path: root/cover.tex

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\begin{document}

\title{{\bf COVER LETTER \\`Complex complex landscapes'}}
%\footnote{{\bf Key-words}: }
%}}

\author{
Jaron Kent-Dobias
 and 
Jorge Kurchan
}

\maketitle



\vspace{1.cm}


The subject of `Complex Landscapes', which started in the spin-glass literature, is concerned with functions (landscapes) of many variables, having a multiplicity of minimums, which are the objects of interest. Apart from its obvious interest for glassy systems, it has found a myriad applications  in  many domains:  Computer Science, Ecology, Economics, Biology \cite{mezard2009information}.

In the last few years, a renewed interest has developed for landscapes for which the variables are complex. There are a few reasons for this: {\em i)} in Computational Physics, there is the main obstacle of the `sign problem', and a strategy has emerged to attack it deforming the sampling space into complex variables. This is a most natural and promising path, and any progress made will have game-changing impact in solid state physics and lattice-QCD \cite{Cristoforetti_2012_New,Scorzato_2016_The}.
{\em ii)} At a more basic level, following the seminal work of E. Witten \cite{Witten_2010_A,Witten_2011_Analytic}, there has been a flurry of activity concerning the very definition of quantum mechanics, which requires also that one move into the complex plane. 

In all these cases, just like in the real case, one needs to know the structure of the `landscape', where are the saddle points and how they are connected, typical questions of `complexity'.
However, to the best of our knowledge, there are no studies extending the methods of the theory of
complexity to
complex variables. 
We believe our paper will open a field that may find 
numerous applications and will widen our theoretical view of complexity in general.


\bibliography{bezout}


\end{document}















\section{The Kipnis-Marchioro-Presutti model}

Consider the following process: 
\begin{itemize}
\item
choose a pair of neighbouring sites and completely
exchange energy between them
\item
if the site is one of the borders, exchange completely energy with the bath.
\end{itemize}
each choice with probability $1/(N+1)$. From here onwards, we shall denote
$\tau$ a large time, sufficient for any two-site thermalisation.

The evolution operator in one step is:
\begin{eqnarray}
U &=& \frac{1}{N+1} \left[ e^{-\tau L_1^*} +  e^{-\tau L_N^*} + \sum_{i=1}^{N-1}   e^{-\tau L^*_{i,i+1}} \right]
\nonumber \\
&=& \frac{1}{N+1} \left[ e^{-2\tau (T_1 K^-_1 + K^o_1 + k) } +  e^{-2\tau(T_L K^-_L + K^o_L +k)  }
 + \sum_{i=1}^{N-1}   e^{ \frac{-\tau}{k}
(K^+_i K^-_{i+1} + K^-_i K^+_{i+1} - 2 K^o_i K^o_{i+1}
+ 2k^2 )} \right] \nonumber \\
~
\end{eqnarray}
and the dynamics after $n$ steps is given by $U^n$.
Because we are considering large $\tau$, the terms in the sums are  in fact projectors
onto the lowest eigenvalues of the exponents. We shall however keep the notation as it is
in order to stress the symmetry of the bulk terms.

Let us now show that - at the level of energies - this dynamics yields the KMP process 
{\em for $k=\frac{1}{2}$, that is $m=2$}.
Consider first a general $m$, and two neighbouring sites of coordinates $x = \{x_\alpha\}_{\alpha=1,\ldots,m}$, 
$y=\{y_\alpha\}_{\alpha=1,\ldots,m}$.
If they are completely thermalised, it means that (cfr (\ref{bb}):
the joint probability density satisfies
\begin{equation}
\left(x_{\alpha}
\frac{\partial}{\partial y_{\beta}} -
y_{\beta}\frac{\partial}{\partial x_{\alpha}}
  \right) p(x,y)=0
\end{equation}
It is easy to see that this may happen if and only if
\begin{equation}
p(x,y)= p[ \sum_\alpha (x_\alpha^2+y_\alpha^2)]
\end{equation}
In particular let us consider the microcanonical measure
\begin{equation}
p(x,y)= \delta[ \sum_\alpha (x_\alpha^2+y_\alpha^2)-\epsilon ]
\end{equation}
Defining new random variables $\epsilon_1$ and $\epsilon_2$ 
as the energies of the neighboring sites
\be
\epsilon_1 = \sum_\alpha x_\alpha^2
\ee
\be
\epsilon_2 = \sum_\alpha y_\alpha^2
\ee
then their joint probability density will be
\begin{equation}
p(\epsilon_1,\epsilon_2) = \frac{S_m^2}{4} \delta(\epsilon_1+\epsilon_2-\epsilon)
\epsilon_1^{\frac{1}{2}-1} \epsilon_2^{\frac{1}{2}-1}
\end{equation}
where $S_m$ denotes the surface of the unit sphere in $m$ dimension
\be
S_m = \frac{m \pi^{m/2}}{\Gamma(\frac{1}{2}+1)}
\ee
{\em This yields a flat distribution for $m=2$, i.e. the KMP model.}




