Fluctuations of the Free Energy of the Spherical Sherrington–Kirkpatrick Model

Abstract

We consider the fluctuations of the free energy for the 2-spin spherical Sherrington–Kirkpatrick model with no magnetic field. We show that the law of the fluctuations converges to the Gaussian distribution when the temperature is above the critical temperature, and to the GOE Tracy–Widom distribution when the temperature is below the critical temperature. The orders of the fluctuations are markedly different in these two regimes. A universality of the limit law is also proved.

Appendix

In this appendix, we evaluate various constants stated in Sect. 3. As a main tool of the evaluation, we use the Stieltjes transform of the measure $\nu$, defined by

$$\begin{aligned} m(z) = \int _{C_-}^{C_+} \frac{{\,\mathrm {d}\nu }(x)}{x-z}, \qquad z \in \mathbb {C}\setminus [C_-, C_+). \end{aligned}$$

(8.1)

Note that due to the square-root decay of $\frac{{\,\mathrm {d}\nu }(x)}{dx}$ at $x=C_+$, $m(C_+)$ is well-defined.

1.1 Wigner Matrix

We only consider real Wigner matrices. The complex case can be evaluated in the same way and we skip the details. The limiting spectral measure for real Wigner matrices is $\frac{{\,\mathrm {d}\nu }(x)}{dx} = \frac{1}{2\pi } \sqrt{4-x^2}$, $-2\le x\le 2$. Hence

$$\begin{aligned} m(z) = \frac{-z+\sqrt{z^2 -4}}{2}, \end{aligned}$$

(8.2)

and we find that

$$\begin{aligned} \beta _c = \frac{1}{2} \int _{C_-}^{C_+} \frac{{\,\mathrm {d}\nu }(x)}{C_+-x} = -\frac{1}{2} m(2)= \frac{1}{2}. \end{aligned}$$

(8.3)

We now evaluate ${\widehat{\gamma }}$ defined in (2.14). This equation is equivalent to the equation $-m_{\nu }({\widehat{\gamma }}) = 2\beta$. Solving this equation, we find that

$$\begin{aligned} {\widehat{\gamma }} = 2\beta + \frac{1}{2\beta }. \end{aligned}$$

(8.4)

To evaluate (2.15) and (2.16), we note that

$$\begin{aligned} \int _{-2}^2 \log (z-x) {\,\mathrm {d}\nu }(x) = \frac{1}{4} z \left( z - \sqrt{z^2 -4} \right) + \log \left( z + \sqrt{z^2 -4} \right) -\log 2 - \frac{1}{2}. \end{aligned}$$

(8.5)

for $z\in \mathbb {C}\setminus (-\infty , 2]$. This follows by noting that

$$\begin{aligned} -m(z)= \frac{\,\mathrm {d}}{\,\mathrm {d}z} \left[ \frac{1}{4} z \left( z - \sqrt{z^2 -4} \right) + \log \left( z + \sqrt{z^2 -4} \right) \right] \end{aligned}$$

(8.6)

and $\int _{-2}^2 \log (z-x) {\,\mathrm {d}\nu }(x)= \log z + O(z^{-1})$ as $z \rightarrow \infty$. Thus, using (8.4), we obtain

$$\begin{aligned} F(\beta )= {\left\{ \begin{array}{ll} \beta ^2 &{} \quad \text { if } \beta < 1/2 \\ 2\beta - \frac{\log (2\beta ) + 3/2}{2} &{} \quad \text { if } \beta > 1/2. \end{array}\right. } \end{aligned}$$

(8.7)

We now evaluate $\ell (\beta )$ and $\sigma ^2(\beta )$ for the case when $\beta <\beta _c$. First, since

$$\begin{aligned} \int _{C_-}^{C_+} \frac{{\,\mathrm {d}\nu }(x)}{(z-x)^2} = m^{\prime }(z) = -\frac{1}{2} + \frac{z}{2\sqrt{z^2-4}}, \qquad z\in \mathbb {C}\setminus [-2,2], \end{aligned}$$

(8.8)

 (2.18) becomes

$$\begin{aligned} \ell _1(\beta ) =\log (2\beta ) - \frac{1}{2} \log \left( m^{\prime }({\widehat{\gamma }}(\beta )) \right) =\frac{1}{2} \log (1-4\beta ^2). \end{aligned}$$

(8.9)

