One-Body Operator Applications

A one-body operator applies the same one-particle operator to each particle and adds the results. If $A$ acts on the one-particle Hilbert space $\mathcal H_1$, its many-body lift is

$$\widehat A = d\Gamma(A) = \sum_{i,j} A_{ij}a_i^\dagger a_j, \qquad A_{ij} = \langle\varphi_i\vert A \vert\varphi_j\rangle.$$

The bilinear $a_i^\dagger a_j$ removes one particle from mode $j$ and places it in mode $i$. Diagonal matrix elements weight occupations; off-diagonal matrix elements mix modes.

The one-body lift derives the equivalence to $\sum_{\alpha=1}^N A^{(\alpha)}$. Here we use it to evaluate probes, exploit basis changes, evolve reduced states, and identify what cannot be learned from one-body information alone.

Helpful background. One-body operators derives the additive lift; occupation-number representation supplies the matrix basis.

Additive observables in a mode basis

Take an ordered orthonormal basis of complete one-particle modes. We distinguish the one-particle matrix $A$ from its Fock-space lift $\widehat A=d\Gamma(A)$ and the many-body state $\rho_{\rm MB}$ from its one-body reduced matrix $\gamma$. The source index $j$ is annihilated and destination $i$ is created. For a Hermitian matrix, $A_{ij}=A_{ji}^*$, so an off-diagonal transfer comes with its conjugate.

The lift vanishes on vacuum, agrees with $A$ in the one-particle sector, and adds one contribution per particle at fixed $N$. In particular,

$$d\Gamma(I)=N,\qquad [N,d\Gamma(A)]=0.$$

It can mix spin or orbital labels while preserving total number. One-body does not mean diagonal, spatially local, or weak: strong external potentials and nonlocal hopping kernels are one-body, while the full system may also have interactions. Self-adjoint unbounded $A$ gives a self-adjoint lift on its natural direct-sum domain; formal Hermitian coefficients alone do not specify that domain.

Zero-, One-, and Two-Body Structure

Operator degree should be classified by how many particles a term acts on, not only by the number of ladder symbols visible after a change of variables.

Structure	Typical form	Meaning
zero-body	$E_0 I$	constant energy or normalization offset
linear source	$\eta_i a_i^\dagger+\eta_i^*a_i$	creates or removes one particle; not an additive one-body lift
one-body	$A_{ij}a_i^\dagger a_j$	acts on one particle at a time and conserves number
pairing quadratic	$\Delta_{ij}a_i^\dagger a_j^\dagger+\text{h.c.}$	quadratic but number nonconserving; not $d\Gamma(A)$
two-body	$V_{ij;kl}a_i^\dagger a_j^\dagger a_l a_k$	acts on particle pairs

The linear source row describes bosons; an ordinary physical fermion Hamiltonian preserves parity and cannot contain such an isolated odd term with a complex-number coefficient. Calling every quadratic expression a one-body operator hides the distinction between number-conserving lifts and pairing or source terms.

Diagonal and Off-Diagonal Parts

Split

$$\widehat A = \sum_i A_{ii}n_i + \sum_{i\ne j} A_{ij}a_i^\dagger a_j.$$

The diagonal part is determined by mode occupations in the chosen basis. The off-diagonal part probes or generates one-body coherence.

Dropping the off-diagonal terms is justified only when:

• $A$ is diagonal in that basis;
• the off-diagonal matrix elements vanish by symmetry;
• or a stated approximation discards them.

A state diagonal in the occupation basis does not make an arbitrary one-body operator diagonal. State structure and operator structure are separate questions.

Basis Covariance

Let a new orthonormal basis be

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

with unitary $U$. The new annihilation operators are

$$b_\alpha = \sum_i U_{i\alpha}^* a_i,$$

and the matrix transforms as

$$A' = U^\dagger A U.$$

The Fock-space operator is invariant:

$$\sum_{\alpha,\beta} A'_{\alpha\beta} b_\alpha^\dagger b_\beta = \sum_{i,j} A_{ij}a_i^\dagger a_j.$$

Matrix elements and mode operators must be transformed together. Rotating only one of them changes the physical operator.

Diagonalizing a Hermitian One-Body Operator

If $A$ is Hermitian, choose eigenmodes

$$A\lvert\chi_\alpha\rangle = a_\alpha \lvert\chi_\alpha\rangle.$$

Then

$$\widehat A = \sum_\alpha a_\alpha b_\alpha^\dagger b_\alpha.$$

For a purely one-body Hamiltonian,

$$H_1 = d\Gamma(h),$$

this makes occupation-number states in the eigenmode basis many-body energy eigenstates:

$$E_{\boldsymbol n} = \sum_\alpha \epsilon_\alpha n_\alpha.$$

Diagonalizing $h$ does not solve an interacting Hamiltonian. The interaction transforms with the basis and generally remains quartic and nondiagonal.

