# Is the $U(1)_A$ axial vector current even under charge conjugation?

The axial current of a Dirac spinor is given by $$j_A^\mu = \bar{\psi} \gamma^5 \gamma^\mu \psi$$. In this book, in the paragraph under equation (2.18) it is stated that the current is even under charge conjugation. I am trying to show this. Charge conjugation of a spinor $$\psi$$ is defined as $$\psi^{(c)} = C \psi^*$$, where $$C$$ is the unitary charge conjugation matrix which satisfies $$C^\dagger \gamma^\mu C = -(\gamma^\mu)^*$$ for all gamma matrices and $$\bar{\psi}=\psi^\dagger \gamma^0$$.

We have

$$(j^\mu_A)^{(c)} = \overline{\psi^{(c)}}\gamma^5 \gamma^\mu \psi^{(c)} \\ =(C\psi^*)^\dagger \gamma^0 \gamma^5\gamma^\mu C \psi^* \\ = (\psi^\dagger)^* (C^\dagger\gamma^0 C)(C^\dagger \gamma^5 C)(C^\dagger \gamma^\mu C) \psi^* \\ = - (\psi^\dagger)^*(\gamma^0)^*(\gamma^5)^* (\gamma^\mu)^*\psi^* \\ =(\psi^\dagger \gamma^0\gamma^5 \gamma^\mu \psi)^* \\ = (\bar{\psi}\gamma^5 \gamma^\mu \psi)^*$$

where going from the 4th to 5th line, pulling out the complex conjugate picks up a minus sign as the components of the spinors are anti-commuting Grassman numbers. Now that book is telling me that the current is even under charge conjugation so I would expect $$\bar{\psi}\gamma^5 \gamma^\mu \psi$$ is real, but I can't show this.

Any help would be greatly appreciated.

You're almost there. We know that the components of the current are complex numbers, and so we can rewrite your second last line as

$$$$(\psi^\dagger \gamma^0 \gamma^5 \gamma^\mu \psi)^* = (\psi^\dagger \gamma^0 \gamma^5 \gamma^\mu \psi)^\dagger.$$$$

We then perform the Hermitian conjugate, and use following the properties of the gamma matrices

\begin{align} (\gamma^0)^\dagger &= \gamma^0, \\ (\gamma^5)^\dagger &= \gamma^5, \\ (\gamma^\mu)^\dagger &= \gamma^0 \gamma^\mu \gamma^0, \\ \{\gamma^5,\gamma^\mu\} &= 0, \\ (\gamma^0)^2 &= \mathbb{I}_{4}, \end{align}

where $$\mathbb{I}_{4}$$ is the identity to proceed as follows:

$$$$\begin{split} (\psi^\dagger \gamma^0 \gamma^5 \gamma^\mu \psi)^\dagger &= \psi^\dagger (\gamma^\mu)^\dagger (\gamma^5)^\dagger (\gamma^0)^\dagger \psi \\ &= \psi^\dagger \gamma^0 \gamma^\mu \gamma^0 \gamma^5 \gamma^0 \psi\\ &= -\bar{\psi} \gamma^\mu \gamma^5 (\gamma^0)^2 \psi \\ &= \bar{\psi} \gamma^5 \gamma^\mu \psi, \end{split}$$$$

as required.