LBan-1 - Revision history

Rosso: /* Back to REM */

2026-01-13T13:27:51Z

Back to REM

← Older revision		Revision as of 15:27, 13 January 2026
Line 181:		Line 181:

	'''Key observations:'''		'''Key observations:'''
	* At zero temperature (<math>\beta=\infty</math>), the ground-state energy is self-averaging. Its leading contribution is deterministic and extensive, <math>F\sim -\sqrt{2 \log 2}\, N</math>.		* At zero temperature (<math>\beta=\infty</math>), the ground-state energy is self-averaging. Its leading contribution is deterministic and extensive, <math>F_N(\beta=\infty)\sim -\sqrt{2 \log 2}\, N</math>.

	* Sample-to-sample fluctuations are subextensive (''N''-independent), with a finite standard deviation <math>\sigma = \sqrt{2 \log 2}</math>. We will later see that this scale coincides with the critical inverse temperature of the model.		* Sample-to-sample fluctuations are subextensive (''N''-independent), with a finite standard deviation <math>\sigma = \sqrt{2 \log 2}</math>. We will later see that this scale coincides with the critical inverse temperature of the model.

Rosso: /* Back to REM */

2026-01-13T13:23:40Z

Back to REM

← Older revision		Revision as of 15:23, 13 January 2026
Line 181:		Line 181:

	'''Key observations:'''		'''Key observations:'''
	* At zero temperature (<math>\beta=\infty</math>), the ground-state energy is self-averaging.		* At zero temperature (<math>\beta=\infty</math>), the ground-state energy is self-averaging. Its leading contribution is deterministic and extensive, <math>F\sim -\sqrt{2 \log 2}\, N</math>.
	Its leading contribution is deterministic and extensive,
	<math>F\sim -\sqrt{2 \log 2}\, N</math>.

	* Sample-to-sample fluctuations are subextensive (''N''-independent),		* Sample-to-sample fluctuations are subextensive (''N''-independent), with a finite standard deviation <math>\sigma = \sqrt{2 \log 2}</math>. We will later see that this scale coincides with the critical inverse temperature of the model.
	with a finite standard deviation
	<math>\sigma = \sqrt{2 \log 2}</math>.
	We will later see that this scale coincides with the critical inverse temperature of the model.

Rosso: /* Part III */

2026-01-13T13:19:51Z

Part III

← Older revision		Revision as of 15:19, 13 January 2026
Line 165:		Line 165:

	<center><math>\pi(z) = \exp(z)\,\exp(-e^{z})</math></center>		<center><math>\pi(z) = \exp(z)\,\exp(-e^{z})</math></center>
			== Back to REM ==

	~~== Back to~~ REM ==		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>
			As a consequence, the minimum energy takes the form

	~~In the REM the variance is~~ <math>\~~sigma^2_M~~ = \frac{\log M}{\log 2} ~~= N~~</math>~~. Then we have:~~		<center><math>
			E_{\min} = a_M + b_M z
			= - \sqrt{2 \log 2}\, N
			+ \frac{1}{2}\, \frac{\log (4 \pi N \log 2)}{\sqrt{2 \log 2}}
			+ \frac{z}{\sqrt{2 \log 2}},
			</math></center>

	~~<center>~~ <math>~~E_{\min} =a_M +b_M z = - \sqrt{2 \log 2}\; N +\frac{1}{2} \frac{ \log (4 \pi N \log2)}{\sqrt{2 \log 2}} + \frac{~~z~~}{\sqrt{2 \log 2}}~~</math> ~~</center>~~		where <math>z</math> is a Gumbel-distributed random variable.

			'''Key observations:'''
			* At zero temperature (<math>\beta=\infty</math>), the ground-state energy is self-averaging.
			Its leading contribution is deterministic and extensive,
			<math>F\sim -\sqrt{2 \log 2}\, N</math>.

