Consider 2 different Majorana fermions $\Psi_{L}, \Psi_{R}$ (physically, neutrinos). In general case I can write the massive part of lagrangian of these fermions in the form $$ L_{m} = (\bar {\Psi}_{1} \bar {\Psi}_{2})\begin{pmatrix} m_{11} & m_{12} \\ m_{21} & m_{22} \end{pmatrix}\begin{pmatrix} \Psi_{1} \\ \Psi_{2} \end{pmatrix} = \bar {\Psi}^{a}\hat {m}_{ab}\Psi^{b}. \qquad (1) $$ Then let's introduce matrix of some gauge interaction $$ \hat {A}_{\mu}^{ab} = A_{\mu} = \begin{pmatrix} A^{11}_{\mu} & A^{12}_{\mu} \\ A^{21}_{\mu} & A^{22}_{\mu} \end{pmatrix}_{ab}. \qquad (2) $$
and corresponding fermion's charge $g$.
The full lagrangian with interaction and mass term takes the form
$$ L = \bar {\Psi}(i\gamma^{\mu}\partial_{\mu} + g\gamma^{\mu}\hat {A}_{\mu})\Psi - \bar {\Psi}\hat {m}\Psi . $$
I want to represent the lagrangian in terms of states with definite masses. Let's call them $\Psi_{L}, \Psi_{R}$. So by introducing mixing unitary matrix
$$ \begin{pmatrix} \Psi_{L} \\ \Psi_{R} \end{pmatrix} = \begin{pmatrix} cos(\theta ) & -sin (\theta ) \\ sin(\theta ) & cos(\theta ) \end{pmatrix}\begin{pmatrix} \Psi_{1} \\ \Psi_{2} \end{pmatrix} $$ I can rewrite the mass term in a form of $$ L_{m} = (\bar {\Psi}_{L} \bar {\Psi}_{R})\begin{pmatrix} m_{1} & 0 \\ 0 & m_{2} \end{pmatrix}\begin{pmatrix} \Psi_{L} \\ \Psi_{R} \end{pmatrix}. $$ But I can't diagonalize the $A$-matrix in general case in this basis. So the $A$-matrix and $m$-matrix don't commute with each other. So there is the question: what is the physical meaning of this result in the case of Majorana fermions?
