What is Nesbet's theorem? 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?
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

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

$\langle \Psi_0|\hat{H}| \Psi_{a}^{r}\rangle = 0$  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.