Number States in Fock Space

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.

A state with definite total number need not have definite occupation of every mode. The phrase “number state” is basis-dependent. It means definite occupation of specified modes, not an intrinsic list of hidden particle labels.

Helpful background. Occupation-Number Basis supplies the labels and Vacuum State supplies the normalized reference vector.

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:

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

The bar denotes closure: a general normalized state can have infinitely many nonzero, square-summable coefficients.

Many-Mode Number States

Choose an ordered orthonormal 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.

Why the factorial normalizes a bosonic state

For $[a,a^\dagger]=1$ and $a\lvert0\rangle=0$, the commutator identity

$$a(a^\dagger)^n\lvert0\rangle =n(a^\dagger)^{n-1}\lvert0\rangle$$

gives the norm recursion $\langle0|a^n(a^\dagger)^n|0\rangle=n\langle0|a^{n-1}(a^\dagger)^{n-1}|0\rangle=n!$. This derives the factor $1/\sqrt{n!}$ without assuming the normalized ladder amplitudes beforehand.

For fermions, the ordered product of distinct orthonormal creation operators has norm one. Repeated creation in the same mode is zero. In either case,

$$\langle m_1,m_2,\ldots|n_1,n_2,\ldots\rangle =\prod_i\delta_{m_i n_i}.$$

The number operators $N_i=a_i^\dagger a_i$ or $c_i^\dagger c_i$ have these vectors as simultaneous eigenstates; Number Operators derives their algebra.

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.

Exercises

1. 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

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

2. 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.

3. **Mean occupation is not a number state.** Compare $\lvert1\rangle$ with $(\lvert0\rangle+\lvert2\rangle)/\sqrt2$ for a bosonic mode. Find $\langle N\rangle$ and $\operatorname{Var}N$ in each.

Solution

Both have mean one. The first has $\langle N^2\rangle=1$, hence variance zero. The second gives outcomes zero and two equally, so $\langle N^2\rangle=2$ and variance one. The mean alone does not specify the distribution. A number-dephased mixture of the second state gives these same number statistics.

References

• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• 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.
• M. O. Scully and M. S. Zubairy, Quantum Optics, Cambridge University Press, 1997.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.