Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

I was reading about the seesaw mechanism in my Lecture notes and got a technical question. See for example page 13.

There it says than in the limit $M_R >> M_D$ and $M_L=0$ then the second mass is $m_2=M_D^2/M_R$. But when I apply the previous limit to the solution $$m_{1/2}=(M_L+M_R)/2 \pm \sqrt{(M_L-M_R)^2/4+M_D^2}$$ I get $m_2=-M_D^2/M_R$. What am I doing wrong here?

share|cite|improve this question
I get the minus sign you are talking about, and so does wikipedia I am thinking about if this matters or not... – DJBunk Apr 12 '13 at 23:05
Scrednicki pg556 states that you can just take $\nu \rightarrow i\nu $ to get ride of the phase. This clearly takes care of the sign of the mass term, and does't change the kinetic term since that will pick up a factor of $(+i)(-i) = +1$. Off hand its not clear to me this doesn't affect any other terms in the Lagrangian or doesn't create any anomaly issues...but apparently its ok. – DJBunk Apr 12 '13 at 23:11
@DJBunk I see. A similar argument is given in my lecture notes in order to rewrite the mass matrix with real Dirac masses, essentially writing $M_D=|M_D|e^{i\phi}$ and then absorbing the phases in the fields. It makes a remark about this procedure when including several families, but I don't understand it due to poor writing. – Barefeg Apr 12 '13 at 23:39
@DJBunk Isn't $\nu \rightarrow i\nu$ just a field redefinition? – twistor59 Apr 13 '13 at 7:25
@twistor59 - I was a little worried that the redefinition might be anomalous. Maybe its not, or maybe it just doesn't matter, or maybe its not anomalous because the gauge group is SU(2). – DJBunk Apr 13 '13 at 13:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.