Physical quantity related to the parity operator

Question

There is a statement in quantum mechanics that for every physical quantity, there exists a Hermitian operator. The converse is also true. So the question is, what physical quantity is related to the parity operator $\hat{P}:$ $\hat{P}\Psi(x)=\Psi(-x)$? It must have a physical quantity, since it's a Hermitian operator.

Answer 1

Yes, parity is a hermitean and unitary operator which does have a (somewhat recondite and marginal) physical interpretation disjoint from its function: It is also an evolution operator of the quantum oscillator for half of its cycle.

I'll be skipping carets of operators to uncluttered formulas and working in optimized natural units ($\hbar=1$, $m=1$, $\omega=1$, to get the drift of formulas friendlier to follow; you know how to repeat all this reinserting the units of your liking).

$$ P x P= -x , \qquad PpP= -p,\\ P=P^{-1}=P^\dagger=\int\!\! dx ~~|\!-\!x\rangle \langle x|, \\ P^2=I. $$

Most importantly, a parity transformation preserves Heisenberg’s/Born's commutation relation, $[x,p]=i$, so it is a quantum canonical transformation.

As you must know, a quantum oscillator Hamiltonian uniformly rotates the operators x and p rigidly in phase space, Condon's (1937) celebrated fractional Fourier transform as a canonical transformation, $$ x\mapsto e^{-iHt} x e^{iHt}, ~~~ p\mapsto e^{-iHt} p e^{iHt}, \\ H= {p^2+ x^2 \over 2}-{1\over 2}. $$ I have shifted away the zero-point energy, since its effect commutes with everything and hence washes out with the inverses of the evolution operators. What's left is the number operator with the standard integer eigenvalues.

The cycle of this quantum phase-space rotation, then, is $t=2\pi$, $e^{i2\pi H}=I$; so the half cycle, $\pi$, just amounts to $$ x\mapsto -x, \qquad p\mapsto -p , ~~~\leadsto \\ \bbox[yellow,5px]{ P= e^{\frac{-i\pi}{2}(p^2+x^2)+{i\pi\over 2}}=P^\dagger }, $$ satisfying the properties posited ($P P^\dagger=I$, etc), as you must check.

• This is a completely general operator canonical transformation, for all systems, and is not predicated on focussing on a quantum oscillator system. It is your prerogative to interpret the exponent as an oscillator hamiltonian or not, for ease of visualization. Again, the specific dynamics of the system specified by its own hamiltonian should not be conflated with the formal action of the parity operator, as you might appreciate from QFT.

This is not my favorite operatorial representation of P,$^\natural$ but it is the one closest to a "physical picture", namely a $t=\pi$ evolution of the quantum oscillator for half its cycle.

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$^\natural$ _{My own favorite representation, much easier to check on eigenstates of x or p, is $$P = \int\!\! dx ~~|\!-\!x\rangle \langle x|=\int\!\! da db ~~ e^{i(a\hat p + b\hat x)}/4\pi .$$}