Is the Hartree-Fock determinant always the minimum energy single Slater determinant solution to a molecular Hamiltonian?

Question

Is it always true that the Hartree-Fock slater determinant is the minimum energy single slater determinant solution to a given molecular Hamiltonian $H$?

By construction, any determinant $|\chi_{i_1}\ldots \chi_{i_N}\rangle$ where the $\chi_{i_j}$ are distinct solutions of the Hartree-Fock equation $f(x) \chi(x) = \epsilon_i \chi(x)$ will have stationary energy $\langle \chi_{i_1}\ldots \chi_{i_N} | H |\chi_{i_1}\ldots \chi_{i_N}\rangle$ (where stationarity is with respect to perturbing the orbitals). The Hartree-Fock slater determinant $|\Phi_0\rangle =|\chi_1\ldots\chi_N\rangle$ is constructed by choosing the lowest $N$ energy solutions of the Hartree-Fock equation.

Now, $|\chi_{i_1}\ldots \chi_{i_N}\rangle$ is an eigenstate of the Hartree-Fock Hamiltonian $H^{HF}$ with energy $\sum_{i=1}^N \epsilon_{i_j}$, so constructing $|\Phi_0\rangle$ from the $N$ lowest energy solutions of the HF equation clearly minimizes the energy of $H^{HF}$. But the molecular Hamiltonian $H$ is given by $$H = H^{HF} + \sum_{i=1}^N\sum_{j>i}^N \frac{1}{r_{ij}} - \sum_{i=1}^N v^{HF}(i)$$ where $v^{HF}$ is the HF potential. This satisfies $$\langle{\Phi_0}|H|\Phi_0\rangle = \sum_{i=1}^N \epsilon_i - \frac{1}{2}\sum_{i=1}^N \sum_{j=1}^N \langle ij|\,|ij\rangle $$ where $$\langle ij|\,|ij\rangle:=\int dx_1 dx_2 \chi_i^*(x_1)\chi_j^*(x_2)\frac{1}{r_{12}}(\chi_i(x_1)\chi_j(x_2) - \chi_j(x_1)\chi_i(x_2))$$ Due to the $\langle ij|\,|ij\rangle$ term, it is not clear to me that constructing $\Phi_0$ from the $N$ lowest energy solutions of the HF equations still results in the lowest energy single slater determinant for $H$. An excited determinant would have a higher energy contribution to the $\sum_i \epsilon_i$ term, but could this is offset by a greater deduction from the $\langle ij|\,|ij\rangle$ term?

Answer 1

By construction, any determinant $|\chi_{i_1}\ldots \chi_{i_N}\rangle$ where the $\chi_{i_j}$ are distinct solutions of the Hartree-Fock equation $f(x) \chi(x) = \epsilon_i \chi(x)$ will have stationary energy $\langle \chi_{i_1}\ldots \chi_{i_N} | H |\chi_{i_1}\ldots \chi_{i_N}\rangle$ (where stationarity is with respect to perturbing the orbitals).

This is not true: only the converged Hartree-Fock determinant $|\Phi_0\rangle = |\chi_{i_1}\ldots \chi_{i_N}\rangle$ has stationary energy with respect to orbital perturbations.

Stationarity of energy requires all orbitals in the determinant to fulfill the Hartree-Fock equations $$ \hat f \chi_i = \varepsilon_i \chi_i $$ where the Fock operator $$\hat{f} = \hat h + \sum_j^N(\hat{J_j} - \hat{K_j})$$ depends on the $N$ occupied orbitals $\{\chi_j\}$ included in the determinant.

The Hartree-Fock determinant $|\Phi_0\rangle$ fulfills this condition and is therefore stationary with respect to orbital perturbation, because the constituting orbitals are converged (in SCF iterations) solutions to these equations, where the Fock operator is calculated from the $N$ orbitals with lowest energies.

Any other (excited) determinant $|\Psi\rangle$ is generally not stationary with respect to perturbing the orbitals. Since other orbitals are occupied in this determinant, it corresponds to a different Fock operator $\hat {f'}$ where the sum is over the orbitals in the determinant $|\Psi\rangle$ which differs from the set of orbitals included in $|\Phi_0\rangle$.

Using this Fock operator $\hat{f'}$ in subsequent SCF cycles, the total energy of the excited determinant is again minimised by tweaking the orbitals. In this course, I suppose, it is quite likely that the excited determinant relaxes back to the ground state Hartree-Fock determinant $|\Phi_0\rangle$. The iterations might also converge to a state with higher energy, but one-determinant representations of excited states are rather inaccurate anyway.

In principle, in should be possible for an excited determinant to converge to a determinant with lower energy if and only if the initial converged solution is merely a local minimum. However, this is very unlikely if using a reasonable initial guess.