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:

$\frac{\sum _{\alpha }{z}_{\alpha }}{z_{{\alpha }_{\min }}}=1+\sum _{\alpha \ne {\alpha }_{\min }}\frac{z_{\alpha }}{z_{{\alpha }_{\min }}}=1+\sum _{\alpha \ne {\alpha }_{\min }}{e}^{-\beta (E_{\alpha }-E_{\min })}\sim 1+\underset{0}{\overset{\mathrm{\infty }}{\int }}dx\frac{d\overline{n(x)}}{dx}{e}^{-\beta x}=1+\underset{0}{\overset{\mathrm{\infty }}{\int }}dx\frac{e^{x/b_M}}{b_M}{e}^{-\beta x}$

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:

$\underset{0}{\overset{\mathrm{\infty }}{\int }}dx{e}^{(1/b_M-\beta )x}/b_M=\frac{1}{\beta b_M-1}=\frac{T}{T_f-T}$

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:

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

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

$⟨\eta (x,t)\eta (x^{\prime },t^{\prime })⟩=2T\delta (x-x^{\prime })\delta (t-t^{\prime })$

The solution can be written in Fourier space as:

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

with Fourier decomposition:

${\hat{h}}_q(t)=\frac{1}{L}\underset{0}{\overset{L}{\int }}{e}^{iqx}h(x,t),h(x,t)=\sum _q{e}^{-iqx}{\hat{h}}_q(t),⟨{\eta }_{q_1}(t^{\prime }){\eta }_{q_2}(t)⟩=\frac{2T}{L}{\delta }_{q_1,-q_2}\delta (t-t^{\prime })$

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

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:

$⟨\mathrm{\Delta }{h}_{\text{flat}}^2⟩=2\frac{T}{L}t+\{\begin{array}{ll}T\sqrt{\frac{2t}{\pi \nu }},& t\ll L^2,\\ \frac{T}{\nu }\frac{L}{12},& t\gg L^2.\end{array}$

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 $T$:

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

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 $⟨\cdot ⟩$ and the initial condition $\bar{\cdot }$:

$\overline{⟨\mathrm{\Delta }{h}^2⟩}=\overline{⟨[h(0,t)-h(0,0){]}^2⟩}=\overline{⟨h^2(0,t)⟩}+\overline{⟨h^2(0,0)⟩}-2\overline{⟨h(0,t)h(0,0)⟩}$

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 $\underset{0}{\overset{L}{\int }}dx{e}^{iqx}=L{\delta }_{q,0}$.

1. Show that:

$\overline{⟨h^2(0,0)⟩}=\frac{T}{\nu L}\sum _{q\ne 0}\frac{1}{q^2},\overline{⟨h(0,t)h(0,0)⟩}=\frac{T}{\nu L}\sum _{q\ne 0}\frac{e^{-\nu q^{2}t}}{q^2}$

1. Show that:

$\overline{⟨h^2(0,t)⟩}=A+⟨\mathrm{\Delta }{h}_{\text{flat}}^2⟩$

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

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

1. Hence write:

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

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:

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

Provide the two asymptotic behaviors of $\overline{⟨\mathrm{\Delta }{h}^2⟩}$.