Free Energy in Spin Glass Models with Conventional Order

Abstract

Recently, Baldwin and Swingle (J Stat Phys 190(7):125, 2023) considered spin glass models with additional conventional order parameters characterizing single-replica properties. These parameters are distinct from the standard order parameter, the overlap, used to measure correlations between replicas. A “min-max” formula for the free energy was prescribed in Baldwin and Swingle (2023). We rigorously verify this prescription in the setting of vector spin glass models featuring additional deterministic spin interactions. Notably, our results can be viewed as a generalization of the Parisi formula for vector spin glass models in Panchenko (Ann Probab 46(2):865–896, 2018), where the order parameter for self-overlap is already present.

Appendix A: Convergence of \(m_N\)

As mentioned in Remark 1.4, we can show that $m_N$ in (1.4) always converges under the Gibbs measure associated with $F^{\textsf{soc}}_N$ in (1.2). When $h=\textbf{s}$, we have that $m_N = \frac{\sigma \sigma ^\intercal }{N}$ is the self-overlap and such a result has been proved in [19, Theorem 1.1 (1) and (2)]. A straightforward modification gives the desired result below.

Proposition A.1

Under conditions (H0)–(H4), if h is bounded and measurable, then $m_N$ in (1.4) satisfies

$$\begin{aligned} \lim _{N\rightarrow \infty } \mathbb {E}\left\langle \left|m_N - \nabla \mathscr {P}^h(0)\right| \right\rangle = 0 \end{aligned}$$

where $\mathscr {P}^h$ is defined in (2.4) and $\left\langle \cdot \right\rangle$ is the Gibbs measure associated with $F^{\textsf{soc}}_N$ in (1.2).

For completeness, we present the proof, which follows from the straightforward combination of the next two lemmas. We assume (H0)–(H4) henceforth.

Lemma A.2

Let $\left\langle \cdot \right\rangle$ be associated with $F^{\textsf{soc}}_N$. If h is bounded and measurable, then

$$\begin{aligned} \lim _{N\rightarrow \infty }\mathbb {E}\left\langle m_N\right\rangle = \nabla \mathscr {P}^h(0). \end{aligned}$$

Proof

Recall $\mathcal {F}_N$ defined in (2.6). Let $y\in \mathbb {R}^d$ and $r>0$. The convexity of $\mathcal {F}_N$ by Lemma 2.3 implies

$$\begin{aligned} \frac{\mathcal {F}_N(0,0)-\mathcal {F}_N(0,-ry)}{r} \leqslant y\cdot \nabla \mathcal {F}_N(0,0) \leqslant \frac{\mathcal {F}_N(0,ry)-\mathcal {F}_N(0,0)}{r}. \end{aligned}$$

Sending $N\rightarrow \infty$ and then $r\rightarrow 0$, and using Lemma 2.1 and the differentiability of $\mathscr {P}^h$ in Lemma 2.2, we get

$$\begin{aligned} \lim _{N\rightarrow \infty } y\cdot \nabla \mathcal {F}_N(0,0) = y\cdot \nabla \mathscr {P}^h(0). \end{aligned}$$

(A.1)

Varying y, we get $\lim _{N\rightarrow \infty } \nabla \mathcal {F}_N(0,0) = \nabla \mathscr {P}^h(0)$ in $\mathbb {R}^d$. Recall from (2.11) that $\nabla \mathcal {F}_N(0,0) = \mathbb {E}\left\langle m_N\right\rangle$ where $\left\langle \cdot \right\rangle$ is associated with $\mathcal {F}_N(0,0)$. The desired result follows from the observation that $\mathcal {F}_N(0,0)=F^{\textsf{soc}}_N$.

Lemma A.3

Let $\left\langle \cdot \right\rangle$ be associated with $F^{\textsf{soc}}_N$. If h is bounded and measurable, then

$$\begin{aligned} \lim _{N\rightarrow \infty }\mathbb {E}\left\langle \left|m_N - \mathbb {E}\left\langle m_N\right\rangle \right| \right\rangle . \end{aligned}$$

Proof

For $x\in \mathbb {R}^d$, we write $\mathcal {F}_N(x)=\mathcal {F}_N(0,x)$ (in (2.6)) for brevity. Let $\left\langle \cdot \right\rangle _x$ be the Gibbs measure associated with $\mathcal {F}_N(x)$. Since $\mathcal {F}_N(0) = F^{\textsf{soc}}_N$, we have $\left\langle \cdot \right\rangle =\left\langle \cdot \right\rangle _0$. Fix any $y\in \mathbb {R}^d$ and set $g(\sigma ) = Ny\cdot m_N = y\cdot \sum _{i=1}^Nh(\sigma _i)$. It suffices to show

$$\begin{aligned} \lim _{N\rightarrow \infty } \frac{1}{N}\mathbb {E}\left\langle \left|g(\sigma ) - \mathbb {E}\left\langle g(\sigma )\right\rangle _0\right| \right\rangle _0 =0. \end{aligned}$$

