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?
$$
\hat {P}\Psi = \pm \Psi , \quad \hat {C} \Psi = \Psi^{*}, \quad \hat {C} \hat {P}\Psi = \pm \Psi^{*} = \hat {P}\hat {C}\Psi = \pm \Psi^{*}.
$$
 A: 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,
\begin{equation} 
\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) 
\end{equation} 
Under parity we have that $ a _{ {\mathbf{p}} } \rightarrow a _{ - {\mathbf{p}} } $ and $ b _{ {\mathbf{p}} } \rightarrow b _{ - {\mathbf{p}} } $ which results in,
\begin{equation} 
P \phi ( t, {\mathbf{x}} )  P = \phi ( t , - {\mathbf{x}} ) 
\end{equation} 
Under complex conjugation we have that $ a _{ {\mathbf{p}} } \leftrightarrow  b _{ {\mathbf{p}} } $ which results in
\begin{equation} 
C \phi ( t , {\mathbf{x}} ) C = \phi ^\ast ( t , {\mathbf{x}} ) 
\end{equation} 
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,
\begin{equation} 
C  P \phi (x) P C = C \phi ( t , - {\mathbf{x}} ) C = \phi ^\ast ( t , - {\mathbf{x}} ) 
\end{equation} 
\begin{equation} 
P C \phi (x) CP = C \phi ^\ast  ( t , {\mathbf{x}} ) C = \phi ^\ast ( t , - {\mathbf{x}} ) 
\end{equation} 
and hence the two operators must commute.
