Occupation Number Representation

Once identical particles cannot be labeled, the natural way to describe a many-body state is not to say "which particle is where" but "how many particles occupy each single-particle level." That list of occupation numbers, $\{n_1,n_2,n_3,\dots\}$ , is the occupation-number representation, and it is the working language of quantum statistical mechanics. A state is specified completely by its occupations, the total particle number is $N=\sum_k n_k$ , and the total energy is $E=\sum_k n_k\epsilon_k$ . Indistinguishability, which made labeled states redundant, is exactly what makes the occupation list a complete and non-redundant description. We take the formalism from the quantum-mechanics track. In aq-qm-occupation-number-fock-space a many-body state is written as "a list of occupation numbers, one per mode, with the total particle number read off by summing them," living in Fock space, and aq-qm-second-quantization builds every such state "from the vacuum with creation operators whose (anti)commutation encodes the statistics," with the fermion cap enforced by " $(c_i^\dagger)^2=0$ ." This article does not re-derive Fock space or the operator algebra; it uses the occupation-number picture to do the statistical bookkeeping: to show that $\{n_k\}\mapsto(N,E)$ is an exact, basis-independent map, and — crucially — that in the grand canonical ensemble the levels become independent, so the grand partition function factorizes, $\Xi=\prod_k\Xi_k$ . That factorization is the launch pad for everything in SM5. Because each level contributes an independent factor, the mean occupation of a level is a single-level calculation, and the sign of the exchange (boson vs fermion) sets whether that factor is a geometric series or a two-term sum. This article sets up the machinery and hands the explicit Bose–Einstein and Fermi–Dirac occupations forward to their dedicated articles — it counts and factorizes, but deliberately stops short of evaluating $\langle n_k\rangle$ .

Key takeaways

A many-body state of identical particles is specified by its occupation list $\{n_k\}$ ; the total number is $N=\sum_k n_k$ and energy $E=\sum_k n_k\epsilon_k$ .
The occupation-number basis and Fock space are owned by aq-qm-occupation-number-fock-space; the creation/annihilation operators encoding statistics by aq-qm-second-quantization.
The map $\{n_k\}\mapsto(N,E)$ is exact and basis-independent; the number operator $\hat N=\sum_k \hat n_k$ reads $N$ off any state.
The number of distinct occupation lists matches the boson/fermion state counts ( $\binom{g+N-1}{N}$ and $\binom{g}{N}$ ).
In the grand canonical ensemble the levels are independent, so $\Xi=\prod_k\Xi_k$ — the setup for the Bose–Einstein and Fermi–Dirac distributions (SM5).
Key idea: labeling particles is redundant; labeling levels by their occupation is complete. Because $E-\mu N$ is a sum over levels, the grand partition function factorizes into independent single-level factors.

The core insight

Describe a state by how many particles occupy each level, $\{n_k\}$ , not by which particle sits where. Then $N=\sum n_k$ , $E=\sum n_k\epsilon_k$ , and — because $E-\mu N=\sum_k(\epsilon_k-\mu)n_k$ is a sum over independent levels — the grand partition function factorizes, $\Xi=\prod_k\Xi_k$ . This factorization is the entire setup for quantum statistics.

Where this lives today

The occupation-number representation is the default language of every many-body calculation. In condensed-matter physics, band occupations $\langle n_{\mathbf k}\rangle$ define the Fermi surface, and the factorized grand partition function yields the electronic specific heat and the Pauli susceptibility. In quantum optics, photon occupation numbers describe Fock states, squeezed light, and the input to boson sampling. In ultracold atoms, quantum-gas microscopes measure site occupations directly, testing the occupation statistics site by site. And in quantum field theory the same occupation-number/Fock-space structure, with creation and annihilation operators, is the arena in which particles are created and destroyed — the formalism owned by aq-qm-second-quantization that this article consumes statistically.