\section{Dual model}


The expectation value of an observable at time $t$, starting from an initial
distribution $|init\rangle$ is:


\begin{equation}
<O> = \langle - | O e^{-Ht} | init \rangle
\end{equation}
where $\langle - |$ is a constant.
Taking the adjoint $ x_i \to x_i$, $\partial_i \to -\partial_i$:
\begin{equation}
<O> = \langle - | O e^{-Ht} | init \rangle= \langle init|  e^{-H^\dag t} O |- \rangle
\end{equation}
where $H^\dag(K^\pm, K^o)=H( K^\pm, -K^o)$ (because of the change of signs of the derivatives)
\begin{eqnarray}
-H^\dag&=& \frac{4}{1} \sum_i \left(
K^+_i K^-_{i+1} + K^-_i K^+_{i+1} - 2 K^o_i K^o_{i+1}
+ \frac{m^2}{8} \right)
\nonumber\\
&+&2 \left(T_1 K^-_1 + K^o_1 + \frac{1}{4}\right)
+2 \left(T_L K^-_L + K^o_L +\frac{1}{4}\right)
\end{eqnarray}
In particular, for the generating function we had chosen
 \begin{equation}
 O |- \rangle = \Pi_i \frac{x_i^{2 \xi_i}}{(2\xi_i -1)!!}|-\rangle=|\xi_1,...,\xi_N\rangle
\end{equation}

Considered as an operator acting on `particle number', as counted by $K^o$, $H^\dag$ does not
conserve the probability.
The trick we used can be expressed as follows: introduce the particle number $\xi_o$ and $\xi_{N+1}$ 
and the operators $A^+_o$ and $A^+_{N+1}$, which create particles in boundary sites with unit rate.
We consider now the {\em enlarged} process generated by
\begin{eqnarray}
-H^{dual}&=& \frac{4}{1} \sum_i \left(
K^+_i K^-_{i+1} + K^-_i K^+_{i+1} - 2 K^o_i K^o_{i+1}
+ \frac{m^2}{8} \right)
\nonumber\\
&+&2 \left(A^+_o K^-_1 + K^o_1 - \frac{1}{4}\right)
+2 \left( A^+_{N+1} K^-_N + K^o_N -\frac{1}{4}\right)
\end{eqnarray}
which conserves ({\it seems}) particle number and probability.
We wish to prove that:

\begin{eqnarray}
<O> &=& \langle init| e^{-H^\dag t} |\xi_1,...,\xi_N \rangle \nonumber \\
&=& \sum_{\xi_o,\xi_{N+1}}
  T_1^{\xi_o} T_{L}^{\xi_{N+1}} \langle \xi_o \xi_{N+1} | \otimes \langle
  init| e^{-H^{dual} t} |\xi_1,...,\xi_N \rangle \otimes
  |\xi_o=0,\xi_{N+1}=0 \rangle \nonumber \\
\label{ggg}
\end{eqnarray}


I think the proof is obvious, because developing the exponential of $H^{dual}$ all the $A^+$ can be
collected because they commute with everything else, and the experctation value
\begin{equation}
\sum_{\xi_o}  T_1^{\xi_o}  \langle \xi_o  |[A^+_o]^r  |\xi_o=0 \rangle = T_1^r
 \end{equation}
just puts back as many $T$'s as necessary.

I do not know exactly how to use (\ref{ggg}) in general, but in the large time limit the evolution
voids the chain of particles


\section{Dual of KMP}

I think that the argument runs through without changes if we use $U$ defined for the KMP model.
We just have to note that each term corresponds to an evolution of two sites (or a site and the bath)
and so in the dual it corresponds to sharing the particles between those two sites, or emptying
the sites at the borders.

{\bf: NOTE by Cristian}

We can check that the duality function chosen in the original paper by KMP
do coincide with the duality function of our process for $m=2$ (and the random
variables are the energies).
Indeed we start from
\be
f(x,\xi) = \prod_i (\sum_{\alpha} x_{i,\alpha}^2)^{\xi}
\ee
When the bath have equal temperature (let's us choose T=1) then the stationary 
measure is 
\be
\pi(x) = \prod_i \frac{1}{(2\pi)^{m/2}} \exp\left(-\sum_{\alpha}\frac{x_{i,\alpha}^2}{2}\right)
\ee
Let us focus on a fixed $i$ (that is in this short computation we write $x$ for $x_i$). 
We have
\begin{eqnarray}
\E(f(x,\xi)) 
&=&
\int dx_1 \cdots \int dx_m (x_1^2+\ldots + x_m^2)^{\xi} \exp-\left(\frac{x_{1}^2}{2}+\ldots+\frac{x_{1}^2}{2}\right) 
\nonumber \\
& = & 
\int dr S_m r^{2\xi} \exp-\left(\frac{r^2}{2}\right)
\nonumber \\
& = &
\frac{\frac{1}{2}\Gamma(\frac{1}{2}+\xi)}{\Gamma(\frac{m}{2}+1)} 2^\xi
\nonumber \\
\end{eqnarray}
Special cases:
\begin{itemize}
\item $m=1$

$$
\E(f(x,\xi)) = (2\xi-1)!!
$$
where one uses that $\Gamma(\frac{1}{2}+\xi)= \frac{\sqrt{\pi}(2\xi-1)!!}{2^{\xi}}$ and $\Gamma(\frac{3}{2}) = \frac{\sqrt{\pi}}{2}$
\item $m=2$

$$
\E(f(x,\xi)) = \xi! 2^\xi
$$
where one uses that $\Gamma(1+\xi)= \xi!$ and $\Gamma(2) = 1$.
Thus, if one defines the energies as 
$$
\epsilon_i = \sum_{\alpha}\frac{x_{i,\alpha}^2}{2}
$$
one recover the choice of KMP for the dual function
$$
O(\epsilon_i,\xi) = \prod_i \frac{\epsilon_i^{\xi_i}}{\xi_i!}
$$
\end{itemize}






