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In classical mechanics, the position vector changes under parity as $\textbf{x}\to -\textbf{x}$. This is the definition of parity in classical mechanics.

Operators (such as $\mathcal{O}$) change under symmetry operators represented by $\mathscr{Q}$ as $$\mathcal{O}\to \mathscr{Q}^{-1}\hat{\mathcal{O}}\mathscr{Q}.$$

Can we take $$\mathscr{P}^{-1}\hat{\textbf{x}}\mathscr{P}=-\hat{\textbf{x}}$$ as the definition of parity in quantum mechanics (where the object $\mathscr{P}$ represents the parity operator)?

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No we cannot, since the only requirement$$\mathscr{P}^{-1}\hat{\textbf{x}}\mathscr{P}=-\hat{\textbf{x}}$$ does not fix the parity operator uniquely even in the simplest case. Further information with the form of added requirements is necessary to fix the parity operator.

The definition of parity operator actually depends on the system you are considering. Let us consider the simplest spin-zero particle in QM.

Its Hilbert space is (isomorphic to) $L^2(\mathbb R^3)$.

Parity is supposed to be a symmetry, so in view of Wigner's theorem, it is an operator $H: L^2(\mathbb R^3) \to L^2(\mathbb R^3)$ which may be either unitary or antiunitary.

Here the parity operator is fixed by a pair of natural requirements, the former is just that in the initial question, the latter added requirement concerns the momentum operators. $$UX_kU^{-1} =-X_k\quad, k=1,2,3 \tag{1}$$ and $$UP_kU^{-1} =-P_k\quad, k=1,2,3 \tag{2}$$ Notice that (2) is independent from (1), we could define operators satisfying (1) but not (2).

First of all, these requirements decide the unitary/antiunitary character. Indeed, from CCR, $$[X_k,P_h] = i\delta_{hk}I\tag{3}$$ we have $$U[X_k,P_h] U^{-1} = \delta_{kh} Ui IU^{-1} = \pm i \delta_{kh}I$$
that is $$[UX_kU^{-1}, UP_hU^{-1}]=\pm i \delta_{kh}I\:,$$ so that, from (1) and (2), $$[X_k,P_h] = \pm i \delta_{kh}I$$ Comparing with (3), this identity rules out the minus sign corresponding to an antiunitary operator. $U$ must be unitary.

Let us prove that (1)-(2) fix $U$ up to a phase. It is not possible to define $U$ more precisely because this arbitrary phase is just the degree of freadom permitted by Wigner's theorem in defining symmetries in terms of unitary or anti unitary operators.

Suppose that, for another unitary operator $V$, we also have $$VX_kV^{-1} =-X_k\quad, k=1,2,3 \tag{1'}$$ and $$VP_kV^{-1} =-P_k\quad, k=1,2,3 \tag{2'}\:.$$ As a consequence of (1) and (2), $$U^{-1}VX_kV^{-1}U =X_k\quad, k=1,2,3 $$ and $$U^{-1}VP_kV^{-1}U = P_k\quad, k=1,2,3 \:.$$ In other words, $L:= U^{-1}V$ satisfies $$L X_k = X_kL\:, \quad L P_k = P_kL\quad, k=1,2,3 \:.$$ Since the system of operators $X_k$ and $P_k$ is irreducible in $L^2(\mathbb R^3)$, Schur's lemma implies that $$L= e^{i\gamma}I$$ for some fixed real $\gamma$. Namely, $$V = e^{i\gamma}U\:.$$

To conclude, it is enough to find an unitary operator satisfying both (1) and (2). Per direct inspection one sees that $$(U\psi)(x) = \psi(-x) \tag{4}$$ does the job. All remaining possibilities are included in the arbitrary phase $e^{i\gamma}$. Choice (4) has a nice further property shared with only the other possibility $$(U\psi)(x) = -\psi(-x) \tag{4'}$$ In both situations (and only for these choices of the phase), $$UU=I\:.$$ Since we already know that $U^{-1}=U^\dagger$, we conclude that $$U= U^{-1}= U^\dagger\:.$$ In other words the said choices of the phase make $U$ an observable, which plays an important role in particle physics (it can be conserved or not depending on the Hamiltonian).

Extending the notion of particle by including the spin, the Hilbert space enlarges to $L^2(\mathbb R^3)\otimes {\mathbb C}^{2s+1}$, where the second factor includes an irreducible representation of $SU(2)$ generated by the three spin operators $S_k$, $k=1,2,3$. An analysis similar to the previous one can be developed by adding to (1) and (2) the further requirement $$U S_kU^{-1} = S_k\:, \quad k =1,2,3,$$ together with spin commutation relations $$[S_k,S_h]= i \sum_{p=1}^3\epsilon_{khp} S_p\:,$$ but I stop here.

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  • $\begingroup$ This answer is confusing me, first you state that "No we cannot, since that only requirement does not define the parity operator even in the simplest case, further information is necessary." Then you go on to demonstrate, that it contrary to what you stated actually is the only requirement necassary: "Let us prove that (1)-(2) fix U up to a phase". $\endgroup$ – pindakaas Jun 7 '18 at 11:59
  • $\begingroup$ The first requirement is Eq. (1) . It is not enough to fix the parity operator uniquely. If adding also the second restriction (2), then the parity operator is fixed up to a sign. Is it clear now? $\endgroup$ – Valter Moretti Jun 7 '18 at 12:36
  • $\begingroup$ Notice that (2) is independent from (1). You can find operators satisfying (1) but not (2). So (1) is not a well-behaved definition of parity operator since it does not determine the parity operator without adding (2). $\endgroup$ – Valter Moretti Jun 7 '18 at 12:46
  • $\begingroup$ One comment: With only (1) being defined, the translation operator $T$ can be shown to anticommute with the parity operator $\{U,T\}=0$. Since momentum operator can be expressed in terms of translation operator, it follows that (1) implies (2). $\endgroup$ – Neoh Oct 14 '18 at 6:26
  • $\begingroup$ I cannot see your proof. Could you be more explicit please? $\endgroup$ – Valter Moretti Oct 14 '18 at 14:28

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