The physical picture

Think of the single-particle energy levels as a set of bins, each labeled by its quantum numbers and energy $\epsilon_k$ . A many-body configuration is just a tally: bin 1 holds $n_1$ particles, bin 2 holds $n_2$ , and so on. For bosons any tally is allowed; for fermions each bin holds 0 or 1. Nothing else is needed — because the particles are indistinguishable, "which particle is in bin 3" is not a physical question, only "how many." The total number and energy are read off by summing the tallies weighted by 1 and by $\epsilon_k$ . The power of this description appears in the grand canonical ensemble, where the system exchanges both energy and particles with a reservoir. The Boltzmann–Gibbs weight is $e^{-\beta(E-\mu N)}$ , and since $E-\mu N=\sum_k(\epsilon_k-\mu)n_k$ is a sum over independent levels, the weight is a product over levels. Summing over all occupation lists then factorizes into a product of independent single-level sums, $\Xi=\prod_k\Xi_k$ . Each level becomes its own tiny grand-canonical system, and the whole many-body problem collapses into "solve one level, then multiply." That is why the occupation-number representation, not the labeled one, is the language in which Bose–Einstein and Fermi–Dirac statistics are actually computed.

Figure

Occupation-number vector on a single-particle spectrum gives total N and E

The punchline: a single occupation vector on the spectrum yields the exact totals $N=6$ and $E=14.5$ — the complete state, with no reference to which particle is which.

Figure

Grand partition function factorizes into a product of independent single-level factors

Because $E-\mu N$ is a sum over levels, the grand partition function is a product of independent single-level factors. Solving one level and multiplying is exactly how the Bose–Einstein and Fermi–Dirac distributions are obtained in SM5.

The core equations

A many-body state is a list of occupation numbers, and the totals are linear functionals of that list — the definition owned by aq-qm-occupation-number-fock-space.

|n_1,n_2,n_3,\dots\rangle,\qquad N=\sum_k n_k,\qquad E=\sum_k n_k\,\epsilon_k \qquad (1)

where: $n_k$ — occupation of single-particle level $k$ ; $\epsilon_k$ — its energy; $N,E$ — total particle number and energy. Physical reading: the occupation list is the complete state; $N$ and $E$ are read off by summing. The allowed occupations depend on the statistics, exactly the ceilings of the previous articles.

n_k\in\{0,1,2,\dots\}\ (\text{bosons}),\qquad n_k\in\{0,1\}\ (\text{fermions}) \qquad (2)

where: the fermion cap $n_k\le1$ is enforced operatorially by $(c_k^\dagger)^2=0$ (owned by aq-qm-second-quantization). Physical reading: the occupation range is the statistical fingerprint of the exchange sign. In the grand canonical ensemble the exponent $E-\mu N$ is a sum over levels, so the grand partition function factorizes into single-level pieces.

\Xi=\sum_{\{n_k\}}e^{-\beta\sum_k(\epsilon_k-\mu)n_k}=\prod_k\Big(\sum_{n_k}z^{n_k}e^{-\beta\epsilon_k n_k}\Big)=\prod_k\Xi_k \qquad (3)

where: $z=e^{\beta\mu}$ — fugacity; $\Xi_k$ — single-level grand partition function ( $1/(1-ze^{-\beta\epsilon_k})$ bosons, $1+ze^{-\beta\epsilon_k}$ fermions). Physical reading: independent levels make the many-body problem a product of one-level problems — the entire setup for quantum statistics. (The mean occupations $\langle n_k\rangle=z\,\partial_z\ln\Xi_k$ are derived in aq-sm-bose-einstein-distribution and aq-sm-fermi-dirac-distribution; here we stop at the factorization.)

Derivations

Derivation AThe occupation list is an exact, basis-independent bijection to (N, E)

The goal is to establish that the occupation-number representation is not just convenient shorthand but an exact and complete description of a many-body state of identical particles, equivalent to the symmetrized wavefunction, with the totals $N$ and $E$ recovered exactly and independent of the choice of single-particle basis. This underwrites everything that follows: if the map were lossy or basis-dependent, factorizing the partition function over occupation lists would be illegitimate.

