Fock state

Inquantum mechanics,aFock stateornumber stateis aquantum statethat is an element of aFock spacewith a well-defined number ofparticles(orquanta). These states are named after theSovietphysicistVladimir Fock.Fock states play an important role in thesecond quantizationformulation of quantum mechanics.

The particle representation was first treated in detail byPaul Diracforbosonsand byPascual JordanandEugene Wignerforfermions.The Fock states of bosons and fermions obey useful relations with respect to the Fock spacecreation and annihilation operators.

Definition

One specifies a multiparticle state of N non-interacting identical particles by writing the state as a sum oftensor productsof N one-particle states. Additionally, depending on the integrality of the particles'spin,the tensor products must bealternating(anti-symmetric) orsymmetric productsof the underlying one-particleHilbert space.Specifically:

• Fermions,having half-integer spin and obeying thePauli exclusion principle,correspond to antisymmetric tensor products.
• Bosons,possessing integer spin (and not governed by the exclusion principle) correspond to symmetric tensor products.

If the number of particles is variable, one constructs theFock spaceas thedirect sumof the tensor product Hilbert spaces for eachparticle number.In the Fock space, 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${\textstyle \left\{\mathbf {k} _{i}\right\}_{i\in I}}$be anorthonormal basisof 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 aFock stateif it is an element of the occupancy number basis.

A Fock state satisfies an important criterion: for eachi,the state is an eigenstate of theparticle number operator${\widehat {N_{{\mathbf {k} }_{i}}}}$corresponding to thei-th elementary statek_i.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 the number of particles in the i-th statek_i,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$n_{{\mathbf {k} }_{i}}$.

Fock states often form the most convenientbasisof a Fock space. Elements of a Fock space that aresuperpositionsof states of differingparticle number(and thus not eigenstates of the number operator) are not Fock states. For this reason, not all elements of a Fock space are referred to as "Fock states".

If we define the aggregate particle number operator${\textstyle {\widehat {N}}}$as

${\widehat {N}}=\sum _{i}{\widehat {N_{{\mathbf {k} }_{i}}}},$

the definition of Fock state ensures that thevarianceof measurement$\operatorname {Var} \left({\widehat {N}}\right)=0$,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$|f\rangle$,any Fock state of two identical particles given by$|1_{\mathbf {k} _{1}},1_{\mathbf {k} _{2}}\rangle$,and anyoperator${\widehat {\mathbb {O} }}$,we have the following condition forindistinguishability:

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

So, we must have$\left\langle f\left|{\widehat {\mathbb {O} }}\right|1_{\mathbf {k} _{1}},1_{\mathbf {k} _{2}}\right\rangle =e^{i\delta }\left\langle f\left|{\widehat {\mathbb {O} }}\right|1_{\mathbf {k} _{2}},1_{\mathbf {k} _{1}}\right\rangle$

where$e^{i\delta }=+1$forbosonsand$-1$forfermions.Since$\langle f|$and${\widehat {\mathbb {O} }}$are arbitrary, we can say,

$\left|1_{\mathbf {k} _{1}},1_{\mathbf {k} _{2}}\right\rangle =+\left|1_{\mathbf {k} _{2}},1_{\mathbf {k} _{1}}\right\rangle$for bosons and

$\left|1_{\mathbf {k} _{1}},1_{\mathbf {k} _{2}}\right\rangle =-\left|1_{\mathbf {k} _{2}},1_{\mathbf {k} _{1}}\right\rangle$for fermions.

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 thecreation and annihilation operators.

Bosonic Fock state

Bosons,which are particles with integer spin, follow a simple rule: their composite eigenstate is symmetricunder operation by anexchange 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 bosoniccreation and annihilation operators,denoted by$b^{\dagger }$and$b$respectively. The action of these operators on a Fock state are given by the following two equations:

• Creation operator${\textstyle b_{{\mathbf {k} }_{l}}^{\dagger }}$:

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

• Annihilation operator${\textstyle b_{{\mathbf {k} }_{l}}}$:

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

The Operation of creation and annihilation operators on Bosonic Fock states.

Non-Hermiticity of creation and annihilation operators

The bosonic Fock state creation and annihilation operators are notHermitian operators.

Proof that creation and annihilation operators are not Hermitian.

