T-I - Revision history

Ros: /* To know more */

2026-03-11T09:53:00Z

To know more

← Older revision		Revision as of 11:53, 11 March 2026
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	* A note on terminology:		* A note on terminology:
	<div style="font-size:89%">		<div style="font-size:89%">
	The terms “quenched” and “annealed” come from metallurgy and refer to the procedure in which you cool a very hot piece of metal: a system is quenched if it is cooled very rapidly (istantaneously changing its environment by putting it into cold water, for instance) and has to adjusts to this new fixed environment; annealed if it is cooled slowly, kept in (quasi)equilibrium with its changing environment at all times. Think now at how you compute the free energy of a disordered system, and at disorder as the environment. In the quenched protocol, you first compute the average over the configurations of the system (with the Boltzmann weight) keeping the disorder (environment) fixed, so the configurations have to adjust to the given disorder. Then you take the log and only afterwards average over the randomness (not even needed, at large <math>N</math>, if the free-energy is self-averaging). In the annealed protocol instead, the disorder (environment) and the configurations are treated on the same footing and adjust to each others, you average over both simultaneously. The quenched case corresponds to equilibrating at fixed disorder, and it is justified when there is a separation of timescales and the disorder evolved much slower than the variables; the annealed one to considering the disorder as a dynamical variable, that co-evolves with the variables.		The terms “quenched” and “annealed” come from metallurgy and refer to the procedure in which you cool a very hot piece of metal: a system is quenched if it is cooled very rapidly (istantaneously changing its environment by putting it into cold water, for instance) and has to adjusts to this new fixed environment; annealed if it is cooled slowly, kept in (quasi)equilibrium with its changing environment at all times. Think now at how you compute the free energy of a disordered system, and at disorder as the environment. In the quenched protocol, you first compute the average over the configurations of the system (with the Boltzmann weight) keeping the disorder (environment) fixed, so the configurations have to adjust to the given disorder. Like the hot metal has to adjust itself to the new, fixed, cold water). Then you take the log and only afterwards average over the randomness, that plays the role of the environment (not even needed, at large <math>N</math>, if the free-energy is self-averaging). In the annealed protocol instead, the disorder (environment) and the configurations are treated on the same footing and adjust to each others, you average over both simultaneously. The quenched case corresponds to equilibrating at fixed disorder, and it is justified when there is a separation of timescales and the disorder evolved much slower than the variables (the cold water changes temperature much slower than the metal you put in it); the annealed one to considering the disorder as a dynamical variable, that co-evolves with the variables.
	</div>		</div>

Ros: /* Problem 1: the energy landscape of the REM */

2026-01-18T20:07:50Z

Problem 1: the energy landscape of the REM

← Older revision		Revision as of 22:07, 18 January 2026
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	<ol start="3">		<ol start="3">
	<li> <em> Rare events.</em> For <math> \|\epsilon\| > \sqrt{2 \, \log 2} </math> the annealed entropy is negative: the average number of configurations with those energy densities is exponentially small in <math> N </math>. This implies that the probability to get configurations with those energy is exponentially small in <math> N </math>: these configurations are rare. Do you have an idea of how to show this, using the expression for <math> \mathbb{E}[\mathcal{N}_N]?</math> What is the typical value of <math> \mathcal{N}_N </math> in this region? Putting everything together, derive the form of the typical value of the entropy density. </li>		<li> <em> Rare events.</em> For <math> \|\epsilon\| > \sqrt{2 \, \log 2} </math> the annealed entropy is negative: the average number of configurations with those energy densities is exponentially small in <math> N </math>. This implies that the probability to get configurations with those energy is exponentially small in <math> N </math>: these configurations are rare. Do you have an idea of how to show this, using the expression for <math> \mathbb{E}(\mathcal{N}_N)?</math> What is the typical value of <math> \mathcal{N}_N </math> in this region? Putting everything together, derive the form of the typical value of the entropy density. </li>
	</ol>		</ol>
	<br>		<br>

