Occupation Numbers

An occupation number counts particles in a specified one-particle mode. A measurement returns an integer, but an average over repeated measurements can be fractional. The central distinction is between a count, its probability law, and the information about a quantum state that those probabilities leave out.

Helpful background. Occupation-number basis states and number operators supply the counting notation; quantum statistics supplies the ensemble comparison.

Mode counters, configurations, and means

A complete mode label includes spatial and internal quantum numbers: for example, momentum and spin $(\mathbf k,\sigma)$. Choose an orthonormal mode basis and let $d_i^\dagger$ create a particle in mode $i$. Then

$$\hat n_i=d_i^\dagger d_i,\qquad \hat N=\sum_i\hat n_i,\qquad [\hat n_i,\hat n_j]=0.$$

The commuting counters have joint occupation-number eigenstates $\lvert\boldsymbol n\rangle=\lvert n_1,n_2,\ldots\rangle$. Each bosonic $n_i$ is a nonnegative integer; each fermionic $n_i$ is zero or one. In a fixed-$N$ sector the allowed configurations additionally satisfy $\sum_i n_i=N$. The tuple names occupied modes, not individually labeled particles.

Object	Notation	Meaning
Mode counter	$\hat n_i$	An observable
Configuration	$\boldsymbol n$	A joint integer outcome
Counting probability	$P(\boldsymbol n)$	Probability of that outcome
Mean occupation	$\overline n_i$	Average count in mode $i$
Natural occupation	$\nu_\alpha$	Eigenvalue of the one-body density matrix

The number-operator algebra implies $\hat n_i^2=\hat n_i$ for fermions. Thus a single complete fermionic mode has $0\leq\overline n_i\leq1$ in every state. A spatial orbital with two spin states contains two modes and can hold two electrons. A bosonic mode counter is not a projector and its mean can exceed one.

The probability law of a count

For a density operator $\rho$, the complete counting law is

$$P(\boldsymbol n)=\langle\boldsymbol n\vert\rho\vert\boldsymbol n\rangle, \qquad P(\boldsymbol n)\geq0,\qquad \sum_{\boldsymbol n}P(\boldsymbol n)=1.$$

For a pure state this becomes $P(\boldsymbol n)=|\langle\boldsymbol n\vert\Psi\rangle|^2$. Only diagonal matrix elements appear. Coherences $\langle\boldsymbol n\vert\rho\vert\boldsymbol m\rangle$ with $\boldsymbol n\neq\boldsymbol m$ can affect other measurements even when all these probabilities are unchanged.

The mean and covariance follow from the joint law:

$$\overline n_i=\sum_{\boldsymbol n}n_iP(\boldsymbol n),\qquad C_{ij}=\langle\hat n_i\hat n_j\rangle-\overline n_i\overline n_j, \qquad \operatorname{Var}(N)=\sum_{ij}C_{ij}.$$

Here $C_{ii}=\operatorname{Var}(n_i)$. These formulas require the indicated moments to exist; a normalized bosonic state need not have finite mean number or variance. The means discard information about fluctuations and correlations. The entire $P(\boldsymbol n)$ determines every function of these counters, but still not the full quantum state.

Independent thermal modes

For the ideal Hamiltonian $H=\sum_i\epsilon_i\hat n_i$, the positive-temperature grand-canonical state has independent occupations. Define $x_i=e^{-\beta(\epsilon_i-\mu)}$, with $\beta=1/(k_{\mathrm B}T)$. The elementary laws are

Statistics	Probability for one mode	Mean and variance
Thermal Bose	$(1-x_i)x_i^n$, $x_i<1$	$f_i=x_i/(1-x_i)$; $\operatorname{Var}=f_i(1+f_i)$
Thermal Fermi	$P(1)=x_i/(1+x_i)$, $P(0)=1/(1+x_i)$	$f_i=x_i/(1+x_i)$; $\operatorname{Var}=f_i(1-f_i)$
Grand-canonical classical ideal gas	$e^{-x_i}x_i^n/n!$	Mean and variance both $x_i$

The full trace must converge when infinitely many modes are included. Under that condition $P(\boldsymbol n)=\prod_iP_i(n_i)$ and $C_{ij}=0$ for $i\neq j$. Derivations are given in Bose–Einstein statistics, Fermi–Dirac statistics, and the Maxwell–Boltzmann limit.

For fermions, the Bernoulli form and variance $f_i(1-f_i)$ hold for the one-mode count in any state; equilibrium determines the value of $f_i$. The geometric Bose law is specifically thermal. A coherent bosonic state instead has Poisson counts. At small $x_i$ all three agree to first order in the probabilities of zero or one particle, but normalized multiparticle coincidences can remain different.

