Outline

Fock space is the direct sum of identical particle Hilbert space for all different particle numbers:

$$\mathcal{F}=\mathcal{H}_0\oplus \mathcal{H}_1 \oplus\cdots$$

Many-body Hilbert space of fermions and bosons is (Anti-)Symmetrized tensor product space:

$n$ .- --body Hilbert space for identical particles $\mathcal{H}_n$ is a subspace of $(\mathcal{H})^{\otimes n}$
$\mathcal{H}_n$ is the image of multi-linear (anti-)symmetrization mapping:

$$\mathcal{S}:(\mathcal{H}_1)^{\otimes n}\rightarrow \mathcal{H}_n$$

The (anti-)symmetrization mapping is defined on a tensor product basis as:

$$\begin{aligned} \mathcal{S}:\ket{1}\cdots\ket{n}\mapsto \frac 1 {\sqrt{n!}}\sum_{\sigma\in S_n} \ket{\sigma(1)}\cdots\ket{\sigma(n)} \ \ & \text{for boson} \\ \mathcal{S}:\ket{1}\cdots\ket{n}\mapsto \frac 1 {\sqrt{n!}}\sum_{\sigma\in S_n} (-1)^{\sigma} \ket{\sigma(1)}\cdots\ket{\sigma(n)} \ \ & \text{for fermion} \end{aligned}$$

Second quantization language and some simple many-body wavefunction

The Fock Space

Permutation Group

[Definition] : permutation is a map : $\sigma=\begin{pmatrix}1 & 2 & \cdots & n \\ \sigma(1) & \sigma(2) & \cdots & \sigma(n)\end{pmatrix}\equiv\{j,\sigma(j)\}$ means replacing $j$ by $\sigma(j)$ , and $\sigma(j)\in\{1,2,\cdots,n\}$
[Definition] : Transposition : $(i,j)$ means exchange $i$ and $j$ while keeping others fixed.
Any permutation can be represented as a product of transpositions of neighbors $(i,i+1)$ < , and the number of transpositions is defined as the parity of the permutation.
The sign(signature) of a permutation : $(-1)^{\sigma}$ , is $+1$ for even one and $-1$ for odd one. It can be rewritten as:

$$(-1)^{\sigma} = \prod_{i\lt j} \text{sgn }(\sigma(j)-\sigma(i))$$

And satisfies $(-1)^{\sigma\cdot \mu} = (-1)^{\sigma}(-1)^{\mu}$
There are only two 1-d representation of permutation group:

the trivial one $R(\sigma)=1$
the alternative one $R(\sigma)=(-1)^{\sigma}$

Traditional Treatment

The configuration of $n$ .- - indentical particles $x=(\bm{x}_1,\bm{x}_2,\cdots,\bm{x}_n)$ has the same physical picture with all permutations of $\bm{x}_1,\cdots,\bm{x}_n$
This means that $n$ .- --body state: $\psi(x)=\psi(\bm{x}_1,\bm{x}_2,\cdots,\bm{x}_n)$ should be 'invariant' under permutations of $\bm{x}_i$ :
the wavefunction may get complex phase, but the density matrix should be the same

[Assumption] : the n-body wavefunction is a 1-d representation of the permutation group $S_n$ , then there are only two possibilities: bosons and fermions.

Being a 1-d representation means, for $\sigma \in S_n$ ,

$$\psi(\bm{x}_{\sigma(1)},\cdots,\bm{x}_{\sigma(n)}) = R(\sigma) \psi(\bm{x}_1,\cdots,\bm{x}_n)$$

Where   $R(\sigma)$   is a complex number of unit modulus.

    
        Note
    
    

the requirement of 1-d representation is come from we believe the wavefunction (with 1 component) can be completely discribe the many-body state

Bosons : correspond to the trivial representation of permutation group, which means any permutation has no effect on the wavefunction. Integer spin particles.
Fermions : correspond to the alternative representation of permutation group, which means pariwise exchage changes the sign of the wavefunction. Half-odd-integer spin particles.
In 2-dimensional space, braiding group instead of permutation group should be considered, there are paticles(anyons) beyond bosons and fermions.

