb 4, 2023 at 20:06
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$\begingroup$ Sure. It has the d.o.f. and gauge symmetries of the SM, but people were smitten by the SM because it was renormalizable, when it emerged. Something happening at scales far above v , so, BSM or GUT-like, could produces it as a "low-energy" effective lagrangian one . $\endgroup$ Commented Feb 4, 2023 at 20:10