	~~'''Key Observations:'''~~		* Sample-to-sample fluctuations are subextensive (''N''-independent),
	* ~~the ground state energy is self~~-~~averaging with an extensive deterministic part <math> \sim~~ -~~\sqrt{2 \log 2} \cdot N = f_\infty(\beta>\beta_c) \cdot N </math>.~~		with a finite standard deviation
	* Its fluctuations are ~~very small~~ (''N'' independent) with a standard deviation <math>\sqrt{2 \log 2} ~~=\beta_c~~ </math>.		<math>\sigma = \sqrt{2 \log 2}</math>.
			We will later see that this scale coincides with the critical inverse temperature of the model.

Rosso: /* Back to REM */

2025-09-15T08:42:17Z

Back to REM

← Older revision		Revision as of 10:42, 15 September 2025
Line 170:		Line 170:
	In the REM the variance is <math>\sigma^2_M = \frac{\log M}{\log 2} = N</math>. Then we have:		In the REM the variance is <math>\sigma^2_M = \frac{\log M}{\log 2} = N</math>. Then we have:

	<center> <math>E_{\min} =a_M +b_M z = - \sqrt{2 \log 2}\; N +\frac{1}{2} \frac{ \log (4 \pi \log2 N)}{\sqrt{2 \log 2}} + \frac{z}{\sqrt{2 \log 2}}</math> </center>		<center> <math>E_{\min} =a_M +b_M z = - \sqrt{2 \log 2}\; N +\frac{1}{2} \frac{ \log (4 \pi N \log2)}{\sqrt{2 \log 2}} + \frac{z}{\sqrt{2 \log 2}}</math> </center>

Rosso: /* Back to canonical ensemble: the freezing transition */

2025-09-15T08:40:35Z

Back to canonical ensemble: the freezing transition

← Older revision		Revision as of 10:40, 15 September 2025
Line 115:		Line 115:
	<center> <math> \overline{Z_N(\beta)} = \int_{-\infty}^{\infty} d\epsilon \; \overline{\mathcal{N}_N(\epsilon)} \, e^{-\beta N \epsilon} = \int_{-\infty}^{\infty} d\epsilon \; \exp \left[N (\ln 2 -\epsilon^2/2 -\beta \epsilon)\right]. </math> </center>		<center> <math> \overline{Z_N(\beta)} = \int_{-\infty}^{\infty} d\epsilon \; \overline{\mathcal{N}_N(\epsilon)} \, e^{-\beta N \epsilon} = \int_{-\infty}^{\infty} d\epsilon \; \exp \left[N (\ln 2 -\epsilon^2/2 -\beta \epsilon)\right]. </math> </center>
	Using the saddle point for large N we find <math> \epsilon_{saddle} =-\beta</math> and thus		Using the saddle point for large N we find <math> \epsilon_{saddle} =-\beta</math> and thus
	<center> <math> f^{\text{ann.}}(\beta) = \ln 2/\beta + \dfrac{\beta}{2}</math> </center>		<center> <math> f^{\text{ann.}}(\beta) = - \ln 2/\beta - \dfrac{\beta}{2}</math> </center>

	The '''quenched partition function''' is obtained replacing the mean with the typical value:		The '''quenched partition function''' is obtained replacing the mean with the typical value:
Line 122:		Line 122:
	* For <math>\beta < \beta_c </math>, <math> \epsilon_{saddle} =-\beta</math> and the annealed calculation works		* For <math>\beta < \beta_c </math>, <math> \epsilon_{saddle} =-\beta</math> and the annealed calculation works
	*For <math>\beta > \beta_c </math>, <math> \epsilon_{saddle} =-\beta_c</math> and the free energy freezes to a temperature independent value. As a result, the quenched average on the free energy density is:		*For <math>\beta > \beta_c </math>, <math> \epsilon_{saddle} =-\beta_c</math> and the free energy freezes to a temperature independent value. As a result, the quenched average on the free energy density is:
	<center> <math> f_\infty(\beta) = \begin{cases} \ln 2/\beta + \dfrac{\beta}{2}, & \text{for } \beta < \beta_c \\ \sqrt{2 \ln 2}, & \text{for } \beta> \beta_c \end{cases} </math> </center>		<center> <math> f_\infty(\beta) = \begin{cases} -\ln 2/\beta - \dfrac{\beta}{2}, & \text{for } \beta < \beta_c \\ -\sqrt{2 \ln 2}, & \text{for } \beta> \beta_c \end{cases} </math> </center>

