On the energy landscape of the mixed even p-spin model

Abstract

We investigate the energy landscape of the mixed even p-spin model with Ising spin configurations. We show that for any given energy level between zero and the maximal energy, with overwhelming probability there exist exponentially many distinct spin configurations such that their energies stay near this energy level. Furthermore, their magnetizations and overlaps are concentrated around some fixed constants. In particular, at the level of maximal energy, we prove that the Hamiltonian exhibits exponentially many orthogonal peaks. This improves the results of Chatterjee (Disorder chaos and multiple valleys in spin glasses, 2009) and Ding et al. (Ann Probab 43(6):3468–3493, 2015), where the former established a logarithmic size of the number of the orthogonal peaks, while the latter proved a polynomial size. Our second main result obtains disorder chaos at zero temperature and at any external field. As a byproduct, this implies that the fluctuation of the maximal energy is superconcentrated when the external field vanishes and obeys a Gaussian limit law when the external field is present.

Appendix

This appendix is devoted to establishing Proposition 2. We prove a priori estimate first:

Lemma 10

Let $0<r_0<r_1<r_2<\infty .$ Suppose that $\kappa _1,\kappa _2\in L^\infty ([r_0 ,r_2]\times \mathbb {R})$ and $g\in L^\infty (\mathbb {R})$ with $\Vert \kappa _i\Vert _\infty \le C_i$ for $i=1,2$ and $\Vert g\Vert _\infty \le C_0$. Assume that u is the classical solution to

$$\begin{aligned} \partial _ru(r,x)=\partial _{xx}u(r,x)+\kappa _1(r,x)\partial _xu(r,x)+k_2(r,x),\,\,\forall (r,x)\in (r_0 ,{r_2} ]\times \mathbb {R} \end{aligned}$$

with initial condition $u(r_0 ,x)=g(x).$ If

$$\begin{aligned} r\mapsto \Vert \partial _xu(r,\cdot )\Vert _\infty \end{aligned}$$

is continuous on $[r_0 ,{r_2} ]$, then there exists a nonnegative continuous function F on $[0,\infty )^3$ depending only on $r_0 ,r_2$ such that

$$\begin{aligned} \sup _{(r,x)\in [r_1 ,r_2]\times \mathbb {R}}|\partial _xu(r,x)|&\le F(C_0,C_1,C_2). \end{aligned}$$

Proof

From the Duhamel principle,

$$\begin{aligned} u(r,x)&=P_{r-r_0 }g(x)+\int _{r_0 }^rP_{r-w}(\kappa _1(w,\cdot )\partial _xu(w,\cdot )+\kappa _2(w,\cdot ))(x)dw, \end{aligned}$$

where for any $f\in L^\infty (\mathbb {R})$,

$$\begin{aligned} P_{a}f(x)\,{:=}\,\frac{1}{\sqrt{4\pi a}}\int _{\mathbb {R}}f(y)e^{-\frac{(x-y)^2}{4a}}dy. \end{aligned}$$

A direct computation leads to

$$\begin{aligned} \partial _xu(r,x)&=\partial _x(P_{r-r_0 }g(x))+\int _{r_0 }^r\partial _x\bigl (P_{r-w}(\kappa _1(w,\cdot )\partial _xu(w,\cdot )+\kappa _2(w,\cdot ))(x)\bigr )dw. \end{aligned}$$

Note that

$$\begin{aligned} \partial _x(P_{r-r_0 }g(x))&=\frac{-1}{2(r-r_0 )\sqrt{4\pi (r-r_0 )}}\int _{\mathbb {R}}(x-y)g(y)e^{-\frac{(x-y)^2}{4(r-r_0 )}}dy, \end{aligned}$$

from which

$$\begin{aligned} |\partial _x(P_{r-r_0 }g(x))|&\le \frac{C_0}{2(r-r_0 )\sqrt{4\pi (r-r_0 )}}\int _{\mathbb {R}}|x-y|e^{-\frac{(x-y)^2}{4(r-r_0 )}}dy=\frac{C_0}{\sqrt{\pi (r-r_0 ) }}. \end{aligned}$$

