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 k_i 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 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 $|n_{{\mathbf{k}}_1},n_{{\mathbf{k}}_2},..n_{{\mathbf{k}}_i}...\rangle$ . In this expression, $n_{{\mathbf{k}}_i}$ denotes number of particles in the i-th state k_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 $n_{{\mathbf{k}}_i}$ .^[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]

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}}...n_{{\mathbf{k}}_{l}},...\rangle

\langle n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}}-1,...|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}}}\langle n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}}-1,...|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}}-1,...\rangle

(\langle n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}},...|b_{{\mathbf{k}}_l}|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}}-1,...\rangle)^*

= \langle n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}}-1...|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}\langle n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}}-1...|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}}+1...\rangle

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.^[5]^:45

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 $\widehat {N_{{{{\mathbf {k}}}_{l}}}}=b_{{{{\mathbf {k}}}_{l}}}^{{\dagger }}b_{{{{\mathbf {k}}}_{l}}}$ , where $\widehat{N_{{\mathbf{k}}_l}}|n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}}...\rangle=n_{{\mathbf{k}}_{l}} |n_{{\mathbf{k}}_{1}}, n_{{\mathbf{k}}_{2}} ,n_{{\mathbf{k}}_{3}}...n_{{\mathbf{k}}_{l}}...\rangle$ ^[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.