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diff --git a/frsb_kac_new.tex b/frsb_kac_new.tex deleted file mode 100644 index 90bf081..0000000 --- a/frsb_kac_new.tex +++ /dev/null @@ -1,955 +0,0 @@ -\documentclass[fleqn]{article} - -\usepackage{fullpage,amsmath,amssymb,latexsym,graphicx} -\usepackage{appendix} - -\begin{document} -\title{Full solution of the Kac--Rice problem for mean-field models.\\ -or Full solution for the counting of saddles of mean-field glass models} -\author{Jaron Kent-Dobias \& Jorge Kurchan} -\maketitle -\begin{abstract} - We derive the general solution for the computation of saddle points - of mean-field complex landscapes. The solution incorporates Parisi's solution for the ground state, as it should. -\end{abstract} -\section{Introduction} - -The computation of the number of metastable states of mean field spin glasses -goes back to the beginning of the field. Over forty years ago, -Bray and Moore \cite{Bray_1980_Metastable} attempted the first calculation for - the Sherrington--Kirkpatrick model, in a paper remarkable for being one of the first applications of a replica symmetry breaking scheme. As became clear when the actual ground-state of the model was computed by Parisi \cite{Parisi_1979_Infinite} with a different scheme, the Bray--Moore result - was not exact, and in fact the problem has been open ever since. -To this date the program of computing the number of saddles of a mean-field -glass has been only carried out for a small subset of models. -These include most notably the (pure) $p$-spin model ($p>2$) \cite{Rieger_1992_The, Crisanti_1995_Thouless-Anderson-Palmer}. -The problem of studying the critical points of these landscapes -has evolved into an active field in probability theory \cite{Auffinger_2012_Random, Auffinger_2013_Complexity, BenArous_2019_Geometry} - -In this paper we present what we argue is the general replica ansatz for the -computation of the number of saddles of generic mean-field models, including the Sherrington--Kirkpatrick model. It reproduces the Parisi result in the limit -of small temperature for the lowest states, as it should. - - -\section{The model} - -Here we consider, for definiteness, the mixed $p$-spin model, itself a particular case -of the `Toy Model' of M\'ezard and Parisi \cite{Mezard_1992_Manifolds} -\begin{equation} - H(s)=\sum_p\frac{a_p^{1/2}}{p!}\sum_{i_1\cdots i_p}J_{i_1\cdots i_p}s_{i_1}\cdots s_{i_p} -\end{equation} -for $\overline{J^2}=p!/2N^{p-1}$. Then -\begin{equation} - \overline{H(s_1)H(s_2)}=Nf\left(\frac{s_1\cdot s_2}N\right) -\end{equation} -for -\begin{equation} - f(q)=\frac12\sum_pa_pq^p -\end{equation} -Can be thought of as a model of generic gaussian functions on the sphere. -To constrain the model to the sphere, we use a Lagrange multiplier $\mu$, with the total energy being -\begin{equation} - H(s)+\frac\mu2(s\cdot s-N) -\end{equation} - -At any critical point, the hessian is -\begin{equation} - \operatorname{Hess}H=\partial\partial H+\mu I -\end{equation} -$\partial\partial H$ is a GOE matrix with variance -\begin{equation} - \overline{(\partial_i\partial_jH)^2}=\frac1Nf''(1)\delta_{ij} -\end{equation} -and therefore its spectrum is given by the Wigner semicircle with radius $\sqrt{4f''(1)}$, or -\begin{equation} - \rho(\lambda)=\frac1{\pi\sqrt{f''(1)}}\sqrt{4f''(1)-\lambda^2} -\end{equation} -and the spectrum of $\operatorname{Hess}H$ is this shifted by $\mu$, or $\rho(\lambda+\mu)$. - -The parameter $\mu$ fixes the spectrum of the hessian. By manipulating it, one -can decide to find the complexity of saddles of a certain macroscopic index, or -of minima with a certain harmonic