\section{ Dual of SEP: here goes an outline of how to proceed for the SSEP}


\be
H=-L_{SEP}^*
\ee
\begin{eqnarray}
L^*_{SEP} &=& \frac{1}{j}
  \sum_i  \left(J^+_i J^-_{i+1} + J^-_i J^+_{i+1} + 2 J^o_i J^o_{i+1}
 - 2 j^2  \right)\\
&+&\alpha (J^-_1 - J^o_1-j) + \gamma (J^+_1 + J^o_1-j)
+ \delta (J^-_L - J^o_L-j) + \beta (J^+_L + J^o_L-j)\nonumber
\end{eqnarray}
The factor $1/j$ is analogous to the factor $1/m$ in (\ref{bb}).
The operators $J^+_i, J^-_i, J^o_i$ act on the Hilbert space
 corresponding to  $0 \le r \le n$  particles per site $\otimes_i |r\rangle_i$
as follows:
\begin{eqnarray}
J^+_i |r\rangle_i &=& (2j-r)  |r+1\rangle_i \nonumber \\
 J^-_i |r\rangle_i &=& r  |r-1\rangle_i \nonumber \\
J^o_i |r\rangle_i &=& (r-j)  |r\rangle_i
\end{eqnarray}

The conjugation properies are as follows. There is an operator $Q$,
{\em diagonal in this basis  } (I give the expression below), such that:
\begin{equation}
[J^+_i]^\dag = Q[J^-_i]Q^{-1} \qquad [J^-_i]^\dag = Q[J^+_i]Q^{-1}
\end{equation}
while $[J^z_i]^\dag=J^z_i= Q[J^z_i]Q^{-1}$.



The expectation value of an observable at time $t$, starting from an initial
distribution $|init\rangle$ is:


\begin{equation}
<O> = \langle - | O e^{-Ht} | init \rangle
\end{equation}
where $\langle - |$ is a constant.
As before:
\begin{eqnarray}
<O> &=& \langle - | O e^{-Ht} | init \rangle=
\langle init|  e^{-H^\dag t} O |- \rangle= \nonumber \\
& & \langle init|Q  e^{-{\bar H} t}  Q^{-1}O   |- \rangle=
\langle init|Q \;  e^{-{\bar H} t}  Q^{-1}O Q  Q^{-1} |- \rangle
\end{eqnarray}


{\em  $ {\bar H}$ is the same operator as $H$ but with
$J^+$ substituted by  $J^-$, and vice-versa.}
Our job is now to make the rotation that will eliminate the $J^+$'s in
the border terms of $ {\bar H}$.




The transformation is of the form
\begin{eqnarray}
e^{\mu J^+} J^+ e^{-\mu J^+}&=&J^+ \nonumber \\
e^{\mu J^+} J^o e^{-\mu J^+} &=&J^o - \mu J^+ \nonumber \\
e^{\mu J^+} J^- e^{-\mu J^+} &=& J^- + 2 \mu J^o - \mu^2 J^+
\end{eqnarray}
for suitable $\mu$.
Putting $\mu=-1$, we get that {\bf the bulk term is left invariant,
precisely because of the SU(2) symmetry}. The boundary terms {\bf of $\bar H$}
transform further into:
\begin{eqnarray}
& e^{\mu J^+_1} \left[ \alpha (J^+_1 - J^o_1-j) + \gamma (J^-_1 + J^o_1-j)
\right] e^{-\mu J^+_1}= \nonumber \\ & \gamma(J^-_1 + 2 \mu J^o_1 - \mu^2
J^+_1 +J^o_1 - \mu J^+_1 -j) + \alpha (J^+_1 - J^o_1 + \mu J^+_1 -j)
= \nonumber \\
& \alpha(- J^o_1 -j) + \gamma (J^-_1 -J^o_1 -j)
\label{trans}
\end{eqnarray}
which is of the same form we have in the $SU(1,1)$ model.
The same can be done in the other boundary term.

We thus get:
\begin{eqnarray}
<O> &=& \langle - | O e^{-Ht} | init \rangle=
\langle init|Q \; e^{-{\bar H} t}  Q^{-1}O Q  Q^{-1} |- \rangle \nonumber\\
&= & \langle init|Q   e^{ \sum_i J^+_i}  e^{-{\bar H_{dual}} t}
 e^{ -\sum_i J^+_i}  Q^{-1}O Q  Q^{-1} |- \rangle \nonumber \\
&= & \langle init|Q   e^{ \sum_i J^+_i}  e^{-{\bar H_{dual}} t}
 e^{ -\sum_i J^+_i}  Q^{-1}O Q e^{ \sum_i J^+_i}
  e^{ -\sum_i J^+_i}  |- \rangle \nonumber \\
 &= & \langle init|Q  Q^{-1}  e^{ \sum_i J^+_i}  e^{-{\bar H_{dual}} t}
 e^{ -\sum_i J^+_i}  Q^{-1}O Q e^{ \sum_i J^+_i}  |-_{dual} \rangle
\end{eqnarray}
where we have defined $H_{dual}$  as the transformed Hamiltonian.

