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?


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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