Letter format and bibliography updates.

1 files changed, 75 insertions, 1344 deletions

diff --git a/cover.tex b/cover.tex
index 586fa38..2d2fc10 100644
--- a/cover.tex
+++ b/cover.tex
@@ -1,1358 +1,89 @@
-
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+\usepackage[utf8]{inputenc} % why not type "Bézout" with unicode?
+\usepackage[T1]{fontenc} % vector fonts plz
+\usepackage{newtxtext,newtxmath} % Times for PR
+\usepackage[
+  colorlinks=true,
+  urlcolor=purple,
+  citecolor=purple,
+  filecolor=purple,
+  linkcolor=purple
+]{hyperref} % ref and cite links with pretty colors
+\usepackage{xcolor}
+
+\signature{
+  Jaron Kent-Dobias \& Jorge Kurchan
 }
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+\address{
+  Laboratoire de Physique \\
+  Ecole Normale Sup\'erieure \\
+  24, rue Lhomond \\ 
+  75005 Paris
 }
 
-\setstretch{1.24}
-
 \begin{document}
-
-\title{{\bf COVER LETTER \\`Complex complex landscapes'}}
-%\footnote{{\bf Key-words}: }
-%}}
-
-\author{
-Jaron Kent-Dobias
- and 
-Jorge Kurchan
+\begin{letter}{
+  Editorial Office\\
+  Physical Review Letters\\
+  1 Research Road\\
+  Ridge, NY 11961
 }
 
-\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.
-
-
+\opening{}
+
+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{Mezard_2009_Information}.
+
+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.
+
+\closing{Sincerely,}
+\end{letter}
+
+\bibliographystyle{unsrt}
 \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}