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| -rw-r--r-- | 2-point.tex | 408 |
1 files changed, 268 insertions, 140 deletions
diff --git a/2-point.tex b/2-point.tex index 3cd8422..5c40278 100644 --- a/2-point.tex +++ b/2-point.tex @@ -20,6 +20,9 @@ ]{biblatex} \usepackage{anyfontsize,authblk} +\usepackage{tikz} +\usetikzlibrary{calc,fadings,decorations.pathreplacing,perspective,3d} + \addbibresource{2-point.bib} \begin{document} @@ -35,6 +38,104 @@ \begin{abstract} \end{abstract} + +\newcommand\pgfmathsinandcos[3]{% + \pgfmathsetmacro#1{sin(#3)}% + \pgfmathsetmacro#2{cos(#3)}% +} +\newcommand\LongitudePlane[3][current plane]{% + \pgfmathsinandcos\sinEl\cosEl{#2} % elevation + \pgfmathsinandcos\sint\cost{#3} % azimuth + \tikzset{#1/.style={cm={\cost,\sint*\sinEl,0,\cosEl,(0,0)}}} +} + +\newcommand\LatitudePlane[3][current plane]{% + \pgfmathsinandcos\sinEl\cosEl{#2} % elevation + \pgfmathsinandcos\sint\cost{#3} % latitude + \pgfmathsetmacro\yshift{\cosEl*\sint} + \tikzset{#1/.style={cm={\cost,0,0,\cost*\sinEl,(0,\yshift)}}} % +} +\newcommand\NewLatitudePlane[4][current plane]{% + \pgfmathsinandcos\sinEl\cosEl{#3} % elevation + \pgfmathsinandcos\sint\cost{#4} % latitude + \pgfmathsetmacro\yshift{#2*\cosEl*\sint} + \tikzset{#1/.style={cm={\cost,0,0,\cost*\sinEl,(0,\yshift)}}} % +} +\newcommand\TangentPlane[5][current plane]{% + \pgfmathsinandcos\sinEl\cosEl{#3} % elevation + \pgfmathsinandcos\sint\cost{#4} % latitude + \pgfmathsinandcos\sinu\cosu{#5} % azimuth + \pgfmathsetmacro\yshift{#2*\cosEl*\sint} + \tikzset{#1/.style={cm={\cost*\cosu,\cost*\sinu*\sinEl,0,\cost*\sinEl*\cosEl,(0,0)}}} % +} +\newcommand\DrawLongitudeCircle[2][1]{ + \LongitudePlane{\angEl}{#2} + \tikzset{current plane/.prefix style={scale=#1}} + % angle of "visibility" + \pgfmathsetmacro\angVis{atan(sin(#2)*cos(\angEl)/sin(\angEl))} % + \draw[current plane] (\angVis:1) arc (\angVis:\angVis+180:1); + \draw[current plane,dashed] (\angVis-180:1) arc (\angVis-180:\angVis:1); +} +\newcommand\DrawLatitudeCircle[2][1]{ + \LatitudePlane{\angEl}{#2} + \tikzset{current plane/.prefix style={scale=#1}} + \pgfmathsetmacro\sinVis{sin(#2)/cos(#2)*sin(\angEl)/cos(\angEl)} + % angle of "visibility" + \pgfmathsetmacro\angVis{asin(min(1,max(\sinVis,-1)))} + \draw[current plane] (\angVis:1) arc (\angVis:-\angVis-180:1); + \draw[current plane,dashed] (180-\angVis:1) arc (180-\angVis:\angVis:1); +} + +%% document-wide tikz options and styles + +\tikzset{% + >=latex, % option for nice arrows + inner sep=0pt,% + outer sep=2pt,% + mark coordinate/.style={inner sep=0pt,outer sep=0pt,minimum size=3pt, + fill=black,circle}% +} + +\begin{tikzpicture} % "THE