# What is Peccei-Quinn transformation?

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?

• Your expressions don't make sense because $\gamma_5$ is a $4 \times 4$ matrix, while $\chi$ and $\xi$ only have two components. To see what's going on, just expand explicitly in components. Jun 23, 2019 at 10:06

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 $$\chi \rightarrow e^{i\alpha}\chi$$, $$\xi \rightarrow e^{i\alpha}\xi$$.