For a Fock state,$|n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},n_{\mathbf {k} _{3}}\dots n_{\mathbf {k} _{l}},\dots \rangle$, $${\begin{aligned}\left\langle n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},n_{\mathbf {k} _{3}}\dots n_{\mathbf {k} _{l}}-1,\dots \left|b_{\mathbf {k} _{l}}\right|n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},n_{\mathbf {k} _{3}}\dots n_{\mathbf {k} _{l}},\dots \right\rangle &={\sqrt {n_{\mathbf {k} _{l}}}}\left\langle n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},n_{\mathbf {k} _{3}}\dots n_{\mathbf {k} _{l}}-1,\dots |n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},n_{\mathbf {k} _{3}}\dots n_{\mathbf {k} _{l}}-1,\dots \right\rangle \\[6pt]\left(\left\langle n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},n_{\mathbf {k} _{3}}\dots n_{\mathbf {k} _{l}},\dots \left|b_{\mathbf {k} _{l}}\right|n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},n_{\mathbf {k} _{3}}\dots n_{\mathbf {k} _{l}}-1,\dots \right\rangle \right)^{*}&=\left\langle n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},n_{\mathbf {k} _{3}}\dots n_{\mathbf {k} _{l}}-1\dots \left|b_{\mathbf {k} _{l}}^{\dagger }\right|n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},n_{\mathbf {k} _{3}}\dots n_{\mathbf {k} _{l}},\dots \right\rangle \\&={\sqrt {n_{\mathbf {k} _{l}}+1}}\left\langle n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},n_{\mathbf {k} _{3}}\dots n_{\mathbf {k} _{l}}-1\dots |n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},n_{\mathbf {k} _{3}}\dots n_{\mathbf {k} _{l}}+1\dots \right\rangle \end{aligned}}$$

Therefore, it is clear that adjoint of creation (annihilation) operator doesn't go into itself. Hence, they are not Hermitian operators.

But adjoint of creation (annihilation) operator is annihilation (creation) operator.

Operator identities

The commutation relations of creation and annihilation operators in abosonic systemare

$\left[b_{i}^{\,},b_{j}^{\dagger }\right]\equiv b_{i}^{\,}b_{j}^{\dagger }-b_{j}^{\dagger }b_{i}^{\,}=\delta _{ij},$

$\left[b_{i}^{\dagger },b_{j}^{\dagger }\right]=\left[b_{i}^{\,},b_{j}^{\,}\right]=0,$

where$[\ \,\ \ ]$is thecommutatorand$\delta _{ij}$is theKronecker delta.

N bosonic basis states

Number of particles (N)	Bosonic basis states
0	$|0,0,0...\rangle$
1	$|1,0,0...\rangle$,$|0,1,0...\rangle$,$|0,0,1...\rangle$,...
2	$|2,0,0...\rangle$,$|1,1,0...\rangle$,$|0,2,0...\rangle$,...
$n$	$|n_{{\mathbf {k} }_{1}},n_{{\mathbf {k} }_{2}},n_{{\mathbf {k} }_{3}}...n_{{\mathbf {k} }_{l}},...\rangle$

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:

  $b_{{\mathbf {k} }_{l}}^{\dagger }|0_{{\mathbf {k} }_{1}},0_{{\mathbf {k} }_{2}},0_{{\mathbf {k} }_{3}}...0_{{\mathbf {k} }_{l}},...\rangle =|0_{{\mathbf {k} }_{1}},0_{{\mathbf {k} }_{2}},0_{{\mathbf {k} }_{3}}...1_{{\mathbf {k} }_{l}},...\rangle$

  and,$b_{\mathbf {k} _{l}}|0_{\mathbf {k} _{1}},0_{\mathbf {k} _{2}},0_{\mathbf {k} _{3}}...0_{\mathbf {k} _{l}},...\rangle =0$.That is, thel-th creation operator creates a particle in thel-th statek_l,and the vacuum state is a fixed point of annihilation operators as there are no particles to annihilate.
• We can generate any Fock state by operating on the vacuum state with an appropriate number ofcreation operators:

  $|n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}}...\rangle ={\frac {\left(b_{\mathbf {k} _{1}}^{\dagger }\right)^{n_{\mathbf {k} _{1}}}}{\sqrt {n_{\mathbf {k} _{1}}!}}}{\frac {\left(b_{\mathbf {k} _{2}}^{\dagger }\right)^{n_{\mathbf {k} _{2}}}}{\sqrt {n_{\mathbf {k} _{2}}!}}}...|0_{\mathbf {k} _{1}},0_{\mathbf {k} _{2}},...\rangle$

