Skip to article frontmatterSkip to article content
Site not loading correctly?

This may be due to an incorrect BASE_URL configuration. See the MyST Documentation for reference.

Hartree–Fock theory

Hartree–Fock equation

In the Hartree–Fock approximation, the many-electron wave function takes the form of a Slater determinant

Ψ=1N!ψ1(r1)ψN(r1)ψ1(rN)ψN(rN)| \Psi \rangle = \frac{1}{\sqrt{N!}} \begin{vmatrix} \psi_{1}(\mathbf{r}_1) & \cdots & \psi_{N}(\mathbf{r}_1) \\ \vdots & \ddots & \vdots \\ \psi_{1}(\mathbf{r}_N) & \cdots & \psi_{N}(\mathbf{r}_N) \\ \end{vmatrix}

where ψi\psi_i are the single-electron wave functions known as spin orbitals. The Hartree–Fock energy and the associated state is found by minimizing the energy functional

EHF=minψE[ψ]E_\mathrm{HF} = \min_{\psi} E[\psi]

under the constraint that the spin orbitals remain orthonormal. Here, ψ\psi collectively refers to the entire set of NN spin orbitals. Such a constrained minimization is conveniently performed by means of the technique of Lagrange multipliers.

Lagrangian

In Hartree–Fock theory, we introduce the real-valued Lagrangian

L[ψ]=E[ψ]i,j=1Nεji(ψiψjδij)L[\psi] = E[\psi] - \sum_{i,j=1}^N \varepsilon_{ji} \big( \langle \psi_i | \psi_j \rangle - \delta_{ij} \big)

and search for the set of spin orbitals, ψ\psi, that results in a first variation that vanishes

δL=0\delta L = 0

Expressing the energy as the expectation value of the electronic Hamiltonian with respect to a Slater determinant and using the general expressions for matrix elements, we arrive at

δL=i=1Nδψih^ψi+i,j=1N(δψiψjg^ψiψjδψiψjg^ψjψiεjiδψiψj)+complex conjugate=i=1Nδψi(f^ψij=1Nεjiψj)+complex conjugate\begin{align*} \delta L & = \sum_{i=1}^N \langle \delta \psi_i | \hat{h} | \psi_i \rangle + \sum_{i,j=1}^N \big( \langle \delta \psi_i \psi_j | \hat{g} | \psi_i \psi_j\rangle - \langle \delta \psi_i \psi_j | \hat{g} | \psi_j \psi_i\rangle - \varepsilon_{ji} \langle \delta \psi_i | \psi_j \rangle \big) + \mbox{complex conjugate} \\ &= \sum_{i=1}^N \langle \delta \psi_i | \big( \hat{f} | \psi_i \rangle - \sum_{j=1}^N \varepsilon_{ji} | \psi_j \rangle \big) + \mbox{complex conjugate} \end{align*}

where we have introduced the one-electron Fock operator

f^=h^+j=1N(J^jK^j)\hat{f} = \hat{h} + \sum_{j=1}^N \big( \hat{J}_j - \hat{K}_j \big)

with

J^jψi=[e2ψj(r)24πε0rrd3r]ψiK^jψi=[e2ψj(r)ψi(r)4πε0rrd3r]ψj\begin{align*} \hat{J}_j | \psi_i \rangle & = \Big[ \int \frac{e^2 |\psi_j(\mathbf{r}')|^2}{4\pi\varepsilon_0 |\mathbf{r} - \mathbf{r}'|} d^3\mathbf{r}' \Big] | \psi_i \rangle \\ % \hat{K}_j | \psi_i \rangle & = \Big[ \int \frac{e^2 \psi_j^\dagger(\mathbf{r}')\psi_i(\mathbf{r}')}{4\pi\varepsilon_0 |\mathbf{r} - \mathbf{r}'|} d^3\mathbf{r}' \Big] | \psi_j \rangle \end{align*}

Since the first-order variation in the Lagrangian is required to vanish for general variations in the spin orbitals, we have shown that the Hartree–Fock solution is given by

f^ψij=1Nεjiψj=0\hat{f} | \psi_i \rangle - \sum_{j=1}^N \varepsilon_{ji} | \psi_j \rangle = 0

