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I am having trouble in diagonalizing analytically the mass matrix associated to the inverse seesaw mechanism with lagrangian: $L= -y_1\bar{L}\tilde{H}N_{R_1} - -y_2\bar{L}\tilde{H}N_{R_2} + N_{R_1}^TM_{R_1}C^\dagger N_{R_1} + N_{R_2}^TM_{R_2}C^\dagger N_{R_2} +\Lambda \bar{N_{R_1}}N_{R_2}^C + h.c. $

Where the mass matrix is given by:

\begin{pmatrix} 0 & y_1 v_H/ \sqrt{2} & y_2 v_H/ \sqrt{2}\\ y_1v_H/\sqrt{2} & M_{R_1} & \Lambda \\ y_2v_H/\sqrt{2} & \Lambda & M_{R_2} \end{pmatrix}

If we take a U(1) where L has charge 1, $N_{R_1}$ has charge 1 and $N_{R_1}$ has charge -1, we will obtain that $y_2v_H/\sqrt{2}, M_{R_1}$ and $M_{R_2}$ have to be small. The resulting eigenvalue for the neutrino mass should be:

$m_{\nu} = \frac{v_H^2}{2(\Lambda^2 - \mu \mu')}[\mu y_1^2 + y_2^2 \mu' -2\Lambda y_2y_1] $ where $\mu$ and $\mu'$ have been used insted of $M_{R_1}$ and $M_{R_2}$ respectively.

I have tried to obtain this result but I can't, even by using approximation for the small values I don't recover the denominator in any way. Can someone tell me what could I be doing wrong in the procedure and if there's some way to get this result?

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  • $\begingroup$ You have cloyed your expression with too many superfluous parameters, and omitted crucial information on their relative sizes, providing irrelevant information instead. You are asking about the smallest eigenvalue of a 3x3 matrix after all. Introduce the μs and parameters a,b for the effective Yukawa couplings, and describe what your difficulty is. Large and small components of your matrix should jump at you and the reader. $\endgroup$
    – Cosmas Zachos
    Commented Jan 5, 2023 at 15:05
  • $\begingroup$ I don't have a reference, these are just notes taken in class. It should have been just an introduction on the topic so no computations were required and therefore it wasn't treated in detail, but I have to do this diagonalization for other reasons $\endgroup$
    – shescientist
    Commented Jan 5, 2023 at 15:55
  • $\begingroup$ Mathematica should provide the eigenvalues... As a warmup, try $y_2=0$. give your variables simple names. $\endgroup$
    – Cosmas Zachos
    Commented Jan 5, 2023 at 16:26
  • $\begingroup$ Yes, but I have to do both the analytic and numerical computation and then compare the two for an assignment $\endgroup$
    – shescientist
    Commented Jan 5, 2023 at 16:29
  • $\begingroup$ Mathematica would provide the analytic one, and confirm it. $\endgroup$
    – Cosmas Zachos
    Commented Jan 5, 2023 at 16:30

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