TBan-III: Difference between revisions

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=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 <math>\sim 1/M</math>. 
* '''Low-Temperature Phase:''' 
: Below a critical freezing temperature <math>T_f</math>, 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, <math>M</math>-independent probabilities.
''' Calculating the Freezing Temperature <math>T_f</math>'''
Thanks to the computation of <math>\overline{n(x)}</math>, we can identify the fingerprints of the glassy phase and calculate <math>T_f</math>. 
Let's compare the weight of the ground state against the weight of all other states:
<center> 
<math>
\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 + \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}
</math>
</center>
=== Behavior in Different Phases:===
* '''High-Temperature Phase (<math> T > T_f= b_M = 1/\sqrt{2 \log2}</math>):''' 
: 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 (<math> T < T_f= b_M = 1/\sqrt{2 \log2}</math>):''' 
: In this regime, the integral is finite: 
<center> 
<math>
\int_0^\infty dx \, e^{ (1/b_M-\beta) x}/b_M = \frac{1}{\beta b_M-1}  = \frac{T}{T_f - T}
</math> 
</center> 
This result implies that below the freezing temperature <math>T_f</math>, 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  =
= Exercise 2: Edwards-Wilkinson Interface with Stationary Initial Condition  =


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'''Questions:'''
'''Questions:'''


# '''Compute''' the ensemble average of the Gaussian initial condition:   
* '''Compute''' the ensemble average of the Gaussian initial condition:   


<center><math>
<center><math>
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</math></center>
</math></center>


*Hint:* Write the integral in terms of Fourier modes and use <math>\int_0^L dx \, e^{iqx} = L \delta_{q,0}</math>.
'''Hint:''' Write the integral in terms of Fourier modes and use <math>\int_0^L dx \, e^{iqx} = L \delta_{q,0}</math>.


# '''Show''' that:
* '''Show''' that:


<center><math>
<center><math>
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</math></center>
</math></center>


# '''Show''' that:
* '''Show''' that:


<center><math>
<center><math>
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</math></center>
</math></center>


# '''Hence''' write:
* '''Hence''' write:


<center><math>
<center><math>
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Estimate <math>C(t)</math> for <math>t \gg L^2</math>.
Estimate <math>C(t)</math> for <math>t \gg L^2</math>.


# Estimate <math>C(t)</math> for <math>t \ll L^2</math> and large <math>L</math>.   
* Estimate <math>C(t)</math> for <math>t \ll L^2</math> and large <math>L</math>.   


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


<center><math>
<center><math>

Latest revision as of 13:36, 16 September 2025

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:

h(x,t)t=ν2h(x,t)+η(x,t)

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

η(x,t)η(x,t)=2Tδ(xx)δ(tt)

The solution can be written in Fourier space as:

h^q(t)=h^q(0)eνq2t+0tdseνq2(ts)ηq(s)

with Fourier decomposition:

h^q(t)=1L0Leiqxh(x,t),h(x,t)=qeiqxh^q(t),ηq1(t)ηq2(t)=2TLδq1,q2δ(tt)

where q=2πn/L,n=,1,0,1,.

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:

Δhflat2=2TLt+{T2tπν,tL2,TνL12,tL2.

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:

Pstat.[h]exp[ν2T0Ldx(xh)2]

For simplicity, set the initial center of mass to zero: 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 and the initial condition :

Δh2=[h(0,t)h(0,0)]2=h2(0,t)+h2(0,0)2h(0,t)h(0,0)

Questions:

  • Compute the ensemble average of the Gaussian initial condition:
h^q1(0)h^q2(0)

Hint: Write the integral in terms of Fourier modes and use 0Ldxeiqx=Lδq,0.

  • Show that:
h2(0,0)=TνLq01q2,h(0,t)h(0,0)=TνLq0eνq2tq2
  • Show that:
h2(0,t)=A+Δhflat2

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

A(t)=TνLq0e2νq2tq2
  • Hence write:
C(t)Δh2Δhflat2=2TνLn=1(1eν(2πn/L)2t)2(2πn/L)2

Estimate C(t) for tL2.

  • Estimate C(t) for tL2 and large L.

Hint: Write the series as an integral using the continuum variable z=2πn/L. It is helpful to know:

0ds(1es2)2s2=π(22)

Provide the two asymptotic behaviors of Δh2.