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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? $$ \hat {P}\Psi = \pm \Psi , \quad \hat {C} \Psi = \Psi^{*}, \quad \hat {C} \hat {P}\Psi = \pm \Psi^{*} = \hat {P}\hat {C}\Psi = \pm \Psi^{*}. $$

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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.

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  • $\begingroup$ Just a remark, but your answer assumes that we're dealing with a Klein-Gordon field. OP's question is about a general scalar particle. $\endgroup$ – Vibert Feb 25 '14 at 5:32
  • $\begingroup$ @Vibert: Good point, I didn't think about any other possibility... $\endgroup$ – JeffDror Feb 25 '14 at 11:08

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