Second, we evaluate (see 3.3) $\tau _\ell (\varphi )= t_\ell ({\widehat{\gamma }}(\beta ))$ for $\ell =0,1,2,$ and 4, where

$$\begin{aligned} t_\ell (z)= \frac{1}{\pi } \int _{-2}^2 \log (z-x) \frac{T_\ell (x/2)}{\sqrt{4-x^2}} {\,\mathrm {d}x}, \qquad z\in \mathbb {C}\setminus (-\infty , 2), \end{aligned}$$

(8.10)

We start by evaluating

$$\begin{aligned} t_\ell ^{\prime }(z)= \frac{1}{\pi } \int _{-2}^2 \frac{T_\ell (x/2)}{(z-x)\sqrt{4-x^2}} {\,\mathrm {d}x}. \end{aligned}$$

(8.11)

Recall the recursions $T_{\ell +2}(y)=2yT_{\ell +1}(y)-T_{\ell }(y)$, $\ell \ge 0$, and the orthogonality relation $\int _{-1}^1 \frac{y^kT_\ell (y)}{\sqrt{1-y^2}} {\,\mathrm {d}y}=0$ for all integers $0\le k<\ell$, for the Chebyshev polynomials. From this we obtain the recursions

$$\begin{aligned} t_{\ell +2}^{\prime }(z)= z t_{\ell +1}^{\prime }(z)-t_{\ell }^{\prime }(z) \end{aligned}$$

(8.12)

for $\ell \ge 0$. Since $T_0(x/2)=1$ and $T_1(x/2)=x/2$, it is easy to check from residue calculus that

$$\begin{aligned} t_{0}^{\prime }(z) = \frac{1}{\sqrt{z^2-4}} , \qquad t_{1}^{\prime }(z) = \frac{z}{2\sqrt{z^2-4}} -\frac{1}{2}. \end{aligned}$$

(8.13)

Hence

$$\begin{aligned} t_{2}^{\prime }(z)= \frac{z^2-2}{2\sqrt{z^2-4}} -\frac{z}{2}, \qquad t_4^{\prime }(z)= \frac{z^4-4z^2+2}{2\sqrt{z^2-4}} -\frac{z^3}{2}+z. \end{aligned}$$

(8.14)

Taking the anti-derivatives and noting from (8.10) that $\tau _0(z)=\log z+ O(1/z)$ and $t_\ell (z)= O(1/z)$, $\ell \ge 1$, as $z\rightarrow \infty$ using the orthogonality relations of the Chebyshev polynomials, we find that

$$\begin{aligned} t_{0}(z)= \log \left( z+ \sqrt{z^2-4} \right) -\log 2, \qquad t_1(z)= \frac{1}{2} \sqrt{z^2-4}-\frac{z}{2}, \end{aligned}$$

(8.15)

and

$$\begin{aligned} t_2(z)= \frac{z}{4} \sqrt{z^2-4}-\frac{z^2}{4} + \frac{1}{2}, \qquad t_4(z)= \frac{1}{8} (z^3-2z)\sqrt{z^2-4}-\frac{z^4}{8} + \frac{z^2}{2} - \frac{1}{4}.\qquad \end{aligned}$$

(8.16)

Hence,

$$\begin{aligned} \tau _{0}(\varphi )= -\log (2\beta ), \quad \tau _{1}(\varphi )= -2\beta , \quad \tau _{2}(\varphi )= -2\beta ^2, \quad \tau _{4}(\varphi )= -4\beta ^4. \end{aligned}$$

(8.17)

Third, $V_{\textit{GOE}}(\varphi )$ (see (3.7)) is evaluated from lemma 8.1 below. Hence we find that

$$\begin{aligned} V_{\textit{GOE}}(\varphi )= \frac{1}{2\pi ^2} L({\widehat{\gamma }}(\beta ), {\widehat{\gamma }}(\beta )) = -2 \log (1-4\beta ^2) \end{aligned}$$

(8.18)

where F(z, w) is defined in (8.20). We also have (see (3.6))

$$\begin{aligned} M_{\textit{GOE}}(\varphi )= \frac{1}{2} \log (1-4\beta ^2). \end{aligned}$$

(8.19)

Therefore, combining with (8.9), (8.17), we obtain (3.12) and (3.13).

It remains to prove following Lemma.