Degeneracy and Basis Freedom

Within a degenerate eigenspace of $A$, the diagonal form does not select a unique basis. One may use that freedom to simplify another commuting symmetry, localize orbitals, preserve real structure, or reduce the complexity of interaction matrix elements.

If two one-particle Hermitian operators commute,

$$[A,B] = 0,$$

they admit a common eigenbasis in a finite-dimensional space. For unbounded operators, formal commutation on a common domain is insufficient: joint spectral resolution requires strong commutation of their spectral measures, and continuous spectra need not have normalizable eigenvectors. Their lifts then commute as well. Degeneracy labels should be included in the complete mode index rather than suppressed when another operator acts inside the degenerate subspace.

Continuum Form

If $A_x$ is the position-space representation of the one-particle operator,

$$\widehat A = \int d^dx\, \psi^\dagger(x) A_x \psi(x).$$

Examples include:

$$\begin{aligned} N &= \int d^dx\, \psi^\dagger\psi, \\ T &= \int d^dx\, \psi^\dagger \left( -\frac{\hbar^2\nabla^2}{2m} \right) \psi, \\ U &= \int d^dx\, U(x)\psi^\dagger\psi. \end{aligned}$$

For differential operators, the domain and boundary conditions matter. Integrating by parts can move derivatives between fields only when the boundary term is controlled.

Field Operators in Many-Body Models explains continuum fields and regularization. Real-Space Representation explains cell normalization, finite differences and boundaries.

Lattice Form

On a lattice,

$$\widehat A = \sum_{i,j} A_{ij}a_i^\dagger a_j.$$

Representative structures are:

$$\begin{aligned} \text{onsite potential:} &\quad \sum_i U_i n_i, \\ \text{hopping:} &\quad \sum_{i,j}t_{ij}a_i^\dagger a_j, \\ \text{polarization:} &\quad \sum_i q\mathbf R_i n_i. \end{aligned}$$

Hopping is one-body even though it couples two sites. It transfers one particle at a time. An intersite density product $n_i n_j$ is two-body because it couples simultaneous occupations.

The current generated by hopping is developed in Density Operators and Current Operators. The Tight-Binding Model develops the dispersion and boundary effects of these hopping matrices.

Complete Internal Labels

For orbital or site label $i$ and internal label $\alpha$,

$$\widehat A = \sum_{i,j} \sum_{\alpha,\beta} A_{i\alpha,j\beta} a_{i\alpha}^\dagger a_{j\beta}.$$

The matrix can contain:

• spin rotations and Zeeman coupling;
• spin-orbit-coupled hopping;
• orbital hybridization;
• interband transitions;
• coherent species conversion;
• sublattice mixing.

For spin-$1/2$, a local spin operator is

$$S_i^a = \frac{\hbar}{2} \sum_{\alpha,\beta} c_{i\alpha}^\dagger (\sigma^a)_{\alpha\beta} c_{i\beta}.$$

It is one-body and preserves total particle number, though it can change $N_\uparrow$ and $N_\downarrow$ separately.

External Probes

A classical probe that couples to a one-particle observable produces

$$H_{\mathrm{probe}}(t) = -f(t)\widehat A.$$

Examples include:

• an electric field coupled to dipole or polarization;
• a magnetic field coupled to spin or magnetic moment;
• a scalar potential coupled to density;
• a vector potential coupled through the current and diamagnetic terms;
• a laser or radio-frequency field coupling internal states.

The perturbation is one-body even when the response is strongly many-body. Interactions change the state, spectrum, matrix elements, and response function.

One-Body Reduced Density Matrix

For a many-body density operator $\rho_{\mathrm{MB}}$, define

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

Then

$$\gamma^\dagger = \gamma, \qquad \gamma \geq 0,$$

and

$$\operatorname{Tr}_{\mathcal H_1}\gamma = \langle N\rangle.$$

For a fixed-$N$ state, the trace-one reduced state is

$$\rho^{(1)} = \frac{\gamma}{N}.$$

We assume finite $\langle N\rangle$ so that $\gamma$ is trace class. For a Hermitian unbounded observable, the expectation also requires the corresponding integrability. The unnormalized convention is often more convenient in many-body theory because its eigenvalues are mean occupations.

