- #1
HeavyMetal
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Homework Statement
[/B]
Taken straight out of Szabo and Ostlund's "Quantum Chemistry" problem 2.1:
Given a set of K orthonormal spatial functions, [itex]\{\psi_{i}^{\alpha}(\mathbf{r})\}[/itex], and another set of K orthonormal functions, [itex]\{\psi_{i}^{\beta}(\mathbf{r})\}[/itex], such that the first set is not orthogonal to the second set, i.e.,
[tex]\int d\mathbf{r}\ \psi_{i}^{\alpha*}(\mathbf{r})\psi_{j}^{\beta}(\mathbf{r}) = S_{ij}[/tex]
where [itex]\mathbf{S}[/itex] is an overlap matrix, show that the set [itex]\{\chi_{i}\}[/itex] of 2K spin orbitals, formed by multiplying [itex]\{\psi_{i}^{\alpha}(\mathbf{r})\}[/itex] by the [itex]\alpha[/itex] spin function and [itex]\{\psi_{i}^{\beta}(\mathbf{r})\}[/itex] by the [itex]\beta[/itex] spin function, i.e.,
[tex]\left.\begin{aligned}
\chi_{2i-1}(\mathbf{x}) = \psi_{i}^{\alpha}(\mathbf{r})\alpha(\omega)\\
\chi_{2i}(\mathbf{x}) = \psi_{i}^{\beta}(\mathbf{r})\beta(\omega)
\end{aligned}
\right\}
\qquad \text{i = 1, 2, ..., k}[/tex]
is an orthonormal set.
Homework Equations
The above given data, and [possibly] these identities:
[tex]\int d\omega\ \alpha^*(\omega)\alpha(\omega) = \langle\alpha\|\alpha\rangle = 1[/tex]
[tex]\int d\omega\ \alpha^*(\omega)\beta(\omega) = \langle\alpha\|\beta\rangle = 0[/tex]
[tex]\int d\mathbf{x}\ \chi_{i}^*(\mathbf{x})\chi_{j}(\mathbf{x}) = \langle\chi_{i}\|\chi_{j}\rangle = \delta_{ij}[/tex]
The Attempt at a Solution
First, I insert two 1's stealthily into the first equation to get:
[tex]\int d\mathbf{r}\ \psi_{i}^{\alpha*}(\mathbf{r})*1*1*\psi_{j}^{\beta}(\mathbf{r})\\
= \int\int d\mathbf{r}\ d\omega\ \psi_{i}^{\alpha*}(\mathbf{r})\alpha^*(\omega)\alpha(\omega)\beta^*(\omega)\beta(\omega)\psi_{j}^{\beta}(\mathbf{r})[/tex]
Then, I insert the definition of the spin orbitals:
[tex]\int \int d\mathbf{x}\ d\omega\ \chi_{i}^{\alpha*}(\mathbf{x}) \alpha(\omega)\beta^*(\omega) \chi_{j}^{\beta}(\mathbf{x})[/tex]
I then get lost and, searching for a solution equal to a Kronecker delta, I say that it rearranges to:
[tex]\int \int d\mathbf{x}\ d\omega\ \chi_{i}^{\alpha*}(\mathbf{x})\chi_{j}^{\beta}(\mathbf{x})\alpha(\omega)\beta^*(\omega)\\
= \int d\omega\ \langle\chi_{i}^{\alpha}\|\chi_{j}^{\beta}\rangle\alpha(\omega)\beta^*(\omega)\\
= \int d\omega\ \delta_{ij}\ \alpha(\omega)\beta^*(\omega)\\
= \delta_{ij}\ * 0[/tex]
But this answer gives me zero, not just the Kronecker delta that I am searching for...
I am very new to index notation, so I am sorry if this is trivial!
As always, thanks in advance!
-HeavyMetal