T-2: Difference between revisions

From Disordered Systems Wiki
Jump to navigation Jump to search
 
(145 intermediate revisions by the same user not shown)
Line 1: Line 1:
In this set of problems, we use the replica method to study the equilibrium properties of a prototypical mean-field toy model of glasses, the spherical <math>p</math>-spin model.
<strong>Goal: </strong> deriving the equilibrium phase diagram of the Random Energy Model (REM). Notion such as: freezing transition, entropy crisis, condensation, overlap distribution. <br>
<strong> Techniques: </strong> saddle point approximation, thermodynamics, probability theory.
<br>




=== The model: spherical p-spin ===
== Problems==
In the spherical <math>p</math>-spin model the configurations <math> \vec{\sigma}=(\sigma_1, \cdots, \sigma_N) </math> that the system can take satisfy the spherical constraint <math> \sum_{i=1}^N \sigma_i^2=N </math>, and the energy associated to each configuration is  
=== Problem 2: the REM: freezing transition, condensation & glassiness ===
 
In this Problem we compute the equilibrium phase diagram of the model, and in particular the quenched free energy density <math>f(\beta):=f_\infty(\beta) </math> which controls the scaling of the typical value of the partition function, <math>Z_N(\beta) \sim e^{-N \beta \, f(\beta) +o(N) } </math>. We show that the free energy equals to
<center><math>
f(\beta) =  
\begin{cases}
&- \left( T \log 2 + \frac{1}{2 T}\right) \quad \text{if} \quad T \geq T_f\\
& - \sqrt{2 \,\log 2} \quad \text{if} \quad T <T_f
\end{cases} \quad \quad T_f= \frac{1}{ \sqrt{2 \, \log 2}}.
</math></center>
At <math> T_f </math> a transition occurs, often called freezing transition: in the whole low-temperature phase, the free-energy is “frozen” at the value that it has at the critical temperature  <math>T= T_f </math>. <br>
We also characterize the overlap distribution of the model: the overlap between two configurations <math> \vec{\sigma}^\alpha, \vec{\sigma}^\beta </math> is
<math>q_{\alpha \, \beta}= N^{-1} \sum_{i=1}^N \sigma_i^\alpha \, \sigma_i^\beta</math>, and its distribution is
<center>
<center>
<math>
<math>
E(\vec{\sigma}) =\sum_{1 \leq i_1 \leq i_2 \leq \cdots i_p \leq N} J_{i_1 \,i_2 \cdots i_p} \sigma_{i_1} \sigma_{i_2} \cdots \sigma_{i_p},
P_{N, \beta}(q)= \sum_{\alpha, \beta} \frac{e^{-\beta E(\vec{\sigma}^\alpha)}}{Z_N}\frac{e^{-\beta E(\vec{\sigma}^\beta)}}{Z_N}\delta(q-q_{\alpha \beta}).
</math></center>
</math>
 
</center>
where the coupling constants <math>J_{i_1 \,i_2 \cdots i_p}</math> are independent random variables with Gaussian distribution with zero mean and variance <math> p!/ (2 N^{p-1}),</math> and <math> p \geq 3</math> is an integer.
 
=== Quenched vs annealed, and the replica method ===
In TD1, we defined the quenched free energy density as the quantity controlling the scaling of the typical value of the partition function <math>Z </math>, which means:
<center><math>
f= -\lim_{N \to \infty} \frac{1}{\beta N} \overline{ \log Z}.
</math></center>


