Boson Coherent State
Definition and Properties
[Definition] : The generic form of the coherent state from a set of boson mode $\hat a_i$ is:
$$ \ket{\bm{z}}=\exp(\bm{z}\cdot \hat {\bm{a}}^\dagger)\ket{0}=\exp(\sum_{i}z_i \hat a_i^\dagger)\ket{0} $$
[Theorem] : The overlap between two coherent states is:
$$ \bra{\bm{z}}\bm{z}'\rangle=\exp(\bm{z}^*\cdot \bm{z}') $$
Proof
With that $\hat b(\bm{z})^\dagger=\bm{z}\cdot \hat {\bm{a}}^\dagger$ , we have: $$ [\hat b(\bm{z}), \hat b(\bm{z}')^\dagger]=\bm{z}^* \cdot \bm{z}' $$ Then: $$ \begin{aligned} \bra{\bm{z}}\bm{z}'\rangle &= \bra{0}\exp\big(\sum_i z_i^* \hat a_i\big)\exp\big(\sum_j z'_j \hat a_j^\dagger\big)\ket{0} \\ &=\bra{0}e^{\hat b(z)} e^{\hat b(z')^\dagger} \ket{0} \\ &=\bra{0} [e^{\hat b(z)},e^{\hat b(z')^\dagger}]\ket{0} \\ &= \exp(\bm{z}^*\cdot \bm{z}') \end{aligned} $$This shows that different coherent states are not orthogonal, they form an over-complete basis.
[Theorem] : (The closure relation), with the integral measure:
$$ \td \mu(\bm{z})= \prod_i \frac {\td (\text{Re}z_i)\td (\text{Im}z_i)} {\pi}\exp\Big(-\sum_i |z_i|^2\Big) $$
one has:
$$ \int \td \mu(\bm{z})\ket{\bm{z}}\bra{\bm{z}}=1 $$
Proof
We need only prove that $\int \td \mu(\bm{z})\ket{\bm{z}}\bra{\bm{z}}\psi\rangle =\ket{\psi}$ . hold for arbitrary $\ket{\psi}$ . For the fact that: $$ \hat a_i^\dagger \ket{0}=\frac {\partial} {\partial z_i} \ket{\bm{z}} \Bigg|_{\bm{z}=0} $$ That means we can construct all many-particle states linearly from the coherent state. So we can directly choose the $\ket{\psi}=\ket{\bm{x}}$ is a coherent state. Then one has: $$\begin{aligned} \int \td \mu(\bm{z}) \ket{\bm{z}}\bra{\bm{z}}\bm{x}\rangle &= \int \td\mu(\bm{z})\ket{\bm{z}}e^{\bm{z}^*\cdot \bm{x}} \\ &= \prod_i \int \frac {\td u_i \td v_i} {\pi} e^{-u_i^2-v_i^2}e^{(u_i-\ti v_i)x_i} e^{(u_i+\ti v_i)\hat a_i^\dagger} \ket{0} \\ &=\prod_i \frac 1 \pi\Big( \int \td u e^{-u^2}e^{u x_i} e^{u \hat a_i^\dagger} \Big)\Big( \int \td u e^{-u^2}e^{-\ti u x_i} e^{\ti u \hat a_i^\dagger} \Big)\ket{0} \\ &= \prod_i \frac 1 \pi \cdot \pi\exp\{\frac {(x_i+\hat a_i^\dagger)^2} 4+\frac {(-\ti x_i+\ti \hat a_i^\dagger)^2} 4\} \ket{0}\\ &= \prod_i e^{x_i \hat a_i^\dagger} \ket{0}\\ &=\ket{\bm{x}} \end{aligned}$$ q.e.d.[Theorem] : The trace of an operator $\hat A$ over the entire Boson Fock space can be written with the coherent state:
$$ \text{Tr}\hat A:=\sum_{n_1,n_2,\cdots} \bra{n_1,n_2,\cdots}\hat A\ket{n_1,n_2,\cdots}=\int \td \mu(\bm{z}) \bra{\bm{z}}\hat A\ket{\bm{z}} $$
Proof
With the expansion of coherent state: $$ \ket{\bm{z}}=\sum_{n=0}^{\infty} \frac 1 {n!} \Big(\sum_i z_i \hat a_i^\dagger\Big)^n \ket{0}=\sum_{n=0}^{\infty}\sum_{k_1+\cdots+k_m=n} \frac {z_1^{k_1}\cdots z_m^{k_m}} {k_1!\cdots k_m!}\prod_{j=1}^m \hat a_j^{\dagger k_j}\ket{0} $$ Where we assume that the number of Boson modes is $m\lt \infty$ . The second equality is caused by the Binomial theorem. Then one can find that component of the $n$ -Boson subspace is proportional to the $n$ -monomial of $\{z_1,\cdots\}$ . To prove the relation of trace, we need to show that the integral will vanish all matrix elements of $\hat A$ between two states of different occupation of Bosons. That is to show: $$ \int \td \mu(\bm{z}) \ \prod_{i=1}^m z_i^{k_i} z_i^{* l_i}=0 $$ If exists $i$ so that $k_i\neq l_i$ . We can directly compute it: $$\begin{aligned} \int \td \mu(\bm{z}) \prod_{i=1}^m z_i^{k_i}z_i^{* l_i} &= \prod_{i=1}^m \int \frac {\td z_i \td z_i^*} {2\pi \ti} e^{-|z_i|^2} z_i^{k_i}z_i^{*l_i} \\ &= \prod_{i=1}^{m} \int \frac {\td u \td v} {\pi} e^{-u^2 -v^2} (u^2+ v^2)^{\min (k_i,l_i)} (u\pm \ti v)^{\Delta_i} \\ &= \prod_{i=1}^m \int \frac {r \td r \td \phi} {\pi} e^{-r^2} r^{2\min(k_i,l_i)} r^{\Delta_i} e^{\pm \ti \Delta_i \phi} \end{aligned}$$ The integral of $\phi$ vanishes if $\Delta_i\neq 0$ . Then we have: $$ \int \td \mu(\bm{z}) \prod_{i=1}^m z_i^{k_i}z_i^{* l_i}=\delta_{k,l}\prod_{i=1}^m k_i! $$ Then we have: $$\begin{aligned} \int \td \mu(\bm{z}) \bra{\bm{z}}\hat A\ket{\bm{z}}&=\sum_{n=0}^{\infty}\sum_{k_1+\cdots+k_m=n} \frac {\bra{0}\Big(\prod_{j=1}^m\hat a_j^{\dagger k_j}\Big)^\dagger\hat A\Big(\prod_{j=1}^m \hat a_j^{\dagger k_j}\Big)\ket{0}} {k_1!\cdots k_m!}\\ &=\sum_{n=0}^{\infty} \sum_{k_1+\cdots+k_m=n} \bra{k_1,\cdots}\hat A \ket{k_1,\cdots} = \text{Tr}\hat A \end{aligned}$$ q.e.d.[Theorem] : One can construct any many body states of Bosons from coherent state by:
