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| -rw-r--r-- | frsb_kac-rice.tex | 31 |
1 files changed, 22 insertions, 9 deletions
diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex index e918176..a7a03cc 100644 --- a/frsb_kac-rice.tex +++ b/frsb_kac-rice.tex @@ -1006,16 +1006,26 @@ At $\hat \beta>\hat \beta_f$ there is a further transition. \subsection{\textit{R} and \textit{D}: response functions} -The matrix field $R$ is related to responses of the stationary points to perturbations of the tensors $J$: +The matrix field $R$ is related to responses of the stationary points to +perturbations of the tensors $J$. Since the only dependence on $J$ lies in the +measure, once the normalization $\mathcal N$ is replicated one finds +\begin{equation} + \begin{aligned} + \frac1{N^p}\sum_{i_1\cdots i_p}\frac{\partial\langle s_{i_1}\cdots s_{i_p}\rangle}{\partial J^{(p)}_{i_1\cdots i_p}} + &=\lim_{n\to0}\frac1{N^p}\sum_{i_1\cdots i_p}\frac\partial{\partial J^{(p)}_{i_1\cdots i_p}} + \int\left(\prod_a^nd\nu(\mathbf s_a)\right)\,s^1_{i_1}\cdots s^1_{i_p} \\ + & =\lim_{n\to0}\int\left(\prod_a^nd\nu(\mathbf s_a)\right)\sum_b^n\left[ + \hat\beta\left(\frac{\mathbf s_1\cdot\mathbf s_b}N\right)^p+ + p\left(-i\frac{\mathbf s_1\cdot\hat{\mathbf s}_b}N\right)\left(\frac{\mathbf s_1\cdot\mathbf s_b}N\right)^{p-1} + \right] + \end{aligned} +\end{equation} +Taking the average of this expression over disorder and averaging over the equivalent replicas in the integral gives, similar to before, \begin{equation} \begin{aligned} \overline{\frac1{N^p}\sum_{i_1\cdots i_p}\frac{\partial\langle s_{i_1}\cdots s_{i_p}\rangle}{\partial J^{(p)}_{i_1\cdots i_p}}} - =\lim_{n\to0}\overline{\int\prod_\alpha^nd\nu(s_\alpha)\,\frac1n\sum_{ab}\left[ - \hat\beta\left(\frac{s_a\cdot s_b}N\right)^p+ - p\left(-i\frac{\hat s_a\cdot s_b}N\right)\left(\frac{s_a\cdot s_b}N\right)^{p-1} - \right]} \\ - =\lim_{n\to0}\frac1n\sum_{ab}(\hat\beta C_{ab}^p+pR_{ab}C_{ab}^{p-1}) - =\hat\beta+pr_d-\int_0^1dx\,c^{p-1}(x)(\hat\beta c(x)+pr(x)) + &=\lim_{n\to0}\int D[C,R,D]\,\frac1n\sum_{ab}^n(\hat\beta C_{ab}^p+pR_{ab}C_{ab}^{p-1})e^{nN\Sigma[C,R,D]}\\ + &=\hat\beta+pr_d-\int_0^1dx\,c^{p-1}(x)(\hat\beta c(x)+pr(x)) \end{aligned} \end{equation} In particular, when the energy is unconstrained ($\hat\beta=0$) and there is replica symmetry, @@ -1029,7 +1039,10 @@ tend to align with a field. The energy constraint has a significant contribution due to the perturbation causing stationary points to move up or down in energy. -The matrix field $D$ is related to the response of the complexity to such perturbations: +The matrix field $D$ is related to the response of the complexity to +perturbations to the variance of the tensors $J$. This can be found by taking +the expression for the complexity and inserting the dependence of $f$ on the +coefficients $a_p$, then differentiating: \begin{equation} \begin{aligned} \frac{\partial\Sigma}{\partial a_p} @@ -1051,7 +1064,7 @@ When the saddle point of the Kac--Rice problem is supersymmetric, \end{equation} and in particular for $p=1$ \begin{equation} - \frac{\partial\Sigma}{a_1} + \frac{\partial\Sigma}{\partial a_1} =\frac{\hat\beta}4\overline{\frac1N\sum_i\frac{\partial\langle s_i\rangle}{\partial J_i^{(1)}}} \end{equation} i.e., the change in complexity due to a linear field is directly related to the |