L1 ICTS - Revision history

Rosso: /* Back to REM */

2026-03-01T14:19:45Z

Back to REM

← Older revision		Revision as of 16:19, 1 March 2026
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	In the REM, the variance of the energies scales with the system size as		In the REM, the variance of the energies scales with the system size as
	~~<center>~~<math>\sigma_M^2 = \frac{\log M}{\log 2} = N.</math~~></center~~>		<math display="block">\sigma_M^2 = \frac{\log M}{\log 2} = N.</math>
	As a consequence, the minimum energy takes the form		As a consequence, the minimum energy takes the form
			<math display="block">
	~~<center>~~<math>

	E_{\min} = a_M + b_M z		E_{\min} = a_M + b_M z
	= - \sqrt{2 \log 2}\, N		= - \sqrt{2 \log 2}\, N
	+ \frac{1}{2}\, \frac{\log (4 \pi N \log 2)}{\sqrt{2 \log 2}}		+ \frac{1}{2}\, \frac{\log (4 \pi N \log 2)}{\sqrt{2 \log 2}}
	+ \frac{z}{\sqrt{2 \log 2}},		+ \frac{z}{\sqrt{2 \log 2}},
	</math>~~</center>~~		</math>

	where <math>z</math> is a Gumbel-distributed random variable.		where <math>z</math> is a Gumbel-distributed random variable.

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	the average number of states within an energy window ''x'' above the ground		the average number of states within an energy window ''x'' above the ground
	state is		state is
	~~<center>~~		<math display="block">
	<math>
	\overline{n(x)} = e^{x/b_M}-1,		\overline{n(x)} = e^{x/b_M}-1,
	</math>		</math>
	~~</center>~~
	where <math>b_M</math> is the scaling parameter of the minimum.		where <math>b_M</math> is the scaling parameter of the minimum.

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	<math>z_\alpha = e^{-\beta E_\alpha}</math> the Boltzmann weights. We compare the total		<math>z_\alpha = e^{-\beta E_\alpha}</math> the Boltzmann weights. We compare the total
	weight of all excited states to that of the ground state:		weight of all excited states to that of the ground state:
	~~<center>~~		<math display="block">
	<math>
	\frac{\sum_\alpha z_\alpha}{z_{\alpha_{\min}}}		\frac{\sum_\alpha z_\alpha}{z_{\alpha_{\min}}}
	= 1 + \sum_{\alpha\neq\alpha_{\min}} e^{-\beta(E_\alpha-E_{\min})}.		= 1 + \sum_{\alpha\neq\alpha_{\min}} e^{-\beta(E_\alpha-E_{\min})}.
	</math>		</math>
	~~</center>~~

	Replacing the discrete sum by an integral over energy differences <math>x=E-E_{\min}</math>,		Replacing the discrete sum by an integral over energy differences <math>x=E-E_{\min}</math>,
	and using the result of Exercise 3, we obtain		and using the result of Exercise 3, we obtain
	~~<center>~~		<math display="block">
	<math>
	\sum_{\alpha\neq\alpha_{\min}} e^{-\beta(E_\alpha-E_{\min})}		\sum_{\alpha\neq\alpha_{\min}} e^{-\beta(E_\alpha-E_{\min})}
	\;\sim\;		\;\sim\;
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	\int_0^\infty dx\,\frac{1}{b_M}e^{x/b_M}e^{-\beta x}.		\int_0^\infty dx\,\frac{1}{b_M}e^{x/b_M}e^{-\beta x}.
	</math>		</math>
	~~</center>~~

	The integral converges if and only if		The integral converges if and only if
	~~<center>~~		<math display="block">
	<math>
	\beta > \beta_f \equiv \frac{1}{b_M}.		\beta > \beta_f \equiv \frac{1}{b_M}.
	</math>		</math>
	~~</center>~~

	== Freezing transition ==		== Freezing transition ==

Rosso: /* Freezing transition */

2026-03-01T14:18:09Z

Freezing transition

← Older revision		Revision as of 16:18, 1 March 2026
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	== Freezing transition ==		== Freezing transition ==

