---
title: Exchange, Slater Determinants, and Hartree-Fock
module: Many-Electron Atoms
moduleNumber: 5
lessonNumber: 3
order: 503
summary: >
  A product wave function ignores that electrons are identical fermions. Enforcing
  antisymmetry writes the state as a Slater determinant, which vanishes whenever
  two electrons share a spin-orbital — the exclusion principle made algebraic.
  The energy of a determinant carries a new term with no classical analogue, the
  exchange integral, nonzero only for parallel spins; it lowers the energy of
  aligned electrons and carves a Fermi hole around each one. Adding the exchange
  operator to the mean field gives the Hartree-Fock equations, and what they still
  miss defines the correlation energy.
topics: [Many-Electron Atoms]
sources:
  - book: Bransden & Joachain
    ref: "Ch. 7 — Many-Electron Atoms; §7.2–7.3 Antisymmetry, Slater Determinants, and the Hartree-Fock Equations"
  - book: Griffiths & Schroeter
    ref: "Ch. 5 — Identical Particles; §5.1–5.2 Two-Particle Systems and Exchange"
  - book: Foot
    ref: "Ch. 4 — The Central-Field Approximation and Exchange Symmetry"
draft: false
---

The [Hartree self-consistent field](/atomic-physics/many-electron-atoms/central-field-self-consistent)
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,
$\vec x_i = (\vec r_i, m_{s_i})$:[^bj-72]

$$
\Psi(\ldots,\vec x_i,\ldots,\vec x_j,\ldots)
= -\,\Psi(\ldots,\vec x_j,\ldots,\vec x_i,\ldots).
$$

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

$$
\Psi_A(\vec x_1,\vec x_2)
= \frac{1}{\sqrt2}\big[\psi_a(\vec x_1)\psi_b(\vec x_2) - \psi_a(\vec x_2)\psi_b(\vec x_1)\big].
$$

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:[^gs-51]

$$
\underbrace{\text{symmetric space}}_{\text{electrons close}} \otimes \text{singlet spin},
\qquad
\underbrace{\text{antisymmetric space}}_{\text{electrons apart}} \otimes \text{triplet spin}.
$$

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](/atomic-physics/many-electron-atoms/hund-rules-ground-terms).

$$
% caption: The two-electron spatial density along the line r₁ = r₂. The
% antisymmetric (triplet) combination is zero on the diagonal — a Fermi hole —
% while the symmetric (singlet) combination peaks there.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (6.4,0) node[right, black!70] {separation};
\draw[->, black] (0,0) -- (0,3.2) node[above, black!70] {pair density};
\draw[black, dashed] (0,0) -- (0,3.0);
\node[black, anchor=south east] at (0.0,3.0) {coincide};
% symmetric: peaks at zero separation
\draw[black, very thick] (0,2.7) .. controls (0.8,2.5) and (1.6,1.7) .. (2.6,1.05)
  .. controls (3.6,0.55) and (4.8,0.32) .. (6.1,0.25);
\node[black, anchor=west] at (2.6,1.25) {singlet (space symmetric)};
% antisymmetric: zero at coincidence, rises then falls
\draw[acc, very thick, dashed] (0,0) .. controls (0.7,0.9) and (1.3,1.75) .. (2.1,1.9)
  .. controls (3.1,2.05) and (4.4,0.9) .. (6.1,0.4);
\node[acc, anchor=west] at (2.3,2.05) {triplet (space antisymmetric)};
\fill[acc] (0,0) circle (1.8pt);
\node[acc, anchor=north west] at (0.05,-0.05) {Fermi hole};
\end{tikzpicture}
$$

## The Slater determinant

For $N$ electrons the antisymmetric combination of $N$ spin-orbitals is a
determinant. Given occupied spin-orbitals $\psi_1,\ldots,\psi_N$, write[^bj-72]

$$
\Psi(\vec x_1,\ldots,\vec x_N)
= \frac{1}{\sqrt{N!}}
\begin{vmatrix}
\psi_1(\vec x_1) & \psi_2(\vec x_1) & \cdots & \psi_N(\vec x_1)\\
\psi_1(\vec x_2) & \psi_2(\vec x_2) & \cdots & \psi_N(\vec x_2)\\
\vdots & \vdots & \ddots & \vdots\\
\psi_1(\vec x_N) & \psi_2(\vec x_N) & \cdots & \psi_N(\vec x_N)
\end{vmatrix}.
$$

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.

