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\begin{document}
\draft
\title{Functional approach to coherent states in non commutative
theories}
 
\author{Musongela Lubo \footnote{E-mail muso@ictp.trieste.it} \\
        The Abdus Salam International Centre for Theoretical Physics P.O.Box
586\\
34100 Trieste, Italy}
  
\maketitle
\begin{abstract}
  In many high dimensional noncommutative theories, no state saturates
simultaneously all the non trivial Heisenberg uncertainty relations. This
differs  from the usual theory where the squeezed states possess this
property. The important role played by these states when recovering classical 
mechanics as a limit of  quantum theory makes  necessary  the
investigation of  the possible generalizations in the noncommutative
context. 

We propose an extension based on a variational
principle. The action considered is the sum  of the squares of the terms
associated to the non trivial Heisenberg  uncertainty relations. We first verify
that our proposal works in the usual theory: we find the known gaussian 
functions and, besides them, other states which can be expressed as products 
of gaussians with specific hypergeometrics.

We illustrate our construction in three models defined on a four dimensional
phase space: two models endowed with a minimal length uncertainty
 and the popular case in which
the  commutators of the positions are constants. In these three models, our proposal leads
to   second order partial differential equations. We find analytical solutions 
in specific cases. We briefly discuss how our method may be applied to the fuzzy sphere.

To emphasize that the recovering of  classical behaviours is not a trivial
question in the non commutative context, we show how the difference of structure between the
 Poisson brackets  and the
commutators in the  theories analyzed here generically leads to a loss of periodicity
for the harmonic oscillator.  



\end{abstract}
 

\pacs{{\bf PACS numbers:03.65 Ca, 0365 Db}} 


\section{Introduction}
    Many works have been devoted to noncommutative quantum theories recently.
One of the main motivations  is the hope that a non trivial
structure of space time at small distances may give birth to theories with 
better ultaviolet behaviors. 

There are two ways of apprehending non commutative theories. The first one
postulates a modification of the commutation relations from the
start. This is the case for example of J.Madore \cite{madore}. The
second one, in the wake of E.Witten and S.Seiberg \cite{witten},
considers non commutative theories as  low energy phenomenological
implications  of string theory. The philosophy of this work mostly relies
on the first approach.

Many non commutative quantum theories have not  fulfilled the initial hope
concerning the behavior of Green functions \cite{phi4}. One of the remarquable
cases wich do not fall in this category has been formulated by
Kempf-Mangano-Mann(KMM) \cite{kmm1,kmm3,kmm4,kmm5}. This could be done thanks to a carefull analysis
of the states physically allowed in the theory. With that idea in mind, it was
suggested that the analysis about the loss of causality may need a more detailed 
treatment \cite{lubo}. 

Some efforts  have been devoted to the understanding of quantum mechanics when
non commutativity sets in. For example, H.Falomir et al. \cite{falomir} have studied
the Bohm-Aharonov effect in this context, obtaining for the deformation
parameter new bounds which are consistent with previous ones \cite{oldbound}. The
noncommutative oscillator in arbitrary dimension has been analyzed by
A.Hatzinikas et al \cite{hatzi} while R.Banarjee \cite{banar} has shown 
its link to dissipation. The relation with usual canonical
variables has been analyzed by A.smailagic \cite{smail}. Some theories which also have
special small structures have been used in the study of  physical processes like
black hole evaporation \cite{brout,muso}.

Our aim is not to solve a specific problem in this framework but to try to
understand the structure of some miningfull sets of states. In two recent
works, K.Bolonek and P.Kosinski \cite{kosi} have shown that all the non trivial
Heisenberg uncertainties can not be satisfied simultaneously in a theory where
the  commutators of the positions are are non vanishing constants. This raises an
 important question
since in the unmodified theory, the states which saturate all the non trivial 
 Heisenberg
uncertainties are also the ones which lead to quasi classical behaviors. In 
these states the mean values of the position and momentum operators
reproduce the behavior of the classical Hamilton solution for the harmonic
oscillator, the relative uncertainties being negligible \cite{cohen}. 


  
Adapting  for our
purposes an idea  recently used by
S.Detournay, Cl.Gabriel and Ph.Spindel  \cite{spindel}, we  construct 
a  functional which attains its minima on the coherent
 states in the usual theory. We then generalize this functional for the non commutative
theories we are interested in.  The functions on which these functionals
attain their  minima are, in our opinion, 
candidates to the status of coherent states. 

This paper is organized as follows. 
The second section is devoted to the impossibility of
saturating the non trivial  Heisenberg uncertainties in three particular
noncommutative theories. 
In the  third section, we show explicitly how our proposal 
works in the usual one dimensional case; we verify that the usual squeezed states satisfy
the second order differential equation which comes from our variationnal
principle.
In the fourth section we apply the same method to the three models studied in
section $2$. We obtain  that the states for which the functionals are extremal
verify  second order partial differential equations. In the fifth section we show that 
for a generic state describing a harmonic oscillator on the commutative plane and in
 the K.M.M theory, the position is not rigorously periodic; altough  
  phenomenologically unobservable, this fact raises an important conceptual question.  
The sixth section is devoted to a discussion of our results and a part of the computations
are put in the appendix.






\section{Examples of high dimensional 
theories with non trivial Heisenberg uncertainties that cannot be satisfied
simultaneously}

In the usual quantum theory, the coherent states satisfy {\em simultaneously}
the relations 

\be
\label{eq1}
\Delta x_i \Delta p_i = \frac{ \vert \langle [\hat x_i, \hat p_i] \rangle 
\vert}{2} = \frac{\hbar}{2} \ee
 for each index $i$. On the
contrary, a product such as $\Delta x_1 \Delta p_2$  can be arbitrary since the
commutator of the  associated variables vanishes. The first type of
uncertainty will be called {\em non trivial}. The aim of this section is to
show that in some  models  the non trivial uncertainties
cannot be fulfilled at the same time. The same reasoning works in higher
dimensions. The states which satisfy Eq.(\ref{eq1}) are in the kernels of 
 operators of the form
\be
\label{eq2}
\left( \hat x - \langle \hat x \rangle + \frac{\langle [\hat x, \hat p] \rangle }{2 (\Delta p)^2}  ( \hat
x - \langle \hat x \rangle )  \right)   \quad ,
\ee 
which is linear in $\hat x$, $\hat p$ and contains a piece proportional to
the identity.


\subsection{The non commutative plane}
  
 Let us consider a two dimensional theory in which the non vanishing
commutation relations are the following \cite{}:
 \be
\label{eq3}
[ \hat x_1 , \hat x_2] = i \, \theta  \quad , \quad [ \hat x_j , \hat p_k] = i
\, \hbar \, \delta_{jk} \quad ; \quad    \theta, \hbar \quad > 0 \quad .
\ee 
This theory admits a representation in which the operators and the scalar
product are given by the following formula
\be
\label{eq4}
\hat x_1 = i \hbar \partial_{p_1} - \frac{1}{2} \frac{\theta}{\hbar} p_2
\quad , \quad \hat x_2 = i \hbar \partial_{p_2} + \frac{1}{2}
\frac{\theta}{\hbar} p_1 \quad  \quad , \quad \hat p_1 = p_1 \quad , \quad
\hat p_2 = p_2  \quad , \quad \quad \langle \phi \vert \psi \rangle = \int
d^2 p \, \phi^*(p) \, \psi(p) \quad . 
\ee   
As  reminded above, the equality in
the Heisenberg relation is attained only for those states which satisfy
Eq.(\ref{eq2}). In our case, 
\be \label{eq5}
\Delta x_1 \Delta x_2 = \frac{\theta}{2}  \Longrightarrow  {\hat a}_1  \vert
\psi \rangle  = 0 \quad , \quad  
\Delta x_1 \Delta p_1 = \frac{\hbar}{2}  \Longrightarrow  {\hat a}_2  \vert
\psi \rangle  = 0 \quad , \quad  
\Delta x_2 \Delta p_2 = \frac{\hbar}{2}  \Longrightarrow  {\hat a}_3  \vert
\psi \rangle  = 0 \quad ;  
\ee
the operators $\hat a_i$ are given by the following expressions
\be
\label{eq6}
{\hat a}_1 =  \hat x_1 + i \lambda_1 \hat x_2 + \mu_1 \quad , \quad 
{\hat a}_2 =  \hat x_1 + i \lambda_2 \hat p_1 + \mu_2 \quad , \quad 
{\hat a}_3 =  \hat x_2 + i \lambda_3 \hat p_2 + \mu_3 \quad , 
\ee
 $\lambda_i$ and $\mu_i$ being respectively real and complex constants. The second
  and the third
equalities in Eq.(\ref{eq5}) can not be satisfied simultaneously. Let us suppose
this is realized by a state $\vert \psi \rangle$. Then this state must obey
the equation
 \be
\label{eq7}
[ \hat a_2 , \hat a_3 ] \vert \psi \rangle = i \theta \vert \psi \rangle = 0 
\ee 
which  admits only the null vector as a solution. A more extensive analysis
of the other equalities can be found in \cite{}.

The constants appearing in the operators $\hat a_i$ in Eq.(\ref{eq6}) do not
play any role in the considerations implying the commutators. We shall discard
them for simplicity in the more complicated model which is considered below.


\subsection{The K.M.M. 1 model}
   The first extension of the one dimensional model endowed with a minimal
length uncertainty was proposed in \cite{kmm2}:  \be
\label{eq8}
[ \hat x_j ,\hat p_k] = i \hbar (1 + \beta  \hat{{\vec p}^2} ) \delta_{j k} 
\quad , \quad [ \hat x_j ,\hat x_k] = 2 i \hbar \beta ( \hat p_j  \hat x_k -
\hat p_k  \hat x_j)  \quad . 
\ee 
It lacks translational  symmetry since the second relation is not invariant
under the transformation $\hat x_k  \rightarrow \hat x_k + \hat a_k $. This 
is also the case of the fuzzy sphere  \cite{}; one can not discard  this
extension  on this sole argument.

