T-2

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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p} -spin model.

The model: spherical p-spin

In the spherical Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p} -spin model the configurations Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{\sigma}=(\sigma_1, \cdots, \sigma_N) } that the system can take satisfy the spherical constraint Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{i=1}^N \sigma_i^2=N } , and the energy associated to each configuration is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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}, }

where the coupling constants Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_{i_1 \,i_2 \cdots i_p}} are independent random variables with Gaussian distribution with zero mean and variance Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p!/ (2 N^{p-1}),} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p \geq 3} 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z } , which means:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f= -\lim_{N \to \infty} \frac{1}{\beta N} \overline{ \log Z}. }

The annealed free energy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_{ ann} } instead controls the scaling of the average value of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z } . It is defined by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_{ann} = -\lim_{N \to \infty} \frac{1}{\beta N} \log \overline{Z}. }

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:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \log x= \lim_{n \to 0} \frac{x^n-1}{n} }

which can be easily shown to be true by Taylor expanding Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^n= e^{n \log(x)}= 1+ n\log x+ O(n^2) } . Applying this to the average of the partition function, we see that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f= -\lim_{N \to \infty} \lim_{n \to 0}\frac{1}{\beta N n} \frac{\overline{Z^n}-1}{n}. }

Therefore, to compute the quenched free-energy we need to compute the moments Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\overline{Z^n}}} and then take the limit Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \to 0} . The annealed one only requires to do the calculation with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n=1} .

Problem 2.1: the annealed free energy

1. Energy correlations. At variance with the REM, in the spherical Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p} -spin the energies at different configurations are correlated. Show that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \overline{E(\vec{\sigma}) E(\vec{\tau})}= N q(\vec{\sigma}, \vec{\tau})^p/2 + o(1) } , where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q(\vec{\sigma}, \vec{\tau})= \frac{1}{N}\sum_{i=1}^N \sigma_i \tau_i } is the overlap between the two configurations. Why can we say that for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p \to \infty } this model converges with the REM discussed in the previous lecture?

2. Energy contribution. Show that computing Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \overline{Z}} boils down to computing the average Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \overline{e^{-\beta J_{i_1 \, \cdots i_p} \sigma_{i_1} \cdots \sigma_{i_p}}}} , which is a Gaussian integral. Compute this average. Hint: if X is a centered Gaussian variable with variance Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sigma^2} , then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \overline{e^{\alpha X}}=e^{\frac{\alpha^2 \sigma^2}{2} }} .

3. Entropy contribution. The volume of a sphere Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_N} of radius Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{N}} in dimension Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N} is given by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N^{\frac{N}{2}} \pi^{\frac{N}{2}}/\left(\frac{N}{2}\right)!} . Use the large-N asymptotic of this to conclude the calculation of the annealed free energy: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_{ann}= - \frac{1}{\beta}\left( \frac{\beta^2}{4}+ \frac{1}{2}\log (2 \pi e)\right). } 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?

Problem 2.2: the replica trick and the quenched free energy

1. Step 1: average over the disorder. By using the same Gaussian integration discussed above, show that the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} -th moment of the partition function is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

   Justify why averaging over the disorder induces a coupling between the replicas.

2. Step 2: identify the order parameter. Using the identity Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1=\int dq_{ab} \delta \left( q({\vec{\sigma}^a, \vec{\sigma}^b})- q_{ab}\right) } , show that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \overline{Z^n}} can be rewritten as an integral over Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n(n-1)/2} variables only, as: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

   In the derivation, you can use the fact that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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)}} , where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S[Q]= n \log (2 \pi e)/2 + (1/2)\log \det Q} . The matrix Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q} is an order parameter: explain the analogy with the magnetisation in the mean-field solution of the Ising model.

3. Step 3: the saddle point (RS). For large N, the integral can be computed with a saddle point approximation for general Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} . The saddle point variables are the matrix elements Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q_{ab}} with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \neq b} . Show that the saddle point equations read Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 }

   To solve these equations and get the free energy, one needs to make an assumption on the structure of the matrix Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q} .

Let us consider the simplest possible ansatz for the structure of the matrix Q, that is the Replica Symmetric (RS) ansatz:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q=\begin{pmatrix} 1 & q_0 &q_0 \cdots& q_0\\ q_0 & 1 &q_0 \cdots &q_0\\ &\cdots& &\\ q_0 & q_0 &q_0 \cdots &1 \end{pmatrix} }

Under this assumption, there is a unique saddle point variable, that is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q_0} .

