My book gives this Lagrangian:

$$ L = -|\partial \phi|^2 -V(\phi) -\bar \psi_L \not \partial \psi_L -\bar \psi_R \not \partial \psi_R -g(\phi \bar \psi_L \psi_R + \phi^* \bar \psi_R \psi_L) $$

It's supposed to have $U(1)$ symmetry, with $\phi$ having charge $+1$, $\psi_L$ having $+1/2$ and $\psi_R$ having $-1/2$, which is the reason for those cubic terms.

However, the book also says that

$$ \psi_{L/R} = \frac{1}{2}(1\pm\gamma_5)\psi $$

so I deduce that

$$ \bar \psi_L \psi_R = \frac{1}{4} \bar \psi (1+\gamma_5)(1-\gamma_5)\psi = \frac{1}{4}\bar \psi (1-\gamma_5^2)\psi = 0 $$

because $\gamma_5^\dagger = \gamma_5$ and $\gamma_5^2 = 1$.

It's can't be zero otherwise the interaction terms given by the book are wrong. What happened?


$$ \bar \psi_L = \psi_L^\dagger \gamma_0 = \frac{1}{2}\psi^\dagger (1+\gamma_5)\gamma_0 = \frac{1}{2}\psi^\dagger \gamma_0 \gamma^0 (1+\gamma_5)\gamma_0 = \frac{1}{2}\bar \psi (1+\gamma_5) $$

using $\gamma_0 \gamma^0 = 1$ and $\gamma^0 \gamma_5 \gamma_0=\gamma_5$.

  • 2
    $\begingroup$ I think you need to flip the sign of the $\gamma_5$ when you move it past a $\gamma_0$ $\endgroup$
    – Ihle
    Commented May 24, 2015 at 19:51
  • $\begingroup$ @Ihle I edited the post to explain how I calculated the Dirac adjoint. $\endgroup$
    – Lelota
    Commented May 24, 2015 at 20:00
  • $\begingroup$ @Lelota look again at what you typed. It is clear that you imagined one more $\gamma^0$ that wasn't there. $\endgroup$
    – Danu
    Commented May 25, 2015 at 9:26

3 Answers 3


@Ihle is right. $\gamma^5\gamma^0=-\gamma^0\gamma^5$. You don't need to insert $\gamma^0\gamma^0$.


Your second last step is incorrect: $\bar{\psi}_L = \frac{1}{2}\psi^\dagger\gamma_0\gamma^0(1+\gamma_5)\gamma_0 = \frac{1}{2}\bar{\psi}(1-\gamma_5)$, since $\{\gamma_5,\gamma_\mu\} = 0 \implies \gamma_5\gamma_0 = -\gamma_0\gamma_5$.


$$\overline{\psi_L}\psi_R = \overline{\frac{1}{2}(1-\gamma^5) \psi} \frac{1}{2}(1+\gamma^5)\psi=\left(\frac{1}{2}(1-\gamma^5) \psi\right)^\dagger \gamma^0 \frac{1}{2}(1+\gamma^5)\psi = \psi^\dagger\frac{1}{2}(1-\gamma^5)\gamma^0\frac{1}{2}(1+\gamma^5)\psi$$ Since $\gamma^5\gamma^0 = -\gamma^0 \gamma^5$, it yields: $$\overline{\psi_L}\psi_R = \psi^\dagger \gamma^0 \frac{1}{2}(1+\gamma^5) \frac{1}{2}(1+\gamma^5)\psi = \overline{\psi}\frac{1}{2}(1+\gamma^5)\psi \ne 0 $$


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