The representation we shall use for this model is the following:
\be 
\label{eq9} 
\hat x_k = i \hbar (1 + \beta {\vec p}^2) \partial_{p_k}  \quad , \quad \hat
p_k = p_k  \quad , \quad \langle \phi \vert \psi \rangle = \int d^2 p \,
\frac{1}{1+ \beta \vec{p}^2} \phi^*(p) \psi(p) \quad . 
\ee 

Similarly to the case analyzed in the previous subsection, saturating
non trivial Heisenberg uncertainties results in analyzing  the kernels of
the first order operators  given in Eq.(\ref{eq2}). The pairs of variables exhibiting 
non trivial commutators are the same than in the previous model. We therefore use 
again the operators defined in Eq.(\ref{eq6}). If a state saturates all the three
 Heisenberg inequalities, it is in the kernel
of all the following  commutators:
\begin{eqnarray}
\label{eq10}
&& [ \hat a_1 , \hat a_2 ] = - \lambda_2 \hbar (1 + \beta \vec{p}^2) - 2 \hbar
\beta  \lambda_1  ( \hat p_2  \hat x_1 -
\hat p_1  \hat x_2)  \quad , \quad 
[ \hat a_1 , \hat a_3 ] =   2 \hbar
\beta  \lambda_2  ( \hat p_2  \hat x_1 -
\hat p_1  \hat x_2) - i \hbar  \lambda_1  \lambda_3 (1 + \beta \vec{p}^2)
\quad , \nonumber\\ &&   [ \hat a_2 , \hat a_3 ]= - 2 i \hbar
\beta    ( \hat p_2  \hat x_1 -
\hat p_1  \hat x_2) \quad .
\end{eqnarray} 
 From the last expression, we see that such a state is in the kernel of the 
  momentum operator $\hat L_z \sim ( \hat
p_2  \hat x_1 - \hat p_1  \hat x_2)$. Combining this with the fact that it is in the
 kernel of the first
and the second commutators one obtains $\lambda_2 = \lambda_1 \lambda_3 =
0$. This is forbiden by the interpretation of the parameters since for example
\be
\label{eq11}
\lambda_2 = \frac{\hbar}{2(\Delta p_1)^2} \langle 1 + \beta \hat{\vec p}^2
 \rangle  \quad  . \ee 


\subsection{The K.M.M. 2 model}

  The second extension of the K.M.M model to higher dimensions is invariant under
rotations and translations. The non trivial commutation relations are  \cite{kmm1}:

\be
\label{eq12}
[ \hat x_j , \hat p_k] = i \hbar \left(  f(\hat{{\vec p}^2})  \delta_{j
k} + g(\hat{{\vec p}^2}) \hat p_j  \hat p_k \right) \quad , \ee 
and the condition 
\be
\label{eq13}
g = \frac{2 f f'}{f - 2 p^2 f'} \quad .
\ee
enforces the commutation of the positions among themselves. The same property holds for 
the momenta.  
The functions $f$ and $g$ are supposed positive. In this model, a position operator
 in one
direction has a non trivial commutation relation with  the momentum associated
to another direction:
\be  
\label{eq14} 
[ \hat x_1 , \hat p_2 ] = i h g(\hat{{\vec p}^2}) \hat p_1  \hat p_2 \quad .
\ee 

A representation of this extension is realized by the following formulas:
\be
\label{eq15}
\hat x_i = i \hbar \left( f' + {\vec p}^2  g'+ \frac{3}{2} g \right) p_i + f
\partial_{p_i} + g p_i p_j  \partial_{p_j} \quad , \quad \hat p_i = p_i \quad
, \quad \langle \phi \vert \psi \rangle = \int d^2 p \phi^*(p) \psi(p) \quad .
\ee 

This theory has four non trivial uncertainty relations concerning the
following couples of variables: $(x_1,p_1),(x_2,p_2),(x_1,p_2),(x_2,p_1)$. The
 corresponding operators are
\be
\label{eq16}
\hat a_1 = \hat x_1 + i \lambda \hat p_1 \quad , \quad 
\hat a_2 = \hat x_2 + i \mu \hat p_2 \quad , \quad 
\hat a_3 = \hat x_1 + i \tau \hat p_2 \quad , \quad 
\hat a_4 = \hat x_2 + i \sigma  \hat p_1 \quad , 
\ee
so that the relevant commutators, are, in this case


\begin{eqnarray}
\label{eq17}
&& [ \hat a_1 , \hat a_2 ] = \hbar g(\hat{{\vec p}^2}) ( - \mu \hat p_1 \hat
p_2 + \lambda \hat p_2 \hat p_1) \quad , \quad 
[ \hat a_1 , \hat a_3 ]= - \hbar \left( g(\hat{{\vec p}^2}) ( \tau \hat p_1
\hat p_2 - \lambda {\hat p_1}^2 ) - \lambda  f(\hat{{\vec p}^2})  \right) 
\quad , \quad  \nonumber\\
&&[ \hat a_1 , \hat a_4 ]= - \hbar \left( g(\hat{{\vec p}^2}) ( - \lambda \hat
p_2 \hat p_1 + \sigma {\hat p_1}^2 ) + \sigma  f(\hat{{\vec p}^2})  \right) 
\quad , \quad 
[ \hat a_2 , \hat a_3 ]= - \hbar \left( g(\hat{{\vec p}^2}) ( - \mu \hat
p_1 \hat p_2 + \tau {\hat p_2}^2 ) + \tau  f(\hat{{\vec p}^2})  \right) 
\quad , \quad  \nonumber\\
&&[ \hat a_2 , \hat a_4 ]= - \hbar \left( g(\hat{{\vec p}^2}) (  \sigma \hat
p_2 \hat p_1 - \mu {\hat p_2}^2 ) - \mu  f(\hat{{\vec p}^2})  \right) 
\quad , \quad 
[ \hat a_3 , \hat a_4 ]= - \hbar \left( g(\hat{{\vec p}^2}) (  \sigma {\hat
p_1}^2 -  \tau {\hat
p_2}^2 ) +  (\sigma - \tau) f(\hat{{\vec p}^2}) \right)  \quad .
\nonumber\\ 
\end{eqnarray} 

From now on, we use the explicit form of the operators given earlier.
The kernel of the first commutator is not empty; the equality $\lambda=
\mu$ allows a well behaved state which saturates simultaneously the
uncertainty relations concerning the couples of variables $(x_1,p_1)$ and
$(x_2,p_2)$. These states are the ones used by A.Kempf \cite{kmm2}, S.Detournay 
Cl.Gabriel and Ph.Spindel 
 \cite{spindel} while studying the existence of a minimal uncertainty
in length. We are interested in a
conceptually different problem: {\em in this work we put all the non trivial
commutation relations on the same footing.}

The
action of the second commutator on a state $\vert \psi \rangle$ vanihes 
only if the equation
\be
\label{eq18}
\left( ( \tau p_1 p_2 - \lambda p_1^2 ) - \lambda \frac{f({\vec p}^2)}{f({\vec
p}^2)} \right) \psi(\vec p) = 0  \ee 
is satisfied. Clearly this is the case only when the expression under
parantheses vanishes. As its second term is a function of $ {\vec p}^2$,
 one should have the same thing for its first term. This is possible
only if $\tau = \lambda=0 $, but  since 
\be  
\label{eq19} 
\lambda = \frac{\hbar}{2(\Delta p_1)^2} \int  d^2 p \psi^*(\vec p)
\left(f({\vec p}^2) + g({\vec
p}^2) p_1^2 \right) \psi(p)  \quad ,
\ee 
and the functions $f,g$ are positive, that is not possible. 

The third, the
fourth and the fifth commutators face a similar situation. The last commutator
can be divided by $g({\vec p}^2)$; the resulting expression 
vanishes if its first term is a function of ${\vec p}^2$, like its second
term. This happens if $\tau = - \sigma$ and then the commutator  reduces to
 to 
\be 
\label{eq20}     \sigma \left(  2 f({\vec
p}^2) + g({\vec p}^2) {\vec p}^2 \right)  \quad .
 \ee 
The expression under parentheses does not vanish because of our restrictions on
the signs of the two functions parameterizing the theory. One is led to
 $\tau = \sigma = 0$ which is not forbiden since the integral
\be
\label{eq21}
\int d^2 p \psi^*(p) g({\vec p}^2) p_1 p_2 \psi(p)
\ee 
can vanish. However, the considered state is then an eigenfunction of the
position operators; it may not be physical since  with very simple
assumptions on the functions $f,g$ which are compatible  with ours, there
is a minimal uncertainty in length \cite{spindel}.

In conclusion, in each of the three models presented, no state 
saturates all the non trivial uncertainties.




\section{The least square variational principle in the usual 1D case}
   
Let us consider a theory whose  operators are generically denoted $\hat
G_i$ and whose non trivial uncertainty relations can not be satisfied
simultaneously. Our proposal is that the states
which minimize the  sum 

\be 
\label{eq22} 
S_n=\sum_{j,k} \left( (\Delta  G_j)^{2 n} (\Delta  G_k)^{2 n} -  \frac{ \vert
\langle [\hat G_j, \hat G_k ] \rangle  \vert^{2 n}}{2^{2 n}} \right)^2 + \lambda
 (\langle \psi \vert \psi \rangle - 1 ) \quad ,
\ee 
for all couples of variables $(j,k)$ whose commutators do not vanish, are a 
possible generalization of the coherent states. The values of the index $n$ 
 in which we
will  be interested are $1/2$ and $1$. We shall come back to this point
later. It is important  that all  choices of $n$ lead to the same
differential equation when the action is varied, as we shall see. 

In the usual theory, $S$ simply
vanishes on the sqeezed states. 
The difficult point in our proposal lies in the fact that the variation of the
sum given in Eq.(\ref{eq22}) leads to a second order differential equation
which in principle admits more solutions.

 Let us show how this happens in the
simplest i.e undeformed theory. The action we consider is
\be
\label{eq23}
S = \left( \Delta x^2  \Delta p^2 - \frac{1}{4}  \right)^2 + \lambda \left(
1 - \langle \psi \vert \psi \rangle \right)  \quad ,
\ee 
where for simplicity we have set $\hbar=1$.
The last term  enforces the normalization of the
states trough the introduction of the Lagrange multiplier $\lambda$.

The definition of the uncertainty
\be 
\label{eq24} 
\Delta x^2 = \frac{\langle \psi \vert \hat x^2  \vert  \psi\rangle}{\langle
\psi \vert   \psi\rangle} - \left( \frac{\langle \psi \vert \hat x  \vert 
\psi\rangle}{\langle \psi \vert   \psi\rangle} \right)^2
\ee 
renders necessary the computation of  derivatives such as
\be
\label{eq25}
\frac{\delta}{\delta \psi^*(p)} \langle \psi \vert  \hat x \psi\rangle =
\frac{\delta}{\delta \psi^*(p)}  \int dq \, \psi^*(q) i \partial_q \psi(q) =
\int dq \, \delta(p-q) i \partial_q \psi(q) = i \partial_p \psi(p)   \quad . \ee 
Performing similar computations, we obtain the following
differential equation when varying the action $S$:

\be
\label{eq26}
 a   \psi^{''}(p) + i b \psi^{'}(p) + ( c p^2 + d p + e) \psi(p) = 0  \quad ,
\ee 
where the real constants $a,b,c,d,e$ are linked to the observables of the
solution by the relations

\begin{eqnarray}
\label{eq27}
a &=& - \left( \Delta x^2  \Delta p^2 - \frac{1}{4}  \right)\Delta p^2  \quad , 
\quad b = -2 \left( \Delta x^2  \Delta p^2 - \frac{1}{4}  \right) \Delta p^2 \langle \hat x \rangle
\quad , 
\quad c = \left( \Delta x^2  \Delta p^2 - \frac{1}{4}  \right)\Delta x^2  \quad , 
\nonumber\\
d &=& -2 \left( \Delta x^2  \Delta p^2 - \frac{1}{4}  \right) \Delta x^2 \langle
\hat p \rangle  \quad , \quad
 e =   \left( \Delta x^2  \Delta p^2 - \frac{1}{4}  \right)
 \left( - 2 \Delta p^2  \Delta x^2  +
\frac{1}{4} \right) + \lambda  \quad .
\end{eqnarray} 
The appearance of the observables of the solution such as $\langle \hat x \rangle,
\cdots $ in this equation is similar to the situation encountered in
Eq.(\ref{eq2}) i.e for usual coherent states.