	=Part III=		=Part III=

Rosso: /* Gaussian Case */

2025-09-13T12:08:23Z

Gaussian Case

← Older revision		Revision as of 14:08, 13 September 2025
Line 142:		Line 142:
	== Gaussian Case ==		== Gaussian Case ==

	We ask you to prove that for a Gaussian distribution with zero mean and variance <math>\(\sigma^2\)</math>, the cumulative can be written as:		We ask you to prove that for a Gaussian distribution with zero mean and variance <math>\sigma^2</math>, the cumulative can be written as:


	<center><math> P(E) = \exp(A(E)), \quad \text{with } A(E)= -\frac{E^2}{2 \sigma^2} - \log\!\!\left(\frac{\sqrt{2 \pi}\, \|E\|}{\sigma}\right)+\ldots, \quad \text{and } A'(E)= -\frac{E}{\sigma^2} +\ldots</math></center>		<center><math> P(E) = \exp(A(E)), \quad \text{with } A(E)= -\frac{E^2}{2 \sigma^2} - \log\!\!\left(\frac{\sqrt{2 \pi}\, \|E\|}{\sigma}\right)+\ldots, \quad \text{and } A'(E)= -\frac{E}{\sigma^2} +\ldots</math></center>

	* '''Typical Minimum''':		* '''Typical Minimum''': From the first relation <math>A(E_{\min}^{\text{typ}}) = - \ln M</math> one obtains, for large \(M\):
	From the first relation

	<math>A(E_{\min}^{\text{typ}}) = - \ln M</math>

	one obtains, for large \(M\):

	<center><math>E_{\min}^{\text{typ}} = -\sigma \sqrt{2 \log M} \left( 1 - \frac{1}{4} \frac{\log (4 \pi \log M)}{\log M} + \ldots \right)</math></center>		<center><math>E_{\min}^{\text{typ}} = -\sigma \sqrt{2 \log M} \left( 1 - \frac{1}{4} \frac{\log (4 \pi \log M)}{\log M} + \ldots \right)</math></center>

	* '''Gumbel Scaling''':		* '''Gumbel Scaling''': From the second relation, we look for a value <math>a_M</math> such that <math>A(a_M) = - \ln M </math>, and then expand <math>A(E)</math> around <math>a_M</math>:
	From the second relation, we look for a value 'a_M' such that \(A(a_M) = - \ln M\), and then expand \(A(E)\) around 'a_M':

	<center><math>Q_M(E) \sim \exp\!\!\Bigl(-M P(E)\Bigr) \sim \exp\!\!\left[-M \exp\!\!\bigl(A(a_M) + A'(a_M)(E-a_M)\bigr)\right]</math></center>		<center><math>Q_M(E) \sim \exp\!\!\Bigl(-M P(E)\Bigr) \sim \exp\!\!\left[-M \exp\!\!\bigl(A(a_M) + A'(a_M)(E-a_M)\bigr)\right]</math></center>
Line 169:		Line 162:
	<center><math>z = \frac{E_{\min} - a_M}{b_M}</math></center>		<center><math>z = \frac{E_{\min} - a_M}{b_M}</math></center>

	is \(M\)-independent and Gumbel distributed:		is <math>M</math>-independent and Gumbel distributed:

	<center><math>\pi(z) = \exp(z)\,\exp(-e^{z})</math></center>		<center><math>\pi(z) = \exp(z)\,\exp(-e^{z})</math></center>

Rosso: /* Gaussian Case */

2025-09-13T12:05:42Z

Gaussian Case

← Older revision		Revision as of 14:05, 13 September 2025
Line 142:		Line 142:
	== Gaussian Case ==		== Gaussian Case ==

