\section{Introduction}	
 Let $H, A, B$ be  smooth classical observables on $\mathbb{R}^{2d}$ in the variables $X=(x,\xi)$.  The Poisson brackets
 is defined as $\{A,B\} =\partial_\xi A\cdot\partial_xB-\partial_xA\cdot\partial_\xi B$. So the classical  time evolution of $A$  determined by the Hamilton equation for $H$ is solution of the equation:
 \begin{eqnarray}}  \def\eea{\end{eqnarray}
 \frac {d}{dt}A(t) &=& \{A(t), H\} \\
 A(0) &=&A. \nonumber
 \eea
 The Weyl quantization $\widehat A$  of $A$  is defined as as the following operator:
  \begin{equation}}   \def\eeq{\end{equation} 
  \widehat A f(x) :={\rm Op}^w_\hbar f(x)=(2\pi\hbar)^{-d} \int_{\mathbb{R}^{2d}} A\!\left(\frac{x+y}{2}, \xi\right) 
{\rm e}^{i\xi\cdot(x-y)/\hbar} f(y) \,dy\, d\xi
\eeq
for any $f\in{\mathcal S}(\mathbb{R}^d)$.  \\
The quantum time evolution of the quantum observable $\widehat A$ must satisfy the Heisenberg equation
\begin{eqnarray}}  \def\eea{\end{eqnarray}
 \frac {d}{dt}\widehat A(t) &=& \frac{i}{\hbar}[ \widehat A(t),  \widehat H]\\
  \widehat A(0) &=&\widehat A
 \eea
 where $[\hat A, \hat B] = \hat A\hat B -\hat B\hat A$.\\
The Moyal bracket  of the observables $A, H$ is defined such that 
\begin{equation}}   \def\eeq{\end{equation}
\frac{i}{\hbar}[ \widehat A,  \widehat H]={\rm Op}^w_\hbar(\{A,H\}_\circledast.
\eeq
Notice that it results from   the Weyl quantization  calculus with a small parameter $\hbar$  that we have
$$
\lim_{\hbar \searrow 0}\{A,H\}_\circledast = \{A,H\}.
$$

There is an equivalent definition  by introducing the Moyal product  $A\circledast B$ (see the next section)
such that 
$$
({\rm Op}^w_\hbar A)({\rm Op}^w_\hbar B) = {\rm Op}^w_\hbar(A\circledast B).
$$
Then we have
$$
\{A,B\}_\circledast  = \frac{i}{\hbar}(A\circledast B-B\circledast A).
$$
These definitions make sense for $A, B\in {\mathcal S}(\mathbb{R}^{2d})$ and can be extended to suitable  classes of symbols with moderate growth. To be more explicite we introduce  the classes
${\mathbb{S}}_\delta^\mu$, for $\delta<1$, $\mu\in\mathbb{R}$. $A\in{\mathbb{S}}_\delta^\mu$ iff  $A\in C^\infty(\mathbb{R}^{2d})$ and   for any multiindex $\gamma\in\mathbb{N}^{2d}$
 we have:
$$
\vert\partial^\gamma_X A(X)\vert \leq C_\gamma\langle X\rangle^{\mu+\delta\vert\gamma\vert}
$$
Using Theorem A.1 in \cite{BR}, we can see that $A\circledast H$ is a smooth symbol if $H\in{\mathbb{S}}_\delta^\mu$ and $A\in{\mathbb{S}}^0_0$.
Our aim here is to prove the following result.
\begin{theorem}\label{MainThm} Assume that  $\hbar$  is fixed ($\hbar =1$). 
Let be $H\in {\mathbb{S}}_\delta^\mu$ for some $\mu\in\mathbb{R}$ and $\delta<1$.  Assume that  for any $A\in{\mathbb{S}}^0_0$  we have 
$ \{A, H\}_\circledast =  \{A, H\}$. Then $H(X)$ is a polynomial in $X=(x,\xi)$ of degree at most 2.  
\end{theorem}
\begin{remark}
It is well known that if $H$ is a polynomial of degree at most 2 then $\{A, H\}_\circledast =  \{A, H\}$  for any $A\in{\mathbb{S}}^\nu_0$. I do not know any  reference  for a proof of the converse statement. The proof given here is a direct consequence of basic properties of the Weyl quantization. 
\end{remark}
\begin{remark}
The usual proofs of the Groenewold-van Hove Theorem on the phase space $\mathbb{R}^{2d}$  concern  more general quantization procedures  but are restricted to polynomial symbols $A, H$.\\
       A quotation from \cite{vHove} p.66-67:\\
       "On \'etablit ensuite qu'une correspondance biunivoque entre grandeurs classiques et quantiques, ayant le  caract\`ere d'un isomorphisme entre alg\`ebres de Lie, existe entre les grandeurs repr\'esent\'ees par des polyn\^omes de degr\'e 0, 1, 2 en les variables $p_1,\cdots p_N,q_1,\cdots q_n$ mais ne peut \^etre \'etendue sans perdre ses propri\'et\'es essentielles \`a l'ensemble de toutes les grandeurs classiques"   \\
       The Theorem of Groenewold-van Hove is detailed  p.76 and the quadratic case p.87 of \cite{vHove}.\\
       Notice that the quadratic case is related with the metaplectic representation \cite{Gotay}.
\end{remark}  

