From 892a57ae59ef90474a5be664d52f369a0bc1f3a9 Mon Sep 17 00:00:00 2001 From: Jaron Kent-Dobias Date: Thu, 24 Feb 2022 21:47:59 +0100 Subject: More futzing with superfield language and notation. --- stokes.tex | 154 +++++++++++++++++++++++++++++-------------------------------- 1 file changed, 72 insertions(+), 82 deletions(-) diff --git a/stokes.tex b/stokes.tex index b6f3e36..54b0b64 100644 --- a/stokes.tex +++ b/stokes.tex @@ -1184,37 +1184,30 @@ trick. Based on the experience from similar problems \cite{Castellani_2005_Spin- expected to be exact wherever the complexity is positive. As in \S\ref{sec:stationary.hessian}, these can be bright into a manifestly complex form using Cauchy--Riemann relations. This gives -\begin{eqnarray} \label{eq:real.kac-rice} +\begin{equation} \label{eq:real.kac-rice} \mathcal N - &=\int dz^*dz\,d\hat z^*d\hat z\,d\eta^*d\eta\,d\gamma^*d\gamma\exp\left\{ - -\operatorname{Re}\left( - \hat z^T\partial\tilde\mathcal S_p+\eta^T\partial\partial\mathcal S_p\gamma - \right) - \right\} \\ - &=\int dz^*dz\,d\hat z^*d\hat z\,d\eta^*d\eta\,d\gamma^*\,d\gamma\exp\left\{ - -\frac12\left( - \hat z^T\partial\tilde\mathcal S_p+\hat z^\dagger(\partial\tilde\mathcal S_p)^*+ - \eta^T\partial\partial\mathcal S_p\gamma+ - \eta^\dagger(\partial\partial\mathcal S_p)^*\gamma^* + =\int dz^*dz\,d\hat z^*d\hat z\,d\eta^*d\eta\,d\gamma^*d\gamma\exp\left\{ + \operatorname{Re}\left( + \hat z^T\partial\tilde\mathcal S_p+\eta^T\partial\partial\tilde\mathcal S_p\gamma \right) \right\} -\end{eqnarray} +\end{equation} where $\eta$ and $\gamma$ are Grassmann variables. This can be more conveniently studied using the method of superfields. Introducing the one-component Grassman variables $\theta$ and $\bar\theta$, define the superfield -\begin{eqnarray} - \zeta(i)&=z+\bar\theta(i)\eta^*+\gamma\theta(i)+\hat z\bar\theta(i)\theta(i) -\end{eqnarray} +\begin{equation} + \phi(1)=z+\bar\theta(1)\eta+\gamma\theta(1)+\hat z\bar\theta(1)\theta(1) +\end{equation} +and its measure $d\phi=dz\,d\hat z\,d\eta\,d\gamma$. Then the expression for the number of stationary points can be written in a very compact form, as \begin{equation} - \mathcal N=\int d\zeta^*d\zeta\,\exp\left\{ - -\frac12\int d1\,\left( - \tilde\mathcal S_p(\zeta(1))+\tilde\mathcal S_p(\zeta^*(1)) - \right) + \mathcal N=\int d\phi^*d\phi\,\exp\left\{ + \int d1\,\operatorname{Re} + \tilde\mathcal S_p(\phi(1)) \right\} \end{equation} -where $d1=d\bar\theta(1)\,d\theta(1)$ is a shorthand for integration over the auxiliary grassman variables. +where $d1=d\bar\theta(1)\,d\theta(1)$ denotes the integration over the grasssman variables. This can be related to the previous expression by expansion with respect to the Grassman variables, recognizing that $\theta^2=\bar\theta^2=0$ restricts the series to two derivatives. @@ -1223,46 +1216,54 @@ From here the process can be treated as usual, averaging over the couplings and replacing bilinear combinations of the fields with their own variables via a Hubbard--Stratonovich transformation. Defining the supermatrix \begin{equation} - A(i,j)=\frac1N\{\zeta(i),\zeta^*(i)\}\otimes\{\zeta(j),\zeta^*(j)\} + Q(1,2)=\frac1N\left[\matrix{ + \phi(1)^T\phi(2)&\phi(1)^T\phi(2)^*\cr + \phi(1)^\dagger\phi(2)&\phi(1)^\dagger\phi(2)^* + }\right] +\end{equation} +the result can be written, neglecting constant factors, +\begin{equation} + \overline\mathcal N\simeq\int dQ\,e^{NS_\mathrm{eff}(Q)} \end{equation} -the result can be written +for an effective action functional of the supermatrix $Q$ \begin{equation} - \Sigma= + S_{\mathrm{eff}}= \int d1\,d2\,\operatorname{Tr}\left( - \frac1{16}\left[ - \matrix{1&1\cr1&1} - \right]A^{(p)}(1,2)+\frac p4\left[ - \matrix{\epsilon&0\cr0&\epsilon^*} - \right](I-A(1,1))\delta(1,2) + \frac14\left[ + \matrix{\frac14&\frac14\cr\frac14&\frac14} + \right]Q^{(p)}(1,2)-\frac p2\left[ + \matrix{\frac\epsilon2&0\cr0&\frac{\epsilon^*}2} + \right](Q(1,1)-I)\delta(1,2) \right) - +\frac12\log\det A + +\frac12\log\det Q \end{equation} where the exponent in parentheses denotes element-wise exponentiation, and \begin{equation} - \delta(i,j)=(\theta(i)-\theta(j))(\bar\theta(i)-\bar\theta(j)) + \delta(1,2)=(\bar\theta(1)-\bar\theta(2))(\theta(1)-\theta(2)) \end{equation} -is the super Dirac-$\delta$, and the determinant is a superdeterminant. This leads to the condition for a saddle point of +is the superspace $\delta$-function, and the determinant is a superdeterminant. +This leads to the condition for a saddle point of \begin{equation} 0 - =\frac{\partial\Sigma}{\partial A(1,2)} - =\frac p{16}A^{(p-1)}(1,2)-\frac p4\left[ - \matrix{\epsilon&0\cr0&\epsilon^*} + =\frac{\partial S_\mathrm{eff}}{\partial Q(1,2)} + =\frac p{16}Q^{(p-1)}(1,2)-\frac p2\left[ + \matrix{\frac\epsilon2&0\cr0&\frac{\epsilon^*}2} \right]\delta(1,2) - +\frac12A^{-1}(1,2) + +\frac12Q^{-1}(1,2) \end{equation} -where $\odot$ denotes element-wise multiplication and the inverse superfield is defined by +where the inverse supermatrix is defined by \begin{equation} - I\delta(1,2)=\int d3\,A^{-1}(1,3)A(3,2) + I\delta(1,2)=\int d3\,Q^{-1}(1,3)Q(3,2) \end{equation} Making such a transformation, we arrive at the saddle point equations \begin{eqnarray} 0 - &=\int d3\,\frac{\partial\Sigma}{\partial A(1,3)}A(3,2) \\ - &=\frac p{16}\int d3A^{(p-1)}(1,3)A(3,2)-\frac p4\left[ - \matrix{\epsilon&0\cr0&\epsilon^*} - \right]A(1,2)+\frac12I\delta(1,2) + &=\int d3\,\frac{\partial S_\mathrm{eff}}{\partial Q(1,3)}Q(3,2) \\ + &=\frac p{16}\int d3\,Q^{(p-1)}(1,3)Q(3,2)-\frac p2\left[ + \matrix{\frac\epsilon2&0\cr0&\frac{\epsilon^*}2} + \right]Q(1,2)+\frac12I\delta(1,2) \end{eqnarray} -When expanded, the supermatrix $A$ contains nine independent bilinear +When expanded, the supermatrix $Q$ contains nine independent bilinear combinations of the original variables: $z^\dagger z$, $\hat z^T z$, $\hat z^\dagger z$, $\hat z^T\hat z$, $\hat z^\dagger\hat z$, $\eta^\dagger\eta$, $\gamma^\dagger\gamma$, $\eta^\dagger\gamma$, and $\eta^T\gamma$. The saddle @@ -1288,69 +1289,59 @@ 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 +To determine this, we perform the same Kac--Rice procedure as in the previous section, but now with two probe points, or