Slater Determinants

Every ab initio electronic structure package — Gaussian, ORCA, Psi4, Q-Chem, VASP for periodic solids — builds its wavefunctions from Slater determinants at the foundational layer, whether the final method is Hartree–Fock itself, density functional theory (which still typically uses a single determinant of Kohn–Sham orbitals to represent the noninteracting reference system), configuration interaction, or coupled-cluster theory (built as an exponential operator acting on a reference determinant). Drug discovery pipelines that use DFT to estimate binding energies of candidate inhibitors, materials-design codes that predict battery electrode voltages or semiconductor band gaps, and spectroscopy software that predicts UV-Vis or NMR spectra all rest on this same determinantal skeleton somewhere beneath the surface. The exchange energy that the determinant generates automatically is also the physical basis of exchange-correlation functionals in DFT (the "exchange" half of "exchange-correlation" is literally named for this effect), of ferromagnetism in solid-state physics (parallel electron spins on adjacent atoms lower exchange energy, favoring aligned magnetic moments — the quantum mechanical seed of the Heisenberg exchange model used to describe magnetic materials), and of the periodic table's shell structure itself, since the Pauli-forced impossibility of double occupation is what forces electrons into successively higher shells as atomic number increases, generating the aufbau pattern that in turn organizes chemical periodicity, valence, and reactivity trends taught from the very first day of general chemistry.

Ask the tutor: How do I count and classify all the distinct Slater determinants for a given electron configuration to build up its atomic term symbols?

Start with the simplest nontrivial case: two electrons, one in orbital $\chi_a$ and one in orbital $\chi_b$, where each $\chi$ denotes a spin-orbital — a spatial orbital multiplied by a spin function $\alpha$ or $\beta$. The naive product wavefunction $\chi_a(1)\chi_b(2)$ assigns electron 1 to orbital $a$ and electron 2 to orbital $b$. But electrons are indistinguishable; no measurement can tell "electron 1" from "electron 2." Assigning them labels this way is a bookkeeping fiction, and the wavefunction must respect that fiction's limits: under the exchange operator $\hat P_{12}$ that swaps all coordinates of electrons 1 and 2, the probability density $|\Psi|^2$ must be unchanged, which allows $\Psi$ to either stay the same (bosons) or flip sign (fermions). Electrons obey the second rule. The wavefunction must satisfy

This is the antisymmetry postulate, and it is more fundamental than — and implies — the exclusion principle. The simple product $\chi_a(1)\chi_b(2)$ fails this test: swapping labels gives $\chi_a(2)\chi_b(1)$, an unrelated function, not $-\chi_a(1)\chi_b(2)$. The fix is to antisymmetrize by hand, forming the combination

which by direct substitution does satisfy $\hat P_{12}\Psi = -\Psi$. This is exactly a $2\times2$ determinant, and the pattern generalizes: for $N$ electrons in $N$ spin-orbitals, the antisymmetrized wavefunction is the determinant of the matrix of spin-orbitals evaluated at each electron's coordinates, divided by $\sqrt{N!}$ for normalization. This is the Slater determinant, named for John C. Slater who introduced the determinantal form in 1929. The determinant structure has two built-in, automatic consequences that are worth seeing pictorially before the algebra. First, if any two spin-orbitals are identical — say $\chi_a=\chi_b$ — two rows (or columns) of the determinant become identical, and a determinant with two identical rows is exactly zero. No two electrons can share all four quantum numbers: this is the Pauli exclusion principle, derived rather than postulated. Second, the off-diagonal exchange term $-\chi_a(2)\chi_b(1)$ in the two-electron wavefunction only survives (does not integrate to zero against itself) when $\chi_a$ and $\chi_b$ share the same spin; this generates a purely quantum-mechanical spatial correlation between same-spin electrons, called Fermi correlation, with no explicit $1/r_{12}$ interaction required to produce it.

The rows of the matrix are electrons, the columns are spin-orbitals (or vice versa, by convention — swapping the convention only changes an overall sign, which normalization removes). Expanding the determinant by cofactors reproduces, term by term, every possible way of assigning the $N$ electrons to the $N$ spin-orbitals, each term carrying the sign of the permutation that generated it. This is precisely what "antisymmetrized sum over all assignments" means, and the determinant notation is simply the compact way of writing that sum with the correct signs pre-attached.

Fixing one electron with spin $\alpha$ at a point in space, the exchange term in the determinant forces the probability of finding a second $\alpha$-spin electron at the same point to vanish exactly — the "Fermi hole." An electron of opposite spin $\beta$ feels no such restriction from exchange (only from ordinary Coulomb repulsion, which is a separate, smaller effect at short range). This is a purely statistical effect of antisymmetry, present even for noninteracting electrons, and it is the reason exchange energy lowers the energy of high-spin configurations — the quantum mechanical seed of Hund's first rule.

For $N$ electrons occupying spin-orbitals $\chi_1,\chi_2,\ldots,\chi_N$, the normalized Slater determinant is

where:

• rows correspond to electrons (labeled $1,\ldots,N$, meaning "the coordinates, space and spin, of electron $i$"),
• columns correspond to the $N$ occupied spin-orbitals $\chi_p$,
• $1/\sqrt{N!}$ normalizes the determinant, assuming the spin-orbitals are themselves orthonormal.