We now have to study $ |-_{dual} \rangle \equiv e^{ -\sum_i J^+_i}
 Q^{-1} |- \rangle$
Because we know that  terms like those proportional to $\gamma$ and $\alpha$
anihilate the measure to the left:
\begin{eqnarray}
& & \langle - | (J^-_i - J^o_i-j) =0\nonumber \\
& & \langle - | (J^+_i + J^o_i-j) =0
\end{eqnarray}
this implies that in the new variables and following all the transformations
(cfr (\ref{trans})):
\begin{eqnarray}
& & (J^-_i -J^o_i -j)e^{ -\sum_i J^+_i} Q^{-1} |- \rangle= 0 \nonumber \\
& & ( -J^o_i -j)e^{ -\sum_i J^+_i} Q^{-1} |- \rangle =0
\end{eqnarray}
which implies that $( J^o_i +j) |-_{dual} \rangle= J^-_i  |-_{dual} \rangle=0$,
and this means that
\begin{equation}
J^o_i |-_{dual} \rangle =-j |-_{dual}  \rangle
\end{equation}
is the vacuum of particles in this base!

All in all we are left with:
\begin{eqnarray}
<O> &=&  \langle init|Q \;  e^{ \sum_i J^+_i}
  e^{-{\bar H_{dual}} t} e^{ -\sum_i J^+_i}  Q^{-1}O Q e^{ \sum_i J^+_i}
  |-_{dual} \rangle \nonumber \\
 &=&  \langle init|Q \;  e^{ \sum_i J^+_i}
  e^{-{\bar H_{dual}} t} {\hat O}
  |-_{dual} \rangle
\end{eqnarray}
where $ {\hat O} \equiv e^{ -\sum_i J^+_i} Q^{-1}O Q e^{ \sum_i
J^+_i}$.  We have to start with the
vacuum $ |-_{dual} \rangle$, then apply $ {\hat O} $, (which creates
particles because it contains many $J^+$'s), and then there is the
dual evolution. The final configuration has to be overlapped with
$\langle f| \equiv  \langle init|Q \;  e^{ \sum_i J^+_i}$.
For large times, there will be no particle left except in the two extra sites
in the borders.

\section{Constructive approach}

Here I would like to say the following: if I have a modle of transport
of which I do not know if it has a Dual one, I can proceed as follows.
I take a small version with no baths and a few sites. I write the
evolution operator and I diagonalise it numerically.  If there is a
non abelian group, the eigenvalues will be in degenerate
multiplets. Hence, if I find multiplets, then very probably there is a
dual model, if I do not, then there cannot be one.  It would be nice
to show it with the KMP model with two or three sites.

Another thing is to consider higher groups. $SU(3)$ has already been studied
for two kinds of particles. We know how to map to a dual in that
case, if it has not been done yet.

\newpage
{\bf THIS PART HAS BEEN WRITTEN BY CRISTIAN} 

The aim of this file is to set notation in the two languages.
Let us focus on duality for the case we already know:
SU(1,1) model with $k=1/4$. To fix ideas let us consider only
the bulk part of the system with periodic boundary conditions.