replicas of the system. The number of stationary 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^*) + \mathcal N + &=\int d\phi_1\,d\phi_2^*\,d\phi_2\,\exp\left\{ + \int d1 \left[ + \tilde\mathcal S_p(\phi_1(1))+\operatorname{Re}\tilde\mathcal S_p(\phi_2(1)) \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)\} + Q(1,2) + =\left[ + \matrix{ + \phi_1(1)^T\phi_1(2) & \phi_1(1)^T\phi_2(2) & \phi_1(1)^T\phi_2(2)^* \cr + \phi_2(1)^T\phi_1(2) & \phi_2(1)^T\phi_2(2) & \phi_2(1)^T\phi_2(2)^* \cr + \phi_2(1)^\dagger\phi_1(2) & \phi_2(1)^\dagger\phi_2(2) & \phi_2(1)^\dagger\phi_2(2)^* + } + \right] \end{equation} \begin{equation} - S + S_\mathrm{eff} =\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[ + \frac14\left[\matrix{1&\frac12&\frac12\cr\frac12&\frac14&\frac14\cr\frac12&\frac14&\frac14}\right]Q^{(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) + Q(1,1)-I \right)\delta(1,2) - \right\}+\frac12\det A + \right\}+\frac12\det Q \end{equation} -where the exponent in parentheses denotes the element-wise power. - \begin{equation} - 0=\frac{\partial S}{\partial A(1,2)} + 0=\frac{\partial S_\mathrm{eff}}{\partial Q(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 p4\left[\matrix{1&\frac12&\frac12\cr\frac12&\frac14&\frac14\cr\frac12&\frac14&\frac14}\right]\odot Q^{(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) + +\frac12Q^{-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) \\ + &=\int d3\,\frac{\partial S_\mathrm{eff}}{\partial Q(1,3)}Q(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) + \left\{\left[\matrix{1&\frac12&\frac12\cr\frac12&\frac14&\frac14\cr\frac12&\frac14&\frac14}\right]\odot Q^{(p-1)}(1,3)\right\}Q(3,2) -\frac p2 \left[ - \matrix{\epsilon_1&0&0\cr0&\frac12\epsilon_2&0\cr0&0&\frac12\epsilon_2^*}\right]A(1,2) + \matrix{\epsilon_1&0&0\cr0&\frac12\epsilon_2&0\cr0&0&\frac12\epsilon_2^*}\right]Q(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. +deal of complexity lies within. The supermatrix $Q$ depends on 35 independent +bilinear products, and when the superfields are expanded produces 48 (not entirely independent) 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 @@ -1359,7 +1350,7 @@ 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. This is related to the influence of +replicas, like $\eta_1^T\eta_2$, vanish. This is related to the influence of the relative position of the two replicas to their spectra, with the vanishing being equivalent to having no influence, e.g., the value of the determinant at each stationary point is exactly what it would be in the one-replica problem @@ -1367,9 +1358,8 @@ with the same invariants, e.g., energy and radius. 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 +This leaves two bilinear products: $z_2^\dagger z_2$ and $z_2^\dagger x_1$, or one real and one complex number. The first is the radius +of the complex saddle, while the other is 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 $\Delta=\Delta z^\dagger\Delta z=\|\Delta z\|^2$ and $\delta=\frac{\Delta z^T\Delta z}{|\Delta z^\dagger\Delta z|}$. Once again -- cgit v1.2.3-70-g09d2