A similar computation also yields that

$$\begin{aligned} \int _{r_0 }^r\partial _x\bigl (P_{r-w}(\kappa _1(w,\cdot )\partial _xu(w,\cdot )+\kappa _2(w,\cdot ))(x)\bigr )dw&\le \int _{r_0 }^r\frac{C_1\Vert \partial _xu(w,\cdot )\Vert _\infty +C_2}{\sqrt{\pi (r-w)}} dw. \end{aligned}$$

As a result,

$$\begin{aligned} \Vert \partial _xu(r,\cdot )\Vert _\infty&\le \phi _0(r)+C_1\int _{r_0 }^r\frac{\Vert \partial _xu(w,\cdot )\Vert _\infty }{\sqrt{\pi (r-w)}} dw, \end{aligned}$$

where

$$\begin{aligned} \phi _0(r):=\frac{C_0}{\sqrt{\pi (r-r_0 )}}+\frac{2C_2\sqrt{r-r_0 }}{\sqrt{\pi }}. \end{aligned}$$

From the Gronwall inequality,

$$\begin{aligned} \Vert \partial _xu(r,\cdot )\Vert _\infty&\le \phi _0(r)+\int _{r_0 }^r\frac{C_1\phi _0(w)}{\sqrt{\pi (r-w)}}\exp \Bigl (\int _{w}^r\frac{dl}{\sqrt{\pi (r-l)}}\Bigr )dw\\&=\phi _0(r)+\int _{r_0 }^r\frac{C_1\phi _0(w)}{\sqrt{\pi (r-w)}}\exp \Bigl (2\sqrt{\frac{r-w}{\pi }}\Bigr )dw\\&\le \phi (r), \end{aligned}$$

where

$$\begin{aligned} \phi (r):=\phi _0(r)+\frac{C_1}{\pi }\exp \Bigl (2\sqrt{\frac{r_2-r_0 }{\pi }}\Bigr )\bigl (\pi C_0+4C_2\sqrt{r_2-r_0 }\bigr ), \end{aligned}$$

and the last inequality was obtained by using $\int _{r_0 }^r1/\sqrt{(r-w)(w-r_0 )}dw =\pi .$ This finishes our proof. $\square$

With the help of Lemma 10, we obtain some controls on the spacial derivatives of $\Phi _\gamma .$

Lemma 11

Let $s_1\in (0,1)$. For any $\gamma \in \mathcal {U}_d,$

$$\begin{aligned} \sup _{(s,x)\in [0,1)\times \mathbb {R}}|\partial _x\Phi _\gamma (s,x)|\le 1 \end{aligned}$$

(62)

and for $k\ge 2$

$$\begin{aligned} \sup _{(s,x)\in [0,s_1]\times \mathbb {R}}|\partial _x^k\Phi _\gamma (s,x)|\le F_k(\gamma (s_1)), \end{aligned}$$

(63)

where $F_k$ is nonnegative continuous on $[0,\infty )$ depending on $s_1$ only and independent of $\gamma .$

Proof

Assume that $\gamma \in \mathcal {U}_d.$ One can explicitly solve $\Phi _\gamma$ in the classical sense by performing the Cole-Hopf transformation. In fact, if $\gamma =\sum _{i=0}^{m}a_i1_{[q_{i},q_{i+1})}$ for some sequences

$$\begin{aligned}&0=q_0<q_1<\cdots<q_m<q_{m+1}=1,\\&0\le a_0\le a_1\le \cdots \le a_{m-1}\le a_{m}<\infty , \end{aligned}$$

then for any $1\le i\le m,$

$$\begin{aligned} \Phi _{\gamma }(s,x)&=\frac{1}{a_i}\log \mathbb {E}\exp a_i\Phi _\gamma (q_{i+1},x+z\sqrt{\xi '(q_{i+1})-\xi '(s)}) \end{aligned}$$

