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authorJaron Kent-Dobias <jaron@kent-dobias.com>2022-02-11 16:54:52 +0100
committerJaron Kent-Dobias <jaron@kent-dobias.com>2022-02-11 16:54:52 +0100
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Started writing up the 2 replica superfield calculation.
-rw-r--r--stokes.tex105
1 files changed, 104 insertions, 1 deletions
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.