Entanglement in Quantum Chemistry

Entanglement can help describe correlations among molecular orbitals, fragments, or local spins. Because electrons are identical fermions, however, antisymmetry alone is not the same as correlation beyond a Hartree–Fock determinant. An orbital entropy also depends on which orbitals define the subsystems. The examples below keep these distinctions explicit before using entropy as a chemical diagnostic.

Helpful background. Slater determinants supplies fermionic signs; spin and spatial wavefunctions explains exchange of complete electron labels; entanglement entropy explains reduced-state spectra.

Electronic Hilbert Space

A one-electron spin-orbital depends on spatial and spin variables,

$$q=(\mathbf r,s),$$

and belongs to a one-particle Hilbert space $\mathcal h$. An $N$-electron wavefunction belongs to the antisymmetric sector

$$\wedge^N\mathcal h.$$

Equivalently, in a chosen spin-orbital basis, one may use fermionic Fock space and occupation-number states. A determinant occupying spin-orbitals $i_1,\ldots,i_N$ is represented by

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

with a fixed ordering convention for signs.

The spin-orbital basis is not unique. Canonical molecular orbitals, localized orbitals, atomic orbitals, natural orbitals, and active-space orbitals can all be useful. Entanglement between orbitals is therefore a statement about a chosen mode decomposition, not an invariant property of electron labels.

Antisymmetry Is Not the Same as Correlation

A single Slater determinant is already antisymmetric. For two occupied spin-orbitals $a,b$,

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

This state is not a product in formal particle slots, but those slots are not physical electron subsystems. In the occupation basis defined by $a,b$, the same determinant is simply

$$\lvert1_a,1_b\rangle.$$

It has definite occupations of those two spin-orbitals. The antisymmetric sign is essential physics, but it is not by itself the same as electron correlation in the quantum-chemistry sense.

Quantum chemistry usually reserves “correlation” for effects missing from the best single-determinant mean-field reference for a chosen electronic Hamiltonian. Exchange is already included in Hartree–Fock through antisymmetry. Correlation beyond the best determinant requires a state that cannot be reduced to one determinant by rotating the orbitals, or another genuinely correlated ansatz. Orbital optimization can improve a chosen determinant but cannot turn the single-determinant class into the full many-electron space. Exchange and Correlation develops the atomic energy, pair-hole, and density-functional accounting behind this distinction.

Electronic Hamiltonian

In an orthonormal spin-orbital basis, the nonrelativistic clamped-nuclei electronic Hamiltonian has the second-quantized form

$$H = \sum_{pq} h_{pq} a_p^\dagger a_q + \frac12 \sum_{pqrs} g_{pqrs} a_p^\dagger a_q^\dagger a_s a_r.$$

The constant nuclear repulsion is omitted here; it must be added for the total molecular energy at the specified geometry. The one-electron integrals contain kinetic energy and electron–nuclear attraction. In atomic units,

$$h_{pq} = \int dq\, \varphi_p^*(q) \left( -\frac12\nabla^2 +v_{\mathrm{nuc}}(\mathbf r) \right) \varphi_q(q).$$

The two-electron integrals contain Coulomb repulsion:

$$g_{pqrs} = \int dq_1\,dq_2\, \varphi_p^*(q_1) \varphi_q^*(q_2) \frac{1}{r_{12}} \varphi_r(q_1) \varphi_s(q_2).$$

The measure $dq$ means a spatial integral and a sum over spin. The displayed $g_{pqrs}$ is not antisymmetrized, which explains the factor $1/2$; using antisymmetrized integrals instead changes that prefactor and convention. Many-particle Hamiltonians develops the operator construction.

Hartree–Fock Reference

Hartree–Fock approximates the electronic ground state by one optimized Slater determinant,

$$\lvert\Phi_0\rangle = a_{i_1}^\dagger \cdots a_{i_N}^\dagger \lvert0\rangle.$$

Within that determinant, exchange symmetry is exact. The approximation lies in describing the many-electron state by independent occupied spin-orbitals moving in a self-consistent mean field.