GLOBE" showcase + \def\R{4 } % sphere radius + \def\angEl{20} % elevation angle + \def\angAz{-20} % azimuth angle + \filldraw[ball color=white] (0,0) circle (\R); + \filldraw[fill=white] (0,0) circle (\R); + + \foreach \t in {0,45} { \DrawLatitudeCircle[\R]{\t} } + \foreach \t in {-120} { \DrawLongitudeCircle[\R]{\t} } + + \pgfmathsetmacro\H{\R*cos(\angEl)} % distance to north pole + \coordinate (O) at (0,0); + \node[circle,draw,black,scale=0.3] at (0,0) {}; + \coordinate (N) at (0,\H); + \draw[left] node at (0,\H){$\pmb\sigma_1$}; + \draw[thick, ->](O)--(N); + + \NewLatitudePlane[planeP]{\R}{\angEl}{45}; + \path[planeP] (-120:\R) coordinate (P); + \draw[left] node at (P){$\mathbf s_1$}; + + \NewLatitudePlane[equator]{\R}{\angEl}{00}; + \path[equator] (-30:\R) coordinate (Pprime); + \draw[right] node at (Pprime){$\pmb\sigma_b$}; + + \NewLatitudePlane[sbplane]{\R}{\angEl}{45}; + \path[sbplane] (20:\R) coordinate (sb); + \draw[right] node at (sb){$\mathbf s_b$}; + + \TangentPlane[tplane]{\R}{\angEl}{45}{-120}; + \draw[tplane,fill=gray,fill opacity=0.3] circle (1); + \draw[shift={(P)},rotate around y=45,canvas is xy plane at z=0,->,thick] (0,0) -> ({sin(10)},{cos(10)}); + + \draw[thick, ->] (O)->(P); + \draw[thick, ->] (O)->(Pprime); + \draw[thick, ->] (O)->(sb); + + +\end{tikzpicture} + \cite{Ros_2020_Distribution, Ros_2019_Complex, Ros_2019_Complexity} \section{Model} @@ -451,14 +552,26 @@ through the determinant of their Hessians, but only on the bulk of the distribution. As we saw, this bulk is unaffected by the conditions of energy and proximity. However, these conditions give rise to small-rank perturbations to the Hessian, which can lead a subextensive number of eigenvalues leaving the -bulk (we will study \emph{one}). - -\cite{Ikeda_2023_Bose-Einstein-like} - +bulk. We study the possibility of \emph{one} stray eigenvalue. + +We use a technique recently developed to find the smallest eigenvalue of a +random matrix \cite{Ikeda_2023_Bose-Einstein-like}. One defines a quadratic +statistical mechanics model with configurations defined on the sphere, whose +interaction tensor is given by the matrix in question. By construction, the +ground state is located in the direction of the eigenvector associated with the +smallest eigenvalue, and the ground state energy is proportional to that +eigenvalue. + +Our matrix is the Hessian evaluated at a stationary point of the mixed $p$-spin +model, conditioned on the relative position, energies, and stabilities +discussed above. We must restrict the artificial spherical model to lie in the +tangent plane of the `real' spherical configuration space at the point of +interest. The