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Diffstat (limited to 'stokes.tex')
| -rw-r--r-- | stokes.tex | 43 |
1 files changed, 27 insertions, 16 deletions
@@ -921,8 +921,8 @@ Green function \cite{Livan_2018_Introduction} gives \end{equation} The average is then made over $J$ and Hubbard--Stratonovich is used to change variables to the replica matrices -$N\alpha_{\alpha\beta}=(\zeta^{(\alpha)})^\dagger\zeta^{(\beta)}$ and -$N\chi_{\alpha\beta}=(\zeta^{(\alpha)})^T\zeta^{(\beta)}$, and a series of +$N\alpha_{\alpha\beta}=(z^{(\alpha)})^\dagger z^{(\beta)}$ and +$N\chi_{\alpha\beta}=(z^{(\alpha)})^Tz^{(\beta)}$, and a series of replica vectors. The replica-symmetric ansatz leaves all replica vectors zero, and $\alpha_{\alpha\beta}=\alpha_0\delta_{\alpha\beta}$, $\chi_{\alpha\beta}=\chi_0\delta_{\alpha\beta}$. The result is @@ -943,7 +943,7 @@ $\chi_{\alpha\beta}=\chi_0\delta_{\alpha\beta}$. The result is \includegraphics{figs/spectra_1.5.pdf} \caption{ - Eigenvalue and singular value spectra of a random matrix $A+\lambda_0I$, where the entries of $A$ are complex-normal distributed with $\overline{|A|^2}=A_0=\sqrt{5/4}$ and $\overline{A^2}=C_0=\frac34e^{-i3\pi/4}$. + Eigenvalue and singular value spectra of a random matrix $A+\lambda_0I$, where the entries of $A$ are complex-normal distributed with $\overline{|A|^2}=A_0=5/4$ and $\overline{A^2}=C_0=\frac34e^{-i3\pi/4}$. The diaginal shifts differ in each plot, with (a) $\lambda_0=0$, (b) $\lambda_0=\frac12|\lambda_{\mathrm{th}}|$, (c) $\lambda_0=|\lambda_{\mathrm{th}}|$, and (d) @@ -1440,23 +1440,29 @@ unfortunately more complicated. They can be simplified somewhat by examination of the real two-replica problem. There, all bilinear products involving fermionic fields from different -replicas, like $\eta_x\cdot\eta_z$, vanish. Making this ansatz, the equations -can be solved for the remaining 5 bilinear products, eliminating all the -fermionic fields. +replicas, like $\eta_x\cdot\eta_z$, vanish. This is related to the influence of +the relative position of the two replicas to their spectra, with the vanishing +being equivalent to having no influence, e.g., the value of the determinant at +each stationary point is exactly what it would be in the one-replica problem +with the same invariants, e.g., energy and radius. Making this ansatz, the +equations can be solved for the remaining 5 bilinear products, eliminating all +the fermionic fields. This leaves three bilinear products: $z^\dagger z$, $z^\dagger x$, and $(z^\dagger x)^*$, or one real and one complex number. The first is the radius of the complex saddle, while the others are a generalization of the overlap in the real case. For us, it will be more convenient to work in terms of the difference $\Delta z=z-x$ and the constants which characterize it, which are -$\Delta=\Delta z^\dagger\Delta z$ and $\delta=\Delta z^T\Delta z$. Once again +$\Delta=\Delta z^\dagger\Delta z=\|\Delta z\|^2$ and $\delta=\frac{\Delta z^T\Delta z}{|\Delta z^\dagger\Delta z|}$. Once again we have one real (and strictly positive) variable $\Delta$ and one complex variable $\delta$. -Though the value of $\delta$ is bounded by $\Delta$ by $|\delta|\leq\Delta$, in -reality this bound is not the relevant one, because we are confined on the -manifold $N=z^2$. This bound is most easily established by returning to a -$2N$-dimensional real problem, with $x=x_1$ and $z=x_2+iy_2$. The constraint gives $x_2^Ty_2=0$, $x_1^Tx_1=1$, and $x_2^Tx_2=1+y_2^Ty_2$. Then +Though the value of $\delta$ is bounded by $|\delta|\leq1$ as a result of the +inequality $\Delta z^T\Delta z\leq\|\Delta z\|^2$, in reality this bound is not +the relevant one, because we are confined on the manifold $N=z^2$. The relevant +bound is most easily established by returning to a $2N$-dimensional real +problem, with $x=x_1$ and $z=x_2+iy_2$. The constraint gives $x_2^Ty_2=0$, +$x_1^Tx_1=1$, and $x_2^Tx_2=1+y_2^Ty_2$. Then, by their definitions, \begin{equation} \Delta=1+x_2^Tx_2+y_2^Ty_2-2x_1^Tx_2=2(1+y_2^Ty_2-x_1^Tx_2) \end{equation} @@ -1464,19 +1470,24 @@ $2N$-dimensional real problem, with $x=x_1$ and $z=x_2+iy_2$. The constraint giv \Delta=2(1+|y_2|^2-\sqrt{1-|y_2|^2}\cos\theta_{xx}) \end{equation} \begin{eqnarray} - \delta&=2-2x_1^Tx_2-2ix_1^Ty_2=2(1-|x_2|\cos\theta_{xx}-i|y_2|\cos\theta_{xy}) \\ + \delta\Delta&=2-2x_1^Tx_2-2ix_1^Ty_2=2(1-|x_2|\cos\theta_{xx}-i|y_2|\cos\theta_{xy}) \\ &=2(1-\sqrt{1-|y_2|^2}\cos\theta_{xx}-i|y_2|\cos\theta_{xy}) \end{eqnarray} -$\cos^2\theta_{xy}\leq1-\cos^2\theta_{xx}$ -These equations along with the inequality produce the required bound on $|\delta|$ as a function of $\Delta$ and $\arg\delta$. +There is also an inequality between the angles $\theta_{xx}$ and $\theta_{xy}$ +between $x_1$ and $x_2$ and $y_2$, respectively, which takes that form +$\cos^2\theta_{xy}+\cos^2\theta_{xx}\leq1$. This results from the fact that +$x_2$ and $y_2$ are orthogonal, a result of the constraint. These equations +along with the inequality produce the required bound on $|\delta|$ as a +function of $\Delta$ and $\arg\delta$. \begin{figure} \includegraphics{figs/bound.pdf} + \includegraphics{figs/example_bound.pdf} \caption{ The line bounding $\delta$ in the complex plane as a function of - $\Delta=1,2,\ldots,6$ (inner to outer). Notice that for $\Delta\leq4$, - $|\delta|=\Delta$ is saturated for positive real $\delta$, but is not for + $\Delta=0,1,2,\ldots,6$ (outer to inner). Notice that for $\Delta\leq4$, + $|\delta|=1$ is saturated for positive real $\delta$, but is not for $\Delta>4$, and $\Delta=4$ has a cusp in the boundary. This is due to $\Delta=4$ corresponding to the maximum distance between any two points on the real sphere. |