stiffness. When $\mu$ is taken to be within -the range $\pm\sqrt{4f''(1)}=\pm\mu_m$, the critical points are constrained to have -index $\mathcal I=\frac12N(1-\mu/\mu_m)$. When $\mu>\mu_m$, the critical -points are minima whose sloppiest eigenvalue is $\mu-\mu_m$. Finally, -when $\mu=\mu_m$, the critical points are marginal minima. - - -The parameter $\mu$ fixes the spectrum of the hessian. When it is an integration variable, -and one restricts the domain of all integrations to compute saddles of a certain macroscopic index, or -of minima with a certain harmonic stiffness, its value is the `softest' mode that adapts to change the Hessian \cite{Fyodorov_2007_Replica}. When it is fixed, then the restriction of the index of saddles is `payed' by the realization of the eigenvalues of the Hessian, usually a -`harder' mode. - -{\tiny NOT SURE WORTHWHILE - -\subsection{What to expect?} - -In order to try to visualize what one should expect, consider two pure p-spin models, with -\begin{equation} - H = H_1 + H_2=\alpha_1 \sum_{ijk} J^1_{ijk} s_i s_j s_k + - \alpha_2 \sum_{ijk} J^2_{ijk} \bar s_i \bar s_j \bar s_k +\epsilon \sum_i s_i \bar s_i -\end{equation} -The complexity of the first and second systems in terms of $H_1$ and of $H_2$ -have, in the absence of coupling, the same dependence, but are stretched to one another: -\begin{equation} - \Sigma_1(H_1)= \Sigma_o(H_1/\alpha_1) \qquad ; \qquad \Sigma_2(H_2)= \Sigma_o(H_2/\alpha_2) -\end{equation} -Each system has a ground state energy $E_{gs}^{1,2}$, a threshold energy $E_{thres}^{1,2}$ (a well-defined notion, since we are considering pure p-spins), the corresponding limit values $X^{1,2}_{gs}=\left. \frac{d \Sigma_1}{dE_{1,2}}\right|_{E^{gs}_{12}}$ -and $X^{1,2}_{thres}=\left. \frac{d \Sigma_1}{dE_{1,2}}\right|_{E^{thres}_{12}}$ -Considering the cartesian product of both systems, we have, in terms of the total energy -$H=H_1+H_2$ three regimes: -\begin{itemize} -\item {\bf Unfrozen}: -\begin{eqnarray} -& & X_1 \equiv \frac{d \Sigma_1}{dE_1}= X_2 \equiv \frac{d \Sigma_2}{dE_2} - \end{eqnarray} -\item {\bf Semi-frozen} -As we go down in energy, one of the systems (say, the first) reaches its frozen phase, - the first system is thus concentrated in a few states of $O(1)$ energy, while the second is not, so that $X_1=X_1^{gs}> X_2$. The lowest energy is reached when systems are frozen. -\item {\bf Semi-threshold } As we go up from the unfrozen upwards in energy, -the second system reaches its threshold $X_2^{thres}$. At higher energies minima are extremely rare, so the minima of the second system remain stuck at its threshold for higher energies. - -\item{\bf Both systems reach their thresholds} There essentially no more minima above that. -\end{itemize} -Consider now two combined vectors $({\bf s},{\bf \hat s})$ and $({\bf s}',{\bf \hat s}')$ -chosen at the same energies.\\ - -$\bullet$ Their normalized overlap is close to one when both subsystems are frozen, -between zero and one in the semifrozen phase, and zero at all higher energies.\\ - -$\bullet$ In phases where one or both systems are stuck in their thresholds (and only in those), the -minima are exponentially subdominant with respect to saddles, because a saddle is found by releasing the constraint of staying on the threshold. - -} - -\section{Equilibrium} - -Here we review the equilibrium solution. \cite{Crisanti_1992_The, Crisanti_1993_The, Crisanti_2004_Spherical, Crisanti_2006_Spherical} -The free energy is well known to take the form -\begin{equation} - \beta F=-1-\log2\pi-\frac12\lim_{n\to0}\frac1n\left(\beta^2\sum_{ab}^nf(Q_{ab})+\log\det Q\right) -\end{equation} -which must be