Parity of spatial wavefunctions

Parity of spatial wavefunctions is a property describing how a wavefunction behaves under inversion, being classified as even (g) if it remains unchanged or odd (u) if it changes sign.

Consider the hydrogenic wavefunction $\psi=R_{n,l}(r)Y_l^{\vert m_l\vert}(\theta,\phi)$ , where $R_{n,l}(r)$ and $Y_l^{\vert m_l\vert}(\theta,\phi)=P_l^{\vert m_l\vert}(cos\theta)e^{im_l\phi}$ are the radial and angular wavefunctions respectively. In spherical coordinates, an inversion ( $\hat{i}=\hat{\sigma}\hat{C_2}$ ) through the origin results in:

- $r\rightarrow r$ (the radius is always positive)
- $\theta\rightarrow \pi-\theta$ (polar angle reflects across the $xy$ -plane)
- $\phi\rightarrow\phi+\pi$ (azimuthal angle rotates by half a circle)

Therefore,

$\hat{i}R_{n,l}(r)Y_l^{\vert m_l\vert}(\theta,\phi)=\hat{\sigma}R_{n,l}(r)P_l^{\vert m_l\vert}(cos\theta)e^{im_l(\phi+\pi)}=\hat{\sigma}R_{n,l}(r)Y_l^{\vert m_l\vert}(\theta,\phi)e^{im_l\pi}$

Since $e^{im_l\pi}=(e^{i\pi})^{m_l}=(-1)^{m_l}$ ,

$\hat{i}R_{n,l}(r)Y_l^{\vert m_l\vert}(\theta,\phi)=(-1)^{m_l}R_{n,l}(r)\hat{\sigma}Y_l^{\vert m_l\vert}(\theta,\phi)$

$P_l^{\vert m_l\vert}(cos\theta)$ are the associated Legendre polynomials given by $P_l^{\vert m_l\vert}(cos\theta)=\frac{1}{2^ll!}(1-cos^2\theta)^{\frac{\vert m_l\vert}{2}}\frac{d^{l+\vert m_l\vert}}{dcos\theta^{l+\vert m_l\vert}}(cos^2\theta-1)^l$ , or equivalently, $P_l^{\vert m_l\vert}(x)=\frac{1}{2^ll!}(1-x^2)^{\frac{\vert m_l\vert}{2}}\frac{d^{l+\vert m_l\vert}}{dx^{l+\vert m_l\vert}}(x^2-1)^l$ , where $x=cos\theta$ . So,

$\hat{\sigma}P_l^{\vert m_l\vert}(x)=P_l^{\vert m_l\vert}(-x)=\frac{1}{2^ll!}[1-(-x)^2]^{\frac{\vert m_l\vert}{2}}\frac{d^{l+\vert m_l\vert}}{d(-x)^{l+\vert m_l\vert}}[(-x)^2-1]^l$

Question

Prove by induction the expression $\frac{d^n}{d(-x)^n}=(-1)^n\frac{d^n}{dx^n}$ .

Answer

For $n=1$ , let $u=-x$ . Applying the chain rule, $\frac{d}{du}=\frac{dx}{du}\frac{d}{dx}=(-1)\frac{d}{dx}$ . So, $\frac{d}{d(-x)}=\frac{d}{du}=(-1)\frac{d}{dx}$ and the expression is valid. Assume the expression holds for some $n$ , i.e. $\frac{d^n}{d(-x)^n}=(-1)^n\frac{d^n}{dx^n}$ . Then for $n+1$ ,

$\frac{d^{n+1}}{d(-x)^{n+1}}=\frac{d}{d(-x)}\frac{d^n}{d(-x)^n}=(-1)^n\frac{d}{d(-x)}\frac{d^n}{dx^n}=(-1)^{n+1}\frac{d^{n+1}}{dx^{n+1}}$

By mathematical induction, $\frac{d^{n}}{d(-x)^{n}}=(-1)^{n}\frac{d^{n}}{dx^{n}}$ .

It follows that

$\hat{\sigma}P_l^{\vert m_l\vert}(x)=(-1)^{l+\vert m_l\vert}\frac{1}{2^ll!}[1-x^2]^{\frac{\vert m_l\vert}{2}}\frac{d^{l+\vert m_l\vert}}{dx^{l+\vert m_l\vert}}(x^2-1)^l=(-1)^{l+\vert m_l\vert}P_l^{\vert m_l\vert}(x)$

and

$\hat{i}\psi=(-1)^{m_l}(-1)^{l+\vert m_l\vert}\psi=(-1)^{l+2\vert m_l\vert}\psi=(-1)^l\psi$

Hence, the parity of hydrogenic wavefunctions is given by $(-1)^l$ , where $\psi$ is even if $l$ is even and odd if $l$ is odd. For an $N$ -electron atom with a wavefunction expressed as a product of hydrogenic orbitals $\Psi(\boldsymbol{\mathit{r}}_1,\cdots,\boldsymbol{\mathit{r}}_N)=\psi_1(\boldsymbol{\mathit{r}}_1)\cdots\psi_N(\boldsymbol{\mathit{r}}_N)$ , its parity is

$\prod^N_i(-1)^{l_i}=(-1)^{\sum^N_il_i}$

because $\hat{i}\Psi(\boldsymbol{\mathit{r}}_1,\cdots,\boldsymbol{\mathit{r}}_N)=\Psi(-\boldsymbol{\mathit{r}}_1,\cdots,-\boldsymbol{\mathit{r}}_N)=\psi_1(-\boldsymbol{\mathit{r}}_1)\cdots\psi_N(-\boldsymbol{\mathit{r}}_N)=\hat{i}\psi_1(\boldsymbol{\mathit{r}}_1)\cdots\hat{i}\psi_N(\boldsymbol{\mathit{r}}_N)$ .

So, the wavefunction is even if $\sum^N_il_i$ is even and odd if $\sum^N_il_i$ is odd.

If the multielectron wavefunction is properly antisymmetrised, such as in a Slater determinant, each term in the determinant is a permutation of the same set of one-electon orbitals. Since parity depends only on the set of occupied orbitals and not their ordering, every permutation acquires the same overall factor $(-1)^{\sum^N_il_i}$ . Therefore, the parity of the Slater determinant remains $(-1)^{\sum^N_il_i}$ .

Parity of spatial wavefunctions

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