Normal Ordering in Many-Body QM

Normal ordering is more than a visual rule that places creation operators to the left. In many-body quantum mechanics it is a way to rewrite an operator around a chosen reference state. The reference may be the empty Fock vacuum, a filled Slater determinant, a coherent condensate, or a Bogoliubov quasiparticle vacuum. Changing that reference changes the contractions and therefore changes which constant, one-body, pairing, and residual interaction terms appear.

The central logical distinction is

$$\text{all induced terms retained} \quad\Longrightarrow\quad \text{exact operator identity},$$

whereas

$$\text{residual terms discarded} \quad\Longrightarrow\quad \text{reference-dependent approximation}.$$

This page develops that distinction for nonrelativistic many-body systems. The creation-left definition and elementary vacuum examples live in Normal Ordering. The systematic sum over contractions is introduced separately in Wick’s Theorem Preview.

Helpful background. Normal ordering supplies the vacuum algebra; two-body operators fixes the antisymmetrized coefficient and prefactor convention.

Particles and holes around a filled reference

Take orthonormal fermionic modes and a Slater determinant $|\Phi\rangle$. Indices $i,j$ denote its occupied orbitals, $a,b$ its unoccupied orbitals, and $p,q$ arbitrary orbitals. Braces $\{O\}_\Phi$ will denote reference normal ordering, not an anticommutator. We first work in finitely many modes; continuum products need the regulator qualification below.

For the empty vacuum, every $a_p$ annihilates the reference and creators are moved to the left. Reordering generates contractions, as in $a_pa_q^\dagger=\delta_{pq}-a_q^\dagger a_p$. The nonconstant vacuum-normal products have zero vacuum expectation, but need not have zero expectation in a filled state. A filled reference reverses the roles of operators on its occupied modes.

Excitations relative to a determinant

Suppose $\lvert\Phi\rangle$ contains occupied fermion orbitals. Then

$$a_i^\dagger\lvert\Phi\rangle = 0$$

for every occupied $i$, while

$$a_a\lvert\Phi\rangle = 0$$

for every unoccupied $a$. Relative to $\lvert\Phi\rangle$:

• $a_a^\dagger$ creates a particle excitation;
• $a_a$ annihilates a particle excitation;
• $a_i$ creates a hole excitation;
• $a_i^\dagger$ annihilates a hole excitation.

Thus an operator already normal ordered relative to $\lvert0\rangle$ need not be normal ordered relative to $\lvert\Phi\rangle$.

Particle–hole vacuum

For a Slater determinant with occupation numbers $n_p\in\{0,1\}$, define quasiparticle annihilators

$$\beta_p = \begin{cases} a_p,&n_p=0,\\ a_p^\dagger,&n_p=1. \end{cases}$$

They satisfy

$$\beta_p\lvert\Phi\rangle = 0.$$

Their adjoints create excitations:

$$\beta_p^\dagger = \begin{cases} a_p^\dagger,&n_p=0,\\ a_p,&n_p=1. \end{cases}$$

The particle-hole transformation preserves the canonical anticommutation relations:

$$\{\beta_p,\beta_q^\dagger\} = \delta_{pq}.$$

Reference normal ordering places every $\beta^\dagger$ to the left of every $\beta$. This definition immediately guarantees

$$\langle\Phi\rvert \{O\}_\Phi \lvert\Phi\rangle = 0$$

for every nonconstant normal-ordered quasiparticle monomial.

Reference Contractions

In the occupation basis of $\lvert\Phi\rangle$,

$$\langle\Phi\rvert a_p^\dagger a_q \lvert\Phi\rangle = n_p\delta_{pq},$$

and

$$\langle\Phi\rvert a_p a_q^\dagger \lvert\Phi\rangle = (1-n_p)\delta_{pq}.$$

Both contractions are needed. A filled mode has a nonzero $a_p^\dagger a_p$ contraction, whereas an empty mode has a nonzero $a_p a_p^\dagger$ contraction.

