Two-Body Operator Applications

A number-conserving two-body operator acts on particle pairs. In an orthonormal one-particle basis, its standard Fock-space form is

$$\widehat V = \frac12 \sum_{i,j,k,l} V_{ij;kl} a_i^\dagger a_j^\dagger a_l a_k.$$

The operator annihilates an incoming pair in modes $k,l$ and creates an outgoing pair in modes $i,j$. The coefficient $V_{ij;kl}$ is a two-particle matrix element, and the prefactor depends on whether those matrix elements are unsymmetrized, symmetrized, antisymmetrized, or summed over a restricted index set.

The foundational equivalence with $\sum_{\alpha<\beta}v^{(\alpha\beta)}$ is derived in Two-Body Operators. This page is the many-body application guide: coefficient conventions, pair scattering, two-body reduced density matrices, basis transformations, contact and Coulomb interactions, lattice terms, symmetry tests, truncation, and computational checks.

Helpful background. Two-body operators derives the pair lift; one-body operator applications introduces the reduced-matrix convention.

Pair interactions and ordered coefficients

Take orthonormal complete modes and a self-adjoint two-particle interaction $v$ that commutes with exchange. We use $\eta=+1$ for bosons and $-1$ for fermions. The semicolon in $V_{ij;kl}$ separates outgoing labels from incoming labels. All four indices are summed unless a restriction is shown.

On the fixed-$N$ sector, the interaction is $\sum_{\alpha<\beta}v^{(\alpha\beta)}$, with one term per unordered particle pair. It vanishes for $N=0,1$. The foundational lift derives its full-index prefactor $1/2$; the useful task here is to translate between coefficient and pair-density conventions without losing that normalization.

Ordered-Product Matrix Elements

Define

$$\begin{aligned} V_{ij;kl} ={}& \langle \varphi_i\otimes\varphi_j \vert v \vert \varphi_k\otimes\varphi_l \rangle. \end{aligned}$$

These are matrix elements on $\mathcal H_1\otimes\mathcal H_1$ before symmetrizing or antisymmetrizing the basis states. With this convention,

$$\widehat V = \frac12 \sum_{i,j,k,l} V_{ij;kl} a_i^\dagger a_j^\dagger a_l a_k.$$

Reading from right to left:

1. $a_k$ removes a particle from mode $k$;
2. $a_l$ removes a particle from mode $l$;
3. $a_j^\dagger$ creates a particle in mode $j$;
4. $a_i^\dagger$ creates a particle in mode $i$.

For fermions, this displayed order is part of the convention. Reordering a string and retaining the same coefficient without the corresponding sign changes the operator.

Body Rank and Polynomial Degree

A number-conserving two-body interaction is quartic in ladder operators because it contains two creations and two annihilations. The two descriptions should not be conflated:

Structure	Representative form	Interpretation
one-body	$A_{ij}a_i^\dagger a_j$	acts additively on individual particles
two-body	$V_{ij;kl}a_i^\dagger a_j^\dagger a_l a_k$	acts on particle pairs
pairing quadratic	$\Delta_{ij}a_i^\dagger a_j^\dagger+\mathrm{h.c.}$	changes particle number by two; not a fixed-$N$ pair interaction
three-body	$W_{ijk;lmn}a_i^\dagger a_j^\dagger a_k^\dagger a_n a_m a_l$	acts on triples

A density product can hide this grading. For example, $n_i^2$ contains both a pair term and a one-body term for bosons. Normal ordering exposes the distinction.

Hermiticity

For a Hermitian two-particle operator $v$,

$$V_{ij;kl} = V_{kl;ij}^*.$$

Indeed,

$$\left( a_i^\dagger a_j^\dagger a_l a_k \right)^\dagger = a_k^\dagger a_l^\dagger a_j a_i.$$

Relabeling the incoming and outgoing index pairs then gives $\widehat V^\dagger=\widehat V$. In numerical work, the residual

$$V_{ij;kl}-V_{kl;ij}^*$$

is a direct diagnostic of integral-generation, storage, or convention errors.