\section{Probabilistic language}
We have two stochastic Markovian process with continuous time.
\begin{itemize}
\item
\underline{The first process $X(t) \in \R^N$} is given by the Fokker-Planck equation:
\be
\frac{dp(x,t)}{dt}  = L^* p(x,t)
\ee
where $p(x,t)$ represents the probability density
for the process $X(t)$, that is
$$
p(x,t)dx = Prob (X(t)\in (x,x+dx))
$$
and
\begin{eqnarray}
L^*p(x,t)
& = &
\sum_i L^*_{i,i+1} p(x,t) \noindent\\
& = &
\sum_i \left(x_i\frac{\partial}{\partial x_{i+1}} -x_{i+1}\frac{\partial}{\partial x_{i}}\right)^2 p(x,t)
\end{eqnarray}
\item
\underline{The second process $\Xi(t) \in \N^N$} is characterized by the master equation
\be
\frac{dP(\xi,t)}{dt} = {\cal L^*} P(\xi,t)
\ee
where $P(\xi,t)$ represents the
probability mass function for the process $\Xi(t)$, that is
$$
P(\xi,t) = Prob (\Xi(t) = \xi)
$$
and
\begin{eqnarray}
{\cal L}^*P(\xi,t)
& = &
\sum_i {\cal L}^*_{i,i+1}P(\xi,t) \nonumber \\
& = &
\sum_i 2\xi_i \left(1+ 2\xi_{i+1}\right) P(\xi^{i,i+1},t)
+ \left(1+2\xi_i\right)2\xi_{i+1} P(\xi^{i+1,i},t) \nonumber\\
& & - 2\left(2\xi_i + \frac{1}{2}\right)\left(2\xi_{i+1} + \frac{1}{2}\right) P(\xi,t)
+ \frac{1}{2}P(\xi,t)
\end{eqnarray}
and $\xi^{i,j}$ denotes the configuration that is obtained by removing one particle
at $i$ and adding one particle at $j$.
\newpage
\item
\underline{In general, Duality means the following}:
there exists functions $O(x,\xi): \R^N \times \N^N \mapsto \R$ such that
the following equality between expectations for the two processes holds
\begin{center}
\fbox{\parbox{9cm}{
\be
\E_x( O(X(t),\xi)) =\E_\xi(O(x,\Xi(t)))
\ee
}}
\end{center}
The subscripts in the expectations denote the initial conditions of the processes:
$X(0) =x$ on the left and $\Xi(0) = \xi$ on the right.
More explicitly we have:
\be
\int dy O(y,\xi) p(y,t; x,0)  = \sum_{\eta} O(x,\eta) P(\eta,t; \xi,0)
\ee
To prove duality it is sufficient to show that
\be
\label{main}
L O(x,\xi) = {\cal L} O(x,\xi)
\ee
where $L$, that is working on $x$, is the adjoint of  $L^*$ and ${\cal L}$, that is working on $\xi$,
is the adjoint of  ${\cal L}^*$.
Indeed we have:
\begin{eqnarray}
\E_x( O(X(t),\xi))
& = &
\int dy O(y,\xi) p(y,t; x,0) \\
& = &
\sum_{\eta} \int dy O(y,\eta) p(y,t; x,0) \delta_{\eta,\xi} \\
& = &
\sum_{\eta} \int dy O(y,\eta) e^{tL^*} \delta(y-x) \delta_{\eta,\xi} \\
& = &
\sum_{\eta} \int dy e^{tL} O(y,\eta)  \delta(y-x) \delta_{\eta,\xi} \\
& = &
\sum_{\eta} \int dy e^{t{\cal L}} O(y,\eta)  \delta(y-x) \delta_{\eta,\xi}  \\
& = &
\sum_{\eta} \int dy O(y,\eta) e^{t{\cal L}^*} \delta(y-x) \delta_{\eta,\xi}  \\
& = &
\sum_{\eta} \int dy O(y,\eta) P(\eta,t;\xi,0) \delta(y-x) \\
& = &
\sum_{\eta}  O(x,\eta) P(\eta,t;\xi,0)  \\
& = &
\E_\xi(O(x,\Xi(t)))
\end{eqnarray}
\newpage
\item
\underline{For the present case, the proper function to be considered are}
\be
\label{Oss}
O(x,\xi) = \prod_{i} \frac{x_i^{2\xi_i}}{(2\xi_i-1)!!}
\ee
Let us check Eq.(\ref{main}) on this choice. We have
\begin{eqnarray*}
&&
L_{i,i+1} O(x,\xi)
=
\left(\prod_{k\not\in\{i,i+1\}}  \frac{x_k^{2\xi_k}}{(2\xi_k -1)!!}\right)
\times
\\
&&\left(2\xi_{i+1}(2\xi_{i+1}-1) \frac{x_i^{2\xi_i+2}}{(2\xi_i -1)!!}\frac{x_{i+1}^{2\xi_{i+1}-2}}{(2\xi_{i+1} -1)!!}
- 2\xi_{i}(2\xi_{i+1}+1) \frac{x_i^{2\xi_i}}{(2\xi_i -1)!!}\frac{x_{i+1}^{2\xi_{i+1}}}{(2\xi_{i+1} -1)!!}
\right.
\\
&&\left.- 2\xi_{i+1}(2\xi_{i}+1) \frac{x_i^{2\xi_i}}{(2\xi_i -1)!!}\frac{x_{i+1}^{2\xi_{i+1}}}{(2\xi_{i+1} -1)!!}
+2\xi_{i}(2\xi_{i}-1) \frac{x_i^{2\xi_i-2}}{(2\xi_i -1)!!}\frac{x_{i+1}^{2\xi_{i+1}+2}}{(2\xi_{i+1} -1)!!}
\right)
\\
\end{eqnarray*}
which implies
\begin{eqnarray*}
L_{i,i+1} O(x,\xi)
& = &
\Big(2\xi_{i+1}(2\xi_{i}+1) [O(x,\xi^{i+1,i})-O(x,\xi)]
\\
&&
\;+\;2\xi_{i}(2\xi_{i+1}+1) [O(x,\xi^{i,i+1})-O(x,\xi)]\Big)
\\
& = &
{\cal L}_{i,i+1} O(x,\xi)
\end{eqnarray*}

