Hartree-Fock Hamiltonian and higher-order terms

Question

I'm diving into Hartree-Fock methods, and I'm confused on why the Hartree-Fock Hamiltonian reduces into a single particle Hamiltonian.

When applying Wick's theorem to the Fermi Sea vacuum, we use the particle-hole picture which transforms the Hamiltonian into

$$ H = \sum_i h_{ii} + \sum_{ij} \frac{1}{2} \left[ \langle ij \rVert V \rVert ij \rangle - \langle ij \rVert V \rVert ji \rangle \right] + \sum_{pq} \left[ h_{pq} + \sum_i \left[ \langle pi \rVert V \rVert qi \rangle - \langle pi \rVert V \rVert iq \rangle \right] \right] : c_p^\dagger c_q : + \frac{1}{2} \sum_{pqrs} \langle pq \rVert V \rVert rs \rangle : c_p^\dagger c_q^\dagger c_s c_r:. $$

where $ : :$ denotes normal ordering in the particle-hole picture in a Fermi sea vacuum.

The first term is the Hartree-Fock energy, which we want to minimize. The second term, which is the Fock matrix, gives the single particle/hole excitations. Variational principle tells us that the self-consistency method will involve diagonalizing the Fock Matrix, so we can expect a full SCF will give a diagonal one-particle term corresponding to the single quasiparticle energy $\varepsilon_p \delta_{pq}$.

The typical literature for Hartree-Fock states that applying mean field to the two-body interaction we reduced the Hamiltonian to a diagonal single particle Hamiltonian, which is true if we look until this order. e.g., http://felix.physics.sunysb.edu/%7Eallen/555-07/Mean-field-theory.pdf. However, I do not understand why are we ignoring the two-particle normal ordered interaction (third term).

Am I missing something in my understanding? Any help on this will be appreciated!

Answer 1

In most cases, the last term cannot be neglected, as it contains the important electron correlation effects. This term is usually taken into account after the mean-field procedure, via e.g. many-body perturbation theory (MBPT). To make the notations easier, let us write the Hamiltonian as $$ H=E_\text{vac}+\sum_{p,q}f^q_p :\hspace{-1mm}a^p_q\hspace{-1mm}:+\frac{1}{4}\sum_{p,q,r,s}\tilde{v}^{qs}_{pr} :\hspace{-1mm}a^{pr}_{qs}\hspace{-1mm}: \ , $$ where $$ :\hspace{-1mm}a^p_q\hspace{-1mm}:=:\hspace{-1mm}a^{{\color{black}\dagger}}_pa^{{\color{white}\dagger}}_q\hspace{-1mm}: \ \ , \ \ :\hspace{-1mm}a^{pr}_{qs}\hspace{-1mm}:=:\hspace{-1mm}a^{{\color{black}\dagger}}_pa^{{\color{black}\dagger}}_ra^{{\color{white}\dagger}}_sa^{{\color{white}\dagger}}_q\hspace{-1mm}: \ , $$ $$ E_\text{vac}=\sum_{i}h^i_i+\frac{1}{2}\sum_{i,j}\tilde{v}^{ij}_{ij} \ , $$ and the Fockian matrix elements are $$ f^q_p=h^q_p+\sum_{i}\tilde{v}^{qi}_{pi} \ \ , \ \ \tilde{v}^{qs}_{pr}=v^{qs}_{pr}-v^{qs}_{rp} \ . $$ Indices $p,q,r,s$ stand for general spin orbital indices, while indices $i,j,k,l$ and $a,b,c,d$ denote occupied and virtual orbitals, respectively (hole and particle states).

Note that you can choose in principle any kind of molecular orbitals in the above equations, as the second-quantized Hamiltonian does not care about the underlying one-particle basis (you are not tied to Hartree-Fock, you could choose Kohn-Sham orbitals, etc.). Of course, you have to build all integrals $h^q_p$, $v^{qs}_{pr}$, $f^q_p$ (and consequently the Fermi vacuum energy $E_\text{vac}$) with the chosen orbitals.

