In quantum mechanics, a Slater determinant is an expression that describes the wave function of a multi-fermionic system. It satisfies anti-symmetry requirements, and consequently the Pauli principle, by changing sign upon exchange of two electrons (or other fermions).^[1] Only a small subset of all possible fermionic wave functions can be written as a single Slater determinant, but those form an important and useful subset because of their simplicity.

The Slater determinant arises from the consideration of a wave function for a collection of electrons, each with a wave function known as the spin-orbital ${\displaystyle \chi (\mathbf {x} )}$, where ${\displaystyle \mathbf {x} }$ denotes the position and spin of a single electron. A Slater determinant containing two electrons with the same spin orbital would correspond to a wave function that is zero everywhere.

The Slater determinant is named for John C. Slater, who introduced the determinant in 1929 as a means of ensuring the antisymmetry of a many-electron wave function,^[2] although the wave function in the determinant form first appeared independently in Heisenberg's^[3] and Dirac's^[4][5] articles three years earlier.

Definition

Example: Matrix elements in a many electron problem

Many properties of the Slater determinant come to life with an example in a non-relativistic many electron problem.^[7]

• The one particle terms of the Hamiltonian will contribute in the same manner as for the simple Hartree product, namely the energy is summed and the states are independent
• The multi-particle terms of the Hamiltonian, i.e. the exchange terms, will introduce a lowering of the energy of the eigenstates

Starting from an Hamiltonian:

$${\displaystyle {\hat {H}}_{\text{tot}}=\sum _{i}{\frac {\mathbf {p} _{i}^{2}}{2m}}+\sum _{I}{\frac {\mathbf {P} _{I}^{2}}{2M_{I}}}+\sum _{i}V_{\text{nucl}}(\mathbf {r_{i}} )+{\frac {1}{2}}\sum _{i\neq j}{\frac {e^{2}}{|\mathbf {r} _{i}-\mathbf {r} _{j}|}}+{\frac {1}{2}}\sum _{I\neq J}{\frac {Z_{I}Z_{J}e^{2}}{|\mathbf {R} _{I}-\mathbf {R} _{J}|}}}$$

where ${\displaystyle \mathbf {r} _{i}}$ are the electrons and ${\displaystyle \mathbf {R} _{I}}$are the nuclei and

${\displaystyle V_{\text{nucl}}(\mathbf {r} )=-\sum _{I}{\frac {Z_{I}e^{2}}{|\mathbf {r} -\mathbf {R} _{I}|}}}$

For simplicity we freeze the nuclei at equilibrium in one position and we remain with a simplified Hamiltonian

${\displaystyle {\hat {H}}_{e}=\sum _{i}^{N}{\hat {h}}(\mathbf {r} _{i})+{\frac {1}{2}}\sum _{i\neq j}^{N}{\frac {e^{2}}{r_{ij}}}}$

where

${\displaystyle {\hat {h}}(\mathbf {r} )={\frac {{\hat {\mathbf {p} }}^{2}}{2m}}+V_{\text{nucl}}(\mathbf {r} )}$

and where we will distinguish in the Hamiltonian between the first set of terms as ${\displaystyle {\hat {G}}_{1}}$ (the "1" particle terms) and the last term ${\displaystyle {\hat {G}}_{2}}$ which is the "2" particle term or exchange term

${\displaystyle {\hat {G}}_{1}=\sum _{i}^{N}{\hat {h}}(\mathbf {r} _{i})}$

${\displaystyle {\hat {G}}_{2}={\frac {1}{2}}\sum _{i\neq j}^{N}{\frac {e^{2}}{r_{ij}}}}$

The two parts will behave differently when they have to interact with a Slater determinant wave function. We start to compute the expectation values

${\displaystyle \langle \Psi _{0}|G_{1}|\Psi _{0}\rangle ={\frac {1}{N!}}\langle \det\{\psi _{1}...\psi _{N}\}|G_{1}|\det\{\psi _{1}...\psi _{N}\}\rangle }$

In the above expression, we can just select the identical permutation in the determinant in the left part, since all the other N! − 1 permutations would give the same result as the selected one. We can thus cancel N! at the denominator

${\displaystyle \langle \Psi _{0}|G_{1}|\Psi _{0}\rangle =\langle \psi _{1}...\psi _{N}|G_{1}|\det\{\psi _{1}...\psi _{N}\}\rangle }$

Because of the orthonormality of spin-orbitals it is also evident that only the identical permutation survives in the determinant on the right part of the above matrix element

${\displaystyle \langle \Psi _{0}|G_{1}|\Psi _{0}\rangle =\langle \psi _{1}...\psi _{N}|G_{1}|\psi _{1}...\psi _{N}\rangle }$

This result shows that the anti-symmetrization of the product does not have any effect for the one particle terms and it behaves as it would do in the case of the simple Hartree product.

