We assume here that we are only interested in the ground state of the system and expand the exact wave function in term of a series of Slater determinants
$$
\vert \Psi_0\rangle = \vert \Phi_0\rangle + \sum_{m=1}^{\infty}C_m\vert \Phi_m\rangle,
$$
where we have assumed that the true ground state is dominated by the
solution of the unperturbed problem, that is
$$
\hat{H}_0\vert \Phi_0\rangle= W_0\vert \Phi_0\rangle.
$$
The state \( \vert \Psi_0\rangle \) is not normalized, rather we have used an intermediate
normalization \( \langle \Phi_0 \vert \Psi_0\rangle=1 \) since we have \( \langle \Phi_0\vert \Phi_0\rangle=1 \).
$$
\hat{H}\vert \Psi_0\rangle = E\vert \Psi_0\rangle,
$$
and multiplying the latter from the left with \( \langle \Phi_0\vert \) gives
$$
\langle \Phi_0\vert \hat{H}\vert \Psi_0\rangle = E\langle \Phi_0\vert \Psi_0\rangle=E,
$$
and subtracting from this equation
$$
\langle \Psi_0\vert \hat{H}_0\vert \Phi_0\rangle= W_0\langle \Psi_0\vert \Phi_0\rangle=W_0,
$$
and using the fact that the both operators \( \hat{H} \) and \( \hat{H}_0 \) are hermitian
results in
$$
\Delta E=E-W_0=\langle \Phi_0\vert \hat{H}_I\vert \Psi_0\rangle,
$$
which is an exact result. We call this quantity the correlation energy.
Here we have assumed that our model space defined by the operator \( \hat{P} \) is one-dimensional, meaning that
$$
\hat{P}= \vert \Phi_0\rangle \langle \Phi_0\vert ,
$$
and
$$
\hat{Q}=\sum_{m=1}^{\infty}\vert \Phi_m\rangle \langle \Phi_m\vert .
$$
$$
\vert \Psi_0\rangle= (\hat{P}+\hat{Q})\vert \Psi_0\rangle=\vert \Phi_0\rangle+\hat{Q}\vert \Psi_0\rangle.
$$
Going back to the Schr\"odinger equation, we can rewrite it as, adding and a subtracting a term \( \omega \vert \Psi_0\rangle \) as
$$
\left(\omega-\hat{H}_0\right)\vert \Psi_0\rangle=\left(\omega-E+\hat{H}_I\right)\vert \Psi_0\rangle,
$$
where \( \omega \) is an energy variable to be specified later.
$$
\left(\omega-\hat{H}_0\right)^{-1}=\frac{1}{\left(\omega-\hat{H}_0\right)}.
$$
We can rewrite Schroedinger's equation as
$$
\vert \Psi_0\rangle=\frac{1}{\omega-\hat{H}_0}\left(\omega-E+\hat{H}_I\right)\vert \Psi_0\rangle,
$$
and multiplying from the left with \( \hat{Q} \) results in
$$
\hat{Q}\vert \Psi_0\rangle=\frac{\hat{Q}}{\omega-\hat{H}_0}\left(\omega-E+\hat{H}_I\right)\vert \Psi_0\rangle,
$$
which is possible since we have defined the operator \( \hat{Q} \) in terms of the eigenfunctions of \( \hat{H} \).
$$
\hat{Q}\frac{1}{\left(\omega-\hat{H}_0\right)}\hat{Q}=\hat{Q}\frac{1}{\left(\omega-\hat{H}_0\right)}=\frac{\hat{Q}}{\left(\omega-\hat{H}_0\right)}.
$$
With these definitions we can in turn define the wave function as
$$
\vert \Psi_0\rangle=\vert \Phi_0\rangle+\frac{\hat{Q}}{\omega-\hat{H}_0}\left(\omega-E+\hat{H}_I\right)\vert \Psi_0\rangle.
$$
This equation is again nothing but a formal rewrite of Schr\"odinger's equation
and does not represent a practical calculational scheme.
It is a non-linear equation in two unknown quantities, the energy \( E \) and the exact
wave function \( \vert \Psi_0\rangle \). We can however start with a guess for \( \vert \Psi_0\rangle \) on the right hand side of the last equation.
$$
\vert \Psi_0\rangle=\sum_{i=0}^{\infty}\left\{\frac{\hat{Q}}{\omega-\hat{H}_0}\left(\omega-E+\hat{H}_I\right)\right\}^i\vert \Phi_0\rangle,
$$
for the wave function and
$$
\Delta E=\sum_{i=0}^{\infty}\langle \Phi_0\vert \hat{H}_I\left\{\frac{\hat{Q}}{\omega-\hat{H}_0}\left(\omega-E+\hat{H}_I\right)\right\}^i\vert \Phi_0\rangle,
$$
which is now a perturbative expansion of the exact energy in terms of the interaction
\( \hat{H}_I \) and the unperturbed wave function \( \vert \Psi_0\rangle \).
