What is Peccei-Quinn transformation?

Question

In Srednicki's textbook Quantum Field Theory, Problem 94.2 considers a massless quark represented by a pair of Weyl fermions $\chi$ and $\xi$. Part a) asks us to show that the lagrangian is invariant under a Peccei-Quinn transformation $\chi \rightarrow e^{i\alpha} \chi$, $\xi \rightarrow e^{i\alpha} \xi$, $\Phi \rightarrow e^{-2i\alpha}\Phi$, where $\Phi$ is a complex scalar field that interacts with $\chi$ and $\xi$ to form the Yukawa interaction $\mathcal{L}_{Yuk} = y \Phi \chi\xi + h.c.$ ... From this I guess that the Peccei-Quinn transformation is a transformation in which the two fermions transform by the same phase factor. However, the answer states that,

If we define a Dirac field $\Psi = \left( \begin{array}{cols} \chi \\ \xi^{\dagger} \end{array} \right)$, then the PQ transformation is $\Psi \rightarrow e^{-i\alpha \gamma_{5}} \Psi$, ...

However, if $\Psi \rightarrow e^{-i\alpha \gamma_{5}} \Psi$, then $\chi \rightarrow e^{-i\alpha\gamma_{5}} \chi$, $\xi \rightarrow e^{i\alpha\gamma_{5}} \xi$; so $\chi$ and $\xi$ do not transform by the same phase factor. I am puzzled. What is Peccei-Quinn transformation?

Answer 1

Using chiral representation \begin{equation} \gamma_{5} = \left( \begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array} \right), \end{equation} we have \begin{equation} e^{-i\alpha\gamma_{5}} = \cos\alpha - i\gamma_{5}\sin\alpha = \left( \begin{array}{cc} \cos\alpha +i\sin\alpha & 0 \\ 0 & \cos\alpha - i\sin\alpha \end{array} \right) = \left( \begin{array}{cc} e^{i\alpha} & 0 \\ 0 & e^{-i\alpha} \end{array} \right). \end{equation} Then \begin{equation} \Psi \rightarrow e^{-i\alpha\gamma_{5}}\Psi = \left( \begin{array}{cc} e^{i\alpha} & 0 \\ 0 & e^{-i\alpha} \end{array} \right)\left( \begin{array}{c} \chi \\ \xi^{\dagger} \end{array} \right) = \left( \begin{array}{c} e^{i\alpha}\chi \\ e^{-i\alpha}\xi^{\dagger} \end{array} \right), \tag{1} \end{equation} \begin{equation} \overline\Psi \rightarrow \overline{\Psi}e^{-i\alpha\gamma_{5}} = \left( \begin{array}{cc}\xi & \chi^{\dagger} \end{array} \right) \left( \begin{array}{cc} e^{i\alpha} & 0 \\ 0 & e^{-i\alpha} \end{array} \right) = \left( \begin{array}{cc} e^{i\alpha}\xi & e^{-i\alpha}\chi^{\dagger} \end{array} \right). \tag{2} \end{equation} So, the transformations $(1)$ and $(2)$ are consistent with the Peccei-Quinn transformation $\chi \rightarrow e^{i\alpha}\chi$, $\xi \rightarrow e^{i\alpha}\xi$.