	We ask you to prove that for a Gaussian distribution with zero mean and variance \(\sigma^2\), the cumulative can be written as:		We ask you to prove that for a Gaussian distribution with zero mean and variance <math>\(\sigma^2\)</math>, the cumulative can be written as:

	~~<math>P(E) = \exp(A(E))</math>~~

	~~with~~		<center><math> P(E) = \exp(A(E)), \quad \text{with } A(E)= -\frac{E^2}{2 \sigma^2} - \log\!\!\left(\frac{\sqrt{2 \pi}\, \|E\|}{\sigma}\right)+\ldots, \quad \text{and } A'(E)= -\frac{E}{\sigma^2} +\ldots</math></center>

	<center><math>A(E)= -\frac{E^2}{2 \sigma^2} - \log\!\!\left(\frac{\sqrt{2 \pi}\, \|E\|}{\sigma}\right)+\ldots~~</math></center>~~

	and

	~~<center><math>~~A'(E)= -\frac{E}{\sigma^2} +\ldots</math></center>

	* '''Typical Minimum''':		* '''Typical Minimum''':
Line 164:		Line 157:

	* '''Gumbel Scaling''':		* '''Gumbel Scaling''':
	From the second relation, we look for a value \(a_M\) such that \(A(a_M) = - \ln M\), and then expand \(A(E)\) around \(a_M\):		From the second relation, we look for a value 'a_M' such that \(A(a_M) = - \ln M\), and then expand \(A(E)\) around 'a_M':

	<center><math>Q_M(E) \sim \exp\!\!\Bigl(-M P(E)\Bigr) \sim \exp\!\!\left[-M \exp\!\!\bigl(A(a_M) + A'(a_M)(E-a_M)\bigr)\right]</math></center>		<center><math>Q_M(E) \sim \exp\!\!\Bigl(-M P(E)\Bigr) \sim \exp\!\!\left[-M \exp\!\!\bigl(A(a_M) + A'(a_M)(E-a_M)\bigr)\right]</math></center>

Rosso: /* Il Caso Gaussiano */

2025-09-13T11:59:42Z

Il Caso Gaussiano

@@ Line 140: / Line 140: @@
 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 for large <math>M</math> and that  allows one to express the random variable <math>E_{\min}</math> in a scaling form: <math>E_{\min} = a_M + b_M z</math>, where <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>.
-== Il Caso Gaussiano ==
+== Gaussian Case ==
 <math>P(E) = \exp(A(E))</math>
-* '''Typical minimum''': From the first relation
+with
- <math> A(E_{\min}^{\text{typ}})=- \ln M</math>
-you get for large M:
+<center><math>A(E)= -\frac{E^2}{2 \sigma^2} - \log\!\!\left(\frac{\sqrt{2 \pi}\, |E|}{\sigma}\right)+\ldots</math></center>
 <center> <math>E_{\min}^{\text{typ}} = -\sigma \sqrt{2 \log M} \left( 1- \frac{1}{4} \frac{\log (4 \pi \log M)}{\log M} + \ldots \right)</math> </center>
-* '''Gumbel Scaling''': From the second relation we need to find an <math>a_M</math> that absorb the 'M' factor and a b_M that absorb the 'M' Dependence. In the Gaussian case it is better to A(a_M)=- \ln M and expand <math>A(E)</math> around this value:
+is \(M\)-independent and Gumbel distributed:
-<center> <math>\pi(z) = \exp(z) \exp(-e^z)  </math> </center>
+<center><math>\pi(z) = \exp(z)\,\exp(-e^{z})</math></center>
 == Back to REM ==

Rosso: /* Back to REM */

2025-09-13T11:57:00Z

Back to REM

← Older revision		Revision as of 13:57, 13 September 2025
Line 140:		Line 140:
	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 for large <math>M</math> and that allows one to express the random variable <math>E_{\min}</math> in a scaling form: <math>E_{\min} = a_M + b_M z</math>, where <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>.		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 for large <math>M</math> and that allows one to express the random variable <math>E_{\min}</math> in a scaling form: <math>E_{\min} = a_M + b_M z</math>, where <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>.