Let us  verify that the well known coherent states 
satisfy  Eq.(\ref{eq26}). We thus look for a gaussian \cite{cohen}:
\be
\label{eq28}
\psi(p) = N \exp{( (z_1+ i z_2) p^2 + (z_3+i z_4) p)} \quad .
\ee 
It is readily seen that this state satisfies  Eq.(\ref{eq26}) if
 
 \be 
\label{eq29} 
z_1= \frac{i \sqrt{c}}{2 \sqrt{a}} \quad , \quad z_2= 0 \quad ; \quad z_3 = 
 \frac{i d}{2 \sqrt{a} \sqrt{c}} \quad , \quad z_4 = - \frac{b}{2 a} \quad ,
  \quad e = - \frac{b^2}{4 a} - 
 i \sqrt{a} \sqrt{c} + \frac{d^2}{4  c} \quad ,
\ee 
or
\be 
\label{eq30} 
z_1= - \frac{i \sqrt{c}}{2 \sqrt{a}} \quad , \quad z_2= 0 \quad ; \quad z_3 = 
- \frac{i d}{2 \sqrt{a} \sqrt{c}} \quad , \quad z_4 = - \frac{b}{2 a} \quad \quad e = - \frac{b^2}{4 a} + 
 i \sqrt{a} \sqrt{c} + \frac{d^2}{4  c} \quad . 
\ee 


Due to Eq.(\ref{eq27}), the $z_i$ are real. The normalisation of the wave function to
$1$  is achieved by an appropriate choice of the prefactor $N$. One finds

 \be
\label{eq31}
\Delta x^2 = \frac{ c^2}{2  a^2} \quad , \quad \Delta p^2 =
\frac{ a^2}{2  c^2} \quad  \Longrightarrow \Delta x \Delta p = \frac{1}{4}
\ee 
so that the action $S$ vanishes on this wave function. We have thus explicitely
verified that the usual coherent states  verify our second 
order differential equation.

Instead of simply looking for the conditions under which the known gaussians
obey the differential equation Eq.(\ref{eq26}), one may look for the
cases in which  this equation factorizes, leading to two first order operators
 $ k_1( k_2(\psi(p)))$. It is not difficult to show that this occurs in two cases:
\be
\label{eq32}
 e = - \frac{b^2}{4 a} + 
 i \sqrt{a} \sqrt{c} + \frac{d^2}{4  c} \quad , \quad
  k_{1,2}(\psi(p))= - \sqrt{a}\, \psi^{\prime}(p) +
   \left(- \frac{i b}{2 \sqrt{a}} \pm 
  \frac{i d}{2 \sqrt{c}} \pm i \sqrt{c} p \right) \psi(p) \quad ,                          
\ee
\be
\label{eq33}
 e = - \frac{b^2}{4 a} - 
 i \sqrt{a} \sqrt{c} + \frac{d^2}{4  c} \quad , \quad
  k_{1,2}(\psi(p))=  \sqrt{a}\, \psi^{\prime}(p) +
   \left( \frac{i b}{2 \sqrt{a}} \pm 
  \frac{i d}{2 \sqrt{c}} \pm i \sqrt{c} p \right) \psi(p) \quad .                          
\ee


The  conditions on the parameters $a,b,c,e$ displayed in the last part of 
Eq.(\ref{eq29},\ref{eq30}) coincide with the ones
obtained in 
the first part of Eq.(\ref{eq32},\ref{eq33}). They are  quite restrictive
and means that we are investigating only a part of the set of solutions. To find the
most general solution to Eq.(\ref{eq26}), let us make the following parametrization
\be 
\label{eq34} 
 \psi(p)= \exp{\left( \frac{- 4 i b p+ (-a)^{1/2} c^{-3/2} (d+2 c p)^2 }{8 a} \right)}
  (d+2 c p) X(q)  \quad ; \quad q= - \frac{1}{4}(- a)^{-1/2} c^{-3/2} (d+2 c p)^2 \quad.
\ee
To simplify the expressions, let us also introduce the quantity 
\be
\label{eq35}
\alpha= \frac{ -(-a)^{1/2} b^2 c +
 a (12 a c^{3/2}+ (-a)^{1/2}(d^2 - 4 c e ) )}{16 a^2 c^{3/2}} \quad .
\ee 
The differential equation now assumes the form
\be
\label{eq36}
X''(q) + \left(1 + \frac{3}{2 q} \right) X'(q) + \frac{\alpha}{q} X(q) = 0 \quad,
\ee
and the general solution can be   written as a sum of hypergeometric functions:
\be
\label{eq37}
X(q) = C_1 \,  q^{-1/2} \, \,   {_1} F_{1} \left( -\frac{1}{2}+ \alpha, \frac{1}{2},- q \right)+
 C_2 \, \, \,  {_1} F_1 \left( \alpha, \frac{3}{2},- q \right) \quad .
\ee

Using for the constant $e$ the expression displayed in Eq.(\ref{eq29}), the
first argument of the two hypergeometrics equals $1/2$. As the second
hypergeometric becomes a constant in this case, one recovers the special
solution given in Eq.(\ref{eq28}) by taking $C_1=0$.

Apart from the states which are known to have a vanishing
value of $S$, the formalism we have used introduces new
states. The fact that the states displayed
in Eq.(\ref{eq28}-\ref{eq30}) exhibit the absolute minimum of the action and form an over
complete set will be enough to discard the other states.


It should be stressed that normalizable solutions do not exist for all the values
of the parameters. In fact, the parameterization 
\be
\label{eq38}
\psi(p) = \exp{\left( - \frac{i b}{2 a} p \right)} u(p)
\ee
leads to the equation
\be
\label{eq39}
u''(p) - V(p) u(p) = 0 ,  \quad \rm{where} \quad 
 V(p) =  \left( \frac{c}{a} p^2 + \frac{d}{a} p + \frac{e}{a} + \frac{b^2}{4 a^2} 
  \right) \quad .
\ee
Integrating by parts and discarding the boundary term, one obtains the relation
\be
\label{eq40}
\int_{-\infty}^{+\infty}  \left( u'(p)^2 + V(p) u(p)^2 \right) = 0 \quad ,
\ee
which  can not be satisfied unless $V(p)$ admits two distinct zeros. This results 
in the following inequality:
\be
\label{eq41}
e >\frac{d^2}{4 c} - \frac{b^2}{4 a} \quad .
\ee
This  reasoning applies when the function $u(p)$ is real. If it is complex,
its real and imaginary parts obey  Eq.(\ref{eq39}) since its parameters are real and 
the same conclusion holds. 

 


To end this section, let us notice that the choice $n=1$ in
Eq.(\ref{eq23}) can be modified without changing drastically the situation. In
fact, trivial relations such as

\be  
\label{eq42}  
  \frac{\delta}{\delta \psi^{*}(p)} \Delta x = \frac{1}{2 \Delta x  }
\frac{\delta}{\delta \psi^{*}(p)} (\Delta x)^2  
\ee 
show that if we work with $n=1/2$, we will end up with  a
differential equation of the same form than  Eq.(\ref{eq26}). The "only" difference will
be encoded in a modified Eq.(\ref{eq27}) which gives the
link between the parameters $a,\cdots,e$ and the observables of the state. The
advantage of the first formula lies in the fact that  one does not have to
worry about a  possible division by zero like in Eq.(\ref{eq33}). In the usual
theory one knows that $\Delta x$ never vanishes on a physical state so that
the second choice does not bring in any new problem. We shall see that in
other theories one should be more carefull.

\section{High dimensional extensions}

\subsection{Generalized coherent states}
    
The generalization of the notion of coherent states in non commutative
    theories has been considered by many authors \cite{coh1,coh2,coh3,coh4}.
 Having a
     deformation of the
     usual commutation relations, one constructs operators
     obeying the relation
\be
\label{eq43}
[  \tilde a, \tilde a^{+} ] = F(\tilde a \tilde a^{+}) \quad .
\ee
The coherent states are then defined to be the eigenstates of the modified 
destruction operators:
\be
\label{eq44}
 \tilde a \vert \xi \rangle = \xi \vert \xi \rangle \quad .
 \ee  
The function $F$ is not arbitrary; when the deformed creation-destruction
operators can be constructed from the usual ones $a, a^{+}$ in the following way:
\be
\label{eq45}
\tilde a = f( \hat n +1) a  \quad , \quad n = a^{+} a \quad ,
\ee
 a link exists between the functions $f$ and $F$.

This approach is mathematically  consistent and useful in the sense that a 
decomposition of the unity is obtained quite easily and  many properties of the usual 
coherent states survive. However, the link with the 
saturation of the Heisenberg uncertainties is far from trivial. 
In this work we shall do the opposite: we will use the uncertainties
 as the primordial ingredients.


\subsection{The Non commutative plane}

 Let us define  the  variables $\hat X_k, \hat P_k$ 
 linked to the ones given in Eq.(\ref{eq3}) by the relations  
\be
\label{eq46}
\hat x_k = \sqrt{\theta} \hat X_k  \quad , \quad  \hat p_k =
\frac{\hbar}{\sqrt{\theta}} \hat P_k  \quad .
 \ee 
 The three non trivial
commutation relations lead to the dimensionless  action
\be 
\label{eq47} 
\tilde S = \left( \Delta X_1 \Delta X_2 - \frac{1}{2}  \right)^2 + \left(
\Delta X_1 \Delta P_1 - \frac{1}{2}  \right)^2 +  \left( \Delta X_2 \Delta P_2
- \frac{1}{2} \right)^2  + \tilde \lambda \left( \langle \psi \vert \psi
\rangle  - 1\right)  \quad . 
\ee

Multiplying by an overall factor, and going back to the former variables, we
obtain the action which will be used in this subsection:
 \be
\label{eq48}
S=  \hbar^2 \left( \Delta x_1 \Delta x_2 - \frac{1}{2} \theta
\right)^2 + \theta^2 \left( \Delta x_1 \Delta p_1 - \frac{1}{2}  \hbar
\right)^2 +  \theta^2 \left( \Delta x_2 \Delta p_2 - \frac{1}{2} 
\hbar \right)^2  + \lambda \left( \langle \psi \vert \psi
\rangle  - 1\right)  \quad .  
\ee 

Two remarks are in order at this point. First, the change of variables
performed in Eq.(\ref{eq46}) is not well defined in the limit $\theta
\longrightarrow 0$ and the action $S$ does not coincide with two copies of the
one dimensional case displayed in Eq.(\ref{eq23}). However, the
differential equation it generates has that good property. We will come back to this
point in the appendix.
The best states would be those which,
being extrema of the action given in Eq.(\ref{eq48}), approach  the well known
coherent states when the deformation parameter goes to zero. The second remark
is that from a dimensional analysis, one could  guess the form of the
terms appearing in Eq.(\ref{eq48}) but not their relative weights. The
introduction of generic dimensionless variables and the adoption of the
democratic rule displayed in Eq.(\ref{eq47}) fixes in an anambigous way the
action displayed in Eq.(\ref{eq48}). Introducing
constant factors to balance the different terms  leads to the same
differential equation.

