Occupation-Number Basis

The occupation-number basis labels a many-particle state by how many particles occupy each one-particle mode. Instead of asking which formal particle slot contains which state, it asks:

$$\text{how many particles are in mode }1, \quad \text{how many in mode }2, \quad \ldots$$

A basis vector is written schematically as

$$\lvert n_1,n_2,n_3,\ldots\rangle,$$

where $n_i$ is the occupation number of mode $i$. This is the natural notation for identical particles because identical particles do not carry observable individual labels.

Single-Particle Modes

Start with a one-particle Hilbert space $\mathcal h$ and choose an orthonormal basis of one-particle modes:

$$\{\lvert\varphi_1\rangle,\lvert\varphi_2\rangle,\ldots\}.$$

A mode may be:

• a spatial orbital in an atom;
• a spin-orbital in quantum chemistry;
• a momentum and spin state in a gas;
• a harmonic-oscillator mode;
• an optical mode with a specified frequency, polarization, and spatial profile;
• a lattice-site orbital in a tight-binding model.

The occupation-number basis depends on this chosen mode basis. Changing modes changes the meaning of the occupation labels, just as changing a one-particle basis changes the components of a vector.

Occupation Numbers

The occupation number $n_i$ records how many particles occupy mode $i$. The total particle number is

$$N = \sum_i n_i.$$

For a fixed-$N$ problem, only occupation strings with the same total $N$ are included. For example,

$$\lvert 2,0,1\rangle$$

has three particles in total: two in mode $1$, none in mode $2$, and one in mode $3$.

Occupation notation does not say “particle 1 is in mode 1.” It says “mode 1 has occupation 2.” That is exactly the shift needed for identical particles.

Bosonic Occupation Numbers

For bosons, each mode may have any nonnegative integer occupation:

$$n_i=0,1,2,\ldots .$$

Thus a bosonic basis vector may look like

$$\lvert 2,0,1,0,\ldots\rangle_B.$$

This means that two bosons occupy mode $1$, one boson occupies mode $3$, and the displayed other modes are empty.

For two bosons in two distinct orthonormal modes $\lvert a\rangle$ and $\lvert b\rangle$, the occupation vector

$$\lvert 1_a,1_b\rangle_B$$

corresponds in slot language to the symmetric state

$$\frac{1}{\sqrt2} \bigl( \lvert a\rangle_1\lvert b\rangle_2 + \lvert b\rangle_1\lvert a\rangle_2 \bigr).$$

For two bosons in the same mode,

$$\lvert 2_a\rangle_B$$

corresponds to the symmetric two-particle state

$$\lvert a\rangle_1\lvert a\rangle_2.$$

The mode occupation already knows the particles are identical; it does not need to list two separate particle names.

Fermionic Occupation Numbers

For fermions, Pauli exclusion restricts each mode to occupation zero or one:

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

A fermionic basis vector such as

$$\lvert 1,0,1,1\rangle_F$$

means that modes $1$, $3$, and $4$ are occupied, while mode $2$ is empty.

For two fermions in two distinct orthonormal modes $\lvert a\rangle$ and $\lvert b\rangle$, the occupation vector

$$\lvert 1_a,1_b\rangle_F$$

corresponds to the antisymmetric state

$$\frac{1}{\sqrt2} \bigl( \lvert a\rangle_1\lvert b\rangle_2 - \lvert b\rangle_1\lvert a\rangle_2 \bigr).$$

The forbidden two-fermion occupation

$$\lvert 2_a\rangle_F$$

does not exist. In fermionic notation, mode $a$ can have occupation $0$ or $1$, not $2$.

For fermions, a fixed ordering of modes is part of the convention. Signs in later creation-operator formulas depend on that ordering. The occupation string itself records which modes are filled; the operator algebra records the signs produced by reordering fermionic operations.

Mapping Examples

Consider three one-particle modes $a,b,c$.

A bosonic state with two particles in $a$ and one in $c$ is

$$\lvert 2_a,0_b,1_c\rangle_B.$$

It has total particle number

$$N=2+0+1=3.$$

A fermionic state with modes $a$ and $c$ occupied is

$$\lvert 1_a,0_b,1_c\rangle_F.$$

It has total particle number $N=2$. The corresponding slot wavefunction is the antisymmetrized combination of one particle in mode $a$ and one in mode $c$.

