Fock state explained

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] The Fock states of bosons and fermions obey useful relations with respect to the Fock space creation and annihilation operators.

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. Additionally, depending on the integrality of the particles' spin, the tensor products must be alternating (anti-symmetric) or symmetric products of the underlying one-particle Hilbert space. Specifically:

• Fermions, having half-integer spin and obeying the Pauli 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 the Fock space as the direct sum of the tensor product Hilbert spaces for each particle 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 $\backslash left\backslash \_$ 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.

\widehat{N _{{k} _{i}}} corresponding to the i-th elementary state k_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 _{{k} _{1},n} _{{k} _{2},..n} _{{k} _{i}...\rangle} . In this expression, n _{{k} _{i}} denotes the number of particles in the i-th state k_i, and the particle number operator for the i-th state, \widehat{N _{{k} _{i}}} , acts on the Fock state in the following way:\widehat{N _{{k} _{i}}|n} _{{k} _{1},n} _{{k} _{2},..n} _{{k} _{i}...\rangle} = n _{{k} _{i}|n} _{{k} _{1},n} _{{k} _{2},..n} _{{k} _{i}...\rangle}

Hence the Fock state is an eigenstate of the number operator with eigenvalue

n _{{k} _{i}} .^[2]

Fock states often form the most convenient basis of a Fock space. Elements of a Fock space that are superpositions of states of differing particle 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 $\backslash widehat$ as

\widehat{N} = \sum _{i} \widehat{N _{{k} _{i}},}

the definition of Fock state ensures that the variance of 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 _{{{k} _{1}},} 1 _{{{k} _{2}}\rangle} , and any operator \widehat{ O } , we have the following condition for indistinguishability:^[3] \left|\left\langle f\left|\widehat{ O }\right|1_, 1_\right\rangle\right|^2 = \left|\left\langle f\left|\widehat\right|1_, 1_\right\rangle\right|^2.

So, we must have

\left\langle f\left|\widehat{ O }\right|1_, 1_\right\rangle = e^\left\langle f\left|\widehat\right|1_, 1_\right\rangle

where

e ^{i\delta} = +1 for bosons and -1 for fermions. Since \langle f| and \widehat{ O } are arbitrary, we can say, \left| 1 _{k _{1}} , 1 _{k _{2}} \right\rangle = + \left| 1 _{k _{2}} , 1 _{k _{1}} \right\rangle for bosons and \left| 1 _{k _{1}} , 1 _{k _{2}} \right\rangle = - \left| 1 _{k _{2}} , 1 _{k _{1}} \right\rangle for fermions.^[3]

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:

• Creation operator $b^\_$:

b ^{\dagger} _{{ k} _{l}|n} _{{k} _{1} }, n_,n_...n_,...\rangle=\sqrt |n_, n_,n_...n_+1,...\rangle

• Annihilation operator $b\_$:

b _{{k} _{l}|n} _{{k} _{1} }, n_,n_...n_,...\rangle=\sqrt |n_, n_, n_...n_-1,...\rangle

Non-Hermiticity of creation and annihilation operators

The bosonic Fock state creation and annihilation operators are not Hermitian operators.^[4]

Operator identities

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

\left[b _{i,} b ^{\dagger} _{j\right]} \equiv b _{i} b ^{\dagger} _{j} - b _{i} = \delta _{i} , ^[4] \left[b ^{\dagger} _{i,} b ^{\dagger} _{j\right]} = \left[b _{i,} b _{j\right]} = 0 , ^[4] where [ , ] is the commutator and \delta _{i} is the Kronecker delta.