extremized over the matrix $Q$. When the solution is a Parisi matrix, this can also be written in a functional form. If $P(q)$ is the probability distribution for elements $q$ in a row of the matrix, then -\begin{equation} - \chi(q)=\int_1^qdq'\,\int_0^{q'}dq''\,P(q'') -\end{equation} -Since it is the double integral of a probability distribution, $\chi$ must be concave, monotonically decreasing and have $\chi(1)=0$, $\chi'(1)=1$. The free energy can be written as a functional over $\chi$ as -\begin{equation} - \beta F=-1-\log2\pi-\frac12\int dq\,\left(\beta^2f''(q)\chi(q)+\frac1{\chi(q)}\right) -\end{equation} - - -We insert a $k$ step RSB ansatz (the steps are standard, and are reviewed in -Appendix A), and obtain $q_0=0$ -\begin{equation} - \begin{aligned} - \beta F= - -1-\log2\pi - -\frac12\left\{\beta^2\left(f(1)+\sum_{i=0}^k(x_i-x_{i+1})f(q_i)\right) - +\frac1{x_1}\log\left[ - 1+\sum_{i=1}^{k}(x_i-x_{i+1})q_i - \right]\right.\\ - \left.+\sum_{j=1}^k(x_{j+1}^{-1}-x_j^{-1})\log\left[ - 1+\sum_{i=j+1}^{k}(x_i-x_{i+1})q_i-x_{j+1}q_j - \right] - \right\} - \end{aligned} -\end{equation} -The zero temperature limit is most easily obtained by putting $x_i=\tilde -x_ix_k$ and $x_k=\tilde\beta/\beta$, $q_k=1-z/\beta$, which ensures the $\tilde x_i$, -$\tilde\beta$, and $z$ have nontrivial limits. Inserting the ansatz and taking the limit -carefully treating the $k$th term in each sum separately from the rest, we get -\begin{equation} - \begin{aligned} - \lim_{\beta\to\infty}\tilde\beta F= - -\frac12z\tilde\beta f'(1)-\frac12\left\{\tilde\beta^2\left(f(1)+\sum_{i=0}^{k-1}(\tilde x_i-\tilde x_{i+1})f(q_i)\right) - +\frac1{\tilde x_1}\log\left[ - \tilde\beta z^{-1}\left(1+\sum_{i=1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i\right)+1 - \right]\right.\\ - \left.+\sum_{j=1}^{k-1}(\tilde x_{j+1}^{-1}-\tilde x_j^{-1})\log\left[ - \tilde\beta z^{-1}\left(1+\sum_{i=j+1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i-\tilde x_{j+1}q_j\right)+1 - \right] - \right\} - \end{aligned} -\end{equation} -This is a $k-1$ RSB ansatz with all eigenvalues scaled by $\tilde\beta z^{-1}$ and shifted by -1, with effective temperature $\tilde\beta$, and an extra term. This can be seen -more clearly by rewriting the result in terms of the matrix $\tilde Q$, a -$(k-1)$-RSB Parisi matrix built from the $\tilde x_1,\ldots,\tilde x_{k-1}$ and -$q_1,\ldots,q_{k-1}$, which gives -\begin{equation} \label{eq:ground.state.free.energy} - \lim_{\beta\to\infty}\tilde\beta F - =-\frac12z\tilde\beta f'(1)-\frac12\lim_{n\to0}\frac1n\left( - \tilde\beta^2\sum_{ab}^nf(\tilde Q_{ab})+\log\det(\tilde\beta z^{-1}\tilde Q+I) - \right) -\end{equation} -In the continuum case, this is -\begin{equation} \label{eq:ground.state.free.energy.cont} - \lim_{\beta\to\infty}\tilde\beta F - =-\frac12z\tilde\beta f'(1)-\frac12\int dq\left( - \tilde\beta^2f''(q)\tilde\chi(q)+\frac1{\tilde\chi(q)+\tilde\beta z^{-1}} - \right) -\end{equation} - -The zero temperature limit of the free energy loses one level of replica -symmetry breaking. Physically, this is a result of the fact that in $k$-RSB, -$q_k$ gives the overlap within a state, e.g., within the basin of a well inside -the energy landscape. At zero temperature, the measure is completely localized -on the bottom of the well, and therefore the overlap with each state becomes -one. We will see that the complexity of low-energy stationary points in -Kac--Rice is also given by a $(k-1)$-RSB anstaz. Heuristicall, this is because -each stationary point also has no width and therefore overlap one with itself. - - - -\section{Kac--Rice} - -\cite{Auffinger_2012_Random, BenArous_2019_Geometry} - -The