From 4812679a26026036d16a408a5b63f4e33d070759 Mon Sep 17 00:00:00 2001 From: Jaron Kent-Dobias Date: Fri, 11 Feb 2022 16:54:52 +0100 Subject: Started writing up the 2 replica superfield calculation. --- stokes.tex | 105 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++- 1 file changed, 104 insertions(+), 1 deletion(-) (limited to 'stokes.tex') diff --git a/stokes.tex b/stokes.tex index 33f2f5b..f2ce388 100644 --- a/stokes.tex +++ b/stokes.tex @@ -1097,7 +1097,7 @@ imaginary energy join. &\simeq(2\pi)^{N/2-1}\int d\lambda\,e^{\lambda N-\frac N2\int d\lambda'\,\rho(\lambda')\log(\beta\lambda'+\lambda)} \\ \end{eqnarray} -\subsection{Pure \textit{p}-spin} +\subsection{Pure \textit{p}-spin: where are the saddles?} Since $H$ is holomorphic, any critical point of $\operatorname{Re}H$ is also one of $\operatorname{Im}H$, and therefore of $H$ itself. Writing $z=x+iy$ for @@ -1354,6 +1354,109 @@ physical dynamics, are a problem we hope to address in future work. } \end{figure} +\subsection{Pure \textit{p}-spin: where are my neighbors?} + +The problem of counting the density of Stokes points in an analytic +continuation of the spherical models is quite challenging, as the problem of +finding dyramic trajectories with endpoints at stationary points is already +difficult, and once made complex the problem has twice the number of fields +squared. + +In this section, we begin to address the problem heuristically by instead +asking: if you are at a stationary point, where are your neighbors? The +stationary points geometrically nearest to a given stationary point should make +up the bulk of its adjacent points in the sense of being susceptible to Stokes +points. The distribution of these near neighbors in the complex plane therefore +gives a sense of whether many Stokes lines should be expected, and when. + +To determine this, we perform the same Kac--Rice produce as in the previous +section, but now with two probe points, or replicas of the system. The number of +critical points with given energies $\epsilon_1$ and $\epsilon_2$ are +\begin{equation} + \mathcal N(\epsilon_1,\epsilon_2) + =\int dx\,dz\,dz^*\,\delta(\partial H(x))\,\delta(\operatorname{Re}\partial H(z))\delta(\operatorname{Im}\partial H(z))|\det\operatorname{Hess}H(z)|^2|\det\operatorname{Hess}H(x)| +\end{equation} + +\begin{eqnarray} + \mathcal N(\epsilon_1,\epsilon_2) + &=\int d\phi\,d\zeta^*d\zeta\exp\left\{ + \int d\bar\theta\,d\theta \left[ + H(\phi)+\operatorname{Re}H(\zeta) + \right] + \right\} \\ + &=\int d\phi\,d\zeta^*d\zeta\exp\left\{ + \int d\bar\theta\,d\theta \left[ + H(\phi)+\frac12H(\zeta)+\frac12H^*(\zeta^*) + \right] + \right\} +\end{eqnarray} +\begin{eqnarray} + \phi(i)&=x+\bar\theta(i)\eta_x^*+\eta_x^*\theta(i)+\hat x\bar\theta(i)\theta(i) \\ + \zeta(i)&=z+\bar\theta(i)\eta_z^*+\eta_z\theta(i)+\hat z\bar\theta(i)\theta(i) \\ + \zeta^*(i)&=z^*+\bar\theta(i)\eta_{z^*}^*+\eta_{z^*}\theta(i)+\hat z^*\bar\theta(i)\theta(i) +\end{eqnarray} +\begin{equation} + A(i,j)=\{\phi(i),\zeta(i),\zeta^*(i)\}\otimes\{\phi(j),\zeta(j),\zeta^*(j)\} +\end{equation} +\begin{equation} + S + =\int d1\,d2\,\operatorname{Tr}\left\{ + \frac14\left[\matrix{1&\frac12&\frac12\cr\frac12&\frac14&\frac14\cr\frac12&\frac14&\frac14}\right]A^{(p)}(1,2) + +\frac p2\left[ + \matrix{\epsilon_1&0&0\cr0&\frac12\epsilon_2&0\cr0&0&\frac12\epsilon_2^*} + \right]\left( + I-A(1,1) + \right)\delta(1,2) + \right\}+\frac12\det A +\end{equation} +where the exponent in parentheses denotes the element-wise power. + +\begin{equation} + 0=\frac{\partial S}{\partial A(1,2)} + = + \frac p4\left[\matrix{1&\frac12&\frac12\cr\frac12&\frac14&\frac14\cr\frac12&\frac14&\frac14}\right]\odot A^{(p-1)}(1,2) + -\frac p2\left[ + \matrix{\epsilon_1&0&0\cr0&\frac12\epsilon_2&0\cr0&0&\frac12\epsilon_2^*}\right]\delta(1,2) + +\frac12A^{-1}(1,2) +\end{equation} +where $\odot$ denotes element-wise multiplication. +\begin{eqnarray} + 0 + &=\int d3\,\frac{\partial S}{\partial A(1,3)}A(3,2) \\ + &=\frac p4\int d3\, + \left\{\left[\matrix{1&\frac12&\frac12\cr\frac12&\frac14&\frac14\cr\frac12&\frac14&\frac14}\right]\odot A^{(p-1)}(1,3)\right\}A(3,2) + -\frac p2 \left[ + \matrix{\epsilon_1&0&0\cr0&\frac12\epsilon_2&0\cr0&0&\frac12\epsilon_2^*}\right]A(1,2) + +\frac12I\delta(1,2) +\end{eqnarray} +Despite being able to pose the saddle point problem in a compact way, a great +deal of complexity lies within. The supermatrix $A$ depends on 35 independent +bilinear products, and when the superfields are expanded produces 48 equations. +These equations can be split into 30 involving bilinear products of the +fermionic fields and 18 without them. The 18 equations without fermionic +bilinear products can be solved with a computer algebra package to eliminate 17 +of the 20 non-fermionic bilinear products. The fermionic equations are +unfortunately more complicated. + +They can be simplified somewhat by examination of the real two-replica problem. +There, all bilinear products involving fermionic fields from different +replicas, like $\eta_x\cdot\eta_z$, vanish. Making this ansatz, the equations +can be solved for the remaining 5 bilinear products, eliminating all the +fermionic fields. + +This leaves three bilinear products: $z^\dagger z$, $z^\dagger x$, and +$(z^\dagger x)^*$, or one real and one complex number. The first is the radius +of the complex saddle, while the others are a generalization of the overlap in +the real case. For us, it will be more convenient to work in terms of the +difference $\Delta z=z-x$ and the constants which characterize it, which are +$\Delta=\Delta z^\dagger\Delta z$ and $\delta=\Delta z^T\Delta z$. Once again +we have one real (and strictly positive) variable $\Delta$ and one complex +variable $\delta$. + +Though the value of $\delta$ is bounded by $\Delta$ by $|\delta|\leq\Delta$, in +reality this bound is not the relevant one, because we are confined on the +manifold $N=z^2$. + \section{The $p$-spin spherical models: numerics} To confirm the presence of Stokes lines under certain processes in the $p$-spin, we studied the problem numerically. -- cgit v1.2.3-70-g09d2