\item \underline{How to find the proper normalization?}
Suppose that we are in the general following situation:
\begin{itemize}
\item We have a generator $L$ of a Markov process $X(t)$.
\item We know its stationary measure $\pi(x)$:
\be
L^* \pi(x) = 0
\ee
\item We have functions $f(x,\xi)$ for which the following holds:
\be
\label{aaa}
L f(x,\xi) = \sum_{\eta} r(\xi,\eta) f(x,\eta)
\ee
with
\be
\label{bbb}
r(\xi,\eta) \ge 0 \qquad  \mbox{if}\quad \xi \neq \eta
\ee
\be
\label{ccc}
r(\xi,\xi)  \le 0 \qquad  \mbox{if}\quad \xi = \eta
\ee
\end{itemize}
The matrix $r$ resembles the generator of a dual Markov process,
but it is not because it does not satisfy the condition
$\sum_{\eta} r(\xi,\eta) = 0$.
In order to find the generator of the dual process we proceed as
follows:
\begin{enumerate}
\item Define
\be
m(\xi) = \int f(x,\xi) \pi(x) dx
\ee
\item Define
\be
q(\xi,\eta)= m(\xi)^{-1} r(\xi,\eta) m(\eta)
\ee
\item Define
\be
O(x,\xi) = m(\xi)^{-1} f(x,\xi)
\ee
\end{enumerate}
Then the matrix $q$ can be seen as the generator of the dual Markov process $\Xi(t)$, that is
\be
L O(x,\xi) = \sum_{\eta} q(\xi,\eta) O(x,\eta)
\ee
with
\be
q(\xi,\eta) \ge 0 \qquad  \mbox{if}\quad \xi \neq \eta
\ee
\be
q(\xi,\xi)  \le 0 \qquad  \mbox{if}\quad \xi = \eta
\ee
\be
\sum_{\eta} q(\xi,\eta) = 0
\ee
Indeed we have:
\begin{eqnarray}
L O(x,\xi)
&=&
L m(\xi)^{-1} f(x,\xi) \nonumber \\
&=&
m(\xi)^{-1} \sum_{\eta} r(\xi,\eta) f(x,\eta) \nonumber \\
&=&
m(\xi)^{-1} \sum_{\eta} m(\xi)q(\xi,\eta) m(\eta)^{-1} m(\eta) O(x,\eta)\nonumber \\
&=&
\sum_{\eta} q(\xi,\eta) O(x,\eta)
\end{eqnarray}
and
\begin{eqnarray}
\sum_{\eta} q(\xi,\eta)
&=&
\sum_{\eta} m(\xi)^{-1} r(\xi,\eta) m(\eta)  \nonumber \\
&=&
m(\xi)^{-1} \sum_{\eta}  r(\xi,\eta) \int f(x,\eta) \pi(x) dx \nonumber \\
&=&
m(\xi)^{-1} \int L f(x,\xi) \pi(x) dx \nonumber \\
&=&
m(\xi)^{-1} \int f(x,\xi) L^* \pi(x) dx \nonumber \\
&=&
0
\end{eqnarray}



\item \underline{Our case}. Among all the invariant measure
of the $X(t)$ process, namely the normalized function with
spherical symmetry $p(x) = p(\sum_i x_i^2)$, a special role is
played by the Gibbs measure
$$
\pi(x)
= \left(\frac{\beta}{2\pi}\right)^{(N/2)} e^{-\beta\sum_i \frac{x_i^2}{2}}
= \left(\frac{\beta}{2\pi}\right)^{(N/2)} \prod_i e^{-\beta\frac{x_i^2}{2}}
$$
which is selected as soon as the system is placed in contact with
thermal bath working at inverse temperature $\beta$.
Moreover: If $Z$ is a centered Gaussian, namely $Z\sim N(0,\sigma^2)$,
then
$$
\E(Z^{2n}) = \sigma^{2n} (2n-1)!!
$$
If one start from
$$
f(x,\xi) = \prod_i x_i^{2\xi}
$$
which satisfy (\ref{aaa}),(\ref{bbb}),(\ref{ccc}) and apply
the previous procedure, one arrives to (\ref{Oss}).

{\bf Remark:} Note that, in applying the procedure, the
dependence on $\beta$ disappear!!!!
\end{itemize}