stationary points of a function can be counted using the Kac--Rice formula, -which integrates a over the function's domain a $\delta$-function containing -the gradient multiplied by the absolute value of the determinant -\cite{Rice_1939_The, Kac_1943_On}. In addition, we insert a $\delta$-function -fixing the energy density $\epsilon$, giving the number of stationary points at -energy $\epsilon$ and radial reaction $\mu$ as -\begin{equation} - \mathcal N(\epsilon, \mu) - =\int ds\,\delta(N\epsilon-H(s))\delta(\partial H(s)+\mu s)|\det(\partial\partial H(s)+\mu I)| -\end{equation} -This number will typically be exponential in $N$. In order to find typical -counts when disorder is averaged, we will want to average its logarithm -instead, which is known as the complexity: -\begin{equation} - \Sigma(\epsilon,\mu)=\lim_{N\to\infty}\frac1N\overline{\log\mathcal N(\epsilon, \mu)} -\end{equation} -The radial reaction $\mu$, which acts like a kind of `mass' term, takes a -fixed value here, which means that the complexity is for a given energy -density and hessian spectrum. This will turn out to be important when we -discriminate between counting all solutions, or selecting those of a given -index, for example minima. The complexity of solutions without fixing $\mu$ is -given by maximizing the complexity as a function of $\mu$. - -If one averages over $\mathcal N$ and afterward takes its logarithm, one arrives at the annealed complexity -\begin{equation} - \Sigma_\mathrm a(\epsilon,\mu) - =\lim_{N\to\infty}\frac1N\log\overline{\mathcal N(\epsilon,\mu)} -\end{equation} -This has been previously computed for the mixed $p$-spin models \cite{BenArous_2019_Geometry}, with the result -\begin{equation} - \begin{aligned} - \Sigma_\mathrm a(\epsilon,\mu) - =-\frac{\epsilon^2(f'(1)+f''(1))+2\epsilon\mu f'(1)+f(1)\mu^2}{2f(1)(f'(1)+f''(1))-2f'(1)^2}-\frac12\log f'(1)\\ - +\operatorname{Re}\left[\frac\mu{\mu+\sqrt{\mu^2-4f''(1)}} - -\log\left(\frac1{2f''(1)}\left(\mu-\sqrt{\mu^2-4f''(1)}\right)\right) - \right] - \end{aligned} -\end{equation} -The annealed complexity is known to equal the actual (quenched) complexity in -circumstances where there is at most one level of RSB. This is the case for the -pure $p$-spin models, or for mixed models where $1/\sqrt{f''(q)}$ is a convex -function. However, it fails dramatically for models with higher replica -symmetry breaking. For instance, when $f(q)=\frac12(q^2+\frac1{16}q^4)$, the -anneal complexity predicts that minima vanish well before the dominant saddles, -a contradiction for any bounded function, as seen in Fig.~\ref{fig:frsb.complexity}. - -\subsection{The replicated problem} - -The replicated Kac--Rice formula was introduced by Ros et al.~\cite{Ros_2019_Complex}, and its -effective action for the mixed $p$-spin model has previously been computed by -Folena et al.~\cite{Folena_2020_Rethinking}. Here we review the derivation. - -In order to average the complexity over disorder properly, the logarithm must be dealt with. We use the standard replica trick, writing -\begin{equation} - \begin{aligned} - \log\mathcal N(\epsilon,\mu) - &=\lim_{n\to0}\frac\partial{\partial n}\mathcal N^n(\epsilon,\mu) \\ - &=\lim_{n\to0}\frac\partial{\partial n}\int\prod_a^n ds_a\,\delta(N\epsilon-H(s_a))\delta(\partial H(s_a)+\mu s_a)|\det(\partial\partial H(s_a)+\mu I)| - \end{aligned} -\end{equation} - -As noted by Bray and Dean \cite{Bray_2007_Statistics}, gradient and Hessian -are independent for a random Gaussian function, and the average over disorder -breaks into a product of two independent averages, one for the gradient factor -and one for the determinant. The