(A.2)

Step 1. We show

$$\begin{aligned} \lim _{N\rightarrow \infty } \frac{1}{N}\mathbb {E}\left\langle \left|g(\sigma ) - \left\langle g(\sigma )\right\rangle _0\right| \right\rangle _0 =0. \end{aligned}$$

(A.3)

We denote by $(\sigma ^l)_{l\in \mathbb {N}}$ independent copies of $\sigma$ under the relevant Gibbs measure. Let $r>0$. Integrating by parts, we get

$$\begin{aligned}&r\mathbb {E}\left\langle \left|g\left(\sigma ^1\right) -g\left(\sigma ^2\right)\right|\right\rangle _0\\&= \int _0^r \mathbb {E}\left\langle \left|g\left(\sigma ^1\right) -g\left(\sigma ^2\right)\right|\right\rangle _{sy} \textrm{d}s -\int _0^r \int _0^t \frac{\textrm{d}}{\textrm{d}s} \mathbb {E}\left\langle \left|g\left(\sigma ^1\right) -g\left(\sigma ^2\right)\right|\right\rangle _{sy}\textrm{d}s \textrm{d}t. \end{aligned}$$

The integrand in the last term can be estimated as follows

$$\begin{aligned} \frac{\textrm{d}}{\textrm{d}s} \mathbb {E}\left\langle \left|g\left(\sigma ^1\right) -g\left(\sigma ^2\right)\right|\right\rangle _{sy} = \mathbb {E}\left\langle \left|g\left(\sigma ^1\right) -g\left(\sigma ^2\right)\right|\left(g\left(\sigma ^1\right)+g\left(\sigma ^2\right)-2g\left(\sigma ^3\right)\right)\right\rangle _{sy} \\ \geqslant -2 \mathbb {E}\left\langle \left|g\left(\sigma ^1\right) -g\left(\sigma ^2\right)\right|^2\right\rangle _{sy}\geqslant -8 \mathbb {E}\left\langle \left|g\left(\sigma \right) -\left\langle g\left(\sigma \right)\right\rangle _{sy}\right|^2\right\rangle _{sy}. \end{aligned}$$

Inserting this into the previous display, we obtain

$$\begin{aligned}&\mathbb {E}\left\langle \left|g\left(\sigma ^1\right) -g\left(\sigma ^2\right)\right|\right\rangle _0\\&\leqslant \frac{1}{r}\int _0^r\mathbb {E}\left\langle \left|g\left(\sigma ^1\right) -g\left(\sigma ^2\right)\right|\right\rangle _{sy} \textrm{d}s + \frac{8}{r}\int _0^r \int _0^t \mathbb {E}\left\langle \left|g\left(\sigma \right) -\left\langle g\left(\sigma \right)\right\rangle _{sy}\right|^2\right\rangle _{sy}\textrm{d}s \textrm{d}t \\&\leqslant \frac{2}{r}\int _0^r\mathbb {E}\left\langle \left|g\left(\sigma \right) -\left\langle g\left(\sigma \right)\right\rangle _{sy}\right|\right\rangle _{sy} \textrm{d}s + 8\int _0^r \mathbb {E}\left\langle \left|g\left(\sigma \right) -\left\langle g\left(\sigma \right)\right\rangle _{sy}\right|^2\right\rangle _{sy}\textrm{d}s . \end{aligned}$$

Setting $\varepsilon _N = \frac{1}{N} \int _0^r \mathbb {E}\left\langle \left|g\left(\sigma \right) -\left\langle g\left(\sigma \right)\right\rangle _{sy}\right|^2\right\rangle _{sy}\textrm{d}s$, we can rewrite the above estimate as

$$\begin{aligned} \frac{1}{N}\mathbb {E}\left\langle \left|g\left(\sigma ^1\right) -g\left(\sigma ^2\right)\right|\right\rangle _0 \leqslant 2 \sqrt{\frac{\varepsilon _N}{rN}}+ 8 \varepsilon _N. \end{aligned}$$

