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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.

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    $\begingroup$ The physical quantity is just called the parity. For example, quarks have a parity of $+1$. $\endgroup$
    – knzhou
    Commented Apr 1, 2018 at 11:16
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    $\begingroup$ "The converse is also true." [citation needed] In fact, this statement is wrong, see this question and its answers. $\endgroup$
    – ACuriousMind
    Commented Apr 1, 2018 at 11:30
  • $\begingroup$ I agree with ACuriousMind, I don’t think that the converse is also true. I don’t like knzhou’s idea that parity itself is a measurable quantity either. To measure parity you would have to measure some sort of quantity in two different places, so it is not really a physical quantity in the usual sense. $\endgroup$
    – Y2H
    Commented Apr 1, 2018 at 13:30
  • $\begingroup$ Linked. $\endgroup$ Commented Mar 13, 2022 at 10:01

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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.


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

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  • $\begingroup$ This is very specific, applying only to the case of a single quantum harmonic oscillator, which is not a context where I think the parity operator is particularly useful. $\endgroup$ Commented Mar 4, 2022 at 17:07
  • $\begingroup$ Formally, it is completely abstract and general. The OP asked for some "physical" interpretation, and I pointed out that, yes, for a very special operator, the oscillator Hamiltonian, this does have some such interpretation. I cannot clarify this in my answer further. The oscillator, creation and annihilation operators, are used in all kinds of contexts, not involving oscillators visibly. I hope you appreciate that Condon's stuff is completely general ! $\endgroup$ Commented Mar 4, 2022 at 17:12
  • $\begingroup$ I stand by my original comment. Appealing that the use of the CCR algebra implies generality is really very weak. $\endgroup$ Commented Mar 7, 2022 at 2:31
  • $\begingroup$ Either you are misunderstanding my evident point, or I yours. The formulas are correct, general, and incontrovertible. Recognizing the operator as the oscillator propagator may be frivolous, but satisfactory to some. $\endgroup$ Commented Mar 7, 2022 at 3:09
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    $\begingroup$ Tu quoque after curt invitations to embarassment??? My point has been firmly that dynamics is irrelevant to the significance of parity. And that the number operator is what counts, which happens to be the oscillator hamiltonian, but need not be the system's hamiltonian. Tertium non datur. $\endgroup$ Commented Mar 8, 2022 at 17:25

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