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| -rw-r--r-- | 2-point.tex | 102 |
1 files changed, 55 insertions, 47 deletions
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 \[ |