Rearranging neutrino expression in see-saw maths

  • #1
venus_in_furs
19
1
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
[itex] \bar{\nu_R} \nu_R^c = \bar{\nu_R^c} \nu_R [/itex]

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

Thank you in advance!
 

Answers and Replies

  • #2
Orodruin
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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
venus_in_furs
19
1
ahhhhh thankyou, I think I have it .....

[itex] \overline{ \nu_R^c} \nu_R + h.c. = \overline{ \nu_R} \nu_R^c + h.c. [/itex]

[itex] (1) + h.c. = (2) + h.c. [/itex]

[itex] (1)^{\dagger} +h.c. = (2) + h.c. [/itex]


Proof

[itex] (1)^{\dagger} = ( \overline{ \nu_R^c} \nu_R )^{\dagger} [/itex]

[itex] = ( - \nu_R^T C ^{\dagger} \nu_R )^{\dagger} [/itex]

[itex] = \nu_R^{\dagger} C \nu_R^{\ast} [/itex]

[itex] = \nu_R^{\dagger}\gamma^0 C \gamma^0 \nu_R^{\ast} [/itex]

[itex] = \overline{\nu_R} C \gamma^0 \nu_R^{\ast} [/itex]

[itex] = \overline{\nu_R} C \overline{\nu_R^{T} } [/itex]

[itex] = \overline{\nu_R} \nu_R^{c} = (2) [/itex]
 

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