 \section{Weyl calculus}
 \subsection{Introduction to the Weyl quantization}
 In this section, we recall some  basic properties of the Weyl calculus (for more details see \cite{ho}).\\
 Weyl quantization start by quantization of exponent of linear forms $L_Y(X) = \sigma(Y,X)=\eta\cdot x-y\cdot\xi$
  with $X=(x,\xi)$, $Y=(y,\eta)$. Apart the usual properties asked for an admissible quantization, Weyl quantization is uniquely determined by imposing
  that the Weyl symbol of ${\rm e}^{i\widehat{L_Y}}$ is ${\rm e}^{iL_Y}$. Recall that $ \widehat T(Y):={\rm e}^{-\frac{i}{\hbar}\widehat{L_Y}}$   is the Weyl-Heisenberg translation operator by $Y$ in the phase space $\mathbb{R}^{2d}$. 
  Then  for any observable $A$,  using a Fourier transform,  the  Weyl quantization $ A$ is defined  for any $\psi\in {\mathcal S}(\mathbb{R}^d)$, as
  $$
   \widehat A\psi = (2\pi)^{-d}\int_{\mathbb{R}^{2d}}{\tilde A}_\sigma(Y) T(Y)\psi dY
  $$
  where ${\tilde A}_\sigma(Y)=\int_{\mathbb{R}^{2d}} A(z){\rm e}^{-i\sigma(Y,z)}dz$ is the symplectic Fourier transform of $A$ (in the sense of distributions).
   So that the family $\{T(Y)\}_{Y\in\mathbb{R}^{2d}}$ is an over-complete   basis for operators   between the Schwartz spaces ${\mathcal S}(\mathbb{R}^d)$ and 
   ${\mathcal S}'(\mathbb{R}^d)$ 
  
 \subsection{The Moyal Product}
We first recall the formal product rule for quantum observables 
with Weyl 
quantization. Let $A, B \in {\mathcal S}(\mathbb{R}^{2d})$. The Moyal product $C:=A\circledast B$ is the  
observable 
$C$ such that $\widehat{A}\cdot\widehat{B} = \widehat{C}$. Some computations with 
the Fourier 
transform give the following well known formula \cite{ho} 
\begin{equation}}   \def\eeq{\end{equation}\label{moy1}
C(x,\xi) = 
\exp\left(\frac{i\hbar}{2}\sigma(D_q,D_p;D_{q^\prime},D_{p^\prime})\right)A(q,p) 
B(q^\prime,p^\prime)\vert_{(q,p)=(q^\prime,p^\prime)=(x,\xi)}, 
\eeq 
where $\sigma$ is the symplectic bilinear form $\sigma((q,p), (q',p'))=p\cdot q'-p'\cdot q$ and 
$D=i^{-1}\hbar\nabla$. 
By expanding the exponential term in a formal power series in $\hbar$ we get 
\begin{equation}}   \def\eeq{\end{equation}\label{prod2} 
C(x,\xi) = 
\sum_{j\geq 
0}\frac{\hbar^j}{j!}\left(\frac{i}{2}\sigma(D_q,D_p;D_{q^\prime},D_{p^\prime}) \right)^j 
A(q,p)B(q^\prime,p^\prime)\vert_{(q,p)=(q^\prime,p^\prime)=(x,\xi)}. 
\eeq 
So that $C(x,\xi)$ is a formal power series in $\hbar$ with coefficients 
given by 
\begin{equation}}   \def\eeq{\end{equation} \label{exprod}
C_j(A,B;x,\xi) = \frac{1}{2^j}\sum_{\vert\alpha+\beta\vert=j} 
\frac{(-1)^{\vert \beta\vert}}{\alpha!\beta!} 
(D^\beta_x\partial^\alpha_\xi A).( D^\alpha_x\partial^\beta_\xi B)(x,\xi). 
\eeq 
\medskip 
Furthermore we need  a remainder estimates for the expansion of the Moyal product.\\
For every $N \geq 1$. we denote
\begin{equation}}   \def\eeq{\end{equation} 
R_N(A,B;z):=
A\circledast B(z) - \sum_{0\leq j \leq N}\hbar^jC_j(z) .
\eeq 
The following estimate is a particular case of Theorem A.1 in \cite{BR} see also Remark A.3.