> **Definition (Slater determinant).** The normalized antisymmetric product of $N$
> orthonormal spin-orbitals, written as the determinant above. It is the simplest
> $N$-electron wave function consistent with fermion antisymmetry, and it vanishes
> identically whenever two of its spin-orbitals coincide.

$$
% caption: A Slater determinant as an N×N array: columns index spin-orbitals,
% rows index electrons. Swapping electrons swaps rows (sign flip); repeating an
% orbital repeats a column (determinant zero).
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth,
  cell/.style={draw, black, minimum width=12mm, minimum height=8mm}]
\definecolor{acc}{HTML}{4A6FA5}
\foreach \c/\lab in {0/{u_1},1/{u_2},2/{u_3}}{
  \node[cell] at (\c*1.3,2.6) {$\lab$};
}
\foreach \r/\lab in {0/1,1/2,2/3}{
  \node[cell] at (-1.5,1.3-\r*0.9) {elec \lab};
}
\foreach \r in {0,1,2}{
  \foreach \c in {0,1,2}{
    \pgfmathtruncatemacro{\orb}{\c+1}
    \pgfmathtruncatemacro{\ele}{\r+1}
    \node[cell] at (\c*1.3,1.3-\r*0.9) {$u_{\orb}(\ele)$};
  }
}
\node[black, anchor=north] at (1.3,-1.9) {rows = electrons, columns = orbitals};
\end{tikzpicture}
$$

## Direct and exchange integrals

The reward for antisymmetry appears in the energy. With a Hamiltonian
$H = \sum_i h(i) + \sum_{i<j} g(i,j)$, $g = e^2/4\pi\varepsilon_0 r_{ij}$,
evaluating $\langle\Psi|H|\Psi\rangle$ on a single Slater determinant gives[^bj-72]

$$
E = \sum_i I_i + \frac12\sum_{i,j}\big(J_{ij} - K_{ij}\big),
$$

where $I_i = \langle\psi_i|h|\psi_i\rangle$ is the one-body energy and the two
two-electron integrals are

$$
J_{ij} = \iint \psi_i^\ast(1)\psi_j^\ast(2)\,g\,\psi_i(1)\psi_j(2)\,\d^3r_1\,\d^3r_2,
$$
$$
K_{ij} = \iint \psi_i^\ast(1)\psi_j^\ast(2)\,g\,\psi_j(1)\psi_i(2)\,\d^3r_1\,\d^3r_2.
$$

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 $\langle\chi_i|\chi_j\rangle$ 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.

$$
% caption: Direct vs exchange. The direct integral J repels two fixed charge
% clouds; the exchange integral K swaps the orbitals between the two electrons and
% survives only for parallel spins.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth]
\definecolor{acc}{HTML}{4A6FA5}
% --- direct ---
\node[black!70] at (1.4,2.6) {direct $J_{ij}$};
\draw[black] (0.4,1.4) circle (7pt); \node[black] at (0.4,1.4) {$i$};
\draw[black] (2.4,1.4) circle (7pt); \node[black] at (2.4,1.4) {$j$};
\draw[black, thick, dashed] (0.65,1.4) -- (2.15,1.4);
\node[black, anchor=north] at (1.4,1.0) {cloud $i$ repels cloud $j$};
% --- exchange ---
\begin{scope}[xshift=5.6cm]
\node[black!70] at (1.4,2.6) {exchange $K_{ij}$};
\fill[acc!25] (0.4,1.4) circle (7pt); \node[acc] at (0.4,1.4) {$i$};
\fill[acc!25] (2.4,1.4) circle (7pt); \node[acc] at (2.4,1.4) {$j$};
\draw[acc, thick, ->] (0.65,1.6) .. controls (1.4,2.2) .. (2.15,1.6);
\draw[acc, thick, ->] (2.15,1.2) .. controls (1.4,0.6) .. (0.65,1.2);
\node[acc, anchor=north] at (1.4,0.55) {orbitals swap; parallel spins only};
\end{scope}
\end{tikzpicture}
$$

## 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:[^bj-73]

$$
\left[h + \sum_j\big(\mathcal J_j - \mathcal K_j\big)\right]\psi_i = \varepsilon_i\,\psi_i,
$$

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

$$
\mathcal J_j\,\psi_i(\vec r) = \left[\int \frac{e^2\,|\psi_j(\vec r')|^2}{4\pi\varepsilon_0\,|\vec r-\vec r'|}\,\d^3r'\right]\psi_i(\vec r),
$$
$$
\mathcal K_j\,\psi_i(\vec r) = \left[\int \frac{e^2\,\psi_j^\ast(\vec r')\,\psi_i(\vec r')}{4\pi\varepsilon_0\,|\vec r-\vec r'|}\,\d^3r'\right]\psi_j(\vec r).
$$

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 $\vec r'$, 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.

