# 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?

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.

• Okay, thanks! I'll look at that paper carefully later, but first, is the answer simpler/different for scalar fields? I'm already confused enough with those. – knzhou Jan 24 '18 at 14:36
• For scalar fields, the situation is trivial. 1) $\lambda = 1$, because the plane wave solutions are scalar. 2) There is no signature ambiguity: $O(3,1)\cong O(1,3)$. 3) The charge conjugation commutation with the parity does not impose any constraint. Thus, by a repeated application of the parity operator we are left with the constraint $\eta^2 = 1$, and the possible values $(+1)$ for scalars and$(-1)$ for pseudoscalars. – David Bar Moshe Jan 24 '18 at 15:16
• Okay, I see. And what about the fact that in the Poincare group, parity and time translation always commute? Why does this not imply that on the Hilbert space, parity and the Hamiltonian always commute? – knzhou Jan 24 '18 at 19:28
• From figure 2 of the article, the parity is not an element of the Lorentz group SO(3,1), but only its combination with time reversal ($PT$). Thus you can talk about a theory invariant under the Poincaré group $=$ Translations $\rtimes$ Lorentz, without having a parity invariant Hamiltonian. In the standard model, with the exception of the Weyl fermions, the parity violating terms are in the interaction terms and not in the free part. Thus far away from interactions, you can talk about conserved intrinsic parity. For Weyl fermions, parity does not even exist as they are spinors and not pinors. – David Bar Moshe Jan 25 '18 at 8:39
• Let me try to rephrase this to see if I understand it: a scalar field theory transforms under a representation of $G_1 = \text{translations} \rtimes \text{Lorentz connected to identity}$. A free scalar field theory transforms under a representation of $G_2 = \text{translations} \rtimes O(3, 1)$, defining a conserved parity operator. If we turn on general parity-violation interactions we can keep using the same definition of parity, but it does not commute with time translation, i.e. we do not have even a projective representation of $G_2$. – knzhou Jan 29 '18 at 17:00

• 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.