# Parity Operator

$$\hat{\mathcal{P}}|\vec{r}\rangle = |-\vec{r}\rangle$$. $$\hat{\mathcal{P}}\Psi(\vec{r}) =\Psi(-\vec{r})$$.

## Properties

• $$\hat{\mathcal{P}}$$ is Hermitian. Check with $$\langle \varphi|\hat{\mathcal{P}}\psi\rangle$$. (One step requires a variable change from $$\vec{r}$$ to $$-\vec{r}$$).
• $$\hat{\mathcal{P}}^2 = \mathbb{I}$$
• $$\hat{\mathcal{P}}^n = \mathbb{I},\hat{\mathcal{P}}$$ for n even or odd respectively.
• Unitary
• Eigenvalues: $$\pm 1$$. Odd and even functions are the eigenvectors
• $$\Psi(\vec{r}) = \alpha\Psi_+(\vec{r}) + \beta\Psi_-(\vec{r})$$
• $$\Psi_{\pm}(\vec{r}) = \frac{1}{2}\left[\Psi(\vec{r})\pm\Psi(\vec{r})\right]$$

## Theorem

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

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

### Examples

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

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

$$\hat{H} = \frac{-\hbar^2}{2m}\frac{d^2}{dx^2} + V(x)$$. 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

$$\mathcal{P}|\alpha\rangle = \Xi_\alpha|\alpha\rangle,\mathcal{P}|\beta\rangle = \Xi_\beta|\beta\rangle$$. $$\Xi_{\alpha,\beta}=\pm 1$$. $$\langle \beta|\vec{R}|\alpha=-\langle \beta|\mathcal{P}\vec{R}\mathcal{P}|\alpha\rangle = -\Xi_\alpha\Xi_\beta\langle \beta|\vec{R}|\alpha\rangle$$. So, $$\Xi_\alpha\Xi_\beta = -1$$ or $$\langle \beta|\vec{R}|\alpha\rangle = 0$$. In other words, if they have the same parity, the interaction dependent on $$\vec{R}$$ cannot get you from $$\alpha$$ to $$\beta$$.

Created: 2023-06-25 Sun 02:32

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