# Fock state

In quantum mechanics, a Fock state or number state is a quantum state that is an element of a Fock space with a well-defined number of particles (or quanta). These states are named after the Soviet physicist Vladimir Fock. Fock states play an important role in the second quantization formulation of quantum mechanics.

The particle representation was first treated in detail by Paul Dirac for bosons and by Pascual Jordan and Eugene Wigner for fermions. [1]:35

## Definition

One specifies a multiparticle state of N non-interacting identical particles by writing the state as a sum of tensor products of N one-particle states. The tensor products must be alternating products or symmetric products of the underlying one-particle Hilbert space according to whether the particles are fermions or bosons. If the number of particles is variable, one constructs Fock space as the direct sum of the tensor product Hilbert spaces for each particle number.

Then it is possible to specify the same state in a new notation, the occupancy number notation, by specifying the number of particles in each possible one-particle state.

Let ki be an orthonormal basis of states in the underlying one-particle Hilbert space. This induces a corresponding basis of the Fock space called the "occupancy number basis". A quantum state in the Fock space is called a Fock state if it is an element of the occupancy number basis.

A Fock state satisfies an important criterion: for each i, the state is an eigenstate of the particle number operator $\widehat{N_{{\mathbf{k}}_i}}$ corresponding to the i-th elementary state ki. The corresponding eigenvalue gives the number of particles in the state. This criterion nearly defines the Fock states (one must in addition select a phase factor).

A given Fock state is denoted by $|n_{{\mathbf{k}}_1},n_{{\mathbf{k}}_2},..n_{{\mathbf{k}}_i}...\rangle$. In this expression, $n_{{\mathbf{k}}_i}$ denotes number of particles in the i-th state ki, and the particle number operator for the i-th state, $\widehat{N_{{\mathbf{k}}_i}}$, acts on the Fock state in the following way:

$\widehat{N_{{\mathbf{k}}_i}}|n_{{\mathbf{k}}_1},n_{{\mathbf{k}}_2},..n_{{\mathbf{k}}_i}...\rangle=n_{{\mathbf{k}}_i}|n_{{\mathbf{k}}_1},n_{{\mathbf{k}}_2},..n_{{\mathbf{k}}_i}...\rangle$

Hence the Fock state is an eigenstate of the number operator with eigenvalue $n_{{\mathbf{k}}_i}$.[2]:478

Fock states form the most convenient basis of a Fock space. Elements of a Fock space which are superpositions of states of differing particle number (and thus not eigenstates of the number operator) are not Fock states. Thus, not all elements of a Fock space are referred to as "Fock states".

The definition of Fock state ensures that ${\rm Var}(\widehat{N})=0$, i.e., measuring the number of particles in a Fock state always returns a definite value with no fluctuation. Here

$\widehat{N}=\sum_i \widehat{N_{{\mathbf{k}}_i}}$.

## Example using two particles

For any final state $|f\rangle$, any Fock state of two identical particles given by $|1_{{\mathbf{{k}}_1}}, 1_{{\mathbf{{k}}_2}}\rangle$, and any operator $\widehat{\mathbb{O}}$, we have the following condition for indistinguishability:[3]:191

$|\langle f|\widehat{\mathbb{O}}|1_{{\mathbf{{k}}_1}}, 1_{{\mathbf{{k}}_2}}\rangle|^2=|\langle f|\widehat{\mathbb{O}}|1_{{\mathbf{{k}}_2}}, 1_{{\mathbf{{k}}_1}}\rangle|^2$.

So, we must have, $\langle f|\widehat{\mathbb{O}}|1_{{\mathbf{{k}}_1}}, 1_{{\mathbf{{k}}_2}}\rangle=e^{i\delta}\langle f|\widehat{\mathbb{O}}|1_{{\mathbf{{k}}_2}}, 1_{{\mathbf{{k}}_1}}\rangle$

where $e^{i\delta}=+1$ for bosons and $-1$ for fermions.