The Structure of Many-body Hilbert Space

[Definition] : Fock space : the Hilbert space of identical particles with indefinite particle number, is the direct sum of 0-partile & 1-particle & … Hilbert space:

$$\mathcal{F}= \mathcal{H}_0 \oplus \mathcal{H}_1 \oplus \mathcal{H}_2 \oplus \cdots$$

[Definition] : 0-particle Hilbert space: noted as $\mathcal{H}_0$ , is a 1-d Hilbert space with only the vacuum state : $\ket{0}$
[Definition] : 1-particle Hilbert space: noted as $\mathcal{H}_1$ s , is the linear space of 1-body wavefunction. can be finite or infinite dimension.

[Definition] : $n(\gt2)$ --particle Hilbert space: noted as $\mathcal{H}_n$ , is a subspace of the tensor product $(\mathcal{H}_1)^{\otimes n} \equiv \bigotimes_{i=1}^n \mathcal{H}_1$ ,, with (anti-)symmetrization between the factor $\mathcal{H}_1$ s s

There shall be a (linear) map, maps the direct product state to the state of $\mathcal{H}_n$ like:

$$\mathcal{S} : (\mathcal{H}_1)^{\otimes n} \rightarrow \mathcal{H}_n ; \ \ \ket{\psi_1}\otimes\ket{\psi_2}\otimes \cdots \otimes \ket{\psi_n} \mapsto \ket{\psi_1,\psi_2,\cdots,\psi_n}$$

And permutations of       $\psi_i$       produce the same n-body state, up to a phase.

Bosons: permutations of $\psi_i$ are trivial:

$$\mathcal{S} : \ket{\psi_1}\cdots\ket{\psi_n} \mapsto \frac 1 {\sqrt{n!}} \sum_{\sigma \in S_n} \ket{\psi_{\sigma(1)}}\cdots\ket{\psi_{\sigma(n)}}$$

And

$$\ket{\psi_{\sigma(1)},\cdots,\psi_{\sigma(n)}} = \ket{\psi_1,\cdots,\psi_n}$$

And the wavefunction is the coefficient on the coordinate eigenstate basis    $\ket{\bm{x}_i}$    , we have:

$$\psi(\bm{x}_1,\cdots,\bm{x}_n) = \frac 1 {\sqrt{n!}} \sum_{\sigma\in S_n}\prod_i \psi_{\sigma(i)}(\bm{x}_i)\equiv \frac 1 {\sqrt{n!}}\text{perm}[\psi_j(\bm{x}_i)]$$

The norm of the wavefunction is:

$$\sqrt{\text{perm}[\langle \psi_i |\psi_j\rangle]} = \sqrt{\sum_{\sigma\in S_n}\prod_i \langle \psi_i|\psi_{\sigma(i)}\rangle}$$

Fermions: permutations of $\psi_i$ produce the factor $(-1)^{\sigma}$ :

$$\mathcal{S} : \ket{\psi_1}\cdots\ket{\psi_n} \mapsto \frac 1 {\sqrt{n!}} \sum_{\sigma \in S_n} (-1)^{\sigma} \ket{\psi_{\sigma(1)}}\cdots\ket{\psi_{\sigma(n)}}$$

And

$$\ket{\psi_{\sigma(1)},\cdots,\psi_{\sigma(n)}} =(-1)^{\sigma} \ket{\psi_1,\cdots,\psi_n}$$

And the wavefunction is the coefficient on the coordinate eigenstate basis    $\ket{\bm{x}_i}$    , we have:

$$\psi(\bm{x}_1,\cdots,\bm{x}_n) = \frac 1 {\sqrt{n!}} \sum_{\sigma\in S_n}(-1)^{\sigma} \prod_i \psi_{\sigma(i)}(\bm{x}_i) = \frac 1 {\sqrt{n!}}\det[\psi_j(\bm{x}_i)]$$

The norm of the wavefunction is:

$$\sqrt{\det[\langle \psi_i|\psi_j\rangle]}$$

Basis of Many-body Hilbert Space

Given a complete orthonormal basis $\{\ket{j}\}_{j=1}^{J}$ , we can construct the basis of $\mathcal{H}_n$ :