Exchange Symmetry of the Interaction

Let $P_{12}$ exchange the two one-particle factors. A physical pair interaction for identical particles satisfies

$$[v,P_{12}]=0.$$

In the ordered product basis, this implies

$$V_{ij;kl} = V_{ji;lk}.$$

This identity exchanges both particles in the bra and ket. It does not generally imply $V_{ij;kl}=V_{ji;kl}$ or $V_{ij;kl}=V_{ij;lk}$ separately.

For a coordinate-diagonal symmetric potential $v(x,x')=v(x',x)$, the same relation follows by interchanging the integration variables.

Statistics-Adapted Coefficients

The ladder algebra projects the coefficient tensor onto the exchange sector appropriate to the particles. Define

$$W_{ij;kl}^{(\eta)} = V_{ij;kl} + \eta V_{ij;lk}.$$

For an exchange-symmetric interaction,

$$W_{ji;kl}^{(\eta)} = \eta W_{ij;kl}^{(\eta)}, \qquad W_{ij;lk}^{(\eta)} = \eta W_{ij;kl}^{(\eta)}.$$

The interaction may then be written

$$\widehat V = \frac14 \sum_{i,j,k,l} W_{ij;kl}^{(\eta)} a_i^\dagger a_j^\dagger a_l a_k.$$

For bosons, $W^{(+)}$ is symmetrized. For fermions, $W^{(-)}$ is antisymmetrized. This $W$ is not a normalized pair-basis matrix element; normalized states with coincident bosonic indices introduce additional square-root factors.

Antisymmetrized fermionic integrals

The common notation $\langle ij\Vert kl\rangle=W_{ij;kl}^{(-)}$ uses the full-sum prefactor $1/4$. With a full sum, using $1/2$ instead doubles the interaction. Compare the definition and the annihilator order together: some texts interchange the final two indices.

Unique normalized pair states

For bosons, label unique pairs by $i\le j$ and define

$$B_{ij}^\dagger=\frac{b_i^\dagger b_j^\dagger}{\sqrt{1+\delta_{ij}}},\qquad v^{\rm pair}_{ij;kl}=\frac{W_{ij;kl}^{(+)}}{\sqrt{(1+\delta_{ij})(1+\delta_{kl})}}.$$

Then $B_{ij}^\dagger|0\rangle$ are normalized two-particle basis states. For fermions use $i<j$, $B_{ij}^\dagger=c_i^\dagger c_j^\dagger$, and $v^{\rm pair}_{ij;kl}=W_{ij;kl}^{(-)}$. In either case,

$$\widehat V=\sum_{\text{unique }(i,j),(k,l)}v^{\rm pair}_{ij;kl}B_{ij}^\dagger B_{kl}.$$

There is no additional prefactor. This is especially convenient for checking the $N=2$ matrix. For example, a bosonic same-mode diagonal pair has $v^{\rm pair}_{ii;ii}=V_{ii;ii}$, not $2V_{ii;ii}$; the square-root normalizations remove that apparent factor.

Action on Bosonic Number States

For four distinct modes,

$$\begin{aligned} &a_i^\dagger a_j^\dagger a_l a_k \lvert\ldots,n_i,\ldots,n_j,\ldots, n_k,\ldots,n_l,\ldots\rangle \\ &\quad= \sqrt{ (n_i+1)(n_j+1)n_k n_l } \times \lvert\ldots,n_i+1,\ldots,n_j+1,\ldots, n_k-1,\ldots,n_l-1,\ldots\rangle. \end{aligned}$$

Coincident indices require sequential ladder action. Two important identities are

$$a_i^\dagger a_i^\dagger a_i a_i = n_i(n_i-1)$$

and

$$(a_i^\dagger)^2a_j^2 \lvert n_i,n_j\rangle ={} \sqrt{(n_i+1)(n_i+2)} \times \sqrt{n_j(n_j-1)} \lvert n_i+2,n_j-2\rangle.$$

The first counts ordered pairs in one mode; the Hamiltonian prefactor turns it into the unordered-pair count $n_i(n_i-1)/2$.

