Recent content by whitewolf91

My reasoning: [(\alpha\cdot \tau)\otimes I, (\epsilon\cdot \tau)\otimes\gamma_5]= =[\alpha\cdot\tau,\epsilon\cdot\tau]\otimes\gamma_5= =\alpha_a\epsilon_b[\tau_a,\tau_b]\otimes\gamma_5= =i\alpha_a\epsilon_b\epsilon_{abc}\tau_c\otimes\gamma_5= =i(\alpha \times...

@The_Duck: thank you, that solves the problem. @samalkhaiat: I follow you until the very end, when you imply that: e^{i(\alpha+\epsilon \gamma_5)\,\cdot\, \tau/2}=e^{i\alpha\,\cdot\, \tau/2}e^{i\epsilon \gamma_5\,\cdot\, \tau/2} Isn't this false since the exponents are non commuting?

An SU(N) matrix is not necessarily hermitian, so it is not guaranteed that it can be diagonalized. If it were hermitian however then P would be unitary and it would be automatic that the square root is in SU(N).

I want to show that the group SU_L(N)\times SU_R(N) is the same as SU_V(N)\times SU_A(N) - i.e. that it is possible to rewrite the transformation: \begin{cases} \psi_L \to \psi'_L=V_L\,\psi_L\\ \psi_R \to \psi'_R=V_R\,\psi_R \end{cases} , where V_L and V_R are N\times N SU(N)...