diff options
| -rw-r--r-- | bezout.bib | 17 | ||||
| -rw-r--r-- | bezout.tex | 2 | ||||
| -rw-r--r-- | cover.tex | 1419 |
3 files changed, 87 insertions, 1351 deletions
@@ -267,6 +267,16 @@ doi = {10.1007/jhep11(2012)023} } +@book{Mezard_2009_Information, + author = {Mézard, Marc and Montanari, Andrea}, + title = {Information, physics, and computation}, + publisher = {Oxford University Press}, + year = {2009}, + address = {Great Clarendon Street, Oxford}, + isbn = {9780198570837}, + series = {Oxford Graduate Texts} +} + @article{Nguyen_2014_The, author = {Nguyen, Hoi H. and O'Rourke, Sean}, title = {The Elliptic Law}, @@ -294,12 +304,7 @@ 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}, @@ -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 \cite{mezard2009information}. The most tractable family of these +most notably constraint satisfaction \cite{Mezard_2009_Information}. 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} @@ -1,1358 +1,89 @@ - -\documentclass[12pt,reqno,a4paper,twoside]{article} -% \ProvidesPackage{makra} -\usepackage{amsmath,amsthm,amstext,amscd,amssymb,euscript} -%,showkeys} -%,times} -\usepackage{epsf} -\usepackage{color} -\usepackage{verbatim} -\usepackage{graphicx} -\usepackage{esint} -\usepackage{tikz} -\usepackage{setspace} -\usepackage{mathrsfs} - -\usepackage{todonotes} - -%\usepackage{natbib} - - - -\usepackage{bm} -\usepackage[normalem]{ulem} - - -\textwidth 6in -\topmargin -0.50in -\oddsidemargin 0in -\evensidemargin 0in -\textheight 9.00in -%\pagestyle{plain} -%%%%%%%%%%%%%%%%%% Macros %%%%%%%%%%% -\def\mybox #1{\fbox{\parbox{5.8in}{#1}}} -\newcommand{\m}[1]{{\marginpar{\scriptsize #1}}} - -\def\mep{\mathbf{mep}_{n}^{\delta}} -\def\r{{\mathbf r}} -\def\O{{\mathcal{O}}} - -\def\I{{\mathcal{I}}} -\def\fee{\mathcal{F}} - -\def\F{{\EuScript{F}}} - -\renewcommand{\phi}{\varphi} -\newcommand{\compose}{\circ} -\renewcommand{\subset}{\subseteq} -\renewcommand{\emptyset}{\varnothing} -\newcommand{\interval}{[\underline\alpha,\overline \alpha]} -\def\liminfn{\liminf_{n\to\infty}} -\def\limsupn{\limsup_{n\to\infty}} -\def\limn{\lim_{n\to\infty}} -\def\disagree{\not\longleftrightarrow} -\newcommand{\Zd}{\mathbb Z^d} -\newcommand{\kk}{\mathbf k} -\renewcommand{\Pr}{\mathbb P} -\newcommand{\dist}{\text{dist}} -\newcommand{\Cal}{\mathcal} -\def\1{ {\mathit{1} \!\!\>\!\! 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+\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 } -% -\newcommand{\graphh}{ -\begin{tikzpicture}[scale=.7] -\draw[gray, thick] (1,-1.5)--(0,0) -- (2,3)--(4,3.4)--(5.0,2)--(4.9,1)--(3.0,-1.2)--(1,-1.5); -\tikzstyle myBG=[line width=3.5pt,opacity=1.0] -% -\draw[white,myBG] (3.2,1.5) -- (5.2,3); -\draw[gray, thick] (3.2,1.5) -- (5.2,3); -% -\draw[gray, thick] (0,0)--(-1.5,0); -\draw[gray, thick] (3,-1.2)--(4,-2); -\draw[gray, thick] (4,3.4)--(4.3,4.6); -\draw[gray, thick] (1,-1.5)--(2,0.2)--(0,0); -\draw[gray, thick] (2,0.2)--(4.9,1); -\draw[gray, thick] (3,-1.2)--(2,0.2)--(3.2,1.5)--(2,3); -\draw[gray, thick] (3.2,1.5)--(4,3.4); -\draw[gray, thick] (3.2,1.5)--(5,2); -\draw[gray, thick] (4,3.4)--(5.2,3)--(5.7,2.2)--(5.7,1.3)--(4.9,1); -\draw[gray, thick] (5.7,2.2)--(5,2); -% -% -% -\filldraw[black!70] (0,0) circle (2pt); -\filldraw[black!70] (1,-1.5) circle (2pt); -\filldraw[black!70] (2,3) circle (2pt); -\filldraw[black!70] (4,3.4) circle (2pt); -\filldraw[black!70] (5,2) circle (2pt); -\filldraw[black!70] (4.9,1) circle (2pt); -\filldraw[black!70] (2,0.2) circle (2pt); -\filldraw[black!70] (3.2,1.5) circle (2pt); -% \filldraw[black!70] (5.7,1.3) circle (2pt); -\filldraw[black!70] (5.7,2.2) circle (2pt); -\filldraw[black!70] (5.2,3) circle (2pt); -\filldraw[black!70] (3,-1.2) circle (2pt); -% -% -\filldraw[black!70] (-1.5,0) circle (2pt) node[anchor=east] {$\Gamma_3,T$}; -\filldraw[white] (-1.5,0) circle (1pt); -\filldraw[black!70] (4,-2) circle (2pt) node[anchor=west] {$\Gamma_2,T$}; -\filldraw[white] (4,-2) circle (1pt); -\filldraw[black!70] (4.3,4.6) circle (2pt) node[anchor=west] {$\Gamma_1,T$}; -\filldraw[white] (4.3,4.6) circle (1pt) ; -\end{tikzpicture} +\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} |