Step 1. State the bijection. The quantum-mechanics owner aq-qm-occupation-number-fock-space records that a Fock-space basis state is "a list of occupation numbers, one per mode, with the total particle number read off by summing them," $N=\sum_k n_k$ . Each symmetrized (boson) or antisymmetrized (fermion) $N$ -particle wavefunction corresponds to exactly one occupation list, and vice versa. Why: indistinguishability makes the ordering of orbitals irrelevant, so the multiset of occupied orbitals — i.e. the occupation list — is the complete invariant.

Step 2. Recover the totals. Given $\{n_k\}$ and the single-particle spectrum $\{\epsilon_k\}$ , the total energy is $E=\sum_k n_k\epsilon_k$ and the number is $N=\sum_k n_k$ , both linear in the occupations. For the example $\{n_k\}=(2,0,1,3)$ on $\{\epsilon_k\}=(0,1,2.5,4)$ this gives $N=6$ and $E=14.5$ . Why: the many-body energy of non-interacting particles is additive over occupied single-particle levels.

Step 3. Match the state counts. The number of occupation lists with $\sum_k n_k=N$ over $g$ levels is $\binom{g+N-1}{N}$ for bosons (any $n_k$ ) and $\binom{g}{N}$ for fermions ( $n_k\in\{0,1\}$ ) — exactly the symmetric/antisymmetric subspace dimensions of aq-sm-symmetrization-postulate-wavefunction-symmetry. Why: counting occupation lists is the same combinatorial problem as counting symmetrized states.

Step 4. Confirm basis independence. The total number operator $\hat N=\sum_k\hat n_k$ has a basis-independent value equal to the trace of the number matrix; a unitary change of single-particle basis preserves that trace. Why: $N$ is a physical observable, invariant under relabeling of the modes. Verified: the $\{n_k\}\mapsto(N=6,E=14.5)$ map, the occupation-list counts matching $\binom{g+N-1}{N}$ and $\binom{g}{N}$ , and the basis-independence of $\mathrm{tr}\,\hat n$ under a unitary are checked in verify_occupation-number-representation.py.

Go deeper · Derivation B · 1 minFactorization of the grand partition function into independent levels

Derivation BFactorization of the grand partition function into independent levels

The goal is to prove the single most useful fact about the occupation-number representation for statistical mechanics: that the grand canonical partition function factorizes into a product of independent single-level factors. This is what reduces the intractable many-body sum to a set of one-level problems and is the exact point from which the Bose–Einstein and Fermi–Dirac distributions are launched in SM5. We build the factorization but deliberately stop before evaluating the mean occupations.

Step 1. Write the grand-canonical weight. Summing over all occupation lists, $\Xi=\sum_{\{n_k\}}e^{-\beta(E-\mu N)}=\sum_{\{n_k\}}e^{-\beta\sum_k(\epsilon_k-\mu)n_k}$ . Why: the grand ensemble weights each configuration by its energy and particle number through $e^{-\beta(E-\mu N)}$ .

Step 2. Recognize the sum-in-the-exponent. Because $E-\mu N=\sum_k(\epsilon_k-\mu)n_k$ is a sum over independent levels, the exponential is a product over levels: $e^{-\beta\sum_k(\ldots)}=\prod_k e^{-\beta(\epsilon_k-\mu)n_k}$ . Why: the exponential of a sum is a product of exponentials, and the levels do not couple for non-interacting particles.

Step 3. Interchange sum and product. Summing over each $n_k$ independently, $\Xi=\prod_k\big(\sum_{n_k}(ze^{-\beta\epsilon_k})^{n_k}\big)=\prod_k\Xi_k$ , where $z=e^{\beta\mu}$ . For fermions each factor is $\Xi_k=1+ze^{-\beta\epsilon_k}$ (sum $n_k=0,1$ ); for bosons $\Xi_k=1/(1-ze^{-\beta\epsilon_k})$ (geometric sum). Why: an unrestricted sum over a product of independent variables factorizes into a product of independent sums.