Reading it: every term in the determinant's expansion is one particular way of assigning each electron to one spin-orbital, with a $+$ or $-$ sign fixed by the permutation parity; the sum over all $N!$ such assignments, with these signs, is exactly what "fully antisymmetrized" requires. Swapping any two rows (relabeling two electrons) swaps two rows of the matrix, which flips the sign of the determinant — antisymmetry is automatic, not imposed afterward. Setting any two columns equal (two electrons in the same spin-orbital) gives a zero determinant — the exclusion principle, likewise automatic. A shorthand notation writes only the diagonal, understanding the rest:

The energy of a single Slater determinant, for a Hamiltonian with one-electron (kinetic + nuclear attraction) terms $\hat h(i)$ and two-electron (repulsion) terms $1/r_{ij}$, is

where:

• $h_{ii} = \langle \chi_i(1)|\hat h(1)|\chi_i(1)\rangle$ is the one-electron energy (kinetic energy plus nuclear attraction) of spin-orbital $\chi_i$,
• $J_{ij} = \displaystyle\int\int |\chi_i(1)|^2 \frac{1}{r_{12}} |\chi_j(2)|^2\, d\tau_1 d\tau_2$ is the Coulomb integral — the ordinary electrostatic repulsion between the charge clouds of electrons in $\chi_i$ and $\chi_j$,
• $K_{ij} = \displaystyle\int\int \chi_i^*(1)\chi_j(1)\frac{1}{r_{12}}\chi_j^*(2)\chi_i(2)\, d\tau_1 d\tau_2$ is the exchange integral, nonzero only when $\chi_i$ and $\chi_j$ share the same spin, arising purely from antisymmetry,
• the $i=j$ terms in the double sum satisfy $J_{ii}=K_{ii}$ exactly, so they cancel and do not contribute a spurious self-interaction.

Reading it: the total electronic energy of an independent-electron (single-determinant) description is the sum of orbital energies, plus the net Coulomb repulsion between every pair of electrons, minus an exchange correction that lowers the energy specifically for pairs of electrons with parallel spin. The exchange term is a pure consequence of the determinant's antisymmetry — no such term exists for a simple (non-antisymmetrized) orbital product.

Example 1 — Foundational: helium ground state as an explicit Slater determinant, and confirming its energy formula. Helium's ground configuration is $1s^2$: both electrons in the spatial orbital $1s$, necessarily with opposite spins by the exclusion principle just derived. The two spin-orbitals are $\chi_1 = 1s(\mathbf r)\alpha(\omega)$ and $\chi_2 = 1s(\mathbf r)\beta(\omega)$, and the Slater determinant is

Because both electrons share the same spatial orbital, the spatial part factors out cleanly; the antisymmetry is carried entirely by the spin singlet function $\big[\alpha(1)\beta(2)-\alpha(2)\beta(1)\big]/\sqrt2$, which is itself antisymmetric under $1\leftrightarrow2$, while $1s(1)1s(2)$ is symmetric — product of symmetric-space and antisymmetric-spin gives overall antisymmetric, exactly as required. Applying the two-electron energy formula derived above, with $\chi_a=1s\alpha$, $\chi_b=1s\beta$:

Since both spin-orbitals share the identical spatial part $1s$, the Coulomb and exchange integrals become numerically identical ($J_{1s,1s}=K_{1s,1s}$, both computed from the same spatial density $|1s|^2$ against itself), but $K_{1s,1s}$ multiplies a spin-orbital overlap that — computed properly with the spin variables — actually corresponds to opposite spins, giving $K=0$ in the spin-orbital sense used for distinct-spin pairs. Concretely, using hydrogenic $1s$ orbitals with effective nuclear charge $Z_{\text{eff}}=27/16$ (the standard variational value refined for electron-electron screening) gives the well-known energy

More directly, the exact first-order (uncorrelated $Z=2$ hydrogenic orbitals) result is

which already captures the qualitative point: the total energy is the sum of two hydrogenic orbital energies ($-2$ hartree each for $Z=2$) plus a positive Coulomb repulsion term ($+5Z/8 = +1.25$ hartree), with exchange contributing zero because the two electrons have opposite spin. This is $37.9\%$ higher (less negative) than the experimental ground-state energy of $-2.9037$ hartree, the difference being electron correlation beyond what a single Slater determinant with hydrogenic orbitals can capture — precisely the missing physics that Hartree–Fock (optimized orbitals within the single-determinant ansatz) and post-Hartree–Fock methods are built to recover. Limiting case check: if the two electrons in $1s^2$ somehow had the same spin, the determinant would require identical spin-orbitals and would vanish outright — there is no triplet $1s^2$ state, consistent with every observed helium spectrum showing $1s^2\,{}^1S_0$ as the only $1s^2$ term.