Expectation Values From the One-Body Matrix

Every one-body expectation is determined by $\gamma$:

$$\begin{aligned} \langle\widehat A\rangle &= \sum_{i,j} A_{ij} \gamma_{ji}, \\ &= \operatorname{Tr}_{\mathcal H_1} \left( A\gamma \right). \end{aligned}$$

If $A$ is diagonal,

$$\langle\widehat A\rangle = \sum_i A_i \langle n_i\rangle.$$

If $A$ is not diagonal, off-diagonal coherences $\gamma_{ji}$ are required.

Two different many-body states can have the same $\gamma$ and therefore agree on every one-body expectation while differing in pair correlations, entanglement, or higher moments. One-body data are complete for one-body observables, not for the full many-body state.

Natural-Orbital Basis

Diagonalizing

$$\gamma \lvert\chi_\alpha^{\mathrm{nat}}\rangle = \nu_\alpha \lvert\chi_\alpha^{\mathrm{nat}}\rangle$$

defines natural orbitals and natural occupations. In that basis,

$$\gamma_{\alpha\beta} = \nu_\alpha\delta_{\alpha\beta}.$$

This basis diagonalizes the state’s one-body coherence, not necessarily the Hamiltonian or the interaction. For fermions, $0\le\nu_\alpha\le1$; for bosons, $\nu_\alpha\ge0$ with the total fixed by $\operatorname{Tr}\gamma$. These constraints are developed in Occupation Numbers. Off-Diagonal Long-Range Order explains the thermodynamic scaling of natural occupations for bosonic condensation and the Pauli obstruction for fermions.

Transition One-Body Density Matrix

For initial and final many-body states, define

$$\gamma_{ij}^{fi} = \langle\Psi_f\vert a_j^\dagger a_i \vert\Psi_i\rangle.$$

Then

$$\langle\Psi_f\vert \widehat A \vert\Psi_i\rangle = \sum_{i,j} A_{ij} \gamma_{ji}^{fi}.$$

This factorization separates the probe or operator matrix $A_{ij}$ from the many-body transition information.

Transition one-body densities appear in spectroscopy, response theory, quantum chemistry, and nuclear structure. They are not ordinary density operators: for $f\ne i$, they need not be positive or Hermitian.

Connectivity of Basis States

A one-body bilinear changes at most one occupied mode into one other mode.

For determinants constructed from one common orthonormal orbital basis, a one-body matrix element can be nonzero only when the determinants:

• are identical;
• or differ by one occupied spin-orbital replacement.

Determinants differing by two or more replacements require a two-body or higher operator at the basis-state level. Fermionic ordering still supplies the sign.

For bosonic number states, a bilinear changes one occupation down by one and another up by one. This sparse connectivity is central in exact diagonalization and configuration-interaction calculations.

Interacting eigenstates are superpositions of basis states, so their full transition matrix elements can connect complicated states even though every contributing bilinear has one-particle connectivity.

Algebra of Bilinears

For canonical bosonic or fermionic modes, ordinary commutators of number-conserving bilinears satisfy

$$[a_i^\dagger a_j,a_k^\dagger a_l] ={} \delta_{jk}a_i^\dagger a_l - \delta_{il}a_k^\dagger a_j.$$

Consequently,

$$[d\Gamma(A),d\Gamma(B)] = d\Gamma([A,B]).$$

The map $d\Gamma$ preserves the Lie algebra of one-particle operators. No statistics-dependent sign remains in the final bilinear commutator.

This identity is useful for:

• lifting symmetry generators;
• deriving equations of motion;
• checking conserved one-body quantities;
• constructing orbital rotations;
• recognizing closed algebras of collective bilinears.

The identity is exact for the finite matrices of a fixed-number, finite-mode calculation. On unrestricted bosonic Fock space, even a finite mode set gives unbounded bilinears; the algebra holds on their common finite-particle core. Continuum applications require corresponding domain care.

Symmetries and Conserved One-Body Quantities

For a purely one-body Hamiltonian

$$H_1 = d\Gamma(h),$$

if

$$[h,A] = 0,$$

then

$$[H_1,\widehat A] = 0.$$

With interactions,

$$H = H_1+V,$$

one must also check

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

A symmetry of the one-particle Hamiltonian need not survive an interaction, boundary condition, external field, or truncation. Conversely, many interactions preserve total number, spin, momentum, or another lifted generator.

Closed Dynamics for a Quadratic Hamiltonian

Let

$$H_1 = \sum_{i,j} h_{ij}a_i^\dagger a_j.$$

The one-body density matrix evolves as

$$i\hbar \frac{d\gamma}{dt} = [h,\gamma].$$

No Gaussian-state assumption is needed. This equation is closed: knowing $\gamma(0)$ and $h(t)$ determines every later one-body expectation for the quadratic, number-conserving system.

