The correlation energy of a system is defined as the difference between the exactly energy and the energy in the Hartree-Fock method: $E_\mathrm{cor} = E - E_\mathrm{HF}$. In the case of an atom or a molecule, is it possible say that $E_\mathrm{cor} < 0$, since the interaction between the electrons is positive?

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    $\begingroup$ Being nit-picky, there are systems in which HF provides the exact answer and thus $E_{\text{cor}}$ is zero and thus not negative. $\endgroup$ – TAR86 Sep 30 '19 at 10:46

The answer to this question rests on the fact that Hartree-Fock is a variational method. A variational method is one where you supply a guess wavefunction ($|\Psi\rangle$) with some parameter(s) ($\alpha$) and then use those parameter(s) to minimize the energy. Doing this, you are guaranteed that your resulting energy is an upper bound to the exact ground state energy. $$ E_{\text{exact}} \leq \frac{\langle\Psi_{\alpha}|\hat{H}|\Psi_{\alpha}\rangle}{\langle\Psi_{\alpha}|\Psi_{\alpha}\rangle} $$

So, if we are talking about the exact correlation energy (such as one given by full CI) then the answer is that the correlation energy must always be negative. This is because Hartree-Fock always provides an upper bound to the exact ground state energy. However, if we use a non-variational method to approximate the correlation energy (such as coupled cluster) then we could find cases where we calculate a positive correlation energy, but this is distinct from the exact correlation energy.

(For an authoritative description of this in the context of quantum chemistry with examples and practice problems, I recommend the first chapter of Szabo and Ostlund's Modern Quantum Chemistry)


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