integration of all variables, including the -disorder in the last factor, may be restricted to the domain such that the -matrix $\partial\partial H(s_a)-\mu I$ has a specified number of negative -eigenvalues (the index $\mathcal I$ of the saddle), (see Fyodorov -\cite{Fyodorov_2007_Replica} for a detailed discussion). In practice, we are -therefore able to write -\begin{equation} - \begin{aligned} - \Sigma(\epsilon, \mu) - &=\lim_{N\to\infty}\frac1N\lim_{n\to0}\frac\partial{\partial n}\int\left(\prod_a^nds_a\right)\,\overline{\prod_a^n \delta(N\epsilon-H(s_a))\delta(\partial H(s_a)+\mu s_a)} - \times - \overline{\prod_a^n |\det(\partial\partial H(s_a)+\mu I)|} - \end{aligned} -\end{equation} -To largest order in $N$, the average over the product of determinants factorizes into the product of averages, each of which is given by the same expression depending only on $\mu$: -\begin{equation} - \begin{aligned} - \mathcal D(\mu) - &=\frac1N\overline{\log|\det(\partial\partial H(s_a)+\mu I)|} - =\int d\lambda\,\rho(\lambda+\mu)\log|\lambda| \\ - &=\operatorname{Re}\left\{ - \frac12\left(1+\frac\mu{2f''(1)}\left(\mu-\sqrt{\mu^2-4f''(1)}\right)\right) - -\log\left(\frac1{2f''(1)}\left(\mu-\sqrt{\mu^2-4f''(1)}\right)\right) - \right\} - \end{aligned} -\end{equation} -The $\delta$-functions are treated by writing them in the Fourier basis, introducing auxiliary fields $\hat s_a$ and $\hat beta$, -\begin{equation} - \prod_a^n\delta(N\epsilon-H(s_a))\delta(\partial H(s_a)+\mu s_a) - =\int \frac{d\hat\beta}{2\pi}\prod_a^n\frac{d\hat s_a}{2\pi} - e^{\hat\beta(N\epsilon-H(s_a))+i\hat s_a\cdot(\partial H(s_a)+\mu s_a)} -\end{equation} -$\hat \beta$ is a parameter conjugate to the state energies, i.e. playing the -role of an inverse temperature for the metastable states. The average over disorder can now be taken, and since everything is Gaussian it gives -\begin{equation} - \begin{aligned} - \overline{ - \exp\left\{ - \sum_a^n(i\hat s_a\cdot\partial_a-\hat\beta)H(s_a) - \right\} - } - &=\exp\left\{ - \frac12\sum_{ab}^n - (i\hat s_a\cdot\partial_a-\hat\beta) - (i\hat s_b\cdot\partial_b-\hat\beta) - \overline{H(s_a)H(s_b)} - \right\} \\ - &=\exp\left\{ - \frac N2\sum_{ab}^n - (i\hat s_a\cdot\partial_a-\hat\beta) - (i\hat s_b\cdot\partial_b-\hat\beta) - f\left(\frac{s_a\cdot s_b}N\right) - \right\} \\ - &\hspace{-14em}=\exp\left\{ - \frac N2\sum_{ab}^n - \left[ - \hat\beta^2f\left(\frac{s_a\cdot s_b}N\right) - -2i\hat\beta\frac{\hat s_a\cdot s_b}Nf'\left(\frac{s_a\cdot s_b}N\right) - -\frac{\hat s_a\cdot \hat s_b}Nf'\left(\frac{s_a\cdot s_b}N\right) - +\left(i\frac{\hat s_a\cdot s_b}N\right)^2f''\left(\frac{s_a\cdot s_b}N\right) - \right] - \right\} - \end{aligned} -\end{equation} - -We introduce new fields -\begin{align} - Q_{ab}=\frac1Ns_a\cdot s_b && - R_{ab}=-i\frac1N\hat s_a\cdot s_b && - D_{ab}=\frac1N\hat s_a\cdot\hat s_b -\end{align} -$Q_{ab}$ is the overlap between spins belonging to different replicas. The -meaning of $R_{ab}$ is that of a response of replica $a$ to a linear field in -replica $b$: -\begin{equation} - R_{ab} = \frac 1 N \sum_i \overline{\frac{\delta s_i^a}{\delta h_i^b}} -\end{equation} -The $D$ may similarly be seen as the variation of the complexity with respect to a random field. - -By substituting these parameters into the expressions above and then making a change of variables in the integration from $s_a$ and $\hat s_a$ to these three matrices, we arrive at the form for the complexity -\begin{equation} - \begin{aligned} - &\Sigma(\epsilon,\mu) - =\mathcal D(\mu)+\hat\beta\epsilon+\\ - &\lim_{n\to0}\frac1n\left( - -\mu\sum_a^nR_{aa} - +\frac12\sum_{ab}\left[ - \hat\beta^2f(Q_{ab})+(2\hat\beta