\begin{lem}\label{lem:Moyalest}
Let be $A\in{\mathbb S}^{\mu_A}_\delta$ and  $B\in{\mathbb S}^{\mu_B}_\delta$  then for any $N\geq 1$, $\gamma\in\mathbb{N}^{2d}$, $M\geq M_0$  there exists $C_{N,\gamma,M}>0$(independent of $(A, B)$)
such that 
\begin{eqnarray}}  \def\eea{\end{eqnarray}\label{remest1} 
\vert\partial_z^\gamma R_N(A,B;z)\vert \leq C_{N,\gamma,M}\hbar^{N+1}\sum_{\substack{\vert\alpha+\beta\vert=N+1\\ \vert\mu+\nu\vert\leq M+\vert\gamma\vert}}\\\sup_{u, v\in\mathbb{R}^{2d}} (1+\vert u\vert^2+\vert v\vert^2)^{(M_0-M)/2}\vert\partial_u^{(\alpha,\beta)+\mu}A(z+u)\vert\vert\partial_v^{(\beta,\alpha)+\nu}B(z+v)\vert
\eea
In particular $ R_N(A,B,z)\in \mathbb S_\delta^{\mu_{AB}}$ for some $\mu_{AB}\geq \mu_A+\mu_B$.
\end{lem} 
For proving  this Lemma  one assume first that $A, B\in{\mathcal{S}}(\mathbb{R}^{2d})$. For the general case we put $A_\varepsilon(X)={\rm e}^{-\epsilon\vert X\vert^2}A(X)$, 
$B_\varepsilon(X)={\rm e}^{-\epsilon\vert X\vert^2}B(X)$ and pass to the limit for $\varepsilon\searrow 0$.
\section{Proof of Theorem(\ref{MainThm})}
 Here $\hbar=1$. It is enough to consider the test observables $ T_Y:={\rm e}^{-iL_Y}$ ($Y\in\mathbb{R}^{2d}$). \\
 We have 
 $$
 \widehat T_Y\widehat H\widehat T_Y^* = [\widehat T_Y, \widehat H]\widehat T_Y^* + \widehat H
 $$
Using the assumption  of  Theorem(\ref{MainThm}) we get 
\begin{equation}}   \def\eeq{\end{equation}\label{MPC1}
\frac{1}{i}(\{T_T^*, H\}\circledast T_Y)(X) = H(X+Y)-H(X), \forall X, Y\in\mathbb{R}^{2d}.
\eeq
Computing the Poisson bracket in \eqref{MPC1} gives
\begin{equation}}   \def\eeq{\end{equation}\label{MPC2}
(((y\cdot\partial_x H +\eta\cdot\partial_\xi H)T_Y^*)\circledast T_Y)(X) = H(X+Y) -H(X),\;\forall X, Y\in\mathbb{R}^{2d}.
\eeq
Our aim is to prove that \eqref{MPC2}  implies that $H(X)$ is a polynomial of degree at most 2. For that purpose we shall compute 
the asymptotic expansion  as $Y\rightarrow 0$ of the left hand side  of \eqref{MPC2} and compare it with the Taylor expansion for  $H(X+Y)$ modulo
 $O(\vert Y\vert^4)$. From that we shall conclude that  all the third  order derivatives of $H$ vanish in $X$ hence the conclusion will follow.
 