\section{Quantum language}


Here we start from a quantum spin chain
$$
H = - 4 \sum_i \left( K^+_iK^-_{i+1} + K^-_iK^+_{i+1} -2 K^0_iK^0_{i+1} + \frac{1}{8}\right)
$$
where the spin $K_i$'s satisfy the SU(1,1) algebra
\begin{eqnarray}
\label{commutatorsSU11}
[K_i^{0},K_i^{\pm}] &=& \pm K_i^{\pm} \nonumber \\
{[}K_{i}^{-},K_{i}^{+}{]} &=& 2K_i^{0}
\end{eqnarray}
We are going to see the Schr\"odinger equation with imaginary time
\begin{equation}
\label{schroedinger}
\frac{d}{dt}|\psi(t) \rangle = -H |\psi(t)\rangle\;.
\end{equation}
as the evolution equation for the probability distribution of
a Markovian stochastic process.
\begin{itemize}
\item
\underline{The Hamiltonian possesses the SU(1,1) invariance}. If we define
\be
K^+ = \sum_{i} K_i^+
\ee
\be
K^- = \sum_{i} K_i^-
\ee
\be
K^0 = \sum_{i} K_i^0
\ee
we find that
\be
[H,K^+] = 0
\ee
\be
[H,K^-] = 0
\ee
\be
[H,K^0] = 0
\ee
\item
\underline{Since $[H,K^+] = 0$} there exist a basis to study the stochastic process associated to
$H$ where \underline{$K^+$ is diagonal}. We might consider the following representation
\begin{eqnarray}
\label{Koper}
K^+_i &=& \frac{1}{2}  x_{i}^2 \nonumber \\
K^-_i &=& \frac{1}{2} \frac{\partial^2}{\partial x_{i}^2} \nonumber \\
K^o_i &=& \frac{1}{4}  \left\{\frac{\partial}{\partial x_{i}} x_{i} +
 x_{i} \frac{\partial}{\partial x_{i}} \right \}
\end{eqnarray}
If we use this representation then
$$
H = -L^*
$$
and the probability density function for the $X(t)$ process is encoded in
the state $|\psi(t)\rangle$, namely
\begin{equation}
|\psi(t) \rangle = \int dx p(x,t) |x\rangle
\end{equation}
where we have introduced the notation $|x\rangle$ to denote a completely
localized state, that is a vector which together with its transposed
$\langle x|$ form a complete basis of a Hilbert space and its dual:
\begin{equation}
\langle x|x' \rangle = \delta(x-x')
\end{equation}
It immediately follows that
\begin{equation}
\langle x|\psi(t) \rangle = p(x,t)
\end{equation}
To compute expectation with respect to the $X(t)$ process
we introduce the flat state
\begin{equation}
\langle - | = \int dx \;\langle x|
\end{equation}
which is such that
\begin{equation}
\langle - | x\rangle = 1
\end{equation}
Then for any observable $A = A(X(t))$ we have that its expectation value
at time $t$ can be written as
\begin{equation}
\langle A(t) \rangle_x = \int dy \,A(y)\, p(y,t;x,0) = \langle -|A|  \psi(t) \rangle_x = \langle -|A e ^{-tH}| x\rangle
\end{equation}
\item
\underline{Since $[H,K^0] = 0$} there exist a basis to study the stochastic process associated to
$H$ where \underline{$K^0$ is diagonal}. We might consider the following representation
\begin{eqnarray}
\label{Koper2}
K^+_i|\xi\rangle &=& \left(\frac{1}{2} + \xi\right)  |\xi+1\rangle\nonumber \\
K^-_i|\xi\rangle &=& \xi  |\xi-1\rangle\nonumber \\
K^o_i|\xi\rangle &=& \left(\xi + \frac{1}{4}\right)  |\xi\rangle
\end{eqnarray}
where $|\xi\rangle$ denotes a vector which together with its transposed
$\langle \xi|$ form a complete basis of a Hilbert space and its dual, that is
\begin{equation}
\langle \xi|\eta \rangle = \delta_{\xi,\eta}
\end{equation}
If we use this representation then
$$
H = -{\cal L}^*
$$
and the probability mass function for the $\Xi(t)$ process is encoded in
the state $|\phi(t)\rangle$, namely
\begin{equation}
|\phi(t) \rangle = \sum_{\xi} P(\xi,t) |\xi\rangle
\end{equation}
It immediately follows that
\begin{equation}
\langle \xi|\phi(t) \rangle = P(\xi,t)
\end{equation}
To compute expectation with respect to the $\Xi(t)$ process
we introduce the flat state
\begin{equation}
\langle -_{dual} | = \sum_{\xi} \;\langle \xi|
\end{equation}
which is such that
\begin{equation}
\langle -_{dual} | \xi\rangle = 1
\end{equation}
Then for any observable $A=A(\Xi(t))$ we have that its expectation value
at time $t$ can be written as
\begin{equation}
\langle A(t) \rangle_\xi = \sum_{\eta}\,A(\eta)\, p(\eta,t;\xi,0) = \langle -_{dual}|A|  \phi(t) \rangle_{\xi} = \langle -_{dual}|A e ^{-tH}| \xi\rangle
\end{equation}
\item
\underline{The claim is the following: Duality, in general, is going from the basis
where}\\
\underline{one generator of the group is diagonal to a basis where another generator of}\\
\underline{ the group is diagonal.}

In our case we change from a basis where $K^+$ is diagonal to the base where $K^0$ is diagonal.