The discussion of section $2$ has shown that  there is no state on which 
the action vanishes. We now look for those on which it is extremal. Varying
the action we obtain the field equation

\be
\label{eq49}
\frac{\delta S}{\delta \psi^*(p)} = b_1 \frac{\delta }{\delta \psi^*(p)}
\Delta x_1 + b_2 \frac{\delta }{\delta \psi^*(p)} \Delta x_2 + b_3 \frac{\delta
}{\delta \psi^*(p)} \Delta p_1 + b_4 \frac{\delta }{\delta \psi^*(p)} \Delta
p_2+ \lambda \frac{\delta }{\delta \psi^*(p)}  \langle \psi \vert \psi
\rangle = 0  \quad ,
\ee 
where the coeficients $b_i$ are given by the following expressions


\begin{eqnarray}
\label{eq50}
b_1 &=& 2 \hbar^2  \left( \Delta x_1 \Delta x_2 - \frac{1}{2} \theta
\right)  \Delta x_2 + 2 \theta^2  \left( \Delta x_1 \Delta p_1 - \frac{1}{2}
\hbar \right)  \Delta p_1  \quad , \nonumber\\
 b_2 &=& 2 \hbar^2  \left( \Delta x_1 \Delta x_2 - \frac{1}{2} \theta
\right)  \Delta x_1 + 2 \theta^2  \left( \Delta x_2 \Delta p_2 - \frac{1}{2}
\hbar \right)  \Delta p_2  \quad , \nonumber\\
b_3 &=& 2 \theta^2  \left( \Delta x_1 \Delta p_1 - \frac{1}{2}
\hbar \right)  \Delta x_1 \quad , \quad  
b_4 = 2 \theta^2  \left( \Delta x_2 \Delta p_2 - \frac{1}{2}
\hbar \right)  \Delta x_2 \quad .   
\end{eqnarray} 

Intermediate results such as
\be
\label{eq51}
\frac{\delta }{\delta \psi^*(p)} \Delta x_1^2 = \left[ - \hbar^2
\partial^2_{p_1} - i ( \theta p_2 + 2 \hbar \langle \hat x_1 \rangle)
\partial_{p_1}  + \left(  \frac{1}{4} \frac{\theta^2}{\hbar^2}  p_2^2 +
\frac{\theta}{\hbar} \langle \hat x_1 \rangle p_2 + \langle \hat x_1 \rangle^2
- (\Delta x_1)^2 \right) \right] \psi(p) \quad ,
\ee 


\be
\label{eq52}
\frac{\delta }{\delta \psi^*(p)} \Delta p_1^2 = \left( p_1^2 - 2 \langle
\hat p_1\rangle  p_1 + \langle
\hat p_1\rangle^2 - (\Delta p_1)^2 \right) \psi(p) \quad ,
\ee 

are necessary to recast  the equation in the following form

\be 
\label{eq53} 
\left[ \bar{a}_1 \partial^2_{p_1} + \bar{a}_2 \partial^2_{p_2} + i  \left( \bar{a}_1
\frac{\theta}{\hbar^2}  p_2 + a_3 \right)
\partial_{p_1}  + i  \left( -\frac{\theta}{\hbar^2} \bar{a}_2 p_1
+ a_4 \right) \partial_{p_2} + ( \bar{a}_5 p_1^2 + \bar{a}_6 p_2^2
+ a_7 p_1 + a_8 p_2 + a_9 ) \right] \psi(p) = 0  \quad . 
\ee  
The real numbers
$a_i$ are linked to the previous coefficients $b_i$ by the following set of
relations

\begin{eqnarray}
\label{eq54}
\bar{a}_1&=& - \hbar^2 \frac{b_1}{2 \Delta x_1} \quad , \quad \bar{a}_2 = - \hbar^2
\frac{b_2}{2 \Delta x_2} \quad , \quad  
a_3 = - \hbar \langle \hat x_1\rangle \frac{b_1}{ \Delta x_1}  \quad ,
\quad a_4 =   - \hbar \langle \hat x_2\rangle \frac{b_2}{ \Delta x_2}
\quad , \nonumber\\   \bar{a}_5 &=& \frac{1}{8} \frac{\theta^2}{\hbar^2}
\frac{b_2}{\Delta x_2} + \frac{b_3}{2 \Delta x_3} \quad , \quad 
\bar{a}_6= \frac{1}{8} \frac{\theta^2}{\hbar^2} \frac{b_1}{\Delta x_1} +
\frac{b_4}{2 \Delta p_2} \quad , \nonumber\\
&a_7& = -\frac{\theta}{\hbar} \langle \hat x_2 \rangle \frac{b_2}{2 \Delta
x_2} - \langle \hat p_1 \rangle \frac{b_3}{2 \Delta p_1} \quad , \quad 
a_8 = -\frac{\theta}{\hbar} \langle \hat x_1 \rangle \frac{b_1}{2 \Delta
x_1} - \langle \hat p_2 \rangle \frac{b_4}{ \Delta p_2} \quad , \nonumber\\
&a_9& = \left( \langle \hat x_1 \rangle^2 - (\Delta x_1)^2  \right)
\frac{b_1}{2 \Delta x_1} + 
\left( \langle \hat x_2 \rangle^2 - (\Delta x_2)^2
 \right) \frac{b_2}{2 \Delta x_2} + 
\left( \langle \hat p_1 \rangle^2 - (\Delta p_1)^2  \right)
\frac{b_3}{2 \Delta p_1}  \nonumber\\
& & +  
\left( \langle \hat p_2 \rangle^2 - (\Delta p_2)^2  \right)
\frac{b_4}{2 \Delta p_2} + \lambda  \quad .
\end{eqnarray} 

To summerize,  states which are  extrema of the action given in
Eq.(\ref{eq48})  satisfy the second order partial differential equation
displayed in Eq.(\ref{eq53}). The real numerical factors appearing in this 
equation are linked to the observables of these state-solutions by the two sets
of relations given in Eqs.(\ref{eq50},\ref{eq54}).

It is obvious that the coefficients $b_i$ are positive. As a consequence, one
has
\be
\label{eq55}
\bar{a}_1 , \bar{a}_2  \leq 0  \quad , \quad \bar{a}_5 , \bar{a}_6  \geq 0  \quad .
\ee 
From  section $2$ we know that $(\bar{a}_1,\bar{a}_2) \neq (0,0)$ i.e at
 least one of
these two quantities must be strictly negative. Similarly, one of the two
quantities $\bar{a}_5,\bar{a}_6$ must be strictly positive.

We don't have the most general solution to the second order partial
differential equation  we have obtained. We shall look for
special cases in which explicit solutions can be found. 

To begin, let us take into account the signs of the coefficients by the following 
parametrization:
\newline  $\bar{a}_1= - a_1^2 , \bar{a}_2= - a_2^2  , 
\bar{a}_5=  a_5^2  , \bar{a}_6=  a_6^2$. To simplify the formulas, 
we  take $\hbar = \theta = 1.$ Introducing the variables $y_1,y_2$ and the function
$\phi$ by the relations
\be
\label{eq56}
p_1 = a_1 y_1 - \frac{a_4}{a_2^2} \quad , \quad p_2 = a_2 y_2 + \frac{a_3}{a_1^2} 
\quad , \quad  \phi(y_1,y_2) = \psi(p_1,p_2) \quad ,
\ee
the differential equation take the simpler form
\be
\label{eq57}
\left[ - \partial^2_{y_1} -  \partial^2_{y_2} +
 i t_1^2 ( - y_2 \partial_{y_1} + y_1 \partial_{y_2}) +
  ( t_2^2  y_1^2 + t_3^2   y_2^2
+ t_4  y_1 + t_5  y_2 + t_6 ) \right] \phi(y) = 0  \quad , 
\ee
where
\begin{eqnarray}
\label{eq58}
t_1^2 &=& a_1 a_2 \quad , \quad t_2^2= a_1^2 a_5^2 \quad , \quad t_3^2= a_2^2 a_6^2 
\quad , \quad t_4 = \left( - 2 \frac{a_5^2}{a_2^2} a_4 + a_7 \right) a_1 \quad ,
\quad  t_5 = \left(  2 \frac{a_6^2}{a_1^2} a_3 + a_8 \right) a_2 \quad , \nonumber\\
t_6&=& a_9 + \left( \frac{a_4 a_5}{a_2^2} \right)^2 -   \frac{a_4 a_7}{a_2^2} +
\left( \frac{a_3 a_6}{a_1^2} \right)^2 +   \frac{a_3 a_8}{a_1^2} \quad .
\end{eqnarray}


\subsubsection{The gaussian solution}
  
Similarly to the one dimensional case, one can, {\em in this case}, postulate a
solution which is the exponential of a quadratic function:
 \be 
\label{eq59}
\psi(p) = N \exp{( c_1 y_1^2 + c_2 y_2^2 +c_3 y_1 y_2 + c_4 y_1 + c_5 y_2)}
\quad .
 \ee 
Plugging this wave function in the differential equation, one ends up with the relations
\begin{eqnarray}  
\label{eq60}  
c_3 &=& \frac{t_1^2}{4(t_2^2-t_3^2)} 
\left( \sqrt{-(t_1^2 - 4 t_2^2)(t_1^2 - 4 t_3^2)} +
 i ( t_1^4 - 2( t_2^2+t_3^2)) \right)  \quad , 
 \quad c_1 = \frac{1}{2} \sqrt{- c_3^2 - i c_3 t_1^2+ t_2^2} \quad , \nonumber\\ 
c_2 &=&  \frac{1}{2} \sqrt{- c_3^2 - i c_3 t_1^2+ t_3^2} \quad , \quad 
c_5= \frac{2 c_3 t_4 + i t_1^2 t_4 - 4 c_1 t_5}{16 c_1 c_2 - 4 c_3^2 - t_1^4}
 \quad , \quad c_4= - \frac{2 c_3 c_5 - i c_5 t_1^2 - t_4}{4 c_1} \quad , \nonumber\\
 & & 2 c_1 + 2 c_2 + c_4^2 + c_5^2 - t_6 = 0 \quad .
\end{eqnarray} 
Each coefficient has been  written solely in terms of those
appearing before it in the list and the first one ($c_3$) depends only on the
 parameters of the equation. The last formula of Eq.(\ref{eq60}) is a constraint
  which has to be satisfied for the
differential equation   to admit a gaussiann solution; this is reminiscent of 
Eqs.(\ref{eq29},\ref{eq30}) in the usual one dimensional case. 

The normalizability of the state  
imposes conditions on the real parts of $c_1,c_2,c_3$.


It should  be noted that when the parameters $\hbar,\theta$ are restored, the 
coefficient   $c_3$ of the term $p_1 p_2$
vanishes in the limit $\theta \rightarrow 0$, so that the  usual separation of
the variables $p_1,p_2$ is recovered.  



\subsubsection{The cylindrically symmetric solution}

In the preceding subsection we saw that a carefully chosen relation between the
coefficients of the differential equation led to an explicit solution.
Altough in the usual one dimensional case  such a choice
was ultimately justified because it led to the absolute minimum of the action, nothing like
that occurs here. One has to ressort to a second order analysis to verify the true
 nature of the critical points represented by the states obtained so far. That will
  not be  done here; we simply find some explicit solutions. 

 There is a set of simple relations between the parameters of the equation
  which allows a cylindrical separation of
  variables. In fact, if 
\be
\label{eq61}
t_3 = t_2 \quad , \quad t_4 = t_5 = 0  \quad  ,
\ee
the parameterization $\phi = R(r) e^{i m \theta} $ in the polar coordinates linked to
the cartesian coordinates $(y_1,y_2)$ leads to the ordinary differential equation

\be
\label{eq62}
R''(r) + \frac{1}{r} R'(r) + \left( \frac{m^2}{r^2} + (t_6 - m t_1^2) + t_2^2 r^2
  \right) R(r) = 0 \quad .
\ee
In the special case where the relation $t_6 = m t_1^2$ holds, the solution can be 
recasted as a combination of two Bessel functions:
\be
\label{eq63}
R(r) = C_1  I_{- \frac{m}{2}} \left(\frac{t_2 r^2}{2} \right) +
 C_2 I_{ \frac{m}{2}} \left(\frac{t_2 r^2}{2}  \right) \quad .
\ee



 
A polar separation of variables in the original coordinates 
$(p_1,p_2)$ requires more relations then Eq.(\ref{eq62}):
\be
\label{eq64}
a_1 = a_2 \quad , \quad  a_3 = a_4 = a_7 = a_8=0 \quad , a_5 = a_6   \quad .
\ee 


\subsubsection{A third explicit solution}
  
  Let us consider the case 
  \be
  \label{eq65}
  (t_1,t_3,t_4,t_5,t_6) = (2,2,0,0,-4) \quad , \quad {\rm with} \quad  t_2 \quad
   {\rm arbitrary} \quad .
  \ee
  It is straightforward to verify that for any integer $m$ and any 
   complex constants $c_{01},c_{11}$, the function
\be
\label{eq66}
\phi(r,\theta) = r^{m} \exp{ \left[ - c_{01} e^{i \theta} r - r^2 \left(1 + 
\frac{1}{4} c_{11} e^{2 i \theta} \right) + i m \theta \right] }
\ee
is a solution of Eq.(\ref{eq57}).