As $\langle f|$ and $\widehat{\mathbb{O}}$ are arbitrary, we can say, $|1_{{\mathbf{{k}}_1}}, 1_{{\mathbf{{k}}_2}}\rangle=+|1_{{\mathbf{{k}}_2}}, 1_{{\mathbf{{k}}_1}}\rangle$ for bosons

and $|1_{{\mathbf{{k}}_1}}, 1_{{\mathbf{{k}}_2}}\rangle=-|1_{{\mathbf{{k}}_2}}, 1_{{\mathbf{{k}}_1}}\rangle$ for fermions. [3]:191

Note that the number operator does not distinguish bosons from fermions; indeed, it just counts particles without regard to their symmetry type. To perceive any difference between them, we need other operators, namely the creation and annihilation operators.

## Bosonic Fock state

Bosons, which are particles with integer spin, follow a simple rule: their composite eigenstate is symmetric[4] under operation by an exchange operator. For example, in a two particle system in the tensor product representation we have $\hat{P}\left|x_1, x_2\right\rangle = \left|x_2, x_1\right\rangle$ .

### Boson Creation and Annihilation operators

We should be able to express the same symmetric property in this new Fock space representation. For this we introduce non-Hermitian bosonic creation and annihilation operators,[4] denoted by $b^{\dagger}$ and $b$ respectively. The action of these operators on a Fock state are given by the following two equations:

$b^{\dagger}_{{\mathbf{k}}_l}|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}},...\rangle=\sqrt{n_{{\mathbf{k}}_{l}} +1 } |n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}}+1 ,...\rangle$ [4]
$b_{{\mathbf{k}}_l}|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}},...\rangle=\sqrt{n_{{\mathbf{k}}_{l}}} |n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}}-1 ,...\rangle$ [4]

#### Hermiticity of Creation and Annihilation operator

Creation and Annihilation operators are not Hermitian operators.[4]

#### Operator Identities

The commutation relations of creation and annihilation operators in a bosonic system are

$[b^{\,}_i, b^\dagger_j] \equiv b^{\,}_i b^\dagger_j - b^\dagger_jb^{\,}_i = \delta_{i j},$ [4]
$[b^\dagger_i, b^\dagger_j] = [b^{\,}_i, b^{\,}_j] = 0,$ [4]

where $[\ \ , \ \ ]$ is the commutator and $\delta_{i j}$ is the Kronecker delta.

### N bosonic basis states $|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}},...\rangle$

Number of Particles (N) Bosonic basis states[6]:11
0 $|0,0,0...\rangle$
1 $|1,0,0...\rangle$,$|0,1,0...\rangle$,$|0,0,1...\rangle$,...
2 $|2,0,0...\rangle$,$|1,1,0...\rangle$,$|0,2,0...\rangle$,...
... ...

### Action on some specific Fock states

• For a vacuum state (no particle is in any state), expressed as $|0_{{\mathbf{k}}_{1}}, 0_{{\mathbf{k}}_{2}} ,0_{{\mathbf{k}}_{3}}...0_{{\mathbf{k}}_{l}},...\rangle$, we have:
$b^{\dagger}_{{\mathbf{k}}_l}|0_{{\mathbf{k}}_{1}}, 0_{{\mathbf{k}}_{2}} ,0_{{\mathbf{k}}_{3}}...0_{{\mathbf{k}}_{l}},...\rangle=|0_{{\mathbf{k}}_{1}}, 0_{{\mathbf{k}}_{2}} ,0_{{\mathbf{k}}_{3}}...1_{{\mathbf{k}}_{l}} ,...\rangle$

and, $b_{{\mathbf{k}}_l}|0_{{\mathbf{k}}_{1}}, 0_{{\mathbf{k}}_{2}} ,0_{{\mathbf{k}}_{3}}...0_{{\mathbf{k}}_{l}},...\rangle=0$ [4]

• We can generate any Fock state by operating on the vacuum state with an appropriate number of creation operators:

$|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ...\rangle=\frac{(b^{\dagger}_{{\mathbf{k}}_1})^{{\mathbf{k}}_1}}{\sqrt{{{\mathbf{k}}_1}!}} \frac{(b^{\dagger}_{{\mathbf{k}}_2})^{{\mathbf{k}}_2}}{\sqrt{{{\mathbf{k}}_2}!}}...|0_{{\mathbf{k}}_{1}}, 0_{{\mathbf{k}}_{2}} ,...\rangle$

• For a single mode Fock state, expressed as, $|n_{{\mathbf{k}}}\rangle$,
$b^{\dagger}_{\mathbf{k}}|n_{{\mathbf{k}}}\rangle=\sqrt{n_{\mathbf{k}} +1} |n_{{\mathbf{k}}}+1\rangle$

and, $b_{\mathbf{k}}|n_{{\mathbf{k}}}\rangle=\sqrt{n_{\mathbf{k}}} |n_{{\mathbf{k}}}-1\rangle$

### Action of Number operator

The number operators for a bosonic system are given by $\widehat{N_{{\mathbf{k}}_l}}=b^{\dagger}_{{\mathbf{k}}_l}b_{{\mathbf{k}}_l}$, where $\widehat{N_{{\mathbf{k}}_l}}|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}}...\rangle=n_{{\mathbf{k}}_{l}} |n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}}...\rangle$ [4]

Number operators are Hermitian operators.

### Symmetric behaviour of Bosonic Fock states

The commutation relations of the creation and annihilation operators ensure that the bosonic Fock states have the appropriate symmetric behaviour under particle exchange. Here, exchange of particles between two states is done by annihilating one particle in one state and creating one in another. If we start with a Fock state, $|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,....n_{{\mathbf{k}}_{m}}...n_{{\mathbf{k}}_{l}}...\rangle$, and want to shift a particle from state $k_l$ to state $k_m$, then we operate the Fock state by $b_{{\mathbf{k}}_{m}}^{\dagger}b_{{\mathbf{k}}_{l}}$ in the following way:

Using the commutation relation we have, $b_{{\mathbf{k}}_{m}}^{\dagger}.b_{{\mathbf{k}}_{l}}=b_{{\mathbf{k}}_{l}}.b_{{\mathbf{k}}_{m}}^{\dagger}$

$b_{{\mathbf{k}}_{m}}^{\dagger}.b_{{\mathbf{k}}_{l}}|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,....n_{{\mathbf{k}}_{m}}...n_{{\mathbf{k}}_{l}}...\rangle=b_{{\mathbf{k}}_{l}}.b_{{\mathbf{k}}_{m}}^{\dagger}|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,....n_{{\mathbf{k}}_{m}}...n_{{\mathbf{k}}_{l}}...\rangle=\sqrt{n_{{\mathbf{k}}_{m}} +1 }\sqrt{n_{{\mathbf{k}}_{l}}}|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,....n_{{\mathbf{k}}_{m}}+1...n_{{\mathbf{k}}_{l}}-1...\rangle$

So, the Bosonic Fock state behaves to be symmetric under operation by Exchange operator.

## Fermionic Fock state

### Fermion Creation and Annihilation operators

To be able to retain the antisymmetric behaviour of fermions, for Fermionic fock states we introduce non-Hermitian Fermion Creation and annihilation operators,[4] defined as, for a Fermionic fock state, $|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}},...\rangle$, Creation operator acts as:

$c^{\dagger}_{{\mathbf{k}}_l}|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}},...\rangle=\sqrt{n_{{\mathbf{k}}_{l}} +1 } |n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}}+1 ,...\rangle$ [4]

and Annihilation operator acts as:

$c_{{\mathbf{k}}_l}|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}},...\rangle=\sqrt{n_{{\mathbf{k}}_{l}}} |n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}}-1 ,...\rangle$ [4]

These two actions are done antisymmetrically, which we shall discuss later.

#### Operator Identities

The anticommutation relations of creation and annihilation operators in a fermionic system are,

$\{c^{\,}_i, c^\dagger_j\} \equiv c^{\,}_i c^\dagger_j + c^\dagger_jc^{\,}_i = \delta_{i j},$ [4]
$\{c^\dagger_i, c^\dagger_j\} = \{c^{\,}_i, c^{\,}_j\} = 0,$ [4]

where ${\ \ , \ \ }$ is the anticommutator and $\delta_{i j}$ is the Kronecker delta. These anticommutation relation will be used to show antisymmetric behaviour of Fermionic Fock states.

