Fock state

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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[edit]

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 |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, 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.

Example using two particles[edit]

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

Operation by a Number operator[edit]

Suppose, for a number operator given by \widehat{N_{{\mathbf{k}}_1}} , we have for bosons, \widehat{N_{{\mathbf{k}}_1}}{|1_{{\mathbf{{k}}_1}}, 1_{{\mathbf{{k}}_2}}\rangle}_{bosonic}=1_{{\mathbf{{k}}_1}}|1_{{\mathbf{{k}}_1}}, 1_{{\mathbf{{k}}_2}}\rangle

and for fermions \widehat{N_{{\mathbf{k}}_1}}{|1_{{\mathbf{{k}}_1}}, 1_{{\mathbf{{k}}_2}}\rangle}_{fermionic}=1_{{\mathbf{{k}}_1}}|1_{{\mathbf{{k}}_1}}, 1_{{\mathbf{{k}}_2}}\rangle

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[edit]

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[edit]

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

Hermiticity of Creation and Annihilation operator[edit]

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

Operator Identities[edit]

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[edit]

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[edit]

  • 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[edit]

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[edit]

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[edit]

Fermion Creation and Annihilation operators[edit]

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[edit]

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[edit]

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[edit]

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 |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:

|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 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 \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[edit]

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[edit]

The Operation of creation and annihilation operators on Fermionic 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 maximum occupation number of any state is 1, more than 1 fermions 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 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 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[edit]

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[edit]

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}

For free Schrodinger Theory,[3]:189

\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[edit]

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[edit]

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[edit]

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[edit]

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 2n'th derivative of the Dirac delta function and therefore not a classical probability distribution.

See also[edit]

References[edit]

  1. ^ Friedrichs, K. O. (1953). Mathematical aspects of the Quantum Theory of Fields. Interscience Publishers. ASIN B0006ATGK4. 
  2. ^ Mandel, Wolf (1995). Optical coherence and quantum optics. Cambridge University Press. ISBN 0521417112. 
  3. ^ a b c d Gross, Franz (1999). Relativistic Quantum Mechanics and Field Theory. Wiley-VCH. ISBN 0471353868. 
  4. ^ a b c d e f g h i j k l m n o "Quantum Mechanics 1 Lecture Notes on Identical Particles, TIFR, Mumbai". 
  5. ^ a b Altland, Simons (2006). Condensed Matter Field Theory. Cambridge University Press. ISBN 0521769752. 
  6. ^ a b Bruus, Flensberg (2003). Many-Body Quantum Theory in Condensed Matter Physics: An Introduction. OUP Oxford. ISBN 0198566336. 
  7. ^ Schwabl, Hilton, Lahee (2008). Advanced Quantum Mechanics. Springer. ISBN 3540850619. 
  8. ^ 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
  9. ^ 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

External links[edit]