Bosons: basis are $\ket{i_1,i_2,\cdots,i_n}$ , for all $1\leq i_1\leq i_2 \leq \cdots \leq i_n\leq m$
These basis are orthogonal , but not normalized . And the number of basis (dimension of $\mathcal{H}_n)$ is $\begin{pmatrix} n+J-1 \\ n \end{pmatrix} = \frac {\Gamma(n+J)} {\Gamma(n+1)\Gamma(J)}$
Fermions: basis are $\ket{i_1,i_2,\cdots,i_n}$ , for all $1\lt i_1\lt i_2 \lt \cdots \lt i_n\lt m$
When $J\leq n$ , there is no many-body state, which is known as Pauli exclusion
These basis are orthonormal . And the number of basis (dimension of $\mathcal{H}_n)$ is $\begin{pmatrix} J \\ n \end{pmatrix} = \frac {\Gamma(1+J)} {\Gamma(n+1)\Gamma(J-n+1)}$

Occupation Number Rep.

denote the above basis $\ket{i_1,i_2,\cdots,i_n}$ as $\ket{1^{n_1},\cdots,m^{n_m}}$ , where $n_j$ is the number of appearance of $i_j$ , For $n-$ pparticle state, $\sum n_i=n$ ..

[Definition] : Occupation basis is the normalized state of this vector:

$$\ket{n_1,\cdots,n_m} = (n_1!)^{-1/2}(n_2!)^{-1/2}\cdots(n_m!)^{-1/2}\ket{1^{n_1},\cdots,m^{n_m}}$$

For fermion , $n_j=0 \text{ or }1$

Second Quantization

Creation & Annihilation Operators

[Definition] : Creation operator is an operator maps vector from $\mathcal{H}_n$ to one in $\mathcal{H}_{n+1}$ :

$$\hat \psi^{\dagger} : \mathcal{H}_n \rightarrow \mathcal{H}_{n+1} \ \ ; \ \hat \psi^{\dagger}\ket{\psi_1,\cdots,\psi_n} = \ket{\psi,\psi_1,\cdots,\psi_n}$$

In occupation basis, it is :

$$\begin{aligned} \text{For bosons: } & \hat a_i^{\dagger} \ket{\cdots,n_i,\cdots} = \sqrt{n_i+1}\ket{\cdots,(n_i+1),\cdots} \\ \text{For fermions: } & \hat a_i^{\dagger} \ket{\cdots,n_i=0,\cdots} = (-1)^{\sum_{j=1}^{i-1} n_j}\ket{\cdots,n_i=1,\cdots} \end{aligned}$$

[Definition] : Annihilation operator is an operator maps vector from $\mathcal{H}_n$ to one in $\mathcal{H}_{n-1}$ :

$$\hat \psi : \mathcal{H}_n \rightarrow \mathcal{H}_{n-1} \ \ ; \ \hat \psi \ket{\psi_1,\cdots,\psi_n} = \sum_{i=1}^n (\nu)^{i-1} \langle \psi |\psi_i \rangle \ket{\psi_1,\cdots,\psi_{i-1},\psi_{i+1},\cdots,\psi_n}$$

Where $\nu=\pm 1$ for bosons $\nu=1$ ,, for fermions $\nu=-1$
In occupation basis, it is:

$$\begin{aligned} \text{For bosons: } & \hat a_i \ket{\cdots,n_i,\cdots} = \sqrt{n_i}\ket{\cdots,(n_i-1),\cdots} \\ \text{For fermions: } & \hat a_i \ket{\cdots,n_i=1,\cdots} = (-1)^{\sum_{j=1}^{i-1} n_j}\ket{\cdots,n_i=0,\cdots} \end{aligned}$$

And:

$$\ket{n_1,n_2,\cdots,n_m} = \frac {1} {\sqrt{\prod_j n_j!}}\prod_{j=1}^m \hat e_j^{\dagger n_j}\ket{0}$$

For vacuum, we have:

$$\begin{aligned} \hat \psi \ket{0}&=0 \text{ for any annihilation operator} \\ \bra{0}\hat \psi^{\dagger} &= 0 \text{ for any creation operator} \end{aligned}$$

[Theorem] : (Anti-)Commutation relations :

$$\begin{aligned} \text{For bosons: } & [\hat\psi,\hat\psi^{\dagger}] = 1 \\ \text{For fermions: } & \{\hat\psi,\hat\psi^{\dagger}\} = 1 \end{aligned}$$

[Definition] : occupation number operator : $\hat n_{\psi} = \hat \psi^{\dagger}\hat \psi $ ::

