From 9f97debabab68e4fcffe5c2c9ec485cc58502b1c Mon Sep 17 00:00:00 2001 From: Jaron Kent-Dobias Date: Tue, 25 Jun 2024 16:02:23 +0200 Subject: Some consistency fixes, including a lot in the GOE example. --- marginal.tex | 191 +++++++++++++++++++++++++++++++++-------------------------- 1 file changed, 108 insertions(+), 83 deletions(-) (limited to 'marginal.tex') diff --git a/marginal.tex b/marginal.tex index 26a88fe..45d65dc 100644 --- a/marginal.tex +++ b/marginal.tex @@ -202,12 +202,14 @@ as the emergence of an imaginary part in the function. As an example, we compute \begin{equation} \label{eq:large.dev} - e^{NG_\lambda^*(\mu)} - =P_{\lambda_\mathrm{min}(B+\mu I)=\lambda^*} - =\overline{\delta\big(N\lambda^*-N\lambda_\mathrm{min}(B+\mu I)\big)} + \begin{aligned} + e^{NG_{\lambda^*}(\mu)} + &=P\big(\lambda_\mathrm{min}(B+\mu I)=\lambda^*\big) \\ + &=\overline{\delta\big(N\lambda^*-N\lambda_\mathrm{min}(B+\mu I)\big)} + \end{aligned} \end{equation} where the overline is the average over $B$, and we have defined the large -deviation function $G_\sigma(\mu)$. +deviation function $G_{\lambda^*}(\mu)$. Using the representation of $\lambda_\mathrm{min}$ defined in \eqref{eq:λmin}, we have \begin{widetext} \begin{equation} @@ -225,10 +227,10 @@ representation, we have =\overline{\lim_{\beta\to\infty}\lim_{m\to0}\int d\hat\lambda\prod_{\alpha=1}^m\left[d\mathbf s^\alpha\,\delta(N-\|\mathbf s^\alpha\|^2)\right] \exp\left\{-\beta\sum_{\alpha=1}^m(\mathbf s^\alpha)^T(B+\mu I)\mathbf s^\alpha+\hat\lambda\left[N\lambda^*-(\mathbf s^1)^T(B+\mu I)\mathbf s^1\right]\right\}} \end{equation} -having introduced the parameter $\hat\lambda$ in the Fourier representation of -the $\delta$-function. The whole expression, so transformed, is a simple +having introduced the auxiliary parameter $\hat\lambda$ in the Fourier representation of +the $\delta$-function. The whole expression, so transformed, is an exponential integral linear in the matrix $B$. Taking the average over $B$, we -have +find \begin{equation} \begin{aligned} &e^{NG_{\lambda^*}(\mu)} @@ -241,27 +243,27 @@ have \end{equation} \end{widetext} We make the Hubbard--Stratonovich transformation to the matrix field -$Q_{ab}=\frac1N\mathbf s_a^T\mathbf s_b$. This gives +$Q^{\alpha\beta}=\frac1N\mathbf s^\alpha\cdot\mathbf s^\beta$. This gives \begin{equation} e^{NG_{\lambda^*}(\mu)} =\lim_{\beta\to\infty}\lim_{m\to0}\int d\hat\lambda\,dQ\, - e^{N\mathcal U_\mathrm{GOE}(\hat\lambda,Q\mid\mu,\lambda^*,\beta)} + e^{N\mathcal U_\mathrm{GOE}(\hat\lambda,Q\mid\beta,\lambda^*,\mu)} \end{equation} where the effective action is given by \begin{equation} \begin{aligned} - &\mathcal U_\textrm{GOE}(\hat\lambda, Q\mid\lambda^*,\mu,\beta) + &\mathcal