Ros: /* Problem 1: the energy landscape of the REM */

2026-01-18T20:06:47Z

Problem 1: the energy landscape of the REM

← Older revision		Revision as of 22:06, 18 January 2026
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	In this problem we study the random variable <math> \mathcal{N}_N(E)dE </math>, that is the number of configurations having energy <math> E_\alpha \in [E, E+dE] </math>. For large <math> N </math> this variable scales exponentially <math>\mathcal{N}_N(E) \sim e^{N S_N\left( E/N\right)}</math>. Let <math> \epsilon=E/N </math>. The goal of this Problem is to show that the asymptotic value of the entropy <math> S_N(\epsilon) </math>, that is self-averaging, is given by (henceforth, <math> S(\epsilon):=S_\infty(\epsilon) </math>):		In this problem we study the random variable <math> \mathcal{N}_N(E)dE </math>, that is the number of configurations having energy <math> E_\alpha \in [E, E+dE] </math>. For large <math> N </math> this variable scales exponentially <math>\mathcal{N}_N(E)dE \sim e^{N S_N\left( E/N\right)}dE</math>. Let <math> \epsilon=E/N </math>. The goal of this Problem is to show that the asymptotic value of the entropy <math> S_N(\epsilon) </math>, that is self-averaging, is given by (henceforth, <math> S(\epsilon):=S_\infty(\epsilon) </math>):

	[[File:Entropy REM.png\|thumb\|left\|x140px\|Entropy of the Random Energy Model]]		[[File:Entropy REM.png\|thumb\|left\|x140px\|Entropy of the Random Energy Model]]

Ros: /* Problem 1: the energy landscape of the REM */

2026-01-18T20:06:10Z

Problem 1: the energy landscape of the REM

← Older revision		Revision as of 22:06, 18 January 2026
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	<ol>		<ol>
	<li> <em> Averages: the annealed entropy.</em> We begin by computing the annealed entropy <math> S_{\text{a}} </math>, which is defined by the average <math> \mathbb{E}[{\mathcal{N}_N(E)} dE]= \text{exp}\left(N S_{\text{a}}\left( E/N \right)+ o(N)\right)dE </math>. Compute this function using the representation <math> \mathcal{N}_N(E)dE= \sum_{\alpha=1}^{2^N} \chi_\alpha(E) dE \;</math> [with <math> \chi_\alpha(E)=1</math> if <math> E_\alpha \in [E, E+dE]</math> and <math> \chi_\alpha(E)=0</math> otherwise]. </li>		<li> <em> Averages: the annealed entropy.</em> We begin by computing the annealed entropy <math> S_{\text{a}} </math>, which is defined by the average <math> \mathbb{E}({\mathcal{N}_N(E)} dE)= \text{exp}\left(N S_{\text{a}}\left( E/N \right)+ o(N)\right)dE </math>. Compute this function using the representation <math> \mathcal{N}_N(E)dE= \sum_{\alpha=1}^{2^N} \chi_\alpha(E) dE \;</math> [with <math> \chi_\alpha(E)=1</math> if <math> E_\alpha \in [E, E+dE]</math> and <math> \chi_\alpha(E)=0</math> otherwise]. </li>
	</ol>		</ol>
	<br>		<br>

	<ol start="2">		<ol start="2">
	<li><em> Self-averaging.</em> For <math> \|\epsilon\| \leq \sqrt{2 \, \log 2} </math> the quantity <math> \mathcal{N}_N </math> is self-averaging: its distribution concentrates around the average value <math> \mathbb{E}[\mathcal{N}_N] </math> when <math> N \to \infty </math>. Show this by computing the second moment <math>\mathbb{E}[\mathcal{N}^2_N]</math>. Deduce that <math> S(\epsilon)= S_a(\epsilon)</math> when <math> \|\epsilon\| \leq \sqrt{2 \, \log 2} </math>. This property of being self-averaging is no longer true in the region where the annealed entropy is negative: why does one expect fluctuations to be relevant in this region?</li>		<li><em> Self-averaging.</em> For <math> \|\epsilon\| \leq \sqrt{2 \, \log 2} </math> the quantity <math> \mathcal{N}_N </math> is self-averaging: its distribution concentrates around the average value <math> \mathbb{E}(\mathcal{N}_N) </math> when <math> N \to \infty </math>. Show this by computing the second moment <math>\mathbb{E}(\mathcal{N}^2_N)</math>. Deduce that <math> S(\epsilon)= S_a(\epsilon)</math> when <math> \|\epsilon\| \leq \sqrt{2 \, \log 2} </math>. This property of being self-averaging is no longer true in the region where the annealed entropy is negative: why does one expect fluctuations to be relevant in this region?</li>
	</ol>		</ol>
	<br>		<br>