\begin{eqnarray}
\langle - |\prod_i\frac{(2K_i^+)^{\xi_i}}{(2\xi_i-1)!!}|\psi(t)\rangle_x
& = &
\int dy \; \langle y |\prod_i\frac{(2K_i^+)^{\xi_i}}{(2\xi_i-1)!!} e^{tL^*}|x\rangle \nonumber \\
& = &
\sum_{\eta} \int dy \; \langle y |\prod_i\frac{(2K_i^+)^{\eta_i}}{(2\eta_i-1)!!}e^{tL^*}|x\rangle \langle \eta|\xi\rangle\nonumber \\
& = &
\sum_{\eta} \int dy \; \langle y| \otimes \langle \eta| \prod_i\frac{(2K_i^+)^{\eta_i}}{(2\eta_i-1)!!} e^{tL^*} | x\rangle \otimes|\xi\rangle\nonumber \\
& = &
\sum_{\eta} \int dy \; \langle x| \otimes \langle \xi | e^{tL} \prod_i\frac{(2K_i^+)^{\eta_i}}{(2\eta_i-1)!!} | y\rangle \otimes|\eta\rangle \nonumber \\
& = &
\sum_{\eta} \int dy \; \langle x| \otimes \langle \xi | e^{tL} \prod_i\frac{y^{2\eta_i}}{(2\eta_i-1)!!} | y\rangle \otimes|\eta \rangle \nonumber \\
& = &
\sum_{\eta} \int dy \; \langle x| \otimes \langle \xi | e^{t{\cal L}} \prod_i\frac{y_i^{2\eta_i}}{(2\eta_i-1)!!} | y\rangle \otimes|\eta \rangle \nonumber \\
& = &
\sum_{\eta} \int dy \; \langle x| \otimes \langle \xi | e^{t{\cal L}} \prod_i\frac{y_i^{K_i^0 -\frac{1}{2}\1}}{(K_i^0 -\frac{3}{2}\1)!!} | y\rangle \otimes|\eta \rangle \nonumber \\
& = &
\sum_{\eta} \int dy \; \langle y| \otimes \langle \eta |\prod_i\frac{y_i^{K_i^0 -\frac{1}{2}\1}}{(K_i^0 -\frac{3}{2}\1)!!} e^{t{\cal L}^*}  | x\rangle \otimes|\xi \rangle\nonumber \\
& = &
\sum_{\eta} \int dy \; \langle \eta | \prod_i\frac{y_i^{K_i^0 -\frac{1}{2}\1}}{(K_i^0 -\frac{3}{2}\1)!!} e^{t{\cal L}^*}  |\xi \rangle   \langle y | x\rangle   \nonumber \\
& = &
\sum_{\eta} \langle \eta | \prod_i\frac{y_i^{K_i^0 -\frac{1}{2}\1}}{(K_i^0 -\frac{3}{2}\1)!!} |\phi(t)\rangle_{\xi}  \nonumber \\
& = &
\langle -_{dual} |\prod_i\frac{y_i^{K_i^0 -\frac{1}{2}\1}}{(K_i^0 -\frac{3}{2}\1)!!}|\phi(t)\rangle_{\xi}
\end{eqnarray}


\end{itemize}

\section{General k}

A convenient $(2j+1)$-dimensional representation of the SU(2) algebra is given by
\begin{eqnarray}
J^+_i |n_i\rangle &=& (2j-n_i) |n_i+1\rangle \nonumber \\
J^-_i |n_i\rangle &=& n_i      |n_i-1\rangle \nonumber \\
J^0_i |n_i\rangle &=& (n_i-j)  |n_i\rangle
\end{eqnarray}
where the quantum numbers $n_i\in\{0,1,\ldots,2j\}$.
{\bf Note that in this representation the adjoint of $J^+_i$ is not
$J^-_i$, UNLESS $j=1/2$}.

A matrix representation is:
$$
J^+ = \left(
\begin{array}{cccc}
 0  &        &        &   \\
 2j & \ddots &        &   \\
    & \ddots & \ddots &   \\
    &        &    1   &  0\\
\end{array}\right)
\qquad
J^- = \left(
\begin{array}{cccc}
 0  &  1     &        &     \\
    & \ddots & \ddots &     \\
    &        & \ddots &  2j \\
    &        &        &  0  \\
\end{array}\right)
\qquad
J^0 = \left(
\begin{array}{cccc}
 -j  &        &        &   \\
     & \ddots &        &   \\
     &        & \ddots &   \\
     &        &        &  j\\
\end{array}\right)
$$

In the SU(1,1) case one can use the infinite dimensional representation
\begin{eqnarray}
\label{newrepresentationsu11}
K^+_i |n_i\rangle &=& (2k+n_i) |n_i+1\rangle \nonumber \\
K^-_i |n_i\rangle &=& n_i      |n_i-1\rangle \nonumber \\
K^0_i |n_i\rangle &=& (n_i+k)  |n_i\rangle
\end{eqnarray}
where the quantum numbers $n_i\in\{0,1,2,\ldots\}$.
A matrix representation is:
$$
K^+ = \left(
\begin{array}{cccc}
 0  &        &        &   \\
 2k & \ddots &        &   \\
    & 2k+1   & \ddots &   \\
    &        & \ddots & \ddots\\
\end{array}\right)
\qquad
K^- = \left(
\begin{array}{cccc}
 0  &  1     &        &     \\
    & \ddots &   2    &     \\
    &        & \ddots &  \ddots \\
    &        &        &  \ddots \\
\end{array}\right)
\qquad
K^0 = \left(
\begin{array}{cccc}
 k   &        &        &   \\
     &  k+1   &        &   \\
     &        & k+2    &   \\
     &        &        &  \ddots\\
\end{array}\right)
$$
Let's check that in this representation the operator is stochastic.
I will do it for the bulk:
\begin{eqnarray}
L_{i,i+1}|n_i,n_{i+1}\rangle
&=&
(2k+n_i) n_{i+1}|n_i +1 ,n_{i+1}-1\rangle \nonumber\\
&+&
n_i(2k+n_{i+1})|n_i -1 ,n_{i+1}+1\rangle \nonumber\\
&+&
(-2(n_i+k)(n_{i+1}+k)+2k^2)|n_i,n_{i+1}\rangle
\end{eqnarray}
The sum of the rates is
$$
(2k+n_i) n_{i+1}+
n_i(2k+n_{i+1})
-2(n_i+k)(n_{i+1}+k)+2k^2 =0
$$









% \end{document}