U_\textrm{GOE}(\hat\lambda, Q\mid\beta,\lambda^*,\mu) =\hat\lambda(\lambda^*-\mu)-m\beta\mu \\ - &+\sigma^2\left[\beta^2\sum_{ab}^mQ_{ab}^2 - +2\beta\hat\lambda\sum_a^mQ_{1a}^2 + &+\sigma^2\left[\beta^2\sum_{\alpha\gamma}^m(Q^{\alpha\gamma})^2 + +2\beta\hat\lambda\sum_\alpha^m(Q^{1\alpha})^2 +\hat\lambda^2 \right]+\frac12\log\det Q \end{aligned} \end{equation} -where $Q_{aa}=1$ because of the spherical constraint. We can evaluate this +and $Q^{\alpha\alpha}=1$ because of the spherical constraint. We can evaluate this integral using the saddle point method. We make a replica symmetric ansatz for $Q$, because this is a 2-spin model, but with the first row singled out because -of its unique coupling with $\hat\lambda$. This gives +of its unique coupling with $\hat\lambda$. The resulting matrix has the form \begin{equation} \label{eq:Q.structure} Q=\begin{bmatrix} 1&\tilde q_0&\tilde q_0&\cdots&\tilde q_0\\ @@ -271,16 +273,17 @@ of its unique coupling with $\hat\lambda$. This gives \tilde q_0&q_0&q_0&\cdots&1 \end{bmatrix} \end{equation} -with $\sum_{ab}Q_{ab}^2=m+2(m-1)\tilde q_0^2+(m-1)(m-2)q_0^2$, $\sum_aQ_{1a}^2=1+(m-1)\tilde q_0^2$, +The relevant expressions in the effective action produce $\sum_{\alpha\beta}(Q^{\alpha\beta})^2=m+2(m-1)\tilde q_0^2+(m-1)(m-2)q_0^2$, $\sum_\alpha(Q^{1\alpha})^2=1+(m-1)\tilde q_0^2$, and \begin{equation} \log\det Q=(m-2)\log(1-q_0)+\log(1+(m-2)q_0-(m-1)\tilde q_0^2) \end{equation} -Inserting these expressions and taking the limit of $m$ to zero, we find +Inserting these expressions into the effective action and taking the limit of +$m$ to zero, we arrive at \begin{equation} e^{NG_{\lambda^*}(\mu)} =\lim_{\beta\to\infty}\int d\hat\lambda\,dq_0\,d\tilde q_0\, - e^{N\mathcal U_\textrm{GOE}(\hat\lambda,q_0,\tilde q_0\mid\mu,\lambda^*,\beta)} + e^{N\mathcal U_\textrm{GOE}(\hat\lambda,q_0,\tilde q_0\mid\beta,\lambda^*,\mu)} \end{equation} with the effective action \begin{equation} @@ -305,32 +308,37 @@ However, taking the limit with $y\neq\tilde y$ results in an expression for the action that diverges with $\beta$. To cure this, we must take $\tilde y=y$. The result is \begin{equation} \begin{aligned} - \mathcal U_\textrm{GOE}(\hat\lambda,y,\Delta z\mid\mu,\lambda^*,\infty) - &=\hat\lambda(\lambda^*-\mu) - +\sigma^2\big[ - \hat\lambda^2-4(y+\Delta z) - \big] \\ - &\qquad+\frac12\log\left(1+\frac{2\Delta z}{y^2}\right) + &\mathcal U_\textrm{GOE}(\hat\lambda,y,\Delta z\mid\infty,\lambda^*,\mu) + =\hat\lambda(\lambda^*-\mu) \\ + &\qquad+\sigma^2\big[ + \hat\lambda^2+4(y+\Delta z) + \big] + +\frac12\log\left(1-\frac{2\Delta z}{y^2}\right) \end{aligned} \end{equation} Extremizing this action over the new parameters $y$, $\Delta z$, and $\hat\lambda$, we have \begin{align} - \hat\lambda=-\frac1\sigma\sqrt{\frac{(\mu+\lambda^*)^2}{(2\sigma)^2}-1} + \hat\lambda&=\frac1\sigma\sqrt{\left(\frac{\mu-\lambda^*}{2\sigma}\right)^2-1} \\ - y=\frac1{2\sigma}\left(\frac{\mu+\lambda^*}{2\sigma}-\sqrt{\frac{(\mu+\lambda^*)^2}{(2\sigma)^2}-1}\right) - &\\ - \Delta z=\frac1{4\sigma^2}\left(1-\frac{\mu+\lambda^*}{2\sigma}\left(\frac{\mu+\lambda^*}{2\sigma}-\sqrt{\frac{(\mu+\lambda^*)^2}{(2\sigma)^2}-1}\right)\right) + y&=\frac1{2\sigma}\left[ + \frac{\mu-\lambda^*}{2\sigma}+\sqrt{\left(\frac{\mu-\lambda^*}{2\sigma}\right)^2-1} + \right]^{-1} + \\ + \Delta z&=\frac1{4\sigma^2}\left[ + \left(\frac{\mu-\lambda^*}{2\sigma}\right)^2-1 + -\frac{\mu-\lambda^*}{2\sigma}\sqrt{\left(\frac{\mu-\lambda^*}{2\sigma}\right)^2-1} + \right] \end{align} Inserting this solution into $\mathcal S_\infty$ we find \begin{equation} \label{eq:goe.large.dev} \begin{aligned} &G_{\lambda^*}(\mu) - =\mathop{\textrm{extremum}}_{y,\Delta z,\hat\lambda} - \mathcal U_\mathrm{GOE}(y,\Delta z,\hat\lambda\mid\mu,\lambda^*,\infty) \\ - &=-\tfrac{\mu+\lambda^*}{2\sigma}\sqrt{\Big(\tfrac{\mu+\lambda^*}{2\sigma}\Big)^2-1} - +\log\left( - \tfrac{\mu+\lambda^*}{2\sigma}+\sqrt{\Big(\tfrac{\mu+\lambda^*}{2\sigma}\Big)^2-1} - \right) + =\mathop{\textrm{extremum}}_{\hat\lambda,y,\Delta z} + \mathcal U_\mathrm{GOE}(\hat\lambda,y,\Delta z\mid\infty,\lambda^*,\mu) \\ + &=-\frac{\mu-\lambda^*}{2\sigma}\sqrt{\left(\frac{\mu-\lambda^*}{2\sigma}\right)^2-1} \\ + &\hspace{5em}-\log\left[ + \frac{\mu-\lambda^*}{2\sigma}-\sqrt{\left(\frac{\mu-\lambda^*}{2\sigma}\right)^2-1} + \right] \end{aligned} \end{equation} This function is plotted in Fig.~\ref{fig:large.dev} for $\lambda^*=0$. For $\mu<2\sigma$ $G_{0}(\mu)$ has an @@ -1376,7 +1384,7 @@ taking the zero-temperature limit, we find for $\alpha=\frac32$ and $f(q)=q^2+q^3$. The ground state energy $E_\mathrm{gs}$ and the threshold energy $E_\mathrm{th}$ are marked on the plot. - } + } \label{fig:ls.complexity} \end{figure} \section{Conclusions} @@ -1572,24 +1580,28 @@ fixing the trace of the Hessian, this gives \end{align} so that the differential form of the symmetry is \begin{equation} - \mathcal D=\bar{\pmb\eta}\frac\partial{\partial\mathbf x} - -i\hat\beta\bar{\pmb\eta}\frac\partial{\partial\hat{\mathbf x}} - -i\hat{\mathbf x}\frac\partial{\partial\pmb\eta} + \mathcal D=\bar{\pmb\eta}\cdot\frac\partial{\partial\mathbf x} + -i\hat\beta\bar{\pmb\eta}\cdot\frac\partial{\partial\hat{\mathbf x}} + -i\hat{\mathbf x}\cdot\frac\partial{\partial\pmb\eta} \end{equation} The Ward identities associated with this symmetry give rise to relationships among the order parameters. These identities are \begin{align} - 0=\frac1N\mathcal D\langle\mathbf x_a^T\pmb\eta_b\rangle - =\frac1N\left[ - \langle\bar{\pmb\eta}_a^T\pmb\eta_b\rangle- - i\langle\mathbf