For a possibly time-dependent one-body observable,

$$\frac{d}{dt} \langle\widehat A\rangle ={} \frac{i}{\hbar} \langle[H,\widehat A]\rangle + \left\langle d\Gamma(\partial_t A) \right\rangle.$$

When $H=H_1$, the commutator closes on another one-body operator:

$$[H_1,\widehat A] = d\Gamma([h,A]).$$

Interactions Open a Hierarchy

For

$$H = H_1+V,$$

with a two-body interaction $V$, the one-body equation has the structure

$$i\hbar \frac{d\gamma}{dt} = [h,\gamma] + \mathcal C[\Gamma^{(2)}],$$

where $\Gamma^{(2)}$ is a two-body reduced density matrix and $\mathcal C$ denotes the interaction contraction. Its index convention, sum rules, and contraction with pair interactions are developed in Two-Body Operators.

The dynamics of $\gamma$ are no longer closed. The two-body density evolves using three-body information, producing the BBGKY-type hierarchy. Mean-field theories close this hierarchy approximately by expressing higher correlations in terms of lower ones.

This is a precise reason interacting many-body dynamics cannot generally be reduced to evolving occupied one-particle orbitals.

Means Versus Fluctuations

For Hermitian $A$, the one-body density matrix determines $\langle\widehat A\rangle$, but generally not $\langle\widehat A^2\rangle$.

Introduce

$$\eta = \begin{cases} +1,&\text{bosons},\\ -1,&\text{fermions}. \end{cases}$$

Using

$$a_j a_k^\dagger = \delta_{jk} + \eta a_k^\dagger a_j,$$

one finds

$$\widehat A^2 ={} \sum_{i,l} (A^2)_{il} a_i^\dagger a_l + \eta \sum_{i,j,k,l} A_{ij}A_{kl} a_i^\dagger a_k^\dagger a_j a_l.$$

The first term is one-body; the second contains two-body information. Therefore the variance

$$\operatorname{Var}(\widehat A) = \langle\widehat A^2\rangle - \langle\widehat A\rangle^2$$

usually requires a two-body reduced density matrix even though $\widehat A$ itself is one-body.

Extensivity and Spectral Bounds

Suppose the one-particle Hermitian operator satisfies

$$a_{\min}I \leq A \leq a_{\max}I.$$

On the fixed-$N$ sector,

$$N a_{\min}I \leq d\Gamma(A) \leq N a_{\max}I.$$

If these one-particle bounds remain independent of system size, the many-body bounds are at most linear in particle number. This does not apply unchanged to an unbounded position or kinetic-energy operator, or to coefficients that grow with system size. The intensive average per particle is

$$\frac{\widehat A}{N}$$

on a fixed nonzero-$N$ sector.

On bosonic Fock space, a nonzero bounded Hermitian $A$ has an unbounded lift because the same orbital can be occupied arbitrarily often. Fermionic exclusion changes the conclusion: a rank-one projector lifts to a bounded occupation operator, even in an infinite mode space. Other fermionic lifts, such as total number, are unbounded when infinitely many modes are available.

Projection and Active-Space Truncation

Let $P_1$ project onto retained one-particle modes. The directly projected operator is

$$A_P = P_1 A P_1,$$

with lift

$$d\Gamma(A_P).$$

This is exact only for the projected model. If $A$ couples retained and discarded modes, the truncated operator omits those matrix elements. Convergence of the Hamiltonian does not automatically guarantee convergence of every observable.

When high-energy modes are eliminated dynamically, let $\mathcal P$ project onto the retained many-body model space and let $S$ generate the decoupling transformation. A consistent effective observable can be

$$\widehat A_{\mathrm{eff}} = \mathcal P e^S \widehat A e^{-S}\mathcal P.$$

Even if $\widehat A$ began as one-body, the transformed effective operator can acquire two-body and higher terms. Effective Hamiltonians and effective observables must be derived at the same level of approximation.

Mean-Field One-Body Operators

Hartree, Hartree–Fock, Kohn–Sham, and related approximations often produce an effective one-body matrix

$$h_{\mathrm{eff}}[\gamma].$$

Its bilinear form does not make the original interacting Hamiltonian one-body. The effective coefficients may:

• depend on the state or density;
• require self-consistency;
• break a symmetry at the solution level;
• omit connected correlations;
• include double-counting subtraction constants;
• depend on frequency or energy in a more general self-energy description.

A mean-field decoupling is an approximation, not an operator identity. Bogoliubov and pairing mean fields also require anomalous averages such as $\langle a_i a_j\rangle$ and a Nambu matrix; $\gamma$ alone does not supply those data.

