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diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex index 0da2e88..741cc80 100644 --- a/frsb_kac-rice.tex +++ b/frsb_kac-rice.tex @@ -1287,40 +1287,51 @@ exponentially many stationary points. One can construct a schematic 2RSB model from two 1RSB models. Consider two independent pure models with $p_1$ and $p_2$ couplings, -respectively, with energies $H_{p_1}({\mathbf s})$ and $H_{p_2}({\mathbf -\sigma})$ of sizes $N$, and couple them weakly with $\varepsilon \; {\mathbf -\sigma} \cdot {\mathbf s}$. The landscape of the pure models is much simpler -than that of the mixed because, in these models, fixing the stability $\mu$ is -equivalent to fixing the energy: $\mu=pE$. This implies that at each energy -level there is only one type of stationary point. Therefore, for the pure -models our formulas for the complexity and its Legendre transforms are -functions of one variable only, $E$, and each instance of $\mu^*$ inside mus be -replaced with $pE$. +respectively, with energies $H_{p_1}({\mathbf s})$ and +$H_{p_2}({\boldsymbol\sigma})$ of sizes $N$, and couple them weakly with +$\varepsilon \; {\boldsymbol \sigma} \cdot {\mathbf s}$. The landscape of the +pure models is much simpler than that of the mixed because, in these models, +fixing the stability $\mu$ is equivalent to fixing the energy: $\mu=pE$. This +implies that at each energy level there is only one type of stationary point. +Therefore, for the pure models our formulas for the complexity and its Legendre +transforms are functions of one variable only, $E$, and each instance of +$\mu^*$ inside must be replaced with $pE$. In the joint model, we wish to fix the total energy, not the energies of the individual two models. Therefore, we insert a $\delta$-function containing $(E_1+E_2)-E$ and integrate over $E_1$ and $E_2$. This results in a joint complexity (and Legendre transform) -\begin{eqnarray} - e^{N\Sigma(E)}&=&\int dE_1\, dE_2\, d\lambda \, e^{N[ \Sigma_1(E_1) + \Sigma_2(E_2) + O(\varepsilon) -\lambda N [(E_1+E_2)-E]}\nonumber \\ - e^{NG(\hat \beta)}&=&\int dE\, dE_1\, dE_2\, d\lambda\, e^{N[-\hat \beta E+ \Sigma_1(E_1) + \Sigma_2(E_2) + O(\varepsilon) -\lambda N [(E_1+E_2)-E]} -\end{eqnarray} -The maximum is given by $\Sigma_1'=\Sigma_2'=\hat \beta$, provided it occurs in the phase -in which both $\Sigma_1$ and $\Sigma_2$ are non-zero. The two systems are `thermalized', -and it is easy to see that, because many points contribute, the overlap between two -global configurations $$\frac1 {2N}({\mathbf s^1},{\mathbf \sigma^1})\cdot ({\mathbf s^2},{\mathbf \sigma^2})=\frac1 {2N}[ {\mathbf s^1}\cdot {\mathbf s^2}+ {\mathbf \sigma^1}\cdot {\mathbf \sigma^2}] =0$$ This is the `annealed' phase of a Kac-Rice calculation. - -Now start going down in energy, or up in $\hat \beta$: there will be a point $e_c$, $\hat \beta_c$ -at which one of the subsystems freezes at its lower energy density, say it is system one, -while system two is not yet frozen. At an even higher $\hat \beta=\hat \beta_f$, both systems are frozen. -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: -$$\frac1 {2N}({\mathbf s^1},{\mathbf \sigma^1})\cdot ({\mathbf s^2},{\mathbf \sigma^2}) =\frac1 {2N}[ {\mathbf s^1}\cdot {\mathbf s^2}+ {\mathbf \sigma^1}\cdot {\mathbf \sigma^2}] = \frac1 {2N} {\mathbf s^1}\cdot {\mathbf s^2} $$ -The distribution of this overlap is the one 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 $\hat \beta$, starting at $x=1$ at $\hat \beta_c$ and decreasing wuth increasing $\hat \beta$. -Globally, the joint Kac-Rice system is $1RSB$, but note that the global overlap between different states is at most $1/2$. -At $\hat \beta>\hat \beta_f$ there is a further transition. +\begin{align} + e^{N\Sigma(E)}&=\int dE_1\, dE_2\, d\lambda \,\exp\left\{N\left[\Sigma_1(E_1) + \Sigma_2(E_2) + O(\varepsilon) -\lambda\Big((E_1+E_2)-E\Big)\right]\right\} \\ + e^{NG(\hat \beta)}&=\int dE\, dE_1\, dE_2\, d\lambda\, \exp\left\{N\left[-\hat\beta E+\Sigma_1(E_1) + \Sigma_2(E_2) + O(\varepsilon) -\lambda\Big((E_1+E_2)-E\Big)\right]\right\} +\end{align} +The saddle point is given by $\Sigma_1'(E_1)=\Sigma_2'(E_2)=\hat\beta$, provided that +both $\Sigma_1(E_1)$ and $\Sigma_2(E_2)$ are non-zero. In this situation, two systems are `thermalized', +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 +\end{equation} +This is the `annealed' phase of a Kac-Rice calculation. + +Now start going down in energy, or up in $\hat \beta$: there will be a point +$E_c$ or $\hat \beta_c$ at which one of the subsystems (say it is system one) +freezes at its lowest energy density, while system two is not yet frozen. At +this point, $\Sigma_1(E_1)=0$ and $E_1$ is the ground state energy. At an even +higher value $\hat \beta=\hat \beta_f$, both systems will become frozen in +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 +\end{equation} +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 $\hat \beta$, +starting at $x=1$ at $\hat \beta_c$ and decreasing with increasing $\hat +\beta$. Globally, the joint complexity of the system is 1RSB, but note that the +global overlap between different states is at most $1/2$. At $\hat \beta>\hat +\beta_f$ there is a further transition. \subsection{\textit{R} and \textit{D}: response functions} |