In an arbitrary basis, define

$$\rho_{pq} \equiv \langle\Phi\rvert a_p^\dagger a_q \lvert\Phi\rangle.$$

This index order is the transpose of the common one-particle matrix $\gamma_{pq}=\langle a_q^\dagger a_p\rangle$. For a Slater determinant, $\rho$ is Hermitian and idempotent:

$$\rho^\dagger = \rho, \qquad \rho^2 = \rho.$$

The complementary contraction follows from the canonical anticommutator:

$$\langle\Phi\rvert a_p a_q^\dagger \lvert\Phi\rangle = \delta_{pq} - \rho_{qp}.$$

One-Body Strings Around a Reference

The simplest reference-normal-ordering identity is

$$a_p^\dagger a_q = \{a_p^\dagger a_q\}_\Phi + \rho_{pq}.$$

In the diagonal occupation basis,

$$a_p^\dagger a_q = \{a_p^\dagger a_q\}_\Phi + n_p\delta_{pq}.$$

For the total number operator,

$$\widehat N = N_\Phi + \sum_p \{a_p^\dagger a_p\}_\Phi,$$

where

$$N_\Phi = \sum_p n_p.$$

The normal-ordered correction counts particles above the reference and holes below it with opposite signs. A one-particle-one-hole excitation has the same total particle number as $\lvert\Phi\rangle$.

Fermionic Four-Operator Identity

For a Slater determinant or a number-conserving mixed Gaussian reference (with mixed-state Wick ordering as explained below),

$$a_p^\dagger a_q^\dagger a_s a_r ={} \{a_p^\dagger a_q^\dagger a_s a_r\}_\Phi + \rho_{pr} \{a_q^\dagger a_s\}_\Phi - \rho_{ps} \{a_q^\dagger a_r\}_\Phi - \rho_{qr} \{a_p^\dagger a_s\}_\Phi + \rho_{qs} \{a_p^\dagger a_r\}_\Phi + \rho_{pr}\rho_{qs} - \rho_{ps}\rho_{qr}.$$

The last line is the double-contraction contribution. Taking the reference expectation gives

$$\begin{aligned} \langle a_p^\dagger a_q^\dagger a_s a_r \rangle_\Phi ={}& \rho_{pr}\rho_{qs} - \rho_{ps}\rho_{qr}. \end{aligned}$$

The minus sign is exchange. It is not a convention that may be dropped.

Deriving the Reference-Normal-Ordered Hamiltonian

Consider a fermionic Hamiltonian with one- and two-body terms:

$$H ={} \sum_{p,q} t_{pq}a_p^\dagger a_q + \frac14 \sum_{p,q,r,s} \overline v_{pq;rs} a_p^\dagger a_q^\dagger a_s a_r.$$

The factor $1/4$ accompanies fully antisymmetrized matrix elements. Insert the one- and two-body normal-ordering identities and collect equal quasiparticle ranks. In the orbital basis where $\lvert\Phi\rangle$ occupies $i,j,\ldots$,

$$H ={} E_\Phi + \sum_{p,q} f_{pq} \{a_p^\dagger a_q\}_\Phi + \frac14 \sum_{p,q,r,s} \overline v_{pq;rs} \{a_p^\dagger a_q^\dagger a_s a_r\}_\Phi.$$

The induced coefficients are

$$E_\Phi = \sum_i t_{ii} + \frac12 \sum_{i,j} \overline v_{ij;ij},$$

and

$$f_{pq} = t_{pq} + \sum_i \overline v_{pi;qi}.$$

No approximation has been made. The same operator has been reorganized into zero-body, one-body, and residual two-body pieces relative to $\lvert\Phi\rangle$.

Meaning of the three pieces

The scalar $E_\Phi=\langle\Phi|H|\Phi\rangle$ is the reference energy. It matters when comparing variational states, binding energies or thermodynamic potentials. An irrelevant fixed energy zero in one evolution problem does not justify dropping a reference-dependent scalar in an energy comparison.

The Fock matrix $f$ contains the original one-body term and the contraction of one particle pair in the interaction. The microscopic interaction remains present in the residual two-body operator. That residual can create double particle–hole excitations from the reference and need not be small.

A two-mode example

Let $H=U n_1n_2$ and take the reference $|\Phi\rangle=|1,0\rangle$. The only nonzero contraction is $\langle n_1\rangle_\Phi=1$. The exact decomposition is

$$H=U n_2+U(n_1-1)n_2.$$

The first term is induced one-body energy; the second is the residual reference-normal quartic term. On $|1,1\rangle$ the first gives the correct interaction energy $U$ and the residual vanishes. On $|0,1\rangle$ the two terms cancel, giving zero. Dropping the residual instead predicts $U$ for that state. The example shows why an exact reorganization is not automatically a good approximation far from its reference.

