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diff --git a/marginal.tex b/marginal.tex index 7df99d5..c461f81 100644 --- a/marginal.tex +++ b/marginal.tex @@ -238,8 +238,10 @@ with the effective action We need to evaluate the integral above using the saddle point method, but in the limit of $\beta\to\infty$. We expect the overlaps to concentrate on one as $\beta$ goes to infinity. We therefore take \begin{align} + \label{eq:q0.limit} q_0&=1-y\beta^{-1}-z\beta^{-2}+O(\beta^{-3}) \\ + \label{eq:q0t.limit} \tilde q_0&=1-\tilde y\beta^{-1}-(z+\Delta z)\beta^{-2}+O(\beta^{-3}) \end{align} However, taking the limit with $y\neq\tilde y$ results in an expression for the @@ -506,7 +508,8 @@ $\mathcal O$ then the average gives \end{equation} The result is an integral that only depends on the many vector variables we have introduced through their scalar products with each other. We therefore make a change of variables in the integration from those vectors to matrices that encode their possible scalar products. These matrices are -\begin{align} +\begin{equation} \label{eq:order.parameters} + \begin{aligned} C_{ab}=\frac1N\mathbf x_a\cdot\mathbf x_b && R_{ab}=-i\frac1N\mathbf x_a\cdot\hat{\mathbf x}_b @@ -520,7 +523,8 @@ have introduced through their scalar products with each other. We therefore make X^c_{ab}=\frac1N\mathbf x_a\cdot\mathbf s_b^c \\ \hat X^c_{ab}=\frac1N\hat{\mathbf x}_a\cdot\mathbf s_b^c -\end{align} + \end{aligned} +\end{equation} Order parameters that mix the normal and Grassmann variables generically vanish in these settings \cite{Kurchan_1992_Supersymmetry}. @@ -843,7 +847,7 @@ When expanded, this supermatrix is constructed of the scalar products of the real and Grassmann vectors that make up $\pmb\phi$. The change of variables to these order parameters again results in the Jacobian of \eqref{eq:coordinate.jacobian}, contributing \begin{equation} - \frac N2\log\det J(C,R,D,G,Q,X,\hat X) + \frac N2\log\det J(C,R,D,Q,X,\hat X)-\frac N2\log\det G^2 \end{equation} Up to this point, the expressions above are general and independent of a given ansatz. However, we expect that the order parameters $X$ and $\hat X$ are zero, @@ -881,56 +885,52 @@ Now we have reduced the problem to an extremal one over the order parameters $\hat\beta$, $\hat\lambda$, $C$, $R$, $D$, $G$, and $Q$, it is time to make an ansatz for the form of order we expect to find. We will focus on a regime where the structure of stationary points is replica symmetric, and further where -typical pairs of stationary points have no overlap. This gives +typical pairs of stationary points have no overlap. This requires that $f(0)=0$, or that there is no constant term in the random polynomials. This gives \begin{align} - C=I && R=r_dI && D = d_dI && G = g_dI + C=I && R=rI && D = dI && G = gI \end{align} We further take a planted replica symmetric structure for the matrix $Q$, -identical to that in \eqref{eq:Q.structure}. +identical to that in \eqref{eq:Q.structure}. The resulting effective action is +the same as if we had made an annealed calculation in the complexity, though +the previous expressions are general. \begin{equation} \begin{aligned} - \mathcal S - =-\frac\alpha2\log\left[ + \mathcal S_\beta + =\hat\beta E-\mu(r+g) + +\frac12\log\frac{d+r^2}{g^2}\frac{1-2q_0+\tilde