The annealed free energy <math>f_{ ann} </math> instead controls the scaling of the average value of  <math>Z </math>. It is defined by
<center><math>
f_{ann} = -\lim_{N \to \infty} \frac{1}{\beta N} \log \overline{Z}.
</math></center>
These formulas differ by the order in which the logarithm and the average over disorder are taken. Computing the average of the logarithm is in general a hard problem, which one can address by using a smart representation of the logarithm, that goes under the name of replica trick: 
<center><math>
\log x= \lim_{n \to 0} \frac{x^n-1}{n}
</math></center>
which can be easily shown to be true by Taylor expanding <math>x^n= e^{n \log(x)}= 1+ n\log x+ O(n^2) </math>. Applying this to the average of the partition function, we see that
<center><math>
f= -\lim_{N \to \infty} \lim_{n \to 0}\frac{1}{\beta N n} \frac{\overline{Z^n}-1}{n}.
</math></center>
Therefore, to compute the quenched free-energy we need to compute the moments <math>{\overline{Z^n}}</math> and then take the limit  <math>n \to 0</math>. The annealed one only requires to do the calculation with <math>n=1</math>.


=== Problem 1: the annealed free energy ===


<ol>
<ol>
<li> <em> Energy correlations.</em> At variance with the REM, in the spherical <math>p</math>-spin the energies at different configurations are correlated. Show that <math>\overline{E(\vec{\sigma}) E(\vec{\tau})}= N q(\vec{\sigma}, \vec{\tau})^p/2 + o(1) </math>, where
<li><em> The freezing transition.</em>  
<center><math>q(\vec{\sigma}, \vec{\tau})= \frac{1}{N}\sum_{i=1}^N \sigma_i \tau_i </math></center>
The partition function the REM reads
is the overlap between the two configurations. Why can we say that for <math>p \to \infty </math> this model converges with the REM discussed in the previous lecture?</li>
<math>
</ol>
Z_N(\beta) = \sum_{\alpha=1}^{2^N} e^{-\beta E_\alpha}= \int dE \, \mathcal{N}_N(E) e^{-\beta E}.
<ol start="2">
</math>  
<li> <em> Energy contribution.</em> Show that computing <math>\overline{Z}</math> boils down to computing the average <math> \overline{e^{-\beta J_{i_1 \, \cdots i_p} \sigma_{i_1} \cdots \sigma_{i_p}}}</math>, which is a Gaussian integral. Compute this average. Hint: if X is a centered Gaussian variable with variance <math>\sigma^2</math>, then <math>\overline{e^{\alpha X}}=e^{\frac{\alpha^2 \sigma^2}{2} }</math>.
Using the behaviour of the typical value of <math> \mathcal{N}_N </math> determined in Problem 1, derive the free energy of the model (hint: perform a saddle point calculation). What is the order of this thermodynamic transition?  
</ol></li>
<ol start="3">
<li> <em> Entropy contribution.</em> The volume of a sphere  <math>S_N</math> of radius <math>\sqrt{N}</math> in dimension <math>N</math> is given by <math>N^{\frac{N}{2}} \pi^{\frac{N}{2}}/\left(\frac{N}{2}\right)!</math>. Use the large-N asymptotic of this to conclude the calculation of the annealed free energy:
<center><math>
f_{ann}= - \frac{1}{\beta}\left( \frac{\beta^2}{4}+ \frac{1}{2}\log (2 \pi e)\right).
</math></center>
This result is only slightly different with respect to the free-energy of the REM in the high-temperature phase: can you identify the source of this difference? </li>
</ol>
</ol>
<br>
=== Problem 2: the replica trick and the quenched free energy ===
<ol>
<li> <em> Step 1: average over the disorder.</em> By using the same Gaussian integration discussed above, show that the <math>n</math>-th moment of the partition function is
<center>
<math>\overline{Z^n}= \int_{S_N} \prod_{a=1}^n d \vec{\sigma}^a e^{\frac{\beta^2}{4} \sum_{a,b=1}^n \left[q({\vec{\sigma}^a, \vec{\sigma}^b})\right]^p}</math></center>
Justify why averaging over the disorder induces a coupling between the replicas.
</li>
</li>
</ol>
<br>
<br>