$$
\Delta E=\sum_{i=0}^{\infty}\langle \Phi_0\vert \hat{H}_I\left\{\frac{\hat{Q}}{\omega-\hat{H}_0}\left(\omega-E+\hat{H}_I\right)\right\}^i\vert \Phi_0\rangle=
$$
$$
\langle \Phi_0\vert \left(\hat{H}_I+\hat{H}_I\frac{\hat{Q}}{E-\hat{H}_0}\hat{H}_I+
\hat{H}_I\frac{\hat{Q}}{E-\hat{H}_0}\hat{H}_I\frac{\hat{Q}}{E-\hat{H}_0}\hat{H}_I+\dots\right)\vert \Phi_0\rangle.
$$
$$
\Delta E=\sum_{i=0}^{\infty}\langle \Phi_0\vert \hat{H}_I\left\{\frac{\hat{Q}}{\omega-\hat{H}_0}\left(\omega-E+\hat{H}_I\right)\right\}^i\vert \Phi_0\rangle=$$
$$
\langle \Phi_0\vert \left(\hat{H}_I+\hat{H}_I\frac{\hat{Q}}{E-\hat{H}_0}\hat{H}_I+
\hat{H}_I\frac{\hat{Q}}{E-\hat{H}_0}\hat{H}_I\frac{\hat{Q}}{E-\hat{H}_0}\hat{H}_I+\dots\right)\vert \Phi_0\rangle.
$$
This expression depends however on the exact energy \( E \) and is again not very convenient from a practical point of view. It can obviously be solved iteratively, by starting with a guess for \( E \) and then solve till some kind of self-consistency criterion has been reached.
Actually, the above expression is nothing but a rewrite again of the full Schr\"odinger equation.
$$
\hat{Q}\frac{1}{\hat{e}-\hat{Q}\hat{H}_I\hat{Q}}=
$$
$$
\hat{Q}\left[\frac{1}{\hat{e}}+\frac{1}{\hat{e}}\hat{Q}\hat{H}_I\hat{Q}
\frac{1}{\hat{e}}+\frac{1}{\hat{e}}\hat{Q}\hat{H}_I\hat{Q}
\frac{1}{\hat{e}}\hat{Q}\hat{H}_I\hat{Q}\frac{1}{\hat{e}}+\dots\right]\hat{Q}.
$$
$$
\Delta E=
\langle \Phi_0\vert \hat{H}_I+\hat{H}_I\hat{Q}\frac{1}{E-\hat{H}_0-\hat{Q}\hat{H}_I\hat{Q}}\hat{Q}\hat{H}_I\vert \Phi_0\rangle.
$$
In RS perturbation theory we set \( \omega = W_0 \) and obtain the following expression for the energy difference
$$
\Delta E=\sum_{i=0}^{\infty}\langle \Phi_0\vert \hat{H}_I\left\{\frac{\hat{Q}}{W_0-\hat{H}_0}\left(\hat{H}_I-\Delta E\right)\right\}^i\vert \Phi_0\rangle=
$$
$$
\langle \Phi_0\vert \left(\hat{H}_I+\hat{H}_I\frac{\hat{Q}}{W_0-\hat{H}_0}(\hat{H}_I-\Delta E)+
\hat{H}_I\frac{\hat{Q}}{W_0-\hat{H}_0}(\hat{H}_I-\Delta E)\frac{\hat{Q}}{W_0-\hat{H}_0}(\hat{H}_I-\Delta E)+\dots\right)\vert \Phi_0\rangle.
$$
$$
\hat{Q}\Delta E\vert \Phi_0\rangle = \hat{Q}\Delta E\vert \hat{Q}\Phi_0\rangle = 0.
$$
Inserting this results in the expression for the energy results in
$$
\Delta E=\langle \Phi_0\vert \left(\hat{H}_I+\hat{H}_I\frac{\hat{Q}}{W_0-\hat{H}_0}\hat{H}_I+
\hat{H}_I\frac{\hat{Q}}{W_0-\hat{H}_0}(\hat{H}_I-\Delta E)\frac{\hat{Q}}{W_0-\hat{H}_0}\hat{H}_I+\dots\right)\vert \Phi_0\rangle.
$$
$$
\Delta E=\sum_{i=1}^{\infty}\Delta E^{(i)}.
$$
We get the following expression for \( \Delta E^{(i)} \)
$$
\Delta E^{(1)}=\langle \Phi_0\vert \hat{H}_I\vert \Phi_0\rangle,
$$
which is just the contribution to first order in perturbation theory,
$$
\Delta E^{(2)}=\langle\Phi_0\vert \hat{H}_I\frac{\hat{Q}}{W_0-\hat{H}_0}\hat{H}_I\vert \Phi_0\rangle,
$$
which is the contribution to second order.
$$
\Delta E^{(3)}=\langle \Phi_0\vert \hat{H}_I\frac{\hat{Q}}{W_0-\hat{H}_0}\hat{H}_I\frac{\hat{Q}}{W_0-\hat{H}_0}\hat{H}_I\Phi_0\rangle-
\langle\Phi_0\vert \hat{H}_I\frac{\hat{Q}}{W_0-\hat{H}_0}\langle \Phi_0\vert \hat{H}_I\vert \Phi_0\rangle\frac{\hat{Q}}{W_0-\hat{H}_0}\hat{H}_I\vert \Phi_0\rangle,
$$
being the third-order contribution.