Action on Fermionic Number States

For fermions, a quartic term is nonzero only when:

• both required incoming modes are occupied;
• the outgoing modes are available after the annihilations;
• no repeated creation or annihilation index violates Pauli exclusion.

The sign is determined by the declared global mode ordering. It is safest to apply the operators from right to left with the same parity rule used for every fermionic basis state.

In one common orthonormal orbital basis, a two-body operator connects Slater determinants that differ by at most two occupied spin-orbital replacements. This is the origin of the double-excitation selection rule in configuration interaction. The value still includes direct, exchange, and ordering signs; connectivity alone does not determine the matrix element.

Basis Covariance

Let a new orthonormal one-particle basis be

$$\lvert\chi_\alpha\rangle = \sum_i \lvert\varphi_i\rangle U_{i\alpha},$$

with unitary $U$. The new creation operators satisfy

$$b_\alpha^\dagger = \sum_i U_{i\alpha}a_i^\dagger.$$

The two-body tensor transforms as

$$V'_{\alpha\beta;\gamma\delta} ={} \sum_{i,j,k,l} U_{i\alpha}^* U_{j\beta}^* V_{ij;kl} \times U_{k\gamma} U_{l\delta}.$$

Using $V'$ with the $b$ operators gives the same many-body operator. A one-particle unitary rotation preserves body rank, Hermiticity, and exchange symmetry, but it need not preserve locality, density-density form, or sparsity.

A real-space onsite interaction, for example, generally becomes a momentum-scattering tensor after Fourier transformation.

Continuum Field Form

For a real symmetric coordinate-diagonal pair potential $v(x,x')$,

$$\widehat V ={} \frac12 \int dx\,dx'\, \psi^\dagger(x) \psi^\dagger(x') \times v(x,x') \psi(x') \psi(x).$$

Here $x$ may include position and a discrete internal label. Expanding

$$\psi(x) = \sum_i \varphi_i(x)a_i$$

gives

$$V_{ij;kl} ={} \int dx\,dx'\, \varphi_i^*(x) \varphi_j^*(x') \times v(x,x') \varphi_k(x) \varphi_l(x').$$

The Field Operators in Many-Body Models page explains distributional products, dimensions and cutoffs. A genuinely nonlocal interaction kernel needs four coordinate arguments, rather than the two displayed here.

Contact interactions

The standard regulated contact terms are

$$\widehat V_B=\frac{g_B}{2}\int d^dr\,\psi^\dagger\psi^\dagger\psi\psi, \qquad \widehat V_F=g_F\int d^dr\,\psi_\uparrow^\dagger\psi_\downarrow^\dagger\psi_\downarrow\psi_\uparrow.$$

The bosonic term counts identical pairs with $1/2$; the fermionic expression lists the opposite-spin pair only once. The same-component fermionic $s$-wave contact vanishes, without excluding finite-range or odd-partial-wave interactions. Field-operator applications distinguishes a bare coupling, a scattering-matched pseudopotential, and the required ultraviolet prescription. Normal ordering alone does not make a three-dimensional delta interaction a well-defined cutoff-independent microscopic potential.

Coulomb Interaction

For particles of charge magnitude $q$ in a homogeneous isotropic medium approximated by a static permittivity $\epsilon$,

$$v_C(\mathbf r,\mathbf r') = \frac{q^2} {4\pi\epsilon \lvert\mathbf r-\mathbf r'\rvert}.$$

The electron-electron term is

$$\widehat V_C ={} \frac12 \sum_{\sigma,\sigma'} \int d^3r\,d^3r'\, \psi_\sigma^\dagger(\mathbf r) \psi_{\sigma'}^\dagger(\mathbf r') \times \frac{q^2} {4\pi\epsilon \lvert\mathbf r-\mathbf r'\rvert} \psi_{\sigma'}(\mathbf r') \psi_\sigma(\mathbf r).$$

Spin is conserved by the interaction but still belongs to every mode label. In electronic-structure language, fixed-nucleus electron-nucleus attraction is one-body, whereas electron-electron repulsion is two-body.

