Dirac Lagrangian in terms of left and right handed fields

Question

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....

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 $\{\gamma^0,\gamma^5\} = 0$