x_a^T\hat{\mathbf x}_b\rangle - \right] - =G_{ab}+R_{ab} \\ - 0=\frac iN\mathcal D\langle\hat{\mathbf x}_a^T\pmb\eta_b\rangle - =\frac1N\left[ - \hat\beta\langle\bar{\pmb\eta}_a^T\pmb\eta_b\rangle - +\langle\hat{\mathbf x}_a^T\hat{\mathbf x}_b\rangle - \right] - =\hat\beta G_{ab}+D_{ab} + \begin{aligned} + 0&=\frac1N\mathcal D\langle\mathbf x_a\cdot\pmb\eta_b\rangle + =\frac1N\left[ + \langle\bar{\pmb\eta}_a\cdot\pmb\eta_b\rangle- + i\langle\mathbf x_a\cdot\hat{\mathbf x}_b\rangle + \right] \\ + &=G_{ab}+R_{ab} + \end{aligned} \\ + \begin{aligned} + 0&=\frac iN\mathcal D\langle\hat{\mathbf x}_a\cdot\pmb\eta_b\rangle + =\frac1N\left[ + \hat\beta\langle\bar{\pmb\eta}_a\cdot\pmb\eta_b\rangle + +\langle\hat{\mathbf x}_a\cdot\hat{\mathbf x}_b\rangle + \right] \\ + &=\hat\beta G_{ab}+D_{ab} + \end{aligned} \end{align} These identities establish $G_{ab}=-R_{ab}$ and $D_{ab}=\hat\beta R_{ab}$, allowing elimination of the matrices $G$ and $D$ in favor of $R$. Fixing the @@ -1600,74 +1612,85 @@ trace to $\mu$ explicitly breaks this symmetry, and the simplification is lost. In this appendix we derive an expression for the asymptotic spectral density in the two-sphere multispherical spin glass that we describe in Section -\ref{sec:multispherical}. \cite{Livan_2018_Introduction} +\ref{sec:multispherical}. We use a typical approach of employing replicas to +compute the resolvent \cite{Livan_2018_Introduction}. The resolvent for the +Hessian of the multispherical model is given by an integral over $\mathbf +y=[\mathbf y^{(1)},\mathbf y^{(2)}]\in\mathbb R^{2N}$ as +\begin{widetext} \begin{equation} \begin{aligned} - &G(\lambda) - =\lim_{n\to0}\int\|\mathbf y_1\|^2\,\prod_{a=1}^nd\mathbf y_a\, + G(\lambda) + &=\lim_{n\to0}\int\|\mathbf y_1\|^2\,\prod_{a=1}^nd\mathbf y_a\, \exp\left\{ - -\frac12\mathbf y_a^T(\operatorname{Hess}H(\mathbf x,\pmb\omega)+\lambda I)\mathbf y_a + -\frac12\mathbf y_a^T(\operatorname{Hess}H(\mathbf x,\pmb\omega)-\lambda I)\mathbf y_a \right\} \\ & - =\lim_{n\to0}\int\big(\|\mathbf y_1^{(1)}\|^2+\|\mathbf y_1^{(2)}\|^2\big)\,\prod_{a=1}^nd\mathbf y_a\, \\ - &\times\exp\left\{ + =\lim_{n\to0}\int\big(\|\mathbf y_1^{(1)}\|^2+\|\mathbf y_1^{(2)}\|^2\big)\,\prod_{a=1}^nd\mathbf y_a\, + \exp\left\{ -\frac12\begin{bmatrix}\mathbf y_a^{(1)}\\\mathbf y_a^{(2)}\end{bmatrix}^T \left( \begin{bmatrix} \operatorname{Hess}H_1(\mathbf x^{(1)},\omega_1) & -\epsilon \\ -\epsilon & \operatorname{Hess}H_2(\mathbf x^{(2)},\omega_2) \end{bmatrix} - +\lambda I + -\lambda I \right)\begin{bmatrix}\mathbf y_a^{(1)}\\\mathbf y_a^{(2)}\end{bmatrix} - \right\} \\ + \right\} \end{aligned} \end{equation} -If $Y_{ab}^{(ik)}=\frac1N\mathbf y_a^{(i)}\cdot\mathbf y_b^{(j)}$ is the matrix -of overlaps of the $\mathbf