For a fixed Hamiltonian and basis, the correlation energy is often defined as

$$E_{\mathrm{corr}} = E_0-E_{\mathrm{HF}},$$

where $E_0$ is the exact ground-state energy in that same model space. Variationally,

$$E_{\mathrm{corr}}\le0.$$

This number is not itself an entanglement measure. It is an energy difference. It can be related to correlation and multiconfigurational structure, but the relationship depends on the Hamiltonian, basis, state, and reference determinant.

Configuration Interaction Preview

A general correlated wavefunction in a finite spin-orbital basis can be expanded as a linear combination of determinants:

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

Relative to a reference determinant $\lvert\Phi_0\rangle$, singly and doubly excited determinants have the schematic form

$$\lvert\Phi_i^a\rangle = a_a^\dagger a_i \lvert\Phi_0\rangle,$$

and

$$\lvert\Phi_{ij}^{ab}\rangle = a_a^\dagger a_b^\dagger a_j a_i \lvert\Phi_0\rangle.$$

Here $i,j$ usually label occupied spin-orbitals in the reference and $a,b$ virtual spin-orbitals. Full configuration interaction in a finite basis includes all determinants consistent with particle number and spin-orbital space. Truncated CI, coupled-cluster theory, perturbation theory, selected CI, and tensor-network methods choose different ways to approximate the same enormous correlated state space.

A determinant expansion alone does not prove correlation beyond one determinant. For example,

$$\frac{\lvert1100\rangle+\lvert1010\rangle}{\sqrt2} =a_1^\dagger\frac{a_2^\dagger+a_3^\dagger}{\sqrt2}\lvert0\rangle$$

is a single determinant in rotated orbitals. In contrast, $(\lvert1100\rangle+\lvert0011\rangle)/\sqrt2$ is not a single determinant: its one-particle density matrix has four occupations $1/2$, instead of two ones and two zeros. The eigenvalues of that one-particle density matrix are invariant under orbital rotation.

Orbital Entanglement

Orbital entanglement treats chosen orbitals or groups of orbitals as subsystems. For a spatial orbital $p$ with spin-up and spin-down modes, the local occupation basis is

$$\lvert0\rangle, \qquad \lvert\uparrow\rangle, \qquad \lvert\downarrow\rangle, \qquad \lvert\uparrow\downarrow\rangle.$$

Given a many-electron state $\rho$, the one-orbital reduced state is

$$\rho_p = \operatorname{Tr}_{\bar p}\rho,$$

and the one-orbital entropy is

$$s_p = -\operatorname{Tr}(\rho_p\log_2\rho_p).$$

For two orbitals $p,q$, define

$$\rho_{pq} = \operatorname{Tr}_{\overline{pq}}\rho, \qquad s_{pq} = -\operatorname{Tr}(\rho_{pq}\log_2\rho_{pq}).$$

The orbital mutual information is

$$I(p:q) = s_p+s_q-s_{pq}.$$

We use base-two logarithms and the full quantum mutual information. Some chemistry papers use natural logarithms or include a factor $1/2$ in their orbital-interaction convention; numerical values must be converted before comparison. For a globally pure electronic state, $s_p$ measures entanglement of this orbital with the rest. For mixed states it also includes mixing, and $I(p:q)$ always measures total correlation rather than entanglement alone. The occupation-factor description assumes a fixed fermionic ordering; local particle-number or parity restrictions can reduce the entanglement accessible by the allowed operations.

These quantities are useful diagnostics. They help identify strongly entangled active orbitals, choose orbital orderings in density-matrix renormalization group calculations, and distinguish nearly single-reference regions from multireference regions. They are also basis dependent. Localized orbitals and canonical orbitals can give different entanglement patterns for the same physical state.