Long-range Coulomb models also require boundary and neutrality conventions. In a periodic calculation, the treatment of the zero-momentum component, ionic background, and finite-size corrections is part of the model definition.

Lattice Density Interactions

A generic lattice density interaction can be written

$$\widehat V_{\mathrm{dd}} = \frac12 \sum_{i,j} U_{ij} a_i^\dagger a_j^\dagger a_j a_i,$$

where $i,j$ are complete lattice-mode labels. If $i\neq j$,

$$a_i^\dagger a_j^\dagger a_j a_i = n_i n_j.$$

When $U_{ij}=U_{ji}$ is real and only distinct modes are included,

$$\widehat V_{\mathrm{dd}} = \frac12 \sum_{i\neq j} U_{ij}n_i n_j.$$

If each unordered bond is listed once, the same interaction is

$$\widehat V_{\mathrm{dd}} = \sum_{\langle i,j\rangle} U_{ij}n_i n_j.$$

The summation symbol therefore carries normalization information.

Beyond Density-Density Terms

Two-body operators need not be diagonal in occupation number. Representative structures include:

$$c_{a\uparrow}^\dagger c_{b\downarrow}^\dagger c_{a\downarrow} c_{b\uparrow}$$

for spin exchange, and

$$c_{a\uparrow}^\dagger c_{a\downarrow}^\dagger c_{b\downarrow} c_{b\uparrow}$$

for pair hopping. Bosonic models can contain $(a_i^\dagger)^2a_j^2$, and projected interactions can generate density-assisted hopping.

Calling every quartic term a density interaction discards physically important scattering channels.

Momentum-Space Representation

For a translationally invariant continuum in volume $\Omega$, choose

$$\psi_\sigma(\mathbf r) = \frac{1}{\sqrt\Omega} \sum_{\mathbf k} e^{i\mathbf k\cdot\mathbf r} a_{\mathbf k\sigma}.$$

If $\widetilde v(\mathbf q)$ is the Fourier transform of $v(\mathbf r-\mathbf r')$, then

$$\widehat V ={} \frac{1}{2\Omega} \sum_{\mathbf k,\mathbf k',\mathbf q} \sum_{\sigma,\sigma'} \widetilde v(\mathbf q) \times a_{\mathbf k+\mathbf q,\sigma}^\dagger a_{\mathbf k'-\mathbf q,\sigma'}^\dagger a_{\mathbf k',\sigma'} a_{\mathbf k,\sigma}.$$

One particle gains wave vector $\mathbf q$ while the other loses it; physical momentum transfer is $\hbar\mathbf q$. The total incoming and outgoing momenta agree term by term:

$$(\mathbf k+\mathbf q) + (\mathbf k'-\mathbf q) = \mathbf k+\mathbf k'.$$

For contact interactions, $\widetilde v(\mathbf q)$ is momentum independent within the effective theory. For the three-dimensional Coulomb kernel, it is proportional to $1/q^2$ away from $\mathbf q=0$.

On a lattice, crystal momentum is conserved modulo a reciprocal lattice vector, so Umklapp processes can occur.

This section records the interaction-specific formula. The finite-box and continuum normalization dictionary, kinetic terms, density modes, and lattice Fourier conventions are developed in Momentum-Space Representation.