	We thus identify the freezing temperature as		We thus identify the freezing temperature as
	~~<center>~~		<math display="block">
	<math>
	T_f = \frac{1}{\beta_f} = b_M.		T_f = \frac{1}{\beta_f} = b_M.
	</math>		</math>
	~~</center>~~

	For <math>T>T_f</math>, the contribution of excited states dominates and the Gibbs measure		For <math>T>T_f</math>, the contribution of excited states dominates and the Gibbs measure
	is spread over an exponential number of configurations.		is spread over an exponential number of configurations.

Rosso: /* Detour: Extreme Value Statistics */

2026-03-01T14:17:37Z

Detour: Extreme Value Statistics

← Older revision		Revision as of 16:17, 1 March 2026
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	This is an estimation of the typical value of the minimum. It is a crucial relation that will be used frequently.		This is an estimation of the typical value of the minimum. It is a crucial relation that will be used frequently.
	* '''Second relation''':		* '''Second relation''':
	<math display="block">Q_M(E) = (1-P(E))^M= e^{M \log(1 - P(E))} \sim \exp\left(-M P(E)\right) </math>		<math display="block">Q_M(E) = (1-P(E))^M= e^{M \log(1 - P(E))} \sim \exp\left(-M P(E)\right) </math>
	The first two steps are exact, but the resulting distribution depends on <math>M</math> and the precise form of <math>p(E)</math>. In contrast, the last step is an approximation, valid when <math>M \times P(E)=O(1)</math> and thus, for large <math>M</math>, when		The first two steps are exact, but the resulting distribution depends on <math>M</math> and the precise form of <math>p(E)</math>. In contrast, the last step is an approximation, valid when <math>M \times P(E)=O(1)</math> and thus, for large <math>M</math>, when
	<math> P(E)\ll 1 </math>. This second relation allows to express the random variable <math>E_{\min}</math> in a scaling form: <math>E_{\min} = a_M + b_M z</math>. The two parameters <math>a_M</math> and <math>b_M</math> are deterministic and <math>M</math>-dependent, while <math>z</math> is a random variable that is independent of <math>M</math>.		<math> P(E)\ll 1 </math>. This second relation allows to express the random variable <math>E_{\min}</math> in a scaling form: <math>E_{\min} = a_M + b_M z</math>. The two parameters <math>a_M</math> and <math>b_M</math> are deterministic and <math>M</math>-dependent, while <math>z</math> is a random variable that is independent of <math>M</math>.

Rosso: /* Detour: Extreme Value Statistics */

2026-03-01T14:16:53Z

Detour: Extreme Value Statistics

Show changes

Rosso: /* Random Energy Model (REM) */

2026-03-01T14:12:57Z

Random Energy Model (REM)

← Older revision		Revision as of 16:12, 1 March 2026
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	The solution of the Sherrington-Kirkpatrick (SK) model is challenging. To make progress, we first study the Random Energy Model (REM), introduced by B. Derrida. This model simplifies the problem by neglecting correlations between the <math>M=2^N</math> configurations and assuming that the energies <math>E_{\alpha}</math> are independent and identically distributed (i.i.d.) random variables. Here, "independent" means that the energy of one configuration does not influence the energy of another, e.g., a configuration identical to the previous one except for a spin flip. "Identically distributed" indicates that all configurations follow the same probability distribution.		The solution of the Sherrington-Kirkpatrick (SK) model is challenging. To make progress, we first study the Random Energy Model (REM), introduced by B. Derrida. This model simplifies the problem by neglecting correlations between the <math>M=2^N</math> configurations and assuming that the energies <math>E_{\alpha}</math> are independent and identically distributed (i.i.d.) random variables. Here, "independent" means that the energy of one configuration does not influence the energy of another, e.g., a configuration identical to the previous one except for a spin flip. "Identically distributed" indicates that all configurations follow the same probability distribution.