	==~~Back to REM~~ ==		== Il Caso Gaussiano ==

	Thi chiediamo di dimostrare che per una distribuzione Gaussiana di media zero e varianza sigma^2 la cumulativa puo' essere scritta come		Thi chiediamo di dimostrare che per una distribuzione Gaussiana di media zero e varianza sigma^2 la cumulativa puo' essere scritta come
Line 148:		Line 148:
	<center> <math> A'(E)= -\frac{E}{\sigma^2} +\ldots </math> </center>		<center> <math> A'(E)= -\frac{E}{\sigma^2} +\ldots </math> </center>

	From the ~~second~~ relation ~~we impose~~ <math> A(E_{\min}^{\text{typ}})=- \ln M</math>~~. For~~ large ~~<math>~~M~~</math> we get~~:		* '''Typical minimum''': From the first relation
			<math> A(E_{\min}^{\text{typ}})=- \ln M</math>
			you get for large M:
	<center> <math>E_{\min}^{\text{typ}} = -\sigma \sqrt{2 \log M} \left( 1- \frac{1}{4} \frac{\log (4 \pi \log M)}{\log M} + \ldots \right)</math> </center>		<center> <math>E_{\min}^{\text{typ}} = -\sigma \sqrt{2 \log M} \left( 1- \frac{1}{4} \frac{\log (4 \pi \log M)}{\log M} + \ldots \right)</math> </center>

			* '''Gumbel Scaling''': From the second relation we need to find an <math>a_M</math> that absorb the 'M' factor and a b_M that absorb the 'M' Dependence. In the Gaussian case it is better to A(a_M)=- \ln M and expand <math>A(E)</math> around this value:
	~~We look for a scaling form of~~ the ~~random variable~~ <math>~~E_{\min} =~~ a_M ~~+ b_M z~~</math> ~~such~~ that ~~<math>a_M</math>, <math>b_M</math> are deterministic and '~~'M'~~' dependent while <math>z</math> is random~~ and ''M'~~' independent~~.


	In the Gaussian case it is better to A(a_M)=- \ln M ~~a_M~~ and expand <math>A(E)</math> around this value:
	<center> <math>Q_M(E) \sim \exp\left(-M P(E)\right) \sim \exp\left(-M e^{A(a_M) +A'(a_M)\cdot (E-a_M)}\right) </math> </center>		<center> <math>Q_M(E) \sim \exp\left(-M P(E)\right) \sim \exp\left(-M e^{A(a_M) +A'(a_M)\cdot (E-a_M)}\right) </math> </center>
	Hence setting		Hence setting
	<center> <math> b_M = \frac{1}{A'(a_M)}= \frac{ \sigma}{\sqrt{2 \log M}}</math> </center>		<center> <math> b_M = \frac{1}{A'(a_M)}= \frac{ \sigma}{\sqrt{2 \log M}}</math> </center>
			The random variable <math>z= (E_{\min} - a_M) /b_M </math> ''M'' independent and Gumbel distributed:

			<center> <math>\pi(z) = \exp(z) \exp(-e^z) </math> </center>

			== Back to REM ==

	In the REM the variance is <math>\sigma^2_M = \frac{\log M}{\log 2} = N</math>. Then we have:		In the REM the variance is <math>\sigma^2_M = \frac{\log M}{\log 2} = N</math>. Then we have:

Rosso: /* Back to REM */

2025-09-13T11:05:22Z

Back to REM

← Older revision		Revision as of 13:05, 13 September 2025
Line 142:		Line 142:
	==Back to REM ==		==Back to REM ==

