When we define the operators $P_{L,R}=\frac{1\mp\gamma_{5}}{2}$, we are able to define right- and left-handed Dirac fields:

$$\psi=(P_{L}\psi+ P_{R}\psi)=:\psi_{L}+\psi_{R}$$

The corresponding Dirac-adjoints are therefore defined through $$\overline{\psi}_{L,R}:=\overline{\psi}P_{L,R}$$

Plugging $\psi=\psi_{L}+\psi_{R}$ and $\overline{\psi}=\overline{\psi}_{L}+\overline{\psi}_{R}$ into the definitions of the Dirac Lagrangian $$\mathcal{L}=i\overline{\psi}\gamma^{\mu}\partial_{\mu}\psi - m\overline{\psi}\psi$$ and using $P_{L}P_{R}=P_{R}P_{L}=0$, one gets

$$\mathcal{L}=i\overline{\psi}_{L}\gamma^{\mu}\partial_{\mu}\psi_{L}+i\overline{\psi}_{R}\gamma^{\mu}\partial_{\mu}\psi_{R} - m(\overline{\psi}_{R}\psi_{L}+\overline{\psi}_{L}\psi_{R})$$

This expression can be found in many QFT books.....

Now to my question...Isn't the mass term zero? Because

$$\overline{\psi}_{R}\psi_{L}=\overline{\psi}\underbrace{P_{R}P_{L}}_{=0}\psi=0$$ and the same for the other term....


1 Answer 1


Your mistake is in saying that $\overline{\psi_{L,R}} = \overline{\psi}P_{L,R}$.

In fact, $$ \overline{\psi_{L,R}} = (P_{L,R}\psi)^† \gamma^0 = \overline{\psi}\gamma^0 P_{L,R}^† \gamma^0$$ $$ = \overline{\psi} P_{R,L}$$

To see why that last line is true we have to expand $P_{L,R} = (I\mp \gamma^5)/2$ and note that $(\gamma^5)^† = \gamma^5$, $(\gamma^0)^2 = I$, and $\{\gamma^0,\gamma^5\} = 0$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.