R_{ab}-D_{ab})f'(Q_{ab}) - +R_{ab}^2f''(Q_{ab}) - \right] - +\frac12\log\det\begin{bmatrix}Q&iR\\iR&D\end{bmatrix} - \right) - \end{aligned} -\end{equation} -where $\hat\beta$, $Q$, $R$ and $D$ must be evaluated at extrema of this -expression. With $Q$, $R$, and $D$ distinct replica matrices, this is -potentially quite challenging. - -{\bf Inserting the expression for $D$ in the action, and writing $R=R_d+\tilde R$ ($\tilde R$ is zero in the diagonal) , the linear term in $\tilde R$ is -$Tr \left[\tilde R (\hat \beta -R_d Q^{-1} f')\right]$ } - -\section{Replica ansatz} - -We shall make the following ansatz for the saddle point: -\begin{align}\label{ansatz} - Q_{ab}= \text{a Parisi matrix} && - R_{ab}=R_d \delta_{ab} && - D_{ab}= D_d \delta_{ab} -\end{align} -From what we have seen above, this means that replica $a$ is insensitive to -a small field applied to replica $b$ if $a \neq b$, a property related to ultrametricity. A similar situation happens in quantum replicated systems, -with time appearing only on the diagonal terms: see Appendix C for details. - -From its very definition, it is easy to see just perturbing the equations -with a field that $R_d$ is the trace of the inverse Hessian, as one expect indeed of a response. -\begin{equation} - R_d = \mathcal D'(\mu) -\end{equation} -Similarly, .... one shows that -\begin{equation} -D_d = \hat \beta R_d -\end{equation} - - -Is it a saddle? - -\begin{equation} - 0=\frac{\partial\Sigma}{\partial R} - =\lim_{n\to0}\frac1n\left(-\mu I+\hat\beta f'(Q)+R\odot f''(Q) - +(QD+R^2)^{-1}R - \right) -\end{equation} -\begin{equation} - 0=\frac{\partial\Sigma}{\partial D} - =\lim_{n\to0}\frac1n\left(-\frac12f'(Q) - +\frac12(QD+R^2)^{-1}Q - \right) -\end{equation} -\begin{equation} - D=f'(Q)^{-1}-RQ^{-1}R -\end{equation} -Diagonal anstaz requires that $f'(Q_{ab})^{-1}=R_d^2Q_{ab}^{-1}$ for $a\neq b$. -\begin{equation} - 0 - =\lim_{n\to0}\frac1n\left(-\mu I+\hat\beta f'(Q)+R\odot f''(Q) - +Q^{-1}Rf'(Q) - \right) -\end{equation} -Diagonal ansatz requires that -\begin{equation} - 0=-\mu I+\hat\beta f'(Q)+R_df''(1)I+R_dQ^{-1}f'(Q) -\end{equation} -or $\hat\beta f'(Q_{ab})=-R_d[Q^{-1}f'(Q)]_{ab}$ for $a\neq b$. -We also have from the first (before the diagonal ansatz) $Df'(Q)=I-RQ^{-1}Rf'(Q)$. Inserting the diagonal, we get $Q^{-1}f'(Q)=\frac1{R_d^2}(I-D_df'(Q))$, and -\begin{equation} - 0=-\mu I+\hat\beta f'(Q)+R_df''(1)I+\frac1{R_d}(I-D_df'(Q)) - =(R_df''(1)+R_d^{-1}-\mu)I+(\hat\beta-D_d/R_d)f'(Q) -\end{equation} - -Always true: -\begin{equation} - 0=-\mu R+\hat\beta Rf'(Q)+R(R\odot f''(Q))+I-Df'(Q) -\end{equation} -Insert $D=D_dI+\delta D$, $R=R_dI+\delta R$, and $D_d=\hat\beta R_d$ -\begin{equation} - 0=-\hat\beta \mu R_d I +\hat\beta R_df'(Q)+R_d^2f''(1)I+I-\hat\beta R_df'(Q) - -\mu\delta R -\end{equation} - -\begin{equation} - 0=(\hat\beta Q+R) f'(Q)+(R\odot f''(Q)-\mu I)Q -\end{equation} - -\begin{equation} - \begin{aligned} - &\Sigma(\epsilon,\mu) - =\frac12+\mathcal D(\mu)+\hat\beta\epsilon+\\ - &\lim_{n\to0}\frac1n\left( - -\frac12\mu\sum_a^nR_{aa} - +\frac12\sum_{ab}\left[ - \hat\beta^2f(Q_{ab})+\hat\beta R_{ab}f'(Q_{ab}) - \right] - +\frac12\log\det(f'(Q)^{-1}Q) - \right) - \end{aligned} -\end{equation} - - -\subsection{Solution} - - -Inserting the diagonal ansatz \eqref{ansatz} one gets -\begin{equation} \label{eq:diagonal.action} - \begin{aligned} - \Sigma(\epsilon,\mu) - =\mathcal D(\mu) - + - \hat\beta\epsilon-\mu R_d - +\frac12(2\hat\beta R_d-D_d)f'(1)+\frac12R_d^2f''(1)+\frac12\log R_d^2 - \\ - +\frac12\lim_{n\to0}\frac1n\left(\hat\beta^2\sum_{ab}f(Q_{ab})+\log\det((D_d/R_d^2)Q+I)\right) - \end{aligned} -\end{equation} -Using