For example, a thermal bosonic mode with $x=1/3$ has $P(0)=2/3$, $P(1)=2/9$, $P(2)=2/27$, mean $1/2$, and variance $3/4$. A mean of one half does not mean “one particle half the time”: the mode can contain two or more particles. Its probability of being nonempty is $1-P(0)=1/3$.

Degenerate levels and sums over modes

An energy level with degeneracy $g_r$ contains $g_r$ distinct modes. For independent ideal grand-canonical modes of energy $\epsilon_r$, the total level population $N_r$ has mean $g_rf_r$. Fermions give a binomial law,

$$P_F(N_r=m)=\binom{g_r}{m}f_r^m(1-f_r)^{g_r-m}, \qquad 0\leq m\leq g_r.$$

Bosons give a negative-binomial law,

$$P_B(N_r=m)=\binom{m+g_r-1}{m}(1-x_r)^{g_r}x_r^m, \qquad m=0,1,\ldots.$$

Their level variances are respectively $g_rf_r(1-f_r)$ and $g_rf_r(1+f_r)$. A large level population does not violate the fermionic single-mode bound.

For a dense spectrum, a mode sum can be approximated by a density-of-states integral:

$$\overline N=\sum_i f(\epsilon_i)\simeq\int d\epsilon\,\nu(\epsilon)f(\epsilon).$$

The density of states $\nu(\epsilon)$ counts modes per unit energy, including each internal degeneracy exactly once. Thus $f(\epsilon)$ is occupation per mode, whereas $\nu(\epsilon)f(\epsilon)$ is particle number per unit energy. Isolated low-energy levels, especially a condensate mode, may need to be separated before replacing the remaining sum by an integral.

Fixed particle number correlates the counts

For an ideal gas in the canonical ensemble,

$$P_N(\boldsymbol n)=Z_N^{-1}e^{-\beta\sum_i\epsilon_i n_i} \delta_{\sum_i n_i,N},$$

with the appropriate Bose or Fermi configurations. The Kronecker delta couples the modes. More generally, any state with exactly $N$ particles obeys

$$\sum_j C_{ij}=\operatorname{Cov}(n_i,N)=0.$$

Positive self-variance must be offset by cross-mode covariances. Individual cross-covariances need not all have the same sign in an arbitrary state.

For one particle shared between two modes, let $P(1,0)=p$ and $P(0,1)=1-p$. Then both variances are $p(1-p)$ and $C_{12}=-p(1-p)$. Their sum gives $\operatorname{Var}(N)=2p(1-p)-2p(1-p)=0$. The negative covariance follows from the fixed total and occurs for a boson as well as a fermion.

Coherence and a change of mode basis

If $\lvert\chi_\alpha\rangle=\sum_iU_{i\alpha}\lvert\varphi_i\rangle$ with unitary $U$, then $b_\alpha=\sum_iU_{i\alpha}^*d_i$. The new counter contains off-diagonal terms:

$$b_\alpha^\dagger b_\alpha =\sum_{ij}U_{i\alpha}U_{j\alpha}^*d_i^\dagger d_j.$$

The old mean counts alone therefore cannot predict the new ones. Introduce the one-body density matrix with the explicit index convention

$$\gamma_{ij}=\langle d_j^\dagger d_i\rangle,\qquad \gamma'=U^\dagger\gamma U,\qquad \operatorname{Tr}\gamma=\overline N.$$

It is Hermitian and positive semidefinite. Its diagonal contains the mean occupations, and its off-diagonal entries contain one-body coherence. See mode occupations for the general basis transformation.

Consider $\lvert\Psi\rangle=(\lvert1,0\rangle+\lvert0,1\rangle)/\sqrt2$ and the equal incoherent mixture of those two configurations. Both give the same original-basis counts, but

$$\gamma_\Psi=\frac12\begin{pmatrix}1&1\\1&1\end{pmatrix}, \qquad \gamma_{\mathrm{mix}}=\frac12\begin{pmatrix}1&0\\0&1\end{pmatrix}.$$

A counter for the symmetric mode $b_+^\dagger=(d_1^\dagger+d_2^\dagger)/\sqrt2$ finds the particle with certainty in the pure state and with probability one half in the mixture. Mixing modes before counting reveals information absent from the original counting law.

Natural occupations and one-body observables

The eigenvectors of $\gamma$ are natural orbitals; its eigenvalues $\nu_\alpha$ are natural occupations. They satisfy $\nu_\alpha\geq0$ and $\sum_\alpha\nu_\alpha=\overline N$. For fermions, $\nu_\alpha\leq1$. A pure Slater determinant has only zero and one as natural occupations. Fractional values rule out a single pure determinant, but can result from an ordinary thermal mixture without interactions; they alone do not diagnose interaction-induced correlations.