$$ \ket{n_1,\cdots}:=\frac 1 {\sqrt{\prod_k n_k!}} \prod_k\hat a_k^{\dagger n_k}\ket{0}=\frac 1 {\sqrt{\prod_k n_k!}}\prod_k \frac {\partial^{n_k}} {\partial z_k^{n_k}}\ket{\bm{z}}\Bigg|_{\bm{z}=\bm{0}} $$
[Theorem] : the average and variance of the particle number $\hat N=\sum_i \hat a_i^\dagger \hat a_i$ is:
$$\begin{aligned} \bar N &:= \frac {\bra{\bm{z}}\hat N\ket{\bm{z}}} {\bra{\bm{z}}\bm{z}\rangle} = \sum_i |z_i|^2 \\ (\Delta N)^2 &= \frac {\bra{\bm{z}}(\hat N-\bar N)^2\ket{\bm{z}}} {\bra{\bm{z}}\bm{z}\rangle} = \bar N \end{aligned}$$
Proof
With the property that: $$ [\hat a_i , e^{\bm{z}\cdot \hat{\bm{a}}^\dagger}]=\sum_{n=1}^{\infty} \frac 1 {n!} [\hat a_i,(\sum_j z_j \hat a_j^\dagger)^n]=\sum_{n=1}^{\infty}\frac n {n!} z_i^n (\bm{z}\cdot \hat {\bm{a}}^\dagger)^{n-1}=z_i e^{\bm{z}\cdot \hat {\bm{a}}^\dagger} $$ Then we have: $$\begin{aligned} \bra{\bm{z}}\hat N\ket{\bm{z}}&=\bra{0}\sum_i (\hat a_i e^{\bm{z}\cdot \hat {\bm{a}}^\dagger})^\dagger(\hat a_i e^{\bm{z}\cdot \hat {\bm{a}}^\dagger}) \ket{0} \\ &= \bra{0}\Big(\sum_i (z_i e^{\bm{z}\cdot \hat {\bm{a}}^\dagger}+e^{\bm{z}\cdot \hat {\bm{a}}^\dagger}\hat a_i)^\dagger(z_i e^{\bm{z}\cdot \hat {\bm{a}}^\dagger}+e^{\bm{z}\cdot \hat {\bm{a}}^\dagger}\hat a_i) \Big)\ket{0} \\ &= \sum_i z_i^* z_i \bra{\bm{z}}\bm{z}\rangle \end{aligned}$$ Then, we have: $$ \frac {\bra{\bm{z}}\hat N\ket{\bm{z}}} {\bra{\bm{z}}\bm{z}\rangle} = \sum_i |z_i|^2 $$ And with that $[\hat a_i^\dagger \hat a_i ,e^{\bm{z}\cdot \hat {\bm{a}}^\dagger}]=z_i\hat a_i^\dagger e^{\bm{z}\cdot \hat {\bm{a}}^\dagger}$ . We have: $$\begin{aligned} \bra{\bm{z}}\hat N^2\ket{\bm{z}}&= \bra{0}(\hat N e^{\bm{z}\cdot \hat {\bm{a}}^\dagger})^\dagger(\hat N e^{\bm{z}\cdot \hat {\bm{a}}^\dagger})\ket{0} \\ &= \bra{0} \big(\bm{z}\cdot \hat {\bm{a}}^\dagger e^{\bm{z}\cdot \hat {\bm{a}}^\dagger} + e^{\bm{z}\cdot \hat {\bm{a}}^\dagger}\hat N\big)^\dagger \big(\bm{z}\cdot \hat {\bm{a}}^\dagger e^{\bm{z}\cdot \hat {\bm{a}}^\dagger}+ e^{\bm{z}\cdot \hat {\bm{a}}^\dagger}\hat N\big)\ket{0} \\ &=\bra{0} \big(\bm{z}\cdot \hat {\bm{a}}^\dagger e^{\bm{z}\cdot \hat {\bm{a}}^\dagger} \big)^\dagger \big(\bm{z}\cdot \hat {\bm{a}}^\dagger e^{\bm{z}\cdot \hat {\bm{a}}^\dagger}\big)\ket{0} \\ &= \bra{\bm{z}}\sum_{m,n}z^*_mz_n \hat a_m \hat a_n^\dagger \ket{\bm{z}} \\ &=\sum_{m,n} z_m^* z_n \bra{\bm{z}}(\delta_{mn}+\hat a_n^\dagger \hat a_m)\ket{\bm{z}} \\ &= \sum_i |z_i|^2 \bra{\bm{z}}\bm{z}\rangle +\sum_{m,n}z_m^* z_n z_n^* z_m \bra{\bm{z}}\bm{z}\rangle \end{aligned}$$ Then: $$ \frac {\bra{\bm{z}}(\hat N-\bar N)^2\ket{\bm{z}}} {\bra{\bm{z}}\bm{z}\rangle} =\frac {\bra{\bm{z}}\hat N^2 \ket{\bm{z}}} {\bra{\bm{z}}\bm{z}\rangle } - \bar N^2 =\sum_i |z_i|^2 = \bar N $$ q.e.d.Coherent State Representation
Like the wavefunction in coordinate representation, we introduce the Coherent State Representation:
For an arbitrary quantum state $\ket{\psi}$ , we can define the "wavefunction" in coherent representation , is the complex function: $\psi(\bm{z}^*)=\bra{\bm{z}}\psi\rangle=\bra{0}e^{\bm{z}^*\cdot \hat {\bm{a}}}\ket{\psi}$ , which satisfies:
$$ \ket{\psi}=\int \td \mu(\bm{z}) \ket{\bm{z}}\bra{\bm{z}}\psi\rangle = \int \td \mu(\bm{z}) \ \psi(\bm{z}^*)\ket{\bm{z}} $$
The representation of operators are:
$$\begin{aligned} \hat a_i \rightarrow \frac {\partial} {\partial z_i^*} \ ; \ \hat a_i^\dagger \rightarrow z_i^* \end{aligned}$$
And the matrix elements of normal ordered operators, which means that all creation operators are at the left side of annihilation operators, is:
$$ \bra{\bm{z}}A(\hat {\bm{a}}^\dagger,\hat {\bm{a}})\ket{\bm{z}'}=A(\bm{z}^*,\bm{z}') \exp(\bm{z}^*\cdot \bm{z}') $$
Proof
The creation/annihilation operator operating on an arbitrary quantum state induces: $$\begin{aligned} \hat a_i \psi(\bm{z}^*) &= \bra{\bm{z}}\hat a_i\ket{\psi} = \bra{0}\frac {\partial} {\partial z_i^*}e^{\bm{z}^*\cdot \hat {\bm{a}}}\ket{\psi} = (\partial_i\psi)(z_1^*,\cdots) \\ \hat a_i^\dagger \psi(\bm{z}^*) &= \bra{\bm{z}}\hat a_i^\dagger \ket{\psi} =z_i^* \psi(\bm{z}^*) \end{aligned}$$ And then the normal ordered matrix elements: $$\begin{aligned} \bra{\bm{z}}A(\hat {\bm{a}}^\dagger,\hat {\bm{a}})\ket{\bm{z}'} &= \sum_{i,j} a_{ij} \bra{\bm{z}}\hat a_1^{i_1\dagger}\cdots \hat a_n^{i_n\dagger} \hat a_1^{j_1}\cdots a_n^{j_n}\ket{\bm{z}'} \\ &=\sum_{i,j}a_{ij}\prod_k (z_k^*)^{i_k} (z'_k)^{j_k} \bra{\bm{z}}\bm{z}'\rangle \\ &=A(\bm{z}^*,\bm{z}') \exp(\bm{z}^*\cdot \bm{z}') \end{aligned}$$ q.e.d.At last, the Schrodinger equation in coherent representation is, with the Hamiltonian is normal ordered as $\hat H=H(\hat{\bm{a}}^\dagger,\hat{\bm{a}},t)$ :
$$ \ti\hbar \frac {\partial \psi(\bm{z}^*,t)} {\partial t}=\int \td \mu(\bm{z}') H(\bm{z}^*,\bm{z}',t)e^{\bm{z}^*\cdot \bm{z'}}\psi(\bm{z}'^*,t) $$
Proof
The generic form of Schrodinger equation is: $$ \ti\hbar \frac {\td} {\td t}\ket{\psi(t)}=\hat H \ket{\psi(t)} $$ Here we let: $$ \ket{\psi(t)}=\int \td \mu(\bm{z})\ket{\bm{z}}\psi(\bm{z}^*,t) $$ Then: $$\begin{aligned} \ti \hbar \frac {\td } {\td t}\ket{\psi(t)} &= \int \td\mu(\bm{z})\ket{\bm{z}} \Big(\ti \hbar \frac {\partial } {\partial t} \psi(\bm{z}^*,t)\Big) \\ &= \int \td \mu(\bm{z}) \int \td \mu(\bm{z}') \ket{\bm{z}}\bra{\bm{z}}\hat H\ket{\bm{z}'} \psi(\bm{z}'^*,t) \\ &= \int \td \mu(\bm{z}) \int \td \mu(\bm{z}')\ket{\bm{z}} H(\bm{z}^*,\bm{z}',t)e^{\bm{z}^*\cdot \bm{z}'} \psi(\bm{z}'^* ,t) \end{aligned}$$ Then compare both sides, we have the Schrodinger equation: $$ \ti\hbar \frac {\partial \psi(\bm{z}^*,t)} {\partial t}=\int \td \mu(\bm{z}') H(\bm{z}^*,\bm{z}',t)e^{\bm{z}^*\cdot \bm{z'}}\psi(\bm{z}'^*,t) $$Fermion Coherent State
Introduction to Grassmann Algebra
[Definition] : Grassmann algebra $\mathcal{G}_N$ is defined by a set of generators $\{\xi_\alpha\} \ ; \ \alpha=1,\cdots,N$ usually over the Complex field ( $\mathbb{C}$ . These generators are called Grassmann variables, which are anti-commutative and are commutative with ordinary complex numbers (c-number) $x\in \mathbb{C}$ :
$$ \xi_\alpha\xi_\beta+\xi_\beta\xi_\alpha=0 \ ; \ x\xi = \xi x $$
A matrix representation of Grassmann variables requires matrices of dimension at least $2^N \times 2^N$ as we can always write $2^N$ distinct product of Grassmann variables, each generators can be included or excluded.
[Definition] : Grassmann numbers forms a vector space naturally isomorphic to $\mathbb{C}^{2^N}$ , each element is the linear combination of $2^N$ products with ordinary complex number:
$$ \big\{1,\ \xi_\alpha,\ \xi_{\alpha_1}\xi_{\alpha_2},\ \cdots,\ \xi_{\alpha_1}\cdots\xi_{\alpha_n}\big\} $$
A Grassmann number can be written as:
$$ g=z_0 + \sum_\alpha z_\alpha \xi_\alpha+\sum_{\alpha_1,\alpha_2}z_{\alpha_1,\alpha_2}\xi_{\alpha_1}\xi_{\alpha_2}+ \cdots $$
Coefficients $z_\alpha,\cdots$ are ordinary complex numbers.
[Definition] : The conjugation operation is defined on the Grassmann algebra $\mathcal{G}_{2N}$ with even number of generators. It is split into two sets: $\{\xi_1,\xi_2,\cdots,\xi_N\}$ and $\{\xi_1^*,\cdots,\xi_N^*\}$ . Define the Conjugation operation as the map : $(\cdot)^*:\mathcal{G}_{2N}\rightarrow \mathcal{G}_{2N}$ , and has such properties:
$$\begin{aligned} (\xi_\alpha)^*&=\xi_\alpha^* \ ; \ \alpha=1,\cdots, N\\ (\xi_\alpha^*)^*&=\xi_\alpha \ ; \ \alpha=1,\cdots,N\\ (\xi_{\alpha_1}\cdots\xi_{\alpha_r})^* &= \xi_{\alpha_r}^*\cdots\xi_{\alpha_1}^* \end{aligned}$$
And it is generalized to the product between Grassmann variables and c-numbers: if $x\in\mathbb{C}$ , then $(x\xi)^*=x^* \xi^*$ , where $x^*$ is the complex conjugation of $x$ .