N bosonic basis states

Number of particles (N)	Bosonic basis states^[5]
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_, n_,n_...n_,...\rangle

Action of number operators

The number operators $\backslash widehat$ for a bosonic system are given by

\widehat{N _{{k} ^{\dagger} ^{{ k} _{l}b} _{{k} _{l}} , where \widehat{N _{{k} _{l}}|n} _{{k} _{1} }, n_,n_...n_...\rangle=n_ |n_, n_,n_...n_...\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 (say, l and m) is done by annihilating a particle in state l and creating one in state m. If we start with a Fock state

|\psi\rangle = \left|n _{k _{1}} , n _{k _{2}} , .... n _{k _{m}} ... n _{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 ^{\dagger} _{k _{m}} b _{k _{l}} in the following way:

Using the commutation relation we have,

b ^{\dagger.b} _{k _{l}} = b _{k _{l}} .b ^{\dagger} _{k _{m}} \begin{align} b ^{\dagger.b} _{k _{l}} \left|n _{k _{1}} , n _{k _{2}} , .... n _{k _{m}} ... n _{k _{l}} ... \right\rangle & = b _{k _{l}} .b ^{\dagger} _{k _{m}} \left|n _{k _{1}} , n _{k _{2}} , .... n _{k _{m}} ... n _{k _{l}} ... \right\rangle \\ & = \sqrt{n _{k _{m}} + 1 }\sqrt{n _{k _{l}} } \left|n_, n_, .... n_ + 1 ... n_ - 1 ...\right\rangle\end

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 for a Fermionic Fock state

|\psi\rangle = |n _{{k} _{1} }, n_,n_...n_,...\rangle as:

• The creation operator

c ^{\dagger} _{{ k} _{l}} acts as:c ^{\dagger} _{{ k} _{l}|n} _{{k} _{1} }, n_,n_...n_,...\rangle=\sqrt |n_, n_,n_...n_+1,...\rangle

• The annihilation operator $c\_$ acts as:

c _{{k} _{l}|n} _{{k} _{1} }, n_,n_...n_,...\rangle=\sqrt |n_, n_,n_...n_-1,...\rangle

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,

\begin{align} \left\{c _{i,} c ^{\dagger} _{j\right\}} \equiv c _{i} c ^{\dagger} _{j} + c _{i} & = \delta _{ij} , \\ \left\{c ^{\dagger} _{i,} c ^{\dagger} _{j\right\}} = \left\{c _{i,} c _{j\right\}} & = 0 , \end{align} ^[4]

where

{\{ , \} } is the anticommutator and \delta _{i} is the Kronecker delta. These anticommutation relations can be used to show antisymmetric behaviour of Fermionic Fock states.

Action of number operators

Number operators $\backslash widehat$ for Fermions are given by

\widehat{N _{{k} ^{\dagger} ^{{ k} _{l}.c} _{{k} _{l}} .\widehat{N _{{k} _{l}}|n} _{{k} _{1} }, n_,n_...n_...\rangle=n_ |n_, n_,n_...n_...\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 _{k _{1}} , n _{k _{2}} , n _{k _{3}} ... n _{k _{l}} ... \right\rangle is obtained by applying a certain sum of permutation operators to the tensor product of eigenkets as follows:\left|n _{k _{1}} , n _{k _{2}} , n _{k _{3}} ... n _{k _{l}} ...\right\rangle = S _{-\left|i} _{1,} i _{2,} i _{3} ... i _{l} ...\right\rangle = (1)/(\sqrt{N!) }\begin \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^[6]

This determinant is called the Slater 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 identical fermions must not occupy the same state (a statement of the Pauli 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 _{{k} _{l}}} must be either 0 or 1.

N fermionic basis states

\left|n _{k _{1}} , n _{k _{2}} , n _{k _{3}} ... n _{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...
...	...