y$, then a short and standard calculation yields +If $Y_{ab}^{(ij)}=\frac1N\mathbf y_a^{(i)}\cdot\mathbf y_b^{(j)}$ is the matrix +of overlaps of the vectors $\mathbf y$, then a short and standard calculation involving the average over $H$ and the change of variables from $\mathbf y$ to $Y$ yields \begin{equation} - G(\lambda)=N\lim_{n\to0}\int dY\,(Y_{11}^{(11)}+Y_{11}^{(22)})\, + \overline{G(\lambda)}=N\lim_{n\to0}\int dY\,\big(Y_{11}^{(11)}+Y_{11}^{(22)}\big)\, e^{nN\mathcal S(Y)} \end{equation} -for +where the effective action $\mathcal S$ is given by \begin{equation} \begin{aligned} &\mathcal S(Y) - =\frac1n\sum_{ab}\left[ + =\lim_{n\to0}\frac1n\left\{ + \frac14\sum_{ab}^n\left[ \sigma_1^2(Y_{ab}^{(11)})^2 +\sigma_2^2(Y_{ab}^{(22)})^2 - \right]+\frac12\log\det\begin{bmatrix} - Y^{(11)}&Y^{(12)}\\Y^{(12)}&Y^{(22)} - \end{bmatrix}\\ - &+\frac1n\sum_a^n\left[ + \right] + +\frac12\sum_a^n\left[ 2\epsilon Y_{aa}^{(12)} - -\omega_1Y_{aa}^{(11)} - -\omega_2Y_{aa}^{(22)} - +\lambda(Y_{aa}^{(11)} - +Y_{aa}^{(22)}) + +(\lambda-\omega_1)Y_{aa}^{(11)} + +(\lambda-\omega_2)Y_{aa}^{(22)} \right] + +\frac12\log\det\begin{bmatrix} + Y^{(11)}&Y^{(12)}\\Y^{(12)}&Y^{(22)} + \end{bmatrix} + \right\} \end{aligned} \end{equation} -Making the replica symmetric ansatz $Y_{ab}^{(ij)}=y^{(ij)}\delta_{ab}$ yields +\end{widetext} +Making the replica symmetric ansatz $Y_{ab}^{(ij)}=y^{(ij)}\delta_{ab}$ for +each of the matrices $Y^{(ij)}$ yields \begin{equation} \begin{aligned} - &\mathcal S(y) - = - \sigma_1^2(y^{(11)})^2 - +\sigma_2^2(y^{(22)})^2 - +\frac12\log( + \mathcal S(y) + &= + \frac14\left[\sigma_1^2(y^{(11)})^2 + +\sigma_2^2(y^{(22)})^2\right]+\epsilon y^{(12)} + \\ + & + \qquad+\frac12\left[(\lambda-\omega_1)y^{(11)} + +(\lambda-\omega_2)y^{(22)}\right] \\ + & + \qquad+\frac12\log( y^{(11)}y^{(22)}-y^{(12)}y^{(12)} - )\\ - &+2\epsilon y^{(12)} - -\omega_1y^{(11)} - -\omega_2y^{(22)} - +\lambda(y^{(11)} - +y^{(22)}) + ) \end{aligned} \end{equation} +while the average resolvent becomes \begin{equation} \overline{G(\lambda)} =N(y^{(11)}+y^{(22)}) \end{equation} +for $y^{(11)}$ and $y^{(22)}$ evaluated at a saddle point of $\mathcal S$. The +spectral density at large $N$ is then given by the discontinuity in its +imaginary point on the real axis, or \begin{equation} \rho(\lambda) =\frac1{i\pi N} @@ -1840,7 +1863,9 @@ $\hat\beta$, $c_0$, $r$, and $r_0$ is \end{aligned} \end{equation} When $f(0)=0$ as in the cases directly studied in this work, this further -simplifies as $c_0=r_0=0$. +simplifies as $c_0=r_0=0$. Extremizing this expression with respect to the +order parameters $\hat\beta$ and $r$ produces the red line of dominant minima +shown in Fig.~\ref{fig:ls.complexity}. \end{widetext} -- cgit v1.2.3-70-g09d2