TBan-III - Revision history

Rosso: /* Exercise 1: Back to REM */

2025-09-16T11:36:54Z

Exercise 1: Back to REM

← Older revision		Revision as of 13:36, 16 September 2025
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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.

	~~==More general REM and systems in Finite dimensions==~~

	In random energy models with i.i.d. random variables, the distribution <math>p(E)</math> determines the dependence of <math>a_M</math> and <math>b_M</math> on ''M'', and consequently their scaling with ''N'', the number of degrees of freedom. It is insightful to consider a more general case where an exponent <math>\omega</math> describes the fluctuations of the ground state energy:
	~~<center> <math>\overline{\left(E_{\min} - \overline{E_{\min}}\right)^2} \sim b_M^2 \propto N^{2\omega}</math> </center>~~

	~~Three distinct scenarios emerge depending on the sign of <math>\omega</math>:~~

	* For <math>\omega < 0</math>: The freezing temperature vanishes with increasing system size, leading to the absence of a freezing transition. This scenario occurs in many low-dimensional systems, such as the Edwards-Anderson model in two dimensions.

	* For <math>\omega = 0</math>: A freezing transition is guaranteed. For the Random Energy Model discussed earlier, <math>T_f = 1/\sqrt{2 \log 2}</math>. An important feature of this transition, as will be explored in the next section, is that condensation does not occur solely in the ground state but also involves a large (albeit not extensive) number of low-energy excitations.

	* For <math>\omega > 0</math>: The freezing temperature grows with the system size, resulting in only the glassy phase. The system condenses entirely into the ground state since the excited states are characterized by prohibitively high energies. This scenario, while less intricate than the <math>\omega = 0</math> case, corresponds to a glassy phase with a single deep ground state.

	The extent to which these scenarios persist in real systems in finite dimensions remains an open question. In these systems, such as the directed polymers, the fluctuations of the ground state energy are characterized by an exponent <math>\theta</math>:
	~~<center> <math>\overline{\left(E_{\min} - \overline{E_{\min}}\right)^2} \sim L^{2\theta}</math> </center>~~

	~~where <math>L</math> is the linear size of the system and <math>N = L^D</math> is the number of degrees of freedom.~~

	At finite temperatures, an analogous exponent is defined by studying the fluctuations of the free energy, <math>F = E - T S</math>. For the directed polymer in low dimwnsion the fluctuations of the ground state exhibit a positive and temperature-independent <math>\theta</math>. In such cases, only the glassy phase exists, aligning with the <math>\omega > 0</math> scenario in REMs.

	On the other hand, for the directed polymer in high dimension, <math>\theta</math> is positive at low temperatures but vanishes at high temperatures. This behavior defines a glass transition mechanism entirely distinct from those observed in mean-field models.

	= Exercise 2: Edwards-Wilkinson Interface with Stationary Initial Condition =		= Exercise 2: Edwards-Wilkinson Interface with Stationary Initial Condition =

Rosso: /* Exercise 2: Edwards-Wilkinson Interface with Stationary Initial Condition */

2025-08-31T15:03:35Z

Exercise 2: Edwards-Wilkinson Interface with Stationary Initial Condition

← Older revision		Revision as of 17:03, 31 August 2025
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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>

Rosso: /* Exercise 1: Back to REM */

2025-08-31T14:33:24Z

Exercise 1: Back to REM

← Older revision		Revision as of 16:33, 31 August 2025
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	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.		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.

			==More general REM and systems in Finite dimensions==

			In random energy models with i.i.d. random variables, the distribution <math>p(E)</math> determines the dependence of <math>a_M</math> and <math>b_M</math> on ''M'', and consequently their scaling with ''N'', the number of degrees of freedom. It is insightful to consider a more general case where an exponent <math>\omega</math> describes the fluctuations of the ground state energy:
			<center> <math>\overline{\left(E_{\min} - \overline{E_{\min}}\right)^2} \sim b_M^2 \propto N^{2\omega}</math> </center>

			Three distinct scenarios emerge depending on the sign of <math>\omega</math>:

			* For <math>\omega < 0</math>: The freezing temperature vanishes with increasing system size, leading to the absence of a freezing transition. This scenario occurs in many low-dimensional systems, such as the Edwards-Anderson model in two dimensions.

			* For <math>\omega = 0</math>: A freezing transition is guaranteed. For the Random Energy Model discussed earlier, <math>T_f = 1/\sqrt{2 \log 2}</math>. An important feature of this transition, as will be explored in the next section, is that condensation does not occur solely in the ground state but also involves a large (albeit not extensive) number of low-energy excitations.

			* For <math>\omega > 0</math>: The freezing temperature grows with the system size, resulting in only the glassy phase. The system condenses entirely into the ground state since the excited states are characterized by prohibitively high energies. This scenario, while less intricate than the <math>\omega = 0</math> case, corresponds to a glassy phase with a single deep ground state.