[Note] : We can’t represent conjugation by taking Hermit adjoint in any finite dimensional matrix representation of $\mathcal{G}_{2N}$ . If we could, that is to say, we have a representation map $R: \mathcal{G}_{2N}\rightarrow \mathbb{C}^{d\times d}$ such that $R(\xi^*)=A^\dagger, R(\xi)=A$ , then according to the anti-commutation relation: $A^\dagger A +A A^\dagger =0\Rightarrow \text{Tr} A^\dagger A =\sum_{i,j}|A_{ij}|^2=0\Rightarrow A=0$ , which is a contradiction.
[Definition] : The analytic functions on Grassmann numbers are Grassmann-number-valued functions whose coefficients are complex analytic.
Analytic functions for low dimension Grassmann algebra has simply forms because of the anti-commutation relation: $\xi\in \mathcal{G}_1\Rightarrow \xi^2=0$ :
$$\begin{aligned} f(\xi) &= f_0+f_1 \xi \\ A(\xi,\xi^*) &= a_0 + a_1 \xi + a'_1 \xi^* + a_{12} \xi^*\xi \end{aligned}$$
[Definition] : In $\mathcal{G}_N$ , define differentiation with respect to a Grassmann variable $\xi_\alpha$ or Derivative, denote as $\partial_\alpha$ or $\partial_{\xi_\alpha}$ . A general function on $\mathcal{G}_N$ is, let $\xi_{\neg \alpha}$ denote the components of $\xi=(\xi_1,\cdots,\xi_N)$ excluding $\xi_\alpha$ , like: $\xi_{\neg \alpha}=(\xi_1,\cdots,\xi_{\alpha-1},\xi_{\alpha+1},\cdots,\xi_N)$ . Then any analytic function of $\mathcal{G}_N$ can be rewritten as:
$$ f=u(\xi_{\neg \alpha}) + \xi_\alpha v(\xi_{\neg \alpha}) $$
Then:
$$ \partial_\alpha f=v(\xi_{\neg \alpha}) $$
The order is important. Note that $\partial_\xi$ and $\partial_{\xi^*}$ is also anti-commutative.
Integral is defined to be a linear mapping with properties mimicking those of ordinary integrals. With the properties:
$$ \int \td \xi \ 1 =\int \td\xi^* \ 1 = 0 \ ; \ \int \td \xi \ \xi=\int \td \xi^* \xi^* = 1 $$
Note that $\int \td \xi^* \ \xi$ is not defined. Also integral operators are anti-commutative. And $\int \td \xi$ can be simply treated as an operator, returns $0$ when applied on $1$ , and returns $1$ when applied on $\xi$
It is obvious that, for Grassmann functions, $\partial_\xi=\int \td\xi$
[Note] : usually we treat the analytic function of Grassmann number as the function of Grassmann variables. In fact the full version of Grassmann function should be written as:
$$ f(g=a_0+\sum_{l=1}^N \sum_{i_1\lt \cdots\lt i_l}a_{i_1,\cdots,i_l}\xi_{i_1}\cdots \xi_{i_l} )=A_0(a)+\sum_{l=1}^N \sum_{i_1\lt \cdots\lt i_l}A_{i_1,\cdots,i_l}(a)\xi_{i_l}\cdots\xi_{i_l} $$
The coefficients are functions of c-numbers. But when we just consider the differentiation and integration for Grassmann variables we do not need consider the form of those c-number functions in detail, that is because the commutation and independence between Grassmann variables and c-number, so they are invariant under the Grassmann differentiation and integration.
[Theorem] : One can define the Dirac function for Grassmann variables: $\delta(\xi,\xi'):=-(\xi-\xi')$ . Then for arbitrary function $f(\xi)$ , one has:
$$ \int \td \xi' \delta(\xi,\xi')f(\xi')=f(\xi) $$
Proof
For arbitrary function, one has: $f(\xi)=f_0+f_1\xi$ . Then: $$\begin{aligned} \int \td \xi' \delta(\xi,\xi')f(\xi') &= \int \td \xi' \Big(-(\xi-\xi')(f_0+f_1\xi')\Big) \\ &= -\int \td \xi' \xi f_0 -\int \td \xi' \xi f_1\xi'+\int \td \xi' \xi' f_0+\int \td \xi' \xi'f_1\xi'\\ &= -\int \td \xi' \xi f_0 +f_1\int \td \xi' \xi' \xi+\int \td \xi' \xi' f_0+\int \td \xi' \xi'f_1\xi'\\ &=0+f_1\xi+f_0+0\\ &=f(\xi) \end{aligned}$$ q.e.d.[Theorem] : Scalar product of Grassmann functions:
$$ (f,g):=\int \td\xi^* \td \xi e^{-\xi^*\xi}f^*(\xi)g(\xi^*)=f_0^* g_0+f_1^*g_1 $$
Proof
With that $f(\xi)=f_0+f_1\xi \ ; \ g(\xi)=g_0+g_1\xi$ , And: $$ e^{-\xi^*\xi}=1-\xi^*\xi+\frac {\xi^*\xi\xi^*\xi} {2!}+\cdots=1-\xi^*\xi $$ Because $\xi^*\xi\xi^*\xi=-\xi^*\xi^*\xi\xi=0$ . Then we have: $$\begin{aligned} (f,g)&= \int \td \xi^*\td \xi e^{-\xi^*\xi}f^*(\xi)g(\xi^*) \\ &= \int \td \xi^* \td \xi (1-\xi^*\xi)(f_0^*+f_1^* \xi)(g_0+g_1\xi^*) \\ &= \int \td \xi^* \td \xi f_1^* g_1 \xi \xi^* - \int \td \xi^* \td \xi \ \xi^*\xi f_0^* g_0\\ &= f_1^*g_1 + f_0^* g_0 \end{aligned}$$ q.e.d.[Theorem] : Linear Change of Grassmann Varibles. Let $\xi=(\xi_1,\cdots,\xi_N)$ generate a Grassmann algebra, let $M$ be an invertible complex $N\times N$ matrix. And $b$ is a $N$ -vector formed by these generators (Note linear combinations of generators are still generators, i.e., following the anti-commute rule). Then one can define a new set of Grassmann generators by $\eta := M \xi+b$ . Then:
$$ \int \td \eta f(\eta) = (\det M)^{-1} \int \td \xi f(M\xi+b) $$
Where $\td \eta=\td \eta_N\cdots\td \eta_1$ , and $\td \xi=\td \xi_N\cdots\td\xi_1$ , with the inverted order.
$$ \int \td \eta \ f(\eta)=\frac 1 {\det M}\int \td \xi f(M\xi+b) $$
Noting this variable substitution rule is different from the case of c-number: $\int \td x \ f(x) = \int \td z \frac {\partial(x)} {\partial(z)} f(x(z))$ i.e. for linear transform for c-numbers $x=M z+b$ , the Jacobian contributes the factor $\det M$ but not $(\det M)^{-1}$ !
