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authorJaron Kent-Dobias <jaron@kent-dobias.com>2020-12-10 11:49:13 +0100
committerJaron Kent-Dobias <jaron@kent-dobias.com>2020-12-10 11:49:13 +0100
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More work on explaining the spectrum.
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@@ -219,30 +219,6 @@ where
\end{widetext}
This leaves a single parameter, $a$, which dictates the magnitude of $z^*\cdot z$, or alternatively the magnitude $y^2$ of the imaginary part. The latter vanishes as $a\to1$, where we should recover known results for the real $p$-spin.
-
-The Hessian of \eqref{eq:constrained.hamiltonian} is $\partial\partial
-H=\partial\partial H_0-p\epsilon I$, or the Hessian of
-\eqref{eq:bare.hamiltonian} with a constant added to its diagonal. The
-eigenvalue distribution $\rho$ of the constrained Hessian is therefore related
-to the eigenvalue distribution $\rho_0$ of the unconstrained one by a similar
-shift, or $\rho(\lambda)=\rho_0(\lambda+p\epsilon)$. The Hessian of
-\eqref{eq:bare.hamiltonian} is
-\begin{equation} \label{eq:bare.hessian}
- \partial_i\partial_jH_0
- =\frac{p(p-1)}{p!}\sum_{k_1\cdots k_{p-2}}^NJ_{ijk_1\cdots k_{p-2}}z_{k_1}\cdots z_{k_{p-2}},
-\end{equation}
-which makes its ensemble that of Gaussian complex symmetric matrices. Given its variances
-$\overline{|\partial_i\partial_j H_0|^2}=p(p-1)a^{p-2}/2N$ and
-$\overline{(\partial_i\partial_j H_0)^2}=p(p-1)\kappa/2N$, $\rho_0(\lambda)$ is constant inside the ellipse
-\begin{equation} \label{eq:ellipse}
- \left(\frac{\mathop{\mathrm{Re}}(\lambda e^{i\theta})}{a^{p-2}+|\kappa|}\right)^2+
- \left(\frac{\mathop{\mathrm{Im}}(\lambda e^{i\theta})}{a^{p-2}-|\kappa|}\right)^2
- <\frac{p(p-1)}{2a^{p-2}}
-\end{equation}
-where $\theta=\frac12\arg\kappa$ \cite{Nguyen_2014_The}. The eigenvalue
-spectrum of $\partial\partial H$ therefore is that of an ellipse whose center
-is shifted by $p\epsilon$.
-
\begin{figure}[htpb]
\centering
@@ -265,6 +241,31 @@ is shifted by $p\epsilon$.
} \label{fig:spectra}
\end{figure}
+
+The Hessian of \eqref{eq:constrained.hamiltonian} is $\partial\partial
+H=\partial\partial H_0-p\epsilon I$, or the Hessian of
+\eqref{eq:bare.hamiltonian} with a constant added to its diagonal. The
+eigenvalue distribution $\rho$ of the constrained Hessian is therefore related
+to the eigenvalue distribution $\rho_0$ of the unconstrained one by a similar
+shift, or $\rho(\lambda)=\rho_0(\lambda+p\epsilon)$. The Hessian of
+\eqref{eq:bare.hamiltonian} is
+\begin{equation} \label{eq:bare.hessian}
+ \partial_i\partial_jH_0
+ =\frac{p(p-1)}{p!}\sum_{k_1\cdots k_{p-2}}^NJ_{ijk_1\cdots k_{p-2}}z_{k_1}\cdots z_{k_{p-2}},
+\end{equation}
+which makes its ensemble that of Gaussian complex symmetric matrices. Given its variances
+$\overline{|\partial_i\partial_j H_0|^2}=p(p-1)a^{p-2}/2N$ and
+$\overline{(\partial_i\partial_j H_0)^2}=p(p-1)\kappa/2N$, $\rho_0(\lambda)$ is constant inside the ellipse
+\begin{equation} \label{eq:ellipse}
+ \left(\frac{\mathop{\mathrm{Re}}(\lambda e^{i\theta})}{a^{p-2}+|\kappa|}\right)^2+
+ \left(\frac{\mathop{\mathrm{Im}}(\lambda e^{i\theta})}{a^{p-2}-|\kappa|}\right)^2
+ <\frac{p(p-1)}{2a^{p-2}}
+\end{equation}
+where $\theta=\frac12\arg\kappa$ \cite{Nguyen_2014_The}. The eigenvalue
+spectrum of $\partial\partial H$ therefore is that of an ellipse in the complex
+plane whose center lies at $-p\epsilon$. Examples of these distributions are
+shown in the insets of Fig.~\ref{fig:spectra}.
+
The eigenvalue spectrum of the Hessian of the real part, or equivalently the
eigenvalue spectrum of $(\partial\partial H)^\dagger\partial\partial H$, is the
singular value spectrum of $\partial\partial H$. When $\kappa=0$ and the
@@ -272,7 +273,9 @@ elements of $J$ are standard complex normal, this corresponds to a complex
Wishart distribution. For $\kappa\neq0$ the problem changes, and to our
knowledge a closed form is not known. We have worked out an implicit form for
this spectrum using the saddle point of a replica symmetric calculation for the
-Green function. Introducing replicas to bring the partition function to
+Green function.
+
+Introducing replicas to bring the partition function to
the numerator gives
\begin{widetext}
\begin{equation}
@@ -296,17 +299,26 @@ the numerator gives
\exp\left\{nN\left[
1+\frac{p(p-1)}{16}a^{p-2}\alpha_0^2-\frac{\alpha_0\sigma}2+\frac12\log(\alpha_0^2-|\chi_0|^2)
+\frac p4\mathop{\mathrm{Re}}\left(\frac{(p-1)}8\kappa^*\chi_0^2-\epsilon^*\chi_0\right)
- \right]\right\}
+ \right]\right\}.
\end{equation}
\end{widetext}
-The argument of the exponential has several saddles, but the one with the smallest value of $\mathop{\mathrm{Re}}\alpha$ gives the correct solution \textcolor{red}{\textbf{we have checked this, but a detailed analysis of the saddle-point integration is still needed to justify it.}}.
+The argument of the exponential has several saddles. The solutions $\alpha_0$
+are the roots of a sixth-order polynomial, but the root with the
+smallest value of $\mathop{\mathrm{Re}}\alpha$ appears gives the correct
+solution. A detailed analysis of the saddle point integration is needed to
+understand why this is so. Given such $\alpha_0$, the density of singular
+values follows from the jump across the cut, or
+\begin{equation}
+ \rho(\sigma)=\frac1{i\pi}\left(\lim_{\mathop{\mathrm{Im}}\sigma\to0^+}G(\sigma)-\lim_{\mathop{\mathrm{Im}}\sigma\to0^-}G(\sigma)\right)
+\end{equation}
The transition from a one-cut to two-cut singular value spectrum naturally
corresponds to the origin leaving the support of the eigenvalue spectrum.
Weyl's theorem requires that the product over the norm of all eigenvalues must
not be greater than the product over all singular values \cite{Weyl_1912_Das}.
Therefore, the absence of zero eigenvalues implies the absence of zero singular
-values. The determination of the threshold energy is therefore reduced to a
+values. The determination of the threshold energy -- the energy at which the
+distribution of singular values becomes gapped -- is therefore reduced to a
geometry problem, and yields
\begin{equation} \label{eq:threshold.energy}
|\epsilon_{\mathrm{th}}|^2