free energy of this model given a point $\mathbf s$ and a +specific realization of the disordered Hamiltonian is \begin{equation} \begin{aligned} - \beta F(\beta\mid\mathbf s) - &=-\frac1N\log\left(\int d\mathbf x\,\delta(\mathbf x\cdot\mathbf s)\delta(N-\mathbf x\cdot\mathbf x)\exp\left\{ + \beta F_H(\beta\mid\mathbf s) + &=-\frac1N\log\left(\int d\mathbf x\,\delta(\mathbf x\cdot\mathbf s)\delta(\|\mathbf x\|^2-N)\exp\left\{ -\beta\frac12\mathbf x^T\partial\partial H(\mathbf s)\mathbf x \right\}\right) \\ &=-\lim_{\ell\to0}\frac1N\frac\partial{\partial\ell}\int\left[\prod_{\alpha=1}^\ell d\mathbf x_\alpha\,\delta(\mathbf x_\alpha^T\mathbf s)\delta(N-\mathbf x_\alpha^T\mathbf x_\alpha)\exp\left\{ @@ -466,23 +579,32 @@ bulk (we will study \emph{one}). \right\}\right] \end{aligned} \end{equation} +where the first $\delta$-function keeps the configurations in the tangent +plane, and the second enforces the spherical constraint. We have anticipated +treated the logarithm with replicas already. We are of course interested in +points $\mathbf s$ that have certain properties: they are stationary points of +$H$ with given energy density and stability, and fixed overlap from a reference +configuration $\pmb\sigma$. We therefore average the free energy above over +such points, giving \begin{equation} \begin{aligned} - F(\beta\mid E_1,\mu_1,q,\pmb\sigma) - &=\int\frac{d\nu(\mathbf s,\omega\mid E_1,\mu_1)\delta(Nq-\pmb\sigma\cdot\mathbf s)}{\int d\nu(\mathbf s',\omega'\mid E_1,\mu_1)\delta(Nq-\pmb\sigma\cdot\mathbf s')}F(\beta\mid\mathbf s) \\ - &=\lim_{n\to0}\int\left[\prod_{a=1}^nd\nu(\mathbf s_a,\omega_a\mid E_1,\mu_1)\,\delta(Nq-\pmb\sigma\cdot\mathbf s_a)\right]F(\beta\mid\mathbf s_1) + F_H(\beta\mid E_1,\mu_1,q,\pmb\sigma) + &=\int\frac{d\nu_H(\mathbf s,\omega\mid E_1,\mu_1)\delta(Nq-\pmb\sigma\cdot\mathbf s)}{\int d\nu_H(\mathbf s',\omega'\mid E_1,\mu_1)\delta(Nq-\pmb\sigma\cdot\mathbf s')}F_H(\beta\mid\mathbf s) \\ + &=\lim_{n\to0}\int\left[\prod_{a=1}^nd\nu_H(\mathbf s_a,\omega_a\mid E_1,\mu_1)\,\delta(Nq-\pmb\sigma\cdot\mathbf s_a)\right]F_H(\beta\mid\mathbf s_1) \end{aligned} \end{equation} +again anticipating the use of replicas. Finally, the reference configuration $\sigma$ should itself be a stationary point of $H$ with its own energy density and stability. Averaging over these conditions gives \begin{equation} \begin{aligned} - F(\beta\mid\epsilon_1,\mu_1,\epsilon_2,\mu_2,q) - &=\int\frac{d\nu(\pmb\sigma,\varsigma\mid E_0,\mu_0)}{\int d\nu(\pmb\sigma',\varsigma'\mid E_0,\mu_0)}\,F(\beta\mid E_1,\mu_1,q,\pmb\sigma) \\ - &=\lim_{m\to0}\int\left[\prod_{a=1}^m d\nu(\pmb\sigma_a,\varsigma_a\mid