Lemma 8.1

Set

$$\begin{aligned} L(z,w)= & {} \int _{-2}^2\int _{-2}^2 \big ( \log (z-x)\log (z-y)\big )\big ( \log (w-x)\nonumber \\&-\log (w-y)\big )Q(x,y) {\,\mathrm {d}x} {\,\mathrm {d}y}, \end{aligned}$$

(8.20)

for $z,w\in \mathbb {C}\setminus (-\infty , 2]$, where

$$\begin{aligned} Q(x,y)= \frac{4-xy}{(x-y)^2\sqrt{4-x^2}\sqrt{4-y^2}}. \end{aligned}$$

(8.21)

Then

$$\begin{aligned} L(z,w)=2\pi ^2 \log \left[ \frac{(z+R(z))(w+R(w))}{2(zw-4+R(z)R(w))} \right] , \qquad R(z)=\sqrt{z^2-4}, \end{aligned}$$

(8.22)

for $z,w\in \mathbb {C}\setminus (-\infty , 2]$, where R(z) is defined with branch cut $[-2,2]$.

Proof of Lemma 8.1

Consider the second derivative

$$\begin{aligned} L_{zw}(z,w)= & {} \int _{-2}^2 \left[ \int _{-2}^2\left( \frac{1}{z-x}-\frac{1}{z-y}\right) \left( \frac{1}{w-x}-\frac{1}{w-y}\right) \right. \nonumber \\&\times \left. \frac{4-xy}{(x-y)^2\sqrt{4-x^2}\sqrt{4-y^2}} {\,\mathrm {d}y} \right] {\,\mathrm {d}x} . \end{aligned}$$

(8.23)

Recall that $R(x)=\sqrt{x^2-4}$ is defined with the branch cut $[-2,2]$. Using this, it is easy to check that

$$\begin{aligned} L_{zw}(z,w)=-\frac{1}{4} \int _{\Sigma } \left[ \int _{\Sigma ^{\prime }} \left( \frac{1}{z-x}-\frac{1}{z-y}\right) \left( \frac{1}{w-x}-\frac{1}{w-y}\right) \frac{4-xy}{(x-y)^2R(x)R(y)} {\,\mathrm {d}y} \right] {\,\mathrm {d}x}\nonumber \\ \end{aligned}$$

(8.24)

where $\Sigma$ and $\Sigma ^{\prime }$ are simple closed contours, oriented positively, that contain the interval $[-2,2]$ in its interior and the points z and w in its exterior, and $\Sigma ^{\prime }$ lies in the interior of $\Sigma$. By residue calculus, we obtain

$$\begin{aligned} L_{zw}(z,w)=\frac{-2\pi ^2}{(z-w)^2} \left( 1+ \frac{4-zw}{R(z)R(w)} \right) . \end{aligned}$$

(8.25)

Integrating with respect to z,

$$\begin{aligned} L_{w}(z,w)=\frac{-2\pi \big ( R(z)-R(w)\big )}{(z-w)R(w)} +C_1(w) \end{aligned}$$

(8.26)

for some function $C_1(w)$ of w. The function $C_1(w)$ is determined by noting that, from (8.26),

$$\begin{aligned} L_{w}(z,w)=\frac{-2\pi }{R(w)} +C_1(w) + O(z^{-1}) \end{aligned}$$

(8.27)

as $z\rightarrow \infty$ with fixed w, while from (8.20)

$$\begin{aligned} L_{w}(z,w)= & {} L(z,w)\nonumber \\= & {} \int _{-2}^2\int _{-2}^2 \left( \log (z-x)-\log (z-y)\right) \left( \frac{1}{w-x}-\frac{1}{w-y}\right) Q(x,y) {\,\mathrm {d}x} {\,\mathrm {d}y},\nonumber \\ \end{aligned}$$

(8.28)

is $O(z^{-1})$. Hence $C_1(w)= \frac{2\pi ^2}{R(w)}$. Integrating with respect to w, we find that

$$\begin{aligned} L(z,w)=2\pi ^2 \log \left[ \frac{(w+R(w))}{2(zw-4+R(z)R(w))} \right] +C_2(z) \end{aligned}$$

(8.29)

for some function $C_2(z)$ of z. By considering the asymptotics as $w\rightarrow \infty$, we find that $C_2(z)=2\pi ^2 \log (z+R(z))$, and this completes the proof of Lemma. $\square$

1.2 Sample Covariance Matrix

We again only consider the real case here. The complex case is similar. The Stieltjes transform of the Marchenko-Pastur distribution ${\,\mathrm {d}\nu }$ in  (3.26) is given by

$$\begin{aligned} m_{\nu }(z) = \frac{-z+\sqrt{C_+C_-}+\sqrt{(z-C_+)(z-C_-)}}{2z}, \qquad z\in \mathbb {C}\setminus [C_-, C_+]. \end{aligned}$$

(8.30)