$[\hat n_{\psi},\hat\psi^{\dagger}]=\hat \psi^{\dagger}$ , namely $\hat \psi^{\dagger}$ increases eigenvalue of $\hat n_{\psi}$ by 1
$[\hat n_{\psi},\hat\psi]=-\hat \psi$ , namely $\hat \psi$ decreases eigenvalue of $\hat n_{\psi}$ by 1
Eigenvalues of $\hat n_{\psi}$ are non-negative integers
Occupation basis $\ket{n_1,\cdots,n_m}$ are eigenstates of $\hat n_i$ , with eigenvalue $n_i$

[Theorem] : General (Anti-)Commutation relations :

$$\begin{aligned} \text{For bosons: } & [\hat\psi,\hat\psi'^{\dagger}] = \langle \psi|\psi'\rangle \ , \ [\hat \psi,\hat \psi']=0=[\hat \psi^{\dagger},\hat \psi'^{\dagger}] \\ \text{For fermions: } & \{\hat\psi,\hat\psi^{\dagger}\} = \langle \psi|\psi'\rangle \ , \ \{\hat \psi,\hat \psi'\}=0=\{\hat \psi^{\dagger},\hat \psi'^{\dagger}\} \end{aligned}$$

[Theorem] : Basis change : $\hat \psi^{\dagger} = \sum_i \langle e_i |\psi \rangle \hat {e_i}^{\dagger}$ , sum is over a complete orthonormal basis

Some Calculation Tricks

[Theorem] : for orthonormal basis of creation(annihilation) operators $\hat e_j^{\dagger}, \hat e_i$ , the commutator:

$$[\hat e_i^{\dagger}\hat e_j , \hat e_k^{\dagger}] = \delta_{jk}\hat e_i^{\dagger}$$

This is true for both bosons and fermions.
The proof can be obtained by the following formula:

$$[\hat A\hat B,\hat C] = \hat A[\hat B,\hat C]+[\hat A,\hat C]\hat B = \hat A\{\hat B,\hat C\}-\{\hat A,\hat C\}\hat B$$

[Theorem] : By the above fact, we have:

$$\hat e_i^{\dagger}\hat e_j \hat e_{i_1}^{\dagger}\cdots\hat e_{i_n}^{\dagger} \ket{0}=\sum_{i'_1,i'_2,\cdots,i'_n} \hat e_{i'_1}^{\dagger}\cdots\hat e_{i'_n}^{\dagger}\ket{0}$$

Where the summation over the sequence $(i'_1,\cdots,i'_n)$ which is $(i_1,\cdots,i_n)$ with one appearance of $j$ replaced by $i$

The Second Quantization

[Rule] : the rule of thumb , to get a many-body Hamiltonian (defined on the Fock space) , simply replace 1-body wavefunction $\psi(x),\psi^*(x)$ in the Energy formula for product states by $\hat \psi(x), \hat \psi^{\dagger}(x)$ , remove the summations over all particles and need some normal ordering

Examples

- `kinetic energy` :
In original way, we write the kinetic energy as: $$T = \sum_i \int \psi^*_i(x)\big(-\frac {\partial_x^2} {2m}\big)\psi_i(x)\td x$$ Then the corresponding many-body term is: $$\hat T = \int \hat \psi^{\dagger}(x)\big(-\frac {\partial_x^2} {2m} \big) \hat \psi(x)\td x= \int \hat \psi^{\dagger}(p)\big(\frac {p^2} {2m} \big) \hat \psi(p)\td p$$ - `1-body potential energy` :
In original way , we write the 1-body potential energy as: $$V_1 = \sum_i \int V(x)\psi^*_i(x)\psi_i(x)\td x$$ And the corresponding many-body term is : $$\hat V_1 = \int V(x)\hat \psi^{\dagger}(x)\hat \psi(x)\td x$$ - `2-body potential energy` :
For 2-body interaction, we have the potential energy in original way: $$V_2 = \frac 1 2 \sum_{i\neq j} \int V(x,x')\psi_i^*(x)\psi_i(x)\psi_j^*(x')\psi_j(x')\td x\td x'$$ The corresponding many-body term is : $$\hat V_2 = \frac 1 2\int V(x,x') \hat \psi^{\dagger}(x)\hat \psi^{\dagger}(x')\hat \psi(x')\hat \psi(x)\td x\td x'$$

Special Many-body States

Fermion Product State(Fermi Sea)