	'''Energy Distribution:''' Show that the energy distribution is given by: <math display="block"> p(E_\alpha) = \frac{1}{\sqrt{2 \pi \sigma_M^2}} \exp\left(-\frac{E_{\alpha}^2}{2 \sigma_M^2}\right) </math>and determine that: <math display="block">\sigma_M^2 = N = \frac{\log M}{\log 2}</math>.		'''Energy Distribution:''' Show that the energy distribution is given by: <math display="block"> p(E_\alpha) = \frac{1}{\sqrt{2 \pi \sigma_M^2}} \exp\left(-\frac{E_{\alpha}^2}{2 \sigma_M^2}\right) </math>and determine that: <math display="block">\sigma_M^2 = N = \frac{\log M}{\log 2}</math>.

	In the following, we present the original solution of the model. Here, we characterize the glassy phase by analyzing the statistical properties of the smallest energy values among the <math>M=2^N</math> configurations. To address this, it is necessary to make a brief detour into the theory of extreme value statistics for i.i.d. random variables.		In the following, we present the original solution of the model. Here, we characterize the glassy phase by analyzing the statistical properties of the smallest energy values among the <math>M=2^N</math> configurations. To address this, it is necessary to make a brief detour into the theory of extreme value statistics for i.i.d. random variables.

Rosso: /* Simpler models */

2026-03-01T14:12:38Z

Simpler models

← Older revision		Revision as of 16:12, 1 March 2026
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	=== Sherrington and Kirkpatrik (SK) Model===		=== Sherrington and Kirkpatrik (SK) Model===
	Sherrington and Kirkpatrik considered the fully connected version of the model with Gaussian couplings:		Sherrington and Kirkpatrik considered the fully connected version of the model with Gaussian couplings:
	~~<center>~~ <math>		<math display="block">
	E= - \sum_{i,j} \frac{J_{ij}}{ \sqrt{N}} \sigma_i \sigma_j		E= - \sum_{i,j} \frac{J_{ij}}{ \sqrt{N}} \sigma_i \sigma_j
	</math~~></center~~>		</math>
	At the inverse temperature <math>		At the inverse temperature <math>
	\beta </math>, the partion function of the model is		\beta </math>, the partion function of the model is
	~~<center>~~ <math>		<math display="block">
	Z= \sum_{\alpha=1}^{2^N} z_{\alpha}, \quad \text{with}\; z_{\alpha}= e^{-\beta E_\alpha}		Z= \sum_{\alpha=1}^{2^N} z_{\alpha}, \quad \text{with}\; z_{\alpha}= e^{-\beta E_\alpha}
	</math~~></center~~>		</math>
	Here <math> E_\alpha </math> is the energy associated to the configuration <math> \alpha </math>.		Here <math> E_\alpha </math> is the energy associated to the configuration <math> \alpha </math>.

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	The solution of the Sherrington-Kirkpatrick (SK) model is challenging. To make progress, we first study the Random Energy Model (REM), introduced by B. Derrida. This model simplifies the problem by neglecting correlations between the <math>M=2^N</math> configurations and assuming that the energies <math>E_{\alpha}</math> are independent and identically distributed (i.i.d.) random variables. Here, "independent" means that the energy of one configuration does not influence the energy of another, e.g., a configuration identical to the previous one except for a spin flip. "Identically distributed" indicates that all configurations follow the same probability distribution.		The solution of the Sherrington-Kirkpatrick (SK) model is challenging. To make progress, we first study the Random Energy Model (REM), introduced by B. Derrida. This model simplifies the problem by neglecting correlations between the <math>M=2^N</math> configurations and assuming that the energies <math>E_{\alpha}</math> are independent and identically distributed (i.i.d.) random variables. Here, "independent" means that the energy of one configuration does not influence the energy of another, e.g., a configuration identical to the previous one except for a spin flip. "Identically distributed" indicates that all configurations follow the same probability distribution.

