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| -rw-r--r-- | frsb_kac-rice.bib | 4 | ||||
| -rw-r--r-- | frsb_kac-rice.tex | 64 |
2 files changed, 43 insertions, 25 deletions
diff --git a/frsb_kac-rice.bib b/frsb_kac-rice.bib index 2d6ed23..f4740ad 100644 --- a/frsb_kac-rice.bib +++ b/frsb_kac-rice.bib @@ -348,14 +348,14 @@ stochastic localization}, @phdthesis{Folena_2020_The, author = {Folena, Giampaolo}, - title = {The mixed $p$-spin model: selecting, following and losing states}, + title = {The mixed {$p$}-spin model: selecting, following and losing states}, year = {2020}, month = {3}, number = {2020UPASS060}, url = {https://tel.archives-ouvertes.fr/tel-02883385}, hal_id = {tel-02883385}, hal_version = {v1}, - school = {Université Paris-Saclay ; Università degli studi La Sapienza (Rome)}, + school = {Université Paris-Saclay \& Università degli studi La Sapienza (Rome)}, type = {Theses} } diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex index 680cc89..f5a259e 100644 --- a/frsb_kac-rice.tex +++ b/frsb_kac-rice.tex @@ -13,7 +13,11 @@ filecolor=MidnightBlue, linkcolor=MidnightBlue ]{hyperref} % ref and cite links with pretty colors -\usepackage[style=phys,eprint=true]{biblatex} +\usepackage[ + style=phys, + eprint=true, + maxnames = 100 +]{biblatex} \addbibresource{frsb_kac-rice.bib} @@ -25,11 +29,11 @@ \author{Jaron Kent-Dobias \& Jorge Kurchan} \maketitle \begin{abstract} - We derive the general solution for counting the stationary points of - mean-field complex landscapes. It incorporates Parisi's solution - for the ground state, as it should. Using this solution, we count the - stationary points of two models: one with multi-step replica symmetry - breaking, and one with full replica symmetry breaking. + We derive the general solution for counting the stationary points of + mean-field complex landscapes. It incorporates Parisi's solution + for the ground state, as it should. Using this solution, we count the + stationary points of two models: one with multi-step replica symmetry + breaking, and one with full replica symmetry breaking. \end{abstract} \section{Introduction} @@ -166,7 +170,8 @@ saddle (see Fyodorov \cite{Fyodorov_2007_Replica} for a detailed discussion). Here we review the equilibrium solution, which has been studied in detail \cite{Crisanti_1992_The, Crisanti_1993_The, Crisanti_2004_Spherical, -Crisanti_2006_Spherical}. The free energy, averaged over disorder, is +Crisanti_2006_Spherical}. For a succinct review, see \cite{Folena_2020_The}. +The free energy, averaged over disorder, is \begin{equation} \beta F = - \overline{\ln \int d\mathbf s \;\delta(\|\mathbf s\|^2-N)\, e^{-\beta H(\mathbf s)}} \end{equation} @@ -1322,7 +1327,9 @@ both $\Sigma_1(E_1)$ and $\Sigma_2(E_2)$ are non-zero. In this situation, two sy and, because many points contribute, the overlap between two global configurations is zero: \begin{equation} - \frac1{2N}({\mathbf s^1},{\boldsymbol\sigma^1})\cdot ({\mathbf s^2},{\boldsymbol\sigma^2})=\frac1 {2N}[ {\mathbf s^1}\cdot {\mathbf s^2}+ {\boldsymbol\sigma^1}\cdot {\boldsymbol\sigma^2}] =0 + \frac1 {2N}\Big\langle({\mathbf s^1},{\boldsymbol\sigma^1})\cdot ({\mathbf s^2},{\boldsymbol\sigma^2})\Big\rangle + =\frac1 {2N}\Big[\langle{\mathbf s^1}\cdot {\mathbf s^2}\rangle+ \langle{\boldsymbol\sigma^1}\cdot {\boldsymbol\sigma^2}\rangle\Big] + =0 \end{equation} This is the `annealed' phase of a Kac-Rice calculation. @@ -1335,8 +1342,12 @@ their ground states. For $\hat \beta_f > \hat \beta> \hat \beta_c$ one system is unfrozen, while the other is, because of coupling, frozen at inverse temperature $\hat \beta_c$. The overlap between two solutions is: \begin{equation} - \frac1 {2N}({\mathbf s^1},{\boldsymbol\sigma^1})\cdot ({\mathbf s^2},{\boldsymbol\sigma^2}) =\frac1 {2N}[ {\mathbf s^1}\cdot {\mathbf s^2}+ {\boldsymbol\sigma^1}\cdot {\boldsymbol\sigma^2}] = \frac1 {2N} {\mathbf s^1}\cdot {\mathbf s^2}\neq0 + \frac1 {2N}\Big\langle({\mathbf s^1},{\boldsymbol\sigma^1})\cdot ({\mathbf s^2},{\boldsymbol\sigma^2})\Big\rangle + =\frac1 {2N}\Big[\langle{\mathbf s^1}\cdot {\mathbf s^2}\rangle+ \langle{\boldsymbol\sigma^1}\cdot {\boldsymbol\sigma^2}\rangle\Big] + = \frac1 {2N}\langle{\mathbf s^1}\cdot {\mathbf s^2}\rangle>0 \end{equation} +nonzero