3
$\begingroup$

I have a question about exercise 2.1 in Szabo and Ostlund's Modern Quantum Chemistry, which concerns the inner product between different spin orbitals: $\require{begingroup}\begingroup\newcommand{\br}{\mathbf{r}}\newcommand{\bx}{\mathbf{x}}$

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?

$\endgroup$
0
4
$\begingroup$

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: $$ \langle \psi_i^{\alpha} \alpha | \psi_j^{\alpha } \alpha \rangle \, , \quad \langle \psi_i^{\alpha} \alpha | \psi_j^{\beta } \beta \rangle \, , \quad \langle \psi_i^{\beta} \beta | \psi_j^{\beta } \beta \rangle \, . $$

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.