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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – knzhou
    Jun 23, 2019 at 10:06

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

  • $\begingroup$ Although it is not asked by the OP, can you please also define the Peccei-Quinn transformation in its entirety/full context. Otherwise, it's just a normal axial transformation. $\endgroup$
    – MadMax
    Jun 25, 2019 at 20:39
  • $\begingroup$ @MadMax - The entirety/full context of the Peccei-Quinn transformation appears only in Problem 94.2 in Srednicki's book, which is basically described in the question. You may look at Problem 94.2 in Srednicki's book if you wish. $\endgroup$
    – Shen
    Jun 26, 2019 at 13:14

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