### Action of Number operator

Number Operator for Fermions is given by $\widehat{N_{{\mathbf{k}}_l}}=c^{\dagger}_{{\mathbf{k}}_l}.c_{{\mathbf{k}}_l}$ and, $\widehat{N_{{\mathbf{k}}_l}}|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}}...\rangle=n_{{\mathbf{k}}_{l}} |n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}}...\rangle$ [4]

#### Maximum Occupation number

The action of the number operator as well as the creation and annihilation operators might seem same as the bosonic ones, but the real twist comes from the maximum occupation number of each state in the fermionic Fock state. Extending the 2-particle fermionic example above, we first must convince ourselves that a fermionic Fock state $|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}}...\rangle$ is obtained by applying a certain sum of permutation operators to the tensor product of eigenkets as follows:

$|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}}...\rangle=S_{-}|i_{1}, i_{2},i_{3}...i_{l}...\rangle=\frac{1}{\sqrt{N!}}\begin{vmatrix} |i_{1}\rangle_{1} & \cdots & |i_{1}\rangle_{N} \\ \vdots & \ddots & \vdots \\ |i_{N}\rangle_{1} & \cdots & |i_{N}\rangle_{N} \end{vmatrix}$ [7]:16

This determinant is called the Slater determinant.[citation needed] If any of the single particle states are the same, two rows of the Slater determinant would be same and hence the determinant would be zero. Hence, two identical fermions must not occupy the same state. Therefore, the occupation number of any single state is either 0 or 1. The eigenvalue associated to the fermionic Fock state $\widehat{N_{{\mathbf{k}}_l}}$ will be either 0 or 1.

### N Fermionic basis states $|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}},...\rangle$

Number of Particles (N) Fermionic basis states[6]:11
0 $|0,0,0...\rangle$
1 $|1,0,0...\rangle$,$|0,1,0...\rangle$,$|0,0,1...\rangle$,...
2 $|1,1,0...\rangle$,$|0,1,1...\rangle$,$|0,1,0,1...\rangle$,$|1,0,1,0...\rangle$...
... ...

### Action on some specific Fock states

• For a single mode fermionic Fock state, expressed as $|0_{{\mathbf{k}}}\rangle$,
$c^{\dagger}_{\mathbf{k}}|0_{{\mathbf{k}}}\rangle=|1_{{\mathbf{k}}}\rangle$

and $c^{\dagger}_{\mathbf{k}}|1_{{\mathbf{k}}}\rangle=0$, as the maximum occupation number of any state is 1, more than 1 fermion cannot occupy the same state.

• For a single mode fermionic Fock state, expressed as $|1_{{\mathbf{k}}}\rangle$,
$c_{\mathbf{k}}|1_{{\mathbf{k}}}\rangle=|0_{{\mathbf{k}}}\rangle$

and $c_{\mathbf{k}}|0_{{\mathbf{k}}}\rangle=0$, as the particle number cannot be less than zero.

• For a multimode fermionic Fock state, expressed as, $|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}},... n_{{\mathbf{k}}_{\beta}},n_{{\mathbf{k}}_{\alpha}},...\rangle$

$c_{{\mathbf{k}}_{\alpha}}|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}},... n_{{\mathbf{k}}_{\beta}},n_{{\mathbf{k}}_{\alpha}},...\rangle= (-1)^{\sum_{\beta < \alpha} n \beta} |n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}},... n_{{\mathbf{k}}_{\beta}},1-n_{{\mathbf{k}}_{\alpha}},...\rangle$,

where $(-1)^{\sum_{\beta < \alpha} n \beta}$ is called the Jordan-Wigner string, which depends on the ordering of the involved single-particle states and adding the fermion occupation numbers of all preceding states.[5]:88