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

Action of number operators

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

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 (say,landm) is done by annihilating a particle in stateland creating one in statem.If we start with a Fock state$|\psi \rangle =\left|n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},....n_{\mathbf {k} _{m}}...n_{\mathbf {k} _{l}}...\right\rangle$,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,$b_{\mathbf {k} _{m}}^{\dagger }.b_{\mathbf {k} _{l}}=b_{\mathbf {k} _{l}}.b_{\mathbf {k} _{m}}^{\dagger }$

${\begin{aligned}b_{\mathbf {k} _{m}}^{\dagger }.b_{\mathbf {k} _{l}}\left|n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},....n_{\mathbf {k} _{m}}...n_{\mathbf {k} _{l}}...\right\rangle &=b_{\mathbf {k} _{l}}.b_{\mathbf {k} _{m}}^{\dagger }\left|n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},....n_{\mathbf {k} _{m}}...n_{\mathbf {k} _{l}}...\right\rangle \\&={\sqrt {n_{\mathbf {k} _{m}}+1}}{\sqrt {n_{\mathbf {k} _{l}}}}\left|n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},....n_{\mathbf {k} _{m}}+1...n_{\mathbf {k} _{l}}-1...\right\rangle \end{aligned}}$

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

• 
  Wigner function of$|0\rangle$
• 
  Wigner function of$|1\rangle$
• 
  Wigner function of$|2\rangle$
• 
  Wigner function of$|3\rangle$
• 
  Wigner function of$|4\rangle$

Fermionic Fock state

Fermion creation and annihilation operators

To be able to retain the antisymmetric behaviour offermions,for Fermionic Fock states we introduce non-Hermitian fermion creation and annihilation operators,defined for a Fermionic Fock state$|\psi \rangle =|n_{{\mathbf {k} }_{1}},n_{{\mathbf {k} }_{2}},n_{{\mathbf {k} }_{3}}...n_{{\mathbf {k} }_{l}},...\rangle$as:

• Thecreation operator$c_{{\mathbf {k} }_{l}}^{\dagger }$acts as:

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

• Theannihilation operator${\textstyle c_{{\mathbf {k} }_{l}}}$acts as:

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

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

Operator identities

The anticommutation relations of creation and annihilation operators in afermionic systemare,

${\begin{aligned}\left\{c_{i}^{\,},c_{j}^{\dagger }\right\}\equiv c_{i}^{\,}c_{j}^{\dagger }+c_{j}^{\dagger }c_{i}^{\,}&=\delta _{ij},\\\left\{c_{i}^{\dagger },c_{j}^{\dagger }\right\}=\left\{c_{i}^{\,},c_{j}^{\,}\right\}&=0,\end{aligned}}$

where${\{\,\ \}}$is theanticommutatorand$\delta _{ij}$is theKronecker delta.These anticommutation relations can be used to show antisymmetric behaviour ofFermionic Fock states.

Action of number operators

Number operators${\textstyle {\widehat {N_{{\mathbf {k} }_{l}}}}}$forFermionsare given by${\widehat {N_{{\mathbf {k} }_{l}}}}=c_{{\mathbf {k} }_{l}}^{\dagger }.c_{{\mathbf {k} }_{l}}$.

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

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$|\psi \rangle =\left|n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},n_{\mathbf {k} _{3}}...n_{\mathbf {k} _{l}}...\right\rangle$is obtained by applying a certain sum of permutation operators to the tensor product of eigenkets as follows:

$\left|n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},n_{\mathbf {k} _{3}}...n_{\mathbf {k} _{l}}...\right\rangle =S_{-}\left|i_{1},i_{2},i_{3}...i_{l}...\right\rangle ={\frac {1}{\sqrt {N!}}}{\begin{vmatrix}\left|i_{1}\right\rangle _{1}&\cdots &\left|i_{1}\right\rangle _{N}\\\vdots &\ddots &\vdots \\\left|i_{N}\right\rangle _{1}&\cdots &\left|i_{N}\right\rangle _{N}\end{vmatrix}}$

This determinant is called theSlater determinant.If any of the single particle states are the same, two rows of the Slater determinant would be the same and hence the determinant would be zero. Hence, two identicalfermionsmust not occupy the same state (a statement of thePauli exclusion principle). Therefore, the occupation number of any single state is either 0 or 1. The eigenvalue associated to the fermionic Fock state${\widehat {N_{{\mathbf {k} }_{l}}}}$must be either 0 or 1.