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 _{k _{1}} , n _{k _{2}} , ... n _{k _{m}} ... n _{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 ^{\dagger} _{k _{m}} .c _{k _{l}} in the following way:

Using the anticommutation relation we have

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

but,

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

Fock states are not energy eigenstates in general

In second quantization theory, the Hamiltonian density function is given by

ak{H} = (1)/(2 m) \nabla _{i} \psi ^{*} (x) \nabla _{i} \psi(x) ^[3]

The total Hamiltonian is given by

\begin{align} l{H} & = \int d ^{3} x ak{H} = \int d ^{3} x \psi ^{*} (x)\left( - (\nabla ^{2})/(2 m) \right)\psi(x) \\ \therefore ak{H} & =- (\nabla ^{2})/(2 m) \end{align}

In free Schrödinger theory,^[3]

ak{H}\psi ^{( + )} _{n} (x) =- (\nabla ^{2})/(2 m) \psi ^{( + )} _{n} (x) = E ^{0} _{n} \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 l{H} = \sum _{n,n'} \int d ^{3} x a ^{\dagger} _{n'} \psi ^{( + ) ^{*}} _{n'} (x) ak{H}a _{n} \psi ^{( + )} _{n} (x)

Only for non-interacting particles do

ak{H} and a _{n} commute; in general they do not commute. For non-interacting particles,l{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, 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 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:

\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

a _{k _{l}} and a ^{\dagger} _{k _{l}} will operate only on \left|n _{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 _{k _{l}} \rangle over all the modes:\left|n _{k _{1}} \right\rangle \left|n _{k _{2}} \right\rangle \left|n _{k _{3}} \right\rangle \ldots \equiv \left|n _{k _{1}} , n _{k _{2}} , n _{k _{3}} ... n _{k _{l}} ... \right\rangle \equiv \left|\{n _{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{align} a _{{k} _{l}} |n _{{k} _{1} }, n_, n_... n_, ...\rangle &= \sqrt |n_, n_, n_... n_-1, ...\rangle \\ a^_ |n_, n_, n_... n_,...\rangle &= \sqrt |n_, n_, n_... n_ + 1, ...\rangle\end

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

\hat{n} _{k} = \sum \hat{n} _{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} _{k} |\{n _{k} \}\rangle = \left( \sum n _{k _{l}} \right) |\{n _{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 _{k} \}\right\rangle = \left( \sum \hbar \omega \left(n _{k _{l}} + (1)/(2) \right)\right) \left|\{n _{k} \}\right\rangle

Source of single photon state

Single photons are routinely generated using single emitters (atoms, ions, molecules, Nitrogen-vacancy center,^[7] Quantum dot^[8]). 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 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) of these states in the representation is a 2 n 'th derivative of the Dirac delta function and therefore not a classical probability distribution.

See also

• Coherent states
• Heisenberg limit
• Nonclassical light

External links

• Vladan Vuletic of MIT has used 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 experiment QuantumLab

Notes and References

1. Book: Friedrichs , K. O. . Mathematical aspects of the Quantum Theory of Fields. Interscience Publishers. 1953. B0006ATGK4 .
2. Book: Mandel , Wolf . Optical coherence and quantum optics. Cambridge University Press. 1995. 0521417112.
3. Book: Gross , Franz . Relativistic Quantum Mechanics and Field Theory. Wiley-VCH. 1999. 0471353868.
4. Web site: Quantum Mechanics 1 Lecture Notes on Identical Particles, TIFR, Mumbai.
5. Book: Bruus , Flensberg . Many-Body Quantum Theory in Condensed Matter Physics: An Introduction . OUP Oxford. 2003. 0198566336.
6. Book: Schwabl, Hilton, Lahee . Advanced Quantum Mechanics . Springer . 2008 . 978-3540850618.
7. C. Kurtsiefer, S. Mayer, P. Zarda, Patrick and H. Weinfurter, (2000), "Stable Solid-State Source of Single Photons", Phys. Rev. Lett. 85 (2) 290--293, doi 10.1103/PhysRevLett.85.290
8. C. Santori, M. Pelton, G. Solomon, Y. Dale and Y. Yamamoto (2001), "Triggered Single Photons from a Quantum Dot", Phys. Rev. Lett. 86 (8):1502--1505 DOI 10.1103/PhysRevLett.86.1502