# Szabo/Ostlund exercise on spin orbital orthonormality


Exercise 2.1. Given a set of $$K$$ orthonormal spatial functions, $$\{\psi_i^\alpha(\br)\}$$, and another set of $$K$$ orthonormal functions, $$\{\psi_j^\beta(\br)\}$$, such that the first set is not orthogonal to the second set, i.e.,

$$\int d\br \, \psi_i^{\alpha*}(\br)\psi_j^\beta(\br) = S_{ij}$$

where $$\mathbf{S}$$ is an overlap matrix, show that the set $$\{\chi_i\}$$ of $$2K$$ spin orbitals, formed by multiplying $$\psi_i^\alpha(\br)$$ by the $$\alpha$$ spin function and $$\psi_j^\beta(\br)$$ by the $$\beta$$ spin function, i.e.,

\begin{align} \chi_{2i-1}(\bx) &= \psi_i^\alpha(\br)\alpha(\omega) \\ \chi_{2i}(\bx) &= \psi_j^\beta(\br)\beta(\omega) \end{align}

with $$i = 1, 2, \ldots, k$$ is an orthonormal set.

I feel quite confused about this question. Consider the spin-orbital integral $$\langle \psi_i^{\alpha} \alpha (\omega) | \psi_j^{\beta } \alpha (\omega) \rangle = \langle \psi_i^{\alpha} | \psi_j^{\beta } \rangle$$ which is not necessarily orthogonal.

The result can only be orthogonal if both ket and bra vectors are in the $$i$$-indexed set. If so, why does the exercise mention a $$j$$-indexed set in the beginning?

Consider the spin-orbital integral $$\langle \psi_i^{\alpha} \alpha (\omega) | \psi_j^{\beta } \alpha (\omega) \rangle = \langle\psi_i^{\alpha} | \psi_j^{\beta } \rangle$$ which is not necessarily orthogonal.
Yes, it is not necessarily orthogonal, but the thing is that you don't need to consider such integrals in the exercise. Read more carefully: spatial orbitals from the first "alpha" set can be multiplied only by the $$\alpha$$ spin function while spatial orbitals from the second "beta" set can be multiplied only by the $$\beta$$ spin function. Thus, there are only three types of integrals you need to consider: $$\langle \psi_i^{\alpha} \alpha | \psi_j^{\alpha } \alpha \rangle \, , \quad \langle \psi_i^{\alpha} \alpha | \psi_j^{\beta } \beta \rangle \, , \quad \langle \psi_i^{\beta} \beta | \psi_j^{\beta } \beta \rangle \, .$$