[Definition] : the Fermion Product State is:

$$|\psi_1,\cdots,\psi_n\rangle = \prod_{i=1}^n \hat \psi_i^{\dagger}\ket{0}$$

Norm of this state is given by the Gram determinant: $\sqrt{\text{det}[\langle \psi_i|\psi_j\rangle]}$ .
If $\psi_i$ are linearly dependent , this state vanishes.
Linearly independent $\psi_i$ span a n-dimensional 1-body Hilbert space. Given a complete orthonormal basis of this sapce $c_i$ , then:

$$\ket{\psi_1,\cdots,\psi_n} = \text{det}[\langle c_i|\psi_j\rangle] \ket{c_1,\cdots,c_n}$$

And this state is the eigenstate of $\hat n=\sum_i \hat c_i^{\dagger}\hat c_i$ with eigenvalue is $n$ .- -.
The parent Hamiltonian of this state is:

$$\hat H = -\sum_{i=1}^n \hat c_i^{\dagger}\hat c_i + \sum_{i=n+1}^m \hat c_i^{\dagger}\hat c_i$$

Namely, the unique ground state of this Hamiltonian is this fermion product state.

[Definition] : Particle-hole Transformation of Fermions of a single fermion mode: formally $\hat c_i \leftrightarrow \hat c_i^{\dagger}$ ..
This will preserve the anti-commutation relations.
This correspongds to a unitary transformation on the Fock space: $\hat U = (\hat c_i+\hat c_i^{\dagger})\cdot (-1)^{\sum_{j\neq i}\hat c_j^{\dagger}\hat c_j}$ , this will let:

$$\hat U^{\dagger}\hat U = \hat 1 \ \ \hat U \hat c_i\hat U^{\dagger} = \hat c_i^{\dagger} \ \ \hat U \hat c_j\hat U^{\dagger} = \hat c_j \text{ for } j\neq i$$

And the unitary transformation on occupation number basis is:

$$\ket{\cdots,n_i=0,\cdots}\leftrightarrow (-1)^{\sum_{j\gt i} n_j} \ket{\cdots,n_i=1,\cdots}$$

And the factor $(-1)^{\sum_{j\gt i}n_j}$ is preserve the matrix elements of $\hat \psi_j , \hat \psi_k^{\dagger}$ for $j\gt i$ ..
In particular, the new “vacuum” is originally $\ket{0,\cdots,n_i=1,\cdots,0}$ ..

Fermion Pairing State(BCS state)

[Definition] : Consider two orthonormal fermion modes $c_1, c_2$ , the pairing state is:

$$\ket{\lambda} = (1+|\lambda|^2)^{-1/2}\exp(\lambda \hat c_1^{\dagger}\hat c_2^{\dagger}) \ket{0}$$

Where $\lambda \in\mathbb{C}$
This state is not an eigenstate of fermion number operator $\hat c_1^{\dagger}\hat c_1+\hat c_2^{\dagger}\hat c_2$ .
[Definition] : Bogoliubov transformation which defines the Boogoliubov quasiparticles :

$$\begin{aligned} \hat \gamma_1 &= u\hat c_1+v \hat c_2^{\dagger} \\ \hat \gamma_2 &= -v\hat c_1^{\dagger}+u\hat c_2 \end{aligned}$$

where $u=(1+|\lambda|^2)^{-1/2} $ and $v=-\lambda(1+|\lambda|^2)^{-1/2}$ .
This two operator satisfy: $\{\hat \gamma_i,\hat \gamma_j^{\dagger}\}=\delta_{ij}$
The pairing state is vanished by $\hat \gamma_i$ , namely $\hat \gamma \ket{\lambda}=0$
And the parent Hamiltonian is

$$\hat H = \hat \gamma_1^{\dagger}\hat \gamma_1 +\hat \gamma_2^{\dagger}\hat \gamma_2$$

[Definition] : Generic fermion pairing state is defined as:

$$\ket{\{f_{ij}\}} \propto \exp(\frac 1 2 \sum_{i,j}f_{ij}\hat c_i^{\dagger}\hat c_j^{\dagger})\ket{0}$$