<ol start="2">
<ol start="2">
<li><em> Step 2: identify the order parameter.</em> Using the identity <math> 1=\int dq_{ab} \delta \left( q({\vec{\sigma}^a, \vec{\sigma}^b})- q_{ab}\right) </math>, show that <math>\overline{Z^n}</math> can be rewritten as an integral over <math>n(n-1)/2</math> variables  only, as:
<li><em> Fluctuations, and back to average vs typical.</em> Similarly to what we did for the entropy, one can define an annealed free energy <math> f_{\text{a}}(\beta) </math> from <math> \overline{Z_N(\beta)}=e^{- N \beta f_{\text{a}}(\beta) + o(N)} </math>: show that in the whole low-temperature phase this is smaller than the quenched free energy obtained above. Putting all the results together, justify why the average of the partition function in the low-T phase is "dominated by rare events".  
 
</ol></li>
<center>
<math>\overline{Z^n}= \int \prod_{a<b} d q_{ab} e^{\frac{N\beta^2}{4} \sum_{a,b=1}^n q_{ab}^p+ \frac{Nn}{2} \log (2 \pi e) + \frac{N}{2} \log \det Q+ o(N)} \equiv \int \prod_{a<b} d q_{ab} e^{N \mathcal{A}[Q]+ o(N)}, \quad \quad Q_{ab} \equiv \begin{cases}
&q_{ab} \text{ if  } a <b\\
&1 \text{ if  } a =b\\
&q_{ba}\text{ if  } a >b
\end{cases}</math></center>
In the derivation, you can use the fact that
<math>
\int_{S_N} \prod_{a=1}^n d \vec{\sigma}^a \, \prod_{a<b}\delta \left(q({\vec{\sigma}^a, \vec{\sigma}^b})- q_{ab} \right)= e^{N S[Q]+ o(N)}</math>, where <math>  S[Q]= n \log (2 \pi e)/2 + (1/2)\log \det Q</math>. The matrix <math> Q</math> is an order parameter: explain the analogy with the magnetisation in the mean-field solution of the Ising model.
 
</li>
</ol>
<br>
<br>


<ol start="3">
<ol start="3">
<li><em> Step 3: the saddle point (RS).</em> For large N, the integral can be computed with a saddle point approximation for general <math>n</math>. The saddle point variables are the matrix elements <math> q_{ab}</math> with <math> a \neq b</math>. Show that the saddle point equations read
<li> <em> Entropy crisis.</em> What happens to the entropy of the model when the critical temperature is reached, and in the low temperature phase? What does this imply for the partition function <math> Z_N</math>?</li>
 
<center><math>
\frac{\partial \mathcal{A}[Q]}{\partial q_{ab}}=\frac{\beta^2}{4}p q_{ab}^{p-1}+ \frac{1}{2} \left(Q^{-1}\right)_{ab}^{-1}=0 \quad \quad \text{for } \quad a \neq b
</math></center>
To solve these equations and get the free energy, one needs to make an assumption on the structure of the matrix <math> Q</math>.
</li>
</ol>
</ol>
<br>
<br>


=== Problem 3: the replica trick and the quenched free energy ===
<ol start="4"><li><em> Overlap distribution and glassiness.</em>  
 
Justify why in the REM the overlap when  <math> N \to \infty</math> typically can take only values zero and one, leading to
<!-- Remember however that to get the free energy, we are interested in the limit <math>n \to 0</math>. A standard way to proceed is: (i) make an ansatz on the structure of the matrix Q, (ii) compute <math>\mathcal{A}[Q]</math> within this ansatz and expand <math>\mathcal{A}[Q]= n\mathcal{A}_0 + O(n^2)</math>, (iii) perform the saddle-point calculation on <math>\mathcal{A}_0</math>. -->
 
Let us consider the simplest possible ansatz for the structure of the matrix Q, that is the Replica Symmetric (RS) ansatz:
 