Ros: /* Large-N disordered systems: a probabilistic dictionary */

2026-01-18T20:05:10Z

Large-N disordered systems: a probabilistic dictionary

← Older revision		Revision as of 22:05, 18 January 2026
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	<math display="block">		<math display="block">
	\lim_{N \to \infty} Y_N =\lim_{N \to \infty} \mathbb{E}(~~\overline~~{Y_N}):= {Y_\infty}, \quad \quad \mathbb{E}({Y_N})=\int \, dy\, P_{Y_N}(y)\, y		\lim_{N \to \infty} Y_N =\lim_{N \to \infty} \mathbb{E}({Y_N}):= {Y_\infty}, \quad \quad \mathbb{E}({Y_N})=\int \, dy\, P_{Y_N}(y)\, y
	</math>		</math>

Ros: /* Large-N disordered systems: a probabilistic dictionary */

2026-01-18T20:04:25Z

Large-N disordered systems: a probabilistic dictionary

← Older revision		Revision as of 22:04, 18 January 2026
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	<math display="block">		<math display="block">
	\lim_{N \to \infty} Y_N =\lim_{N \to \infty} \mathbb{E}({Y_N}):= {Y_\infty}, \quad \quad \mathbb{E}({Y_N})=\int \, dy\, P_{Y_N}(y)\, y		\lim_{N \to \infty} Y_N =\lim_{N \to \infty} \mathbb{E}(\overline{Y_N}):= {Y_\infty}, \quad \quad \mathbb{E}({Y_N})=\int \, dy\, P_{Y_N}(y)\, y
	</math>		</math>

Ros: /* To know more */

2026-01-18T18:26:53Z

To know more

← Older revision		Revision as of 20:26, 18 January 2026
Line 146:		Line 146:
	* A note on terminology:		* A note on terminology:
	<div style="font-size:89%">		<div style="font-size:89%">
	The terms “quenched” and “annealed” come from metallurgy and refer to the procedure in which you cool a very hot piece of metal: a system is quenched if it is cooled very rapidly (istantaneously changing its environment by putting it into cold water, for instance) and has to adjusts to this new fixed environment; annealed if it is cooled slowly, kept in (quasi)equilibrium with its changing environment at all times. Think now at how you compute the free energy of a disordered system, and at disorder as the environment. In the quenched protocol, you first compute the average over the configurations of the system (with the Boltzmann weight) keeping the disorder (environment) fixed, so the configurations have to adjust to the given disorder. Then you take the log and only afterwards average over the randomness (not even needed, at large <math>N</math>, if the free-energy is self-averaging). In the annealed protocol instead, the disorder (environment) and the configurations are treated on the same footing and adjust to each others, you average over both simultaneously. The quenched case corresponds to ~~keeping the environment~~ fixed and ~~looking at how configurations adjust to~~ it, the ~~annealedone~~ to ~~changing~~ the ~~environment fast~~.		The terms “quenched” and “annealed” come from metallurgy and refer to the procedure in which you cool a very hot piece of metal: a system is quenched if it is cooled very rapidly (istantaneously changing its environment by putting it into cold water, for instance) and has to adjusts to this new fixed environment; annealed if it is cooled slowly, kept in (quasi)equilibrium with its changing environment at all times. Think now at how you compute the free energy of a disordered system, and at disorder as the environment. In the quenched protocol, you first compute the average over the configurations of the system (with the Boltzmann weight) keeping the disorder (environment) fixed, so the configurations have to adjust to the given disorder. Then you take the log and only afterwards average over the randomness (not even needed, at large <math>N</math>, if the free-energy is self-averaging). In the annealed protocol instead, the disorder (environment) and the configurations are treated on the same footing and adjust to each others, you average over both simultaneously. The quenched case corresponds to equilibrating at fixed disorder, and it is justified when there is a separation of timescales and the disorder evolved much slower than the variables; the annealed one to considering the disorder as a dynamical variable, that co-evolves with the variables.
	</div>		</div>

Ros: /* Problem 1: the energy landscape of the REM */

2026-01-18T18:24:46Z

Problem 1: the energy landscape of the REM

← Older revision		Revision as of 20:24, 18 January 2026
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	<li><em> Self-averaging.</em> For <math> \|\epsilon\| \leq \sqrt{2 \, \log 2} </math> the quantity <math> \mathcal{N}_N </math> is self-averaging: its distribution concentrates around the average value <math> \mathbb{E}[\mathcal{N}_N] </math> when <math> N \to \infty </math>. Show this by computing the second moment <math>\mathbb{E}[\mathcal{N}^2_N]</math>. Deduce that <math> S(\epsilon)= S_a(\epsilon)</math> when <math> \|\epsilon\| \leq \sqrt{2 \, \log 2} </math>. This property of being self-averaging is no longer true in the region where the annealed entropy is negative: why does one expect fluctuations to be relevant in this region?</li>		<li><em> Self-averaging.</em> For <math> \|\epsilon\| \leq \sqrt{2 \, \log 2} </math> the quantity <math> \mathcal{N}_N </math> is self-averaging: its distribution concentrates around the average value <math> \mathbb{E}[\mathcal{N}_N] </math> when <math> N \to \infty </math>. Show this by computing the second moment <math>\mathbb{E}[\mathcal{N}^2_N]</math>. Deduce that <math> S(\epsilon)= S_a(\epsilon)</math> when <math> \|\epsilon\| \leq \sqrt{2 \, \log 2} </math>. This property of being self-averaging is no longer true in the region where the annealed entropy is negative: why does one expect fluctuations to be relevant in this region?</li>
	</ol>		</ol>
			<br>

