Slater Determinants

A Slater determinant is the standard compact wavefunction for $N$ identical fermions occupying $N$ one-particle spin-orbitals. It builds antisymmetry into the wavefunction by using a determinant:

$$\Psi(q_1,\ldots,q_N) = \frac{1}{\sqrt{N!}} \det \begin{pmatrix} \varphi_1(q_1) & \varphi_2(q_1) & \cdots & \varphi_N(q_1)\\ \varphi_1(q_2) & \varphi_2(q_2) & \cdots & \varphi_N(q_2)\\ \vdots & \vdots & \ddots & \vdots\\ \varphi_1(q_N) & \varphi_2(q_N) & \cdots & \varphi_N(q_N) \end{pmatrix}.$$

Here $q=(\mathbf x,s)$ denotes all one-particle degrees of freedom, including spin. The orbitals $\varphi_j(q)$ are usually chosen orthonormal. 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.

Definition

Let $\varphi_1,\ldots,\varphi_N$ be normalized one-particle states in a Hilbert space $\mathcal h$. The corresponding antisymmetrized slot state is

$$\lvert\varphi_1\cdots\varphi_N\rangle_A = \frac{1}{\sqrt{N!}} \sum_{\pi\in S_N} \operatorname{sgn}(\pi)\, \lvert\varphi_{\pi(1)}\rangle_1 \cdots \lvert\varphi_{\pi(N)}\rangle_N.$$

In the coordinate-spin representation, this is the determinant above:

$$\Psi(q_1,\ldots,q_N) = \frac{1}{\sqrt{N!}} \det \bigl[ \varphi_j(q_i) \bigr]_{i,j=1}^{N}.$$

The rows correspond to particle slots $q_i$. The columns correspond to occupied one-particle spin-orbitals $\varphi_j$.

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 $-1$. The occupied set of spin-orbitals is the same physical occupation; the sign must simply be handled consistently.

Antisymmetry from Determinants

Exchange two particle slots, say $q_r$ and $q_s$. In the determinant, this swaps rows $r$ and $s$. A determinant changes sign under a row swap, so

$$\Psi(\ldots,q_s,\ldots,q_r,\ldots) = -\Psi(\ldots,q_r,\ldots,q_s,\ldots).$$

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 $N$.

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:

$$\varphi_a=\varphi_b \quad \Longrightarrow \quad \det \bigl[ \varphi_j(q_i) \bigr] = 0.$$

More generally, if the occupied one-particle states are linearly dependent, the determinant is zero. This is Pauli exclusion in determinant form: an $N$-fermion determinant needs $N$ 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

Assume

$$\langle\varphi_i\vert\varphi_j\rangle = \delta_{ij}.$$

Using the permutation form,

$$\lvert\Psi\rangle = \frac{1}{\sqrt{N!}} \sum_{\pi\in S_N} \operatorname{sgn}(\pi) \lvert\varphi_{\pi(1)}\rangle_1 \cdots \lvert\varphi_{\pi(N)}\rangle_N.$$

The norm is

$$\langle\Psi\vert\Psi\rangle = \frac{1}{N!} \sum_{\sigma,\pi\in S_N} \operatorname{sgn}(\sigma) \operatorname{sgn}(\pi) \prod_{k=1}^{N} \langle\varphi_{\sigma(k)}\vert \varphi_{\pi(k)}\rangle.$$

Orthonormality makes the product vanish unless $\sigma(k)=\pi(k)$ for every $k$, which means $\sigma=\pi$. The surviving terms each contribute $1$, and there are $N!$ of them. Therefore

$$\langle\Psi\vert\Psi\rangle = \frac{N!}{N!} = 1.$$

This is why the standard Slater determinant carries the prefactor $1/\sqrt{N!}$.

Nonorthogonal Orbitals

Slater determinants are most often built from orthonormal spin-orbitals. If nonorthogonal orbitals are used, the overlap matrix

$$S_{ij} = \langle\varphi_i\vert\varphi_j\rangle$$

enters the normalization. The determinant with the standard $1/\sqrt{N!}$ prefactor has norm squared

$$\det S.$$

Thus, when $\det S>0$, a normalized nonorthogonal determinant can be written as

$$\Psi(q_1,\ldots,q_N) = \frac{ \det[\varphi_j(q_i)]_{i,j=1}^{N} }{ \sqrt{N!\det S} }.$$

If $\det S=0$, the orbitals are linearly dependent and the antisymmetrized state vanishes.

Two-Electron Example

For two orthonormal spin-orbitals $\varphi_a$ and $\varphi_b$, the Slater determinant is

$$\Psi(q_1,q_2) = \frac{1}{\sqrt2} \det \begin{pmatrix} \varphi_a(q_1) & \varphi_b(q_1)\\ \varphi_a(q_2) & \varphi_b(q_2) \end{pmatrix}.$$

Expanding gives

$$\Psi(q_1,q_2) = \frac{1}{\sqrt2} \bigl[ \varphi_a(q_1)\varphi_b(q_2) - \varphi_b(q_1)\varphi_a(q_2) \bigr].$$

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 $\psi(\mathbf x)$:

$$\varphi_a(q)=\psi(\mathbf x)\alpha(s), \qquad \varphi_b(q)=\psi(\mathbf x)\beta(s),$$

where $\alpha$ and $\beta$ denote orthonormal spin-up and spin-down states. Then

$$\Psi(q_1,q_2) = \psi(\mathbf x_1)\psi(\mathbf x_2) \frac{1}{\sqrt2} \bigl[ \alpha(s_1)\beta(s_2) - \beta(s_1)\alpha(s_2) \bigr].$$

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.