E_0,\mu_0)\right]\,F(\beta\mid E_1,\mu_1,q,\pmb\sigma_1) + F_H(\beta\mid\epsilon_1,\mu_1,\epsilon_2,\mu_2,q) + &=\int\frac{d\nu_H(\pmb\sigma,\varsigma\mid E_0,\mu_0)}{\int d\nu_H(\pmb\sigma',\varsigma'\mid E_0,\mu_0)}\,F_H(\beta\mid E_1,\mu_1,q,\pmb\sigma) \\ + &=\lim_{m\to0}\int\left[\prod_{a=1}^m d\nu_H(\pmb\sigma_a,\varsigma_a\mid E_0,\mu_0)\right]\,F_H(\beta\mid E_1,\mu_1,q,\pmb\sigma_1) \end{aligned} \end{equation} -The minimum eigenvalue of the conditioned Hessian is then given by twice the ground state energy, or +This formidable expression is now ready to be averaged over the disordered Hamiltonians $H$. Once averaged, +the minimum eigenvalue of the conditioned Hessian is then given by twice the ground state energy, or \begin{equation} - \lambda_\text{min}=2\lim_{\beta\to\infty}\frac\partial{\partial\beta}\overline{F(\beta\mid\epsilon_1,\mu_1,\epsilon_2,\mu_2,q)} + \lambda_\text{min}=2\lim_{\beta\to\infty}\overline{F_H(\beta\mid\epsilon_1,\mu_1,\epsilon_2,\mu_2,q)} \end{equation} For this calculation, there are three different sets of replicated variables. Note that, as for the computation of the complexity, the $\pmb\sigma_1$ and @@ -491,6 +613,7 @@ $\sigma$ replicas constrained to lie at fixed overlap with \emph{all} the $\mathbf s$ replicas, and the second is the only of the $\mathbf s$ replicas at which the Hessian is evaluated. +Using the same methodology as above, the disorder-dependant terms are captured in the linear operator \begin{equation} \mathcal O(\mathbf t)= \sum_a^m\delta(\mathbf t-\pmb\sigma_a)(i\hat{\pmb\sigma}_a\cdot\partial_\mathbf t-\hat\beta_0) @@ -499,80 +622,145 @@ which the Hessian is evaluated. -\frac12 \delta(\mathbf t-\mathbf s_1)\beta\sum_c^\ell(\mathbf x_c\cdot\partial_{\mathbf t})^2 \end{equation} +\begin{align} + A_{ab}=\frac1N\mathbf x_a\cdot\mathbf x_b + && + X^0_{ab}=\frac1N\pmb\sigma_a\cdot\mathbf x_b + && + \hat X^0_{ab}=-i\frac1N\hat{\pmb\sigma}_a\cdot\mathbf x_b + && + X^1_{ab}=\frac1N\mathbf s_a\cdot\mathbf x_b + && + \hat X^1_{ab}=-i\frac1N\hat{\mathbf s}_a\cdot\mathbf x_b +\end{align} +\begin{equation} + \lambda_\mathrm{min} + =-2\lim_{\beta\to\infty} + \lim_{\substack{\ell\to0\\m\to0\\n\to0}}\frac\partial{\partial\ell}\frac1{\beta N} + \int d\mathcal Q\,d\mathcal X\, + e^{N[ + m\mathcal S_0(\mathcal Q_{00}) + +n\mathcal S_1(\mathcal Q_{11},\mathcal Q_{01}\mid\mathcal Q_{00}) + +\ell\mathcal S_x(\mathcal X\mid\mathcal Q_{00},\mathcal Q_{01},\mathcal Q_{11}) + ]} +\end{equation} \begin{equation} \begin{aligned} - &\frac18\beta^2\int d\mathbf t\,d\mathbf t'\,\delta(\mathbf t-\mathbf s_1)\delta(\mathbf t'-\mathbf s_1)\sum_{ab}^\ell(\mathbf x_a\cdot\partial_{\mathbf t})^2(\mathbf