Occupation-Number Representation

The occupation-number representation describes a many-particle state by the population of each chosen one-particle mode. A basis vector has the form

$$\lvert \boldsymbol n\rangle = \lvert n_1,n_2,\ldots\rangle,$$

where $n_i$ counts particles in mode $i$. This replaces artificial particle-slot labels by physically meaningful mode populations.

For bosons,

$$n_i \in \{0,1,2,\ldots\},$$

whereas for fermions,

$$n_i \in \{0,1\}.$$

That compact difference carries Bose enhancement, Pauli exclusion, distinct Hilbert-space dimensions, and different operator signs.

The Many-Body Hilbert Spaces Overview explains when a fixed-number sector or Fock space is the appropriate model space. The foundational definitions of Fock space and number states live in Fock Space and Occupation Number and the Occupation-Number Basis. This page develops their many-body use: choosing a finite mode set, constructing symmetry sectors, translating fixed-$N$ wavefunctions, and building sparse Hamiltonian matrices.

Helpful background. Occupation-number basis defines mode populations; creation and annihilation operators supplies the ladder actions used to assemble matrices.

Modes and many-particle configurations

For identical particles, labels such as “particle 1” and “particle 2” are coordinate slots, not persistent physical identities. A first-quantized many-body wavefunction

$$\Psi(x_1,\ldots,x_N)$$

must therefore be symmetric for bosons or antisymmetric for fermions. Here $x$ may collect position, spin, and any other one-particle coordinate.

Occupation notation builds the exchange symmetry into the basis itself. Instead of storing exchange-related coordinate coefficients separately, it records how many particles occupy each mode; repeated bosonic assignments would already make some permutations identical. This has three practical advantages:

• exchange symmetry is automatic;
• fixed-particle-number and other conserved sectors are easy to isolate;
• few-body operators connect relatively few configurations when the many-body sector is large.

The representation does not introduce new physics. It is a change of coordinates on the appropriate symmetric or antisymmetric many-particle Hilbert space.

Choose the Mode Basis First

Let the retained one-particle modes be an orthonormal set

$$\mathcal B_1 = \left\{ \lvert\varphi_1\rangle, \ldots, \lvert\varphi_M\rangle \right\}.$$

A mode can be a momentum and spin state, atomic spin-orbital, lattice-site orbital, trap eigenstate, Wannier orbital, or another one-particle basis state.

The occupation tuple

$$\boldsymbol n = (n_1,\ldots,n_M)$$

has meaning only relative to this ordered basis. In particular:

• a spin-up and spin-down orbital at the same site are different modes;
• two crystal momenta in the same band are different modes;
• a localized orbital and a momentum orbital belong to different mode bases;
• fermionic signs require a fixed ordering of the modes.

The sum

$$N(\boldsymbol n) = \sum_{i=1}^{M}n_i$$

is the total particle number of the configuration.

Fixed-N Sectors and Fock Space

A fixed-$N$ calculation retains only configurations satisfying

$$\sum_{i=1}^{M}n_i = N.$$

The corresponding sector is the symmetric or antisymmetric $N$-particle space built from the retained modes.

Fock space collects all allowed particle-number sectors:

$$\mathcal F_{\pm}(\mathcal h_M) = \bigoplus_N \mathcal H_N^{(\pm)}.$$

The plus sign denotes bosons and the minus sign denotes fermions. For $M$ fermionic modes, $N$ stops at $M$. For bosons, $N$ is unbounded unless a particle-number or local-occupation cutoff is imposed.

A number-conserving Hamiltonian can be diagonalized within one fixed-$N$ sector. Grand-canonical calculations and particle-nonconserving effective Hamiltonians require several sectors.

Constructing a Finite Many-Body Basis

To assemble a matrix, first enumerate the states on which it acts:

1. Choose an ordered set of $M$ orthonormal one-particle modes.
2. Choose bosonic or fermionic statistics.
3. Specify the particle-number sector or sectors.
4. Impose any local occupation cutoff for bosons.
5. Impose compatible conserved quantum numbers, such as momentum or spin projection.
6. Enumerate every occupation tuple satisfying all constraints exactly once.
7. Assign a stable index to each tuple.