Three-Electron Example

For three occupied spin-orbitals $a,b,c$, the determinant is

$$\Psi(q_1,q_2,q_3) = \frac{1}{\sqrt{6}} \det \begin{pmatrix} \varphi_a(q_1) & \varphi_b(q_1) & \varphi_c(q_1)\\ \varphi_a(q_2) & \varphi_b(q_2) & \varphi_c(q_2)\\ \varphi_a(q_3) & \varphi_b(q_3) & \varphi_c(q_3) \end{pmatrix}.$$

Expanding the determinant gives six signed product terms. Occupation-number notation compresses the same information into the statement that the three spin-orbitals $a,b,c$ are occupied once.

Occupation-Number Connection

Choose an ordered orthonormal spin-orbital basis

$$\varphi_1,\varphi_2,\ldots .$$

A Slater determinant occupying spin-orbitals

$$\varphi_{i_1},\ldots,\varphi_{i_N}$$

corresponds to a fermionic occupation vector with

$$n_i= \begin{cases} 1, & i\in\{i_1,\ldots,i_N\},\\ 0, & \text{otherwise}. \end{cases}$$

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

$$c_{i_1}^\dagger c_{i_2}^\dagger \cdots c_{i_N}^\dagger \lvert0\rangle,$$

with a fixed mode ordering used to control signs.

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 $N$-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:

$$\lvert\Psi\rangle = \sum_I C_I\lvert\Phi_I\rangle,$$

where each $\lvert\Phi_I\rangle$ 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-$M_L,M_S$ determinants are transformed into configuration-state functions carrying term labels.

Common Mistakes

• Building a determinant from spatial orbitals only when spin is part of the one-particle state.
• Forgetting the factor $1/\sqrt{N!}$ for orthonormal spin-orbitals.
• Thinking a column swap changes the physical occupation rather than only the sign convention.
• Treating a single Slater determinant as the most general fermionic wavefunction.
• Calling the determinant’s antisymmetry a repulsive force; it is a constraint on the state space.
• Confusing electron correlation with antisymmetry. A single determinant is antisymmetric but usually not fully correlated.

Cross-Links

• Symmetric and Antisymmetric Wavefunctions
• Fermions
• Exchange Operators
• Permanents
• Identical-Particle Entanglement Cautions
• Spin and Spatial Wavefunctions
• Pauli Exclusion Principle
• Symmetrization Postulate
• Identical Particle Exercises
• Occupation-Number Basis
• Fermionic Fock Space
• Creation and Annihilation Operators
• Fermionic Anticommutation Relations
• Entanglement in Quantum Chemistry
• Variational Estimate for the Helium Atom
• Slater Determinants in Atoms
• Formula Sheet

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• 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.
• I. N. Levine, Quantum Chemistry, 7th ed., Pearson, 2013.
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, McGraw-Hill, 1971.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.

Exercises

1. Show that exchanging $q_1$ and $q_2$ changes the sign of the two-electron Slater determinant.

Solution

The determinant is

$$\Psi(q_1,q_2) = \frac{1}{\sqrt2} \det \begin{pmatrix} \varphi_a(q_1) & \varphi_b(q_1)\\ \varphi_a(q_2) & \varphi_b(q_2) \end{pmatrix}.$$

Exchanging $q_1$ and $q_2$ swaps the two rows. A determinant changes sign when two rows are swapped, so

$$\Psi(q_2,q_1) = -\Psi(q_1,q_2).$$

2. 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.

3. Prove the normalization of a Slater determinant built from orthonormal spin-orbitals.

Solution

Use the permutation expansion:

$$\lvert\Psi\rangle = \frac{1}{\sqrt{N!}} \sum_{\pi\in S_N} \operatorname{sgn}(\pi) \lvert\varphi_{\pi(1)}\rangle_1 \cdots \lvert\varphi_{\pi(N)}\rangle_N.$$

In the inner product, orthonormality kills all terms except those where the two permutations are identical. There are $N!$ surviving terms, each equal to $1$. The prefactor contributes $1/N!$, so the norm is $1$.

4. Factor the two-electron determinant built from $\psi(\mathbf x)\alpha(s)$ and $\psi(\mathbf x)\beta(s)$ into spatial and spin parts.

Solution

Start with

$$\frac{1}{\sqrt2} \bigl[ \psi(\mathbf x_1)\alpha(s_1) \psi(\mathbf x_2)\beta(s_2) - \psi(\mathbf x_1)\beta(s_1) \psi(\mathbf x_2)\alpha(s_2) \bigr].$$

Factor out the common spatial product:

$$\psi(\mathbf x_1)\psi(\mathbf x_2) \frac{1}{\sqrt2} \bigl[ \alpha(s_1)\beta(s_2) - \beta(s_1)\alpha(s_2) \bigr].$$

The spatial factor is symmetric and the spin factor is antisymmetric.

5. In a spin-orbital basis ordered as $1,2,3,4$, what occupation vector corresponds to the determinant occupying orbitals $1$, $3$, and $4$?

Solution

The occupied orbitals have occupation $1$, and the unoccupied orbital has occupation $0$. The occupation vector is

$$\lvert1,0,1,1\rangle_F.$$