Prove Brillouin's theorem; that is, show that hamiltonian matrix elements between the HF wavefunction \(\Phi_{0}\) and singly excited determinants are identically zero. Hint. Use the Condon-Slater rules.

In this step, the Hartree-Fock (HF) wave function \( \Phi_{0} \) and the singly excited determinants \( \Phi_{a}^{r} \) are defined. Here, \( \Phi_{0} \) is a determinant where each molecular orbital is occupied by two electrons, while \( \Phi_{a}^{r} \) represents the determinant where one electron is promoted from a molecular orbital a to a higher molecular orbital r.

According to the Condon-Slater rules, the integral of an electron-nuclear attraction energy between two determinants is zero unless the two determinants differ by at most a single spin-orbital, both in the placement of electrons and in the spatial parts of the spin orbitals. Hence, it is clear that the integral over a one-electron operator between \( \Phi_{0} \) and \( \Phi_{a}^{r} \) is zero, because the two determinants differ in two orbitals: a and r. Similarly, the integral over the two-electron operator between \( \Phi_{0} \) and \( \Phi_{a}^{r} \) is zero because the two determinants differ in two orbitals. Therefore, by the Condon-Slater rules, the matrix element of the Hamiltonian between \( \Phi_{0} \) and \( \Phi_{a}^{r} \) is zero.

Since the matrix element of the Hamiltonian between \( \Phi_{0} \) and \( \Phi_{a}^{r} \) has been proven to be zero using the Condon-Slater rules, this implies that the Hamiltonian matrix elements between the HF wave function and singly excited determinants are identically zero, as stated by Brillouin's theorem. This concludes the proof of Brillouin's theorem.