Hartree–Fock Stationarity

Let

$$\lvert\Phi_i^a\rangle = a_a^\dagger a_i \lvert\Phi\rangle$$

be a one-particle-one-hole excitation. The Hamiltonian matrix element is

$$\langle\Phi_i^a\rvert H \lvert\Phi\rangle = f_{ai}.$$

For a self-consistent Hartree–Fock reference, stationarity under occupied-unoccupied orbital rotations gives the Brillouin condition

$$f_{ai} = 0.$$

This removes the direct coupling from the reference to single excitations. It does not remove double excitations or make the residual two-body term vanish.

Exact Reorganization Versus Mean Field

The identity

$$H = E_\Phi + H_\Phi^{[1]} + H_\Phi^{[2]}$$

is exact for a Hamiltonian containing at most two-body interactions. A mean-field approximation begins only after one replaces the full problem by selected terms, for example

$$H \longrightarrow E_\Phi + H_\Phi^{[1]}.$$

Dropping $H_\Phi^{[2]}$ discards residual correlations. Conversely, perturbation theory, coupled-cluster theory, configuration interaction, and Green-function methods retain its effects in different controlled or approximate ways.

Normal ordering and mean-field decoupling are related, but they are not synonyms.

Three-Body Operators and Induced Lower Ranks

Let a fully antisymmetrized three-body operator be

$$W ={} \frac1{36} \sum_{p,q,r,s,t,u} w_{pqr;stu} \times a_p^\dagger a_q^\dagger a_r^\dagger a_u a_t a_s.$$

Normal ordering relative to a Slater determinant produces ranks from zero through three:

$$W = W_\Phi^{[0]} + W_\Phi^{[1]} + W_\Phi^{[2]} + W_\Phi^{[3]}.$$

With the displayed antisymmetrization convention,

$$W_\Phi^{[0]} = \frac16 \sum_{i,j,k} w_{ijk;ijk},$$ $$W_\Phi^{[1]} = \sum_{p,q} \left( \frac12 \sum_{i,j} w_{pij;qij} \right) \{a_p^\dagger a_q\}_\Phi,$$

and

$$W_\Phi^{[2]} ={} \frac14 \sum_{p,q,r,s} \left( \sum_i w_{pqi;rsi} \right) \times \{a_p^\dagger a_q^\dagger a_s a_r\}_\Phi.$$

The residual term is

$$W_\Phi^{[3]} ={} \frac1{36} \sum_{p,q,r,s,t,u} w_{pqr;stu} \times \{a_p^\dagger a_q^\dagger a_r^\dagger a_u a_t a_s\}_\Phi.$$

The normal-ordered two-body approximation retains $W_\Phi^{[0]}$, $W_\Phi^{[1]}$, and $W_\Phi^{[2]}$ but discards $W_\Phi^{[3]}$. That truncation can be accurate for suitable references and observables, but it is not an operator identity.

Pure and mixed Gaussian references

A Gaussian state is specified by its first moments and two-point covariance; higher connected cumulants vanish. Slater determinants, coherent states, squeezed vacua and unprojected BCS vacua are examples. A stable quadratic bosonic Hamiltonian or a gapped quadratic fermionic Hamiltonian has a Gaussian quasiparticle ground state. If the ground space is degenerate, an arbitrary superposition within it need not be Gaussian.

A normalizable thermal state of a quadratic Hamiltonian is also Gaussian when traced over the full Fock space. A full-rank mixed state is not a vector annihilated by quasiparticle annihilators. Its Wick ordering is instead defined algebraically by subtracting its contractions; the four-operator identity still applies, with fractional occupations. Restricting the ensemble to a fixed total number generally destroys this factorization. For a coherent reference with nonzero mean, use centered fluctuation operators before applying the pair-contraction rule.

The Wick theorem gives the complete expansion. Its application requires specifying the reference and whether the product is equal-time, time ordered or contour ordered; contractions in these different products are not interchangeable.