Where $c_i$ are some orthonormal basis, and $f_{ij}=-f_{ji}$ are some complex numbers.
By a orthogonal transformation $\hat c_i^{\dagger} \rightarrow O_{ij}\hat {(c')}_j^{\dagger}$ , where $\bm{O}$ is an orthogonal matrix, the antisymmetric matrix $\bm{f}$ can be brought into a standard form:

$$\bm{O}^T \bm{f}\bm{O} = \begin{bmatrix} 0 & \lambda_1 & 0 & 0 & \cdots \\ -\lambda_1 & 0&0&0&\cdots \\ 0&0&0& \lambda_2&\cdots \\ 0&0& -\lambda_2&0&\cdots \\ \vdots&\vdots&\vdots&\vdots&\ddots\end{bmatrix}$$

the state becomes $\propto \exp(\lambda_1 \hat{c'}_1^{\dagger} \hat{c'}_2^{\dagger}+\lambda \hat{c'}_3^{\dagger}\hat {c'}_4^{\dagger}+\cdots)\ket{0}$ .
Bogoliubov transformations can be defined on $\hat {c'}_{2i-1}$ & $\hat {c'}_{2i}$ ..

Boson Coherent State

[Definition] : The coherent state from a single boson mode $\hat b$ is:

$$\ket{z} = e^{-|z|^2/2}e^{z\hat b^{\dagger}} \ket{0}$$

where $z\in \mathbb{C}$ .
This state is not an eigenstate of boson number $\hat b^{\dagger}\hat b$ , but $\hat b\ket{z}=z\ket{z}$ , therefor this state vanished by $\hat b'=\hat b-z$
The parent Hamiltonian is thus $\hat H = (\hat b-z)^{\dagger}(\hat b-z)$
The inner product between two Boson Coherent state is:

$$\langle z'|z\rangle = e^{-(|z|^2+|z'|^2)/2} e^{z(z')^* } $$

Boson Pairing State

Consider two orthonormal boson modes $\hat b_1,\hat b_2$ , the boson pairing state is:

$$\ket{\lambda} = (1-|\lambda|^2)^{1/2}\exp(\lambda \hat b_1^{\dagger}\hat b_2^{\dagger})\ket{0}$$

Where $\lambda \in \mathbb{C} , |\lambda|\lt 1$ .
This state is not an eigenstate of boson number : $\hat b_1^{\dagger}\hat b_1+\hat b_2^{\dagger}\hat b_2$
Bogoliubov transformation:

$$\begin{aligned} \hat \gamma_1 &= u\hat b_1+v\hat b_2^{\dagger} \\ \hat \gamma_2 &= u\hat b_2+v\hat b_1^{\dagger} \end{aligned}$$

where $u=(1-|\lambda|^2)^{-1/2} , v=-\lambda(1-|\lambda|^2)^{-1/2}$ .. Then $\hat \gamma_i\ket{\lambda}=0$
The parent Hamiltonian is:

$$\hat H =\hat \gamma_1^{\dagger}\hat \gamma_1+\hat \gamma_2^{\dagger}\hat \gamma_2$$

The Wick Expansion

Let $\ket{0}$ be the single-particle vacuum , Let $\hat A_i (i=1,\cdots,2n)$ be a set of single-particle operators, namely linear cominations of annihilation and creation operators, then $\bra{0}\hat A_1\hat A_2\cdots\hat A_{2n}\ket{0}$ is the Hafnian(Pfaffian) of matrix $\bra{0}\hat A_i\hat A_j\ket{0}$ for bosons(fermions) .

[Definition] : Hafnian of $2n\times 2n$ symmetric matrix $M_{ij}$ is :

$$\text{Hf}(\bm{M}) = \frac 1 {n!} \sum_{\sigma\in S_{2n}, \sigma(1)\lt \sigma(2), \sigma(3)\lt \sigma(4),\cdots} M_{\sigma(1)\sigma(2)} M_{\sigma(3)\sigma(4)}\cdots M_{\sigma(2n-1)\sigma(2n)}$$

[Definition] : Pfaffian of $2n\times 2n$ anti-symmetric matrix $M_{ij}$ is :

$$\text{Pf}(\bm{M}) = \frac 1 {n!} \sum_{\sigma\in S_{2n}, \sigma(1)\lt \sigma(2), \sigma(3)\lt \sigma(4),\cdots} (-1)^{\sigma} M_{\sigma(1)\sigma(2)} M_{\sigma(3)\sigma(4)}\cdots M_{\sigma(2n-1)\sigma(2n)}$$

[Notes] : this is true only for such free particle states $\ket{0}$ , the matrix $\bra{0}\hat A_i\hat A_j\ket{0}$ may not be symmetric or anti-symmetric, but the above definition of Hafnian/Pfaffian still works for the Wick expansion

Leonard