Does the parity operator always square to 1, in relativistic quantum field theory? I'm a bit confused over whether the parity operator should always square to one. My impression was that if $W(\Lambda, a)$ was a representation of the Poincare group, then the parity operator $\hat{P}$ was defined to be $W(P,0)$ where $P$ is the improper parity Lorentz transformation. Since $P^2 = 1$, this automatically implies $\hat{P}^2 = 1$. Using the group multiplication law, one can prove everything else, e.g. that $\hat{P}$ flips momentum but not spin, and that $\hat{P}$ commutes with $\hat{H}$.
On the other hand, I've also heard that parity can act on a field by multiplying it by a phase $\eta$, called its intrinsic parity, which can be any phase, not just $\pm 1$. I've also heard that parity can be violated, so it doesn't necessarily commute with $\hat{H}$.
Weinberg defines $\hat{P}$ exactly as I do, but when he comes to these phases he gives the cryptic remark

It is easy to say that space inversion $P$ has the group multiplication law $P^2 = 1$; however, the parity operator that is conserved may not be this one, but rather may differ from it by a phase transformation of some sort.

Similarly, Schwartz's textbook gives an even more cryptic remark:

You might expect that the action of $P$ and $T$ should be determined from representation theory. However, recall that technically spinors do not transform under the Lorentz group $O(1, 3)$, only its universal cover $SL(2, \mathbb{C})$, so we are not guaranteed that $T$ and $P$ will act in any nice way. In fact they do not. [...] In any representation, we should have $P^2 = T^2 = 1$.

Both Weinberg and Schwartz seem to dance around the issue of actually defining what parity is, and Schwartz's passage seems almost self-contradictory.
For simplicity, let's ignore the case of spinor fields and focus on scalar fields, so there are no issues with projective representations. Then my questions are:


*

*How exactly is parity defined? If it's defined the way I said above, how can you avoid having $\hat{P}$ commute with $\hat{H}$? If it isn't, is the definition unique and how do you recover familiar results, like how $\hat{P}$ flips momentum but not spin?

*In general, what's wrong with postulating the particles to form a representation of the Poincare group with improper Lorentz transformations?

 A: First let me refer you to the seminal review  by  Berg, DeWitt-Morette, Gwo and Kramer where virtually every known fact about intrinsic parities is derived or quoted. They derive constraints on the values of the intrinsic parities from their action on fermionic free fields:
$$\Psi(x) = \sum_{p,s}\left (a(\mathbf{p}, s) u(p,s) e^{-ipx}+  b^{\dagger}(\mathbf{p}, s) v(p,s) e^{ipx}\right)$$
For a pinor field, the intrinsic parity is given as (equations (38, 39) in the reference):
$$U_P \Psi(x)U_P^{-1} =  (\frac{\eta}{\lambda})\Lambda_P \Psi(Px), \quad \Lambda_P = \Gamma_0,$$ 
where $\eta$ is the phase in the action of the parity operator on the particle's annihilation operator:
$$U_P a(\mathbf{p},s) U_P^{-1} =  \eta a(P\mathbf{p},s),$$
and $\lambda$ is the eigenvalue of the parity matrix action on the positive energy pinor
$$\Lambda_p u(p) = \lambda u(Pp)$$
The intrinsic parity depends only on the ratio $(\frac{\eta}{\lambda})$. By requiring the  antiparticles intrinsic parities  to be the negative complex conjugates of the corresponding particles intrinsic parities and  the parity to commute with the charge conjugation in the specific representation, they reach the constraint:
$$(\frac{\eta}{\lambda})^4=1$$
Thus the intrinsic parities can assume the values$(1, -1, i, -i)$.
The main ambiguity in the values of the intrinsic parities of elementary particles is related to which pin group they belong, please see figure 2 in the reference:
The group , covering the Lorentz group, which supports both spinor representations and distinct parity and time reversal elements is the Pin group which is the double cover of $O(3,1)$. This is why elementary fermions must be pinors rather than spinors because the parity and the charge conjugation are not elements of the spin group but only their combination. The main problem is that Pin(3,1) is NOT isomorphic to Pin(1,3)!, and we do not know which type of pinors the elementary particles are.
A: David Bar Moshe's answer is good, and for future readers I'm just adding a few more subtleties that tripped me up when I was thinking about this.


*

*Since we're dealing with projective representations, we need to take universal covers. The issue is that there are two distinct universal covers of $O(3, 1)$, the two Pin groups.

*In one of these groups, $\hat{P}^2 = 1$. In the other, $\hat{P}^2 = (-1)^F$ where $F = 1$ for half-integer spin and $F = 1$ for integer spin. 

*The subtlety is that for computations one can always redefine parity by multiplying it by $e^{i \alpha Q}$ for some conserved charge $Q$. This will lead to the exact same selection rules.

*In the case of the Standard Model, the conserved charges of electric charge, baryon number, and lepton number are sufficient to define $\hat{P}^2 = 1$ in both cases, so there is no observable difference. This is why books say $\hat{P}^2 = 1$ in one sentence and $\hat{P}^2 \neq 1$ in the next.

*Another way of saying this is that conservation laws are so strict that the stricter conservation law provided in the case $\hat{P}^2 = (-1)^F$ actually adds nothing at all.

*In the case of parity nonconservation, the free Hamiltonian still carries a representation of
$$\text{Pin group} \rtimes \text{spacetime translations}$$
with the exception of theories with chiral fermions. Therefore one can define intrinsic parity of asymptotic states, i.e. the states described by the $S$-matrix. This is why we can speak of a weak decay turning some parity into another.


I'm still figuring this out, so some of the points above may be wrong.