\subsubsection{Factorizability}

 As in the usual case, we can look for the particular conditions  under which the
 second order differential operator can be written as a product. This is found to 
 occur  when

\be
\label{eq67}
t_2 = t_3 = \frac{t_1^2}{2} \quad , \quad t_6 =  t_1^2  \quad , \quad t_4 = t_5 = 0 
\quad .
\ee
The operators $k_1,k_2$ assume the forms
\be
\label{eq68}
k_1 = \left(  i \partial_{y_1} + \partial_{y_2} - 
\frac{i}{2}  t_1^2 (y_1 - i y_2 ) \right)  \quad , 
\quad k_2 = \left(  i \partial_{y_1} - \partial_{y_2} + 
\frac{i}{2}  t_1^2 (y_1 + i y_2 ) \right)  \quad .
\ee

This case of reducibility forms a particular subset which is at the intersection of the
ones corresponding to gaussian solutions and the ones displaying  polar symmetry; it 
does not bring in new solutions.  


\subsection{The K.M.M 1 model}

\subsubsection{The equation}
  
   A reasoning similar to the one used in the previous section leads to the
introduction of the dimensionless phase space variables $\hat X_k, \hat P_k$
given by the relations

\be
\label{eq69}
\hat x_j = \sqrt{2 \beta} \hbar \hat X_j  \quad ; \quad  \hat p_k =
\frac{1}{\sqrt{2 \beta}} \hat P_k \quad .
 \ee 

The action built from the non trivial uncertainty relations, with the choice
$n=1$, multiplied by an overall factor and rewritten in the former variables
reads

\begin{eqnarray}
\label{eq70}
S &=& \frac{1}{16} \left( \Delta x_1^2 \Delta x_2^2 - \hbar^2 \beta^2  \vert
\langle \hat p_1 \hat x_2 - \hat p_2 \hat x_1  \rangle \vert^2  \right)^2
+ \left( \hbar^2 \beta^2 \Delta x_1^2 \Delta p_1^2 - \frac{1}{4} \hbar^4
\beta^2 \langle 1 + \beta \hat{\vec{p}}^2 \rangle^2  \right)^2  \nonumber\\
&+& \left( \hbar^2 \beta^2 \Delta x_2^2 \Delta p_2^2 - \frac{1}{4} \hbar^4
\beta^2 \langle 1 + \beta \hat{\vec{p}}^2 \rangle^2  \right)^2 
+ \lambda (-1 + \langle \psi \vert \psi \rangle) \quad .
\end{eqnarray} 

It should be noticed that our prescription led to a factor $1/16$ in the first
term. An important difference with the first theory we studied is the non
trivial measure needed for the scalar product and the form of
the operators given in Eq.(\ref{eq9}). For example, the equalities

\be
\label{eq71}
\langle \psi \vert \psi \rangle = \int d^2 q  \frac{\psi^*(q) \, 
\psi(q)}{1+\beta \vec{q}^2} \quad {\rm and } \quad \langle \psi \vert \hat
x_1^2 \vert \psi \rangle = -\hbar^2 \int d^2 q \, \psi^{*}(q) \partial_{q_1}
(1+ \beta \vec{q}^2 )\partial_{q_1} \psi(q)  \ee 
lead to the functional derivatives
\be
\label{eq72}
\frac{\delta}{\delta \psi^*(p)} \langle \psi \vert \psi \rangle = \frac{\psi(p)}{1+\beta
\vec{p}^2} \quad , \quad
\frac{\delta}{\delta \psi^*(p)} \langle \psi \vert \hat x_1^2 \vert \psi
\rangle = - \hbar^2 (1 + \beta \vec{p}^2) \partial^2_{p_1} \psi(p) - 2 \beta
\hbar^2 p_1 \partial{p_1} \psi(p) \quad .
 \ee 
If one makes the choice $n=1/2$, one has to compute the derivative of
the absolute value of the commutator of $\hat x_1$ and $\hat x_2$. To get rid of
this absolute value, we can use a trick such as
 \be 
\label{eq73}
 \frac{\delta}{\delta \psi^\star(p)} \vert \langle \hat p_1 \hat x_2 - \hat p_2 \hat
x_1 \rangle \vert = \frac{1}{\vert \langle \hat p_1 \hat x_2 - \hat p_2 \hat
x_1 \rangle \vert} \frac{\delta}{\delta \psi^\star(p)} \left( \langle \hat p_1 \hat
x_2 - \hat p_2 \hat x_1 \rangle  \langle \hat p_1 \hat x_2 - \hat p_2 \hat x_1
\rangle^{*} \right)  \quad ,
\ee 
but then we have to restrict {\em from the start} the set of states on which we
are working to avoid a possible division by zero. Our choice rather leads  to

\be 
\label{eq74}
 \frac{\delta}{\delta \psi^\star(p)} \vert \langle \hat p_1 \hat x_2 - \hat p_2 \hat
x_1 \rangle \vert^2 = \frac{\delta}{\delta \psi^\star(p)} \left( \langle \hat p_1 \hat
x_2 - \hat p_2 \hat x_1 \rangle  \langle \hat p_1 \hat x_2 - \hat p_2 \hat x_1
\rangle^{*} \right)  \quad .
\ee 
We give the details of the derivation in the appendix. The differential
equation associated  to this model reads:
 \begin{eqnarray}
\label{eq75}
 & ( & 1+ \beta \vec{p}^2 )^2 \left( \bar a_1 \partial^2_{p_1} + \bar a_2
\partial^2_{p_2} \right) \psi(p) + (1+ \beta \vec{p}^2 ) \left[ ( 2 \beta \bar a_1
p_1 + i a_3 p_2 + i a_4 ) \partial_{p_1} +  ( 2 \beta \bar a_2 p_2 - i a_3 p_1 + i
a_5 ) \partial_{p_2} \right] \psi(p) \nonumber\\
&+& (a_6 p_1^2 +a_7 p_2^2 + a_8 p_1 + a_9 p_2 + a_{10}) \psi(p) = 0 \quad ,
\end{eqnarray} 
the $a_i$ are real constants. Only two signs are fixed: $\bar a_1, \bar a_2 \leq 0$. To simplify
future formulas, we introduce two positive numbers $(a_1,a_2)$ by the  relations 
$\bar a_1 = - a_1^2 , \bar a_2 = - a_2^2 $.

Let us briefly go back to the one dimensional model from which this one is built. This 
is achieved by discarding the variable $p_2$ everywhere; for commodity we afterwards
rename $p_1$ as $p$. The remaining coefficients are $ a_1,a_4,a_6,a_8$ and $a_{10}$.
 For the
particular choice  
\be 
\label{eq76} 
a_6 = a_8 = 0 \quad , \quad a_4 =  2  a_1^2  \xi \quad , 
\quad a_{10}=  a_1^2( -1 + \xi^2) 
\quad ,
\ee 
one recaptures the solution
\be
\label{eq77}
\psi(p) = \frac{1}{\sqrt{1+p^2}} e^{ i \xi \arctan{p}}
\ee
which was obtained using the usual first order equation \cite{kmm2}. With different relations
between the parameters, one obtains hypergeometric functions.

\subsubsection{The symmetric case}
   
It can be easily verified that if the relations
 $a_1 = a_2 ,a_6=a_7,a_3=a_4=a_5=a_8=a_9=0$ hold, the function 
\be
\label{eq78}
\psi(p) = (p_1^2+p_2^2)^{\frac{c_{00}}{2}} (1+p_1^2+p_2^2)^{\frac{- c_{00}+c_{20}}{2}}
\cos{ \left[ k - c_{00} \arccos{ \left( \frac{p_1}{\sqrt{p_1^2+p_2^2}} \right)}  \right]}
\ee
is a solution of Eq.(\ref{eq75}) the quantities $c_{00},c_{20}$ being given by

\begin{eqnarray}
\label{eq79}
 c_{00} &=& \mp \frac{a_1^2(a_{10} - 2 a_6) + \sqrt{ - a_1^2 ( 2 a_1^2 + a_{10})^2 (a_{10} - a_6)}} 
{ 2 a_1^2 (2 a_1^2 +a_{10})(a_{10}^2+a_{10}-a_6)} \quad , \nonumber\\
c_{20} &=& \frac{( - 2 a_1^4 - a_1^2 a_{10} \pm \sqrt{ - a_1^2 ( 2 a_1^2 + a_{10})^2 (a_{10} - a_6) })
(a_{10} - 2 a_6)}{2 a_1^2 (2 a_1^2+a_{10})(a_1^2+a_{10} - a_6)} \quad , \nonumber\\
\end{eqnarray} 
$k$ being arbitrary. This state is interesting when $c_{20}$ is real and negative enough.

\subsubsection{Factorizability}

The conditions under which the second order operator appearing 
in Eq.(\ref{eq75}) can be written as the product of two first order operators are the
 following:
 \begin{eqnarray}
 \label{eq80}
 a_6 &=& \frac{( 2 a_1 a_2 + a_3 ) (- 2 a_2^3 + a_1 a_3 ) }{4 a_1 a_2^2}
  \quad , \quad a_7= \frac{( 2 a_1 a_2 + a_3 ) (- 2 a_1^3 + a_2 a_3 ) }{4 a_1^2 a_2} 
  \quad ,  \nonumber\\
a_8 &=& - \frac{(2 a_1 a_2 + a_3) a_5}{2 a_2^2} \quad , \quad
a_9 = \frac{(2 a_1 a_2 + a_3) a_4}{2 a_1^2} \quad , \quad
a_{10} = \frac{ a_2( - 2 a_1 (a_1^2 + a_2^2 )( 2 a_1 a_2 + a_3 ) + a_2 a_4^2 ) +
 a_1^2 a_5^2 }{4 a_1^2 a_2^2} \quad .
\end{eqnarray}

The two operators assume the following  forms:
\be
\label{eq81}
k_1 = a_1 (1+p_1^2+p_2^2) \partial_{p_1} + Z_1 \partial_{p_2} + Z_2 \quad , \quad
k_2 = a_2 (1+p_1^2+p_2^2) \partial_{p_1} + Z_3 \partial_{p_2} + Z_4 \quad ,
\ee
where
\begin{eqnarray}
\label{eq82}
Z_1 &=& i a_2 (1+p_1^2+p_2^2) \quad , \quad Z_3 = - Z_1 \quad , \nonumber\\
Z_2 &=& \frac{1}{2 a_1^2 a_2^2} (- i a_2^2 ( a_4 - 2 i a_1^2 p_1 + a_3 p_2 ) a_1 +
 a_1^2 (a_5 - a_3 p_1 - 2 i  a_2^2 p_2 ) a_2 ) \quad , \nonumber\\
 Z_4 &=& \frac{1}{2 a_1^2 a_2^2} (- i a_2^2 ( a_4 + 2 i  a_1^2 p_1 + a_3 p_2 ) a_1 +
 a_1^2 (- a_5 + a_3 p_1 - 2 i  a_2^2 p_2 ) a_2 ) \quad .
\end{eqnarray}