q_0^2}{(1-q_0)^2} + -\frac\alpha2\log\left(\frac{1-f'(1)(2\beta(1-q_0)+\hat\lambda-(1-2q_0+\tilde q_0^2)\beta(\beta+\hat\lambda)f'(1))}{(1-(1-q_0)\beta f'(1))^2}\right) + \\ + -\frac12\mu\hat\lambda+\hat\lambda\lambda^*-\frac\alpha2\log\left[ \frac{ - (f'(1)d-\hat\beta-f''(1)(r^2-g^2+q_0^2\beta^2-\tilde q_0^2\beta(\beta+\hat\lambda)+\beta\hat\lambda+\frac12\hat\lambda^2)))(f(1)-f(0))+(1-rf'(1))^2) + \big[f'(1)d-\hat\beta-f''(1)(r^2-g^2+q_0^2\beta^2-\tilde q_0^2\beta(\beta+\hat\lambda)+\beta\hat\lambda+\frac12\hat\lambda^2)\big]f(1)+(1-rf'(1))^2 }{ (1+gf'(1))^2 } - \right] \\ - +\frac{\alpha f(0)}2\frac{ - \hat\beta-df'(1)+(r^2-g^2+q_0^2\beta^2-\tilde q_0^2\beta(\beta+\lambda)+\lambda\beta+\frac12\lambda^2)f''(1) - }{ - (f'(1)d-\hat\beta- (r^2-g^2+q_0^2\beta^2-\tilde q_0^2\beta(\beta+\hat\lambda)+\beta\hat\lambda+\frac12\hat\lambda^2)f''(1))(f(1)-f(0))+(1-rf'(1))^2 - } - \\ - -\frac\alpha2\log\left(\frac{1-f'(1)(2\beta(1-q_0)+\lambda-(1-2q_0+\tilde q_0^2)\beta(\beta+\lambda)f'(1))}{(1-(1-q_0)\beta f'(1))^2}\right) - +\frac12\log\frac{d+r^2}{g^2}\frac{1-2q_0+\tilde q_0^2}{(1-q_0)^2} - -\mu(r+g)-\frac12\mu\hat\lambda+\hat\beta E+\hat\lambda\lambda^* + \right] \end{aligned} \end{equation} +We expect as before the limits of $q_0$ and $\tilde q_0$ as $\beta$ goes to +infinity to approach one, defining their asymptotic expansion as in +\eqref{eq:q0.limit} and \eqref{eq:q0t.limit}. Upon making this substitution and +taking the zero-temperature limit, we find \begin{equation} \begin{aligned} \mathcal S_\infty - =-\frac\alpha2\log\left[ - \frac{ - (f'(1)d-\hat\beta-f''(1)(r^2-g^2+2y_0\hat\lambda+\Delta z+\frac12\hat\lambda^2)))(f(1)-f(0))+(1-rf'(1))^2) - }{ - (1+gf'(1))^2 - } - \right] \\ - +\frac{\alpha f(0)}2\frac{ - \hat\beta-df'(1)+(r^2-g^2+2y_0\hat\lambda+\Delta z+\frac12\lambda^2)f''(1) - }{ - (f'(1)d-\hat\beta- (r^2-g^2+2y_0\hat\lambda+\Delta z+\frac12\lambda^2)f''(1))(f(1)-f(0))+(1-rf'(1))^2 - } - \\ + =\hat\beta E-\mu(r+g) + +\frac12\log\frac{d+r^2}{g^2}\frac{y_0^2-\Delta z}{y_0^2} -\frac\alpha2\log\left( \frac{ 1-(2y_0+\hat\lambda)f'(1)+(y_0^2-\Delta z)f'(1)^2 }{(1-y_0f'(1))^2} \right) - +\frac12\log\frac{d+r^2}{g^2}\frac{y_0^2-\Delta z}{y_0^2} - -\mu(r+g)-\frac12\mu\hat\lambda+\hat\beta E+\hat\lambda\lambda^* + \\ + -\frac12\mu\hat\lambda+\hat\lambda\lambda^*-\frac\alpha2\log\left[ + \frac{ + \big[f'(1)d-\hat\beta-f''(1)(r^2-g^2+2y_0\hat\lambda+\Delta z+\frac12\hat\lambda^2)\big]f(1)+\big[1-rf'(1)\big]^2 + }{ + (1+gf'(1))^2 + } + \right] \end{aligned} \end{equation} \begin{equation} @@ -950,43 +950,163 @@ replica symmetric structure, formulas for the effective action are generic to any structure and provide a starting point for analyzing the challenging full-RSB setting. +Using the $\mathbb R^{N|2}$ superfields +\begin{equation} + \pmb\phi_a(1)=\mathbf x+\bar\theta_1\pmb\eta+\bar{\pmb\eta}\theta_1+\bar\theta_1\theta_1\hat{\mathbf x}, +\end{equation} +the replicated count of stationary points can be written +\begin{equation} + \begin{aligned} + &\mathcal N(E,\mu)^n + =\int\prod_{a=1}^nd\hat\beta_a\,d\pmb\phi_a\, + \\ + &\qquad\times\exp\left[ + \hat\beta_a E-\frac12\int d1\,B_a(1)\sum_{k=1}^MV^k(\pmb\phi_a(1))^2 + \right] + \end{aligned} +\end{equation} +for $B_a(1)=1-\hat\beta_a\bar\theta_1\theta_1$. +The derivation of the complexity follows from here