because there are few low-energy stationary points in system one, and +there is a nonvanishing probability of selecting one of them twice. The distribution of this overlap is one-half the overlap distribution of a frozen spin-glass at temperature $\hat \beta$, a 1RSB system like the Random Energy Model. The value of $x$ corresponding to it depends on $\hat \beta$, @@ -1345,35 +1356,44 @@ starting at $x=1$ at $\hat \beta_c$ and decreasing with increasing $\hat global overlap between different states is at most $1/2$. At $\hat \beta>\hat \beta_f$ there is a further transition. +This schematic example provides a metaphor for considering what happens in +ordinary models when replica symmetry is broken. At some point certain degree +of freedom `freeze' onto a subextensive number of possible states, while the +remainder are effectively unconstrained. The overlap measures something in the +competition between the number of these unconstrained subregions and their +size. + \subsection{\textit{R} and \textit{D}: response functions} The matrix field $R$ is related to responses of the stationary points to perturbations of the tensors $J$. One adds to the Hamiltonian a random term -$\varepsilon \tilde H_p = \varepsilon \sum_{i_1,...,i_p} \tilde J_{i_1,...,i_p} +$\varepsilon_p \tilde H_p = -\varepsilon_p \sum_{i_1,...,i_p} \tilde J_{i_1,...,i_p} s_{i_1}...s_{i_p}$, where the $\tilde J$ are random Gaussian uncorrelated with -the $J$'s. ] The response to these is: +the $J$'s and having variance $\overline{(\tilde J)^2}=p!/2N^{p-1}$. The response to these is: \begin{equation} - \begin{aligned} - & \overline{ \frac{\partial \langle \tilde H_p \rangle_{\tilde J} } {\partial \varepsilon} } - % \frac1{N^p}\sum_{i_1\cdots i_p}\frac{\partial\langle s_{i_1}\cdots s_{i_p}\rangle}{\partial J^{(p)}_{i_1\cdots i_p}} - % &=\lim_{n\to0}\frac1{N^p}\sum_{i_1\cdots i_p}\frac\partial{\partial J^{(p)}_{i_1\cdots i_p}} - % \int\left(\prod_a^nd\nu(\mathbf s_a)\right)\,s^1_{i_1}\cdots s^1_{i_p} \\ - & =\lim_{n\to0}\int\left(\prod_a^nd\nu(\mathbf s_a)\right)\sum_b^n\left[ +\frac1N\overline{\frac{\partial \langle \tilde H_p \rangle } {\partial \varepsilon_p}} + =\lim_{n\to0}\int\left(\prod_a^nd\nu(\mathbf s_a)\right)\sum_b^n\left[ \hat\beta\left(\frac{\mathbf s_1\cdot\mathbf s_b}N\right)^p+ p\left(-i\frac{\mathbf s_1\cdot\hat{\mathbf s}_b}N\right)\left(\frac{\mathbf s_1\cdot\mathbf s_b}N\right)^{p-1} \right] - \end{aligned} \end{equation} Taking the average of this expression over disorder and averaging over the equivalent replicas in the integral gives, similar to before, \begin{equation} \begin{aligned} - \overline{ \frac{\partial \langle \tilde H_p \rangle_{\tilde J} } {\partial \varepsilon} } + \frac1N\overline{\frac{\partial\langle \tilde H_p \rangle } {\partial \varepsilon_p}} % \overline{\frac1{N^p}\sum_{i_1\cdots i_p}\frac{\partial\langle s_{i_1}\cdots s_{i_p}\rangle}{\partial J^{(p)}_{i_1\cdots i_p}}} &=\lim_{n\to0}\int D[C,R,D]\,\frac1n\sum_{ab}^n(\hat\beta C_{ab}^p+pR_{ab}C_{ab}^{p-1})e^{nN\Sigma[C,R,D]}\\ &=\hat\beta+pr_d-\int_0^1dx\,c^{p-1}(x)(\hat\beta c(x)+pr(x)) \end{aligned} \end{equation} -In particular, when the energy is unconstrained ($\hat\beta=0$) and there is replica symmetry, +The responses as defined by this average perturbation in the pure $p$-spin +energy can be directly related to responses in the tensor polarization of the +stationary points: +\begin{equation} + \frac1{N^p}\overline{\sum_{i_1\cdots i_p}\frac{\partial\langle s_{i_1}\cdots s_{i_p}\rangle}{\partial J^{(p)}_{i_1\cdots i_p}}} + =\frac1N\overline{\frac{\partial\langle \tilde H_p \rangle } {\partial \varepsilon_p}} +\end{equation} +In particular, when the energy is unconstrained ($\hat\beta=0$) and there is replica symmetry, the above formulas imply that \begin{equation} \frac1N\sum_i\frac{\partial\langle s_i\rangle}{\partial J_i^{(1)}} =r_d @@ -1415,10 +1435,8 @@ and in particular for $p=1$ i.e., the change in complexity due to a linear field is directly related to the resulting magnetization of the stationary points. - - - \section{Conclusion} +\label{se:conclusion} We have constructed a replica solution for the general problem of finding saddles of random mean-field landscapes, including systems with many steps of |