Suppose we have a parametrized molecular Hamiltonian $H(R)$, where $R$ are the nuclear coordinates. If we let $B(R)$ be some chosen atomic basis set (like STO-3G) centred at $R$, then using $B(R)$, we let $C(R)$ denote the transition matrix from atomic to molecular orbitals at $R$.

In general, there are many different choices for $C(R)$ at a given $R$, as HF can converge to different unique solutions. However, I want $C(R)$ to vary continuously over some range of $R$. In general, is it possible to enforce this condition when performing Hartree-Fock? If so, how?

I tried starting at a set of parameters $R_0$, computed $C(R_0)$, and then computed $C(R_0 + \Delta)$, $C(R_0 + 2\Delta), ...$ iteratively by using the density matrix yielded by the previous Hartree-Fock step, but this didn't seem to work: I'm still getting some large jump discontinuities in $C(R)$ at certain values of $R$.

  • $\begingroup$ Are the molecular orbitals canonical orbitals, or are they allowed to become non-canonical? $\endgroup$
    – wzkchem5
    Jul 26, 2022 at 15:54

1 Answer 1


First, notes on terminology: ${\bf C}$ are molecular orbital coefficients, not transition matrices. In general one has many nuclei so you have a set of nuclear coordinates $\{{\bf R}_A\}$ instead of a single vector ${\bf R}$; atomic basis sets are typically (but not always!) centered on the individual atoms ${\bf R}_A$.

The actual question does not in fact depend on the basis set at all; instead, the question is whether the Hartree-Fock orbitals are continuous across geometries. The answer is negative: some solutions disappear when the molecular geometry is changed.

Tracking the solutions individually can be done e.g. using holomorphic Hartree-Fock theory, which is reviewed in e.g. the Q-Chem manual.

  • $\begingroup$ +10 this answer thoroughly addresses several points, and yet it's still brief and concise. An all around very well-written answer! $\endgroup$ Jul 26, 2022 at 20:10
  • $\begingroup$ Thank you, this helps a lot! $\endgroup$ Jul 27, 2022 at 14:50

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