Quasiparticle Caution

A Hamiltonian diagonal in quasiparticle occupations,

$$H_{\mathrm{qp}} = E_0 + \sum_\alpha E_\alpha \beta_\alpha^\dagger\beta_\alpha,$$

is one-body in the quasiparticle operators. Its expression in the original particle operators may include pairing terms, constants, and a nontrivial vacuum.

The phrase one-body must therefore identify the degrees of freedom. Bare particles, projected orbitals, collective modes, and Bogoliubov quasiparticles need not define the same operator grading.

Exercises

Two-mode basis rotation

Let

$$A = \begin{pmatrix} \epsilon & t\\ t & \epsilon \end{pmatrix},$$

with real $t$. Use

$$b_\pm = \frac{a_1\pm a_2}{\sqrt2}$$

to diagonalize $\widehat A$.

Solution

The symmetric and antisymmetric one-particle vectors are eigenvectors with eigenvalues

$$a_+ = \epsilon+t, \qquad a_- = \epsilon-t.$$

Therefore

$$\widehat A = (\epsilon+t)b_+^\dagger b_+ + (\epsilon-t)b_-^\dagger b_-.$$

The result holds for bosons or fermions. Statistics changes the allowed occupations, not the one-particle diagonalization.

Contract complex one-body coherences

For

$$A=\begin{pmatrix}2&i\\-i&-1\end{pmatrix},\qquad \gamma=\begin{pmatrix}3/4&i/4\\-i/4&1/4\end{pmatrix},$$

compute $\langle\widehat A\rangle$. How would the interpretation change if the supplied matrix were a trace-one reduced state of an $N$-particle system?

Solution

Contract reversed indices:

$$\operatorname{Tr}(A\gamma)=2(3/4)-(1/4)+i(-i/4)+(-i)(i/4)=7/4.$$

The positive matrix has trace one, so with the unnormalized convention it describes mean particle number one. If it were instead $\rho^{(1)}$ for fixed $N$, the additive observable would have expectation $7N/4$; $7/4$ would be its mean per particle. Using $\sum_{ij}A_{ij}\gamma_{ij}$ would reverse the coherence contribution and give the wrong answer.

Closed one-body dynamics

Let

$$H_1 = d\Gamma(h).$$

Use the bilinear commutator algebra to derive

$$i\hbar\dot\gamma = [h,\gamma].$$

Solution

For

$$\gamma_{ij} = \langle a_j^\dagger a_i\rangle,$$

the Heisenberg equation gives

$$i\hbar \frac{d}{dt} \langle a_j^\dagger a_i\rangle = \left\langle [a_j^\dagger a_i,H_1] \right\rangle.$$

Using the bilinear commutator,

$$i\hbar\dot\gamma_{ij} ={} \sum_k h_{ik}\gamma_{kj} - \sum_k \gamma_{ik}h_{kj}.$$

Therefore

$$i\hbar\dot\gamma = h\gamma-\gamma h = [h,\gamma].$$

No higher reduced density matrix appears because the Hamiltonian is one-body.

Equal one-body data, different fluctuations

Compare two bosonic states $|1,1\rangle$ and $(|2,0\rangle+|0,2\rangle)/\sqrt2$. Find their one-body matrix and the variance of $n_1$.

Solution

Both have $\langle n_1\rangle=\langle n_2\rangle=1$ and zero $\langle b_1^\dagger b_2\rangle$, so $\gamma=I_2$. Every additive one-body mean agrees. In $|1,1\rangle$, $n_1$ is sharp and its variance is zero. In the superposition, $n_1$ is either 0 or 2 with equal probabilities, giving $\langle n_1^2\rangle=2$ and variance 1. Measuring a fluctuation accesses more than the one-body means.

References

• J.-P. Blaizot and G. Ripka, Quantum Theory of Finite Systems, MIT Press (1986).
• H. Bruus and K. Flensberg, Many-Body Quantum Theory in Condensed Matter Physics, Oxford University Press (2004).
• A. J. Coleman, “Structure of fermion density matrices,” Reviews of Modern Physics 35, 668–686 (1963), doi:10.1103/RevModPhys.35.668.
• P. Coleman, Introduction to Many-Body Physics, Cambridge University Press (2015).
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, Dover (2003).
• T. Helgaker, P. Jørgensen, and J. Olsen, Molecular Electronic-Structure Theory, Wiley (2000).
• J. W. Negele and H. Orland, Quantum Many-Particle Systems, Westview Press (1998).
• P. Ring and P. Schuck, The Nuclear Many-Body Problem, Springer (1980).
• A. Szabo and N. S. Ostlund, Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory, Dover (1996).