 We  have  
 $$
 \partial_x^\alpha\partial_\xi^\beta T_Y = i^{-\vert\alpha+\beta\vert}\eta^\alpha y^\beta T_Y
 $$ 
 Let us denote by $C(X,Y)$ the left hand side in \eqref{MPC2}. 
 So using Lemma(\ref{lem:Moyalest})   we have 
 $$
 C(X,Y) = \sum_{0\leq j\leq 2}(C_j(X,Y) + O(\vert Y\vert^4),
 $$
 where 
 \begin{eqnarray}}  \def\eea{\end{eqnarray}
 C_0(X,Y) &= & Y\cdot\nabla_XH(X)\\
 C_1(X,Y) &=& \frac{1}{2}Y\cdot \nabla_X^2H(X)Y,
 \eea
 where $\nabla_X^2H(X)$ is the Hessian matrix of $H$.\\
 Let us compute now $C_2(X,Y)$, which is   an homogeneous polynomials of degree 3 in $Y$.\\
 For simplicity let us consider the 1-D case. The same computation can clearly  be done for $d>1$.\\
 Using \eqref{prod2} we get with $Y=(y,\eta)$, 
 \begin{equation}}   \def\eeq{\end{equation}\label{C2}
 C_2(X,Y) = \frac{1}{8}\left(y^3\partial_x^3H +\eta^3\partial_\eta^3H -y^2\eta\partial_\xi\partial_x^2H - y\eta^2\partial_\xi^2\partial_xH\right).
 \eeq
 According \eqref{MPC2}, $C_2(X,Y)$ must  coincide with the term of order 3  in $Y$ of the Taylor expansion in $X$ for $H(X+Y)-H(X)$.
 But this is possible only if $\partial_x^3H=\partial_\eta^3H =\partial_\xi\partial_x^2H =\partial_\xi^2\partial_xH=0$   for any $(x,\xi)\in \mathbb{R}^2$.
 So $H$ must be a polynomial of degree $\leq 2$. $\square$
 
 \section{Extension  to polynomials of arbitrary degree }
 The asymptotic expansion in $\hbar$ in the Moyal product suggest to introduce the following semi-classical approximations of the Moyal bracket:
 $$
 \{ A,B\}_{\circledast,m} = \{A,B\} +\hbar^2\{A,B\}_3 +\cdots +\hbar^{2m}\{A,B\}_{2m+1},
 $$
 where $\{A,B\}_j = \frac{i}{\hbar}(C_j(A,B)-C_j(B,A))$  (notation of  \eqref{exprod}). 
 Notice that $\{A,B\}_j =0$ for $j$ even.\\
 It is clear that if $H$ is a polynomial of degree at most $2m+2$ then we have $ \{ A,H\}_{\circledast,m} = \{ A,H\}_{\circledast}$  
 for any $A$.  Conversely we have
\begin{theorem}\label{thm:orderm} Assume $\hbar =1$. If for any $A\in {\mathbb S}^0_0$ we have $ \{ A,H\}_{\circledast,m} = \{ A,H\}_{\circledast}$
 then $H$  must be  a polynomial of degree at most $2m+2$.
 \end{theorem}
 {\em Proof}. Here we give a proof different from the case $m=0$, without connection with the Taylor   formula,  for simpler computations.\\
 Using Lemma \ref{lem:Moyalest} we have 
 \begin{equation}}   \def\eeq{\end{equation}\label{comp1}
 T_Y^*\left( \{T_Y,H\}_{\circledast} - \{T_Y,H\}_{\circledast,m} \right)= {\mathcal O}(\vert Y\vert^{2m+3}),\; Y\rightarrow 0.
 \eeq
 Moreover from \eqref{exprod} we get:
 \begin{equation}}   \def\eeq{\end{equation}\label{comp2} 
  T_Y^*\{T_Y,H\}_{2j+1} =\frac{1}{2^{j+1}}\sum_{\vert\alpha+\beta=2j+1}\frac{y^\alpha\eta^\beta}{\alpha!\beta!}\partial_x^\alpha\partial_\xi^\beta H  
   \eeq
Using the assumption of Theorem \ref{thm:orderm}. and \eqref{comp1}  we get that 
$$
 T_Y^*\{T_Y,H\}_{2m+3} = {\mathcal O}(\vert Y\vert^{2m+5})
 $$
 But  $T_Y^*\{T_Y,H\}_{2m+3} $ is an homogeneous polynomial of degree $2m+3$ in $Y$ so we get that this polynomial is 0 and from
 \eqref{comp2} we get that $ \partial_x^\alpha\partial_\xi^\beta H =0 $   for $\vert\alpha+\beta\vert=2m+3$. Then we can conclude that $H(X)$ is a polynomial of degree at most $2m+2$ in $X\in\mathbb{R}^{2d}$. $\square$.
 
	