standard manipulations (Appendix B), one finds also a continuous version -\begin{equation} \label{eq:functional.action} - \begin{aligned} - \Sigma(\epsilon,\mu) - =\mathcal D(\mu) - + - \hat\beta\epsilon-\mu R_d - +\frac12(2\hat\beta R_d-D_d)f'(1)+\frac12R_d^2f''(1)+\frac12\log R_d^2 - \\ - +\frac12\int_0^1dq\,\left( - \hat\beta^2f''(q)\chi(q)+\frac1{\chi(q)+R_d^2/D_d} - \right) - \end{aligned} -\end{equation} -where $\chi(q)$ is defined with respect to $Q$ exactly as in the equilibrium case. -Note the close similarity of this action to the equilibrium replica one, at finite temperature. -\begin{equation} - \beta F=-\frac12\lim_{n\to0}\frac1n\left(\beta^2\sum_{ab}f(Q_{ab})+\log\det Q\right)-1-\log2\pi -\end{equation} - - -\subsubsection{Saddles} -\label{sec:counting.saddles} - -The dominant stationary points are given by maximizing the action with respect -to $\mu$. This gives -\begin{equation} \label{eq:mu.saddle} - 0=\frac{\partial\Sigma}{\partial\mu}=\mathcal D'(\mu)-R_d -\end{equation} -as expected. -To take the derivative, we must resolve the real part inside the definition of -$\mathcal D$. When saddles dominate, $\mu<\mu_m$, and -\begin{equation} - \mathcal D(\mu)=\frac12+\frac12\log f''(1)+\frac{\mu^2}{4f''(1)} -\end{equation} -It follows that the dominant saddles have $\mu=2f''(1)R_d$. Their index -is thus $\mathcal I=\frac12N(1-R_d\sqrt{f''(1)})$. If -$R_d\sqrt{f''(1)}>1$ then we were wrong to assume that saddles dominate, and -the most numerous saddles will be found just above $\mu=\mu_m$. - -\subsubsection{Minima} -\label{sec:counting.minima} - -When minima dominate, $\mu>\mu_m$ and all the roots inside $\mathcal D(\mu)$ are real. Therefore $\mathcal D(\mu)$ is given by its former expression with the real part dropped, and -\begin{equation} - \mathcal D'(\mu)=\frac{2}{\mu+\sqrt{\mu^2-4f''(1)}} -\end{equation} -\begin{equation} - \mu=\frac1{R_d}+R_df''(1) -\end{equation} - -\subsubsection{Recovering the equilibrium ground state} - -The ground state energy corresponds to that where the complexity of dominant -stationary points becomes zero. If the most common stationary points vanish, -then there cannot be any stationary points. In this section, we will show that -it reproduces the ground state produced by taking the zero-temperature limit in -the equilibrium case. - -Consider the extremum problem of \eqref{eq:diagonal.action} with respect to $R_d$ and $D_d$. This gives the equations -\begin{align} - 0 - &=\frac{\partial\Sigma}{\partial D_d} - =-\frac12f'(1)+\frac12\frac1{R_d^2}\lim_{n\to0}\frac1n\operatorname{Tr}((D_d/R_d^2)Q+I)^{-1}Q \label{eq:saddle.d}\\ - 0 - &=\frac{\partial\Sigma}{\partial R_d} - =-\mu+\hat\beta f'(1)+R_df''(1)+\frac1{R_d}-\frac{D_d}{R_d^3}\lim_{n\to0}\frac1n\operatorname{Tr}((D_d/R_d^2)Q+I)^{-1}Q \label{eq:saddle.r} -\end{align} -Adding $2(D_d/R_d)$ times \eqref{eq:saddle.d} to \eqref{eq:saddle.r} and multiplying by $R_d$ gives -\begin{equation} - 0=-R_d\mu+1+R_d^2f''(1)+f'(1)(R_d\hat\beta-D_d) -\end{equation} -At the ground state, minima will always dominate (even if marginal). We can -therefore use the optimal $\mu$ from \S\ref{sec:counting.minima}, which gives -\begin{equation} - 0=f'(1)(R_d\hat\beta-D_d) -\end{equation} -In order to satisfy this equation we must have -$D_d=R_d\hat\beta$. This relationship holds for the most common minima whenever -they dominante, including in the ground state. - -Substituting this into the action, and also substituting the optimal $\mu$ for -minima, and taking $\Sigma(\epsilon_0,\mu^*)=0$, gives -\begin{equation} - \hat\beta\epsilon_0 - =-\frac12R_d\hat\beta f'(1)-\frac12\lim_{n\to0}\frac1n\left( - \hat\beta^2\sum_{ab}^nf(Q_{ab}) - +\log\det(\hat\beta