### Antisymmetric behaviour of Fermionic Fock state

Antisymmetric behaviour of Fermionic states under Exchange operator is taken care of the anticommutation relations. Here, exchange of particles between two states is done by annihilating one particle in one state and creating one in other. If we start with a Fock state, $|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,....n_{{\mathbf{k}}_{m}}...n_{{\mathbf{k}}_{l}}...\rangle$, and want to shift a particle from state $k_l$ to state $k_m$, then we operate the Fock state by $c_{{\mathbf{k}}_{m}}^{\dagger}.c_{{\mathbf{k}}_{l}}$ in the following way:

Using the anticommutation relation we have, $c_{{\mathbf{k}}_{m}}^{\dagger}.c_{{\mathbf{k}}_{l}}=-c_{{\mathbf{k}}_{l}}.c_{{\mathbf{k}}_{m}}^{\dagger}$

$c_{{\mathbf{k}}_{m}}^{\dagger}.c_{{\mathbf{k}}_{l}}|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,....n_{{\mathbf{k}}_{m}}...n_{{\mathbf{k}}_{l}}...\rangle=\sqrt{n_{{\mathbf{k}}_{m}} +1 }\sqrt{n_{{\mathbf{k}}_{l}}}|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,....n_{{\mathbf{k}}_{m}}+1...n_{{\mathbf{k}}_{l}}-1...\rangle$

but, $c_{{\mathbf{k}}_{l}}.c_{{\mathbf{k}}_{m}}^{\dagger}|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,....n_{{\mathbf{k}}_{m}}...n_{{\mathbf{k}}_{l}}...\rangle=-c_{{\mathbf{k}}_{m}}^{\dagger}.c_{{\mathbf{k}}_{l}}|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,....n_{{\mathbf{k}}_{m}}...n_{{\mathbf{k}}_{l}}...\rangle=-\sqrt{n_{{\mathbf{k}}_{m}} +1 }\sqrt{n_{{\mathbf{k}}_{l}}}|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,....n_{{\mathbf{k}}_{m}}+1...n_{{\mathbf{k}}_{l}}-1...\rangle$

So, the Fermionic Fock state behaves to be antisymmetric under operation by Exchange operator.

## Fock states are not Energy eigenstates in general

In Second quantization theory, Hamiltonian density function is given by

$\mathfrak{H}=\frac {1}{2m} \nabla_{i}\psi^{*}(x)\nabla_{i}\psi(x)$ [3]:189

Total Hamiltonian is given by

$\mathcal{H}=\int d^{3}x\,\mathfrak{H}=\int d^{3}x \psi^{*}(x)\left(-\tfrac{\nabla^2}{2m}\right)\psi(x)$

$\therefore \mathfrak{H}=-\tfrac{\nabla^2}{2m}$

$\mathfrak{H}\psi_{n}^{(+)}(x)=-\tfrac{\nabla^2}{2m}\psi_{n}^{(+)}(x)=E_{n}^{0}\psi_{n}^{(+)}(x)$

and

$\int d^{3}x\, \psi_{n}^{(+)^{*}}(x)\psi_{n'}^{(+)}(x)=\delta_{nn'}$

and

$\psi(x)=\sum_n a_n \psi_{n}^{(+)}(x)$,

where $a_n$ is the annihilation operator.

$\therefore \mathcal{H}=\sum_{n,n'}\int d^{3}x\, a^{\dagger}_{n'}\psi_{n'}^{(+)^{*}}(x)\mathfrak{H}a_n \psi_{n}^{(+)}(x)$

Only for non-interacting particles $\mathfrak{H}$ and $a_n$ commute; but in general they do not commute. For non-interacting particles, $\mathcal{H}=\sum_{n,n'}\int d^{3}x\, a^{\dagger}_{n'}\psi_{n'}^{(+)^{*}}(x)E^{0}_{n}\psi_{n}^{(+)}(x)a_n =\sum_{n,n'}E^{0}_{n}a^{\dagger}_{n'}a_n\delta_{nn'}=\sum_{n}E^{0}_{n}a^{\dagger}_{n}a_n=\sum_{n}E^{0}_{n}\widehat{N}$

If they do not commute, Hamiltonian will not have the above expression. Therefore, in general, fock states are not energy eigenstates of a system.

