# Is the parity operator an observable?

I'm trying to justify whether the parity operator is an observable in quantum mechanics, and if so, why. I'm at a loss here, any advice on how to tackle this problem?

According to N. Zettili's "Quantum Mechanics: Concepts and applications" book, he takes as foundation of the theory five postulates, and the second one of them is the following: "To every phisically measurable quantity $$A$$, called an observable or dynamical variable, there corresponds a linear Hermitian operator $$\hat{A}$$ whose eigenvectors form a complete basis".

This would make the parity operator an observable, for it is a linear, hermitian operator, i.e. $$\mathcal{P}=\mathcal{P}^{\dagger}$$. Note also that the parity operator is defined up to a phase of choice, of unitary magnitude, the usual choice for this phase is made so the operator is an observable.

Maybe you can use a deeper explanation given here.

• Bit backward, perhaps - He's not saying that for every linear hermitian operator, there corresponds a physically measurable quantity. Jun 12, 2020 at 8:45
• What you say is totally true, there is no reason in advance to take the statement as true in the opposite direction, but we really talk about observables when some quantity is represented by a linear hermitian operator. On the other hand, parity is indeed a measurable quantity, and each fundamental particle has an intrinsic parity. @Slereah Jun 12, 2020 at 9:32

Yes, parity can be thought of as an observable. It's eigenvalues are $$1$$, corresponding to a state that is symmetric under a parity transformation, and $$-1$$, corresponding to a state that is anti-symmetric under a parity transformation. When you measure parity, you are measuring whether your state has one symmetry property or the other.

Not sure from a practical standpoint how you would build a machine that measures parity, but if you did, you would get the same behavior as any other observable. All states can be written as linear combinations of states with definite parity (eigenstates of $$P$$). You will measure $$\pm1$$ with probabilities that depend on your initial state, and after measurement, the state will collapse into a state with definite parity.

The intrinsic parity of a proton is listed as +1 and of an antiproton as -1, but this is purely a convention. Fermion and antifermion must have opposite parity, but we could equally well take the proton to have parity -1 and the antiproton as +1, provided we did this consistently. You can't measure the (absolute) parity of a proton, so this provides a case where parity is not an observable - under any sensible definition of the word observable.

• You can't measure "absolute" position either. Under one choice of origin, a particle may be located at $x=1$, while under another choice, the particle may be located at $x=-1$. That doesn't mean position is not an observable in quantum mechanics. Dec 31, 2020 at 15:17