For a fixed-$N$ sector, write the configuration set as

$$\mathcal C_N = \left\{ \boldsymbol n: \sum_i n_i=N \right\},$$

with the statistical restrictions on $n_i$ understood.

A general pure state is then

$$\lvert\Psi_N\rangle = \sum_{\boldsymbol n\in\mathcal C_N} C_{\boldsymbol n} \lvert\boldsymbol n\rangle,$$

and normalization requires

$$\sum_{\boldsymbol n\in\mathcal C_N} \left\lvert C_{\boldsymbol n}\right\rvert^2 = 1.$$

Counting configurations

For $M$ complete modes, the standard counts are:

Space	Dimension
N bosons, no local cutoff	$\binom{N+M-1}{N}$
N fermions	$\binom MN$, zero when $N>M$
All fermionic number sectors	$2^M$
Bosonic local cutoff, all sectors	$(n_{\max}+1)^M$

For bosons with both fixed $N$ and $0\le n_i\le n_{\max}$, the count is the coefficient of $x^N$ in $(1+x+\cdots+x^{n_{\max}})^M$. A cutoff $n_{\max}\ge N$ removes no state in that sector. Without any number cutoff, bosonic Fock space is infinite dimensional even for one mode. Hilbert-space scaling derives these counts.

Example: Two Bosons in Three Modes

For $N=2$ and $M=3$,

$$\dim\mathcal H_{2,3}^{(B)} = \binom{4}{2} = 6.$$

One convenient ordered basis is

$$\mathcal B_{2,3}^{(B)} = \bigl\{ \lvert2,0,0\rangle, \lvert1,1,0\rangle, \lvert1,0,1\rangle, \lvert0,2,0\rangle, \lvert0,1,1\rangle, \lvert0,0,2\rangle \bigr\}.$$

The tuple order is conventional, but it must remain fixed when state vectors and matrices are assembled.

Example: Two Fermions in Four Modes

For $N=2$ and $M=4$,

$$\dim\mathcal H_{2,4}^{(F)} = \binom{4}{2} = 6.$$

Using the mode order $1<2<3<4$, a basis is

$$\mathcal B_{2,4}^{(F)} = \bigl\{ \lvert1,1,0,0\rangle, \lvert1,0,1,0\rangle, \lvert1,0,0,1\rangle, \lvert0,1,1,0\rangle, \lvert0,1,0,1\rangle, \lvert0,0,1,1\rangle \bigr\}.$$

For a two-site spin-$1/2$ lattice model, one possible mode order is

$$(1\uparrow,1\downarrow,2\uparrow,2\downarrow).$$

In that convention, $\lvert1,1,0,0\rangle$ has double occupation of site $1$ by opposite spins, while $\lvert1,0,1,0\rangle$ has one spin-up fermion on each site.

A constrained basis may remove every tuple with $n_{i\uparrow}n_{i\downarrow}=1$ before any matrix is built. The t–J Model Preview develops this no-double-occupancy space and its projected hopping operators.

Basis normalization and ordering

We use normalized number states, with each bosonic creation product divided by $\sqrt{\prod_i n_i!}$ and fermionic creators ordered by increasing mode index. Then $\langle\boldsymbol m\vert\boldsymbol n\rangle=\delta_{\boldsymbol m,\boldsymbol n}$. Number states derives the normalization, and fermionic Fock space fixes the ordering signs.

Translation from coordinate amplitudes

Let $\Phi_{\boldsymbol n}(x_1,\ldots,x_N)$ be the normalized permanent or determinant built from the occupied orthonormal modes. Permanents and Slater determinants derive their factorial normalizations. The basis transformation is

$$\Psi(x_1,\ldots,x_N)=\sum_{\boldsymbol n}C_{\boldsymbol n}\Phi_{\boldsymbol n}(x_1,\ldots,x_N),\qquad C_{\boldsymbol n}=\int dx^N\,\Phi_{\boldsymbol n}^*(x_1,\ldots,x_N)\Psi(x_1,\ldots,x_N).$$

It is unitary between the occupation and coordinate descriptions of the same retained one-particle space. A complete infinite mode basis represents the whole continuum sector; a finite retained basis represents its projected sector. The example below illustrates how repeated bosonic assignments combine.

