# 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 important role in second quantization formulation of quantum mechanics.

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

## Definition

Instead of specifying a multiparticle state of N non-interacting particles by writing the state as a tensor product of N one-particle states, it is possible to specify the same state in a new notation, the Fock space representation, by specifying the number of particles in each possible one-particle state. However, this notation loses the ordering of tensor products, which is an important part of the specification of quantum states. To retain the same information in the multiparticle state, one constructs Fock space as the direct sum of Hilbert spaces for different particle numbers.

A quantum state is called a Fock state if it satisfies two criteria:

(i) the state belongs to a Fock space.

(ii) the state is an eigenstate of the particle number operator. The particle number operator operating on a Fock state gives the number of particles in that particular state.

A given Fock state is denoted by ${\displaystyle |n_{{\mathbf {k} }_{1}},n_{{\mathbf {k} }_{2}},..n_{{\mathbf {k} }_{i}}...\rangle }$. In this expression, ${\displaystyle n_{{\mathbf {k} }_{i}}}$ denotes number of particles in the i-th state, and the particle number operator for the i-th state, ${\displaystyle {\widehat {N_{{\mathbf {k} }_{i}}}}}$ acts on the Fock state in the following way:

Hence the Fock state is an eigenstate of the number operator with eigenvalue ${\displaystyle n_{{\mathbf {k} }_{i}}}$.[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 ${\displaystyle {\rm {Var}}({\widehat {N}})=0}$, i.e., measuring the number of particles in a Fock state always returns a definite value with no fluctuation.

## Example using two particles

For any final state ${\displaystyle |f\rangle }$, any Fock state of two identical particles given by ${\displaystyle |1_{\mathbf {k} _{1}},1_{\mathbf {k} _{2}}\rangle }$, and any operator ${\displaystyle {\widehat {\mathbb {O} }}}$, we have the following condition for indistinguishability:[3]:191

### Operation by a Number operator

Hence, using number operator on a state, we can never identify the state whether it is a bosonic one or a fermionic one. For that we need different algebra, which follows in the next sections.

## 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 ${\displaystyle {\hat {P}}\left|x_{1},x_{2}\right\rangle =\left|x_{2},x_{1}\right\rangle }$ .

### 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 ${\displaystyle b^{\dagger }}$ and ${\displaystyle b}$ respectively. The action of these operators on a Fock state are given by the following two equations:

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

${\displaystyle [b_{i}^{\,},b_{j}^{\dagger }]\equiv b_{i}^{\,}b_{j}^{\dagger }-b_{j}^{\dagger }b_{i}^{\,}=\delta _{ij},}$ [4]
${\displaystyle [b_{i}^{\dagger },b_{j}^{\dagger }]=[b_{i}^{\,},b_{j}^{\,}]=0,}$ [4]

### Action on some specific Fock states

${\displaystyle b_{{\mathbf {k} }_{l}}^{\dagger }|0_{{\mathbf {k} }_{1}},0_{{\mathbf {k} }_{2}},0_{{\mathbf {k} }_{3}}...0_{{\mathbf {k} }_{l}},...\rangle =|0_{{\mathbf {k} }_{1}},0_{{\mathbf {k} }_{2}},0_{{\mathbf {k} }_{3}}...1_{{\mathbf {k} }_{l}},...\rangle }$
• We can generate any Fock state by operating on the vacuum state with an appropriate number of creation operators:
${\displaystyle b_{\mathbf {k} }^{\dagger }|n_{\mathbf {k} }\rangle ={\sqrt {n_{\mathbf {k} }+1}}|n_{\mathbf {k} }+1\rangle }$

### Action of Number operator

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, ${\displaystyle |n_{{\mathbf {k} }_{1}},n_{{\mathbf {k} }_{2}},....n_{{\mathbf {k} }_{m}}...n_{{\mathbf {k} }_{l}}...\rangle }$, and want to shift a particle from state ${\displaystyle k_{l}}$ to state ${\displaystyle k_{m}}$, then we operate the Fock state by ${\displaystyle b_{{\mathbf {k} }_{m}}^{\dagger }b_{{\mathbf {k} }_{l}}}$ in the following way:

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, ${\displaystyle |n_{{\mathbf {k} }_{1}},n_{{\mathbf {k} }_{2}},n_{{\mathbf {k} }_{3}}...n_{{\mathbf {k} }_{l}},...\rangle }$, Creation operator acts as:

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

and Annihilation operator acts as:

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

${\displaystyle \{c_{i}^{\,},c_{j}^{\dagger }\}\equiv c_{i}^{\,}c_{j}^{\dagger }+c_{j}^{\dagger }c_{i}^{\,}=\delta _{ij},}$ [4]
${\displaystyle \{c_{i}^{\dagger },c_{j}^{\dagger }\}=\{c_{i}^{\,},c_{j}^{\,}\}=0,}$ [4]

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

### Action of Number operator

#### Maximum Occupation number

Action of Number operator, as well as, 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. Suppose, a Fermionic Fock state ${\displaystyle |n_{{\mathbf {k} }_{1}},n_{{\mathbf {k} }_{2}},n_{{\mathbf {k} }_{3}}...n_{{\mathbf {k} }_{l}}...\rangle }$ be obtained by using some operator from the tensor product of eigenkets as follows:

This determinant is called Slater determinant. If any of the single particle states are 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, occupation number of any single state is either 0 or 1. Eigenvalue of fermionic Fock state ${\displaystyle {\widehat {N_{{\mathbf {k} }_{l}}}}}$ will be either 0 or 1.

### Action on some specific Fock states

${\displaystyle c_{\mathbf {k} }^{\dagger }|0_{\mathbf {k} }\rangle =|1_{\mathbf {k} }\rangle }$

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

${\displaystyle c_{\mathbf {k} }|1_{\mathbf {k} }\rangle =|0_{\mathbf {k} }\rangle }$

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

where, ${\displaystyle (-1)^{\sum _{\beta <\alpha }n\beta }}$ is called Jordan-Wigner String, which depends on the ordering of the involved single-particle states and counting of fermion occupation number 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, ${\displaystyle |n_{{\mathbf {k} }_{1}},n_{{\mathbf {k} }_{2}},....n_{{\mathbf {k} }_{m}}...n_{{\mathbf {k} }_{l}}...\rangle }$, and want to shift a particle from state ${\displaystyle k_{l}}$ to state ${\displaystyle k_{m}}$, then we operate the Fock state by ${\displaystyle c_{{\mathbf {k} }_{m}}^{\dagger }.c_{{\mathbf {k} }_{l}}}$ in the following way:

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

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

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

Total Hamiltonian is given by

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

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

and

${\displaystyle \int d^{3}x\,\psi _{n}^{(+)^{*}}(x)\psi _{n'}^{(+)}(x)=\delta _{nn'}}$

and

${\displaystyle \psi (x)=\sum _{n}a_{n}\psi _{n}^{(+)}(x)}$,

where ${\displaystyle a_{n}}$ is the annihilation operator.

Only for non-interacting particles ${\displaystyle {\mathfrak {H}}}$ and ${\displaystyle a_{n}}$ commute; but in general they do not commute. For non-interacting particles, ${\displaystyle {\mathcal {H}}=\sum _{n,n'}\int d^{3}x\,a_{n'}^{\dagger }\psi _{n'}^{(+)^{*}}(x)E_{n}^{0}\psi _{n}^{(+)}(x)a_{n}=\sum _{n,n'}E_{n}^{0}a_{n'}^{\dagger }a_{n}\delta _{nn'}=\sum _{n}E_{n}^{0}a_{n}^{\dagger }a_{n}=\sum _{n}E_{n}^{0}{\widehat {N}}}$

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

${\displaystyle a|0\rangle =0=\langle 0|a^{\dagger }}$

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

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

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

${\displaystyle \langle 0|F|0\rangle =0}$

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 ${\displaystyle a_{{\mathbf {k} }_{l}}}$ and ${\displaystyle a_{{\mathbf {k} }_{l}}^{\dagger }}$ will operate only on ${\displaystyle |n_{{\mathbf {k} }_{l}}\rangle }$. Since operators corresponding to different modes operate in different sub-spaces of the Hilbert space, the entire field is a direct product of ${\displaystyle |n_{{\mathbf {k} }_{l}}\rangle }$ over all the modes:

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

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

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

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

${\displaystyle {\hat {n}}_{\mathbf {k} }=\sum {\hat {n}}_{{\mathbf {k} }_{l}}}$

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

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

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

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

## 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 ${\displaystyle \scriptstyle \varphi (\alpha )\,}$ of these states in the representation is a ${\displaystyle 2n}$'th derivative of the Dirac delta function and therefore not a classical probability distribution.