Correlated References and Density Cumulants

An interacting reference need not be Gaussian. First consider a number-invariant reference, with no anomalous pair mean. Define its two-body reduced density matrix by

$$\Gamma_{pq;rs} \equiv \langle\Phi\rvert a_p^\dagger a_q^\dagger a_s a_r \lvert\Phi\rangle.$$

The connected two-body cumulant is

$$\lambda_{pq;rs} = \Gamma_{pq;rs} - \left( \rho_{pr}\rho_{qs} - \rho_{ps}\rho_{qr} \right).$$

For a Slater determinant,

$$\lambda_{pq;rs} = 0.$$

For a non-Gaussian reference it is generally nonzero. Pairwise Wick factorization then fails, and higher density cumulants enter generalized expansions. If the reference has an anomalous mean $\kappa_{pq}=\langle a_pa_q\rangle$, the Gaussian four-point expression also contains $+\kappa_{pq}^*\kappa_{rs}$. That term must be subtracted before calling the remainder a connected fourth cumulant; its presence alone does not imply a non-Gaussian state.

Replacing a correlated reference by a Gaussian one with the same one-body density discards $\lambda$ and higher connected information. That replacement is an approximation, not a change of notation.

Generalized Normal Ordering

Generalized normal ordering can be defined relative to multiconfigurational or correlated references so that reference expectations of non-scalar generalized-normal products vanish. The corresponding generalized Wick theorem contains irreducible density matrices, not only pair contractions.

This machinery is valuable in multireference electronic-structure and nuclear many-body methods. The additional information includes:

• one-body contractions need not be projectors;
• two- and higher-body cumulants survive;
• products of generalized-normal strings generate cumulant terms;
• truncations must specify which cumulant ranks are retained.

Ordinary Slater-determinant formulas must not be imported unchanged into a correlated-reference calculation.

Bosonic Displaced Reference

For a bosonic coherent reference $\lvert\boldsymbol\alpha\rangle$, define fluctuation operators

$$b_i = a_i-\alpha_i,$$

so that

$$b_i \lvert\boldsymbol\alpha\rangle = 0.$$

Then

$$a_i = \alpha_i+b_i.$$

Even a one-body number operator separates into several fluctuation ranks:

$$a_i^\dagger a_i ={} |\alpha_i|^2 + \alpha_i^*b_i + \alpha_i b_i^\dagger + b_i^\dagger b_i.$$

An interacting Hamiltonian becomes

$$H = H^{[0]} + H^{[1]} + H^{[2]} + H^{[3]} + H^{[4]} +\cdots$$

in powers of $b,b^\dagger$. Stationarity of the appropriate mean-field functional, including chemical-potential or other constraints when required, cancels the linear term. Retaining only the quadratic part is a Bogoliubov approximation; retaining every power is the original Hamiltonian expressed in shifted variables.

The c-number substitution $a_i\to\alpha_i$ is therefore not an exact operator identity.

Number-Conserving Bosonic References

A condensate number state is not the same as a coherent state. For a sharp-$N$ state,

$$\langle a_i\rangle = 0$$

by the number selection rule, even when one mode has macroscopic occupation. Number-conserving Bogoliubov methods organize fluctuations without assuming a coherent superposition of total particle numbers.

Normal ordering around a coherent state is often efficient, but its broken-$U(1)$ language and its thermodynamic-limit interpretation should be stated. The dilute-gas setting is introduced in Weakly Interacting Bose Gas Preview, and Bogoliubov theory derives the quadratic diagonalization.

Bogoliubov Quasiparticle Vacuum

A fermionic Bogoliubov transformation mixes particle creation and annihilation operators:

$$\beta_\mu = \sum_p \left( U_{p\mu}^*c_p + V_{p\mu}^*c_p^\dagger \right).$$

The matrices $U,V$ must satisfy the conditions that preserve the canonical anticommutation relations.

For a quasiparticle vacuum $\lvert\Phi_B\rangle$,

$$\beta_\mu \lvert\Phi_B\rangle = 0.$$

Normal ordering relative to $\lvert\Phi_B\rangle$ places $\beta^\dagger$ operators left of $\beta$ operators. In the original particle basis, contractions include both a normal density

$$\rho_{pq} = \langle\Phi_B\rvert c_p^\dagger c_q \lvert\Phi_B\rangle$$

and an anomalous density

$$\kappa_{pq} = \langle\Phi_B\rvert c_p c_q \lvert\Phi_B\rangle.$$

Fermionic antisymmetry requires

$$\kappa_{pq} = -\kappa_{qp}.$$

A parity-even Hamiltonian can then contain quasiparticle terms of several even degrees. At a self-consistent Hartree–Fock–Bogoliubov stationary point, the terms that create two quasiparticles directly from the reference vanish. Diagonalizing the quadratic part gives the mean-field quasiparticle spectrum; residual interactions can shift or broaden these excitations.