Spin Entanglement

Spin entanglement in chemistry is rarely about labeled electrons. It is usually about spins associated with orbitals, sites, fragments, centers, or spatial regions. The two-electron singlet,

$$\lvert S\rangle = \frac{1}{\sqrt2} \left( \lvert\uparrow\downarrow\rangle - \lvert\downarrow\uparrow\rangle \right),$$

is maximally entangled across two distinguishable spin degrees of freedom when those spins are assigned to two physical modes or regions. For electrons, the complete state must also satisfy fermionic antisymmetry. Thus spin entanglement must be discussed together with the spatial or orbital part that identifies the parties.

This distinction matters in molecules. A closed-shell determinant can contain spin-singlet pairing due to the exchange-symmetry structure, while an effective model of two separated radical centers can contain operational spin entanglement between centers. The symbols may look similar, but the subsystem split is different.

Open-shell molecules add another layer: a single determinant may not be an eigenstate of total spin. Spin-adapted configuration state functions build proper spin symmetry from determinant combinations. A spin-adapted determinant combination enforces total-spin symmetry; it should not be identified with a particular amount of entanglement without specifying the orbital or fragment split.

Bonding and Correlation

The hydrogen molecule is the standard warning against making single-determinant intuition too absolute. In a simple Heitler–London picture with localized orbitals $a$ and $b$, the covalent spatial factor has the schematic symmetric form

$$\psi_+(\mathbf r_1,\mathbf r_2) \propto a(\mathbf r_1)b(\mathbf r_2) + b(\mathbf r_1)a(\mathbf r_2),$$

paired with a spin singlet to make an antisymmetric two-electron state. Molecular-orbital pictures instead form bonding and antibonding orbitals and often start from a closed-shell determinant.

Near equilibrium, a single-reference description may be a good starting point. During bond stretching or bond breaking, however, near-degenerate configurations become important. A correlated state may require a superposition of determinants to avoid unphysical ionic contributions and to recover the correct separated-atom behavior.

A small model shows why entropy is not a universal scale of chemical correlation. Let $L,R$ be orthonormal localized spatial orbitals and $g=(L+R)/\sqrt2$. In the ordered spin-orbital basis $(L\uparrow,L\downarrow,R\uparrow,R\downarrow)$, the closed-shell determinant is

$$g_\uparrow^\dagger g_\downarrow^\dagger\lvert0\rangle =\frac12(\lvert1100\rangle+\lvert1001\rangle-\lvert0110\rangle+\lvert0011\rangle).$$

Tracing out $R$ gives $\rho_L=I_4/4$, so $s_L=2$ bits, even though the state is one determinant. In the bonding/antibonding orbital split its occupations are definite and its orbital entropy is zero. The covalent singlet $(\lvert1001\rangle-\lvert0110\rangle)/\sqrt2$ instead has $s_L=1$ bit and no empty/doubly occupied fragment fluctuations. Its smaller localized-orbital entropy does not mean that it is closer to a single determinant.

Orbital diagnostics are most useful when compared within a stated basis, active space, and state family. Valence-bond theory develops the chemical meaning of the covalent and ionic components.

Active Spaces and Tensor Networks

In practical electronic-structure calculations, one often chooses an active space: a subset of orbitals treated with high-level correlation while the rest are kept inactive, weakly correlated, or perturbative. Entanglement diagnostics can help choose that active set.

For a globally pure state, large $s_p$ means substantial entanglement with the remaining orbitals in the chosen basis; it need not mean correlation beyond a single determinant, as the example above shows. If $I(p:q)$ is large, orbitals $p$ and $q$ share strong total correlation. These facts can guide active-space construction, orbital localization, and ordering in tensor-network algorithms.

Matrix product state methods represent the many-electron coefficient tensor as a chain of orbital tensors. As in many-body physics, an exact representation with bond dimension $\chi$ across a single chain cut obeys

$$S_A\le\log_2\chi.$$

The same warning applies here as in the many-body bridge: a tensor network is efficient when the chosen ordering and orbital basis keep the relevant entanglement manageable.

