# Dirac Lagrangian in terms of left and right handed fields

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....

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