Fock state

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:

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

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

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

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

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

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

Proof that creation and annihilation operators are not Hermitian.

For a Fock state, ${\displaystyle |n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},n_{\mathbf {k} _{3}}\dots n_{\mathbf {k} _{l}},\dots \rangle }$,

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

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.[5]^: 45

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

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

${\displaystyle \left[b_{i}^{\dagger },b_{j}^{\dagger }\right]=\left[b_{i}^{\,},b_{j}^{\,}\right]=0,}$[4]

where ${\displaystyle [\ \ ,\ \ ]}$ is the commutator and ${\displaystyle \delta _{ij}}$ is the Kronecker delta.

The number operators ${\textstyle {\widehat {N_{{\mathbf {k} }_{l}}}}}$ for a bosonic system are given by ${\displaystyle {\widehat {N_{{\mathbf {k} }_{l}}}}=b_{{\mathbf {k} }_{l}}^{\dagger }b_{{\mathbf {k} }_{l}}}$, where ${\displaystyle {\widehat {N_{{\mathbf {k} }_{l}}}}|n_{{\mathbf {k} }_{1}},n_{{\mathbf {k} }_{2}},n_{{\mathbf {k} }_{3}}...n_{{\mathbf {k} }_{l}}...\rangle =n_{{\mathbf {k} }_{l}}|n_{{\mathbf {k} }_{1}},n_{{\mathbf {k} }_{2}},n_{{\mathbf {k} }_{3}}...n_{{\mathbf {k} }_{l}}...\rangle }$[4]

Number operators are Hermitian operators.

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 ${\displaystyle |\psi \rangle =\left|n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},....n_{\mathbf {k} _{m}}...n_{\mathbf {k} _{l}}...\right\rangle }$, and want to shift a particle from state ${\displaystyle k_{l}}$ 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:

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

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

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

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

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 ${\displaystyle |\psi \rangle =|n_{{\mathbf {k} }_{1}},n_{{\mathbf {k} }_{2}},n_{{\mathbf {k} }_{3}}...n_{{\mathbf {k} }_{l}},...\rangle }$ as:[4]

• The creation operator ${\displaystyle c_{{\mathbf {k} }_{l}}^{\dagger }}$ 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]

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

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

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

The anticommutation relations of creation and annihilation operators in a fermionic system are,

${\displaystyle {\begin{aligned}\left\{c_{i}^{\,},c_{j}^{\dagger }\right\}\equiv c_{i}^{\,}c_{j}^{\dagger }+c_{j}^{\dagger }c_{i}^{\,}&=\delta _{ij},\\\left\{c_{i}^{\dagger },c_{j}^{\dagger }\right\}=\left\{c_{i}^{\,},c_{j}^{\,}\right\}&=0,\end{aligned}}}$ [4]

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

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

${\displaystyle {\widehat {N_{{\mathbf {k} }_{l}}}}|n_{{\mathbf {k} }_{1}},n_{{\mathbf {k} }_{2}},n_{{\mathbf {k} }_{3}}...n_{{\mathbf {k} }_{l}}...\rangle =n_{{\mathbf {k} }_{l}}|n_{{\mathbf {k} }_{1}},n_{{\mathbf {k} }_{2}},n_{{\mathbf {k} }_{3}}...n_{{\mathbf {k} }_{l}}...\rangle }$[4]

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

${\displaystyle \left|n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},n_{\mathbf {k} _{3}}...n_{\mathbf {k} _{l}}...\right\rangle =S_{-}\left|i_{1},i_{2},i_{3}...i_{l}...\right\rangle ={\frac {1}{\sqrt {N!}}}{\begin{vmatrix}\left|i_{1}\right\rangle _{1}&\cdots &\left|i_{1}\right\rangle _{N}\\\vdots &\ddots &\vdots \\\left|i_{N}\right\rangle _{1}&\cdots &\left|i_{N}\right\rangle _{N}\end{vmatrix}}}$[7]^: 16

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 ${\displaystyle {\widehat {N_{{\mathbf {k} }_{l}}}}}$ must be either 0 or 1.

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 ${\displaystyle |\psi \rangle =\left|n_{\mathbf {k} _{1}},n_{\mathbf {k} _{2}},...n_{\mathbf {k} _{m}}...n_{\mathbf {k} _{l}}...\right\rangle }$ and want to shift a particle from state ${\displaystyle k_{l}}$ 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

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

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

but,

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

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