Casimir's Trick in QED

In Griffith's Introduction into Particle Physics (p. 251, eq. 7.125) we derive Casimir's trick

$$\sum_{s_1,s_2}[\bar{v}(s_1,p_1)\Gamma_1 v(s_2,p_2)][\bar{v}(s_a,p_a)\Gamma_2v(s_b,p_b)]^* = \text{Tr}[\Gamma_1(\gamma_\mu p_b^\mu-m_bc)\gamma^0\Gamma_2^\dagger\gamma^0(\gamma_\nu p_a^\nu-m_ac)]$$

wherein $\bar{v}=v^\dagger\gamma^0$ is the Dirac adjoint spinor and $\Gamma\in\mathbb{C}^{4\times4}$. The theorem is useful if we want to evaluate the average of spin configurations.

I have some questions with regard to the derivation used in the book.

1. I read $$\bar{v} \Gamma v=\langle v\gamma^0,\Gamma v\rangle=\langle v,\gamma^0\Gamma v\rangle$$ as the complex scalar product. However if I do so, it does not make sense to use the complex conjugate $*$ on the second square bracket in Casimir's trick equation as the scalar product is real anyway. Am I wrong to interpret $\bar{v}\Gamma v$ as scalar product?

2. In the derivation Griffith uses $\bar{v}=v\gamma^0$ but usually the dirac adjoint is defined as $\bar{v}=\gamma^0 v^\dagger$. Do $\gamma^0,v^\dagger$ commute?

• You need to clean up your notation a bit. None of the expressions in your second question make sense, because $\bar{v}$ is a row vector, $v$ is a column vector, and $\gamma^0$ is a matrix. – knzhou Feb 24 '18 at 15:25
• @knzhou I agree that the notation is problematic (and this is also one reason for this question), however that is the notation used in the referenced book (and in our lecture). – bodokaiser Feb 24 '18 at 15:31
• Correct definition of Dirac adjoint is $\bar \psi = \psi^{\dagger}\gamma^0$. – Blazej Feb 24 '18 at 22:40