Parity Operator

$\hat{P} | \vec{r} ⟩ = | - \vec{r} ⟩$ . $\hat{P} Ψ (\vec{r}) = Ψ (- \vec{r})$ .

Properties

$\hat{P}$ is Hermitian. Check with $⟨ φ | \hat{P} ψ ⟩$ . (One step requires a variable change from $\vec{r}$ to $- \vec{r}$ ).
${\hat{P}}^{2} = I$
${\hat{P}}^{n} = I, \hat{P}$ for n even or odd respectively.
Unitary
Eigenvalues: $\pm 1$ . Odd and even functions are the eigenvectors
$Ψ (\vec{r}) = α Ψ_{+} (\vec{r}) + β Ψ_{-} (\vec{r})$
$Ψ_{\pm} (\vec{r}) = \frac{1}{2} [Ψ (\vec{r}) \pm Ψ (\vec{r})]$

Theorem

$\hat{A}$ is even if $\hat{P} A \hat{P} = \hat{A}$ . $[\hat{A}, \hat{P}] = 0$

$\hat{A}$ is odd if $\hat{P} A \hat{P} = - \hat{A}$ . ${\hat{A}, \hat{P}} = 0$

Examples

$\hat{\vec{R}}$ . $\hat{\vec{R}} | \vec{r} ⟩ = \vec{r} | \vec{r} ⟩$ .

$\hat{P} \hat{\vec{R}} | \vec{r} ⟩ = \vec{r} | \vec{r} ⟩ = \vec{r} | - \vec{r} ⟩$ . $\hat{\vec{R}} \hat{P} | \vec{r} ⟩ = \hat{\vec{R}} | - \vec{r} ⟩ = - \vec{r} | - \vec{r} ⟩$ . Thus, the position operator is odd.

$\hat{H} = \frac{- ℏ^{2}}{2 m} \frac{d^{2}}{d x^{2}} + V (x)$ . The momentum term in this is even. If $V (x) = V (- x)$ then $H (- x) = H (x)$ . I.e. if $V$ is even then $H$ is even. Thus, the Hamiltonian and the parity operator commute. And thus the eigenstates of the Hamiltonian are either even or odd since the Hamiltonian shares an eigenbasis with the parity operator.

Parity Selection Rules

$P | α ⟩ = Ξ_{α} | α ⟩, P | β ⟩ = Ξ_{β} | β ⟩$ . $Ξ_{α, β} = \pm 1$ . $⟨ β | \vec{R} | α = - ⟨ β | P \vec{R} P | α ⟩ = - Ξ_{α} Ξ_{β} ⟨ β | \vec{R} | α ⟩$ . So, $Ξ_{α} Ξ_{β} = - 1$ or $⟨ β | \vec{R} | α ⟩ = 0$ . In other words, if they have the same parity, the interaction dependent on $\vec{R}$ cannot get you from $α$ to $β$ .