			The extent to which these scenarios persist in real systems in finite dimensions remains an open question. In these systems, such as the directed polymers, the fluctuations of the ground state energy are characterized by an exponent <math>\theta</math>:
			<center> <math>\overline{\left(E_{\min} - \overline{E_{\min}}\right)^2} \sim L^{2\theta}</math> </center>

			where <math>L</math> is the linear size of the system and <math>N = L^D</math> is the number of degrees of freedom.

			At finite temperatures, an analogous exponent is defined by studying the fluctuations of the free energy, <math>F = E - T S</math>. For the directed polymer in low dimwnsion the fluctuations of the ground state exhibit a positive and temperature-independent <math>\theta</math>. In such cases, only the glassy phase exists, aligning with the <math>\omega > 0</math> scenario in REMs.

			On the other hand, for the directed polymer in high dimension, <math>\theta</math> is positive at low temperatures but vanishes at high temperatures. This behavior defines a glass transition mechanism entirely distinct from those observed in mean-field models.

	= Exercise 2: Edwards-Wilkinson Interface with Stationary Initial Condition =		= Exercise 2: Edwards-Wilkinson Interface with Stationary Initial Condition =

Rosso at 14:29, 31 August 2025

2025-08-31T14:29:01Z

← Older revision		Revision as of 16:29, 31 August 2025
Line 1:		Line 1:
			=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 =

Rosso: /* Exercise 2: Edwards-Wilkinson Interface with Stationary Initial Condition (7.5 points) */

2025-08-31T13:52:52Z

Exercise 2: Edwards-Wilkinson Interface with Stationary Initial Condition (7.5 points)

← Older revision		Revision as of 15:52, 31 August 2025
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	= Exercise 2: Edwards-Wilkinson Interface with Stationary Initial Condition ~~(7.5 points)~~ =		= Exercise 2: Edwards-Wilkinson Interface with Stationary Initial Condition =

	Consider an Edwards-Wilkinson interface in 1+1 dimensions, at temperature <math>T</math>, and of length <math>L</math> with periodic boundary conditions:		Consider an Edwards-Wilkinson interface in 1+1 dimensions, at temperature <math>T</math>, and of length <math>L</math> with periodic boundary conditions:
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	where <math>q = 2 \pi n / L, \; n = \ldots, -1, 0, 1, \ldots</math>.		where <math>q = 2 \pi n / L, \; n = \ldots, -1, 0, 1, \ldots</math>.

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

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

Rosso: Created page with "= Exercise 2: Edwards-Wilkinson Interface with Stationary Initial Condition (7.5 points) = 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: $\langle \eta(x,t) \eta(x',..."$

2025-08-31T13:50:12Z

Created page with "= Exercise 2: Edwards-Wilkinson Interface with Stationary Initial Condition (7.5 points) = Consider an Edwards-Wilkinson interface in 1+1 dimensions, at temperature <math>T</math>, and of length <math>L</math> with periodic boundary conditions: <center><math> \frac{\partial h(x,t)}{\partial t} = \nu \nabla^2 h(x,t) + \eta(x,t) </math></center> where <math>\eta(x,t)</math> is a Gaussian white noise with zero mean and variance: <center><math> \langle \eta(x,t) \eta(x',..."

New page

= Exercise 2: Edwards-Wilkinson Interface with Stationary Initial Condition (7.5 points) =

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

<center><math>
\frac{\partial h(x,t)}{\partial t} = \nu \nabla^2 h(x,t) + \eta(x,t)
</math></center>

where <math>\eta(x,t)</math> is a Gaussian white noise with zero mean and variance:

<center><math>
\langle \eta(x,t) \eta(x',t') \rangle = 2T \delta(x-x') \delta(t-t')
</math></center>

The solution can be written in Fourier space as:

<center><math>
\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)
</math></center>

with Fourier decomposition:

<center><math>
\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')
</math></center>

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

In class, you computed the mean square displacement of a point <math>h(x,t)</math> starting from a flat interface at <math>t=0</math>, i.e., <math>h(x,0) = 0</math>. The result was:

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

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 <math>h(x,0)</math> is drawn from the equilibrium distribution at temperature <math>T</math>:

<center><math>
P_{\text{stat.}}[h] \propto \exp\left[-\frac{\nu}{2T} \int_0^L dx (\partial_x h)^2 \right]
</math></center>

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

<center><math>
\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}
</math></center>

'''Questions:'''

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

<center><math>
\overline{\hat h_{q_1}(0) \hat h_{q_2}(0)}
</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>.

# '''Show''' that:

<center><math>
\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}
</math></center>

# '''Show''' that:

<center><math>
\overline{\langle h^2(0,t) \rangle} = A + \langle \Delta h_{\text{flat}}^2 \rangle
</math></center>

where the term <math>A</math> depends only on the initial condition. Show that:

<center><math>
A(t) = \frac{T}{\nu L} \sum_{q \neq 0} \frac{e^{-2 \nu q^2 t}}{q^2}
</math></center>

# '''Hence''' write:

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

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>.

*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>
\int_0^\infty ds \frac{(1 - e^{-s^2})^2}{s^2} = \sqrt{\pi}(2 - \sqrt{2})
</math></center>

Provide the two asymptotic behaviors of <math>\overline{\langle \Delta h^2 \rangle}</math>.