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 \( \langle \Phi_0\vert \Phi_0\rangle=1 \).

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 $$ \hat{Q}=\sum_{m=1}^{\infty}\vert \Phi_m\rangle \langle \Phi_m\vert . $$

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 \( \hat{H} \).

Actually, the above expression is nothing but a rewrite again of the full Schr\"odinger equation.

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.

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.

- MBPT introduces order-by-order specific correlations and we make comparisons with exact calculations like FCI
- At every order, we can calculate all contributions since they are well-known and either tabulated or calculated on the fly.
- MBPT is a non-variational theory and there is no guarantee that higher orders will improve the convergence.
- However, since FCI calculations are limited by the size of the Hamiltonian matrices to diagonalize (today's most efficient codes can attach dimensionalities of ten billion basis states, MBPT can function as an approximative method which gives a straightforward (but tedious) calculation recipe.
- MBPT has been widely used to compute effective interactions for the nuclear shell-model.
- But there are better methods which sum to infinite order important correlations. Coupled cluster theory is one of these methods.