N fermionic basis states$\left|n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},n_{\mathbf {k} _{3}}...n_{\mathbf {k} _{l}},...\right\rangle$

Number of particles (N)	Fermionic basis states
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

The operation of creation and annihilation operators on Fermionic Fock states.

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

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

  and$c_{\mathbf {k} }^{\dagger }\left|1_{\mathbf {k} }\right\rangle =0$,as the maximum occupation number of any state is 1. No more than 1 fermion can occupy the same state, as stated in thePauli exclusion principle.
• For a single mode fermionic Fock state, expressed as$\left|1_{\mathbf {k} }\right\rangle$,

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

  and$c_{\mathbf {k} }\left|0_{\mathbf {k} }\right\rangle =0$,as the particle number cannot be less than zero.
• For a multimode fermionic Fock state, expressed as,$\left|n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},...n_{\mathbf {k} _{\beta }},n_{\mathbf {k} _{\alpha }},...\right\rangle$

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

  where$(-1)^{\sum _{\beta <\alpha }n_{\beta }}$is called theJordan–Wigner string,which depends on the ordering of the involved single-particle states and adding the fermion occupation numbers of all preceding states.

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$|\psi \rangle =\left|n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},...n_{\mathbf {k} _{m}}...n_{\mathbf {k} _{l}}...\right\rangle$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

$c_{\mathbf {k} _{m}}^{\dagger }.c_{\mathbf {k} _{l}}=-c_{\mathbf {k} _{l}}.c_{\mathbf {k} _{m}}^{\dagger }$

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

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

Thus, fermionic Fock states are antisymmetric under operation by particle exchange operators.

Fock states are not energy eigenstates in general

Insecond quantizationtheory, theHamiltonian densityfunction is given by

${\mathfrak {H}}={\frac {1}{2m}}\nabla _{i}\psi ^{*}(x)\,\nabla _{i}\psi (x)$

The totalHamiltonianis given by

${\begin{aligned}{\mathcal {H}}&=\int d^{3}x\,{\mathfrak {H}}=\int d^{3}x\psi ^{*}(x)\left(-{\frac {\nabla ^{2}}{2m}}\right)\psi (x)\\\therefore {\mathfrak {H}}&=-{\frac {\nabla ^{2}}{2m}}\end{aligned}}$

In free Schrödinger theory,

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

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.

$\therefore {\mathcal {H}}=\sum _{n,n'}\int d^{3}x\,a_{n'}^{\dagger }\psi _{n'}^{(+)^{*}}(x)\,{\mathfrak {H}}a_{n}\psi _{n}^{(+)}(x)$

Only for non-interacting particles do${\mathfrak {H}}$and$a_{n}$commute; in general they do not commute. For non-interacting particles,

${\mathcal {H}}=\sum _{n,n'}\int d^{3}x\,a_{n'}^{\dagger }\psi _{n'}^{(+)^{*}}(x)\,E_{n}^{0}\psi _{n}^{(+)}(x)a_{n}=\sum _{n,n'}E_{n}^{0}a_{n'}^{\dagger }a_{n}\delta _{nn'}=\sum _{n}E_{n}^{0}a_{n}^{\dagger }a_{n}=\sum _{n}E_{n}^{0}{\widehat {N}}$

If they do not commute, the 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:

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

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

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

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

$\langle 0|F|0\rangle =0$

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 theLamb shiftin 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$a_{\mathbf {k} _{l}}^{\dagger }$will operate only on$\left|n_{\mathbf {k} _{l}}\right\rangle$.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:

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

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

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

We also define the totalnumber operatorfor 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

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

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

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

Source of single photon state

Single photons are routinely generated using single emitters (atoms, ions, molecules,Nitrogen-vacancy center,Quantum dot). However, these sources are not always very efficient, often presenting a 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 non-linearity of some materials like periodically poledLithium niobate(Spontaneous parametric down-conversion), or silicon (spontaneousFour-wave mixing) for example.

Non-classical behaviour

TheGlauber–Sudarshan P-representationof Fock states shows that these states are purely quantum mechanical and have no classical counterpart. The$\scriptstyle \varphi (\alpha )\,$of these states in the representation is a$2n$'th derivative of theDirac delta functionand therefore not a classical probability distribution.

See also

• Coherent states
• Heisenberg limit
• Nonclassical light

External links

• Vladan Vuletic ofMIThasused an ensemble of atoms to produce a Fock state (a.k.a. single photon) source(PDF)
• Produce and measure a single photon state (Fock state) with an interactive experimentQuantumLab