Exchange, Slater Determinants, and Hartree-Fock

The Hartree self-consistent field wrote the atom as a product of orbitals, $\Psi =u_1(1)u_2(2)\cdots u_N(N)$. A product labels the electrons — it says electron $1$ is in orbital $u_1$ — and identical electrons carry no labels. The Hartree ansatz therefore violates the one exact symmetry of the problem, and the physics it drops, exchange, is not a small correction: it splits the helium spectrum by an electronvolt, sets the sign of Hund's rules, and underlies ferromagnetism. This lesson rebuilds the mean field with antisymmetry enforced from the start.

Antisymmetry and the exchange operator

The total wave function of $N$ electrons must change sign under the exchange of any two, where exchange swaps both the spatial and spin coordinates, ${\vec{x}}_i=({\vec{r}}_i,m_{s_i})$:¹

For two electrons the requirement factors cleanly. A product of two spin-orbitals is neither symmetric nor antisymmetric, but the antisymmetric combination is:

Because space and spin separate for a spin-independent Hamiltonian, the spatial part is symmetric or antisymmetric and the spin part carries the opposite symmetry. Two electrons have four spin states: a symmetric triplet ($S=1$, three $m_S$) and an antisymmetric singlet ($S=0$). Antisymmetry of the whole pairs them oppositely:²

The spatial correlation is the physical content. In the triplet the spatial wave function vanishes when ${\vec{r}}_1={\vec{r}}_2$, so parallel-spin electrons keep apart; in the singlet they may sit together. Since keeping apart lowers the Coulomb repulsion, the triplet lies lower for a fixed spatial configuration — the seed of the first Hund rule.

The Slater determinant

For $N$ electrons the antisymmetric combination of $N$ spin-orbitals is a determinant. Given occupied spin-orbitals ${\psi }_1,\dots ,{\psi }_N$, write¹

The determinant does the bookkeeping automatically. Swapping two electrons exchanges two rows, which flips the sign — antisymmetry is built in. Putting two electrons in the same spin-orbital makes two columns equal, and a determinant with two equal columns is zero — the Pauli exclusion principle is now a theorem of linear algebra, not an extra rule.

Direct and exchange integrals

The reward for antisymmetry appears in the energy. With a Hamiltonian $H={\sum }_{i}h(i)+{\sum }_{i<j}g(i,j)$, $g=e^2/4\pi {\epsilon }_0{r}_{ij}$, evaluating $⟨\Psi |H|\Psi ⟩$ on a single Slater determinant gives¹

where $I_i=⟨{\psi }_i|h|{\psi }_i⟩$ is the one-body energy and the two two-electron integrals are

The direct (Coulomb) integral $J_{ij}$ is the electrostatic repulsion between the two charge clouds $|{\psi }_i{|}^2$ and $|{\psi }_j{|}^2$ — the classical, intuitive term already present in the Hartree method. The exchange integral $K_{ij}$ has the orbitals swapped between the two factors on the right; it has no classical interpretation, since it involves the product ${\psi }_i^{\ast }(1){\psi }_i(2)$ that mixes the electron at $1$ with the electron at $2$. Two properties fix its role:

• Spin selectivity. Because $\psi =(\text{spatial})\times (\text{spin})$, the exchange integral carries a factor $⟨{\chi }_i|{\chi }_j⟩$ from the spins in the swapped positions. It is nonzero only when electrons $i$ and $j$ have parallel spins. The direct integral has no such restriction.
• No self-interaction. For $i=j$ the two integrals are equal, $J_{ii}=K_{ii}$, so the diagonal terms cancel in $J_{ij}-K_{ij}$. A determinant contains no spurious repulsion of an electron with itself — a flaw the Hartree method carries and Hartree-Fock removes.

The Hartree-Fock equations

Making $E$ stationary with respect to each orbital, subject to orthonormality, produces the Hartree-Fock equations. They differ from the Hartree equations by one operator, the exchange term:³

with the direct and exchange operators defined by their action on any orbital:

The direct operator ${\mathcal{J}}_j$ is local: it multiplies ${\psi }_i(\vec{r})$ by a potential evaluated at the same point $\vec{r}$. The exchange operator ${\mathcal{K}}_j$ is nonlocal: its value at $\vec{r}$ depends on ${\psi }_i$ at all other points ${\vec{r}}^{\prime }$, because the orbital being acted on appears inside the integral. That nonlocality is what makes Hartree-Fock harder to solve than Hartree, and it is the exact price of antisymmetry.

By Koopmans' theorem, the orbital eigenvalue gives the ionization energy from that orbital in the frozen-orbital approximation, ${\text{IE}}_i\approx -{\epsilon }_i$.

The exchange hole

Exchange reshapes the joint probability of finding two electrons. For two parallel spins the pair density $P(\vec{r},{\vec{r}}^{\prime })$ vanishes as ${\vec{r}}^{\prime }\to \vec{r}$: no two same-spin electrons occupy the same point. Around any electron, the density of other parallel-spin electrons is depleted in a region called the Fermi hole (or exchange hole). Integrating the depletion over all space removes exactly one electron's worth of charge:

The hole is the physical meaning of exchange: an electron pushes same-spin neighbours aside, and the reduced repulsion lowers the energy. Antiparallel electrons feel no such hole in Hartree-Fock — the theory's central omission.

Correlation energy

Hartree-Fock is the best single-determinant approximation, but a single determinant cannot describe the instantaneous avoidance of antiparallel electrons, whose repulsion is treated only in the mean. The difference between the exact non-relativistic energy and the Hartree-Fock energy defines the correlation energy:³

It is always negative — the exact state finds extra ways to keep electrons apart that a determinant cannot represent. In atoms $E_{\text{corr}}$ is a few percent of the total energy but often comparable to the chemical energies of interest, so quantitative work adds correlation on top of Hartree-Fock through configuration interaction (a sum of many determinants) or perturbation theory.

The chain of approximation is now complete: the central field gives orbitals and $\ell$-dependent energies; Hartree-Fock adds exchange and the Fermi hole; correlation is the residual the mean field cannot reach. The next lesson applies all three to the simplest nontrivial case, the two electrons of helium.