Step 4. Stop at the setup. Each level is now an independent grand-canonical subsystem, and the mean occupation follows from $\langle n_k\rangle=z\,\partial_z\ln\Xi_k$ — but that step is owned by aq-sm-bose-einstein-distribution and aq-sm-fermi-dirac-distribution. Here the deliverable is the factorization itself. Why: the de-overlap doctrine assigns the explicit distributions to their dedicated SM5 articles; this article provides the machinery. Verified: the 3-mode fermion factorization $\sum_{\{n_k\}}=\prod_k\Xi_k$ (symbolic), the boson single-mode geometric sum $\to1/(1-x)$ , and the $z^N$ -coefficient of $(1+z)^g$ equalling $\binom{g}{N}$ are checked in verify_occupation-number-representation.py.

Limiting cases

Regime	Occupation-number behavior	Consequence
Bosons	$n_k\in\{0,1,2,\dots\}$	$\Xi_k=1/(1-ze^{-\beta\epsilon_k})$ (geometric)
Fermions	$n_k\in\{0,1\}$	$\Xi_k=1+ze^{-\beta\epsilon_k}$ (two terms)
Independent levels	$E-\mu N=\sum_k(\ldots)n_k$	$\Xi=\prod_k\Xi_k$
Dilute $z\to0$	$\langle n_k\rangle\ll1$	classical Maxwell–Boltzmann
Interacting particles	$E$ not a sum over levels	factorization fails
Fixed $N$ (canonical)	occupations correlated	factorization is only grand-canonical

Worked examples

Example 1 — Read the totals off an occupation list.

Step 1. Setup. Single-particle spectrum $\epsilon_k=(0,1,2.5,4)$ (in units of some $\epsilon_0$ ); occupation list $\{n_k\}=(2,0,1,3)$ .

Step 2. Total number. $N=\sum_k n_k=2+0+1+3=\mathbf{6}$ .

Step 3. Total energy. $E=\sum_k n_k\epsilon_k=2(0)+0(1)+1(2.5)+3(4)=0+0+2.5+12=\mathbf{14.5}\,\epsilon_0$ .

Step 4. Statistics check. For fermions this list is illegal ( $n_1=2>1$ , $n_4=3>1$ ); for bosons it is a valid state. The list is the complete state — no reference to which particle sits where.

Example 2 — Count occupation lists and factorize.

Step 1. Count fermion lists. Distinct fermion occupation lists with $\sum_k n_k=2$ over $g=4$ levels: $\binom{4}{2}=\mathbf{6}$ .

Step 2. Count boson lists. Distinct boson lists with $\sum_k n_k=2$ over $g=4$ : $\binom{4+2-1}{2}=\binom{5}{2}=\mathbf{10}$ .

Step 3. Verify via the fugacity expansion. For four fermion levels at $\epsilon_k=0$ , $\Xi=(1+z)^4=1+4z+6z^2+4z^3+z^4$ . The coefficient of $z^2$ is $\mathbf{6}$ , matching the two-fermion count $\binom{4}{2}$ .

Step 4. The point. The grand partition function's power-series coefficients count the $N$ -particle states, and the whole $\Xi$ factorizes as $(1+z)^4=\prod_{k=1}^4(1+z)$ — four independent single-level factors. This is the factorization that makes SM5 tractable.

Origins · 1 minEnrico Fermi and how this was discovered

Enrico Fermi

1901–1954 · In 1926 Fermi formulated the statistics of particles obeying the exclusion principle, giving the occupation of levels for a fermion gas its quantitative form. The occupation-number bookkeeping he pioneered — counting how many particles sit in each level — is the natural home of the Fermi–Dirac distribution that bears his name.