Two-Body Reduced Density Matrix

For a many-body density operator $\rho_{\mathrm{MB}}$, define the unnormalized two-body reduced density matrix by

$$\Gamma_{ij;kl}^{(2)} = \operatorname{Tr} \left( \rho_{\mathrm{MB}} a_k^\dagger a_l^\dagger a_j a_i \right).$$

This convention mirrors

$$\gamma_{ij} = \operatorname{Tr} \left( \rho_{\mathrm{MB}}a_j^\dagger a_i \right)$$

for the one-body reduced density matrix. We assume finite $\langle N(N-1)\rangle$; finite mean number alone is not sufficient. The two incoming indices appear first in $\Gamma^{(2)}$, while the creation indices appear after the semicolon.

Other communities reverse the pair order or divide by $N(N-1)$. A quoted two-body density matrix is incomplete unless its operator order and normalization are stated.

Off-Diagonal Long-Range Order uses these conventions to formulate Yang’s extensive-eigenvalue criterion for fermion-pair condensation. The distinction between ordered coordinate pairs and unique normalized pair modes matters for that eigenvalue criterion.

Expectation Values from the Two-Body Density Matrix

With the convention above,

$$\langle\widehat V\rangle = \frac12 \sum_{i,j,k,l} V_{ij;kl} \Gamma_{kl;ij}^{(2)}.$$

The reversed pair order is the matrix-trace contraction on the two-particle product space:

$$\langle\widehat V\rangle = \frac12 \operatorname{Tr}_{\mathcal H_1^{\otimes2}} \left( V\Gamma^{(2)} \right).$$

For antisymmetrized fermionic integrals,

$$\langle\widehat V\rangle = \frac14 \sum_{i,j,k,l} \langle ij\Vert kl\rangle \Gamma_{kl;ij}^{(2)}.$$

The one-body density matrix determines expectations of all one-body operators. The two-body density matrix is the additional object needed for all number-conserving pair interactions.

Exchange and Hermiticity of the Two-Body Density Matrix

The ladder algebra gives

$$\begin{aligned} \Gamma_{ji;kl}^{(2)} &= \eta\Gamma_{ij;kl}^{(2)}, \\ \Gamma_{ij;lk}^{(2)} &= \eta\Gamma_{ij;kl}^{(2)}, \\ \Gamma_{ij;kl}^{(2)*} &= \Gamma_{kl;ij}^{(2)}. \end{aligned}$$

For fermions, repeated indices within either pair vanish. For bosons, coincident indices encode same-mode pair occupation and need not vanish.

These identities are useful storage reductions and stringent numerical checks.

Trace and Contraction Sum Rules

The trace of the unnormalized two-body density matrix is

$$\sum_{i,j} \Gamma_{ij;ij}^{(2)} = \langle N(N-1)\rangle.$$

It counts ordered pairs. In a fixed-$N$ sector,

$$\sum_{i,j} \Gamma_{ij;ij}^{(2)} = N(N-1).$$

Partial contraction returns the one-body density matrix:

$$\sum_j \Gamma_{ij;kj}^{(2)} = (N-1)\gamma_{ik}$$

for fixed $N$. In a variable-particle-number state, one must retain the number-operator correlation; replacing $N$ by its mean generally does not give the exact contraction.

In the unique normalized pair convention above, set $C_{ij;kl}=\langle B_{kl}^\dagger B_{ij}\rangle$. Its trace is $\langle N(N-1)\rangle/2$. The ordered-space $\Gamma^{(2)}$, represented in an orthonormal symmetric or antisymmetric pair basis, has matrix $2C$; therefore its nonzero eigenvalues are twice those of $C$. If a source uses instead $M_{ij;kl}=\langle B_{ij}^\dagger B_{kl}\rangle$, then $M=C^T$ and eigenvectors are complex conjugated, while eigenvalues agree.

For fixed $N\ge2$, the trace-one two-particle reduced state is $\Gamma^{(2)}/[N(N-1)]$. Positivity, exchange symmetry, trace, and contraction are necessary consistency checks, but they are not by themselves sufficient to guarantee that an arbitrary tensor is the two-body marginal of a valid many-fermion or many-boson state. This is the many-body representability problem.