	<ol start="3">		<ol start="3">

Ros: /* Problem 1: the energy landscape of the REM */

2026-01-18T18:24:33Z

Problem 1: the energy landscape of the REM

← Older revision		Revision as of 20:24, 18 January 2026
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	\end{cases}		\end{cases}
	</math>		</math>
			<br>

	<ol>		<ol>
	<li> <em> Averages: the annealed entropy.</em> We begin by computing the annealed entropy <math> S_{\text{a}} </math>, which is defined by the average <math> \mathbb{E}[{\mathcal{N}_N(E)} dE]= \text{exp}\left(N S_{\text{a}}\left( E/N \right)+ o(N)\right)dE </math>. Compute this function using the representation <math> \mathcal{N}_N(E)dE= \sum_{\alpha=1}^{2^N} \chi_\alpha(E) dE \;</math> [with <math> \chi_\alpha(E)=1</math> if <math> E_\alpha \in [E, E+dE]</math> and <math> \chi_\alpha(E)=0</math> otherwise]. </li>		<li> <em> Averages: the annealed entropy.</em> We begin by computing the annealed entropy <math> S_{\text{a}} </math>, which is defined by the average <math> \mathbb{E}[{\mathcal{N}_N(E)} dE]= \text{exp}\left(N S_{\text{a}}\left( E/N \right)+ o(N)\right)dE </math>. Compute this function using the representation <math> \mathcal{N}_N(E)dE= \sum_{\alpha=1}^{2^N} \chi_\alpha(E) dE \;</math> [with <math> \chi_\alpha(E)=1</math> if <math> E_\alpha \in [E, E+dE]</math> and <math> \chi_\alpha(E)=0</math> otherwise]. </li>
	</ol>		</ol>
			<br>

	<ol start="2">		<ol start="2">
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	<li> <em> Rare events.</em> For <math> \|\epsilon\| > \sqrt{2 \, \log 2} </math> the annealed entropy is negative: the average number of configurations with those energy densities is exponentially small in <math> N </math>. This implies that the probability to get configurations with those energy is exponentially small in <math> N </math>: these configurations are rare. Do you have an idea of how to show this, using the expression for <math> \mathbb{E}[\mathcal{N}_N]?</math> What is the typical value of <math> \mathcal{N}_N </math> in this region? Putting everything together, derive the form of the typical value of the entropy density. </li>		<li> <em> Rare events.</em> For <math> \|\epsilon\| > \sqrt{2 \, \log 2} </math> the annealed entropy is negative: the average number of configurations with those energy densities is exponentially small in <math> N </math>. This implies that the probability to get configurations with those energy is exponentially small in <math> N </math>: these configurations are rare. Do you have an idea of how to show this, using the expression for <math> \mathbb{E}[\mathcal{N}_N]?</math> What is the typical value of <math> \mathcal{N}_N </math> in this region? Putting everything together, derive the form of the typical value of the entropy density. </li>
	</ol>		</ol>
			<br>

	<ol start="3">		<ol start="3">

Ros: /* Problem 1: the energy landscape of the REM */

2026-01-18T18:24:05Z

Problem 1: the energy landscape of the REM

← Older revision		Revision as of 20:24, 18 January 2026
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	<ol start="3">		<ol start="3">
	<li> <em> The energy landscape.</em> When <math>N </math> is large, if one extract one configuration at random, what will be its energy? What is the typical value of the ground state energy? Deduce this from the entropy calculation: is this consistent with Alberto's lecture? </li>		<li> <em> The energy landscape.</em> When <math>N </math> is large, if one extract one configuration at random (with a uniform measure over all configurations), what will be its energy? What is the equilibrium energy density of the system at infinite temperature? What is the typical value of the ground state energy? Deduce this from the entropy calculation: is this consistent with Alberto's lecture? </li>
	</ol>		</ol>