When the factorizability conditions apply, one can find a family of solutions by 
solving the first order equation  $k_2 \psi(p) = 0$. Going to the variables $v_1,v_2$
defined by
\be
\label{eq83}
v_1 = \frac{2}{a_1^2 - a_2^2} ( a_1 p_1 - i a_2 p_2 ) \quad , \quad
v_2 = \frac{p_1}{a_1}  - i \frac{p_2}{a_2}  \quad ,
\ee
this differential equation reduces to 
\be
\label{eq84}
\frac{\partial \psi}{\partial v_1} + 
\left( \frac{\gamma + \delta v_1}{\epsilon v_1^2 + \nu} \right) \psi = 0 \quad ,
\ee
where $\gamma,\delta$ and $\epsilon$ are constants while $\nu$ is quadratic in $u_2$:
 \be
 \label{eq85}
 \gamma = -  \frac{a_1 a_5 + i a_2 a_4}{2 a_1 a_2} \quad , \quad 
 \delta = \frac{(a_1^2 - a_2^2)(2 a_1 a_2 + a_3)}{4 a_1 a_2} \quad , \quad 
 \epsilon = \frac{1}{2}(a_1^2-a_2^2) \quad ,\quad
 \nu = 2 + \frac{2 a_1^2 a_2^2 v_2^2}{-a_1^2+a_2^2} \quad .
 \ee
The solution to the differential equation can formally be written as
\be
\label{eq86}
\psi = \exp{ (g(v_2)) + \left[ - \frac{i \gamma}{2 \sqrt{\epsilon \nu}}  \left( \log{\left( v_1 + \frac{i}{\sqrt{\epsilon}} \sqrt{\nu}  \right)} - 
 \log{\left( v_1 - \frac{i}{\sqrt{\epsilon}} \sqrt{\nu}  \right)} \right)
 - \frac{\delta}{2 \epsilon} \log{\left( v_1^2 + \frac{\nu}{\epsilon} \right)} \right] } \quad .
\ee
Let us now go back to the original variables and let us also separate the real and
 imaginary part of this function. From now on, we impose $a_1 > a_2$; a similar treatment
 can be performed in the case $a_1 < a_2$. The final result can be recorded as
\be
\label{eq87}
\psi(p) = \exp{\left( g\left(\frac{p_1}{a_1} - i \frac{p_2}{a_2} \right) \right)}
 \left(\frac{L_1}{L_2}\right)^{X_1} L_3^{X_3}  
 \exp{( - X_2 (A_1-A_2))}
 \exp{ \left[i \left(  X_1 ( A_1 - A_2 )+ X_2 \log{\left(\frac{L_1}{L_2}\right)} +
  X_3 A_3 \right) \right] } \quad ,
\ee
with
\begin{eqnarray}
\label{eq88}
L_{1,2} &=& \sqrt{( ( \mp B (a_1^2 - a_2^2) + 2 a_1 p_1 )^2 +  
( \pm A( - a_1^2 + a_2^2) + 2 a_2 p_2 )^2)}
\quad , \quad
A_{1,2} = \arctan{ \left( \frac{A(- a_1^2 + a_2^2) \pm 2 a_2 p_2}{B(a_1^2 - a_2^2) 
\mp 2 a_1 p_1} \right)} \quad , 
\nonumber\\
L_3 &=& \frac{ 
 (2 A B (a_1^2 - a_2^2)^2 - 8 a_1 a_2 p_1 p_2 )^2 + 
( (A^2-B^2) (a_1^2-a_2^2)^2 + 4 (a_1^2 p_1^2 - a_2^2 p_2^2)  )^2 }{a_1^2-a_2^2} 
\quad , \nonumber\\
A_3 &=& \arctan{\left( \frac{ (2 A B (a_1^2 - a_2^2)^2 - 8 a_1 a_2 p_1 p_2 )}{(A^2-B^2)
 (a_1^2-a_2^2)^2 + 4 (a_1^2 p_1^2 - a_2^2 p_2^2)}\right) } \quad , \nonumber\\
X_1 &=& \frac{- A a_2 a_4+ B a_1 a_5}{4 a_1 a_2 (a_1^2-a_2^2)(A^2+B^2)} \quad , \quad
X_2 = \frac{A a_1 a_5+ B a_2 a_4}{4 a_1 a_2 (a_1^2-a_2^2)(A^2+B^2)} \quad ,
X_3 = - \frac{2 a_1 a_2+ a_3}{8 a_1 a_2} \quad ;
\end{eqnarray}
$A$ and $B$, being the real and imaginary part of $\sqrt{\frac{\nu}{\epsilon}}$,
can be expressed as
\begin{eqnarray}
\label{eq89}
A &=& \frac{y}{\sqrt{2} \sqrt{- x + \sqrt{x^2+y^2}}} \quad , \quad  
B = \frac{1}{\sqrt{2}} \sqrt{- x + \sqrt{x^2+y^2}} \quad    , \nonumber\\
\quad {\rm with} \quad x &=& 4  \frac{a_1^2 - a_2^2 -  a_2^2 p_1^2+ a_1^2 p_2^2}{(a_1^2-a_2^2)^2} \quad ,
\quad y= \frac{8 a_1 a_2 p_1 p_2}{(a_1^2-a_2^2)^2} \quad .
\end{eqnarray}
There exist a  parameter space and choices of the arbitrary function $g$
 for which the obtained states are  viable. 

\subsection{The K.M.M 2 model}

We now restrict to the second extension of the KMM model corresponding to the following
choice of the functions appearing in Eq.(\ref{eq12}): 
\be
\label{eq90}
 g(\vec{p}^2) = \beta \quad , \quad  f(\vec{p}^2) = \frac{ \beta \vec{p}^2}{-1+
  \sqrt{1+2 
  \beta \vec{p}^2}} \quad .
\ee
There exists, in this case , a representation in which the operators look much simpler
 than in Eq.(\ref{eq15}):
 \be
 \label{91}
 \hat x_i = i \hbar \partial_{\rho_i} \quad , \quad p_i =
  \frac{\rho_i}{1-\frac{1}{2} \beta \vec{\rho}^2} \quad ;
 \ee
 the same scalar product is the usual one, but now it is defined on the disc of radius
 $\sqrt{\frac{2}{\beta}}$.
 
The couples wich enter the action $S$ are now $(x_1,p_1),(x_1,p_2),(x_2,p_1)$
and $(x_2,p_2)$. The reasoning we have used in the preceding sections lead to the
final equation
\begin{eqnarray}
\label{eq92}
0&= & \left[ \left( 1- \frac{1}{2} \beta \vec{\rho}^2 \right)^2 ( -a_1^2 \partial_{\rho_1}^2- 
a_2^2 \partial_{\rho_2}^2) + \frac{2 i}{\hbar} 
\left( 1- \frac{1}{2} \beta \vec{\rho}^2 \right)
(- a_1^2 \langle \hat x_1\rangle \partial_{\rho_1}  -
 a_2^2 \langle \hat x_2\rangle \partial_{\rho_2}) \right. \nonumber\\
&+& \left.  a_3 \rho_1^2 + a_4 \rho_2^2 + a_5 \rho_1 \rho_2 + 
(a_6 \rho_1+a_7 \rho_2) \left( 1- \frac{1}{2} \beta \vec{\rho}^2 \right)+
a_8 \left( 1- \frac{1}{2} \beta \vec{\rho}^2 \right)^2 + \lambda \right] \psi \quad .
\end{eqnarray}

 
The symmetric case $ a_2 = a_1 \quad , \quad \langle x_1 \rangle =  \langle x_2 \rangle = 0 \quad , \quad
a_3 = a_4 = a_5 = a_6 = a_7 = \lambda = 0 $ admits a solution which
is a combination of Bessel functions multiplied by phases:
 \be
\label{eq93}
\psi(\rho) = \left( C_1 I_{-m} \left(\frac{\sqrt{- a_8}}{a_1} \rho \right)+ 
 C_2 I_{m} \left(\frac{\sqrt{- a_8}}{a_1} \rho \right) \right) e^{i m \theta} \quad .
\ee 


\section{Periodicity of the harmonic oscillator}

Coherent states, in the usual case, are not only the states which minimize the
the action $S$ of Eq.(\ref{eq23}); they also reproduce the classical behavior
of the harmonic oscillator. In this section we will interest ourselves to  the mean values of the positions and momenta of
a harmonic oscillator in the theories under study. In the usual quantum theory, any 
state which is a solution of the Schrodinger equation for the harmonic oscillator
displays periodic positions and momenta; moreover the period coincides exactly with the
one found in classical mechanics \cite{cohen}. We will see that this is generically not the case
when non commutativity sets in.

\subsection{The non commutative plane}

Working in the
 Heisenberg picture, the states are time independent while the operators vary according
 to the evolution  equation 
\be
\label{eq94}
\left(
 \begin{array}{c}
  \dot{\hat x}_1   \\ \dot{\hat x}_2 \\ \dot{\hat p}_1 \\  \dot{\hat p}_2  
\end{array}  
\right) = 
\left(
 \begin{array}{cccc}
  0                         & \frac{k \theta}{\hbar} & \frac{1}{m} & 0 \\
  - \frac{k \theta}{\hbar}  &            0           &          0  & \frac{1}{m} \\
  - k                       &            0           &          0  &    0        \\
  0                         &           - k          &          0  &    0       
 \end{array}
\right)
\left(
 \begin{array}{c}
 {\hat x}_1  \\ {\hat x}_2 \\ {\hat p}_1 \\  {\hat p}_2   
\end{array} \right) \quad .
\ee 
This system of differential equations is easily solved using the exponential method.
One obtains for example

\be
\label{eq95}
 \hat x_1(t)   = M_{11}(t)   \hat{x}_1(0)  +
  M_{12}(t)  \hat{x}_2(0)  + 
 M_{13}(t)  \hat{p}_1(0)  +  M_{14}(t)  \hat{p}_2(0)  \quad ,
\ee 
with
\begin{eqnarray} 
\label{eq96} 
M_{11}(t) & = & \frac{1}{\lambda_1 + \lambda_2} 
( \lambda_1 \cos{\lambda_1 t} +  \lambda_2 \cos{\lambda_2 t} ) \quad , \quad
M_{12}(t)  =  \frac{1}{\lambda_1 + \lambda_2} 
(- \lambda_1 \sin{\lambda_1 t} +  \lambda_2 \sin{\lambda_2 t} ) \nonumber \\
M_{13}(t) & = & \frac{\lambda_1 \lambda_2}{\lambda_1^2 - \lambda_2^2} \frac{\theta}{\hbar}
( - \sin{\lambda_1 t} - \sin{\lambda_2 t} ) \quad , \quad
M_{14}(t) = \frac{\lambda_1 \lambda_2}{\lambda_1^2 - \lambda_2^2} \frac{\theta}{\hbar}
( - \cos{\lambda_1 t} + \cos{\lambda_2 t} ) \quad .
\end{eqnarray} 
The quantities $\lambda_1 {\rm and } \, \lambda_2$ are the eigenstates of the matrix appearing
in Eq.(\ref{eq94}):

\be
\label{eq97}
\lambda_{1,2} = \pm \sqrt{ \frac{k}{m} + \frac{k^2 \theta^2}{2 \hbar^2} \pm 
            \frac{k^{3/2} \theta}{2 \hbar^2 \sqrt{m}}  \sqrt{4 \hbar^2 + k m \theta^2}}
	    \quad  .
\ee 
The operators evaluated at the initial time are given by the 
expressions displayed in Eq.(\ref{eq4}). The mean value of the first position in any 
state $\psi(p_1,p_2)$ is thus found to be
\be
\label{eq98}
\langle  \hat{x}_1(t) \rangle  =  \frac{1}{\hbar (\lambda_1 - \lambda_2) 
(\lambda_1 + \lambda_2)} ( c_1 \cos{\lambda_1 t} + s_1 \sin{\lambda_1 t} +
 c_2 \cos{\lambda_2 t} + s_2 \sin{\lambda_2 t} ) \quad ,
\ee
with
\begin{eqnarray}
\label{eq99}
 c_1 &=&   \hbar \langle \hat x_1(0) \rangle \lambda_1^2 - \lambda_1 \lambda_2
   ( \hbar \langle \hat x_1(0) \rangle - \theta \langle \hat p_2(0) \rangle ) 
   \quad , \quad
s_1 =   - \hbar \langle \hat x_2(0) \rangle \lambda_1^2 + \lambda_1 \lambda_2
   ( \hbar \langle \hat x_2(0) \rangle - \theta \langle \hat p_1(0) \rangle ) \quad ,
    \nonumber\\    
c_2 &= &  \hbar \langle \hat x_1(0) \rangle \lambda_2^2 - \lambda_1 \lambda_2
   ( \hbar \langle \hat x_1(0) \rangle + \theta \langle \hat p_2(0) \rangle ) 
  \quad , \quad   
s_2 =  \hbar \langle \hat x_2(0) \rangle \lambda_2^2 + \lambda_1 \lambda_2
   (- \hbar \langle \hat x_2(0) \rangle + \theta \langle \hat p_1(0) \rangle ) 
   \quad .
\end{eqnarray}  

From Eq.(\ref{eq98}), one sees that the position mean value will be periodic for all states
if the ratio of the two frequencies $\lambda_1 , \lambda_2$ is rational. This requires 
a fine tuning of the parameters and so is not generic. 