Transition Two-Body Density Matrices

For initial and final many-body states, define

$$\Gamma_{ij;kl}^{fi} = \langle\Psi_f\vert a_k^\dagger a_l^\dagger a_j a_i \vert\Psi_i\rangle.$$

Then

$$\langle\Psi_f\vert \widehat V \vert\Psi_i\rangle = \frac12 \sum_{i,j,k,l} V_{ij;kl} \Gamma_{kl;ij}^{fi}.$$

Transition two-body densities encode pair-transfer and double-replacement amplitudes. They are not positive density operators and need not be Hermitian when $f\neq i$.

Pair Density and Pair Correlations

In the continuum, the diagonal pair density is

$$n^{(2)}(x,x') = \langle \psi^\dagger(x) \psi^\dagger(x') \psi(x') \psi(x) \rangle.$$

Its normalization is

$$\int dx\,dx'\, n^{(2)}(x,x') = \langle N(N-1)\rangle.$$

Where the one-body densities are nonzero, a normalized second-order coherence is

$$g^{(2)}(x,x') = \frac{n^{(2)}(x,x')} {n(x)n(x')}.$$

Here $n(x)=\langle\psi^\dagger(x)\psi(x)\rangle$ is the mean one-body density.

The interaction energy for a coordinate-diagonal pair kernel is

$$\langle\widehat V\rangle = \frac12 \int dx\,dx'\, v(x,x')n^{(2)}(x,x').$$

Mean density alone does not determine this energy in a correlated state. Pair avoidance, bunching, antibunching, and short-range correlation holes live in $n^{(2)}$ or the full $\Gamma^{(2)}$.

Slater Determinants and Direct Minus Exchange

For a fermionic Slater determinant, Wick factorization gives

$$\Gamma_{ij;kl}^{(2)} = \gamma_{ik}\gamma_{jl} - \gamma_{il}\gamma_{jk}.$$

In an occupied-orbital basis,

$$\langle\widehat V\rangle = \frac12 \sum_{p,q\in\mathrm{occ}} \left( V_{pq;pq} - V_{pq;qp} \right).$$

The first term is direct; the second is exchange. Exchange follows from fermionic antisymmetry even when $v$ itself is spin independent.

For opposite-spin orbitals, the exchange integral often vanishes because the spin functions are orthogonal. The direct interaction can remain nonzero.

For a spin-independent interaction and orbitals with definite orthogonal spin factors, that opposite-spin exchange is exactly zero; generic spinor orbitals need not obey the simplification. An interacting exact state generally does not obey this factorization. The difference between its $\Gamma^{(2)}$ and the antisymmetrized product of $\gamma$ carries genuine two-particle correlation information.

What the Two-Body Density Matrix Does Not Determine

Together, $\gamma$ and $\Gamma^{(2)}$ determine every expectation value of number-conserving operators with body rank at most two. At fixed $N\ge2$, the contraction sum rule recovers $\gamma$ from $\Gamma^{(2)}$. At $N=1$, every two-body marginal vanishes and carries no information about the one-particle state. The two-body density matrix does not generally determine:

• the full many-body wavefunction or density operator;
• all three-body and higher observables;
• every entanglement property;
• real-time evolution under a two-body Hamiltonian without higher reduced densities.

The last point is important. The equation of motion for $\Gamma^{(2)}$ under a two-body Hamiltonian generally couples to a three-body reduced density matrix. This is the next level of the reduced-density hierarchy.

Symmetries and Conserved Quantities

Every term in the standard pair interaction has two creations and two annihilations, so

$$[N,\widehat V] = 0.$$

For another additive generator $Q=d\Gamma(q)$, the first-quantized criterion is

$$\left[ q\otimes I + I\otimes q, v \right] = 0.$$

When it holds,

$$[Q,\widehat V] = 0.$$

Examples include total momentum for a translationally invariant interaction and total spin for a spin-independent rotationally invariant interaction. A pair interaction can preserve the total generator while changing each particle or mode contribution separately.