	~~Let us analyze in detail the case of a Gaussian distribution with~~ zero ~~mean and variance~~ <math>\~~sigma^2~~</math>~~. Using integration by parts, we can write :~~		Thi chiediamo di dimostrare che per una distribuzione Gaussiana di media zero e varianza sigma^2 la cumulativa puo' essere scritta come
	<center> <math>P(E) = ~~\int_{~~-~~\infty}^E \frac{dx}{\sqrt{2 \pi \sigma^2}} \, e^{-\frac{x^2}{2 \sigma^2}} = \frac{1}{2\sqrt{\pi}} \int_{~~\frac{E^2}{2 \sigma^2}~~}^{~~\~~infty}~~ \frac~~{dt}{\sqrt{t}} e^{-t}= \frac{\sigma}~~{\sqrt{2 \pi} \|E\|} ~~e^{-\frac{E^2}{2 \sigma^2}} - \frac{1}{4\sqrt{\pi}} \int_~~{~~\frac{E^2}{2~~ \sigma~~^2}}^{\infty~~} \~~frac{dt}{t} e^{-t}~~ </math> </center>		<math>P(E) = \exp(A(E))</math>
	~~The asymptotic expansion for <math>E \to -\infty</math> is :~~		con
	<center> <math>P(E) ~~\approx \frac{\sigma}{\sqrt{2 \pi} \|E\|} e^{~~-\frac{E^2}{2 \sigma^2}} + O(\~~frac{e^{-\frac{E^2}{2 \sigma^2}}}{ \|E\|^2} )~~ </math> </center>		<center> <math> A(E)= -\frac{E^2}{2 \sigma^2} - \log(\frac{\sqrt{2 \pi} \|E\|}{\sigma})+\ldots </math> </center>
			<center> <math> A'(E)= -\frac{E}{\sigma^2} +\ldots </math> </center>

	~~In this case it is convenient to introduce the function <math>A(E)</math> defined as <math>P(E) = \exp(A(E))</math>. hence we have:~~		From the second relation we impose <math> A(E_{\min}^{\text{typ}})=- \ln M</math>. For large <math>M</math> we get:
	~~<center> <math> A_{\text{gauss}}(E)= -\frac{E^2}{2 \sigma^2} - \log(\frac{\sqrt{2 \pi} \|E\|}{\sigma})+\ldots </math> </center>~~
	From the second relation we impose <math> ~~A_{\text{gauss}}~~(E_{\min}^{\text{typ}})=- \ln M</math>. For large <math>M</math> we get:
	<center> <math>E_{\min}^{\text{typ}} = -\sigma \sqrt{2 \log M} \left( 1- \frac{1}{4} \frac{\log (4 \pi \log M)}{\log M} + \ldots \right)</math> </center>		<center> <math>E_{\min}^{\text{typ}} = -\sigma \sqrt{2 \log M} \left( 1- \frac{1}{4} \frac{\log (4 \pi \log M)}{\log M} + \ldots \right)</math> </center>

Line 156:		Line 155:


	In the Gaussian case~~, we start from the third relation introduced earlier~~ and expand <math>A(E)</math> around ~~<math>a_M</math>~~:		In the Gaussian case it is better to A(a_M)=- \ln M a_M and expand <math>A(E)</math> around this value:
	<center> <math>Q_M(E) \sim \exp\left(-M P(E)\right) \sim \exp\left(-M e^{A(a_M) +A'(a_M)\cdot (E-a_M)}\right) </math> </center>		<center> <math>Q_M(E) \sim \exp\left(-M P(E)\right) \sim \exp\left(-M e^{A(a_M) +A'(a_M)\cdot (E-a_M)}\right) </math> </center>
	~~Our goal is obtained by~~ setting		Hence setting
	<center> <math>~~a_M = E_{\min}^{\text{typ}}=-\sigma \sqrt{2 \log M} + \ldots \quad \text{and} \quad~~ b_M = \frac{1}{A'(a_M)}= \frac{ \sigma}{\sqrt{2 \log M}}</math> </center>		<center> <math> b_M = \frac{1}{A'(a_M)}= \frac{ \sigma}{\sqrt{2 \log M}}</math> </center>