T-6

Goal: Complete the characterisation of the energy landscape of the spherical $p$-spin.
Techniques: saddle point, random matrix theory.

Problems

Problem 6: the Hessian at the stationary points, and random matrix theory

This is a continuation of problem 5. To get the complexity of the spherical $p$-spin, it remains to compute the expectation value of the determinant of the Hessian matrix: this is the goal of this problem. We will do this exploiting results from random matrix theory discussion in the Tutorial and Exercise 4 .

1. Gaussian Random matrices. Show that the matrix $M$, defined in Problem 5, is a GOE matrix, i.e. a matrix taken from the Gaussian Orthogonal Ensemble, meaning that it is a symmetric matrix with distribution $P_N(M)=Z_N^{-1}\text{exp}(-\frac{N}{4{\sigma }^2}\text{Tr}{M}^2)$ where $Z_N$ is a normalization. What is the value of ${\sigma }^2$?

2. Eigenvalue density and concentration. Let ${\lambda }_{\alpha }$ be the eigenvalues of the matrix $M$. Show that the following identity holds: $$\mathbb{𝔼}[|\text{det}(M-pϵ\mathbb{𝕀})|]=\mathbb{𝔼}\left[\text{exp}((N-1)\int d\lambda {\rho }_{N-1}(\lambda )\log |\lambda -pϵ|)\right],{\rho }_{N-1}(\lambda )=\frac{1}{N-1}\sum _{\alpha =1}^{N-1}\delta (\lambda -{\lambda }_{\alpha })$$ where ${\rho }_{N-1}(\lambda )$ is the empirical eigenvalue distribution. It can be shown that if $M$ is a GOE matrix, the distribution of the empirical distribution has a large deviation form with speed $N^2$, meaning that $P_N[\rho ]=e^{-N^{2}g[\rho ]}$ where now $g[\cdot ]$ is a functional. Using a saddle point argument, show that this implies $$\mathbb{𝔼}\left[\text{exp}((N-1)\int d\lambda {\rho }_{N-1}(\lambda )\log |\lambda -pϵ|)\right]=\text{exp}[N\int d\lambda {\rho }_{\mathrm{\infty }}(\lambda +pϵ)\log |\lambda |+o(N)]$$ where ${\rho }_{\mathrm{\infty }}(\lambda )$ is the typical value of the eigenvalue density, which satisfies $g[{\rho }_{\mathrm{\infty }}]=0$.

3. The semicircle and the complexity. The eigenvalue density of GOE matrices is self-averaging, and it equals to $$\underset{N\to \mathrm{\infty }}{\lim }{\rho }_N(\lambda )=\underset{N\to \mathrm{\infty }}{\lim }\mathbb{𝔼}[{\rho }_N(\lambda )]={\rho }_{\mathrm{\infty }}(\lambda )=\frac{1}{2\pi {\sigma }^2}\sqrt{4{\sigma }^2-{\lambda }^2}$$
   Combining all the results, show that the annealed complexity is $${\mathrm{\Sigma }}_{\text{a}}(ϵ)=\frac{1}{2}\log [4e(p-1)]-\frac{{ϵ}^2}{2}+I_p(ϵ),I_p(ϵ)=\frac{2}{\pi }\int dx\sqrt{1-{(x-\frac{ϵ}{{ϵ}_{\text{th}}})}^2}\log |x|,{ϵ}_{\text{th}}=-2\sqrt{\frac{p-1}{p}}.$$ The integral $I_p(ϵ)$ can be computed explicitly, and one finds: $$I_p(ϵ)=\{\begin{array}{ll}& \frac{{ϵ}^2}{{ϵ}_{\text{th}}^2}-\frac{1}{2}-\frac{ϵ}{{ϵ}_{\text{th}}}\sqrt{\frac{{ϵ}^2}{{ϵ}_{\text{th}}^2}-1}+\log (\frac{ϵ}{{ϵ}_{\text{th}}}+\sqrt{\frac{{ϵ}^2}{{ϵ}_{\text{th}}^2}-1})-\log 2\text{if}ϵ\le {ϵ}_{\text{th}}\\ & \frac{{ϵ}^2}{{ϵ}_{\text{th}}^2}-\frac{1}{2}-\log 2\text{if}ϵ>{ϵ}_{\text{th}}\end{array}$$ Plot the annealed complexity, and determine numerically where it vanishes: why is this a lower bound or the ground state energy density?

4. The threshold and the stability. Sketch ${\rho }_{\mathrm{\infty }}(\lambda +pϵ)$ for different values of $ϵ$; recalling that the Hessian encodes for the stability of the stationary points, show that there is a transition in the stability of the stationary points at the critical value of the energy density ${ϵ}_{\text{th}}=-2\sqrt{(p-1)/p}.$ When are the critical points stable local minima? When are they saddles? Why the stationary points at $ϵ={ϵ}_{\text{th}}$ are called marginally stable ?

Back to dynamics: quenches, and dynamical transitions

Non-rugged vs rugged energy landscapes.

