I'm having some hard time trying to see why the left-handed lagrangian for fermions $\psi$,

$$\mathcal{L} := G\overline{\psi}_{1L}\gamma^\mu\psi_{2L}\overline{\psi}_{3L}\gamma_\mu\psi_{4L}$$

is not invariant under parity transformation. Let $\mathcal{P}$ be the parity operator then we define $$\mathcal{P}\psi := \gamma^0\psi$$

So we have that

$$\mathcal{P}\psi_{L} = \mathcal{P}P_L\psi = P_L\gamma^0\psi$$


$$\mathcal{P} \overline \psi_{L} = \mathcal{P}P_L \psi^\dagger \gamma^0 = P_L(\gamma^0\psi)^\dagger\gamma^0 = P_L\psi^\dagger$$

where we used that $(\gamma^0)^2=1$ and $P_L := (1-\gamma^5)/2$. These are the gamma matrices.

Question: The above calculations are correct?

With this we get that

$$\mathcal{P}(\overline{\psi}_{1L}\gamma^\mu\psi_{2L}) = \overline \psi _{1L}(\gamma^\mu)^\dagger \psi_{2R}$$

where we used that $(\gamma^\mu)^\dagger = \gamma^0 \gamma^\mu \gamma^0$.

Question: With this result can I say that the lagrangian is not invariant. Seems that it will not be the same mostly because it turned right-handed in some fermions, but is this correct?


1 Answer 1


Yes it is correct. Parity maps a left-handed spinor into a right-handed one and vice versa. Since the Fermi Lagrangian you wrote only contains left-handed spinors, it is trivially non invariant under parity.


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