And finally we remain with the trace over the one particle Hamiltonians

${\displaystyle \langle \Psi _{0}|G_{1}|\Psi _{0}\rangle =\sum _{i}\langle \psi _{i}|h|\psi _{i}\rangle }$

Which tells us that to the extent of the one particle terms the wave functions of the electrons are independent of each other and the energy is given by the sum of energies of the single particles.

For the exchange part instead

${\displaystyle \langle \Psi _{0}|G_{2}|\Psi _{0}\rangle ={\frac {1}{N!}}\langle \det\{\psi _{1}...\psi _{N}\}|G_{2}|\det\{\psi _{1}...\psi _{N}\}\rangle =\langle \psi _{1}...\psi _{N}|G_{2}|\det\{\psi _{1}...\psi _{N}\}\rangle }$

If we see the action of one exchange term it will select only the exchanged wavefunctions

${\displaystyle \langle \psi _{1}(r_{1},\sigma _{1})...\psi _{N}(r_{N},\sigma _{N})|{\frac {e^{2}}{r_{12}}}|\mathrm {det} \{\psi _{1}(r_{1},\sigma _{1})...\psi _{N}(r_{N},\sigma _{N})\}\rangle =\langle \psi _{1}\psi _{2}|{\frac {e^{2}}{r_{12}}}|\psi _{1}\psi _{2}\rangle -\langle \psi _{1}\psi _{2}|{\frac {e^{2}}{r_{12}}}|\psi _{2}\psi _{1}\rangle }$

And finally

$${\displaystyle \langle \Psi _{0}|G_{2}|\Psi _{0}\rangle ={\frac {1}{2}}\sum _{ij}\left[\langle \psi _{i}\psi _{j}|{\frac {e^{2}}{r_{ij}}}|\psi _{i}\psi _{j}\rangle -\langle \psi _{i}\psi _{j}|{\frac {e^{2}}{r_{ij}}}|\psi _{j}\psi _{i}\rangle \right]}$$

which instead is a mixing term, the first contribution is called the "coulomb" term and the second is the "exchange" term which can be written using ${\textstyle \sum _{ij}}$ or ${\textstyle \sum _{i\neq j}}$, since the Coulomb and exchange contributions exactly cancel each other for ${\displaystyle i=j}$.

It is important to notice explicitly that the electron-electron repulsive energy ${\displaystyle \langle \Psi _{0}|G_{2}|\Psi _{0}\rangle }$ on the antisymmetrized product of spin-orbitals is always lower than the electron-electron repulsive energy on the simple Hartree product of the same spin-orbitals. The difference is just represented by the second term in the right-hand side without the self-interaction terms ${\displaystyle i=j}$. Since exchange bielectronic integrals are positive quantities, different from zero only for spin-orbitals with parallel spins, we link the decrease in energy with the physical fact that electrons with parallel spin are kept apart in real space in Slater determinant states.

As an approximation

Most fermionic wavefunctions cannot be represented as a Slater determinant. The best Slater approximation to a given fermionic wave function can be defined to be the one that maximizes the overlap between the Slater determinant and the target wave function.^[8] The maximal overlap is a geometric measure of entanglement between the fermions.

A single Slater determinant is used as an approximation to the electronic wavefunction in Hartree–Fock theory. In more accurate theories (such as configuration interaction and MCSCF), a linear combination of Slater determinants is needed.

Discussion

The word "detor" was proposed by S. F. Boys to refer to a Slater determinant of orthonormal orbitals,^[9] but this term is rarely used.

Unlike fermions that are subject to the Pauli exclusion principle, two or more bosons can occupy the same single-particle quantum state. Wavefunctions describing systems of identical bosons are symmetric under the exchange of particles and can be expanded in terms of permanents.

See also

• Antisymmetrizer
• Electron orbital
• Fock space
• Quantum electrodynamics
• Quantum mechanics
• Physical chemistry
• Hund's rule
• Hartree–Fock method

References

External links

• Many-Electron States in E. Pavarini, E. Koch, and U. Schollwöck: Emergent Phenomena in Correlated Matter, Jülich 2013, ISBN 978-3-89336-884-6