This equation is known as the Hartree–Fock equation and it to be solved for the spin orbitals and the associated Lagrange multipliers. We note that the matrix elements of the Fock operator equal the multipliers

fki=ψkf^ψi=j=1Nεjiψkψj=εkif_{ki} = \langle \psi_k | \hat{f} | \psi_i \rangle = \sum_{j=1}^N \varepsilon_{ji} \langle \psi_k | \psi_j \rangle = \varepsilon_{ki}

Canonical form

Apart from a trivial overall phase factor, unitary transformations among the occupied orbitals are shown to leave the Hartree–Fock wave function unchanged. We introduce a unitary transformation that diagonalizes the Hermitian Fock matrix

f=ψf^ψ=Uψf^ψU=UfU\mathbf{f}' = \langle \overline{\psi}' | \hat{f} | \overline{\psi}' \rangle = \mathbf{U}^\dagger \langle \overline{\psi} | \hat{f} | \overline{\psi} \rangle \mathbf{U} = \mathbf{U}^\dagger \mathbf{f} \, \mathbf{U}

We have here adopted the compact overline notation of orbitals. In this basis of canonical spin orbitals, the Hartree–Fock equation takes the form

f^ψi=εiψi\hat{f} | \psi_i \rangle = \varepsilon_{i} | \psi_i \rangle

which we recognize as an eigenvalue equation introducing the orbital energies, εi\varepsilon_{i}, as the eigenvalues of the Fock operator. With an infinite number of solutions to the Hartree–Fock equation, the Hartree–Fock ground state is given by employing the NN spin orbitals with lowest orbital energies in the Slater determinant.

In AO basis

The spatial parts of the spin orbitals, or molecular orbitals (MOs), are expanded as linear combination of atomic orbitals (LCAO). In the basis of spin atomic orbitals, the Fock matrix becomes block diagonal

F=(Fαα00Fββ)\mathbf{F} = \begin{pmatrix} \mathbf{F}^{\alpha\alpha} & \mathbf{0} \\ \mathbf{0} & \mathbf{F}^{\beta\beta} \end{pmatrix}

Using the bar notation to distinguish α\alpha- and β\beta-spin atomic orbitals, we get

Fμν=Fμναα=hμν+γδ(Dγδ(μνγδ)Dγδα(μδγν))Fμˉνˉ=Fμνββ=hμν+γδ(Dγδ(μνγδ)Dγδβ(μδγν))Fμνˉ=Fμˉν=0\begin{align*} F_{\mu\nu} & = F^{\alpha\alpha}_{\mu\nu} = h_{\mu\nu} + \sum_{\gamma\delta} \Big( D_{\gamma\delta}(\mu\nu|\gamma\delta) - D^\alpha_{\gamma\delta}(\mu\delta|\gamma\nu) \Big) \\ F_{\bar{\mu}\bar{\nu}} & = F^{\beta\beta}_{\mu\nu} = h_{\mu\nu} + \sum_{\gamma\delta} \Big( D_{\gamma\delta}(\mu\nu|\gamma\delta) - D^\beta_{\gamma\delta}(\mu\delta|\gamma\nu) \Big) \\ F_{\mu\bar{\nu}} & = F_{\bar{\mu}\nu} = 0 \end{align*}

where

Dγδ=Dγδα+DγδβDγδα=j=1Nα[cγjα]cδjα;Dγδβ=j=1Nβ[cγjβ]cδjβ\begin{align*} D_{\gamma\delta} &= D^\alpha_{\gamma\delta} + D^\beta_{\gamma\delta} \\ D^\alpha_{\gamma\delta}& = \sum_{j=1}^{N_\alpha} \big[c_{\gamma j}^\alpha\big]^* c_{\delta j}^\alpha ; \quad D^\beta_{\gamma\delta} = \sum_{j=1}^{N_\beta} \big[c_{\gamma j}^\beta\big]^* c_{\delta j}^\beta \\ \end{align*}

The canonical Hartree–Fock equation thereby takes the form

FC=SCε\mathbf{F C} = \mathbf{S C} \boldsymbol{\varepsilon} \,

where S\mathbf{S} is the overlap matrix and ε\boldsymbol{\varepsilon} is a diagonal matrix collecting the orbital energies.

