From 293dc60999ef32c0bd4efd915fa85cd0df426a9d Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Tue, 16 May 2023 11:02:41 +0200
Subject: Formatting changes and work.

---
 2-point.tex | 102 ++++++++++++++++++++++++++++++++----------------------------
 1 file changed, 55 insertions(+), 47 deletions(-)

(limited to '2-point.tex')

diff --git a/2-point.tex b/2-point.tex
index 5c4f58b..e8a7a57 100644
--- a/2-point.tex
+++ b/2-point.tex
@@ -446,28 +446,31 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case,
 
 \section{Isolated eigenvalue}
 
-
-\begin{align*}
-  \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\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\{
-    -\beta\frac12\mathbf x^T_\alpha\partial\partial H(\mathbf s)\mathbf x_\alpha
-  \right\}\right]
-\end{align*}
-\begin{align*}
-  F(\beta\mid E_1,\mu_1,q,\pmb\sigma)
-  &=\int\frac{d\nu(\mathbf s\mid E_1,\mu_1)\delta(Nq-\pmb\sigma\cdot\mathbf s)}{\int d\nu(\mathbf s'\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\mid E_1,\mu_1)\,\delta(Nq-\pmb\sigma\cdot\mathbf s_a)\right]F(\beta\mid\mathbf s_1)
-\end{align*}
-\[
+\begin{equation}
+  \begin{aligned}
+    \beta F(\beta\mid\mathbf s)
+    &=-\frac1N\log\left(\int d\mathbf x\,\delta(\mathbf x\cdot\mathbf s)\delta(N-\mathbf x\cdot\mathbf x)\exp\left\{
+      -\beta\frac12\mathbf x^T\partial\partial H(\mathbf s)\mathbf x
+    \right\}\right) \\
+    &=-\lim_{\ell\to0}\frac1N\frac\partial{\partial\ell}\int\left[\prod_{\alpha=1}^\ell d\mathbf x_\alpha\,\delta(\mathbf x_\alpha^T\mathbf s)\delta(N-\mathbf x_\alpha^T\mathbf x_\alpha)\exp\left\{
+      -\beta\frac12\mathbf x^T_\alpha\partial\partial H(\mathbf s)\mathbf x_\alpha
+    \right\}\right]
+  \end{aligned}
+\end{equation}
+\begin{equation}
+  \begin{aligned}
+    F(\beta\mid E_1,\mu_1,q,\pmb\sigma)
+    &=\int\frac{d\nu(\mathbf s,\omega\mid E_1,\mu_1)\delta(Nq-\pmb\sigma\cdot\mathbf s)}{\int d\nu(\mathbf s',\omega'\mid E_1,\mu_1)\delta(Nq-\pmb\sigma\cdot\mathbf s')}F(\beta\mid\mathbf s) \\
+    &=\lim_{n\to0}\int\left[\prod_{a=1}^nd\nu(\mathbf s_a,\omega_a\mid E_1,\mu_1)\,\delta(Nq-\pmb\sigma\cdot\mathbf s_a)\right]F(\beta\mid\mathbf s_1)
+  \end{aligned}
+\end{equation}
+\begin{equation}
   \begin{aligned}
     F(\beta\mid\epsilon_1,\mu_1,\epsilon_2,\mu_2,q)
-    &=\int\frac{d\nu(\pmb\sigma\mid E_0,\mu_0)}{\int d\nu(\pmb\sigma'\mid E_0,\mu_0)}\,F(\beta\mid E_1,\mu_1,q,\pmb\sigma) \\
-    &=\lim_{m\to0}\int\left[\prod_{a=1}^m d\nu(\pmb\sigma_a\mid E_0,\mu_0)\right]\,F(\beta\mid E_1,\mu_1,q,\pmb\sigma_1)
+    &=\int\frac{d\nu(\pmb\sigma,\varsigma\mid E_0,\mu_0)}{\int d\nu(\pmb\sigma',\varsigma'\mid E_0,\mu_0)}\,F(\beta\mid E_1,\mu_1,q,\pmb\sigma) \\
+    &=\lim_{m\to0}\int\left[\prod_{a=1}^m d\nu(\pmb\sigma_a,\varsigma_a\mid E_0,\mu_0)\right]\,F(\beta\mid E_1,\mu_1,q,\pmb\sigma_1)
   \end{aligned}
-\]
+\end{equation}
 \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)
@@ -476,32 +479,36 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case,
   -\frac12
   \delta(\mathbf t-\mathbf s_1)\beta\sum_c^\ell(\mathbf x_c\cdot\partial_{\mathbf t})^2
 \end{equation}
-\begin{align*}
-  &\sum_{ab}^\ell(\mathbf x_a\cdot\partial_{\mathbf s_1})^2(\mathbf x_b\cdot\partial_{\mathbf s_1'})^2\overline{H(\mathbf s_1)H(\mathbf s_1')}\\
-  &=(\mathbf x_a\cdot\mathbf s_1)^2(\mathbf x_b\cdot\mathbf s_1)^2f''''(1)
-  +2(\mathbf x_a\cdot\mathbf s_1)(\mathbf x_b\cdot\mathbf s_1)(\mathbf x_a\cdot\mathbf x_b)f'''(1)
-  +(\mathbf x_a\cdot\mathbf x_b)^2f''(1) \\
-  &=\frac14\beta^2f''(1)\sum_{ab}^\ell A_{ab}^2
-  =\frac14\beta^2f''(1)(1-a_0^2)
-\end{align*}
-\begin{align*}
-  &\sum_{a}^\ell\sum_b^n(i\hat{\mathbf s}_{\mathbf s_b}\cdot\partial_b-\hat\beta_1)(\mathbf x_a\cdot\partial_{\mathbf s_1})^2\overline{H(\mathbf s_1)H(\mathbf s_b)}\\
-  &=-\hat\beta_1(\mathbf x_a\cdot\mathbf s_b)^2f''(C^{11}_{1b})
-  +i(\hat{\mathbf s}_b\cdot\mathbf s_1)(\mathbf x_a\cdot\mathbf s_b)^2f'''(C^{11}_{1b})
-  +2i(\hat{\mathbf s}_b\cdot\mathbf x_a)(\mathbf x_a\cdot\mathbf s_b)f''(C^{11}_{1b}) \\
-  &=
-  \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]
-\end{align*}
+\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)
+  \end{aligned}
+\end{equation}
+\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] \\
+    &=
+    \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]
+  \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})
@@ -818,8 +825,9 @@ which gives
 \]
 so
 \[
-  E_{gs}=\lim_{\beta\to\infty}(\mathcal S/\beta)=\frac12\left(y+\frac1yf''(1)\right)+\frac12X^T(B-yC)X
-  =\frac12\left(y+\frac1yf''(1)\right)
+  E_{gs}=-\lim_{\beta\to\infty}\frac{\partial\mathcal S}{\partial\beta}
+  =-\frac12\left(y+\frac1yf''(1)\right)-\frac12X^T(B-yC)X
+  =-\frac12\left(y+\frac1yf''(1)\right)
 \]
 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
 \[
-- 
cgit v1.2.3-70-g09d2

