# Rearranging neutrino expression in see-saw maths

1. Dec 14, 2014

### venus_in_furs

Hello

I'm working through the see-saw mechanism.
This is currently from the one-generation section.
(I haven't got to the three generation workings yet... although i'm guessing I'll come across something very similar in the three generation model)

Could someone please tell me if it is correct or not that
$\bar{\nu_R} \nu_R^c = \bar{\nu_R^c} \nu_R$

and if it is, could I please have a hint as to why / how to prove this.

2. Dec 14, 2014

### Orodruin

Staff Emeritus
It is not true, the fields on the right are not conjugated while those on the left are. You might want to add the hermitian conjugate on both sides.

3. Dec 14, 2014

### venus_in_furs

ahhhhh thankyou, I think I have it .....

$\overline{ \nu_R^c} \nu_R + h.c. = \overline{ \nu_R} \nu_R^c + h.c.$

$(1) + h.c. = (2) + h.c.$

$(1)^{\dagger} +h.c. = (2) + h.c.$

Proof

$(1)^{\dagger} = ( \overline{ \nu_R^c} \nu_R )^{\dagger}$

$= ( - \nu_R^T C ^{\dagger} \nu_R )^{\dagger}$

$= \nu_R^{\dagger} C \nu_R^{\ast}$

$= \nu_R^{\dagger}\gamma^0 C \gamma^0 \nu_R^{\ast}$

$= \overline{\nu_R} C \gamma^0 \nu_R^{\ast}$

$= \overline{\nu_R} C \overline{\nu_R^{T} }$

$= \overline{\nu_R} \nu_R^{c} = (2)$