For bosons the eigenvalues can exceed one. The Penrose–Onsager condensation criterion requires a natural occupation proportional to particle number, with a nonzero limiting fraction $\nu_{\max}/N$. Merely being bounded above by a constant times $N$ is insufficient. The criterion is independent of the original mode basis.

A natural occupation such as $0.37$ is a mean in the corresponding natural orbital, not an allowed fractional count. For

$$\hat A=\sum_{ij}A_{ij}d_i^\dagger d_j, \qquad \langle\hat A\rangle=\sum_{ij}A_{ij}\gamma_{ji} =\operatorname{Tr}(A\gamma).$$

Only when $A$ is diagonal can one replace this by $\sum_iA_{ii}\overline n_i$. The one-body operator derivation explains the lift from one-particle matrices to Fock space.

Higher moments and counting measurements

Two-body information enters $\langle n_in_j\rangle$ and factorial moments. For nonzero mean define

$$g_i^{(2)}=\frac{\langle\hat n_i(\hat n_i-1)\rangle}{\overline n_i^{\,2}}.$$

It equals two for a single thermal bosonic mode, one for Poisson counts, and zero for a single fermionic mode. These are statements about counting laws. The correlation-function chapter develops spatial and field-operator versions.

For finitely many measured modes, the characteristic function

$$\chi(\boldsymbol\theta)=\left\langle e^{i\sum_j\theta_j\hat n_j}\right\rangle =\sum_{\boldsymbol n}P(\boldsymbol n)e^{i\boldsymbol\theta\cdot\boldsymbol n}$$

generates moments by derivatives at zero when those moments exist. Derivatives of $\ln\chi$ generate cumulants. Independent modes give a product of characteristic functions and a sum of cumulant-generating functions.

An ideal complete count uses projectors $\lvert\boldsymbol n\rangle\langle\boldsymbol n\rvert$. A detector that counts a set of orthogonal modes $R$ instead measures $N_R=\sum_{i\in R}n_i$; sum the probabilities of configurations with that total. A sharp spatial-region counter has this form in a basis adapted to the region, but in a general orbital basis it is $\int_R\psi^\dagger(\mathbf r)\psi(\mathbf r)\,d\mathbf r$, with off-diagonal matrix elements. Finite efficiency, loss, and resolution further change the observed distribution through the measurement model.

Interactions and quasiparticle counts

Occupation states remain a basis when interactions are present, but generally cease to be energy eigenstates. Pair-scattering terms mix configurations, and the equilibrium state need not factorize in bare modes. Applying a Bose or Fermi occupation factor directly to each bare energy then requires an additional approximation.

An ordinary mode rotation mixes annihilation operators only and preserves total particle number. A Bogoliubov transformation mixes creation and annihilation operators, schematically $\beta_\alpha=\sum_i(u_{\alpha i}d_i+v_{\alpha i}d_i^\dagger)$, subject to the canonical commutation or anticommutation conditions. Its vacuum can contain bare particles. “Zero quasiparticles” therefore does not generally mean “zero microscopic particles”; the final exercise makes this explicit for one bosonic mode.

The quasiparticle discussion explains when these excitations behave as long-lived particles. A nonconserved excitation number has no independent equilibrium chemical potential. If quasiparticle energies already diagonalize $H-\mu N$, the microscopic chemical potential must not be subtracted from them again.

Exercises

Fermionic projector identity

Use $\hat n^2=\hat n$ for a single fermionic mode to derive its variance in an arbitrary state. At which mean occupation is the variance largest?

Solution

The projector identity implies

$$\langle\hat n^2\rangle = \langle\hat n\rangle = \overline n.$$

Therefore

$$\operatorname{Var}(n) = \overline n - \overline n^{\,2} = \overline n(1-\overline n).$$

Differentiating with respect to $\overline n$ gives

$$\frac{d}{d\overline n} \operatorname{Var}(n) = 1-2\overline n.$$

The maximum occurs at

$$\overline n = \frac12, \qquad \operatorname{Var}_{\max} = \frac14.$$

This result uses only the binary fermionic spectrum, not thermal equilibrium.

Rotate the mode basis

For the one-particle state

$$\lvert\Psi_\phi\rangle = \frac{ \lvert1,0\rangle + e^{i\phi} \lvert0,1\rangle }{ \sqrt2 },$$

define

$$b_\pm^\dagger = \frac{ d_1^\dagger \pm d_2^\dagger }{ \sqrt2 }.$$

Find $\overline n_+$ and $\overline n_-$. Show that the original means do not determine the answer.