x_b\cdot\partial_{\mathbf t'})^2f\left(\frac{\mathbf t\cdot\mathbf t'}N\right)\\ - &=\frac18\beta^2\sum_{ab}^\ell\left[(\mathbf x_a\cdot\mathbf s_1)^2(\mathbf x_b\cdot\mathbf s_1)^2f''''(1) - +4(\mathbf x_a\cdot\mathbf s_1)(\mathbf x_b\cdot\mathbf s_1)(\mathbf x_a\cdot\mathbf x_b)f'''(1) - +2(\mathbf x_a\cdot\mathbf x_b)^2f''(1)\right] \\ - &=\frac14\beta^2f''(1)\sum_{ab}^\ell A_{ab}^2 - =\frac14\beta^2f''(1)(1-a_0^2) + \ell\mathcal S_x(\mathcal X\mid\mathcal Q) + = + \frac12\beta\sum_b^\ell\bigg\{ + \frac12\beta&f''(1)\sum_a^lA_{ab}^2\\ + &+\sum_a^m\left[ + \big(\hat\beta_0f''(C^{01}_{a1})+R^{10}_{a1}f'''(C^{01}_{a1})\big)(X^0_{ab})^2 + +2f''(C^{01}_{a1})X^0_{ab}\hat X^0_{ab} + \right] \\ + &+\sum_a^n\left[ + \big(\hat\beta_1f''(C^{11}_{a1})+R^{11}_{a1}f'''(C^{11}_{a1})\big)(X^1_{ab})^2 + +2f''(C^{11}_{a1})X^1_{ab}\hat X^1_{ab} + \right] + \bigg\}\\ + &+\frac12\log\det\left( + A- + \begin{bmatrix} + X^0\\\hat X^0\\X^1\\\hat X^1 + \end{bmatrix}^T + \begin{bmatrix} + C^{00}&iR^{00}&C^{01}&iR^{01}\\ + iR^{00}&D^{00}&iR^{10}&D^{01}\\ + (C^{01})^T&(iR^{10})^T&C^{11}&iR^{11}\\ + (iR^{01})^T&(D^{01})^T&iR^{11}&D^{11}\\ + \end{bmatrix}^{-1} + \begin{bmatrix} + X^0\\\hat X^0\\X^1\\\hat X^1 + \end{bmatrix} + \right) \end{aligned} \end{equation} + +\begin{align} + X^0 + = + \begin{subarray}{l} + \hphantom{[}\begin{array}{ccc}\leftarrow&\ell&\rightarrow\end{array}\hphantom{\Bigg]}\\ + \left[ + \begin{array}{ccc} + x_0&\cdots&x_0\\ + 0&\cdots&0\\ + \vdots&\ddots&\vdots\\ + 0&\cdots&0 + \end{array} + \right]\begin{array}{c} + \\\uparrow\\m-1\\\downarrow + \end{array}\\ + \vphantom{\begin{array}{c}n\end{array}} + \end{subarray} + && + \hat X^0 + = + \left[ + \begin{array}{ccc} + \hat x_0&\cdots&\hat x_0\\ + 0&\cdots&0\\ + \vdots&\ddots&\vdots\\ + 0&\cdots&0 + \end{array} + \right] + && + X^1 + = + \begin{subarray}{l} + \hphantom{[}\begin{array}{ccc}\leftarrow&\ell&\rightarrow\end{array}\hphantom{\Bigg]}\\ + \left[ + \begin{array}{ccc} + 0&\cdots&0\\ + x_1&\cdots&x_1\\ + \vdots&\ddots&\vdots\\ + x_1&\cdots&x_1 + \end{array} + \right]\begin{array}{c} + \\\uparrow\\n-1\\\downarrow + \end{array}\\ + \vphantom{\begin{array}{c}n\end{array}} + \end{subarray} + && + \hat X^1 + =\begin{bmatrix} + \hat x_1^0&\cdots&\hat x_1^0\\ + \hat x_1^1&\cdots&\hat x_1^1\\ + \vdots&\ddots&\vdots\\ + \hat x_1^1&\cdots&\hat x_1^1 + \end{bmatrix} +\end{align} \begin{equation} \begin{aligned} - &-\frac12\beta\int d\mathbf t\,d\mathbf t'\,\delta(\mathbf t-\mathbf s_1)\delta(\mathbf t'-\mathbf s_b)\sum_{a}^\ell\sum_b^n(i\hat{\mathbf s}_b\cdot\partial_{\mathbf t'}-\hat\beta_1)(\mathbf x_a\cdot\partial_{\mathbf t})^2f\left(\frac{\mathbf t\cdot\mathbf