T-I

The REM: the energy landscape

To characterize the energy landscape of the REM, we determine the number ${𝒩}(E)dE$ of configurations having energy $E_{\alpha }\in [E,E+dE]$. This quantity is a random variable. For large $N$, its typical value is given by

${𝒩}(E)=e^{N\mathrm{\Sigma }\left(\frac{E}{N}\right)+o(N)},\mathrm{\Sigma }(ϵ)=\{\begin{array}{ll}\log 2-{ϵ}^2& \text{ if }|ϵ|\le \sqrt{\log 2}\\ 0& \text{ if }|ϵ|>\sqrt{\log 2}\end{array}.$

The function $\mathrm{\Sigma }(ϵ)$ is the entropy of the model, and it is sketched in Fig. X. The point where the entropy vanishes, $ϵ=-\sqrt{\log 2}$, is the energy density of the ground state, consistently with what we obtained with extreme values statistics. The entropy is maximal at $ϵ=0$: the highest number of configurations have vanishing energy density.

• The annealed entropy. The annealed entropy ${\mathrm{\Sigma }}^A$ is a function that controls the behaviour of the average number of configurations at a given energy, $\overline{{𝒩}(E)}=e^{N{\mathrm{\Sigma }}^A\left(\frac{E}{N}\right)+o(N)}$. To compute it, write ${𝒩}(E)dE=\sum _{\alpha =1}^{2^N}{\chi }_{\alpha }(E)dE$ with ${\chi }_{\alpha }(E)=1$ if $E_{\alpha }\in [E,E+dE]$ and ${\chi }_{\alpha }(E)=0$ otherwise. Use this together with $p(E)$ to obtain ${\mathrm{\Sigma }}^A$ : when does this function coincide with the entropy defined above?

• Self-averaging quantities. For $|ϵ|\le \sqrt{\log 2}$ the quantity ${𝒩}(E)$ is self-averaging. This means that its distribution concentrates around the average value $\overline{{𝒩}}(E)$ when $N\to \mathrm{\infty }$. Show that this is the case by computing the second moment $\overline{{{𝒩}}^2}$ and using the central limit theorem. Show that this is no longer true in the region where the annealed entropy is negative.

• Average vs typical number. For $|ϵ|>\sqrt{\log 2}$ the annealed entropy is negative, meaning that the average number of configurations with those energy densities is exponentially small in $N$. This implies that configurations with those energy are exponentially rare: do you have an idea of how to show this, using the expression for $\overline{{𝒩}(E)}$? Why is the entropy $\mathrm{\Sigma }(ϵ)$, controlling the typical value of ${𝒩}(E)$, zero in this region? Why the point where the entropy vanishes coincides with the ground state energy of the model?

this will be responsible of the fact that the partition function $Z$ is not self-averaging in the low-T phase, as we discuss below.

The REM: the free energy and the freezing transition

We now compute the equilibrium phase diagram of the model, and in particular the free energy density $f$. The partition function reads

$Z=\sum _{\alpha =1}^{2^N}{e}^{-\beta E_{\alpha }}=\int dE{𝒩}(E)e^{-\beta E}\equiv e^{-\beta Nf+o(N)}.$

We have determined above the behaviour of the typical value of ${𝒩}$ for large $N$. The typical value of the partition function is therefore

$Z=\int dE{𝒩}(E)e^{-\beta E}=\int dϵe^{N[\mathrm{\Sigma }(ϵ)-\beta ϵ]+o(N)}.$

• The critical temperature. In the limit of large $N$, the integral defining $Z$ can be computed with the saddle point method; show that a transition occurs at a critical temperature $T_c=(2\sqrt{\log 2}{)}^{-1}$, and that the free energy density reads

$f=-\frac{1}{\beta }\underset{N\to \mathrm{\infty }}{\lim }\frac{\log Z}{N}=\{\begin{array}{ll}& -(T\log 2+\frac{1}{4T})\text{if}T\ge T_c\\ & -\sqrt{\log 2}\text{if}T<T_c\end{array}$

• Freezing. The transition occurring at $T_c$ is called a freezing transition.