<center>
<center>
<math>
<math>
Q=\begin{pmatrix}
P_\beta(q)= \lim_{N \to \infty} \overline{P_{N, \beta}(q)}= \overline{I_2} \, \delta(q-1)+ (1-\overline{I_2}) \, \delta(q), \quad \quad I_2= \lim_{N \to \infty} \frac{\sum_\alpha z_\alpha^2}{\left(\sum_\alpha z_\alpha\right)^2}, \quad \quad z_\alpha=e^{-\beta E_\alpha}
1 & q &q \cdots& q\\
q & 1 &q \cdots &q\\
&\cdots& &\\
q & q &q \cdots &1
\end{pmatrix}
</math>
</math>
</center>
</center>
Under this assumption, there is a unique saddle point variable, that is <math>q</math>.
Why <math> I_2 </math> can be interpreted as a probability? Show that
 
 
<ol>
<li>
Check that the inverse of the overlap matrix is
<center>
<center>
<math>
<math>
Q^{-1}=\begin{pmatrix}
\overline{I_2}  
\alpha & \beta &\beta \cdots& \beta\\
=\begin{cases}
\beta & \alpha &\beta \cdots &\beta\\
0 \quad &\text{if} \quad T>T_f\\
&\cdots& &\\
1-\frac{T}{T_f} \quad &\text{if} \quad T \leq T_f
\beta & \beta &\beta \cdots &\alpha
\end{cases}
\end{pmatrix}
\quad  
\quad
\text{with}
\quad
\alpha= \frac{1}{1-q}
\quad
\text{and}
\quad
\beta=\frac{-1}{(1-q)[1+(n-1)q]}
</math>
</math>
</center>
</center>
Compute the saddle point equation for  <math>q</math> in the limit  <math>n \to 0</math>, and show that this equation admits always the solution <math>q^*= 0</math>.
and interpret the result. In particular, why is this consistent with the entropy crisis? <br>
</li>
<em> Hint: </em> For <math> T<T_f </math>, use that <math> \overline{I_2}=\frac{T_f}{T} \frac{d}{d\mu}\log  \int_{0}^\infty (1-e^{-u -\mu u^2})u^{-\frac{T}{T_f}-1} du\Big|_{\mu=0}</math></li> </ol>
</ol>
<br>
<br>


<ul>
== Comments==
    <li>3.a) Show linear expansion. Use determinant </li>
 
    <li>3.b) Saddle point equation for q. Solution matching annealed: why is this the case? </li>
[[File:Suscept-experiment.png|thumb|left|x140px| Experimental measurements of the magnetic Field Cooled (top) and Zero-Field Cooled  (bottom) susceptibility in a CuMn spin glass. Figure taken from C. Djurberg, K. Jonason, P. Nordblad, https://arxiv.org/abs/cond-mat/9810314]]
    <li>3.c) Another solution [graphically]</li>
 
</ul>
<ol>
<li><em> Glassiness.</em> The low-T phase of the REM is a frozen phase, characterized by the fact that the free energy is temperature independent, and that the typical value of the partition function is very different from the average value. In fact, the low-T phase is also <em> a glass phase </em> with <math> q_{EA}=1</math>. This is also a phase  where a peculiar symmetry, the so called replica symmetry, is broken. We go back to this concepts in the next sets of problems.


<br>
<br>
<li><em> The physics: susceptibility and experiments.</em> The magnetic susceptibility of the REM can be computed adding a field to the energy function:  <math> E(\vec{\sigma}^\alpha) \to E_\alpha + H \sum_i \sigma^\alpha_i</math>. Redoing the thermodynamics calculation in presence of the field (do its!!) one can show that in the REM
<center><math>
\chi(\beta) = \frac{\partial m(\beta, H) }{\partial H}\Big|_{H=0}=
\begin{cases}
&\frac{\partial \tanh(\beta H) }{\partial H}\Big|_{H=0}= \beta \quad \text{if} \quad T \geq T_f\\
& \frac{\partial \tanh(\beta_f H) }{\partial H}\Big|_{H=0}= \beta_f \quad \text{if} \quad T <T_f
\end{cases}.
</math></center>
where <math>m(\beta, H)</math> is the equilibrium magnetization. Therefore, the susceptibility does not diverge at the transition, as it happens in standard ferromagnets, but becomes flat (recall that what diverges at the spin-glass transition are non-linear susceptibilities, as discussed in Lecture 1 by Alberto). A similar behavior is found in measurements of the Field-Cooled susceptibility in spin glass systems, and it is captured by the Parisi's solution of the SK model with replicas, which therefore captures the physics probed in experiments.