In the cases where the ratio 
is not rational, there are states  for which the coefficients  are such that the 
components of one of the frequencies vanish. For example, the component of frequency
 $\lambda_1$ 
disappears in $ \langle \hat x_1(t) \rangle$  if the following conditions are satisfied:
\be
\label{eq100}
\theta \lambda_2  \langle \hat p_{1}(0) \rangle + \hbar (\lambda_1 - \lambda_2 ) 
 \lambda_2  \langle \hat x_{2}(0) \rangle = 0 \quad , \quad 
 \theta \lambda_2 \langle \hat p_{2}(0) \rangle + \hbar (- \lambda_1 + \lambda_2 ) 
 \lambda_2 \langle \hat x_{1}(0) \rangle = 0 \quad .
\ee 
Moreover, these two conditions also suppress the same components in the mean values
of the other variables
$x_2,p_1,p_2$. The above conditions translate into the vanishing of the following
integrals:
\be
\label{eq101}
\int d^2 p \psi^\ast(p) \left( i \hbar^2 (\lambda_1 - \lambda_2) \partial_{p_2} + 
\frac{1}{2} \theta (\lambda_1 + \lambda_2 ) p_1  \right) \psi(p) \quad , \quad
\int d^2 p \psi^\ast(p) \left( i \hbar^2 (\lambda_1 - \lambda_2) \partial_{p_1} - 
\frac{1}{2} \theta (\lambda_1 + \lambda_2 ) p_2  \right) \psi(p) \quad .
\ee
It is readily found that the operators contained in the parentheses do not have a 
common zero eigenvalue so that the states we are looking for cannot be found by solving two 
first order differential equations. However, one can show that the wave function
\be
\label{eq102}
\psi(p) = \exp{( (- a_1^2+i a_2) p_1^2 + ( - b_1^2+i b_2) p_2^2+ (c_1+i c_2) p_1 +
(d_1 + i d_2) p_2)}
\ee
satisfies the two conditions provided that the following relations hold:
\be
\label{eq103}
c_2 = - \frac{a_2 c_1}{a_1^2} - \frac{1}{4}  \frac{\theta}{ \hbar^2} 
\frac{\lambda_1+\lambda_2}{\lambda_1-\lambda_2} \frac{d_1}{b_1^2} \quad , \quad
d_2 = - \frac{b_2 d_1}{b_1^2} + \frac{1}{4}  \frac{\theta}{ \hbar^2} 
\frac{\lambda_1+\lambda_2}{\lambda_1-\lambda_2} \frac{c_1}{a_1^2} \quad .
\ee
These states can be obtained as solutions of the second order equation given in 
Eq.(\ref{eq53}).

It is readily seen from Eqs.(\ref{eq97},\ref{eq98}) that the usual theory is recovered when the 
deformation parameter goes to zero.

\subsection{The model possessing a minimal length uncertainty}

For the K.M.M theories, the situation is more complicated and we will treat only the 
one dimensional case. The analysis in the Heisenberg picture is not straightforward 
because the evolution of the operators leads to a non linear equation:
\be
\label{eq104} 
\dot{\hat{x}} = \frac{1}{m} ( 1+ \beta \hat{p}^2 ) \hat p \quad . 
\ee
We will rather use the Schrodinger picture and the knowledge of the spectrum of the 
harmonic oscillator \cite{kmm2}. The operators are now fixed and the time dependence is
carried by the state. The energy eigenstates form a basis of the Fock space and so an 
arbitrary state can be expanded as $ \vert \psi \rangle = \sigma^n \vert n \rangle. $
The mean value of the position in this state reads
\be
\label{eq105}
\langle \hat x(t) \rangle = \sum_{m,n} \sigma^n (\sigma^*)^m 
\langle m \vert \hat x \vert n \rangle  \exp{\left(\frac{i}{\hbar} 
( E_m - E_n ) t \right) } \quad ;
\ee 
the time dependence is entirely contained in the exponentials. In the usual case the matrix element
 $ \langle m \vert \hat x \vert n \rangle$ is non vanishing only for integers $m,n$
  which differ by one unit. As the energy spectrum is linear in this case, the arguments of 
  the  exponentials are all equal if we discard the signs. Collecting them, one ends up
  with the usual sine and cosine solutions. In the K.M.M theory, the spectrum of the 
  harmonic oscillator is quadratic \cite{kmm2}: $E_n = a n^2 + b n + c $. This leads us to 
  the following expression for the position mean value:
  \be
  \label{eq106}
  \langle \hat x(t) \rangle = \sum_{m,n} \tau_{m,n} ( \cos{\omega_{m n}t} + 
  i \sin{\omega_{m n}t} ) \quad , \quad \omega_{m n} = 
  \frac{1}{\hbar} ( m - n) ( a ( m+n) + b ) \quad .
  \ee
This mean value will be periodic only if all the frequency ratios 
$\omega_{m_1 n_1}/\omega_{m_2 n_2}$ are rational numbers. From that one  deduces that for 
any two couples $ (m_1,n_1) , (m_2,n_2) $, such that
 $ \langle m_i \vert \hat x \vert n_i \rangle $ do not vanish, there must exist a rational number $r_{12}$
such that
\be
\label{eq107}
r_{12} = \frac{a(m_1+n_1) +b}{a(m_2+n_2) +b} \quad .
\ee
From this one infers that $b/a$ must then be rational; replacing the quantities $a$ 
and $b$ in terms of the quantum parameters \cite{kmm2}, this amounts to ask that
\be
\label{eq108}
\frac{\beta m \hbar \omega}{ \beta m \hbar \omega + \sqrt{4 + \beta m \hbar \omega}}
\ee
is rational. When this condition is not satisfied, periodicity is lost.


\section{A look at the fuzzy sphere}

The fuzzy sphere is a matrix model defined by  the following relations:
\be
\label{eq109}
[ \hat x_k , \hat x_l ] = \frac{i \alpha}{\sqrt{j(j+1)}} \epsilon_{klm} \hat x_m \quad ; \quad 
\hat x_1^2 + \hat x_2^2 + \hat x_3^2 = 1  \quad , \quad
{\rm with} \quad $j$ \quad  {\rm half-integer} \quad .
\ee 
The saturation of the uncertainties related to the pairs of non commuting variables
translate into the formulas  $ m_{jk} \vert \psi \rangle = 0 $, with
\be
\label{eq110}
m_{12} = \hat x_1 + i a \hat x_2 + (d+ i e ) \quad , \quad
m_{23} = \hat x_2 + i b \hat x_3 + (f+ g e ) \quad , \ {\rm and} \quad
m_{31} = \hat x_3 + i c \hat x_1 + (h+ i k ) \quad ,
\ee
where $a,b,c, \cdots$ are real. Considering the following combinations of these equations
\be
\label{eq111}
(c [ m_{12}, m_{23} ] - i  [ m_{23}, m_{31} ]) \vert \psi \rangle = 0 \, , \, 
(- i [ m_{12}, m_{23} ] + b [ m_{31}, m_{12} ]) \vert \psi \rangle = 0 \, , \,
(- a [ m_{23}, m_{31} ] + i  [ m_{31}, m_{12} ]) \vert \psi \rangle = 0 \, , \,  
\ee
one obtains( in units where $\alpha=1$) 
\be
\label{eq112}
(1-i a b c) \hat x_1 \vert \psi \rangle = 0 \quad , \quad
(1-i a b c) \hat x_3 \vert \psi \rangle = 0 \quad , \quad
(1-i a b c) \hat x_2 \vert \psi \rangle = 0 \quad .
\ee
Can the  three Heisenberg inequalities  be saturated  simultaneously? Only two cases
may lead to that situation: 
\begin{itemize}
 \item The first possibility is
        \be
        \label{eq113}
          \hat x_1 \vert \psi \rangle = \hat x_2 \vert \psi \rangle = 
          \hat x_3 \vert \psi \rangle = 0 \quad ,
        \ee
     but then the second part of Eq.(\ref{eq109}) is violated \quad .
 \item The remaining possibility 
       \be
       \label{eq114}
          1 - i a b c = 0 
       \ee 
 can be rewritten as  
 \be
 \label{eq115}
 \frac{ \langle \hat x_1 \rangle \langle \hat x_2 \rangle  \langle \hat x_3 \rangle}
 {(\Delta x_1)^2 (\Delta x_2)^2 (\Delta x_3)^2 } = - 8 i \quad ;
 \ee
 this is not possible since
 all the quantities on the left side  are real.
\end{itemize} 

  The method proposed here can in principle be applied to  the fuzzy
 sphere. The appropriate action is
 \be
 \label{eq116}
 S = \left[ \left( \Delta x_1^2 \Delta x_2^2 - \frac{1}{2} \langle x_3 \rangle^2 \right)^2 +
  {\rm permutations} \right] + \lambda ( \langle \psi \vert \psi \rangle - 1) \quad .
 \ee
 
  One important feature which distinguishes the fuzzy sphere from the 
 models we have studied so far is the fact that its Fock space is finite dimensional. This 
 results in the fact that the action $S$ is now a function rather than a functional.
 