t'}N\right)\\ - &=\frac12\beta\sum_a^\ell\sum_b^n\left[\hat\beta_1(\mathbf x_a\cdot\mathbf s_b)^2f''(C^{11}_{1b}) - -i(\mathbf s_1\cdot\hat{\mathbf s}_b)(\mathbf x_a\cdot\mathbf s_b)^2f'''(C^{11}_{1b}) - -2i(\mathbf x_a\cdot\hat{\mathbf s}_b)(\mathbf x_a\cdot\mathbf s_b)f''(C^{11}_{1b})\right] \\ + \frac2\beta\lim_{\ell\to0}\mathcal S_x(\mathcal X\mid\mathcal Q) &= - \frac12\beta\sum_{a=1}^\ell\sum_{b=2}^n\left[ - \hat\beta_1(X^1_{ab})^2f''(C^{11}_{1b})+(X^1_{ab})^2R^{11}_{1b}f'''(C^{11}_{1b}) - +2X^1_{ab}\hat X_{ab}f''(C^{11}_{1b}) - \right] \\ - &=\frac12\beta\ell\left[\hat\beta_1x_1^2\sum_{b=2}^nf''(C^{11}_{1b}) - +x_1^2\sum_{b=2}^nR^{11}_{1b}f'''(C^{11}_{1b}) - +2x_1\hat x_1\sum_{b=2}^nf''(C^{11}_{1b}) - \right] \\ - &=-\frac12\beta\left[ - (\hat\beta_1f''(q^{11}_0)+r^{11}_0f'''(q^{11}_0))x_1^2+2f''(q^{11}_0)x_1\hat x_1 - \right] + \frac12\beta f''(1)(1-a_0^2) + +\big(\hat\beta_0f''(q)+r_{10}f'''(q)\big)x_0^2 + +2f''(q)x_0\hat x_0 \\ + &- + \big(\hat\beta_1f''(q^{11}_0)+r^{11}_0f'''(q^{11}_0)\big)x_1^2 + -2f''(q^{11}_0)x_1\hat x_1^1 \\ + &+\lim_{\ell\to0}\frac1\ell\frac1\beta\log\det\left( + A- + \begin{bmatrix} + X^0\\\hat X^0\\X^1\\\hat X^1 + \end{bmatrix}^T + \begin{bmatrix} + C^{00}&iR^{00}&C^{01}&iR^{01}\\ + iR^{00}&D^{00}&iR^{10}&D^{01}\\ + (C^{01})^T&(iR^{10})^T&C^{11}&iR^{11}\\ + (iR^{01})^T&(D^{01})^T&iR^{11}&D^{11}\\ + \end{bmatrix}^{-1} + \begin{bmatrix} + X^0\\\hat X^0\\X^1\\\hat X^1 + \end{bmatrix} + \right) \end{aligned} \end{equation} -\begin{align*} - &\sum_{a}^\ell\sum_b^m(i\hat{\pmb\sigma}_b\cdot\partial_b-\hat\beta_0)(\mathbf x_a\cdot\partial_{\mathbf s_1})^2\overline{H(\mathbf s_1)H(\pmb\sigma_b)}\\ - &=-\hat\beta_0(\mathbf x_a\cdot\pmb \sigma_b)^2f''(C^{01}_{1b}) - +i(\hat{\pmb \sigma}_b\cdot\pmb \sigma_1)(\mathbf x_a\cdot\pmb \sigma_b)^2f'''(C^{01}_{1b}) - +2i(\hat{\pmb \sigma}_b\cdot\mathbf x_a)(\mathbf x_a\cdot\pmb \sigma_b)f''(C^{01}_{1b}) \\ - &= - \frac12\beta\sum_{a=1}^\ell\sum_{b=1}^m\left[ - \hat\beta_0(X^0_{ab})^2f''(C^{01}_{b1}) - +(X^0_{ab})^2R^{10}_{b1}f'''(C^{01}_{b1}) - +2X^0_{ab}\hat X^0_{ab}f''(C^{01}_{b1}) - \right] \\ - &=\frac12\beta\left[\big(\hat\beta_0f''(q) - +r_{10}f'''(q)\big)x_0^2 - +2f''(q)x_0\hat x_0 - \right] -\end{align*} -\begin{align} - &\log\det - \begin{bmatrix} - C^{00}&iR^{00}&C^{01}&iR^{01}&X_0\\ - iR^{00}&D^{00}&iR^{10}&D^{01}&\hat X_0\\ - (C^{01})^T&(iR^{10})^T&C^{11}&iR^{11}&X_1\\ - (iR^{01})^T&(D^{10})^T&iR^{11}&D^{11}&\hat X_1\\ - X_0^T&\hat X_0^T&X_1^T&\hat X_1^T&A - \end{bmatrix}\\ - &=\log\det\left( - A- - \begin{bmatrix} - X_0\\\hat X_0\\X_1\\\hat X_1 - \end{bmatrix}^T - \begin{bmatrix} - C^{00}&iR^{00}&C^{01}&iR^{01}\\ - iR^{00}&D^{00}&iR^{10}&D^{01}\\ - (C^{01})^T&(iR^{10})^T&C^{11}&iR^{11}\\ - (iR^{01})^T&(D^{10})^T&iR^{11}&D^{11}\\ - \end{bmatrix}^{-1} - \begin{bmatrix} - X_0\\\hat X_0\\X_1\\\hat X_1 - \end{bmatrix} - \right) -\end{align} \begin{equation} \begin{bmatrix} C^{00}&iR^{00}&C^{01}&iR^{01}\\ @@ -720,66 +908,6 @@ which is \emph{not} a hierarchical matrix! But, all these new hat variables are \end{bmatrix} \end{equation} -\begin{align} - X_0 - = - \begin{subarray}{l} - \hphantom{[}\begin{array}{ccc}\leftarrow&\ell&\rightarrow\end{array}\hphantom{\Bigg]}\\ - \left[ - \begin{array}{ccc} - x_0&\cdots&x_0\\ - 0&\cdots&0\\ - \vdots&\ddots&\vdots\\ - 0&\cdots&0 - \end{array} - \right]\begin{array}{c} - \\\uparrow\\m-1\\\downarrow - \end{array}\\ - \vphantom{\begin{array}{c}n\end{array}} - \end{subarray} - && - \hat X_0 - = - \begin{subarray}{l} - \hphantom{[}\begin{array}{ccc}\leftarrow&\ell&\rightarrow\end{array}\hphantom{\Bigg]}\\ - \left[ - \begin{array}{ccc} - \hat x_0&\cdots&\hat x_0\\ - 0&\cdots&0\\ - \vdots&\ddots&\vdots\\ - 0&\cdots&0 - \end{array} - \right]\begin{array}{c} - \\\uparrow\\m-1\\\downarrow - \end{array}\\ - \vphantom{\begin{array}{c}n\end{array}} - \end{subarray} - && - X_1 - = - \begin{subarray}{l} - \hphantom{[}\begin{array}{ccc}\leftarrow&\ell&\rightarrow\end{array}\hphantom{\Bigg]}\\ - \left[ - \begin{array}{ccc} - 0&\cdots&0\\ - x_1&\cdots&x_1\\ - \vdots&\ddots&\vdots\\ - x_1&\cdots&x_1 - \end{array} - \right]\begin{array}{c} - \\\uparrow\\n-1\\\downarrow - \end{array}\\ - \vphantom{\begin{array}{c}n\end{array}} - \end{subarray} - && - \hat X_1 - =\begin{bmatrix} - \hat x_1^0&\cdots&\hat x_1^0\\ - \hat x_1^1&\cdots&\hat x_1^1\\ - \vdots&\ddots&\vdots\\ - \hat x_1^1&\cdots&\hat x_1^1 - \end{bmatrix} -\end{align} \[ \log\det(A-c)=\log\det A-\frac{c}{\sum_{i=0}^k(a_{i+1}-a_i)x_{i+1}} @@ -824,13 +952,13 @@ which gives \] so \[ - E_{gs}=-\lim_{\beta\to\infty}\frac{\partial\mathcal S}{\partial\beta} - =-\frac12\left(y+\frac1yf''(1)\right)-\frac12X^T(B-yC)X - =-\frac12\left(y+\frac1yf''(1)\right) + \lambda_\mathrm{min}=-2\lim_{\beta\to\infty}\frac{\partial\mathcal S}{\partial\beta} + =-\left(y+\frac1yf''(1)\right)-X^T(B-yC)X + =-\left(y+\frac1yf''(1)\right) \] assuming the last equation is satisfied. The trivial solution, which gives the bottom of the semicircle, is for $X=0$, so the first equation is $y^2=f''(1)$, and \[ - E_{gs}=-\sqrt{f''(1)} + \lambda_\mathrm{min}=-\sqrt{4f''(1)} \] as expected. We need to first the nontrivial solutions with nonzero $X$, but because the coefficients are so nasty this will be a numeric problem... Specifically, we are good for $y$ where one of the eigenvalues of $B-yC$ is zero. |