# Intrinsic parity of particle and antiparticle with spin zero

I need to prove that the intrinsic parities of a particle and antiparticle with spin zero are the same. Can I prove that by an argument that operator of $P$-inversion commutes with charge conjugation operator for the spin-zero particle? $$\hat {P}\Psi = \pm \Psi , \quad \hat {C} \Psi = \Psi^{*}, \quad \hat {C} \hat {P}\Psi = \pm \Psi^{*} = \hat {P}\hat {C}\Psi = \pm \Psi^{*}.$$

I slightly deviate from your notation and use $\phi$ to denote the scalar field as its more standard. Also I should point out that quantum fields are operators and thus under a transformation they get acted on from both the left and the right.
The complex scalar field is given by, $$\phi (x) = \int \frac{ \,d^3p }{ (2\pi)^3 } \frac{1}{ \sqrt{ 2E _{ {\mathbf{p}} }}} \left( a _{ {\mathbf{p}} } e ^{ - i p \cdot x } + b ^\dagger _{ {\mathbf{p}} } e ^{ i p \cdot x } \right)$$ Under parity we have that $a _{ {\mathbf{p}} } \rightarrow a _{ - {\mathbf{p}} }$ and $b _{ {\mathbf{p}} } \rightarrow b _{ - {\mathbf{p}} }$ which results in, $$P \phi ( t, {\mathbf{x}} ) P = \phi ( t , - {\mathbf{x}} )$$ Under complex conjugation we have that $a _{ {\mathbf{p}} } \leftrightarrow b _{ {\mathbf{p}} }$ which results in $$C \phi ( t , {\mathbf{x}} ) C = \phi ^\ast ( t , {\mathbf{x}} )$$
The commuting nature of $C$ and $P$ is then quite trivial. Complex conjugation has nothing to do with what position the field is at. Its easy to see that, $$C P \phi (x) P C = C \phi ( t , - {\mathbf{x}} ) C = \phi ^\ast ( t , - {\mathbf{x}} )$$ $$P C \phi (x) CP = C \phi ^\ast ( t , {\mathbf{x}} ) C = \phi ^\ast ( t , - {\mathbf{x}} )$$ and hence the two operators must commute.