</li>
</li>
</ol>
</ol>
<br>
 
== Check out: key concepts ==
 
Freezing transition, entropy crisis, condensation, overlap distribution.
 
== To know more ==
* Derrida. Random-energy model: limit of a family of disordered models [https://hal.science/hal-03285940v1/document]
* Derrida. Random-energy model: An exactly solvable model of disordered systems [http://www.lps.ens.fr/~derrida/PAPIERS/1981/prb81.pdf]

Latest revision as of 19:42, 9 February 2025

Goal: deriving the equilibrium phase diagram of the Random Energy Model (REM). Notion such as: freezing transition, entropy crisis, condensation, overlap distribution.
Techniques: saddle point approximation, thermodynamics, probability theory.


Problems

Problem 2: the REM: freezing transition, condensation & glassiness

In this Problem we compute the equilibrium phase diagram of the model, and in particular the quenched free energy density which controls the scaling of the typical value of the partition function, . We show that the free energy equals to

At a transition occurs, often called freezing transition: in the whole low-temperature phase, the free-energy is “frozen” at the value that it has at the critical temperature .
We also characterize the overlap distribution of the model: the overlap between two configurations is , and its distribution is


  1. The freezing transition. The partition function the REM reads Using the behaviour of the typical value of determined in Problem 1, derive the free energy of the model (hint: perform a saddle point calculation). What is the order of this thermodynamic transition?


  1. Fluctuations, and back to average vs typical. Similarly to what we did for the entropy, one can define an annealed free energy from : show that in the whole low-temperature phase this is smaller than the quenched free energy obtained above. Putting all the results together, justify why the average of the partition function in the low-T phase is "dominated by rare events".


  1. Entropy crisis. What happens to the entropy of the model when the critical temperature is reached, and in the low temperature phase? What does this imply for the partition function ?


  1. Overlap distribution and glassiness. Justify why in the REM the overlap when typically can take only values zero and one, leading to

    Why can be interpreted as a probability? Show that

    and interpret the result. In particular, why is this consistent with the entropy crisis?

    Hint: For , use that


Comments

Experimental measurements of the magnetic Field Cooled (top) and Zero-Field Cooled (bottom) susceptibility in a CuMn spin glass. Figure taken from C. Djurberg, K. Jonason, P. Nordblad, https://arxiv.org/abs/cond-mat/9810314
  1. Glassiness. The low-T phase of the REM is a frozen phase, characterized by the fact that the free energy is temperature independent, and that the typical value of the partition function is very different from the average value. In fact, the low-T phase is also a glass phase with . This is also a phase where a peculiar symmetry, the so called replica symmetry, is broken. We go back to this concepts in the next sets of problems.

  2. The physics: susceptibility and experiments. The magnetic susceptibility of the REM can be computed adding a field to the energy function: . Redoing the thermodynamics calculation in presence of the field (do its!!) one can show that in the REM

    where is the equilibrium magnetization. Therefore, the susceptibility does not diverge at the transition, as it happens in standard ferromagnets, but becomes flat (recall that what diverges at the spin-glass transition are non-linear susceptibilities, as discussed in Lecture 1 by Alberto). A similar behavior is found in measurements of the Field-Cooled susceptibility in spin glass systems, and it is captured by the Parisi's solution of the SK model with replicas, which therefore captures the physics probed in experiments.

Check out: key concepts

Freezing transition, entropy crisis, condensation, overlap distribution.

To know more

  • Derrida. Random-energy model: limit of a family of disordered models [1]
  • Derrida. Random-energy model: An exactly solvable model of disordered systems [2]