Proof
First, we prove a simple conclusion: shift symmetry of Grassmann integral, with $b\in\mathbb{C}^N$ , one has: $$ \int \td \xi \ f(\xi+b)=\int \td \xi \ f(\xi) $$ With the mathematical induction, when $N=1$ , it is: $$ \int \td \xi \ f(\xi+b)=\int \td \xi (u+v\xi+vb)=v=\int \td f(\xi) $$ Then assume the conclusion works for the case of $N=k$ , so we have for $N=k+1$ : $$\begin{aligned} \int \td\xi f(\xi+b) &= \int \td \xi_{k+1}\td\xi_k\cdots\td \xi_1 \Big(u(\xi_{\neg 1}+b_{\neg 1})+(\xi_1+b_1) v(\xi_{\neg 1}+b_{\neg 1})\Big) \\ &=\int \td \xi_{\neg 1} v(\xi_{\neg 1}+b_{\neg 1}) \\ &= \int \td \xi_{\neg 1} v(\xi_{\neg 1}) \\ &=\int \td \xi\Big(u(\xi_{\neg 1})+\xi_1 v(\xi_{\neg 1})\Big) \\ &=\int \td \xi f(\xi) \end{aligned}$$ q.e.d. So we need only consider the case of $b=0$ . Then we expand the function $f$ as: $$ f(\eta)=A_0+A_1\eta_1+\cdots + A_N\eta_1\cdots\eta_N $$ Coefficients $A_0\cdots A_N$ are c-numbers. The integral on the right side kills all terms except those that contain every $\xi_i$ in a product, or the term involves all $N$ variables: $$\begin{aligned} \int \td \xi \eta_1\cdots\eta_N &= \int \td \xi_N\cdots\td \xi_1 \prod_{i=1}^N \Big(\sum_{j=1}^N M_{ij}\xi_j\Big) \\ &=\int \td \xi_N\cdots\td \xi_1 \sum_{\sigma \in S_N} \prod_{i=1}^N M_{i,\sigma(i)}\xi_{\sigma(i)} \\ &=\int \td \xi_N \cdots \td \xi_1 \xi_1\cdots\xi_N\sum_{\sigma\in S_N} (-1)^\sigma \prod_{i=1}^N M_{i,\sigma(i)} \\ &= \det M \end{aligned}$$ In which $\sigma$ are elements of $N$ -permutation group $S_N$ . Then we have the solution: $$ \int \td \ \xi f(M\xi+b)=\det M \cdot \int \td \eta f(\eta) $$ q.e.d.[Theorem] : The Gaussian integral for Grassmann variables: (the $\prod_i$ ’s order is from $i=1$ to $i=N$ : $\prod_i\xi_i=\xi_1\xi_2\cdots\xi_N$ , same in the following statements.)
$$ \int \prod_{i} \td \eta_i^*\td \eta_i \ \exp\Big(-\eta^\dagger H\eta+J^\dagger \eta +\eta^\dagger J\Big)=(\det H) e^{-J^\dagger H^{-1}J} $$
Where $\{\eta_i,\eta_i^*\} , \{J_i,J_i^*\} \ ; \ i=1,\cdots,N$ are Grassmann variables, and $H$ is an invertible $2N\times 2N$ Hermitian matrix.
Proof
With the linear variables change, let: $\eta \rightarrow \eta+H^{-1}J \ ; \ \eta^\dagger\rightarrow \eta^\dagger+J^\dagger H^{-1}$ , and the integral should be invariant: $$ I=\int \prod_i \td \eta_i^* \td \eta_i \exp\Big(-\eta^\dagger H\eta-J^\dagger H^{-1} J\Big) $$ For the exponents terms are sums of even numbers of Grassmann generators, so they are commutative with each other. So the exponential function can be written as a product of two. And with the diagonalizaiton of $H$ and variables change, note the determinant is $1$ for the transformation, so we have: $$\begin{aligned} I&= e^{-J^\dagger H^{-1} J } \int \prod_i \td \eta_i^* \td \eta_i e^{-\sum_i \lambda_i \eta_i^*\eta} \\ &= e^{-J^\dagger H^{-1} J } \prod_i \int \td \eta_i^* \td \eta_i \Big(1- \lambda_i \eta_i^* \eta_i\Big)\\ &=e^{-J^\dagger H^{-1} J } \prod_i \lambda_i \\ &=\det H e^{-J^\dagger H^{-1} J } \end{aligned}$$ q.e.d.Fermion Coherent States
[Definition] : The generic form of the coherent state from from a set of Fermion mode $\hat a_i$ is:
$$ \ket{\xi}=\exp\big(-\sum_i \xi_i \hat a_i^\dagger\big)\ket{0}=\prod_i (1-\xi_i \hat a_i^\dagger)\ket{0} $$
Where $\xi_i$ are Grassmann generators, and the Fermion Fock space must be enlarged to define a coherent state. These Grassmann numbers and creation/annihilation operators are anti-commutative: $\xi\hat a+\hat a \xi=0$ . And $(\xi\hat a)^\dagger = \hat a^\dagger \xi^*$ . This is much different from the c-numbers and operators.