for all $(s,x)\in [q_i,q_{i+1})\times \mathbb {R},$ where z is a standard normal random variable. Using the initial condition $\Phi _\gamma (1,x)=|x|$ and an iteration argument, it can be easily checked that for any $k\ge 0$, $\partial _x^k\Phi \in C([0,1)\times \mathbb {R})$ and for any $k\ge 0$ and $s_0\in (0,1)$, $\partial _s\partial _x^k\Phi (s,\cdot ) \in C(\mathbb {R})$. Furthermore, it can be verified that (62) holds, and for any $s_0\in (0,1)$, there exists a constant C such that

$$\begin{aligned} \sup _{x\in \mathbb {R}}|\partial _{x}^k\Phi _\gamma (s,x)-\partial _x^k\Phi _\gamma (s',x)|\le C|s-s'|,\,\,\forall s,s'\in [0,1),\,x\in \mathbb {R},\,\,k\ge 2. \end{aligned}$$

This inequality implies that $s\mapsto \Vert \partial _x^k\Phi _\gamma (s,\cdot )\Vert _\infty$ is a continuous function for $s\in [0,1).$ Now define $\zeta (s)=(\xi '(1)-\xi '(s))/2$. Fix $0<s_1<s_0<1$. Set $r_0=\zeta (s_0)$, $r_1=\zeta (s_1)$, and $r_2=\zeta (0)$. Evidently, the function $u(r,x):=\partial _{x}\Phi _\gamma (\zeta ^{-1}(r),x)$ satisfies

$$\begin{aligned} \partial _ru(r,x)&=\partial _{xx}u(r,x)+\kappa _1(r,x)\partial _xu(r,x)+\kappa _2(r,x) \end{aligned}$$

for $(r,x)\in [r_0,r_2]\times \mathbb {R}$ with initial condition $\partial _xu(r_0,x)=\partial _x\Psi _\gamma (\zeta ^{-1}(r_0),x),$ where

$$\begin{aligned} \kappa _1(r,x)&:=\gamma (\zeta ^{-1}(r))\partial _xu(r,x),\\ \kappa _2(r,x)&:=0. \end{aligned}$$

Since $\Vert \kappa _1\Vert _\infty \le \gamma (\zeta ^{-1}(r_0))<\infty$, we can apply Lemma 10 to get that

$$\begin{aligned} \sup _{(s,x)\in [0,s_1]\times \mathbb {R}}|\partial _x\Phi _\gamma (s,x)|&=\sup _{(r,x)\in [r_1,r_2]\times \mathbb {R}}|\partial _xu(r,x)|\\&\le F(1,\gamma (\zeta ^{-1}(r_1)),0)\\&= F(1,\gamma (s_1),0), \end{aligned}$$

where F is a nonnegative continuous function on $[0,\infty )^3$ depending only on $s_1.$ Letting $F_1(y)=F(1,y,0)$ gives (63) with $k=1.$ For $k\ge 2,$ note that

$$\begin{aligned}&\partial _t\bigl (\partial _x^k\Phi _\gamma (s,x)\bigr )\\ {}&\quad =-\frac{\xi ''(s)}{2}\Bigl (\partial _{xx}\bigl (\partial _x^k\Phi _\gamma (s,x)\bigr )+2\gamma (s)\partial _x\Phi _\gamma (s,x)\bigl (\partial _x^{k+1}\Phi _\gamma (s,x)\bigr ) +K(s,x)\Bigr ), \end{aligned}$$

where K(s, x) is the sum of products of spatial derivatives of $\Phi _\gamma$ of order at most k. One may use a similar argument as the case $k=1$ together with an induction procedure to obtain the proof of (63) for all $k\ge 2$. $\square$

Following a similar definition of the weak solution for the original Parisi PDE in Jagannath and Tobasco [33], we define the weak solution of the Parisi PDE (5) as follows.