References

• K. Boguslawski, P. Tecmer, Ö. Legeza, and M. Reiher, “Entanglement measures for single- and multireference correlation effects,” Journal of Physical Chemistry Letters 3, 3129–3135 (2012), doi:10.1021/jz301319v.
• G. K.-L. Chan and S. Sharma, “The density matrix renormalization group in quantum chemistry,” Annual Review of Physical Chemistry 62, 465–481 (2011), doi:10.1146/annurev-physchem-032210-103338.
• T. Helgaker, P. Jørgensen, and J. Olsen, Molecular Electronic-Structure Theory, Wiley (2000).
• J. Rissler, R. M. Noack, and S. R. White, “Measuring orbital interaction using quantum information theory,” Chemical Physics 323, 519–531 (2006), doi:10.1016/j.chemphys.2005.10.018.
• A. Szabo and N. S. Ostlund, Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory, Dover (1996).

Exercises

1. Single determinant in its own spin-orbital basis. Consider a determinant occupying spin-orbitals $1$ and $2$ in a four-spin-orbital basis:

$$\lvert\Phi\rangle = \lvert1100\rangle.$$

What is the entropy of spin-orbital mode $1$?

Solution

Mode $1$ has definite occupation $n_1=1$. Its reduced state is the pure one-mode state

$$\rho_1 = \lvert1\rangle\langle1\rvert.$$

Therefore

$$S(\rho_1)=0.$$

The determinant is antisymmetric as a fermionic state, but in this chosen occupation basis each spin-orbital has definite occupation.

2. A two-configuration superposition. Let

$$\lvert\Psi\rangle = \frac{ \lvert1100\rangle + \lvert0011\rangle }{\sqrt2}.$$

Treat modes $1,2$ as subsystem $A$ and modes $3,4$ as subsystem $B$. Compute $S_A$ in bits.

Solution

The state is already in Schmidt form across $A\vert B$:

$$\lvert\Psi\rangle = \frac{ \lvert11\rangle_A\lvert00\rangle_B + \lvert00\rangle_A\lvert11\rangle_B }{\sqrt2}.$$

Thus

$$\rho_A = \frac12\lvert11\rangle\langle11\rvert + \frac12\lvert00\rangle\langle00\rvert.$$

The nonzero eigenvalues are $1/2$ and $1/2$, so

$$S_A = 1 \quad \text{bit}.$$

3. Exchange versus correlation. Explain why a single Slater determinant can include exchange but still miss electron correlation.

Solution

The determinant is antisymmetric, so it exactly includes the exchange constraint required for fermions. However, it restricts the many-electron state to one antisymmetrized product of spin-orbitals. Correlated electronic motion generally requires superpositions of determinants, changed occupation patterns, or more flexible ansatz structure beyond one mean-field determinant.

4. One-orbital entropy. A spatial orbital has reduced density matrix diagonal in the basis $\lvert0\rangle,\lvert\uparrow\rangle,\lvert\downarrow\rangle,\lvert\uparrow\downarrow\rangle$ with probabilities $1/2,0,0,1/2$. Compute its entropy in bits.

Solution

Only two eigenvalues are nonzero, both equal to $1/2$. Therefore

$$s_p = -2 \left( \frac12\log_2\frac12 \right) = 1.$$

5. A determinant with orbital entropy. Use the localized-orbital expansion of $g_\uparrow^\dagger g_\downarrow^\dagger\lvert0\rangle$ above to find $s_L$, $s_R$, and $I(L:R)$. Compare the covalent singlet.

Solution

The four $L$ occupations are paired with four orthogonal $R$ occupations, all with probability $1/4$. Thus $s_L=s_R=2$ bits. The total state is pure, so $I(L:R)=4$ bits. The covalent singlet has only two equal nonzero Schmidt probabilities, giving $s_L=s_R=1$ and $I(L:R)=2$ bits. Both calculations use the same localized-orbital partition; neither number is a correlation energy or a basis-independent count of determinants.