Hartree–Fock energy

For a given density D\mathbf{D}, the Hartree–Fock energy becomes equal to

EHF=12tr[(h+F)D]+VnnE_\mathrm{HF} = \frac{1}{2} \mathrm{tr} \big[ (\mathbf{h} + \mathbf{F}) \mathbf{D} \big] + V^\mathrm{n-n} \,

where VnnV^\mathrm{n-n} is the nuclear repulsion energy.

Koopmans theorem

The orbital energies of occupied and unoccupied orbitals, respectively, equal

εi=ψif^ψi=ψih^ψi+jiN(ψiJ^jψiψiK^jψi)εa=ψaf^ψa=ψah^ψa+j=1N(ψaJ^jψaψaK^jψa)\begin{align*} \varepsilon_i & = \langle \psi_i |\hat{f} | \psi_i \rangle = \langle \psi_i |\hat{h} | \psi_i \rangle + \sum_{j\neq i}^N \big( \langle \psi_i | \hat{J}_j | \psi_i \rangle - \langle \psi_i | \hat{K}_j | \psi_i \rangle \big) \\ \varepsilon_a & = \langle \psi_a |\hat{f} | \psi_a \rangle = \langle \psi_a |\hat{h} | \psi_a \rangle + \sum_{j=1}^N \big( \langle \psi_a | \hat{J}_j | \psi_a \rangle - \langle \psi_a | \hat{K}_j | \psi_a \rangle \big) \end{align*}

where the cancellation between Coulomb and exchange terms for j=ij=i has been used in the former case. It thus appears as if εi\varepsilon_i relates to the energy of an electron interacting with (N1)(N-1) other electrons, whereas εa\varepsilon_a relates to the energy of an electron interacting with NN other electrons. In accordance with these observations, it is readily shown from the expressions for matrix elements that the ionization energy (IE) and electron affinity (EA) become

IE=EiN1EHFN=εiEA=EHFNEaN+1=εa\begin{align*} \mathrm{IE} &= E_i^{N-1} - E_\mathrm{HF}^N = - \varepsilon_i \\ \mathrm{EA} &= E_\mathrm{HF}^N - E_a^{N+1} = - \varepsilon_a \\ \end{align*}

where, in the frozen orbital approximation, EiN1E_i^{N-1} is the energy of the system after the removal of the electron in spin orbital ii and EaN+1E_a^{N+1} is the energy of the system after the addition of an electron in spin orbital aa.

Brillouin theorem

Based on the expressions for matrix elements, we find

ΨHFH^Ψia=ψif^ψa=0\langle \Psi_\mathrm{HF} | \hat{H} | \Psi_i^a \rangle = \langle \psi_i | \hat{f} | \psi_a \rangle = 0

which shows that there is no coupling between the Hartree–Fock ground state and single excited determinants. This result is known as the Brillouin theorem.

SCF procedure

Due to the summation over occupied spin orbitals that expresses the effective electron interactions, the Fock operator depends on its eigenfunctions and the canonical Hartree–Fock equation is therefore solved iteratively by means of a self-consistent field (SCF) procedure, such as the Roothaan–Hall approach:

Roothaan–Hall scheme

In the following, we will consider the spin-restricted formulation where α\alpha- and β\beta-spin orbitals have identical spatial parts. We also restrict the situation to the common case of a closed-shell system such that

Nα=Nβ=12NDγδα=Dγδβ=12Dγδ=j=1N/2cγjcδj\begin{align*} N_\alpha & = N_\beta = \frac{1}{2} N \\ D^\alpha_{\gamma\delta} & = D^\beta_{\gamma\delta} = \frac{1}{2} D_{\gamma\delta} = \sum_{j=1}^{N/2} c_{\gamma j}^* c_{\delta j} \end{align*}