 To illustrate how our approach applies to the fuzzy sphere, let us take the very 
 simple case $j=1$. A unitary transformation is performed to go from the usual 
 representation to a more symmetric 
 one in which the non vanishing elements of the operators are
 \be
 \label{eq117}
 (X_1)_{23} = (X_1)_{32} = 1 \quad , \quad  (X_2)_{13} = - (X_2)_{31} = -i \quad , \quad
 (X_3)_{12} = (X_3)_{21} = 1 \quad .
 \ee
 The Fock space in this trivial case is six dimensional on the real scalars.
 Any state can be written as
 \be
 \label{eq118}
   \vert \psi \rangle = ( z_1, z_2 ,  z_3 , z_4, 
  z_5, z_6 ) \quad .
  \ee  
 The mean values we need to compute the action can be written as
 \be
 \label{eq119}
 \langle x_k \rangle = \frac{N_k}{G} \quad , \quad 
  \langle x_k^2 \rangle = \frac{K_k}{G} \quad ,
 \ee
 where
 \begin{eqnarray}
 \label{eq120}
 N_1 &=& 2 (z_3 z_5 + z_4 z_6) \quad , \quad N_2 = 2 (- z_2 z_5 + z_1 z_6 ) \quad , 
 \quad N_3 = 2 ( z_1 z_3 + z_2 z_4 ) \quad , \nonumber\\
 K_1 &=& z_3^2 + z_4^2 + z_5^2 + z_6^2 \quad , 
 \quad K_2 = z_1^2 + z_2^2 + z_5^2 + z_6^2
  \quad , \quad K_3 = z_1^2 + z_2^2 + z_3^2 + z_4^2 \quad , \nonumber \\
 G &=& z_1^2 + z_2^2 + z_3^2 + z_4^2 + z_5^2 + z_6^2 \quad . 
 \end{eqnarray}
 The interest of the representation displayed by Eq.(\ref{eq117}) is that it makes the 
 symmetries of the problem more transparent. One easily verifies that the 
 transformation 
 \be
 \label{eq121}
 \tau( z_1, z_2 , z_3, z_4, 
  z_5, z_6 ) = ( z_3 , z_4 , z_5 , z_6 , -z_2,z_1) 
 \ee
generates a circular permutation(discarding the signs) of the functions $N_k$. The same
 holds for the functions $K_k$ while $G$ remains unchanged so that $\tau$ is a 
 symmetry of the action and, in addition, it is idempotent: $\tau^6=- 1$.
 
 Finding the extrema of the action $S$ of Eq.(\ref{eq116}) is not an easy task. 
 We give here two such extrema:
 \begin{eqnarray}
 \label{eq122}
 \vec z &=& (-1,0,0,1,1,0) \quad {\rm with } \quad  S = \frac{16}{27} \quad {\rm and} \nonumber\\ 
 \vec z &=& (0.631646...,0.315353...,0.002528...,0.016494...,-0.316359...,0.633415...)
  \quad  {\rm with} 
\quad  S = 1.19 \, 10^{-8} \quad .
 \end{eqnarray}
 Applying the transformation $\tau$ to these states, one generates states  having the same 
 values of $S$.
\section{conclusions}

In this work, we have suggested an approach towards coherent states which
relies on a functional method involving non trivial Heisenberg uncertainty relations 
and generalizes in a natural way the squeezed states of usual quantum mechanics. 
We have found special solutions to the second order differential equations obtained in three different 
non commutative theories. We have briefly outlined how our method can be applied to
 finite dimensional matrix models like the fuzzy sphere. The problem for the first three
 theories we have studied is that there are too many solutions to the relevant
  equations. On the contrary, for matrix models like the fuzzy sphere, the situation
   is different; the action
is a function of a finite number of variables but its form is non trivial and 
makes the search for solutions non trivial.
  
One of the crucial points which remain to be addressed is the nature of the
critical points found here. To know if these states  are maxima or minima of the
 action, one  to ressort to a second order analysis. However, as the most general
solution of the second order partial differential equations involved are not known, such
a computation cannot tell us by itself if we are in front of an absolute minimum. One
can also, develop the differential equations we obtained order by order 
in the extra parameter $\theta $; this breaks the symmetry with $\hbar$ and makes any
statement about the nature of the solution much more difficult.

Another crucial question about the  states found by the
 method presented here is the  other  properties of the usual  coherent states 
they may possess, like  completeness. If this was the case, they might be legitimate
candidates for the definition of a physically meaningful star product \cite{voros}. 

We did not pay much attention to some subtle questions linked to the self-adjoint 
extensions of the symmetric operators defined in KMM models; a recent and useful
treatment can be found in
\cite{spindel}.    



\underline{Acknowledgements}
 
It is a pleasure to thank S.Detournay, Cl.Gabriel and Ph.Spindel for useful
 discussions in Mons(Belgium) in the early stages of this work.  

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\appendix

\section{Variation of the action in the K.M.M.1 model}



Varying the action of Eq.(\ref{eq70}) one obtains

\begin{eqnarray}
\label{eqa1}
\frac{\delta S}{\delta \psi^*(p)} &=& b_1 \frac{\delta }{\delta \psi^*(p)}
\Delta x_1^2 + b_2 \frac{\delta }{\delta \psi^*(p)} \Delta x_2^2 + b_3
\frac{\delta }{\delta \psi^*(p)} \Delta p_1^2 + b_4 \frac{\delta }{\delta
\psi^*(p)} \Delta p_2^2+ \nonumber\\
&+& b_5 \frac{\delta }{\delta
\psi^*(p)} \langle 1+ \beta \hat{\vec{p}^2} \rangle^2 + b_6 \frac{\delta
}{\delta \psi^*(p)} \vert \langle \hat p_1 \hat x_2 - \hat p_2 \hat x_2
\rangle \vert^2 + \lambda \frac{\delta }{\delta \psi^*(p)} 
\langle \psi \vert \psi \rangle = 0  \quad ,
\end{eqnarray} 
where
\begin{eqnarray}
\label{eqa2}
b_1 &=& \frac{1}{8} \left( \Delta x_1^2 \Delta
x_2^2 - \hbar^2 \beta^2  \vert \langle \hat p_1 \hat x_2 - \hat p_2 \hat x_1 
\rangle \vert^2  \right) \Delta x_2^2  + 2 \left( \hbar^2 \beta^2 \Delta x_1^2
\Delta p_1^2 - \frac{1}{4} \hbar^4 \beta^2 \langle 1 + \beta \hat{\vec{p}}^2
\rangle^2  \right) \hbar^2 \beta^2 \Delta p_1^2  \quad, \nonumber\\
b_3 &=& 2 \left( \hbar^2 \beta^2 \Delta x_1^2
\Delta p_1^2 - \frac{1}{4} \hbar^4 \beta^2 \langle 1 + \beta \hat{\vec{p}}^2
\rangle^2  \right) \hbar^2 \beta^2 \Delta x_1^2  \quad , \nonumber\\
b_5 &=& -\frac{1}{2} \hbar^4 \beta^2  \left( \hbar^2 \beta^2 \Delta x_2^2
\Delta p_2^2 - \frac{1}{4} \hbar^4 \beta^2 \langle 1 + \beta \hat{\vec{p}}^2 \rangle^2 +
 (1 \leftrightarrow 2)  \right)^2  \quad , \nonumber\\
b_6 &=& -\frac{1}{8} \hbar^2 \beta^2 \left( \Delta x_1^2 \Delta x_2^2 - \hbar^2
\beta^2  \vert \langle \hat p_1 \hat x_2 - \hat p_2 \hat x_1  \rangle \vert^2 
\right) \quad .
\end{eqnarray} 
The coefficient $b_2$ is derived from $b_1$ by an exhange of
the indices $1,2$. The same holds for $b_4$ and $b_3$. It is readily seen that $b_5,b_6$ are negative while the
other $b_i$ are positive.

The only derivative which is not computed exactly like the others is the
following

\begin{eqnarray}
\label{eqa3}
\frac{\delta}{\delta \psi^*(p)} \langle  \psi \vert \hat p_1 \hat x_2 - \hat
p_2 \hat x_1 \vert \psi \rangle^* &= i \hbar &  \frac{\delta}{\delta \psi^*(p)}
\int \psi(q) ( - q_1 \partial_{q_2} \psi^*(q) + q_2 \partial_{q_1} \psi^*(q)  )
=i \hbar \int \psi(q) ( - q_1 \partial_{q_2} \delta(p-q) + q_2 \partial_{q_1}
\delta(p-q) )  \nonumber\\
&=& i \hbar (  p_1 \partial_{p_2} \psi(p) - p_2 \partial_{p_1} \psi(p)  )
\quad .
 \end{eqnarray}

The parameters of Eq.(\ref{eq75}) are given by the following formulas

\begin{eqnarray}
\label{eqa4}
a_1 &=& - \hbar^2 b_1 \quad , \quad a_2 = - \hbar^2 b_1 \quad , \quad a_3 =
- 2 \hbar b_6 Re (\langle \hat p_1 \hat x_2 - \hat p_2 \hat x_1 \rangle) \quad
, \quad  a_4 = - 2 \hbar b_1 \langle \hat x_1 \rangle \quad , \quad  a_5 = - 2
\hbar b_2 \langle \hat x_2 \rangle  \quad , \nonumber\\
a_6 &=& b_3 +\beta b_5 \quad , \quad a_7 = b_4 +\beta b_5 \quad , \quad a_8 =
- 2 b_3 \langle \hat p_1 \rangle \quad , \quad  a_9 =
- 2 b_4 \langle \hat p_2 \rangle  \quad , \nonumber\\
a_{10} &=& b_1 ( - \Delta  x_{1}^2 + \langle \hat x_1 \rangle^2 ) 
+ b_2 ( - \Delta  x_{2}^2 + \langle \hat x_2 \rangle^2 ) 
+ b_3 ( - \Delta  p_{1}^2 + \langle \hat p_1 \rangle^2 )
+ b_4 ( - \Delta  p_{2}^2 + \langle \hat p_2 \rangle^2 ) - \beta b_5 \langle
\hat p_1^2 + \hat p_2^2 \rangle \nonumber\\
&-& b_6  \vert \langle \hat p_1 \hat
x_2 - \hat p_2 \hat x_1 \rangle  \vert^2  + \lambda \quad ,
\end{eqnarray} 
from which the signs of $a_1,a_2$ are deduced.


\section{Commutative limit}
  
   Altough the actions we considered as the starting points of our analysis( Eq.(\ref{eq47},\ref{eq70})
    and the one leading to Eq.(\ref{eq92}))
do not tend to two separate copies of the action associated to the one dimensional 
case (Eq.(\ref{eq23})), the equation obtained displays this property. As an illustration, let us 
consider the KMM1 model i.e Eq.(\ref{eq75}). We replace the quantities $a_i$ by their expressions 
in terms of the observables thanks to Eqs.(\ref{eqa2},\ref{eqa4}). We then expand the equation in 
terms of the parameter $\gamma = \frac{\gamma}{\hbar} $:
\be
\label{eqb1}
( O_0 + \gamma O_1+ \gamma^2 O_2 + \cdots) \psi = 0 \quad ,
\ee  
\be
\label{eqb2}
O_0 = \left( - \frac{\hbar^2}{8} \Delta x_1^2 (\Delta x_2^2)^2             \partial_{p_1^2} - 
\frac{i \hbar}{4} \langle \hat  x_1 \rangle \Delta x_1^2 (\Delta x_2^2)^2 \partial_{p_1}  +
( 1 \leftrightarrow 2) \right) + \left( \frac{1}{8} \Delta x_1^2 \Delta x_2^2 ( \langle x_1
 \rangle^2 - \Delta x_1^2) + ( 1 \leftrightarrow 2) + \lambda 
  \right) \, , 
\ee
\be
\label{eqb3}
O_1 = \left( - \frac{\hbar}{4}    \Delta x_1^2 (\Delta x_2^2)^2 ( p_1^2+p_2^2)  \partial_{p_1^2}
-  \frac{1}{4}   \Delta x_1^2 (\Delta x_2^2)^2   ( \hbar p_1 + i 
 \langle \hat x_1 \rangle ) ( p_1^2+p_2^2 ) + ( 1 \leftrightarrow 2) \right) \quad .
\ee
The appearance of the perturbation in the kinetic term can be handled order by order as in 
\cite{muso}.



\end{document}