In the shell-model lectures we showed that we could rewrite the exact state function for say the ground state, as a linear expansion in terms of all possible Slater determinants. That is, we define the ansatz for the ground state as
$$
|\Phi_0\rangle = \left(\prod_{i\le F}\hat{a}_{i}^{\dagger}\right)|0\rangle,
$$
where the index \( i \) defines different single-particle states up to the Fermi level. We have assumed that we have \( N \) fermions.
A given one-particle-one-hole (\( 1p1h \)) state can be written as
$$
|\Phi_i^a\rangle = \hat{a}_{a}^{\dagger}\hat{a}_i|\Phi_0\rangle,
$$
while a \( 2p2h \) state can be written as
$$
|\Phi_{ij}^{ab}\rangle = \hat{a}_{a}^{\dagger}\hat{a}_{b}^{\dagger}\hat{a}_j\hat{a}_i|\Phi_0\rangle,
$$
and a general \( ApAh \) state as
$$
|\Phi_{ijk\dots}^{abc\dots}\rangle = \hat{a}_{a}^{\dagger}\hat{a}_{b}^{\dagger}\hat{a}_{c}^{\dagger}\dots\hat{a}_k\hat{a}_j\hat{a}_i|\Phi_0\rangle.
$$
We can then expand our exact state function for the ground state as
$$
|\Psi_0\rangle=C_0|\Phi_0\rangle+\sum_{ai}C_i^a|\Phi_i^a\rangle+\sum_{abij}C_{ij}^{ab}|\Phi_{ij}^{ab}\rangle+\dots
=(C_0+\hat{C})|\Phi_0\rangle,
$$
where we have introduced the so-called correlation operator
$$
\hat{C}=\sum_{ai}C_i^a\hat{a}_{a}^{\dagger}\hat{a}_i +\sum_{abij}C_{ij}^{ab}\hat{a}_{a}^{\dagger}\hat{a}_{b}^{\dagger}\hat{a}_j\hat{a}_i+\dots
$$
Since the normalization of \( \Psi_0 \) is at our disposal and since \( C_0 \) is by hypothesis non-zero, we may arbitrarily set \( C_0=1 \) with
corresponding proportional changes in all other coefficients. Using this so-called intermediate normalization we have
$$
\langle \Psi_0 | \Phi_0 \rangle = \langle \Phi_0 | \Phi_0 \rangle = 1,
$$
resulting in
$$
|\Psi_0\rangle=(1+\hat{C})|\Phi_0\rangle.
$$
How can we use perturbation theory to determine the same coefficients? Let us study the contributions to second order in the interaction, namely
$$
\Delta E^{(2)}=\langle\Phi_0\vert \hat{H}_I\frac{\hat{Q}}{W_0-\hat{H}_0}\hat{H}_I\vert \Phi_0\rangle.
$$
The intermediate states given by \( \hat{Q} \) can at most be of a \( 2p-2h \) nature if we have a two-body Hamiltonian. This means that second order in the perturbation theory can have \( 1p-1h \) and \( 2p-2h \) at most as intermediate states. When we diagonalize, these contributions are included to infinite order. This means that higher-orders in perturbation theory bring in more complicated correlations.
$$
\Delta E^{(2)}=\frac{1}{4}\sum_{abij}\langle ij\vert \hat{v}\vert ab\rangle \frac{\langle ab\vert \hat{v}\vert ij\rangle}{\epsilon_i+\epsilon_j-\epsilon_a-\epsilon_b}.
$$
$$
E-E_0 =\Delta E=
\sum_{abij}\langle ij | \hat{v}| ab \rangle C_{ij}^{ab},
$$
where the energy \( E_0 \) is the reference energy and \( \Delta E \) defines the so-called correlation energy.
We see that if we set
$$
C_{ij}^{ab} =\frac{1}{4}\frac{\langle ab \vert \hat{v} \vert ij \rangle}{\epsilon_i+\epsilon_j-\epsilon_a-\epsilon_b},
$$
we have a perfect agreement between FCI and MBPT. However, FCI includes such \( 2p-2h \) correlations to infinite order. In order to make a meaningful comparison we would at least need to sum such correlations to infinite order in perturbation theory.
Codes for computing effective Hamiltonians for the shell model as well as many other codes can be obtained from the CENS website.