Using (2.8), we have

$$\begin{aligned} \varepsilon _N&= \int _0^r \frac{\textrm{d}^2}{\textrm{d}s^2}\mathcal {F}_N(sy) \textrm{d}s = y\cdot \nabla \mathcal {F}_N(ry) - y \cdot \nabla \mathcal {F}_N(0)\\&\leqslant \frac{\mathcal {F}_N((r+t)y) - \mathcal {F}_N(ry)}{t} - \frac{\mathcal {F}_N(0) - \mathcal {F}_N(-t y)}{t} \end{aligned}$$

for any $t>0$, where the last inequality follows from the convexity of $\mathcal {F}_N$ given by Lemma 2.3. Combining the above two displays, using Lemma 2.1, and noticing $\sup _N\varepsilon _N<\infty$ (due to (2.5)), we obtain

$$\begin{aligned} \limsup _{N\rightarrow \infty } \frac{1}{8N}\mathbb {E}\left\langle \left|g\left(\sigma ^1\right) -g\left(\sigma ^2\right)\right|\right\rangle _0 \leqslant \frac{\mathscr {P}^h((r+t)y) - \mathscr {P}^h(ry)}{t} - \frac{\mathscr {P}^h(0) - \mathscr {P}^h(-t y)}{t}. \end{aligned}$$

We first send $r\rightarrow 0$ and then $t\rightarrow 0$. Since $\mathscr {P}^h$ is differentiable by Lemma 2.2, the right-hand side vanishes, which yields (A.3).

Step 2. We show

$$\begin{aligned} \lim _{N\rightarrow \infty } \frac{1}{N}\mathbb {E}\left| \left\langle g(\sigma )\right\rangle _0 - \mathbb {E}\left\langle g(\sigma )\right\rangle _0\right| =0. \end{aligned}$$

(A.4)

Recall ${\widetilde{\mathcal {F}}}_N$ below (2.6) and we write ${\widetilde{\mathcal {F}}}_N(\cdot )= {\widetilde{\mathcal {F}}}_N(0,\cdot )$ for brevity. We can use (2.11) to rewrite

$$\begin{aligned} \frac{1}{N}\mathbb {E}\left| \left\langle g(\sigma )\right\rangle _0 - \mathbb {E}\left\langle g(\sigma )\right\rangle _0\right| = \mathbb {E}\left| y\cdot \nabla {\widetilde{\mathcal {F}}}_N(0) - y\cdot \nabla \mathcal {F}_N(0)\right|. \end{aligned}$$

(A.5)

For $r\in (0,1]$, we define

$$\begin{aligned} \delta _N(r) = \left|{\widetilde{\mathcal {F}}}_N(-ry)-\mathcal {F}_N(-ry)\right| + \left|{\widetilde{\mathcal {F}}}_N(0)-\mathcal {F}_N(0)\right| + \left|{\widetilde{\mathcal {F}}}_N(ry)-\mathcal {F}_N(ry)\right|. \end{aligned}$$

In view of (2.12), ${\widetilde{\mathcal {F}}}_N$ is convex, which implies that $y\cdot \nabla {\widetilde{\mathcal {F}}}_N(0) - y\cdot \nabla \mathcal {F}_N(0)$ is bounded from above by

$$\begin{aligned} \frac{{\widetilde{\mathcal {F}}}_N(ry) - {\widetilde{\mathcal {F}}}_N(0)}{r} - y\cdot \nabla \mathcal {F}_N(0) \leqslant \frac{\mathcal {F}_N(ry)-\mathcal {F}_N(0)}{r}-y\cdot \nabla \mathcal {F}_N(0)+ \frac{\delta _N(r)}{r} \end{aligned}$$

and from below by

$$\begin{aligned} \frac{{\widetilde{\mathcal {F}}}_N(0) - {\widetilde{\mathcal {F}}}_N(-ry)}{r} - y\cdot \nabla \mathcal {F}_N(0) \geqslant \frac{\mathcal {F}_N(0)-\mathcal {F}_N(-ry)}{r}-y\cdot \nabla \mathcal {F}_N(0)- \frac{\delta _N(r)}{r}. \end{aligned}$$

By the standard concentration result (e.g. see [40, Theorem 1.2]), there is a constant $C>0$ such that $\sup _{r\in (0,1]}\mathbb {E}\delta _N(r)\leqslant CN^{-\frac{1}{2}}$. This along with Lemma 2.1 and (A.1) gives

$$\begin{aligned}&\limsup _{N\rightarrow \infty } \mathbb {E}\left|y\cdot \nabla {\widetilde{\mathcal {F}}}_N(0) - y\cdot \nabla \mathcal {F}_N(0)\right| \\&\leqslant \left|\frac{\mathscr {P}^h(ry)-\mathscr {P}^h(0)}{r}-y\cdot \nabla \mathscr {P}^h(0)\right| + \left|\frac{\mathscr {P}^h(0)-\mathscr {P}^h(-ry)}{r}-y\cdot \nabla \mathscr {P}^h(0)\right|. \end{aligned}$$

Sending $r\rightarrow \infty$, using the differentiability of $\mathscr {P}^h$, and inserting this to (A.5), we get (A.4).

In conclusion, (A.2) follows from (A.3) and (A.4) and thus the proof is complete.