The notation also handles superpositions. For one particle in two modes,

$$\frac{1}{\sqrt2} \bigl( \lvert 1_a,0_b\rangle + \lvert 0_a,1_b\rangle \bigr)$$

is a one-particle state delocalized across two modes. It is not a two-particle state; each basis vector in the superposition has total occupation $1$.

The normalized basis vectors with definite occupations are developed in Number States.

Why Occupation Notation Scales Better

Slot wavefunctions become unwieldy as particle number grows. A three-boson state with one particle in each of three modes is a sum over $3!$ slot permutations. An $N$-fermion Slater determinant contains a signed sum over $N!$ permutations.

Occupation notation compresses this information:

$$\lvert 1_a,1_b,1_c\rangle$$

records the same mode content without writing every permutation. The symmetry or antisymmetry is carried by the bosonic or fermionic state space and, later, by the creation and annihilation operator algebra.

This is why many-particle quantum mechanics quickly moves from slot-labeled wavefunctions to Fock-space notation. The physics is the same, but the bookkeeping becomes manageable.

Basis Dependence

Occupation is always occupation of a chosen set of modes. If a one-particle mode basis is changed, a state with definite occupation in the old basis may become a superposition of occupation states in the new basis.

For example, a one-particle state occupying mode $a$ can be rewritten in a new basis

$$\lvert+\rangle = \frac{1}{\sqrt2} \bigl( \lvert a\rangle+\lvert b\rangle \bigr), \qquad \lvert-\rangle = \frac{1}{\sqrt2} \bigl( \lvert a\rangle-\lvert b\rangle \bigr).$$

Then

$$\lvert a\rangle = \frac{1}{\sqrt2} \bigl( \lvert+\rangle+\lvert-\rangle \bigr).$$

The phrase “the occupation of mode $a$” is meaningful only after mode $a$ has been specified.

Common Mistakes

• Treating occupation numbers as hidden particle labels.
• Forgetting that occupation depends on a chosen mode basis.
• Allowing fermionic occupations larger than one.
• Thinking a superposition of one-particle occupation states is automatically a many-particle state.
• Confusing a spatial orbital with a spin-orbital when applying Pauli exclusion.
• Ignoring the mode-ordering convention needed for fermionic signs.

Cross-Links

• Indistinguishability
• Symmetrization Postulate
• Bosons
• Fermions
• Pauli Exclusion Principle
• Slater Determinants
• Permanents
• Identical Particle Exercises
• Fock Space Exercises
• Vacuum State
• Number States
• Mode Occupations
• Particle-Number Superselection Preview
• Fock Space Examples
• Bosonic Fock Space
• Fermionic Fock Space
• Creation and Annihilation Operators
• Occupation-Number Representation in Many-Body Models
• Tensor Products of Hilbert Spaces
• Notation and Subsystem Labels
• Entanglement Depends on a Decomposition
• Identical-Particle Entanglement Cautions
• Reference Bridge: Second Quantization

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, McGraw-Hill, 1971.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.

Exercises

1. How many particles are represented by the bosonic occupation vector $\lvert 2,0,1,4\rangle_B$?

Solution

The total particle number is the sum of the occupations:

$$N=2+0+1+4=7.$$

2. List all allowed two-particle fermionic occupation vectors for three modes.

Solution

Each mode has occupation $0$ or $1$, and the total occupation must be $2$. The allowed vectors are

$$\lvert 1,1,0\rangle_F, \qquad \lvert 1,0,1\rangle_F, \qquad \lvert 0,1,1\rangle_F.$$

The vector $\lvert 2,0,0\rangle_F$ is not allowed for fermions.

3. Write the slot-language state corresponding to two bosons occupying distinct orthonormal modes $a$ and $b$.

Solution

The state is symmetric under exchange:

$$\frac{1}{\sqrt2} \bigl( \lvert a\rangle_1\lvert b\rangle_2 + \lvert b\rangle_1\lvert a\rangle_2 \bigr).$$

4. Why is $\lvert 1_a,0_b\rangle$ basis-dependent notation?

Solution

It says that the chosen mode $a$ is occupied and the chosen mode $b$ is empty. If one changes to a different one-particle mode basis, the same physical one-particle state may become a superposition of occupation states in the new basis. Occupation numbers are therefore meaningful only relative to a specified mode basis.