TBan-III

Exercise 1: Back to REM

The Random Energy Model (REM) exhibits two distinct phases:

• High-Temperature Phase:

At high temperatures, the system is in a paramagnetic phase where the entropy is extensive, and the occupation probability of a configuration is approximately $\sim 1/M$.

• Low-Temperature Phase:

Below a critical freezing temperature $T_{f}$, the system transitions into a glassy phase. In this phase, the entropy becomes subextensive (i.e., the extensive contribution vanishes), and only a few configurations are visited with finite, $M$-independent probabilities.

Calculating the Freezing Temperature $T_{f}$

Thanks to the computation of ${\overline {n(x)}}$, we can identify the fingerprints of the glassy phase and calculate $T_{f}$. Let's compare the weight of the ground state against the weight of all other states:

${\frac {\sum _{\alpha }z_{\alpha }}{z_{\alpha _{\min }}}}=1+\sum _{\alpha \neq \alpha _{\min }}{\frac {z_{\alpha }}{z_{\alpha _{\min }}}}=1+\sum _{\alpha \neq \alpha _{\min }}e^{-\beta (E_{\alpha }-E_{\min })}\sim 1+\int _{0}^{\infty }dx\,{\frac {d{\overline {n(x)}}}{dx}}\,e^{-\beta x}=1+\int _{0}^{\infty }dx\,{\frac {e^{x/b_{M}}}{b_{M}}}\,e^{-\beta x}$

Behavior in Different Phases:

• High-Temperature Phase ($T>T_{f}=b_{M}=1/{\sqrt {2\log 2}}$):

In this regime, the weight of the excited states diverges. This indicates that the ground state is not deep enough to render the system glassy.

• Low-Temperature Phase ($T<T_{f}=b_{M}=1/{\sqrt {2\log 2}}$):

In this regime, the integral is finite:

$\int _{0}^{\infty }dx\,e^{(1/b_{M}-\beta )x}/b_{M}={\frac {1}{\beta b_{M}-1}}={\frac {T}{T_{f}-T}}$

This result implies that below the freezing temperature $T_{f}$, the weight of all excited states is of the same order as the weight of the ground state. Consequently, the ground state is occupied with a finite probability, reminiscent of Bose-Einstein condensation.

Exercise 2: Edwards-Wilkinson Interface with Stationary Initial Condition

Consider an Edwards-Wilkinson interface in 1+1 dimensions, at temperature $T$, and of length $L$ with periodic boundary conditions:

${\frac {\partial h(x,t)}{\partial t}}=\nu \nabla ^{2}h(x,t)+\eta (x,t)$

where $\eta (x,t)$ is a Gaussian white noise with zero mean and variance:

$\langle \eta (x,t)\eta (x',t')\rangle =2T\delta (x-x')\delta (t-t')$

The solution can be written in Fourier space as:

${\hat {h}}_{q}(t)={\hat {h}}_{q}(0)e^{-\nu q^{2}t}+\int _{0}^{t}ds\,e^{-\nu q^{2}(t-s)}\eta _{q}(s)$

with Fourier decomposition:

${\hat {h}}_{q}(t)={\frac {1}{L}}\int _{0}^{L}e^{iqx}h(x,t),\quad h(x,t)=\sum _{q}e^{-iqx}{\hat {h}}_{q}(t),\quad \langle \eta _{q_{1}}(t')\eta _{q_{2}}(t)\rangle ={\frac {2T}{L}}\delta _{q_{1},-q_{2}}\delta (t-t')$

where $q=2\pi n/L,\;n=\ldots ,-1,0,1,\ldots$.

In class, we computed the width of the interface starting from a flat interface at $t=0$, i.e., $h(x,0)=0$. The mean square displacement of a point $h(x,t)$ is similar but includes also the contribution of the zero mode. The result is:

$\langle \Delta h_{\text{flat}}^{2}\rangle =2{\frac {T}{L}}t+{\begin{cases}T{\sqrt {\frac {2t}{\pi \nu }}},&t\ll L^{2},\\[2mm]{\frac {T}{\nu }}{\frac {L}{12}},&t\gg L^{2}.\end{cases}}$

The first term describes the diffusion of the center of mass, while the second comes from the non-zero Fourier modes.

Now consider the case where the initial interface $h(x,0)$ is drawn from the equilibrium distribution at temperature $T$:

$P_{\text{stat.}}[h]\propto \exp \left[-{\frac {\nu }{2T}}\int _{0}^{L}dx(\partial _{x}h)^{2}\right]$

For simplicity, set the initial center of mass to zero: ${\hat {h}}_{q=0}(0)=0$. We consider the mean square displacement of the point $h(x=0,t)$. The average is performed over both the thermal noise $\langle \cdot \rangle$ and the initial condition ${\overline {\cdot }}$:

${\overline {\langle \Delta h^{2}\rangle }}={\overline {\langle [h(0,t)-h(0,0)]^{2}\rangle }}={\overline {\langle h^{2}(0,t)\rangle }}+{\overline {\langle h^{2}(0,0)\rangle }}-2{\overline {\langle h(0,t)h(0,0)\rangle }}$

Questions:

1. Compute the ensemble average of the Gaussian initial condition:

${\overline {{\hat {h}}_{q_{1}}(0){\hat {h}}_{q_{2}}(0)}}$

• Hint:* Write the integral in terms of Fourier modes and use $\int _{0}^{L}dx\,e^{iqx}=L\delta _{q,0}$.

1. Show that:

${\overline {\langle h^{2}(0,0)\rangle }}={\frac {T}{\nu L}}\sum _{q\neq 0}{\frac {1}{q^{2}}},\quad {\overline {\langle h(0,t)h(0,0)\rangle }}={\frac {T}{\nu L}}\sum _{q\neq 0}{\frac {e^{-\nu q^{2}t}}{q^{2}}}$

1. Show that:

${\overline {\langle h^{2}(0,t)\rangle }}=A+\langle \Delta h_{\text{flat}}^{2}\rangle$

where the term $A$ depends only on the initial condition. Show that:

$A(t)={\frac {T}{\nu L}}\sum _{q\neq 0}{\frac {e^{-2\nu q^{2}t}}{q^{2}}}$

1. Hence write:

$C(t)\equiv {\overline {\langle \Delta h^{2}\rangle }}-\langle \Delta h_{\text{flat}}^{2}\rangle ={\frac {2T}{\nu L}}\sum _{n=1}^{\infty }{\frac {(1-e^{-\nu (2\pi n/L)^{2}t})^{2}}{(2\pi n/L)^{2}}}$

Estimate $C(t)$ for $t\gg L^{2}$.

1. Estimate $C(t)$ for $t\ll L^{2}$ and large $L$.

• Hint:* Write the series as an integral using the continuum variable $z=2\pi n/L$. It is helpful to know:

$\int _{0}^{\infty }ds{\frac {(1-e^{-s^{2}})^{2}}{s^{2}}}={\sqrt {\pi }}(2-{\sqrt {2}})$

Provide the two asymptotic behaviors of ${\overline {\langle \Delta h^{2}\rangle }}$.