Worked Translation: Two in One Mode, One in Another

For three bosons with occupations

$$n_a=2, \qquad n_b=1,$$

the occupation vector $\lvert2_a,1_b\rangle_B$ corresponds to

$$\Phi_{2_a,1_b}^{(B)} ={} \frac{1}{\sqrt3} \Bigl[ \varphi_a(x_1) \varphi_a(x_2) \varphi_b(x_3) + \varphi_a(x_1) \varphi_b(x_2) \varphi_a(x_3) + \varphi_b(x_1) \varphi_a(x_2) \varphi_a(x_3) \Bigr].$$

The three distinct assignments occur with equal amplitude. Orthogonality of $\varphi_a$ and $\varphi_b$ makes the three products mutually orthogonal, so the factor $1/\sqrt3$ normalizes the state.

Probabilities, Mean Occupations, and Coherence

For a pure state expanded in an orthonormal occupation basis,

$$p(\boldsymbol n) = \left\lvert C_{\boldsymbol n}\right\rvert^2$$

is the probability of obtaining the complete occupation pattern $\boldsymbol n$ in a projective measurement of all mode numbers.

The mean occupation of mode $i$ is

$$\langle N_i\rangle = \sum_{\boldsymbol n} \left\lvert C_{\boldsymbol n}\right\rvert^2 n_i.$$

Similarly,

$$\langle N_iN_j\rangle = \sum_{\boldsymbol n} \left\lvert C_{\boldsymbol n}\right\rvert^2 n_i n_j.$$

These diagonal observables depend only on occupation probabilities. Phase coherence between configurations matters for operators with off-diagonal matrix elements, such as hopping terms.

For a mixed state,

$$\rho = \sum_{\boldsymbol m,\boldsymbol n} \rho_{\boldsymbol m\boldsymbol n} \lvert\boldsymbol m\rangle \langle\boldsymbol n\rvert.$$

The diagonal entries are occupation probabilities. The off-diagonal entries are coherences in the chosen mode basis.

Hamiltonian Matrices in Occupation Space

An ideal diagonal one-body Hamiltonian has the form

$$H_0 = \sum_i \epsilon_i N_i.$$

Every occupation vector is an eigenstate:

$$H_0 \lvert\boldsymbol n\rangle = \left( \sum_i\epsilon_i n_i \right) \lvert\boldsymbol n\rangle.$$

This diagonalization is the starting point for ideal Bose and Fermi gases. Grand-canonical probabilities factorize mode by mode; a fixed total number still couples the allowed occupations statistically.

A general one-body operator contains transfers between modes,

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

Let $\boldsymbol e_i$ denote the unit occupation vector for mode $i$. For bosons and $i\neq j$,

$$b_i^\dagger b_j \lvert\boldsymbol n\rangle ={} \sqrt{ (n_i+1)n_j } \times \lvert \boldsymbol n + \boldsymbol e_i - \boldsymbol e_j \rangle.$$

The result is zero when $n_j=0$. Fermionic transfers obey the same occupation change when the source is occupied and the destination is empty, but acquire a sign fixed by mode ordering.

A number-conserving two-body term removes at most two particles and inserts at most two: occupations can change in as many as four distinct modes. Consequently, few-body Hamiltonians are usually sparse in a large occupation basis.

Worked Matrix: Two-Site Bosons

For the many-site model and its physical regimes, see Bose–Hubbard Model.

Consider the two-site Bose–Hubbard Hamiltonian

$$H ={} -J \left( b_1^\dagger b_2 + b_2^\dagger b_1 \right) + \frac{U}{2} \sum_{i=1}^{2} N_i(N_i-1).$$

In the fixed-$N=2$ basis

$$\mathcal B = \left( \lvert2,0\rangle, \lvert1,1\rangle, \lvert0,2\rangle \right),$$

the matrix is

$$[H]_{\mathcal B} = \begin{pmatrix} U & -\sqrt2J & 0 \\ -\sqrt2J & 0 & -\sqrt2J \\ 0 & -\sqrt2J & U \end{pmatrix}.$$

The interaction is diagonal. Hopping moves one boson and produces the factors $\sqrt2$. This three-state example already displays the general pattern used in exact diagonalization: enumerate configurations, apply each operator term, and accumulate sparse matrix elements.

