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| -rw-r--r-- | 2-point.tex | 54 |
1 files changed, 38 insertions, 16 deletions
diff --git a/2-point.tex b/2-point.tex index 7e3c818..5c4f58b 100644 --- a/2-point.tex +++ b/2-point.tex @@ -482,6 +482,7 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case, +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)}\\ @@ -496,6 +497,9 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case, &=\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{align*} @@ -509,9 +513,9 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case, +(X^0_{ab})^2R^{10}_{b1}f'''(C^{01}_{b1}) +2X^0_{ab}\hat X^0_{ab}f''(C^{01}_{b1}) \right] \\ - &=\frac12\beta\ell\left[\hat\beta_0x_0^2f''(q) - +x_0^2r_{10}f'''(q) - +2x_0\hat x_0f''(q) + &=\frac12\beta\left[\big(\hat\beta_0f''(q) + +r_{10}f'''(q)\big)x_0^2 + +2f''(q)x_0\hat x_0 \right] \end{align*} @@ -718,7 +722,7 @@ which is \emph{not} a hierarchical matrix! But, all these new hat variables are X_1 = \begin{subarray}{l} - \hphantom{[}\begin{array}{ccc}\leftarrow&n&\rightarrow\end{array}\hphantom{\Bigg]}\\ + \hphantom{[}\begin{array}{ccc}\leftarrow&\ell&\rightarrow\end{array}\hphantom{\Bigg]}\\ \left[ \begin{array}{ccc} 0&\cdots&0\\ @@ -727,14 +731,14 @@ which is \emph{not} a hierarchical matrix! But, all these new hat variables are x_1&\cdots&x_1 \end{array} \right]\begin{array}{c} - \\\uparrow\\m-1\\\downarrow + \\\uparrow\\n-1\\\downarrow \end{array}\\ \vphantom{\begin{array}{c}n\end{array}} \end{subarray} && \hat X_1 =\begin{bmatrix} - \hat x_1^0&\cdots&\hat x_1^0\\ + 0&\cdots&0\\ \hat x_1^1&\cdots&\hat x_1^1\\ \vdots&\ddots&\vdots\\ \hat x_1^1&\cdots&\hat x_1^1 @@ -778,32 +782,50 @@ all a constant $\ell\times\ell$ matrix. where $a_{k+1}=1$ and $x_{k+1}=1$. So the basic form of the action is (for replica symmetric $A$) \[ - \beta_x x_1+\frac14\beta^2f''(1)(1-a_0^2)+\frac12\log(1-a_0)+\frac12\frac{a_0}{1-a_0}+\frac12\beta\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}^TB\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}-\frac1{1-a_0}\frac12\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}^TC\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix} + \frac14\beta^2f''(1)(1-a_0^2)+\frac12\log(1-a_0)+\frac12\frac{a_0}{1-a_0}+\frac12\beta\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}^TB\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}-\frac1{1-a_0}\frac12\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}^TC\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix} +\] +for +\[ + B=\begin{bmatrix} + \hat\beta_0f''(q)+r_{10}f'''(q)&f''(q)&0&0\\ + f''(q)&0&0&0\\ + 0&0&-\hat\beta_1f''(q^{11}_0)-r^{11}_0f'''(q^{11}_0)&-f''(q_0^{11})\\ + 0&0&-f''(q_0^{11})&0 + \end{bmatrix} \] Use $X$ for the big vector. Then \[ - 0=-\frac12\beta^2f''(1)a_0+\frac12\frac{a_0-X^TCX}{(1-a_0)^2} + 0=-\beta^2f''(1)a_0+\frac{a_0-X^TCX}{(1-a_0)^2} +\] +\[ + 0=\bigg(\beta B-\frac1{1-a_0}C\bigg)X \] +For a non-trivial solution, we require the following: change of basis in $X$ to diagonalize the matrix $\beta B-C/(1-a_0)$. Then, all of the new basis vectors of $X$ are zero except one, and the second equation is satisfied by tuning $a_0$ to make the coefficient zero. Finally, the first equation is satisfied by choice of the magnitude of this basis vector. +Suppose $a_0=1-1/(y\beta)$, $X\sim X+O(1/\beta)$. Then \[ - 0=\beta BX-\frac1{1-a_0}CX+\beta_X + 0=-f''(1)(1-(y\beta)^{-1})+y^2\left(1-(y\beta)^{-1}-X^TCX\right) \] +For large $\beta$ \[ - X=\left(\frac1{1-a_0}C-\beta B\right)^{-1}\beta_X + 0=-f''(1)+y^2(1-X^TCX) \] -Suppose $a_0=1-y/\beta$ \[ - X=\beta^{-1}\left(\frac1y C-B\right)^{-1}\beta_X + 0=(B-yC)X \] +which gives \[ - 0=-\frac12\beta f''(1)(\beta-y)+\beta\frac12\frac{\beta-y-\beta X^TCX}{y^2} + \mathcal S=\frac14\beta^2f''(1)\big(2(y\beta)^{-1}-(y\beta)^{-2}\big)-\frac12\log(y\beta)+\frac12y\beta(1-(y\beta)^{-1})+\frac12\beta X^TBX-\frac12y\beta X^TCX \] -For large $\beta$ +so \[ - 0=-\frac12\beta^2 f''(1)+\beta^2\frac12\frac{1-X^TCX}{y^2} + E_{gs}=\lim_{\beta\to\infty}(\mathcal S/\beta)=\frac12\left(y+\frac1yf''(1)\right)+\frac12X^T(B-yC)X + =\frac12\left(y+\frac1yf''(1)\right) \] +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 \[ - y^2=\frac{1-X^TCX}{f''(1)} + E_{gs}=-\sqrt{f''(1)} \] +as expected. We need to first the nontrivial solutions with nonzero $X$, but because the coefficients are so nasty this will be a numeric problem... Specifically, we are good for $y$ where one of the eigenvalues of $B-yC$ is zero. \paragraph{Acknowledgements} |