	'''Energy Distribution:''' Show that the energy distribution is given by: ~~<center>~~<math> p(E_\alpha) = \frac{1}{\sqrt{2 \pi \sigma_M^2}} \exp\left(-\frac{E_{\alpha}^2}{2 \sigma_M^2}\right) </math~~></center~~> and determine that: ~~<center>~~<math>\sigma_M^2 = N = \frac{\log M}{\log 2}</math~~></center~~>.		'''Energy Distribution:''' Show that the energy distribution is given by: <math display="block"> p(E_\alpha) = \frac{1}{\sqrt{2 \pi \sigma_M^2}} \exp\left(-\frac{E_{\alpha}^2}{2 \sigma_M^2}\right) </math>and determine that: <math display="block">\sigma_M^2 = N = \frac{\log M}{\log 2}</math>.

	In the following, we present the original solution of the model. Here, we characterize the glassy phase by analyzing the statistical properties of the smallest energy values among the <math>M=2^N</math> configurations. To address this, it is necessary to make a brief detour into the theory of extreme value statistics for i.i.d. random variables.		In the following, we present the original solution of the model. Here, we characterize the glassy phase by analyzing the statistical properties of the smallest energy values among the <math>M=2^N</math> configurations. To address this, it is necessary to make a brief detour into the theory of extreme value statistics for i.i.d. random variables.

Rosso: /* Glass Transition: the Edwards Anderson order parameter */

2026-03-01T14:11:41Z

Glass Transition: the Edwards Anderson order parameter

← Older revision		Revision as of 16:11, 1 March 2026
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	The order parameter for this phase is:		The order parameter for this phase is:

	~~<center>~~<math>		<math display="block">
	q_{EA} = \lim_{t \to \infty} \lim_{N \to \infty} \frac{1}{N} \sum_{i} \sigma_i(0)\,\sigma_i(t).		q_{EA} = \lim_{t \to \infty} \lim_{N \to \infty} \frac{1}{N} \sum_{i} \sigma_i(0)\,\sigma_i(t).
	</math~~></center~~>		</math>

	The quantity <math>q_{EA}</math> measures the overlap of the spin configuration with itself after a long time.		The quantity <math>q_{EA}</math> measures the overlap of the spin configuration with itself after a long time.
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	It can be shown that the susceptibility associated with <math>q_{EA}</math> corresponds to the nonlinear susceptibility:		It can be shown that the susceptibility associated with <math>q_{EA}</math> corresponds to the nonlinear susceptibility:
			<math display="block">
	~~<center>~~<math>
	\frac{M}{H} = \chi + a_3 H^2 + a_5 H^4 + \ldots		\frac{M}{H} = \chi + a_3 H^2 + a_5 H^4 + \ldots
	</math~~></center~~>		</math>

	Here <math>\chi</math> is the linear susceptibility, while <math>a_3, a_5, \ldots</math> are higher-order coefficients.		Here <math>\chi</math> is the linear susceptibility, while <math>a_3, a_5, \ldots</math> are higher-order coefficients.

Rosso: /* Deterministic observables */

2026-03-01T14:11:12Z

Deterministic observables

← Older revision		Revision as of 16:11, 1 March 2026
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	From a more mathematical point of view, it means that the free energy <math> F_N(\beta)=N f_N(\beta)</math> and its derivatives (magnetization, specific heat, susceptibility, etc.), in the limit <math> N \to \infty </math>, these random quantities concentrates around a well defined value. These observables are called self-averaging. This means that,		From a more mathematical point of view, it means that the free energy <math> F_N(\beta)=N f_N(\beta)</math> and its derivatives (magnetization, specific heat, susceptibility, etc.), in the limit <math> N \to \infty </math>, these random quantities concentrates around a well defined value. These observables are called self-averaging. This means that,
	~~<center>~~		<math display="block">
	<math>
	\lim_{N \to \infty} f_N (\beta)= \lim_{N \to \infty} f_N^{\text{typ}}(\beta) =\lim_{N \to \infty} \overline{f_N(\beta)} =f_\infty(\beta)		\lim_{N \to \infty} f_N (\beta)= \lim_{N \to \infty} f_N^{\text{typ}}(\beta) =\lim_{N \to \infty} \overline{f_N(\beta)} =f_\infty(\beta)
	</math>		</math>
	~~</center>~~
	Hence <math> f_N(\beta) </math> becomes effectively deterministic and its sample-to sample fluctuations vanish in relative terms:		Hence <math> f_N(\beta) </math> becomes effectively deterministic and its sample-to sample fluctuations vanish in relative terms:
	~~<center>~~		<math display="block">
	<math>
	\lim_{N \to \infty} \frac{\overline{f_N^2(\beta)}}{(\overline{f_N(\beta)})^2}=1.		\lim_{N \to \infty} \frac{\overline{f_N^2(\beta)}}{(\overline{f_N(\beta)})^2}=1.
	</math>		</math>
	~~</center>~~