## Vacuum fluctuations

The vacuum state or $|0\rangle$ is the state of lowest energy and the expectation values of $a$ and $a^{\dagger}$ vanish in this state:

$a|0\rangle = 0 = \langle0|a^{\dagger}$

The electrical and magnetic fields and the vector potential have the mode expansion of the same general form:

$F(\vec{r},t) = \varepsilon a e^{i\vec{k}x-\omega t} + h.c$

Thus it is easy to see that the expectation values of these field operators vanishes in the vacuum state:

$\langle0|F|0\rangle = 0$

However, it can be shown that the expectation values of the square of these field operators is non-zero. Thus there are fluctuations in the field about the zero ensemble average. These vacuum fluctuations are responsible for many interesting phenomenon including the Lamb shift in quantum optics.

## Multi-mode Fock states

In a multi-mode field each creation and annihilation operator operates on its own mode. So $a_{{\mathbf{k}}_{l}}$ and $a^{\dagger}_{{\mathbf{k}}_{l}}$ will operate only on $|n_{{\mathbf{k}}_{l}}\rangle$. Since operators corresponding to different modes operate in different sub-spaces of the Hilbert space, the entire field is a direct product of $|n_{{\mathbf{k}}_l}\rangle$ over all the modes:

$|n_{{\mathbf{k}}_{1}}\rangle |n_{{\mathbf{k}}_{2}}\rangle |n_{{\mathbf{k}}_{3}}\rangle... \equiv |n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}}...\rangle \equiv |\{n_{\mathbf{k}}\}\rangle$

The creation and annihilation operators operate on the multi-mode state by only raising or lowering the number state of their own mode:

$a_{{\mathbf{k}}_l} |n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}},...\rangle = \sqrt{n_{{\mathbf{k}}_{l}}} |n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}}-1 ,...\rangle$
$a^{\dagger}_{{\mathbf{k}}_l} |n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}},...\rangle = \sqrt{n_{{\mathbf{k}}_{l}} +1 } |n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}}+1 ,...\rangle$

We also define the total number operator for the field which is a sum of number operators of each mode:

$\hat{n}_{\mathbf{k}} = \sum \hat{n}_{\mathbf{k}_l}$

The multi-mode Fock state is an eigenvector of the total number operator whose eigenvalue is the total occupation number of all the modes

$\hat{n}_{\mathbf{k}} |\{n_{\mathbf{k}}\}\rangle = \bigg( \sum n_{\mathbf{k}_l} \bigg) |\{n_{\mathbf{k}}\}\rangle$

In case of non-interacting particles, number operator and Hamiltonian commute with each other and hence multi-mode Fock states become eigenstates of the multi-mode Hamiltonian

$\hat{H} |\{n_{\mathbf{k}}\}\rangle = \bigg( \sum \hbar \omega \big(n_{\mathbf{k}_l} + \frac{1}{2} \big)\bigg) |\{n_{\mathbf{k}}\}\rangle$

## Source of single photon state

Single photons are routinely generated using single emitters (atoms, Nitrogen-vacancy center ,[8] Quantum dot [9]). However, these sources are not always very efficient (low probability of actually getting a single photon on demand) and often complex and unsuitable out of a laboratory environment. Other sources are commonly used that overcome these issues at the expense of a nondeterministic behavior. Heralded single photon sources are probabilistic two-photon sources from whom the pair is split and the detection of one photon heralds the presence of the remaining one. These sources usually rely on the optical nonlinearity of some materials like periodically poled Lithium niobate (Spontaneous parametric down-conversion), or silicon (spontaneous Four-wave mixing) for example.

## Non-classical behaviour

The Glauber-Sudarshan P-representation of Fock states shows that these states are purely quantum mechanical and have no classical counterpart. The $\scriptstyle\varphi(\alpha) \,$[clarification needed] of these states in the representation is a $2n$'th derivative of the Dirac delta function and therefore not a classical probability distribution.