Here's a simple argument that every scalar field theory is parity symmetric.
- Consider the group $G = O(3, 1) \rtimes \mathbb{R}^4$ of improper Lorentz transformations $\Lambda$ and spacetime translations by $a^\mu$. Denote the element of this group comprised of $\Lambda$ followed by $a^\mu$ by $(\Lambda, a^\mu)$.
- In quantum mechanics, a group $G$ of physical transformations is represented on our Hilbert space by some unitary representation $R$. I'm not assuming any symmetry here; it's just that something has to happen when we boost, translate, etc. our system, and that must be represented by operators $R[(\Lambda, a^\mu)]$. For example, every system carries a (possibly projective) $SO(3)$ representation whether it's rotationally symmetric or not, just because you can rotate it.
- Parity is defined as $(P, 0)$, and by definition the parity operator is $$\hat{P} = R[(P, 0)].$$
- Time translation is defined as $(I, e_t)$ and by definition the time translation operator is $$\hat{U} = R[(I, e_t)].$$ The Hamiltonian $\hat{H}$ is by definition proportional to $\log \hat{U}$, e.g. as in the Wightman axioms.
- To avoid the complications of projective representations, we consider a theory with only scalar fields. We know that $(P, 0)$ and $(I, e_t)$ commute. Using nothing but the definition of a representation, $R(g) R(g') = R(gg')$, it's straightforward to show that $\hat{P}$ and $\hat{H}$ commute.
Clearly I'm doing something physically wrong, because there are scalar field theories that break parity. Exactly what step is invalid?