> **Definition (Fock operator).** The effective one-electron operator
> $f = h + \sum_j(\mathcal J_j - \mathcal K_j)$ whose eigenfunctions are the
> Hartree-Fock orbitals. It is solved self-consistently: the operator depends on
> the occupied orbitals, so one iterates guess-solve-rebuild exactly as in the
> Hartree cycle, now including the nonlocal exchange term.

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

## The exchange hole

Exchange reshapes the joint probability of finding two electrons. For two parallel
spins the pair density $P(\vec r,\vec r')$ vanishes as $\vec r'\to\vec r$: 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:

$$
\int \big[\rho_{\text{parallel}}(\vec r') - \rho(\vec r')\big]\,\d^3r' = -1.
$$

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.

$$
% caption: The Fermi (exchange) hole. Around a reference electron the density of
% parallel-spin electrons is depleted; the missing charge integrates to exactly
% one electron.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (-3.0,0) -- (3.2,0) node[right, black!70] {distance from electron};
\draw[->, black] (0,0) -- (0,3.0) node[above, black!70] {parallel density};
\draw[black, dashed] (-3.0,2.4) -- (3.0,2.4);
\node[black, anchor=south east] at (3.0,2.4) {mean};
% depleted near center (hole), rising to mean far out
\draw[black, very thick]
  (-3.0,2.35) .. controls (-2.2,2.25) and (-1.5,1.9) .. (-0.9,1.0)
  .. controls (-0.4,0.35) and (0,0.15) .. (0,0.12)
  .. controls (0,0.15) and (0.4,0.35) .. (0.9,1.0)
  .. controls (1.5,1.9) and (2.2,2.25) .. (3.0,2.35);
\fill[acc] (0,0.12) circle (2.0pt);
\node[acc, anchor=south] at (0,0.25) {Fermi hole};
\end{tikzpicture}
$$

## 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**:[^bj-73]

$$
E_{\text{corr}} = E_{\text{exact}} - E_{\text{HF}} < 0.
$$

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.

$$
% caption: Energy budget. Hartree-Fock captures the mean field and exchange but
% overestimates the energy; the exact state lies lower by the correlation energy.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (0,3.4) node[above, black!70] {energy};
% Hartree-Fock level (higher)
\draw[black, very thick] (0.6,2.6) -- (2.6,2.6);
\node[black, anchor=west] at (2.7,2.6) {Hartree-Fock};
% exact level (lower)
\draw[black, very thick, dashed] (0.6,1.1) -- (2.6,1.1);
\node[black, anchor=west] at (2.7,1.1) {exact};
% gap
\draw[acc, <->] (1.6,2.55) -- (1.6,1.15);
\node[acc, anchor=west] at (1.7,1.85) {correlation energy};
\end{tikzpicture}
$$

The chain of approximation is now complete: the
[central field](/atomic-physics/many-electron-atoms/central-field-self-consistent)
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](/atomic-physics/many-electron-atoms/helium-two-electron-atom).

[^bj-72]: **Bransden & Joachain**, _Physics of Atoms and Molecules_, 2nd ed., §7.2 — antisymmetry under combined space-spin exchange, the singlet/triplet spatial-spin pairing, the Slater determinant and its row/column properties, and the direct and exchange integrals in $\langle\Psi|H|\Psi\rangle$. <https://www.pearson.com/en-gb/subject-catalog/p/physics-of-atoms-and-molecules/P200000005386>
[^gs-51]: **Griffiths & Schroeter**, _Introduction to Quantum Mechanics_, 3rd ed., §5.1 — two-particle systems, symmetric and antisymmetric combinations, and the spatial correlation that keeps antisymmetric-space electrons apart.
[^bj-73]: **Bransden & Joachain**, _Physics of Atoms and Molecules_, 2nd ed., §7.3 — the variational derivation of the Hartree-Fock equations, the local direct and nonlocal exchange operators, the Fock operator, Koopmans' theorem, the Fermi hole, and the definition of correlation energy $E_{\text{exact}}-E_{\text{HF}}$.