Update on Overleaf.

3 files changed, 1367 insertions, 4 deletions

diff --git a/bezout.bib b/bezout.bib
index c654568..8688ed2 100644
--- a/bezout.bib
+++ b/bezout.bib
@@ -294,7 +294,12 @@
  url = {https://doi.org/10.2307%2F2371510},
  doi = {10.2307/2371510}
 }
-
+@book{mezard2009information,
+  title={Information, physics, and computation},
+  author={Mezard, Marc and Montanari, Andrea},
+  year={2009},
+  publisher={Oxford University Press}
+}
 @inproceedings{Scorzato_2016_The,
  author = {Scorzato, Luigi},
  title = {The {Lefschetz} thimble and the sign problem},
diff --git a/bezout.tex b/bezout.tex
index 9dffe65..96ad257 100644
--- a/bezout.tex
+++ b/bezout.tex
@@ -44,7 +44,7 @@
 
 Spin-glasses have long been considered the paradigm of many variable `complex
 landscapes,' a subject that includes neural networks and optimization problems,
-most notably constraint satisfaction.  The most tractable family of these
+most notably constraint satisfaction \cite{mezard2009information}.  The most tractable family of these
 are the mean-field spherical $p$-spin models \cite{Crisanti_1992_The} (for a
 review see \cite{Castellani_2005_Spin-glass}) defined by the energy
 \begin{equation} \label{eq:bare.hamiltonian}
@@ -447,8 +447,8 @@ dynamics, are a problem we hope to address in future work.
 
  This paper provides a first step towards the study of a complex landscape with
  complex variables. The next obvious one is to study the topology of the
- critical points, their basins of attraction following  gradient ascent (the
- Lefschetz thimbles), and descent (the anti-thimbles) \cite{Witten_2010_A,
+ critical points, the sets reached following  gradient descent (the
+ Lefschetz thimbles), and ascent (the anti-thimbles) \cite{Witten_2010_A,
  Witten_2011_Analytic, Cristoforetti_2012_New, Behtash_2017_Toward,
  Scorzato_2016_The}, which act as constant-phase integrating `contours.'
  Locating and counting the saddles that are joined by gradient lines---the
diff --git a/cover.tex b/cover.tex
new file mode 100644
index 0000000..586fa38
--- /dev/null
+++ b/cover.tex
@@ -0,0 +1,1358 @@
+
+\documentclass[12pt,reqno,a4paper,twoside]{article}
+% \ProvidesPackage{makra}
+\usepackage{amsmath,amsthm,amstext,amscd,amssymb,euscript}
+%,showkeys}
+%,times}
+\usepackage{epsf}
+\usepackage{color}
+\usepackage{verbatim}
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+\usepackage{esint}
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+\usepackage{setspace}
+\usepackage{mathrsfs}
+
+\usepackage{todonotes}
+
+%\usepackage{natbib}
+
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+
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+          12:\ifnum\time<10 0\fi\number\time am
+          \else
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+
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+}
+% 
+
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+\filldraw[black!70] (5.2,3) circle (2pt);
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+}
+
+\setstretch{1.24}
+
+\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}