Fermionic Signs and Ordered Modes

For ordered fermionic modes, define

$$Q_j(\boldsymbol n) = \sum_{i<j}n_i.$$

Then

$$\begin{aligned} c_j\lvert\boldsymbol n\rangle &= (-1)^{Q_j(\boldsymbol n)} n_j \lvert \boldsymbol n-\boldsymbol e_j \rangle, \\ c_j^\dagger\lvert\boldsymbol n\rangle &= (-1)^{Q_j(\boldsymbol n)} (1-n_j) \lvert \boldsymbol n+\boldsymbol e_j \rangle. \end{aligned}$$

The parity counts occupied modes crossed while bringing $c_j$ or $c_j^\dagger$ to its ordered position. Changing the mode order changes intermediate signs and basis phases, but not physical predictions when every operator and state is transformed consistently.

A fermionic bitstring is therefore more than a set of occupied labels: its ordered convention is part of the representation.

Basis Indexing in Computation

Numerical many-body calculations store a state as a coefficient vector indexed by configurations. Common encodings include:

• integer tuples for bosonic occupations;
• bitstrings for fermionic occupations;
• combinatorial ranks within a fixed-$N$ sector;
• symmetry-resolved tuples carrying momentum, spin projection, or parity.

For fermions, one possible bit-mask convention is

$$B(\boldsymbol n) = \sum_{i=1}^{M} n_i2^{i-1}.$$

This is an indexing convention, not a physical observable. Software must document whether mode $1$ is stored in the least-significant or most-significant bit and must use the same order for parity signs.

For bosons, a lookup table from occupation tuples to basis indices is often simpler than packing variable-width occupations into one integer.

Symmetry-Resolved Sectors

Particle number is only the first possible restriction. If a Hamiltonian conserves an additive quantum number

$$Q = \sum_i q_iN_i,$$

then each occupation vector has eigenvalue

$$Q(\boldsymbol n) = \sum_i q_i n_i.$$

One may retain only configurations with a chosen value of $Q$. Examples include:

• fixed spin projection $S_z$;
• fixed crystal momentum modulo a reciprocal lattice vector;
• fixed species populations;
• fixed parity or point-group sector after an appropriate basis construction.

If $[H,Q]=0$, different $Q$ sectors do not mix. Working within one block reduces memory use and prevents numerical eigenvectors from combining unrelated sectors.

Not every symmetry is diagonal in a chosen occupation basis. Spatial symmetries may require symmetry-adapted superpositions of occupation configurations rather than a simple tuple filter.

Mode Changes and Truncation

Let two orthonormal mode bases be related by a unitary matrix $U$:

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

A single occupation vector in the $a$ basis generally becomes a superposition of occupation vectors in the $d$ basis. Occupation is therefore basis-dependent even though the physical state is not.

A unitary rotation within the same retained one-particle space is exact at fixed $N$, whether that space is finite or infinite dimensional. Independently chosen mode cutoffs can select different physical subspaces. A local bosonic occupation cutoff can also fail to be invariant under mode rotation: two particles limited to one per mode become a superposition containing double occupation in a rotated basis. Basis choice then affects convergence. Useful choices include:

• eigenmodes of the dominant one-body Hamiltonian;
• localized orbitals for short-range lattice interactions;
• natural orbitals adapted to the one-body density matrix;
• momentum modes when translation symmetry is central.

A compact basis is not automatically accurate. Convergence must be tested by increasing the mode set or occupation cutoff and monitoring the observables of interest.

Hilbert-Space Growth Remains Severe

Occupation notation removes redundant particle permutations, but it does not eliminate genuine many-body complexity.

For ten bosons in twenty modes,

$$\binom{29}{10} = 20{,}030{,}010$$

fixed-$N$ configurations remain. For twenty fermions in forty modes,

$$\binom{40}{20} = 137{,}846{,}528{,}820.$$

These dimensions explain why exact diagonalization requires modest systems, strong symmetry reduction, or both. Larger problems need approximation methods that exploit additional structure rather than merely a better label for the same exponentially large space.

Statistical Ensembles in This Representation

In an ideal gas whose Hamiltonian is diagonal in the selected modes, equilibrium probabilities factorize over occupation numbers in the grand-canonical ensemble. That special simplification underlies the Bose–Einstein and Fermi–Dirac distributions.