Symmetry should be checked against the full coefficient tensor, boundary conditions, and truncation. A formally invariant continuum kernel can lose a symmetry after projection onto an asymmetric basis.

Normal Ordering Relative to a Reference State

Normal ordering relative to a filled Slater determinant or correlated reference is a different operation. Contractions can decompose a two-body interaction schematically as

$$\widehat V = E_{\mathrm{ref}} + \widehat F_{\mathrm{ref}} + \widehat V_{\mathrm{res}},$$

where the three terms are zero-body, one-body, and residual two-body pieces relative to that reference.

This decomposition does not change the original physical body rank. It reorganizes the operator around a chosen state and underlies mean-field and many post-mean-field methods. The reference, contraction convention, and residual term must be stated together.

The explicit fermionic signs, Slater-determinant coefficients, correlated-reference cumulants, and normal-ordered rank truncations are developed in Normal Ordering in Many-Body QM.

Projection and Effective Interactions

Let $P_1$ project onto a retained one-particle subspace. Direct projection gives

$$v_P = (P_1\otimes P_1) v (P_1\otimes P_1),$$

followed by the ordinary two-body lift of $v_P$. This defines the interaction of the projected model.

It need not reproduce low-energy observables of the full model when discarded modes participate virtually. A dynamical decoupling transformation can renormalize the two-body coefficients and induce three-body and higher operators even if the microscopic Hamiltonian began with only pair interactions.

Dropping induced many-body terms is an approximation whose accuracy depends on scale separation, density, and the observable. Effective Hamiltonians in Many-Body Systems develops this issue through projected Hubbard and Anderson reductions.

Mean-Field Reduction Is Not an Identity

Hartree, Fock, Hartree–Fock, Bogoliubov, and related approximations replace part of a quartic interaction by state-dependent lower-degree operators. Schematically,

$$a_i^\dagger a_j^\dagger a_l a_k \longrightarrow \langle a_i^\dagger a_k\rangle a_j^\dagger a_l + \cdots.$$

The omitted terms and subtraction terms depend on the decoupling channel. A resulting bilinear or pairing Hamiltonian is effective and state dependent; it does not prove that the original interaction was one-body.

Different decouplings emphasize density, exchange, pairing, or other order parameters. Comparing their energies requires consistent constants and double-counting corrections.

Computational Scaling and Structure

For $M$ one-particle modes, a dense unsymmetrized tensor has $M^4$ entries. Practical calculations exploit:

• Hermiticity and exchange symmetry;
• particle-number, spin, point-group, and momentum blocks;
• locality and finite interaction range;
• sparse determinant connectivity;
• low-rank, density-fitting, or Cholesky representations where justified;
• on-the-fly matrix-vector products instead of storing the full many-body matrix.

The formal $M^4$ count is not a universal runtime law. The useful complexity depends on the integral structure, basis, symmetry sectors, and numerical method.

Exercises

Statistics-adapted prefactor

Let

$$W_{ij;kl}^{(\eta)} = V_{ij;kl} + \eta V_{ij;lk}.$$

Show that

$$\frac14 \sum_{i,j,k,l} W_{ij;kl}^{(\eta)} a_i^\dagger a_j^\dagger a_l a_k = \frac12 \sum_{i,j,k,l} V_{ij;kl} a_i^\dagger a_j^\dagger a_l a_k.$$

Solution

Write the contribution from the exchanged coefficient as

$$S = \sum_{i,j,k,l} \eta V_{ij;lk} a_i^\dagger a_j^\dagger a_l a_k.$$

Relabel $k\leftrightarrow l$:

$$S = \sum_{i,j,k,l} \eta V_{ij;kl} a_i^\dagger a_j^\dagger a_k a_l.$$

For bosons or fermions,

$$a_k a_l = \eta a_l a_k$$

when the coincident fermionic case is understood to vanish. Hence

$$S = \eta^2 \sum_{i,j,k,l} V_{ij;kl} a_i^\dagger a_j^\dagger a_l a_k.$$

Because $\eta^2=1$, the exchanged term equals the original term. The sum containing $W^{(\eta)}$ is therefore twice the unsymmetrized sum, and the prefactor changes from $1/2$ to $1/4$.

