Problem evaluating $C^{-1}M^\dagger C$

How can I show the following?

$$\overline{\psi_L}M^\dagger (\psi_L)^c=\overline{\psi_L}CM^\dagger\overline{\psi_L}^T$$

where $\psi^c=C\overline{\psi}^T$ and $C=i\gamma^2\gamma^0=-C^T=-C^\dagger=-C^{-1}$. In particular, what is $C^{-1}M^\dagger C$? I'm stuck at this point. If I can show, $$C^{-1}M^\dagger C=M^\dagger$$, the job is done.

see equation 24 of this review

• is $M$ a matrix? if so, in what space? in what space is $C$ a matrix? won't their commutation be trivial? or did i miss something? Mar 4, 2014 at 15:16
• @innisfree, you are right. $M$ is a matrix on flavor space while $C$ only acts on the spinors. Mar 4, 2014 at 15:33
• @JeffDror yes that is exactly what I was implying :) Mar 4, 2014 at 15:41
• So what is the resolution? It is true that M is a matrix on flavor space while C only acts on the spinors, but $\psi_L$ is a vector in the flavor space (in this case), then how do you define $\psi_L^c$?
– SRS
Mar 4, 2014 at 16:31
• Are you ultimately trying to show that $M$ must be symmetric? Mar 4, 2014 at 17:39

As mentioned in the comments, $C$ and $M$ are matrices in different spaces. Explicitly showing the flavor space matrices, in the lepton sector perhaps, we have, \begin{align} \overline{\psi} _L M ^\dagger \left( \psi _L \right) ^c & = \overline{\psi} _L M ^\dagger C\bar{ \psi _L} ^T \\ & = \left( \begin{array}{ccc}\overline{ \psi _{e, L} } & \overline{ \psi _{ \mu , L } } & \overline{ \psi _{ \tau , L}}\end{array} \right) \left( \begin{array}{ccc} M _{1,1} & M _{1,2} & M _{1,3} \\ M _{2,1} & M _{2,2} & M _{2,3} \\ M _{3,1} & M _{3,2} & M _{3,3} \end{array} \right) ^\dagger\left( \begin{array}{ccc} C & 0 & 0 \\ 0 & C & 0 \\ 0 & 0 & C \end{array} \right) \left( \begin{array}{c} \gamma _0 \overline{ \psi _{e, L} }^T \\ \gamma _0 \overline{ \psi _{\mu ,L} } ^T \\ \gamma _0 \overline{ \psi _{\tau, L} } ^T \end{array} \right) \end{align}
• The Dirac mass term is written as $-\mathcal{L}_D=\bar\nu_R m_D\nu_L + \bar\nu_L m_D^\dagger \nu_R$. How can I get the second term by taking the hermitian conjugate of the first term? In case of single fermionic species $m_D$ is number. Right? And the proof is trivial. What about the proof when we have n-fermionic species.
• Its also trivial in the case of $n$ species because $\gamma_0$ commutes with $m_D$ (they act on different spaces) Mar 10, 2014 at 8:44