[Theorem] : The overlap between two coherent states is:
$$ \bra{\xi}\xi'\rangle = e^{\xi^*\cdot \xi'} $$
Proof
With the definition of Fermion coherent state and the orthogonality of Fermion product state, if: $$ \ket{\xi}=\ket{0}+\sum_i c_i(\xi) \hat a_i^\dagger \ket{0} + \sum_{i,j} c_{ij}(\xi) \hat a_i^\dagger \hat a_j^\dagger \ket{0}+\cdots $$ then: $$ \begin{aligned} \braket{\xi|\xi'} &= \braket{0|0} + \sum_{i, j} \braket{0|\hat a_i c_i(\xi)^* c_j(\xi')\hat a_j^\dagger|0} \\ &\indent + \cdots + \sum_{i, j} \braket{0|\hat a_{i_1}\hat a_{i_2}\cdots \hat a_{i_n} c_{i_1}(\xi)^*c_{i_2}(\xi)^*\cdots c_{i_n}(\xi)^* c_{j_1}(\xi')\cdots c_{j_n}(\xi')\hat a_{j_1}^\dagger\cdots \hat a_{j_n}^\dagger|0} + \cdots\\ &= 1 + \sum_i c_i(\xi)^* c_i(\xi') \braket{0|\hat a_i \hat a_i^\dagger|0} \\ &\indent + \cdots + \sum_{i} c_{i_1}(\xi)^*\cdots c_{i_n}(\xi)^* c_{i_1}(\xi')\cdots c_{i_n}(\xi') \braket{0|\hat a_{i_1}\hat a_{i_2}\cdots \hat a_{i_n} \hat a_{i_1}^\dagger\cdots \hat a_{i_n}^\dagger|0} \\ &=1 + \sum_i c_i(\xi)^* c_i(\xi') + \sum_{i_1, i_2}c_{i_1,i_2}(\xi)^* c_{i_1,i_2}(\xi')+\cdots \end{aligned} $$ The coefficients: $$ c_i(\xi)=-\xi_i \ ; \ c_{ij}(\xi)=- \xi_i\xi_j \ ; \ \cdots \ ; \ c_{i_1,\cdots,i_n}(\xi)=(-1)^{n(n+1)/2} \xi_{i_1}\cdots \xi_{i_n} $$ Used the property that $$ [\xi_i \hat a_i , \xi_j\hat a_j^\dagger]=\xi_i\hat a_i \xi_j\hat a_j^\dagger - \xi_j\hat a_j^\dagger \xi_i\hat a_i=-\xi_i\xi_j\{\hat a_i,\hat a_j^\dagger\}=-\xi_i\xi_j\delta_{ij}=0 $$ Because $\xi_i\xi_i=0$ . And $[\xi_i\hat a_i^\dagger,\hat \xi_j\hat a_j^\dagger]=0$ Then: $$\begin{aligned} \bra{\xi}\xi'\rangle &=1+\sum_{n=1}^{\infty} \sum_{i_1,\cdots,i_n} (-1)^{n(n+1)}\xi_{i_n}^* \xi_{i_{n-1}}^*\cdots \xi_{1}^* \xi'_1\cdots \xi'_{i_n} \\ &=1+\sum_{n=1}^{\infty} \sum_{i_1,\cdots,i_n} (-1)^{n(n+1)}\prod_{k=1}^n \xi_{i_k}^*\xi'_{i_k}\\ &=1+\sum_{n=1}^{\infty} \frac 1 {n!} \Big(\sum_{i}\xi_i^*\xi'_i\Big)^n \\ &=\exp(\xi^*\cdot \xi') \end{aligned}$$ Where we used the property that product of two Grassmann numbers are commutative: $\xi_i\xi_j\xi_k\xi_l=\xi_k\xi_l\xi_i\xi_j$ . This property shows that a Grassmann number looks like a half of a c-number.[Theorem] : The closure relation for Fermion coherent state:
$$ \int \prod_i \td \xi_i^* \td \xi_i e^{-\xi^*\cdot \xi}\ket{\xi}\bra{\xi}=1 $$
Proof
Because Fermion coherent states cover all $n$ -particle states of Fermion: $$ \hat a_{i_1}^\dagger \cdots \hat a_{i_n}^\dagger \ket{0}=(-1)^{n(n+1)/2} \partial_{\xi_{i_1}}\cdots \partial_{\xi_{i_n}}\ket{\xi} $$ we need only check the closure relation on an arbitrary coherent state: $$\begin{aligned} \int \prod_i \td \xi_i^* \td \xi_i e^{-\xi^*\cdot \xi}\ket{\xi}\bra{\xi}\xi'\rangle &= \int \prod_i \td \xi_i^* \td \xi_i e^{-\xi^*\cdot \xi}\ket{\xi} e^{\xi^*\cdot \xi'} \\ &= \int \prod_i \td \xi_i^*\td \xi_i e^{-\xi^*\cdot (\xi-\xi')} \ket{\xi} \\ &= \prod_i -\int \td \xi_i \partial_{\xi_i^*} e^{-\xi^*\cdot (\xi-\xi')}\ket{\xi} \end{aligned}$$ The derivative of $\xi^*$ can be computed independently: $$ \partial_{\xi_i^*} e^{-\xi^*\cdot (\xi-\xi')}=\partial_{\xi_i^*}\prod_{j}\Big(1-\xi_j^*(\xi_j-\xi'_j)\Big)=-(\xi_j-\xi'_j)=-\delta(\xi'_j,\xi_j) $$ Then we have: $$ \int \prod_i \td \xi_i^* \td \xi_i e^{-\xi^*\cdot \xi}\ket{\xi}\bra{\xi}\xi'\rangle=\int \prod_i \td \xi_i \delta(\xi',\xi)\ket{\xi}=\ket{\xi'} $$ q.e.d.[Theorem] : The trace of an operator(Fermion operator, should have even numbers of Fermion field operators) is:
$$ \text{Tr} \hat A:=\sum_{n_1,\cdots}\bra{n_1,\cdots}\hat A \ket{n_1,\cdots}=\int \prod_i \td \xi_i^* \td \xi_i e^{-\xi^* \cdot \xi} \bra{-\xi}\hat A\ket{\xi} $$
Proof