For an interacting Hamiltonian, an arbitrary occupation vector is generally not an energy eigenstate. The thermal density operator

$$\rho_\beta = \frac{e^{-\beta H}}{Z}$$

need not be diagonal in the occupation basis, even though that basis remains convenient for representing $H$ and $\rho_\beta$.

Likewise, a fixed mean occupation

$$\langle N_i\rangle$$

does not specify a unique many-body state. Many pure and mixed states can share the same one-mode averages while differing in fluctuations, coherences, and correlations.

Occupation Products and Entanglement

A single occupation vector is a product of mode-number states once a mode decomposition and ordering have been chosen. A superposition such as

$$\frac{1}{\sqrt2} \left( \lvert2,0\rangle + \lvert0,2\rangle \right)$$

can be entangled across the two mode algebras.

For fermions, the mode-factor description is a graded, order-dependent identification. It must not be treated as an ordinary tensor product of distinguishable particle subsystems; fermionic parity and the chosen observable algebras remain part of the physical statement.

For identical particles, statements about entanglement require a declared subsystem structure, operational algebra, and any relevant superselection restrictions. Particle labels should not be reintroduced as if they defined distinguishable subsystems.

Exercises

Add a site-energy bias

Add

$$H_\Delta = \frac{\Delta}{2} (N_1-N_2)$$

to the two-site, two-boson Hamiltonian. Find the matrix in the ordered basis $(\lvert2,0\rangle,\lvert1,1\rangle,\lvert0,2\rangle)$.

Solution

The bias eigenvalues are $\Delta$, $0$, and $-\Delta$ on the three basis states. It does not change the hopping matrix elements. Thus

$$[H+H_\Delta]_{\mathcal B} = \begin{pmatrix} U+\Delta & -\sqrt2J & 0 \\ -\sqrt2J & 0 & -\sqrt2J \\ 0 & -\sqrt2J & U-\Delta \end{pmatrix}.$$

The bias breaks the site-exchange symmetry unless $\Delta=0$.

Compute occupation fluctuations

For two bosonic modes, consider

$$\lvert\Psi\rangle = \frac{1}{2}\lvert2,0\rangle + \frac{1}{\sqrt2}\lvert1,1\rangle + \frac{1}{2}\lvert0,2\rangle.$$

Find $\langle N_1\rangle$, $\operatorname{Var}(N_1)$, and $\operatorname{Cov}(N_1,N_2)$.

Solution

The configuration probabilities are $1/4$, $1/2$, and $1/4$. Therefore

$$\langle N_1\rangle = \frac14(2) + \frac12(1) + \frac14(0) = 1.$$

Also,

$$\langle N_1^2\rangle = \frac14(4) + \frac12(1) = \frac32,$$

so

$$\operatorname{Var}(N_1) = \frac32-1 = \frac12.$$

By symmetry, $\langle N_2\rangle=1$. Only $\lvert1,1\rangle$ contributes to $N_1N_2$, giving

$$\langle N_1N_2\rangle = \frac12.$$

Hence

$$\operatorname{Cov}(N_1,N_2) = \frac12-1 = -\frac12.$$

The negative covariance ensures zero fluctuation of the fixed total $N_1+N_2=2$.

Track a fermionic parity sign

Use the ordered modes $1<2<3<4$ and the state

$$\lvert1,0,1,1\rangle = c_1^\dagger c_3^\dagger c_4^\dagger \lvert0\rangle.$$

Evaluate $c_2^\dagger\lvert1,0,1,1\rangle$ and $c_3\lvert1,0,1,1\rangle$.

Solution

For $c_2^\dagger$, one occupied mode lies to the left:

$$Q_2 = n_1 = 1.$$

Mode $2$ is empty, so creation is allowed and

$$c_2^\dagger \lvert1,0,1,1\rangle = -\lvert1,1,1,1\rangle.$$

For $c_3$, the occupations to the left are $n_1=1$ and $n_2=0$, so

$$Q_3 = 1.$$

Mode $3$ is occupied, and therefore

$$c_3 \lvert1,0,1,1\rangle = -\lvert1,0,0,1\rangle.$$

Both minus signs follow from the declared mode ordering.

References

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• P. Coleman, Introduction to Many-Body Physics, Cambridge University Press (2015).
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• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, Dover (2003).
• G. D. Mahan, Many-Particle Physics, 3rd ed., Springer (2000).
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