Slater Determinants
A Slater determinant is the standard compact wavefunction for identical fermions occupying one-particle spin-orbitals. It builds antisymmetry into the wavefunction by using a determinant:
Here denotes all one-particle degrees of freedom, including spin. The displayed prefactor normalizes the state when the orbitals are orthonormal; nonorthogonal orbitals require the Gram determinant below. In quantum chemistry they are often called spin-orbitals.
Slater determinants are the workhorse representation for many-electron wavefunctions. They encode fermionic antisymmetry exactly, while leaving the choice of one-particle orbitals to physics, approximation, or computation.
Helpful background. Symmetric and Antisymmetric Wavefunctions gives the two-particle case. Occupation-Number Basis helps interpret the filled orbitals.
Determinants of occupied spin-orbitals
Section titled “Determinants of occupied spin-orbitals”Let be normalized one-particle states in a Hilbert space . The corresponding antisymmetrized vector, normalized only when the orbitals are orthonormal, is
In the coordinate-spin representation, this is the determinant above:
The rows correspond to particle slots . The columns correspond to occupied one-particle spin-orbitals .
The ordering of the columns fixes a sign convention. Swapping two columns changes the determinant by a minus sign, which changes the state vector by an overall phase . The occupied set of spin-orbitals is the same physical occupation; the sign must simply be handled consistently.
Antisymmetry from Determinants
Section titled “Antisymmetry from Determinants”Exchange two particle slots, say and . In the determinant, this swaps rows and . A determinant changes sign under a row swap, so
Therefore the Slater determinant lies in the antisymmetric fermionic sector.
This is the same exchange rule used in Symmetric and Antisymmetric Wavefunctions, now written in a form that scales to arbitrary .
Pauli Exclusion in Determinant Form
Section titled “Pauli Exclusion in Determinant Form”If two occupied one-particle spin-orbitals are identical, two columns of the determinant are identical. The determinant then vanishes:
More generally, if the occupied one-particle states are linearly dependent, the determinant is zero. This is Pauli exclusion in determinant form: an -fermion determinant needs linearly independent complete one-particle states.
For electrons, “complete” includes spin. Two electrons may share a spatial orbital only if they occupy different spin-orbitals.
Normalization for Orthonormal Spin-Orbitals
Section titled “Normalization for Orthonormal Spin-Orbitals”Assume
Using the permutation form,
The norm is
Orthonormality makes the product vanish unless for every , which means . The surviving terms each contribute , and there are of them. Therefore
This is why the standard Slater determinant carries the prefactor .
The occupied subspace matters more than a particular orbital basis within it. Replacing the occupied orbitals by , with unitary, multiplies the determinant by , a phase. It therefore leaves the physical state unchanged. Rotating occupied and unoccupied orbitals into one another generally changes the state.
Nonorthogonal Orbitals
Section titled “Nonorthogonal Orbitals”Slater determinants are most often built from orthonormal spin-orbitals. If nonorthogonal orbitals are used, the overlap matrix
enters the normalization. The determinant with the standard prefactor has norm squared
Thus, when , a normalized nonorthogonal determinant can be written as
If , the orbitals are linearly dependent and the antisymmetrized state vanishes.
Two-Electron Example
Section titled “Two-Electron Example”For two orthonormal spin-orbitals and , the Slater determinant is
Expanding gives
This is exactly the antisymmetric two-particle wavefunction.
For a closed-shell two-electron example, take two spin-orbitals built from the same spatial orbital :
where and denote orthonormal spin-up and spin-down states. Then
The spatial part is symmetric and the spin part is the antisymmetric singlet. This is the determinant form of the elementary helium ground-state approximation.
Occupation-Number Connection
Section titled “Occupation-Number Connection”Choose an ordered orthonormal spin-orbital basis
A Slater determinant occupying spin-orbitals
corresponds to a fermionic occupation vector with
The determinant is the first-quantized wavefunction representation of that occupation state. The occupation notation does not list particle labels; it lists filled modes. Later, creation operators write the same state as
with a fixed mode ordering used to control signs.
Hartree-Fock and Quantum Chemistry Preview
Section titled “Hartree-Fock and Quantum Chemistry Preview”A single Slater determinant is the simplest antisymmetric many-electron ansatz. Hartree-Fock theory varies the occupied spin-orbitals to minimize the energy expectation value within the family of single determinants.
This is a powerful approximation because it enforces fermionic antisymmetry exactly. It also produces exchange terms that have no classical counterpart. However, a single determinant is not the most general correlated -electron state. Variational Many-Body States compares determinant expansions with Jastrow, paired, projected, tensor-network, and neural families. General fermionic states may require linear combinations of determinants:
where each is a determinant built from a chosen spin-orbital basis. Configuration interaction, coupled-cluster theory, and many-body perturbation theory build on this idea. Those methods belong to later atoms, molecules, and computational-quantum-mechanics material.
Slater Determinants in Atoms applies the construction to atomic configurations and shows how fixed- determinants are transformed into configuration-state functions carrying term labels.
Exercises
Section titled “Exercises”- Show that exchanging and changes the sign of the two-electron Slater determinant.
Solution
The determinant is
Exchanging and swaps the two rows. A determinant changes sign when two rows are swapped, so
- Explain why a determinant with two identical spin-orbitals vanishes.
Solution
If two spin-orbitals are identical, two columns of the Slater matrix are identical. A determinant with two identical columns is zero. Therefore the antisymmetrized fermionic state vanishes, which is Pauli exclusion in determinant language.
- Prove the normalization of a Slater determinant built from orthonormal spin-orbitals.
Solution
Use the permutation expansion:
In the inner product, orthonormality kills all terms except those where the two permutations are identical. There are surviving terms, each equal to . The prefactor contributes , so the norm is .
- Factor the two-electron determinant built from and into spatial and spin parts.
Solution
Start with
Factor out the common spatial product:
The spatial factor is symmetric and the spin factor is antisymmetric.
- In a spin-orbital basis ordered as , what occupation vector corresponds to the determinant occupying orbitals , , and ?
Solution
The occupied orbitals have occupation , and the unoccupied orbital has occupation . The occupation vector is
References
Section titled “References”- P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
- A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, McGraw-Hill, 1971.
- I. N. Levine, Quantum Chemistry, 7th ed., Pearson, 2013.
- R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
- J. C. Slater, “The Theory of Complex Spectra”, Physical Review 34, 1293–1322 (1929).
- A. Szabo and N. S. Ostlund, Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory, Dover, 1996.