Number States

A number state is a Fock-space state with definite occupation numbers in a chosen mode basis. For one mode, it is written

$$\lvert n\rangle,$$

and it satisfies

$$N\lvert n\rangle = n\lvert n\rangle.$$

For many modes, a number state is written

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

and each $n_i$ is the occupation of mode $i$. Number states are also called occupation-number states or Fock states.

The phrase “number state” is basis-dependent. It means definite occupation of specified modes, not an intrinsic list of hidden particle labels.

One-Mode Number States

For a single bosonic mode, the allowed occupations are

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

The state $\lvert0\rangle$ is the vacuum of that mode. The state $\lvert1\rangle$ has one quantum in the mode, $\lvert2\rangle$ has two, and so on.

The one-mode number operator $N$ has eigenstates

$$N\lvert n\rangle = n\lvert n\rangle.$$

The number states are orthonormal:

$$\langle m\vert n\rangle = \delta_{mn}.$$

They form a basis for the single-mode bosonic Fock space:

$$\mathcal F_{\mathrm{one\ mode}} = \operatorname{span} \{\lvert0\rangle,\lvert1\rangle,\lvert2\rangle,\ldots\}.$$

Many-Mode Number States

Choose an ordered mode basis

$$\varphi_1,\varphi_2,\varphi_3,\ldots .$$

A many-mode number state is

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

It is a simultaneous eigenstate of the mode number operators $N_i$:

$$N_i \lvert n_1,n_2,\ldots\rangle = n_i \lvert n_1,n_2,\ldots\rangle.$$

The total number operator is

$$N_{\mathrm{tot}} = \sum_i N_i,$$

so

$$N_{\mathrm{tot}} \lvert n_1,n_2,\ldots\rangle = \left(\sum_i n_i\right) \lvert n_1,n_2,\ldots\rangle.$$

Only finitely many $n_i$ are nonzero in ordinary finite-particle basis states. Infinite-mode Fock spaces require Hilbert-space completion, but the finite-occupation states are the starting basis.

Bosonic Number States

For bosons,

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

The normalized one-mode bosonic number state can be generated from the vacuum by

$$\lvert n\rangle_B = \frac{(a^\dagger)^n}{\sqrt{n!}} \lvert0\rangle.$$

For many bosonic modes,

$$\lvert n_1,n_2,\ldots\rangle_B = \prod_i \frac{(a_i^\dagger)^{n_i}}{\sqrt{n_i!}} \lvert0\rangle.$$

The product is harmlessly ordered by any fixed convention because distinct bosonic creation operators commute. The factorials normalize repeated occupation of the same mode.

For two modes $a$ and $b$, examples are

$$\lvert2_a,0_b\rangle_B, \qquad \lvert1_a,1_b\rangle_B, \qquad \lvert0_a,2_b\rangle_B.$$

All three are valid two-boson states.

Fermionic Number States

For fermions,

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

A single fermionic mode has only two number states:

$$\lvert0\rangle_F, \qquad \lvert1\rangle_F = c^\dagger\lvert0\rangle_F.$$

There is no state $\lvert2\rangle_F$ for one fermionic mode.

For many fermionic modes, fix the mode ordering once and write

$$\lvert n_1,n_2,\ldots,n_M\rangle_F = (c_1^\dagger)^{n_1} (c_2^\dagger)^{n_2} \cdots (c_M^\dagger)^{n_M} \lvert0\rangle.$$

The order matters. Changing the order of fermionic creation operators can introduce minus signs. The occupation string records which modes are filled; the ordering convention records the sign.

For three fermionic modes, examples of two-particle number states are

$$\lvert1,1,0\rangle_F, \qquad \lvert1,0,1\rangle_F, \qquad \lvert0,1,1\rangle_F.$$

The state $\lvert2,0,0\rangle_F$ is not allowed.

Normalization

For bosons, the factorial in

$$\lvert n\rangle_B = \frac{(a^\dagger)^n}{\sqrt{n!}}\lvert0\rangle$$

ensures

$$\langle n\vert n\rangle=1.$$

The reason is that repeated creation in the same bosonic mode produces factors

$$\sqrt1,\sqrt2,\ldots,\sqrt n,$$

whose product is $\sqrt{n!}$.

For fermions, normalization is simpler but signs are subtler. Since $n_i$ is only $0$ or $1$, no factorial appears. With a fixed mode order and standard anticommutation normalization, the occupation states are orthonormal:

$${}_F\langle m_1,m_2,\ldots \vert n_1,n_2,\ldots\rangle_F = \prod_i \delta_{m_i n_i}.$$

The sign convention affects how operators act, not the norm of a basis state.