Definition 1

(Weak solution) Let $\gamma \in \mathcal {U}.$ Let $\Phi$ be a continuous function on $[0,1]\times \mathbb {R}$ with essentially bounded weak derivative $\partial _x\Phi .$ We say that $\Phi$ is a weak solution to

$$\begin{aligned} \partial _s\Phi (s,x)&=-\frac{\xi ''(s)}{2}\bigl (\partial _{xx}\Phi (s,x)+\gamma (s)\bigl (\partial _x\Phi (s,x)\bigr )^2\bigr ) \end{aligned}$$

on $[0,1)\times \mathbb {R}$ with $\Phi (1,x)=|x|$ if

$$\begin{aligned} \int _0^1\int _{\mathbb {R}}\Bigl (-\Phi \partial _s \phi +\frac{\xi ''(s)}{2}\Bigl (\Phi \partial _{xx}\phi +\gamma (s)\bigl (\partial _{x}\Phi \bigr )^2\phi \Bigr )\Bigr )dxds+\int _{\mathbb {R}}\phi (1,x)|x|dx=0 \end{aligned}$$

for all smooth $\phi$ on $(0,1]\times \mathbb {R}$ with compact support.

Proof of Proposition 2

Let us pick any $(\gamma _n)_n\subset \mathcal {U}_d$ with weak limit $\gamma .$ From the estimates in Lemma 11, one readily sees that for any $k\ge 0,$ the sequence $(\partial _x^k\Phi _{\gamma _n})_{n\ge 1}$ is equicontinuous and uniformly bounded on $[0,s_1)\times \mathbb {R}$ for all $s_1\in (0,1)$. Thus, from the Arzela–Ascoli theorem combined with a diagonal process, on any $[0,s_1]\times [-M,M],$ $(\partial _x^k\Phi _{\gamma _n})_{n\ge 1}$ converges uniformly on any compact subset of $[0,1)\times \mathbb {R}$ for any $k\ge 0$. Here, without loss of generality, we use the same sequence $(\gamma _n)_{n\ge 1}$ instead of adapting a subsequence for notational convenience. The above discussion makes the following well-defined,

$$\begin{aligned} \Phi _{\gamma }(s,x):=\lim _{n\rightarrow \infty }\Phi _{\gamma _n}(s,x) \end{aligned}$$

for $(s,x)\in [0,1]\times \mathbb {R}.$ Note that for $s\in [0,1)$, $\Phi _{\gamma }(s,x)$ is differentiable in $x\in \mathbb {R}$ and satisfies $\Vert \partial _x\Phi _\gamma (s,\cdot )\Vert _\infty <\infty$. Since $\Phi _{\gamma _n}$ satisfies

$$\begin{aligned} \partial _s\Phi _{\gamma _n}(s,x)&=-\frac{\xi ''(s)}{2}\bigl (\partial _{xx}\Phi _{\gamma _n}(s,x)+\gamma _n(s)\bigl (\partial _x\Phi _{\gamma _n}(s,x)\bigr )^2\bigr ) \end{aligned}$$

for $(s,x)\in [0,1)\times \mathbb {R}$ with boundary condition $\Phi _{\gamma _n}(1,x)=|x|,$ passing to the limit gives

$$\begin{aligned} \int _0^1\int _{\mathbb {R}}\Bigl (-\Phi _{\gamma }\partial _s \phi +\frac{\xi ''(s)}{2}\Bigl (\Phi _{\gamma }\partial _{xx}\phi +\gamma (s)\bigl (\partial _{x}\Phi _\gamma \bigr )^2\phi \Bigr )\Bigr )dxds+\int _{\mathbb {R}}\phi (1,x)|x|dx=0 \end{aligned}$$

for all smooth functions $\phi$ on $(0,1]\times \mathbb {R}$ with compact support. Therefore, $\Phi _\gamma$ exists in the weak sense and the fact that it satisfies the Lipschitz property (7) follows by applying (9) and noting that |x| has Lipschitz constant 1. Note that this Lipschitz property also implies that the definition of $\Phi _\gamma$ is independent of the choice of the sequence $(\gamma _n)_{n\ge 1}$. To see the uniqueness of $\Phi _\gamma$, it can be obtained via a fixed point argument identical to [33, Lemma 13]. These together give (i). The above discussion and Lemma 11 imply (ii). Since one can also pick this sequence $(\gamma _n)$ from $\mathcal {U}_c$ and perform a similar procedure as above, we also obtain (iii). $\square$