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-rw-r--r--2-point.tex54
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}