Solution

The one-body density matrix in the original basis is

$$\gamma = \frac12 \begin{pmatrix} 1 & e^{-i\phi}\\ e^{i\phi} & 1 \end{pmatrix}.$$

Using the symmetric and antisymmetric mode vectors gives

$$\overline n_+ = \frac{ 1+\cos\phi }{2},$$

and

$$\overline n_- = \frac{ 1-\cos\phi }{2}.$$

For every $\phi$, the original means are

$$\overline n_1 = \overline n_2 = \frac12.$$

The phase appears only in the off-diagonal entries of $\gamma$, so the old diagonal occupations are insufficient for predicting the new ones.

Degenerate fermionic level

A level contains $g=4$ independent fermionic modes, each with mean occupation $f=1/3$ in the grand-canonical ideal gas. Find the mean, variance, and probability of exactly two particles in the level.

Solution

The level population is binomial:

$$N_r \sim \operatorname{Binomial} \left( 4,\frac13 \right).$$

Thus

$$\overline N_r = 4f = \frac43,$$

and

$$\operatorname{Var}(N_r) = 4f(1-f) = \frac89.$$

The probability of exactly two particles is

$$\begin{aligned} P(N_r=2) &= \binom42 \left( \frac13 \right)^2 \left( \frac23 \right)^2 \\ &= \frac{8}{27}. \end{aligned}$$

No mode has more than one fermion, but the fourfold-degenerate level can contain up to four.

One-body observable from the density matrix

In a two-mode basis, let

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

and

$$A = \begin{pmatrix} \epsilon & it\\ -it & 2\epsilon \end{pmatrix},$$

where $\epsilon$ and $t$ are real. Compute $\langle\hat A\rangle=\operatorname{Tr}(A\gamma)$. What would be missed by using only the mean occupations?

Solution

Matrix multiplication gives

$$\begin{aligned} \operatorname{Tr}(A\gamma) &= \epsilon \left( \frac34 \right) + 2\epsilon \left( \frac14 \right) + (it) \left( -\frac{i}{4} \right) + (-it) \left( \frac{i}{4} \right) \\ &= \frac{5\epsilon}{4} + \frac{t}{2}. \end{aligned}$$

Using only the diagonal occupations would give $5\epsilon/4$ and miss the coherence contribution $t/2$.

Natural occupations

Find the natural occupations of

$$\gamma = \begin{pmatrix} 1/2 & q\\ q^* & 1/2 \end{pmatrix}.$$

What conditions on $q$ follow for a one-particle fermionic density matrix?

Solution

The characteristic polynomial gives eigenvalues

$$\nu_\pm = \frac12 \pm |q|.$$

Positivity requires

$$|q| \leq \frac12.$$

The trace is one, as required for one particle. The fermionic upper bound $\nu_\pm\leq1$ gives the same restriction.

At $|q|=1/2$, the eigenvalues are $(1,0)$ and the one-body state is pure. At $q=0$, they are $(1/2,1/2)$ and there is no one-body coherence in this basis.

Bare particles in a bosonic quasiparticle vacuum

For one bosonic mode, consider

$$\beta = u a + v a^\dagger,$$

with

$$|u|^2-|v|^2=1.$$

If $\beta\lvert\Omega\rangle=0$, show that the bare-particle mean is

$$\langle\Omega| a^\dagger a |\Omega\rangle = |v|^2.$$

Solution

The inverse transformation can be written

$$a = u^*\beta - v\beta^\dagger.$$

Therefore

$$a^\dagger = u\beta^\dagger - v^*\beta.$$

Multiplying and taking the quasiparticle-vacuum expectation value leaves only the term containing $\beta\beta^\dagger$:

$$\begin{aligned} \langle a^\dagger a\rangle &= |v|^2 \langle\Omega| \beta\beta^\dagger |\Omega\rangle \\ &= |v|^2. \end{aligned}$$

The quasiparticle vacuum contains no $\beta$ quanta but has a nonzero mean occupation of the original $a$ particles.

References

• P. Coleman, Introduction to Many-Body Physics, Cambridge University Press (2015) — modern many-body notation and quasiparticle reasoning.
• A. J. Coleman, “Structure of Fermion Density Matrices”, Reviews of Modern Physics 35, 668–686 (1963) — foundational analysis of fermionic reduced density matrices and occupation constraints.
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, Dover (2003) — second-quantized many-body formalism and one-body density matrices.
• K. Huang, Statistical Mechanics, 2nd ed., Wiley (1987) — standard treatment of Bose, Fermi, and classical occupation laws.
• J. W. Negele and H. Orland, Quantum Many-Particle Systems, CRC Press (1998) — occupation-number methods, ensembles, and correlation functions.
• R. K. Pathria and P. D. Beale, Statistical Mechanics, 4th ed., Academic Press (2021) — quantum ideal gases, ensembles, and occupation-number distributions.
• O. Penrose and L. Onsager, “Bose–Einstein Condensation and Liquid Helium”, Physical Review 104, 576–584 (1956) — basis-independent condensation criterion from the one-body density matrix.