Number Operator Preview

For bosons, the mode number operator is

$$N_i = a_i^\dagger a_i.$$

For fermions, it is

$$N_i = c_i^\dagger c_i.$$

In both cases,

$$N_i\lvert n_1,n_2,\ldots\rangle = n_i\lvert n_1,n_2,\ldots\rangle.$$

The total number operator

$$N_{\mathrm{tot}} = \sum_i N_i$$

has eigenvalue equal to the total occupation. A number state is therefore a definite-particle-number state when the sum $\sum_i n_i$ is fixed.

Detailed commutators, anticommutators, and number-operator identities belong to the creation-operator and Number Operators pages. Here the key point is that number states diagonalize occupation.

Number States Versus General Fock States

Number states are a basis, not the only possible states. A single-mode bosonic state can be a superposition

$$\lvert\psi\rangle = \sum_{n=0}^{\infty} c_n\lvert n\rangle, \qquad \sum_n\lvert c_n\rvert^2=1.$$

This state has a definite particle number only if exactly one coefficient $c_n$ is nonzero. Otherwise, number measurements have a distribution.

For many modes, superpositions can also involve different occupation patterns with the same total particle number:

$$\frac{1}{\sqrt2} \bigl( \lvert1_a,0_b\rangle + \lvert0_a,1_b\rangle \bigr).$$

This is a one-particle state delocalized over two modes, not a two-particle state.

Physical Examples

In quantum optics, $\lvert n\rangle$ often means $n$ photons in one specified optical mode. In the harmonic-oscillator-to-field bridge, the same algebra turns oscillator excitation number into mode occupation number.

In lattice many-body physics, $\lvert n_1,\ldots,n_L\rangle$ can record occupation of lattice sites. For bosons, each $n_i$ may be any nonnegative integer. For spinless fermions, each $n_i$ is $0$ or $1$.

In quantum chemistry, fermionic number states are usually occupations of spin-orbitals. A determinant occupying spin-orbitals $1$, $3$, and $4$ corresponds to the number state

$$\lvert1,0,1,1\rangle_F$$

in a four-mode basis.

Common Mistakes

• Treating number states as particle labels rather than mode occupations.
• Forgetting that number states depend on the chosen mode basis.
• Using bosonic occupations $n_i=2,3,\ldots$ for fermionic modes.
• Dropping the bosonic factorial normalization for repeated occupation.
• Thinking every Fock-space state has a definite particle number.
• Confusing a one-particle superposition across modes with a many-particle state.

Cross-Links

• Occupation-Number Basis
• Vacuum State
• Mode Occupations
• Particle-Number Superselection Preview
• Fock Space Examples
• Bosons
• Fermions
• Bosonic Fock Space
• Fermionic Fock Space
• Creation and Annihilation Operators
• Number Operators
• Occupation-Number Representation in Many-Body Models
• Fock Space Exercises
• Bosonic Commutation Relations
• Fermionic Anticommutation Relations
• Slater Determinants
• Ladder-Operator Solution
• Harmonic Oscillator to Fields
• Formula Sheet

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.
• M. O. Scully and M. S. Zubairy, Quantum Optics, Cambridge University Press, 1997.

Exercises

1. What is the total occupation of $\lvert2,0,1,3\rangle_B$?

Solution

The total occupation is

$$N_{\mathrm{tot}} = 2+0+1+3 = 6.$$

2. List the two-mode bosonic number states with total occupation $N=2$.

Solution

The allowed nonnegative occupations are

$$\lvert2,0\rangle_B, \qquad \lvert1,1\rangle_B, \qquad \lvert0,2\rangle_B.$$

3. List the two-mode fermionic number states with total occupation $N=2$.

Solution

Each fermionic mode can be occupied at most once. With two modes and total occupation $2$, the only state is

$$\lvert1,1\rangle_F.$$

4. Show why the normalized bosonic state with two quanta in one mode is $(a^\dagger)^2\lvert0\rangle/\sqrt2$.

Solution

Starting from normalized number states,

$$a^\dagger\lvert0\rangle = \lvert1\rangle,$$

and

$$a^\dagger\lvert1\rangle = \sqrt2\,\lvert2\rangle.$$

Therefore

$$(a^\dagger)^2\lvert0\rangle = \sqrt2\,\lvert2\rangle,$$

so

$$\lvert2\rangle = \frac{(a^\dagger)^2}{\sqrt2}\lvert0\rangle.$$

5. Is

$$\frac{1}{\sqrt2} \bigl( \lvert1_a,0_b\rangle + \lvert0_a,1_b\rangle \bigr)$$

a two-particle state?

Solution

No. Each basis vector in the superposition has total occupation $1$. The state is a one-particle state delocalized over two modes. It is not a state with two particles.