Here $C_+=(\sqrt{d}+1)^2$ and $C_-=(\sqrt{d}-1)^2$. Hence, for real sample covariance matrices, we have

$$\begin{aligned} \beta _c = \frac{1}{2} \int _{C_-}^{C_+} \frac{{\,\mathrm {d}\nu }(x)}{C_+ -x} = -\frac{m_{\nu }(C_+)}{2} = \frac{1}{2 (\sqrt{d} +1)}. \end{aligned}$$

(8.31)

In the high temperature case, $\beta < \beta _c$, we get from the definition of ${\widehat{\gamma }}$ that

$$\begin{aligned} -m_{\nu }({\widehat{\gamma }}) = 2\beta . \end{aligned}$$

(8.32)

Solving the equation, we find that

$$\begin{aligned} {\widehat{\gamma }} = \frac{d}{1-2\beta }+ \frac{1}{2\beta }. \end{aligned}$$

(8.33)

We note that ${\widehat{\gamma }}$ is a decreasing function in $\beta$ and satisfies ${\widehat{\gamma }}>C+$ for $0<\beta < \beta _c$. In order to evaluate $L(\beta )$ in (2.15), note that

$$\begin{aligned} \frac{\,\mathrm {d}}{\,\mathrm {d}\beta } \int _{-2}^2 \log ({\widehat{\gamma }}-\lambda ) {\,\mathrm {d}\nu }(\lambda )= & {} \left( \frac{\,\mathrm {d}{\widehat{\gamma }}}{\,\mathrm {d}\beta } \right) \frac{\,\mathrm {d}}{\,\mathrm {d}{\widehat{\gamma }}} \int _{-2}^2 \log ({\widehat{\gamma }}-\lambda ) {\,\mathrm {d}\nu }(\lambda ) \nonumber \\= & {} -m_{\nu }({\widehat{\gamma }}) \left( \frac{\,\mathrm {d}{\widehat{\gamma }}}{\,\mathrm {d}\beta } \right) = 2\beta \left( \frac{2d}{(1-2\beta )^2} - \frac{1}{2\beta ^2} \right) \end{aligned}$$

(8.34)

using (8.32) and (8.33). Hence, by taking the anti-derivative,

$$\begin{aligned} \int _{-2}^2 \log ({\widehat{\gamma }}(\beta )-\lambda ) {\,\mathrm {d}\nu }(\lambda )= & {} \frac{2\beta }{1-2\beta }d + d \log (1-2\beta ) - \log (2\beta ), \end{aligned}$$

(8.35)

where the constant of integration is determined by the asymptotics that the left hand side behaves as

$$\begin{aligned} \log {\widehat{\gamma }} + O({\widehat{\gamma }}^{-1}) = -\log (2\beta ) + O(\beta ) \end{aligned}$$

as ${\widehat{\gamma }} \rightarrow \infty$, or $\beta \rightarrow 0$. Hence

$$\begin{aligned} L(\beta )= - \frac{d}{2\beta } \log (1-2\beta ) \end{aligned}$$

for $\beta <\beta _c$.

The limit of the free energy per particle $L(\beta )$ for $\beta >\beta _c$ is obtained easily using (8.35) and noting that $C_+={\widehat{\gamma }}(\beta _c)$.

We now evaluate $\ell$ and $\sigma ^2$ for $\beta <\beta _c$. From (2.18), we find that

$$\begin{aligned} \ell _1 = \log (2\beta ) - \frac{1}{2} \log (m^{\prime }_{\nu }({\widehat{\gamma }})) = \frac{1}{2} \log (1-4B^2) \end{aligned}$$

(8.36)

where we set

$$\begin{aligned} B= \frac{\beta \sqrt{d}}{1-2\beta }. \end{aligned}$$

(8.37)

On the other hand, in order to evaluate $M(\varphi )$ and $V(\varphi )$ for $\varphi (x)= \log ({\widehat{\gamma }} -x)$, we observe that (3.27) implies that

$$\begin{aligned} \Phi (x)= \log (\sqrt{d}) +\psi (x), \qquad \psi (x)= \log \left( 2B+\frac{1}{2B}-x \right) \end{aligned}$$

(8.38)

where B is given above. Note that $\psi (x)$ is the same function as $\varphi (x)=\log ({\widehat{\gamma }} -x)$ for Wigner matrices with the change that $\beta$ is replaced by B. Moreover, we have $\tau _\ell (c+f)=\tau _{\ell }(f)$ for $\ell >0$ and $\tau _0(c+f)=c+\tau _0(f)$ for a constant c and function f, from the definition of $\tau _\ell$ and the orthogonality of Chebyshev polynomials. From this and the results obtained for Wigner matrices, we can easily find $M(\varphi )$ and $V(\varphi )$ for sample covariance matrices.