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Maths - writing neutrino states in different forms

  1. Nov 3, 2014 #1
    Hello

    I'm trying to work through a see-saw model derivation and I've become a bit stuck. I've tried lots of sources but the difference in conventions doesn't fill me with confidence when combining these sources.

    I need to get from

    ## \overline{ \nu_L^c } \nu_R^c + h.c ##

    to

    ## \overline{ \nu_R } \nu_L + h.c. ##


    I know the +h.c. allows me to take the h.c. at any point.
    And I know the identities for charge conjugation:

    ## \nu_L^c = C \overline{ \nu_L }^T ##

    ## \overline{ \nu_L^c } = - \nu_L^T C^\dagger ##

    but I still can't work it out!


    Either this is possible, in which case i'd love someone to give me a hint/identity, or they are not equal and I have made a mistake somewhere. Eitherway any input would much appreciated.

    Thanks
     
    Last edited by a moderator: Nov 4, 2014
  2. jcsd
  3. Nov 4, 2014 #2

    Orodruin

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    You should not have to take the hermitian conjugate. In both expressions, you have the conjugate of ##\nu_R## and you have ##\nu_L## without conjugate. I suggest transposing the gamma matrix structure and using that neutrino fields are fermionic.
     
  4. Nov 4, 2014 #3
    Hello

    Thanks for the reply and sorry to bother you again, but I still can't get there.

    I'll state the relations I have been using incase I have made a mistake or muddled conventions up.
    I am assuming all these formulae are the same for LH and RM so please correct me if that is not true for any.

    1. ## \overline{ \nu_L } = \nu_L^{\dagger} \gamma^0 ##

    2. ## \nu_L^c = C \overline{ \nu_L }^T ## which implies (checked in a book) ## \overline{ \nu_L^c } = - \nu_L^T C^{\dagger} ##

    3. ## C C^{\dagger} = 1 ##



    So I start with
    ## \overline{ \nu_L^c } \nu_R^c ## and use (2) to get

    ## \overline{ \nu_L^c } C \overline{ \nu_R }^T ## then use (1) to get

    ## \overline{ \nu_L^c } C ( \nu_R^{\dagger} \gamma^0 )^T ## and use (2) to get

    ## - \nu_L^T C^{\dagger} C ( \nu_R^{\dagger} \gamma^0 )^T ## and use (3)

    ## - \nu_L^T ( \nu_R^{\dagger} \gamma^0 )^T ## and rearrange

    ## - ( \nu_R^{\dagger} \gamma^0 \nu_L )^T ## and use (1) to get

    ## - (\overline{\nu_R} \nu_L )^T ## ... but what I am trying to get to is ## \overline{ \nu_R} \nu_L ## and I'm not sure if this is even possible

    Any comments would be much appreciated.

    Thanks
     
  5. Nov 4, 2014 #4

    Orodruin

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    You are almost there. The point is that the transpose you have is simply the transpose of a 1x1 matrix. I would suggest you think a bit extra about the following step:
    It is fine with regards to the matrix structure, but there may be a sign hidden here somewhere.
     
  6. Nov 4, 2014 #5
    Thanks for the reply. I understand the transpose of a 1D matrix part... but I really can't see the minus.

    ## - \nu_L^T ( \nu_R^{\dagger} \gamma^0 )^T ## using (1)

    ## = -\nu_L^T ( \overline{ \nu_R } )^T ##

    ## = - ( \overline{ \nu_R } \nu_L )^T ##

    which, as you pointed out ## = - \overline{ \nu_R } \nu_L ##

    Am I missing something to do with the gamma matrix? ## \gamma^0 ## is hermitian, right? and it is real ... so ## (\gamma^{0} )^T = \gamma^0 ## . Or is this where my understanding breaks down?

    Thanks again.
     
  7. Nov 4, 2014 #6

    Orodruin

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    Neutrinos are fermions so the field values are Grassman numbers.
     
  8. Nov 4, 2014 #7
    Ahhh, of course, how obvious! Sorry, it's been a while since I've done any of this. Thank you
     
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