Configuration Interaction

Configuration interaction (CI) diagonalizes the electronic Hamiltonian in a space of Slater determinants or spin-adapted combinations. Full CI includes every determinant in a specified one-particle basis and electron sector. It is exact for that represented Hamiltonian, not for the continuum molecule, and its exponential size limits its use.

Helpful background. Hartree–Fock computation supplies orbitals; sparse eigensolvers supplies partial diagonalization.

Determinants and Hamiltonian actions

For $N$ electrons in $M$ orthonormal spin orbitals, the unrestricted determinant count is $\binom{M}{N}$. Fixing spin projection or spatial symmetry reduces it. Specify a canonical ordering of creation operators so that bitstring occupation states carry reproducible fermionic signs.

Using antisymmetrized two-electron integrals,

$$H=\sum_{pq}h_{pq}a_p^\dagger a_q +\frac14\sum_{pqrs}\langle pq\Vert rs\rangle a_p^\dagger a_q^\dagger a_s a_r+E_{\rm nuc}.$$

Each one-body term changes at most one occupied orbital, and each two-body term at most two. Consequently determinants differing by more than a double excitation have zero Hamiltonian matrix element. Generate connected determinants and their parity signs rather than testing every pair. A matrix-free action can contract amplitudes directly without storing all nonzero entries.

Check Hermiticity, particle number, spin-sector preservation where applicable, and the action on hand-computed determinants. A consistent integral-index convention is essential: chemists’ spatial integrals and antisymmetrized spin-orbital integrals have different prefactors and permutations.

Full and truncated CI

Write $|\Psi\rangle=\sum_I c_I|D_I\rangle$. The projected equation is $Hc=Ec$, so Hermitian CI is variational in its chosen determinant space. Nested enlargements lower the ground-state energy when integrals and orbitals are fixed. Singles-and-doubles CI (CISD) restricts the excitation rank relative to a reference determinant; it is not full CI except in sufficiently small electron spaces.

Full CI is invariant under unitary rotations spanning the same complete orbital space. Truncated CI generally is not invariant under rotations that mix occupied and virtual reference orbitals. An active-space calculation is full CI only within its active sector; frozen orbitals and excluded virtuals remain approximations.

A useful convergence record distinguishes orbital-basis enlargement, active-space enlargement, determinant selection, and eigensolver residual. These are four different controls. For selected CI, perturbative corrections can estimate omitted contributions but do not preserve a strict variational upper-bound interpretation.

Why truncated CI is not size-consistent

Take two noninteracting fragments whose correlated states each contain a reference and a double excitation:

$$|\Psi_A\rangle=(1+t_A D_A)|\Phi_A\rangle,\qquad |\Psi_B\rangle=(1+t_B D_B)|\Phi_B\rangle.$$

Their product contains $t_At_BD_AD_B|\Phi_A\Phi_B\rangle$, a quadruple excitation relative to the combined reference. Combined-system CISD excludes this term even though each separate fragment calculation includes all its required doubles. Thus the product of the separate approximate states does not generally belong to the combined truncated space, and $E_{AB}=E_A+E_B$ need not hold.

This is a structural error, independent of the eigensolver tolerance. Coupled cluster treats disconnected excitation products differently through an exponential ansatz.

A minimal correlation benchmark

For a two-determinant block,

$$H=\begin{pmatrix}E_{\rm ref}&v\\v&E_{\rm ex}\end{pmatrix}, \quad E_{\rm ex}>E_{\rm ref},$$

with real coupling $v$, the lower CI energy is

$$E_-=\frac{E_{\rm ref}+E_{\rm ex}}2 -\sqrt{\left(\frac{E_{\rm ex}-E_{\rm ref}}2\right)^2+v^2}.$$

For nonzero $v$ it lies below $E_{\rm ref}$ and approaches $E_{\rm ref}-v^2/(E_{\rm ex}-E_{\rm ref})$ for weak mixing. Compare computed coefficients, energy, normalization, and residual against this expression. Near degeneracy the perturbative denominator fails while diagonalization remains well-defined.

Exercises

How many determinants describe two electrons in four spin orbitals, and how many remain if there are two spatial orbitals with one up-spin and one down-spin electron?

Solution

The full fixed-particle-number space has $\binom42=6$ determinants. Fixing one electron of each spin gives $\binom21\binom21=4$. Fixing spin projection does not by itself select a definite total-spin eigenvalue.

References

• Bartlett, R. J., and M. Musiał, “Coupled-cluster theory in quantum chemistry,” Reviews of Modern Physics 79, 291–352 (2007), doi:10.1103/RevModPhys.79.291 — comparison of linear and exponential correlation expansions.
• Szabo, A., and N. S. Ostlund, Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory, Dover, 1996 — Slater–Condon rules, CI, and size consistency.