Number Symmetry and Anomalous Contractions

The anomalous density carries particle-number charge $-2$ because

$$[\widehat N,c_p c_q] = -2c_p c_q.$$

It vanishes in an exact particle-number eigenstate. A nonzero $\kappa$ therefore signals an unprojected number-breaking reference, a phase reference, or an enlarged relational description.

Pairing correlations do not require a nonzero anomalous average. Neutral quantities such as

$$\langle c_p^\dagger c_q^\dagger c_s c_r \rangle$$

can reveal pairing in a fixed-number state. Number projection generally converts an unprojected quasiparticle vacuum into a non-Gaussian state, so the simple quasiparticle Wick theorem must then be reconsidered. Special projections yielding a vacuum or a single Slater determinant are exceptions.

Basis Transformations

A unitary rotation among annihilation operators,

$$b_\alpha = \sum_p U_{p\alpha}^*a_p,$$

preserves the empty vacuum and does not mix creation with annihilation. Vacuum-normal-ordered polynomials remain vacuum normal ordered after transforming all coefficients consistently.

A particle-hole or Bogoliubov transformation does mix the original creation-annihilation split. It changes the natural reference vacuum and the contractions. The operator is unchanged, but its normal-ordered decomposition is not.

Selecting a different one-particle space can change contractions. Simply projecting a number-conserving one- plus two-body Hamiltonian into the Fock space built from retained orbitals does not itself generate higher-body interactions. Eliminating virtual transitions through discarded states can do so. Basis rotation, reference change, plain projection and effective-Hamiltonian construction are distinct operations.

Continuum Fields and Coincident Points

For equal-time fields,

$$\psi(\mathbf x) \psi^\dagger(\mathbf y) = \delta(\mathbf x-\mathbf y) \pm \psi^\dagger(\mathbf y) \psi(\mathbf x),$$

where the upper sign is bosonic and the lower sign is fermionic. Reordering density products produces contact terms. For either statistics, with the operators ordered consistently,

$$n(\mathbf x)n(\mathbf y) ={} \psi^\dagger(\mathbf x) \psi^\dagger(\mathbf y) \psi(\mathbf y) \psi(\mathbf x) + \delta(\mathbf x-\mathbf y) \psi^\dagger(\mathbf x) \psi(\mathbf y).$$

At coincident points the symbolic term $\delta(\mathbf0)$ warns that operator-valued distributions are being multiplied. A lattice spacing, momentum cutoff, point splitting, smearing, or another regulator may be required.

Normal ordering subtracts contractions associated with a chosen reference. It does not by itself define every singular continuum composite operator or replace renormalization.

Normal Ordering Versus Time Ordering

Normal ordering and time ordering answer different questions.

Ordering	Organizing rule	Main use
normal	creators left of reference annihilators	reference vacuum and Hamiltonian decompositions
time	later times left, with fermionic permutation signs	real-time propagators and perturbation theory
imaginary time	larger imaginary time left, with fermionic permutation signs	equilibrium Matsubara theory
symmetric	average over operator permutations	phase-space and measurement conventions

A time-ordered product is not generally normal ordered. Wick’s theorem relates the two for suitable Gaussian references by adding contractions. In relativistic QFT, those time-ordered contractions become propagators, not merely equal-time Kronecker or Dirac deltas.

Normal Ordering and Measured Correlations

Some measurement models naturally select normal-ordered correlations. For example, idealized particle-counting or photodetection expressions involve

$$G^{(2)}(1,2) = \langle \psi^\dagger(1) \psi^\dagger(2) \psi(2) \psi(1) \rangle.$$

The ordering is part of the observable model. It should not be changed merely because another ordering is algebraically convenient. Ordinary, normal-ordered, symmetrized, retarded, and time-ordered correlators generally encode different physical questions.

The broader taxonomy is developed in Correlation Functions Overview.