nearly identically to that +in Appendix A.2 of \citeauthor{Fyodorov_2022_Optimization} with superoperations +replacing standard ones \cite{Fyodorov_2022_Optimization}. First we insert +Dirac $\delta$ functions to fix each of the $M$ energies $V^k(\pmb\phi_a(1))$ as +\begin{equation} \label{eq:Vv.delta} + \begin{aligned} + &\int dv^k_a\,\delta\big(V^k(\pmb\phi_a(1))-v^k_a(1)\big) + \\ + &\quad=\int dv^k_a\,d\hat v^k_a\,\exp\left[i\int d1\,\hat v^k_a(1)\big(V^k(\pmb\phi_a(1))-v^k_a(1)\big)\right] + \end{aligned} +\end{equation} +The squared $V^k$ appearing in the energy can now be replaced by the variables +$v^k$, leaving the only remaining dependence on the disordered $V$ in the +contribution of \eqref{eq:Vv.delta}, which is linear. The average over the +disorder can then be computed, which yields +\begin{equation} + \begin{aligned} + &\overline{\sum_{k=1}^M\sum_{a=1}^n\exp\left[i\int d1\,\hat v^k_a(1)V^k(\pmb\phi_a(1))\right]} + \\ + & + =\exp\left[ + -\frac12\sum_{k=1}^M\sum_{ab=1}^n\int d1\,d2\,\hat v_a^k(1)f\left(\frac{\pmb\phi_a(1)^T\pmb\phi_b(2)}N\right)\hat v_b^k(2) + \right] + \end{aligned} +\end{equation} +The result is factorized in the indices $k$ and Gaussian in the superfields $v$ +and $\hat v$ with kernel +\begin{equation} + \begin{bmatrix} + B_a(1)\delta_{ab}\delta(1,2) & i\delta_{ab}\delta(1,2) \\ + i\delta_{ab}\delta(1,2) & f\left(\frac{\pmb\phi_a(1)^T\pmb\phi_b(2)}N\right) + \end{bmatrix} +\end{equation} +where $\delta(1,2)=(\bar\theta_1-\bar\theta_2)(\theta_1-\theta_2)$ is the +identity operator for convolutions with $d1$ or $d2$. +Making the $M$ independent Gaussian integrals, we therefore have +\begin{equation} + \begin{aligned} + &\mathcal N(E,\mu)^n + =\int\left(\prod_{a=1}^nd\hat\beta_a\,d\pmb\phi_a\right) + \exp\bigg[ + \sum_a^n\hat\beta_aE \\ + &\qquad-\frac M2\log\operatorname{sdet}\left( + \delta_{ab}\delta(1,2)+B_a(1)f\left(\frac{\pmb\phi_a(1)^T\pmb\phi_b(2)}N\right) + \right) + \bigg] + \end{aligned} +\end{equation} +We make a change of variables from the fields $\pmb\phi$ to matrices $\mathbb Q_{ab}(1,2)=\frac1N\pmb\phi_a(1)^T\pmb\phi_b(2)$. This transformation results in a change of measure of the form +\begin{equation} + \prod_{a=1}^n d\pmb\phi_a=d\mathbb Q\,(\operatorname{sdet}\mathbb Q)^\frac N2 + =d\mathbb Q\,\exp\left[\frac N2\log\operatorname{sdet}\mathbb Q\right] +\end{equation} +We therefore have +\begin{equation} + \begin{aligned} + &\mathcal N(E,\mu)^n + =\int\left(\prod_{a=1}^nd\hat\beta_a\right)\,d\mathbb Q\, + \exp\bigg[ + \sum_a^n\hat\beta_aE + +\frac N2\log\operatorname{sdet}\mathbb Q + \\ + &\qquad-\frac M2\log\operatorname{sdet}\left( + \delta_{ab}\delta(1,2)+B_a(1)f(\mathbb Q_{ab}(1,2)) + \right) + \bigg] + \end{aligned} +\end{equation} +We now need to blow up our supermatrices into our physical order parameters. We have that +\begin{equation} + \begin{aligned} + &\mathbb Q_{ab}(1,2) + =C_{ab}-G_{ab}(\bar\theta_1\theta_2+\bar\theta_2\theta_1) \\ + &\qquad-R_{ab}(\bar\theta_1\theta_1+\bar\theta_2\theta_2) + -D_{ab}\bar\theta_1\theta_2\bar\theta_2\theta_2 + \end{aligned} +\end{equation} +where $C$, $R$, $D$, and $G$ are the matrices defined in +\eqref{eq:order.parameters}. Other possible combinations involving scalar +products between fermionic and bosonic variables do not contribute at physical +saddle points \cite{Kurchan_1992_Supersymmetry}. Inserting this expansion into +the expression above and evaluating the superdeterminants, we find +\begin{equation} + \mathcal N(E,\mu)^n=\int d\hat\beta\,dC\,dR\,dD\,dG\,e^{nN\mathcal S_\mathrm{KR}(\hat\beta,C,R,D,G)} +\end{equation} +where the effective action is given by \begin{widetext} \begin{equation} \begin{aligned} - &\mathcal S - =-\frac1n\frac\alpha2\left\{\log\det\left[ - \hat\beta f(C)+\Big( - f'(C)\odot D+(G\odot G-R\odot R)\odot f''(C) + &\mathcal S_\mathrm{KR}(\hat\beta,C,R,D,G) + =\hat\beta E-\frac1n\operatorname{Tr}(G+R)\mu + +\frac1n\frac12\Big(\log\det(CD+R^2)-\log\det G^2\Big) + \\ + &-\frac1n\frac\alpha2\left\{\log\det\left[ + \Big( + f'(C)\odot D-\hat\beta I+(G\odot G-R\odot R)\odot f''(C) \Big)f(C) - +(I+R\odot f'(C))^2 - \right]-\log\det(I+G\odot f'(C))^2\right\} \\ - &+\frac1n\frac12\Big(\log\det(CD+R^2)-\log\det G^2\Big) - +\hat\beta E+(g_d-r_d)\mu + +(I-R\odot f'(C))^2 + \right]-\log\det(I+G\odot f'(C))^2\right\} \end{aligned} \end{equation} where $\odot$ gives the Hadamard or componentwise product between the matrices, while other products and powers are matrix products and powers. +In the case where $\mu$ is not specified, the model has a BRST symmetry whose +Ward identities give $D=\hat\beta R$ and $G=-R$ +\cite{Annibale_2004_Coexistence, Kent-Dobias_2023_How}. Using these relations, +the effective action becomes particularly simple: +\begin{equation} + \mathcal S(\hat\beta, C, R) + = + \hat\beta E + +\lim_{n\to0}\frac1n\left[ + -\frac\alpha2\log\det\left[ + I-\hat\beta f(C)(I-R\odot f'(C))^{-1} + \right] + +\frac12\log\det(I+\hat\beta CR^{-1}) + \right] +\end{equation} +This effective action is general for arbitrary matrices $C$ and $R$. When using +a replica symmetric ansatz of $C_{ab}=\delta_{ab}+c_0(1-\delta_{ab})$ and +$R_{ab}=r\delta_{ab}+r_0(1-\delta_{ab})$, the resulting function of +$\hat\beta$, $c_0$, $r$, and $r_0$ is \begin{equation} \begin{aligned} - &\hat\beta E+\mu(g_d-r_d)+\frac12\log\frac{d_d+r_d^2}{g_d^2} \\ - &-\frac\alpha2\log\left[ - 1+\hat\beta\big(f(1)-f(0)\big) - \Big(d_d\big(f(1)-f(0)\big)+r_d\big(2+r_df'(1)\big)\Big)f'(1) - +(g_d^2-r_d^2)\big(f(1)-f(0)\big)f''(1) + \mathcal S= + \hat\beta E + -\frac\alpha 2\left[ + \log\left(1-\frac{\hat\beta\big(f(1)-f(c_0)\big)}{1-rf'(1)+r_0f'(c_0)}\right) + -\frac{\hat\beta f(c_0)+r_0f'(c_0)}{ + 1-\hat\beta\big(f(1)-f(c_0)\big)-rf'(1)+rf'(c_0) + }+\frac{r_0f'(c_0)}{1-rf'(1)+r_0f'(c_0)} \right] \\ - &-\alpha f(0)\left( - \big(f(1)-f(0)\big)+\frac{1+r_d\big(2+r_df'(1)\big)f'(1)}{\hat\beta+d_df'(1)+(g_d^2-r_d^2)f''(1)} - \right)^{-1} + +\frac12\left[ + \log\left(1+\frac{\hat\beta(1-c_0)}{r-r_0}\right) + +\frac{\hat\beta c_0+r_0}{\hat\beta(1-c_0)+r-r_0} + -\frac{r_0}{r-r_0} + \right] \end{aligned} \end{equation} - -In the case where $\mu$ is not specified, in which the model is supersymmetric, $D=\hat\beta R$ and the effective action becomes particularly simple: -\begin{equation} - \hat\beta e - -\frac12\frac{\alpha f(0)}{1+\hat\beta\big(f(1)-f(0)\big)+r_df'(1)} - -\frac\alpha2\log\left(1+\frac{\hat\beta\big(f(1)-f(0)\big)}{1+r_df'(1)}\right) - +\frac12\log\frac{\hat\beta+r_d}{r_d} -\end{equation} +When $f(0)=0$ as in the cases directly studied in this work, this further +simplifies as $c_0=r_0=0$. \end{widetext} |