Intrinsic parity

Question

When we apply parity on a field two times, we demand that we should get back the same field. This gives us, $P^{2} =1$, which implies, $ P \psi = e^{i \theta} \psi$ . This extra phase factor is called intrinsic parity. Now, this is not just a phase factor. It is important because it's value decided whether a spin 0 field is a scalar or Pseudo-scalar. But what exactly is intrinsic parity ? On what properties of field it depends?

Answer 1

You are viewing this the wrong way. If you apply parity operation to a scalar field (both real and complex field), you have: $$ P\phi (x) P^{-1} = n_P\phi (\tilde{x}) \tag{1} $$ where, $n_P$ is the intrinsic parity,

and the parity operator effects the coordinate transformation from $x$ to $\tilde{x} = (t, -\boldsymbol{x})$.

Now, if you apply the parity operation again, $$ P^2\phi (x) P^{-2} = n_P P\phi (\tilde{x}) P^{-1} = n_P^2 \phi (x) $$ We should obtain the original field back. Thus, we see that $n_P^2 = 1$, which can only imply that $n_P = \pm 1$.

We cannot determine intrinsic parity with free Lagrangian; for that, we need interactions. Suppose you have an interaction Lagrangian of the form, $$ L_{int} = - \mu \phi^3 - \lambda \phi^4 $$ Applying the parity transformation on the interaction Lagrangian, and subsequently using Eqn. (1) for each $\phi$, we see that parity invariance would require that for the first term, $n_P^3 = 1$, and for the second term $n_P^4 = 1$. Both of these equations are satisfied simultaneously only if $n_P = 1$. Hence, for this case, $\phi$ is a scalar field.