	== Glass Transition: the Edwards Anderson order parameter==		== Glass Transition: the Edwards Anderson order parameter==

Rosso: /* Random energy landascape */

2026-03-01T14:10:37Z

Random energy landascape

← Older revision		Revision as of 16:10, 1 March 2026
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	In the presence of disorder, the energy associated with a given configuration becomes a random quantity. For instance, in the Edwards-Anderson model:		In the presence of disorder, the energy associated with a given configuration becomes a random quantity. For instance, in the Edwards-Anderson model:
	~~<center>~~<math> E = - \sum_{\langle i, j \rangle} J_{ij} \sigma_i \sigma_j, </math~~></center~~>		<math display="block"> E = - \sum_{\langle i, j \rangle} J_{ij} \sigma_i \sigma_j, </math>

	where the sum runs over nearest neighbors <math>\langle i, j \rangle</math>, and the couplings <math>J_{ij}</math> are independent and identically distributed (i.i.d.) Gaussian random variables with zero mean and unit variance.		where the sum runs over nearest neighbors <math>\langle i, j \rangle</math>, and the couplings <math>J_{ij}</math> are independent and identically distributed (i.i.d.) Gaussian random variables with zero mean and unit variance.
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	The energy of this configuration is given by the sum of all the couplings between neighboring spins:		The energy of this configuration is given by the sum of all the couplings between neighboring spins:

	~~<center>~~<math>		<math display="block">
	E[\sigma_1=1,\sigma_2=1,\ldots] = - \sum_{\langle i, j \rangle} J_{ij}.		E[\sigma_1=1,\sigma_2=1,\ldots] = - \sum_{\langle i, j \rangle} J_{ij}.
	</math~~></center~~>		</math>

	Since the couplings are random, the energy associated with this particular configuration		Since the couplings are random, the energy associated with this particular configuration

Rosso: /* Edwards Anderson model */

2026-03-01T14:10:03Z

Edwards Anderson model

← Older revision		Revision as of 16:10, 1 March 2026
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	Edwards and Anderson proposed studying this model with couplings <math>J_{ij}</math> that are independent and identically distributed (i.i.d.) random variables with a zero mean.		Edwards and Anderson proposed studying this model with couplings <math>J_{ij}</math> that are independent and identically distributed (i.i.d.) random variables with a zero mean.
	The coupling distribution is denoted by <math>\pi(J)</math>, and the average over the couplings, referred to as the disorder average, is indicated by an overline:		The coupling distribution is denoted by <math>\pi(J)</math>, and the average over the couplings, referred to as the disorder average, is indicated by an overline:
	~~<center>~~<math> \overline{J} \equiv \int dJ \, J \, \pi(J) = 0. </math~~></center~~>		<math display="block"> \overline{J} \equiv \int dJ \, J \, \pi(J) = 0. </math>

	In the following we will consider Gaussian couplings: <math>\pi(J) = \exp\left(-J^2 / 2\right) / \sqrt{2 \pi}</math>.		In the following we will consider Gaussian couplings: <math>\pi(J) = \exp\left(-J^2 / 2\right) / \sqrt{2 \pi}</math>.