With the definition of Coherent state, we can expand it with the occupation basis of Fermions, assume there are $N$ different fermion modes. $$\begin{aligned} \ket{\xi}&=e^{-\sum_i \xi_i \hat a_i^\dagger}\ket{0}=\ket{0}+\sum_{n=1}^N \frac {(-1)^n} {n!}\Big(\sum_{i=1}^N \xi_i \hat a_i^\dagger\Big)^n\ket{0}\\ &=\sum_{n=1}^N (-1)^n \sum_{k_1+\cdots+k_N=n,k_i=0,1}(\xi_1\hat a_1^\dagger)^{k_1}\cdots(\xi_N \hat a_N^\dagger)^{k_N}\ket{0}\\ &=\sum_{n=1}^N (-1)^n \sum_{k_1+\cdots+k_N=n,k_i=0,1} \xi_N^{k_N}\cdots \xi_1^{k_1}\ket{k_1,\cdots,k_N} \end{aligned}$$ Where we used the property that $\xi_i \hat a_i$ is commutative with any operators and Grassman generators. Then we have, with the fact that $\hat A$ has even number of Fermion field operators so that it is commutative with any Grassmann generators: $$\begin{aligned} \int \prod_i \td \xi_i^* \td\xi_i e^{-\xi^*\cdot \xi} \bra{-\xi}\hat A\ket{\xi}&= \int \prod_i \td \xi_i^* \td\xi \prod_i (1-\xi_i^*\xi_i) \sum_{m,n}(-1)^{m+n} \\ &\indent \cdot \sum_{k,l}' (-1)^{n} (\xi_1^*)^{k_1}\cdots (\xi_N^*)^{k_N} \xi_N^{l_N}\cdots \xi_1^{k_1}\bra{k_1,\cdots,k_N}\hat A\ket{l_1,\cdots,l_N} \end{aligned}$$ Where $\sum_{k,l}'=\sum_{k_1+\cdots+k_N=n,k_i=0,1}\sum_{l_1+\cdots+l_N=m,l_i=0,1}$ . Then we need to compute: $$\begin{aligned} &\int \prod_i \td \xi_i^* \td \xi_i \prod_i (1-\xi_i^*\xi_i) (\xi_1^*)^{k_1}\cdots (\xi_N^*)^{k_N} \xi_N^{l_N}\cdots \xi_1^{l_1} \\ =&\int \prod_{i=2}^N \td \xi_i^*\td\xi_i \Big(\int \td\xi_1^*\td\xi_1 (1-\xi_1^*\xi_1)\xi_1^{*k_1}\xi_1^{l_1}\Big) \prod_{i=2}^N (1-\xi_i^*\xi_i) \xi_2^{*k_2}\cdots\xi_N^{*k_N}\xi_N^{l_N}\cdots\xi_2^{l_2}(-1)^{l_1(l_2+\cdots+l_N+k_N+\cdots+k_2)}\\ =& (\delta_{k_1,0}\delta_{l_1,0}-\delta_{k_1,1}\delta_{l_1,1}) (-1)^{2n-l_1-k_1} \int \prod_{i=2}^N \td \xi_i^*\td\xi_i\prod_{i=2}^N (1-\xi_i^*\xi_i) \xi_2^{*k_2}\cdots\xi_N^{*k_N}\xi_N^{l_N}\cdots\xi_2^{l_2} \td \xi_i^*\td\xi_i\\ =&\prod_i (\delta_{k_i,0}\delta_{l_i,0}-\delta_{k_i,1}\delta_{l_i,1})\\ =&\prod_i ((-1)^{k_i}\delta_{k_i,l_i}) \\ =& (-1)^{n}\delta_{k,l} \end{aligned}$$ Where we used that $1-\xi_i^*\xi$ is commutative with any other Grassmann generators. And if $l_i=k_i$ , $l_i+k_i$ will be always an even number. Then we have: $$\begin{aligned} \int \prod_i \td \xi_i^* \td\xi_i e^{-\xi^*\cdot \xi} \bra{-\xi}\hat A\ket{\xi}&= \int \prod_i \td \xi_i^* \td\xi \prod_i (1-\xi_i^*\xi_i) \sum_{m,n}(-1)^{m+n} \\ &\indent \cdot \sum_{k,l}' (-1)^{n} (\xi_1^*)^{k_1}\cdots (\xi_N^*)^{k_N} \xi_N^{l_N}\cdots \xi_1^{k_1}\bra{k_1,\cdots,k_N}\hat A\ket{l_1,\cdots,l_N}\\ &=\sum_{m,n}(-1)^{m+n} \sum_{k,l}' (-1)^{n+n} \delta_{k,l} \bra{k_1,\cdots,k_N}\hat A\ket{l_1,\cdots,l_N}\\ &=\sum_{n} \sum_{k_1+\cdots+k_N=n,k_i=0,1} \bra{k_1,\cdots,k_N}\hat A\ket{k_1,\cdots,k_N}\\ &=\text{Tr}\hat A \end{aligned}$$ Q.E.D.[Theorem] : One can construct any many body states of Fermions from coherent state by:
$$ \ket{n_1,\cdots}:=\prod_{k} \hat a_k^{\dagger n_k}\ket{0} =(-1)^{n_1+\cdots}\prod_i \partial_{\xi_i}^{n_i}\ket{0} $$
[Theorem] : The average and variance of the particle number $\hat N=\sum_i \hat a_i^\dagger \hat a_i$ is:
$$\begin{aligned} \bar N&:= \frac {\bra{\xi}\hat N\ket{\xi}} {\bra{\xi}\xi\rangle} = \sum_i \xi_i^*\xi_i \\ (\Delta N)^2 & := \frac {\bra{\xi}(\hat N-\bar N)^2\ket{\xi}} {\bra{\xi}\xi\rangle} = \bar N \end{aligned}$$
Proof
With the commutator: $$ [\hat a_i , \prod_j (1-\xi_j\hat a_j^\dagger)]=\prod_{j\lt i}(1-\xi_j\hat a_j^\dagger)[\hat a_i,1-\xi_i\hat a_i^\dagger]\prod_{j\gt i}(1-\xi_j\hat a_j^\dagger)=\prod_{j\lt i}(1-\xi_j\hat a_j^\dagger)\xi_i\prod_{j\gt i}(1-\xi_j\hat a_j^\dagger) $$ Where we used the property that $1-\xi_j\hat a_j^\dagger$ is commutative with $a_i$ when $i\neq j$ , and $\hat a_i\xi_i=-\xi_i\hat a_i$ , the anti-commutation relation. Then: $$ \bra{\xi}\hat a_i^\dagger \hat a_i\ket{\xi}=\bra{0}\Big(\prod_{j\neq i} (1-\xi_j\hat a_j^\dagger)\Big)^\dagger \xi_i^*\xi_i \prod_{j\neq i} (1-\xi_j\hat a_j^\dagger)\ket{0}=\xi_i^*\xi_i \exp(\sum_{j\neq i}\xi_j^*\xi_j) $$ Then: $$ \bar N=\frac {\bra{\xi}\hat N\ket{\xi}} {\bra{\xi}\xi\rangle}=\frac {\sum_i \xi_i^*\xi_i