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Rearranging neutrino expression in see-saw maths

  1. Dec 14, 2014 #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!
     
  2. jcsd
  3. Dec 14, 2014 #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.
     
  4. Dec 14, 2014 #3
    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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