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I'm reading McQuarrie's Physical chemistry textbook [1]. On p. 281, the author explains:

If we use a more flexible trial function of the form in which $\psi(\mathbf{r_1},\mathbf{r_2})$ is a product of one-electron functions, or orbitals,

$$\psi(\mathbf{r_1},\mathbf{r_2})= \phi(\mathbf{r_1})\phi(\mathbf{r_2})\tag{8.15}$$

and allow $\phi(r)$ to be completely general, then we reach a limit that is both practical and theoretical. […]

This limiting value is the best value of the energy that can be obtained using trial function of the form of a product of one-electron wave equations (Equation 8.15). This limit is called the Hartree-Fock limit.

But why does Hartree–Fock limit exist? If we increase the number of parameters of the trial function we use, we may expect better calculation results. Nevertheless, why is the wave function multiplied by the two trial functions limited to the accuracy that can be achieved?

My guess is that the functional form does not properly consider the interelectronic correction term by treating the two trial functions independently. But can't this difference be overcome by using more parameters? I wonder if I can theoretically explain the existence of the Hartree–Fock limit.

In addition, if this limit exists, the function will have to be altered to produce better calculation results. I wonder what theory computational chemistry has developed about this.

Reference

McQuarrie, D. A.; Simon, J. D. Physical Chemistry: A Molecular Approach; University Science Books: Sausalito, California, 1997. ISBN 978-0-935702-99-6.

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  • $\begingroup$ This question might be very well suited for mattermodeling.stackexchange.com. $\endgroup$
    – schneiderfelipe
    Commented Jul 5, 2020 at 15:23
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    $\begingroup$ @Felipe S. S. Schneider I agree that mattermodeling is more specialized community related to the concept of my question. But first of all, because the book that I was asking was related to physical chemistry, I put it up on here. (It seems to span the boundary.) If I can't receive any informative answers, I'd post it on there. Thanks for letting me know such website :) $\endgroup$
    – Arete
    Commented Jul 5, 2020 at 16:41
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    $\begingroup$ HF is variational; that ensures that the energy is higher of equal to the exact energy of the system. Couple this with the fact that HF cannot reproduce all of the correlation energy, electron-electron interactions are treated in a mean field approximation, you can see, that the energy of HF does converge to a higher value than the exact energy. Advances are, to name but a few, perturbation theory, configuration interaction, and other so called post-HF methods. That is a broad field though and a more detailed text on computational chemistry so be more helpful than this site. $\endgroup$
    – Martin - マーチン ♦
    Commented Jul 5, 2020 at 18:19

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The issue is: what parameters can we introduce? As for changing the form of the trial wavefunction to something else to achieve a lower energy, this is prohibited by the variational principle, which HF theory satisfies.

In short (and within the context of Hartree-Fock), the variational principle guarantees that any given trial function will produce an energy equal to or greater than the solution provided by Hartree-Fock.

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