What is Nesbet's theorem?

Question

In a talk by Frank Neese there was a slide with title "Nesbet's theorem" and an equation indicating something like "if your wavefunction has the double excitation amplitudes correct, then you can get the correct FCI energy for free". The caveat being that to get the doubles exactly correct you need the triples and quadruples, and to get the quadruples you need the pentuples and hexuples, etc.

Unfortunately I am not able to find this theorem stated precisely anywhere with a Google search. I have been told that the theorem can be found in the book Molecular Electronic Structure Theory by Helgaker, Jorgensen, and Olsen, however I do not have the book and need the precise statement of this theorem.

Does anyone know the precise statement of Nesbet's theorem and the original source?

Answer 1

Section 4.1.1 (p 237) of Szabo and Ostlund's Modern Quantum Chemistry also describes this issue.

Using intermediate normalisation the correlation energy can be written as

$$\begin{align} \langle \Psi_0|\hat{H}-E_0| \Phi_0\rangle &= \langle \Psi_0|\hat{H}-E_0 \left(| \Psi_0\rangle + \sum_{ar}c_a^r |\Psi_a^r \rangle + \sum_{abrs}c_{ab}^{rs}|\Psi_{ab}^{rs} \rangle +.... \right) \\ &= \sum_{abrs}c_{ab}^{rs}\langle \Psi_0|\hat{H}| \Psi_{ab}^{rs}\rangle \end{align}$$

$\langle \Psi_0|\hat{H}| \Psi_{a}^{r}\rangle = 0$ according to Brillouin's theorem. $$\langle \Psi_0|\hat{H}| \Psi_{abc}^{rst}\rangle = \langle \Psi_0|\hat{H}| \Psi_{ab}^{rs}\rangle \langle \phi_c | \phi_t \rangle = 0$$

and similarly for higher excitations, because $\hat{H}$ is a 2-electron operator.

Answer 2

It's Nesbet's theorem, and you can find it in equation (41) of

Nesbet, R. K. Electronic Correlation in Atoms and Molecules. In Advances in Chemical Physics; Prigogine, I., Ed.; Wiley: New York, 1965; Vol. 9, pp 321–363. DOI: 10.1002/9780470143551.ch4.

$$E_\text{corr} = \frac{1}{4}\sum_{ijab} c^{ab}_{ij}\langle ij || ab \rangle$$

Note Nesbet uses the more general two-body operator $R$ and does not have the prefactor. See slide 7 on this talk by Neese (PDF).

Answer 3

The other answers were very helpful, especially in telling me the correct spelling of Nesbet! With the correct spelling a quick search gave this paper: ChemPhysChem 2017, 18 (23), 3478 which I found to be the most helpful (for me) since it also defines all the notation:

Now let us build an analogous definition for $e_\mathrm c(\mathbf{r})$ in the case of a CI wavefunction $\Psi$ represented by a linear combination of the ground-state determinant $\Phi_0$ and excited determinants $\Phi_i^a$, $\Phi_{ij}^{ab}$, etc.:

$$\Psi = C_0\Phi_0 + \sum_i^{\text{occ}}\sum_a^{\text{virt}}C_i^a\Phi_i^a + \sum_{i<j}^{\text{occ}}\sum_{a<b}^{\text{virt}}C_i^{ab}\Phi_{ij}^{ab} + \cdots$$

In this case, the correlation energy is conveniently provided by Nesbet's theorem:

$$E_c^{\text{CI}} = \sum_{i<j}^{\text{occ}}\sum_{a<b}^{\text{virt}} c_{ij}^{ab} \left< \Phi_0 \middle| \hat{H} \middle| \Phi_{ij}^{ab} \right>$$

where intermediate normalisation is assumed: $c_{ij}^{ab} = C_{ij}^{ab}/C_0$.

So if you have the correct doubles amplitudes of an FCI wavefunction, you can calculate the correlation energy easily using the integrals corresponding to the Hartree–Fock determinant and the double-excitation determinants.