Are there cases where Hartree-Fock is not a good approximation to compute equilibrium geometry when the molecule is in a non bond-breaking condition?

  • $\begingroup$ I would image that cases where some bonds are much stronger than other (i.e. protein folding) would present some special challenges, but I haven't every tried a case like that (I learned Hartree-Fock in a nuclear context where those issues do not arise). $\endgroup$ – dmckee --- ex-moderator kitten Feb 19 '12 at 22:41
  • $\begingroup$ Hartree-Fock should be just fine for proteins except for the computational cost. $\endgroup$ – Jeff Hammond Apr 13 '13 at 20:31

No, there are several cases where the approximation becomes unphysical and inaccurate. To name a few I'm aware of:

  • excited molecule states, basis functions are typically optimized to describe ground states. Configuration Interaction (CI) methods are rather used here. HF covers ground states best.

  • electron correlation, especially if correlation changes with internuclear separation. HF assumes independent electrons. There are Post-HF-Methods for taking this correlation into account based on ordinary HF, e.g. Møller-Plesset many-body perturbation theory

For not small to medium-sized molecules (>20 Atoms) less time-consuming semi-empirical or hybrid (HF+DFT) instead of pure HF methods are used.


Every van-der-Waals bonded molecule like H_2 is not covered by Hartree Fock Theory. Electron Correlation is not considered. So HF is just a good starting point for post HF methods like Møller-Plesset perturbation theory, coupled cluster etc.

Statistically the best available method for the ground state in modern quantum chemistry programs is CCSDT, coupled cluster with single, doublet and triplet exitation.This methodes are time consuming and scale with $N^7$ to $N^8$ where $N$ is the number of gaussian basis functions (GTOs)

Explicit correlation (F12) methods are even better, but they scale horrible.

In HF the ground state has to be given by a single Slater determinat. So even the ground state may not be computeable with Hartree Fock theory.

You must have a good guess for the HF starting point. Test your program with e.g. ozone. It has a broken symmetry unrestricted HF singlet wavefunction. Most likely you converge to a higher state in HF.


Hartree Fock will be unreliable for multiconfigurational systems e.g. involving transition metals or when dispersion interactions are significant (as noted by Alex1167623).


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.