Through Problems 5 and 6, we have shown that the energy landscape of the spherical $p$-spin model has exponentially many stationary points , and that there is a transition at the energy density ${ϵ}_{\mathrm{t}\mathrm{h}}$: for $ϵ>{ϵ}_{\mathrm{t}\mathrm{h}}$ the stationary points are saddles, for $ϵ\le {ϵ}_{\mathrm{t}\mathrm{h}}$ they are local minima. Let us try to deduce something on the systems's dynamics out of this.

Gradient descent dynamics. The local minima are dynamically stable: if I do gradient descent, I get stuck in a local minimum and I exert a small perturbation to the configuration, gradient descent brings me back to the local minimum. In this sense, these configurations are trapping. Therefore, if I try to optimize the landscape, i.e. to reach the ground state, with gradient descent dynamics, I expect that I will not be able to reach the ground state easily, as I will be trapped by these local minima. In fact, for the spherical $p$-spin model it can be shown that starting the gradient descent dynamics from random initial conditions and evolving the configuration with gradient descent (possibly with infinitesimal noise), $$\underset{t\to \mathrm{\infty }}{\lim }\underset{N\to \mathrm{\infty }}{\lim }\frac{E(\overrightarrow{\sigma }(t))}{N}={ϵ}_{\mathrm{t}\mathrm{h}}\ne {ϵ}_{\mathrm{g}\mathrm{s}}.$$ The system gets stuck at the energy density level where local minima start to appear, and does not reach the deeper local minima.

Quenches. We can generalize this protocol to higher $T$: we extract randomly the initial condition of the dynamics, and then we evolve the configuration with Langevin dynamics (gradient descent + noise): $$\frac{d\overrightarrow{\sigma }(t)}{dt}=-{\mathrm{\nabla }}_{\perp }E(\overrightarrow{\sigma })+{\overrightarrow{\eta }}_{\perp }(t),⟨{\eta }_i(t){\eta }_j(t^{\prime })⟩=2T{\delta }_{ij}\delta (t-t^{\prime })$$ In Langevin dynamics, ${\overrightarrow{\eta }}_{\perp }(t)$ a Gaussian vector at each time $t$, uncorrelated from the vectors at other times $t^{\prime }\ne t$, with zero average and some constant variance proportional to temperature. They represent effectively the action of a bath on the system. This dynamical protocol is called a temperature quench. The question we can ask is: does the system equilibrate with the bath under this dynamics? If yes, we should see that $$\underset{t\to \mathrm{\infty }}{\lim }\underset{N\to \mathrm{\infty }}{\lim }\frac{E(\overrightarrow{\sigma }(t))}{N}={ϵ}_{\mathrm{e}\mathrm{q}}(T),$$ where ${ϵ}_{\mathrm{e}\mathrm{q}}(T)$ is the equilibrium energy density (that we can compute using Boltzmann). Equilibrating with the bath would indeed imply that at large time the system visits uniformly the equilibrium energy shell.

Dynamical transition. Now, in the spherical $p$-spin we know that if ${ϵ}_{\mathrm{e}\mathrm{q}}(T)>{ϵ}_{\mathrm{t}\mathrm{h}}$, the energy shell has many stationary points, but they are all unstable saddles and do not trap the dynamics. We expect that this energy shell is relatively easy to explore dynamically, and that equilibration takes place. On the other hand, if ${ϵ}_{\mathrm{e}\mathrm{q}}(T)<{ϵ}_{\mathrm{t}\mathrm{h}}$, in the equilibrium energy shell and at higher energy, there are exponentially many local minima that trap the dynamics, and we expect that reaching equilibrium configurations will be difficult. This tells us that there exists a critical $T_d$, defined by $${ϵ}_{\mathrm{e}\mathrm{q}}(T_d)={ϵ}_{\mathrm{t}\mathrm{h}},$$ such that for $T<T_d$ $$\underset{t\to \mathrm{\infty }}{\lim }\underset{N\to \mathrm{\infty }}{\lim }\frac{E(\overrightarrow{\sigma }(t))}{N}\ne {ϵ}_{\mathrm{e}\mathrm{q}}(T).$$ This is called the <emph>dynamical transition temperature</emph>.

Equilibration timescales. Does it mean that when $T<T_d$, the system <emph>never</emph> equilibrates? This is true only in the limit $N\to \mathrm{\infty }$. When $N$ is finite, there is a timescale ${\tau }_{\mathrm{e}\mathrm{q}}(T,N)$ beyond which the system equilibrates. So, the dynamical transition is a transition only for $N\to \mathrm{\infty }$, and a crossover for finite $N$. However, this equilibration timescale in the spherical $p$-spin scales as $${\tau }_{\mathrm{e}\mathrm{q}}(T<T_d,N)\sim e^N.$$ This is again due to the presence of many local minima/ metastable states, that are separated by <emph>extensive</emph> energy barriers. ^[*] .

[*] - This quantity looks similar to the entropy $S(ϵ)$ we computed for the REM in Problem 1. However, while the entropy counts all configurations at a given energy density, the complexity $\mathrm{\Sigma }(ϵ)$ accounts only for the stationary points.

Check out: key concepts

Metastable states, Hessian matrices, random matrix theory, landscape’s complexity.