Let us introduce the MBPT partitioning $H=H^{(0)}+W$, where $$ H^{(0)}=E_\text{vac}+\sum_{p}f^p_p :\hspace{-1mm}a^p_p\hspace{-1mm}: \ \ , \ \ W=\sum_{p\neq q}f^q_p :\hspace{-1mm}a^p_q\hspace{-1mm}:+\frac{1}{4}\sum_{p,q,r,s}\tilde{v}^{qs}_{pr} :\hspace{-1mm}a^{pr}_{qs}\hspace{-1mm}: \ . $$ The eigenvalue problem of the one-body operator $H^{(0)}$ can be solved using the Wick theorem and the definition of normal ordering: $$ \begin{aligned} H^{(0)}|\Phi\rangle&=E_\text{vac}|\Phi\rangle \ , \\ H^{(0)}:\hspace{-1mm}a^a_i\hspace{-1mm}:|\Phi\rangle&=\Big(E_\text{vac}+f^a_a-f^i_i\Big):\hspace{-1mm}a^a_i\hspace{-1mm}:|\Phi\rangle \ , \\ H^{(0)}:\hspace{-1mm}a^{ab}_{ij}\hspace{-1mm}:|\Phi\rangle&=\Big(E_\text{vac}+f^a_a+f^b_b-f^i_i-f^j_j\Big):\hspace{-1mm}a^{ab}_{ij}\hspace{-1mm}:|\Phi\rangle \ , \end{aligned} $$ and so on, where $|\Phi\rangle$ is the Fermi vacuum that is used to define the normal ordering. If canonical Hartree-Fock orbitals are used, then $f^q_p=\delta^q_p\varepsilon_p$, and the zeroth-order excited state energies take the probably more familiar form expressed with HF orbital energies; also, in that case, the first term of the perturbation operator vanishes.

The zeroth-order energy in the Rayleigh-Schrödinger PT is just $E^{(0)}=\langle\Phi|H^{(0)}|\Phi\rangle=E_\text{vac}$, while the first-order energy correction vanishes because of the normal ordering: $E^{(1)}=\langle\Phi|W|\Phi\rangle=0$. The second-order correction can be found with the Wick theorem: $$ \begin{aligned} E^{(2)}&=-\sum_{i,a}\frac{\langle\Phi|W:\hspace{-1mm}a^a_i\hspace{-1mm}:|\Phi\rangle\langle\Phi|:\hspace{-1mm}a^i_a\hspace{-1mm}:W|\Phi\rangle}{E_\text{vac}+f^a_a-f^i_i-E_\text{vac}} -\frac{1}{4}\sum_{i,j,a,b}\frac{\langle\Phi|W:\hspace{-1mm}a^{ab}_{ij}\hspace{-1mm}:|\Phi\rangle\langle\Phi|:\hspace{-1mm}a^{ij}_{ab}\hspace{-1mm}:W|\Phi\rangle}{E_\text{vac}+f^a_a+f^b_b-f^i_i-f^j_j-E_\text{vac}} \\ &= -\sum_{i,a}\frac{f^a_if^i_a}{f^a_a-f^i_i} -\frac{1}{4}\sum_{i,j,a,b}\frac{\tilde{v}^{ab}_{ij}\tilde{v}^{ij}_{ab}}{f^a_a+f^b_b-f^i_i-f^j_j} \ , \end{aligned} $$ and so on for higher orders. For HF orbitals, $E_\text{vac}=E_\text{HF}$ and $f^i_a=0$, and we get the usual Moller-Plesset formula: $$ E=E_\text{HF}-\frac{1}{4}\sum_{i,j,a,b}\frac{\tilde{v}^{ab}_{ij}\tilde{v}^{ij}_{ab}}{\varepsilon_a+\varepsilon_b-\varepsilon_i-\varepsilon_j}+ \, \text{higher orders} \ . $$

For a more detailed discussion on diagrammatic MBPT and other many-body methods (e.g. MBPT resummation methods like Coupled Cluster), see Many-Body Methods in Chemistry and Physics by Shavitt & Bartlett.