Midpoint parity operator ($R_a\psi (x)=\psi (a-x)$)

On $L^2(0,a)$, the midpoint parity operator is $(R_a\psi )(x)=\psi (a-x)$. Substitution $y=a-x$ gives $⟨f,R_{a}g⟩=⟨R_{a}f,g⟩$ and $\Vert R_{a}f\Vert =\Vert f\Vert$, so it is a bounded self-adjoint operator and a unitary operator, with $R_a^2=I$. It is the parity operator in the centered coordinate $x-a/2$.

For an infinite square well, ${\psi }_n=\sqrt{2/a}\sin (n\pi x/a)$ obeys $R_a{\psi }_n=(-1{)}^{n+1}{\psi }_n$. Reflection preserves the Dirichlet boundary conditions and commutes with the second derivative, so $R_a$ commutes with the Hamiltonian operator on its operator domain. By contrast, differentiating once gives $P{R}_a=-R_{a}P$ for the momentum operator. This distinction permits conserved midpoint parity with changing mean momentum.