Spin-orbital

A spin-orbital $\phi(x)$ expresses the state of an electron as a complete wavefunction, which comprises of a spatial component and a spin component.

It is defined as

$\phi(x)=\left\{\begin{matrix}\chi(\boldsymbol{\mathit{ r}})\alpha(\omega)\\\chi(\boldsymbol{\mathit{ r}})\beta(\omega) \end{matrix}\right.$

where $x$ is a composite coordinate consisting of three spatial coordinates $\boldsymbol{\mathit{ r}}$ and one spin coordinate $\omega$ , while $\alpha$ and $\beta$ are spin wavefunctions describing the two possible spin states of an electron.

In the derivation of the canonical Hartree-Fock equations, $\phi_t^{'}(x)=\chi_t^{'}(\boldsymbol{\mathit{ r}})\sigma(\omega)$ , where $\sigma=\alpha\;or\;\beta$ . When we substitute the spin-orbitals in their explicit forms into eq109, we have

$\biggr[\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)+\sum_{t\neq r}\int\int\chi_t^{'*}(\boldsymbol{\mathit{ r}}')\sigma^*(\omega^{'}) \frac{(1-P_{rt})\chi_t^{'}(\boldsymbol{\mathit{ r}}')\sigma(\omega^{'})}{\vert\boldsymbol{\mathit{ r}}-\boldsymbol{\mathit{ r}}'\vert}d\boldsymbol{\mathit{ r}}'d\omega'\biggr]$ $\chi_r^{'}(\boldsymbol{\mathit{ r}})\sigma(\omega)=\lambda_r^{'}\chi_r^{'}(\boldsymbol{\mathit{ r}}')\sigma(\omega^{'})$
Multiplying throughout by $\sigma^{'}(\omega)$ and integrating with respect to the spin coordinates, while noting that $\int\sigma^{'}(\omega)\sigma(\omega)=\delta_{\sigma^{'}\sigma}$ because of spin orthogonality, we have

$\biggr[\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)+\sum_{t\neq r}\int\chi_t^{'*}(\boldsymbol{\mathit{ r}}')\frac{(1-P_{rt})\chi_t^{'}(\boldsymbol{\mathit{ r}}')}{\vert\boldsymbol{\mathit{ r}}-\boldsymbol{\mathit{ r}}'\vert}d\boldsymbol{\mathit{ r}}'\biggr]\chi_r^{'}(\boldsymbol{\mathit{ r}})=\lambda_r^{'}\chi_r^{'}(\boldsymbol{\mathit{ r}}')$

Question

Why are the spin components of the total wavefunction orthogonal?

Answer

From the article on Hermitian operators, we know that two eigenfunctions of a Hermitian operator that correspond to different eigenvalues are orthogonal. Since the spin components of the total wavefunction $\alpha$ and $\beta$ correspond to different eigenvalues of the spin Hermitian operator $\hat S_z$ , they are orthogonal.

Hence the spin orbitals reduce to spatial orbitals. This implies that we only need to work with spatial orbitals, such as Slater-type orbitals, when we use the Hartree-Fock method to estimate the ground state energies of atoms.

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