# What's wrong with this argument that parity is always conserved?

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?

• What is wrong is the existence of the unitary representation of the whole $O(3,1) \times \mathbb R^4$. When a quantum physical system breaks parity, this just means that the representation which exists is only of $SO(3,1) \times \mathbb R^4$. Existence of the whole representation is decided by Nature not by our mathematical models. Jan 26 '18 at 12:50
• @ValterMoretti I think this what's really confusing me. As a simpler example, there are plenty of systems that are not rotationally invariant, but their Hilbert spaces still carry a representation of $SU(2)$, corresponding to the physical act of rotating the system. (i.e. something has to actually happen when you pick it up and spin it!) Right? So why doesn't everything have to carry an $O(3, 1) \rtimes \mathbb{R}^4$ representation because we can boost, reflect, rotate, translate, etc. everything? Jan 26 '18 at 12:55
• @ValterMoretti Moreover, I thought the parity operator was defined to be $R[(P, 0)]$. If it's not defined as that, then what even is it? Jan 26 '18 at 12:55
• It seems to me that you are confound here dynamical symmetries and symmetries. If a potential has the form $U(x,y,z,) = x^2+y^2+ 3z^2$ the corresponding Hamiltonian is not invariant under rotation, even if there is the standard representation $(U_R\psi)(x) = \psi(R^{-1}x)$. Jan 26 '18 at 13:01
• You may have different notions of symmetry breaking. You may have a representation of a symmetry at the level of algebra of observables, but there is no unitary implementation of it. Or there is but dynamics is not invariant. Finally you may have a unitary implementation but no invariant state... Jan 26 '18 at 13:03