Example 2 — Applied: exchange energy and Hund's rule ordering in the excited configuration $1s^12s^1$ of helium. Consider the excited configuration $1s^12s^1$, with one electron in $1s$ and one in $2s$, now with the freedom for the two electrons to be either spin-paired (singlet) or spin-parallel (triplet). Write $\chi_a = 1s\,\sigma_a$, $\chi_b=2s\,\sigma_b$. Using the two-electron energy formula,

Since $K_{1s,2s}>0$ always (it is a self-repulsion-type integral of an overlap density with itself through $1/r_{12}$, guaranteed positive), the triplet state lies lower in energy than the singlet by exactly $K_{1s,2s}$:

Using tabulated values for helium's $1s2s$ configuration, $K_{1s,2s}\approx 0.044$ hartree $\approx 1.20$ eV. Experimentally, the ${}^3S_1$ (triplet, parahelium's "orthohelium" partner) term lies at $19.82$ eV above the ground state while the ${}^1S_0$ singlet term lies at $20.62$ eV above ground — a splitting of $0.80$ eV, comparable in magnitude (same order, not identical, because real helium orbitals differ somewhat between the two states due to differing screening, and higher-order correlation is not captured by the bare exchange integral) to the exchange-integral estimate. The qualitative statement is exact and is Hund's first rule in its most microscopic form: the state of maximum spin multiplicity is stabilized specifically by exchange energy that a single Slater determinant already contains with no extra assumption. Limiting case check: as the $2s$ orbital is pushed to be spatially identical to $1s$ (a fictitious limit, since they are orthogonal, but useful as a check), $J_{1s,2s}\to J_{1s,1s}$ and $K_{1s,2s}\to J_{1s,1s}$ as well, so $E(\text{triplet}) \to h_{1s}+h_{2s}$, exactly recovering the earlier rule that two electrons cannot share one spatial+spin combination — the triplet built from a common spatial orbital for both electrons cannot exist, and indeed the singlet-triplet gap $K_{1s,2s}$ diverges to the full Coulomb self-energy as the orbitals coalesce, signaling the determinant is trying (and failing) to describe a forbidden state.

Example 3 — Competition level: expanding a three-electron Slater determinant explicitly and verifying antisymmetry under a non-adjacent transposition. Consider three electrons in spin-orbitals $\chi_1,\chi_2,\chi_3$ (lithium's $1s\alpha,1s\beta,2s\alpha$, say, though the labels are kept generic here). The determinant is

Expanding by cofactors along the first row:

This expands into the $3!=6$ terms, one per permutation of $(1,2,3)$ assigned to $(\chi_1,\chi_2,\chi_3)$, each carrying the sign of that permutation:

Now test antisymmetry explicitly under the non-adjacent transposition $1\leftrightarrow3$ (swap electrons 1 and 3, not neighbors in the electron ordering, a stronger check than swapping adjacent labels). Relabel every "$(1)$" as "$(3)$" and every "$(3)$" as "$(1)$" throughout the six-term expansion above:

Reorder each term alphabetically by spin-orbital index (using commutativity of ordinary multiplication of these functions) and compare term by term against the original six terms of $\sqrt6\,\Psi$. Term 1 of $\Psi'$, $\chi_1(3)\chi_2(2)\chi_3(1)$, rewritten as $\chi_3(1)\chi_2(2)\chi_1(3)$, matches term $-\chi_3(1)\chi_2(2)\chi_1(3)$ of the original list with an overall sign flip; carrying out the same matching for all six terms shows every single term of $\Psi'$ equals minus the corresponding term of $\Psi$. Hence

confirming antisymmetry under an arbitrary (not just adjacent) transposition, exactly as guaranteed by the general determinant identity that swapping any two rows multiplies the determinant by $-1$ — here demonstrated by explicit term-by-term bookkeeping rather than by invoking the determinant property abstractly. Limiting case check: setting $\chi_3 \to \chi_1$ (a forbidden repeated spin-orbital) in the six-term expansion causes the terms to cancel in pairs — term 1 $(\chi_1(1)\chi_2(2)\chi_1(3))$ cancels against term 6's negative $(-\chi_1(1)\chi_2(2)\chi_1(3))$ after the substitution, and similarly for the remaining two pairs — collapsing the entire sum to zero, consistent with the exclusion principle demonstrated algebraically in Example 1's two-electron case, now confirmed to hold at three-electron order as well.

John Clarke Slater (1900–1976) was an American physicist whose 1929 paper "The Theory of Complex Spectra" introduced the determinantal wavefunction that now bears his name, in the course of trying to systematize the messy business of assigning term symbols to multi-electron atoms. Slater's broader career ranged remarkably widely: he worked on the theory of chemical bonding (Slater-type orbitals, still a staple of quantum chemistry basis sets), on band theory of solids, on microwave electronics during World War II (he led the fundamental research division at the MIT Radiation Laboratory), and later on molecular and solid-state theory at MIT and the University of Florida. He was notoriously prolific and combative in defending the primacy of physical, quantitative reasoning over what he regarded as excessive formalism, and he pushed hard for treating quantum chemistry as continuous with atomic and solid-state physics rather than as a separate discipline — a stance the determinant, equally at home describing an atom, a molecule, or a solid's Bloch bands, exemplifies perfectly.