Two-body density-matrix sum rules

For a fixed-$N$ state, prove

$$\sum_{i,j} \Gamma_{ij;ij}^{(2)} = N(N-1)$$

and

$$\sum_j \Gamma_{ij;kj}^{(2)} = (N-1)\gamma_{ik}.$$

Solution

For the trace,

$$\begin{aligned} \sum_{i,j} \Gamma_{ij;ij}^{(2)} &= \left\langle \sum_{i,j} a_i^\dagger a_j^\dagger a_j a_i \right\rangle \\ &= \langle N(N-1)\rangle. \end{aligned}$$

The operator counts an ordered choice of two distinct particles. On a fixed-$N$ sector, its eigenvalue is $N(N-1)$.

For the partial contraction,

$$\begin{aligned} \sum_j \Gamma_{ij;kj}^{(2)} &= \left\langle a_k^\dagger \left( \sum_j a_j^\dagger a_j \right) a_i \right\rangle \\ &= \langle a_k^\dagger N a_i\rangle. \end{aligned}$$

Acting on a fixed-$N$ state, $a_i$ first produces the $(N-1)$ sector. Therefore

$$\langle a_k^\dagger N a_i\rangle = (N-1) \langle a_k^\dagger a_i\rangle = (N-1)\gamma_{ik}.$$

Direct and exchange energy

For a fermionic Slater determinant with occupied spin-orbitals $p,q$, use

$$\Gamma_{ij;kl}^{(2)} = \gamma_{ik}\gamma_{jl} - \gamma_{il}\gamma_{jk}$$

to derive its interaction energy.

Solution

In the occupied-orbital basis,

$$\gamma_{ij} = \delta_{ij}n_i, \qquad n_i\in\{0,1\}.$$

Insert the Slater two-body density into

$$\langle\widehat V\rangle = \frac12 \sum_{i,j,k,l} V_{ij;kl} \Gamma_{kl;ij}^{(2)}.$$

The first product of one-body densities sets $k=i$ and $l=j$; the second sets $k=j$ and $l=i$. Only occupied $i,j$ remain:

$$\langle\widehat V\rangle = \frac12 \sum_{p,q\in\mathrm{occ}} \left( V_{pq;pq} - V_{pq;qp} \right).$$

The two contributions are the direct and exchange integrals. For $p=q$, they cancel, excluding fermionic self-interaction.

Additive symmetry test

Let $Q=d\Gamma(q)$ and suppose

$$\left[ q\otimes I + I\otimes q, v \right] = 0.$$

Explain why $[Q,\widehat V]=0$, and apply the statement to a translationally invariant pair potential.

Solution

On a fixed-$N$ sector,

$$Q^{(N)} = \sum_{\alpha=1}^N q^{(\alpha)}, \qquad \widehat V^{(N)} = \sum_{\alpha<\beta} v^{(\alpha\beta)}.$$

For a given pair $(\alpha,\beta)$, all $q^{(\gamma)}$ with $\gamma$ outside the pair commute with $v^{(\alpha\beta)}$. The remaining commutator is

$$\left[ q^{(\alpha)}+q^{(\beta)}, v^{(\alpha\beta)} \right],$$

which vanishes by the two-particle assumption. Summing over pairs gives

$$[Q^{(N)},\widehat V^{(N)}] = 0$$

in every sector, hence $[Q,\widehat V]=0$ on Fock space.

For translations, $q$ is one-particle momentum and a kernel $v(\mathbf r-\mathbf r')$ is invariant under simultaneous translation of both coordinates. Therefore total momentum commutes with the interaction, even though each particle can exchange momentum with the other.

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