History

1924

— Bose introduces occupation-based counting of photon states

1926

— Fermi and Dirac give the occupation statistics of exclusion-obeying particles

1927

— Jordan and Klein/Wigner develop the second-quantized creation/annihilation formalism

1928

— Jordan–Wigner encode fermion occupation with anticommuting operators

1932

— Fock formalizes the space of variable particle number now called Fock space

1957

— The occupation-number language becomes standard for many-body theory (BCS, Landau)

Going further

Texts: Pathria & Beale, Statistical Mechanics (3rd ed.), Ch. 6; Kardar, Statistical Physics of Particles, §7.3; Fetter & Walecka, Quantum Theory of Many-Particle Systems, Ch. 1.
Microscopic owners: the occupation-number basis and Fock space are owned by aq-qm-occupation-number-fock-space; the creation/annihilation operators encoding statistics by aq-qm-second-quantization. This article uses them for statistical bookkeeping.
Leads to: the grand canonical ensemble for ideal quantum gases and the Bose–Einstein and Fermi–Dirac distributions (SM5), where the factorized $\Xi_k$ is evaluated for $\langle n_k\rangle$ .
Provenance: AllQFT `sm` export (Pathria & Beale), aq-sm-occupation-number-representation, ref STAT.QSTAT.OCCN.1203. All bookkeeping, counting, and factorization claims python-verified (15 checks, all PASS).

Why is labeling levels better than labeling particles?

Because identical particles carry no labels: "particle 3 is in level 2" is not a physical statement, only "level 2 has $n_2$ particles" is. Labeling particles produces redundant states related by permutation (the source of the $N!$ overcounting); labeling levels by occupation gives each physical state exactly once. The occupation list is complete and non-redundant, which is why it is the working language of quantum statistics.

Does this article derive the Bose–Einstein or Fermi–Dirac distribution?

No. It sets up the factorized grand partition function $\Xi=\prod_k\Xi_k$ and the single-level factors ( $1/(1-ze^{-\beta\epsilon_k})$ for bosons, $1+ze^{-\beta\epsilon_k}$ for fermions), but the explicit mean occupations $\langle n_k\rangle=1/(e^{\beta(\epsilon_k-\mu)}\mp1)$ are derived in the dedicated articles aq-sm-bose-einstein-distribution and aq-sm-fermi-dirac-distribution (SM5). This article deliberately stops at the machinery.

Why does the grand partition function factorize but the canonical one not?

Factorization requires the levels to be independent, which needs both non-interacting particles and an unconstrained total number. In the grand ensemble $N$ floats, so each level fills independently and $\Xi=\prod_k\Xi_k$ . In the canonical ensemble the constraint $\sum_k n_k=N$ ties the levels together: adding a particle to one level removes it from the pool for others. Trading the fixed $N$ for a chemical potential $\mu$ removes the constraint and decouples the levels.

Is the occupation-number description exact or approximate?

Exact, for identical particles. Each symmetrized (boson) or antisymmetrized (fermion) many-body state corresponds to exactly one occupation list, and vice versa — a bijection, not an approximation. The totals $N=\sum n_k$ and $E=\sum n_k\epsilon_k$ are exact for non-interacting particles. What is approximate is applying it with interactions, where $E$ is no longer a simple sum over single-particle occupations and the levels couple.

What is Fock space and do I need it here?

Fock space is the direct sum of all fixed-particle-number Hilbert spaces, including the vacuum $|0\rangle$ ; it is the arena where the occupation lists live and where particle number can vary (owned by aq-qm-occupation-number-fock-space). For statistical mechanics you need it conceptually — the grand ensemble varies $N$ , so states of different particle number coexist — but the actual computation only uses the occupation lists and the factorized partition function, not the full operator machinery of second quantization.

How does the fermion cap show up operatorially?

Through the creation operators of second quantization: for fermions $(c_k^\dagger)^2=0$ (owned by aq-qm-second-quantization), so you cannot create two particles in the same mode, enforcing $n_k\in\{0,1\}$ . For bosons the creation operators commute and can be applied any number of times, allowing $n_k=0,1,2,\dots$ . In the occupation-number representation this appears simply as the range of each $n_k$ , but its operator origin is the (anti)commutation of the creation operators.