# Intrinsic parity

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?

• I would like to add a question to this. If I consider an interaction , say $l_{int}=\bar\psi \psi \phi$, it can be shown that it is invariant under parity operation, given that scalar is intrinsic scalar. Now, does it mean that a pseudo scalar can not interact through this lagrangian term. OR it can interact but will not be preserved under parity? Commented Sep 4, 2015 at 8:26
• More on intrinsic parity. Commented Oct 26, 2016 at 20:08
• "does it mean that a pseudo scalar can not interact through this lagrangian term": pseudo scalar Yukawa-type interaction is perfectly legit, see here: physics.stackexchange.com/questions/459043/… Commented Nov 20, 2020 at 17:13

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.