Exercises

Zero- and one-body terms from a pair force

Starting from

$$V = \frac14 \sum_{p,q,r,s} \overline v_{pq;rs} a_p^\dagger a_q^\dagger a_s a_r,$$

show that a Slater reference induces

$$V_\Phi^{[0]} = \frac12 \sum_{i,j} \overline v_{ij;ij}$$

and

$$V_\Phi^{[1]} = \sum_{p,q} \left( \sum_i \overline v_{pi;qi} \right) \{a_p^\dagger a_q\}_\Phi.$$

Solution

Insert the four-operator reference-normal-ordering identity. The double contractions give

$$\frac14 \sum_{p,q,r,s} \overline v_{pq;rs} \left( \rho_{pr}\rho_{qs} - \rho_{ps}\rho_{qr} \right).$$

In the occupation basis, $\rho_{pr}=n_p\delta_{pr}$. Using antisymmetry of $\overline v$ makes the two contraction patterns equal after their exchange sign is included. Hence

$$V_\Phi^{[0]} = \frac12 \sum_{i,j} \overline v_{ij;ij}.$$

The four single contractions similarly combine into

$$V_\Phi^{[1]} = \sum_{p,q,i} \overline v_{pi;qi} \{a_p^\dagger a_q\}_\Phi.$$

The remaining uncontracted term is the residual normal-ordered two-body operator.

Brillouin condition

Show that

$$\langle\Phi_i^a\rvert H \lvert\Phi\rangle = f_{ai}$$

for $\lvert\Phi_i^a\rangle=a_a^\dagger a_i\lvert\Phi\rangle$. Why does a Hartree–Fock stationary point set this matrix element to zero?

Solution

In the reference-normal-ordered Hamiltonian, the zero-body term cannot connect states with different quasiparticle content. The residual two-body normal product cannot connect the reference directly to a single particle-hole excitation. The one-body term gives

$$\begin{aligned} \langle\Phi_i^a\rvert H \lvert\Phi\rangle ={}& \sum_{p,q} f_{pq} \langle\Phi_i^a\rvert \{a_p^\dagger a_q\}_\Phi \lvert\Phi\rangle \\ ={}& f_{ai}. \end{aligned}$$

An infinitesimal occupied-unoccupied orbital rotation changes the determinant in the direction of $\lvert\Phi_i^a\rangle$. Stationarity of the Hartree–Fock energy under every such rotation therefore requires $f_{ai}=0$.

Three-body combinatorics

For the antisymmetrized three-body convention used above, explain the factors $1/6$, $1/2$, and $1$ multiplying the induced zero-, one-, and two-body coefficients.

Solution

The original operator carries $1/(3!)^2=1/36$. A zero-body term contracts all three creation operators with all three annihilation operators. There are $3!=6$ signed complete pairings that become equal after antisymmetry, giving

$$\frac{6}{36} = \frac16.$$

For a one-body remainder, choose one uncontracted creator and one uncontracted annihilator in $3\times3=9$ ways. The two contracted pairs have $2!=2$ equivalent pairings. Thus

$$\frac{9\times2}{36} = \frac12.$$

For a two-body remainder, choose one contracted creation-annihilation pair in $3\times3=9$ ways. Relative to the standard two-body prefactor $1/4$, the coefficient satisfies

$$\frac9{36} = \frac14,$$

so the induced antisymmetrized two-body matrix element carries no extra factor beyond the sum over the occupied index.

A fixed-number state beyond pairwise Wick factorization

In four fermionic modes, take $|\Psi\rangle=(|1100\rangle+|0011\rangle)/\sqrt2$. Find the one-body density and $\langle n_1n_2\rangle$. What would a number-conserving Gaussian state with the same one-body density predict?

Solution

Every occupation has mean $1/2$, and a one-body operator cannot connect the two determinants, which differ by two occupied orbitals. Thus $\rho=I/2$. Direct counting gives $\langle n_1n_2\rangle=1/2$, whereas the Gaussian contraction gives $\rho_{11}\rho_{22}-\rho_{12}\rho_{21}=1/4$. The difference $\lambda_{12;12}=1/4$ is a nonzero cumulant. Such a Gaussian comparison state is mixed and has number fluctuations; it does not share the exact fixed-$N=2$ constraint. Agreement of the one-body density therefore misses both pair correlations and the global constraint.

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