e^{\xi^*\cdot \xi-\xi_i^*\xi_i}} {e^{\xi^*\cdot \xi}}=\sum_i \xi_i^*\xi_i e^{-\xi_i^*\xi_i}=\sum_i \xi_i^*\xi_i $$ The last identity is caused by $\xi^*\xi e^{-\xi^*\xi}=\xi^*\xi-\xi^*\xi\xi^*\xi=\xi^*\xi$ . And with: $$\begin{aligned}[] [\hat N,\prod_j(1-\xi_j\hat a_j^\dagger)]&=\sum_i \hat a_i^\dagger \xi_i \prod_{j\neq i} (1-\xi_j\hat a_j^\dagger)\\ &=\sum_i e^{\sum_j -\xi_j\hat a_j^\dagger}-\sum_i\prod_{j\neq i}(1-\xi_j\hat a_j^\dagger)\\ &=e^{-\xi\cdot \hat{\bm{a}}^\dagger}\sum_i (1-e^{\xi_i\hat a_i^\dagger})\\ &=\sum_i (-\xi_i\hat a_i^\dagger) e^{-\xi\cdot \hat{\bm{a}}^\dagger} \end{aligned}$$ Then: $$\begin{aligned} \bra{\xi}\hat N^2\ket{\xi}&=\bra{\xi}\sum_{i,j}\hat a_i \xi_i^* \xi_j \hat a_j^\dagger \ket{\xi}\\ &=\bra{\xi}\sum_{i,j}\xi_i^*\xi_j (-\hat a_j^\dagger \hat a_i+\delta_{ij})\ket{\xi}\\ &=-\sum_{i,j}\xi_i^*\xi_j \bra{\xi}\hat a_j^\dagger \hat a_i\ket{\xi} +\sum_{i}\xi_i^*\xi_i \bra{\xi}\xi\rangle \\ &=-\sum_{i,j}\xi_i^*\xi_j \xi_j^*\xi_i \bra{\xi}\xi\rangle + \bar N \bra{\xi}\xi\rangle\\ &=\bra{\xi}\xi\rangle \Big(\sum_{i,j}\xi_i^*\xi_j^*\xi_j\xi_i+\bar N\Big)=\bra{\xi}\xi\rangle \Big(\bar N^2+\bar N\Big) \end{aligned}$$ Then with that $\bar N$ is the summation of pairs of Grassmann generators, so it is commutative with any single generators or creation/annihilation operators. Then: $$\begin{aligned} (\Delta N)^2 & := \frac {\bra{\xi}(\hat N-\bar N)^2\ket{\xi}} {\bra{\xi}\xi\rangle}\\ &=\frac {\bra{\xi}\hat N^2\ket{\xi}-\bar N^2} {\bra{\xi}\xi\rangle} \\ &=\bar N \end{aligned}$$ q.e.d.Coherent State Representation
Like the wavefunction in coordinate representation, we introduce the Coherent State Representation
For an arbitrary quantum state $\ket{\psi}$ , we can define the "wavefunction" in coherent representation , is the Grassmann function: $\psi(\xi^*)=\bra{\xi}\psi\rangle = \bra{0}e^{-\hat {\bm{a}} \cdot \xi^*}\ket{\psi}$ , which satisfies:
$$ \ket{\psi}=\int \prod_i \td \xi_i^*\td \xi_i e^{-\xi^*\cdot \xi} \ket{\xi}\bra{\xi}\psi\rangle=\int \prod_i \td \xi_i^*\td \xi_i e^{-\xi^*\cdot \xi} \ket{\xi}\psi(\xi^*) $$
The representation of operators are:
$$ \hat a_i \rightarrow \partial_{\xi_i^*} \ ; \ \hat a_i^\dagger \rightarrow \xi_i^* $$
And the matrix elements of normal ordered operators, which means that all creation operators are at the left side of annihilation operators, is:
$$ \bra{\xi}A(\hat {\bm{a}}^\dagger,\hat {\bm{a}})\ket{\xi'}=e^{\xi^*\cdot \xi'}A(\xi^*,\xi') $$
Proof
The creation/annihilation operator operating on an arbitrary quantum state induces: $$\begin{aligned} \hat a_i\psi(\xi^*) &= \bra{\xi}\hat a_i\ket{\psi} = \bra{0}\partial_{\xi_i^*} e^{\hat {\bm{a}} \cdot \xi^*}\ket{\psi} = (\partial_{\xi_i^*} \psi)(\xi)\\ \hat a_i^\dagger \psi(\xi^*) &= \bra{\xi}\hat a_i^\dagger \ket{\psi} = \xi_i^*\psi(\xi) \end{aligned}$$ Where we used the commutation relation $[\hat a_i , \prod_j (1-\xi_j\hat a_j^\dagger)]=\prod_{j\lt i}(1-\xi_j\hat a_j^\dagger)\xi_i\prod_{j\gt i}(1-\xi_j\hat a_j^\dagger)$ , so we have: $$ \hat a_i \ket{\xi}=\xi_i\ket{\xi} $$ And the operator: $$\begin{aligned} \bra{\xi}A(\hat {\bm{a}}^\dagger , \hat {\bm{a}}) \ket{\xi'}&=\sum_{i,j} a_{ij} \bra{\xi}\hat a_1^{\dagger i_1}\cdots \hat a_N^{\dagger i_N} \hat a_N^{j_N}\cdots \hat a_1^{j_1}\ket{\xi'}\\ &=\sum_{i,j}a_{ij} (\xi_1^*)^{i_1}\cdots (\xi_N^*)^{i_N}\bra{\xi}\xi'\rangle(\xi'_N)^{j_N}\cdots (\xi'_1)^{j_1}\\ &=e^{\xi^*\cdot \xi'} \sum_{i,j} a_{ij} (\xi_1^*)^{i_1}\cdots (\xi_N^*)^{i_N}(\xi'_N)^{j_N}\cdots (\xi'_1)^{j_1}\\ &= e^{\xi^*\cdot \xi} A(\xi^*,\xi') \end{aligned}$$ q.e.d.At last, the Schrodinger equation in coherent representation is, with the Hamiltonian is normal ordered as $\hat H=H(\hat {\bm{a}}^\dagger,\hat{\bm{a}},t)$ :
$$ \ti \hbar \frac {\partial \psi(\xi^*,t)} {\partial t}=\int \prod_{i} \td \xi_i'^*\td \xi_i' e^{\xi'^*\cdot \xi'} H(\xi^*,\xi',t) \psi(\xi'^*,t) $$