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authorJaron Kent-Dobias <jaron@kent-dobias.com>2023-05-16 11:02:41 +0200
committerJaron Kent-Dobias <jaron@kent-dobias.com>2023-05-16 11:02:41 +0200
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@@ -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
\[