Maths - writing neutrino states in different forms

[Soph_the_Oaf]

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

 

[Orodruin]

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.

 

[Soph_the_Oaf]

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

 

[Orodruin]

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:

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

$- ( \nu_R^{\dagger} \gamma^0 \nu_L )^T$

It is fine with regards to the matrix structure, but there may be a sign hidden here somewhere.

 

[Soph_the_Oaf]

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.

 

[Orodruin]

Neutrinos are fermions so the field values are Grassman numbers.

 

[Soph_the_Oaf]

Ahhh, of course, how obvious! Sorry, it's been a while since I've done any of this. Thank you