# Is it necessary that spin $0$ particles have intrinsic parity $\pm 1$?

For simplicity let's consider a simple example of real scalar free quantum field $$\phi(x) = \int \mathrm{d}\mu(p) \left( a(p) e^{-ipx} + a^{\dagger}(p)e^{ipx}\right),$$ where $\mathrm{d} \mu(p)$ is Lorentz invariant volume element on mass shell $p^2 =m^2,$ $p^0>0$. Under parity this field transforms to $$\phi_P(x) = P \phi(x)P^{\dagger} = \int \mathrm{d}\mu(p) \left( \eta^* a(p) e^{-ipx} + \eta a^{\dagger}(p) e^{ipx}\right),$$ where $\eta$ is intrinsic parity of the particle created and annihilated by $\phi$, $|\eta|=1$. It is easy to see that in order for $\phi_P$ to commute with $\phi$ for spacelike related points one needs to have $\eta = \pm 1$. This is the argument which is invoked in Weinberg's textbook to conclude that this must be the case and all spin $0$ particles have parity $\pm 1$. In this case $\phi_P=\eta \phi$. It seems that this assumption is made in most texts on QFT. Therefore I ask the question: Is there any fundamental reason why we should assume particularly simple transformation rule of quantum fields under parity?

• What happens if you transform twice? – Emilio Pisanty Aug 10 '16 at 17:37
• If $P^2=1$ is assumed then we also obtain the conclusion that $\eta = \pm 1$. However as far as I know it is possible to construct quantum field theories where parity operator exists but nevertheless it doesn't satisfy $P^2=1$ and it is not possible to redefine it such that it satisfies this equation. – Blazej Aug 10 '16 at 17:40
• I mean parity operator, i.e. operator which implements Lorentz transformation $(t, \vec x) \mapsto (t,- \vec x)$, not the operator which exchanges particles. – Blazej Aug 10 '16 at 17:47
• hah. right you are. – Emilio Pisanty Aug 10 '16 at 17:48