1. Check that the inverse of the overlap matrix is

   Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q^{-1}=\begin{pmatrix} \alpha & \beta &\beta \cdots& \beta\\ \beta & \alpha &\beta \cdots &\beta\\ &\cdots& &\\ \beta & \beta &\beta \cdots &\alpha \end{pmatrix} \quad \quad \text{with} \quad \alpha= \frac{1}{1-q_0} \quad \text{and} \quad \beta=\frac{-1}{(1-q_0)[1+(n-1)q_0]} }

   Compute the saddle point equation for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q_0} in the limit $n\to 0$, and show that this equation admits always the solution $q_{0}=0$: why is this called the paramagnetic solution?

2. Compute the free energy corresponding to the solution $q_{0}=0$, and show that it reproduces the annealed free energy. Do you have an interpretation for this?

1. Overlpa interpretation

Problem 4: the 1-RSB (Replica Symmetry Broken) calculation

In the previous problem, we have chosen a certain parametrization of the overlap matrix $Q$, which corresponds to assuming that typically all the copies of the systems fall into configurations that are at overlap $q_{0}$ with each others, no matter what is the pair of replicas considered. This assumption is however not the good one at low temperature. We now assume a different parametrisation, that corresponds to breaking the symmetry between replicas: in particular, we assume that typically the $n$ replicas fall into configurations that are organized in $n/m$ groups of size $m$; pairs of replicas in the same group are more strongly correlated and have overlap $q_{1}$, while pairs of replicas belonging to different groups have a smaller overlap $q_{0}<q_{1}$. This corresponds to the following block structure for the overlap matrix:

$Q={\begin{pmatrix}1&q_{1}&q_{1}&q_{0}&q_{0}\cdots &q_{0}\\q_{1}&1&q_{1}&q_{0}&q_{0}\cdots &q_{0}\\q_{1}&q_{1}&1&q_{0}&q_{0}\cdots &q_{0}\\\cdots \\\cdots \\\cdots \\q_{0}&q_{0}\cdots &q_{0}&1&q_{1}&q_{1}\\q_{0}&q_{0}\cdots &q_{0}&q_{1}&1&q_{1}\\q_{0}&q_{0}\cdots &q_{0}&q_{1}&q_{1}&1\\\end{pmatrix}}$

Here we have three parameters: $m,q_{0},q_{1}$ (in the formula above, $m=3$).

1. Using that

   $\log \det Q=n{\frac {m-1}{m}}\log(1-q_{1})+{\frac {n-m}{m}}\log[m(q_{1}-q_{0})+1-q_{1}]+\log \left[nq_{0}+m(q_{1}-q_{0})+1-q_{1}\right]$

   show that the free energy now becomes:

   $f_{1RSB}=-{\frac {1}{2\beta }}\left[{\frac {\beta ^{2}}{2}}\left(1+(m-1)q_{1}^{p}-mq_{0}^{p}\right)+{\frac {m-1}{m}}\log(1-q_{1})+{\frac {1}{m}}\log[m(q_{1}-q_{0})+1-q_{1}]+{\frac {q_{0}}{m(q_{1}-q_{0})+1-q_{1}}}\right]$

   Under which limit this reduces to the replica symmetric expression?

2. Compute the saddle point equations with respect to the parameter $q_{0},q_{1}$ and $m$ are. Check that $q_{0}=0$ is again a valid solution of these equations, and that for $q_{0}=0$ the remaining equations reduce to:

   $(m-1)\left[{\frac {\beta ^{2}}{q}}pq_{1}^{p-1}-{\frac {1}{m}}{\frac {1}{1-q_{1}}}+{\frac {1}{m}}{\frac {1}{1+(m-1)q_{1}}}\right]=0,\quad \quad {\frac {\beta ^{2}}{2}}q_{1}^{p}+{\frac {1}{m^{2}}}\log \left({\frac {1-q_{1}}{1+(m-1)q_{1}}}\right)+{\frac {q_{1}}{m[1+(m-1)q_{1}]}}=0$

   How does one recover the paramagnetic solution?

3. We now look for a solution different from the paramagnetic one. To begin with, we set $m=1$ to satisfy the first equation, and look for a solution of

   ${\frac {\beta ^{2}}{2}}q_{1}^{p}+\log \left(1-q_{1}\right)+q_{1}=0$

   Plot this function for $p=3$ and different values of $\beta$, and show that there is a critical temperature $T_{c}$ where a solution $q_{1}\neq 0$ appears: what is the value of this temperature (determined numerically)?

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