Week 2 - Revisiting Parisi's Equations and the Replica Method

Antares Chen University of Chicago

Abstract

We revisit our geometric interpretation of the Parisi equations via a few interactive demos. We then discuss the historical origins of the Parisi Variational Principle via the Replica method. We demonstrate calculations leading to the action integral form of the replicated free energy, setting us up to introduce Replica Symmetry Breaking next time.

Date
Feb 8, 2022 10:30 AM — 12:00 PM
Event
Algorithmic Methods in Statistical Physics
Location
JCL 390

Recap of the Meeting


We spent most of our time discussing geometric intuition for the dynamics described by the Parisi’s PDEs.

  • The equations are given as $$ \begin{cases} \Phi_t(t, x) = - \frac{\beta^2}{2} \cdot \big( \Phi_{xx}(t, x) + \mu(t) \cdot \lvert \Phi_x(t, x) \rvert^2 \big) \\ \Phi(1, x) = \log \big( 2 \cdot \cosh x \big) \end{cases} $$

  • When a change of variable is performed $t(s) = 1 - s$, the PDE becomes $$ \begin{cases} \Phi_s(s, x) = \frac{\beta^2}{2} \cdot \big( \Phi_{xx}(s, x) + \mu( 1 - s ) \cdot \lvert \Phi_x(s, x) \rvert^2 \big) \\ \Phi(0, x) = \log \big( 2 \cdot \cosh x \big) \end{cases} $$

  • The behavior of this dynamic can be interpreted as follows. At time $s = 0$, we begin with the soft absolute value function $\Phi(0, x) = \log \big( 2 \cdot \cosh x \big)$. As $s$ progresses from $0 \rightarrow 1$, the evolution of $\Phi(s, x)$ is always affected by heat diffusion dynamics $$- \frac{\beta^2}{2} \cdot \Phi_{xx}(t, x)$$

  • On the other hand, $\mu(1 - s)$ is non-increasing on the interval $[0, 1]$. The function $\mu(s)$ itself is the cumulative density function of some measure supported on $[0, 1]$. Thus, the evolution of $\Phi(s, x)$ is also affected by a “gradient descent” dynamic $$\frac{\beta^2}{2} \cdot \lvert \Phi_x(s, x) \rvert^2$$ whose weight is dampened as $s$ increases towards $s = 1$.

This picture was helpful to have. It plots the evolution of $\Phi(s, x)$ from $s = 0 \rightarrow 1$

where, on the right, the red contour corresponds to $s = 0$, the navy contour corresponds to $s = 1$, and $\mu(t)$ is given by the CDF of a Gaussian centered at $\frac{1}{2}$ with variance $\frac{1}{100}$

When thinking of $\mu(1 - s)$, consider the above picture as “backwards” i.e. think of $t$ progressing from $1 \rightarrow 0$.

Material


  • [pdf] Scanned presentation notes

  • [demo] [source] Plotting Parisi’s equations

  • [link] Scribe notes – Juspreet Singh Sandhu (see Day 2)