R_d^{-1} Q+I) - \right) -\end{equation} -which is precisely \eqref{eq:ground.state.free.energy} with $R_d=z$, -$\hat\beta=\tilde\beta$, and $Q=\tilde Q$. - -{\em We arrive at one of the main results of our paper: a $(k-1)$-RSB ansatz in -Kac--Rice will predict the correct ground state energy for a model whose -equilibrium state at small temperatures is $k$-RSB } Moreover, there is an -exact correspondance between the saddle parameters of each. If the equilibrium -is given by a Parisi matrix with parameters $x_1,\ldots,x_k$ and -$q_1,\ldots,q_k$, then the parameters $\hat\beta$, $R_d$, $D_d$, $\tilde -x_1,\ldots,\tilde x_{k-1}$, and $\tilde q_1,\ldots,\tilde q_{k-1}$ for the -complexity in the ground state are -\begin{align} - \hat\beta=\lim_{\beta\to\infty}\beta x_k - && - \tilde x_i=\lim_{\beta\to\infty}\frac{x_i}{x_k} - && - \tilde q_i=\lim_{\beta\to\infty}q_i - && - R_d=\lim_{\beta\to\infty}\beta(1-q_k) - && - D_d=\hat\beta R_d -\end{align} - -\section{Examples} - -\subsection{1RSB complexity} - -It is known that by choosing a covariance $f$ as the sum of polynomials with -well-separated powers, one develops 2RSB in equilibrium. This should correspond -to 1RSB in Kac--Rice. For this example, we take -\begin{equation} - f(q)=\frac12\left(q^3+\frac1{16}q^{16}\right) -\end{equation} -With this covariance, the model sees a RS to 1RSB transition at -$\beta_1=1.70615\ldots$ and a 1RSB to 2RSB transition at $\beta_2=6.02198\ldots$. At these points, the average energies are $\langle E\rangle_1=-0.906391\ldots$ and $\langle E\rangle_2=-1.19553\ldots$, and the ground state energy is $E_0=-1.2876055305\ldots$. - -In this model, the RS complexity gives an inconsistent answer for the -complexity of the ground state, predicting that the complexity of minima -vanishes at a higher energy than the complexity of saddles, with both at a -lower energy than the equilibrium ground state. The 1RSB complexity resolves -these problems, predicting the same ground state as equilibrium and that the -complexity of marginal minima (and therefore all saddles) vanishes at -$E_m=-1.2876055265\ldots$, which is very slightly greater than $E_0$. Saddles -become dominant over minima at a higher energy $E_s=-1.287605716\ldots$. -Finally, the 1RSB complexity transitions to a RS description at an energy -$E_1=-1.27135996\ldots$. All these complexities can be seen plotted in -Fig.~\ref{fig:2rsb.complexity}. - -All of the landmark energies associated with the complexity are a great deal -smaller than their equilibrium counterparts, e.g., comparing $E_1$ and $\langle -E\rangle_2$. - -\begin{figure} - \centering - \includegraphics{figs/316_complexity.pdf} - - \caption{ - Complexity of dominant saddles (blue), marginal minima (yellow), and - dominant minima (green) of the $3+16$ model. Solid lines show the result of - the 1RSB ansatz, while the dashed lines show that of a RS ansatz. The - complexity of marginal minima is always below that of dominant critical - points except at the red dot, where they are dominant. - The inset shows a region around the ground state and the fate of the RS solution. - } \label{fig:2rsb.complexity} -\end{figure} - -\begin{figure} - \centering - \begin{minipage}{0.7\textwidth} - \includegraphics{figs/316_comparison_q.pdf} - \hspace{1em} - \includegraphics{figs/316_comparison_x.pdf} \vspace{1em}\\ - \includegraphics{figs/316_comparison_b.pdf} - \hspace{1em} - \includegraphics{figs/316_comparison_R.pdf} - \end{minipage} - \includegraphics{figs/316_comparison_legend.pdf} - - \caption{ - Comparisons between the saddle parameters of the equilibrium solution to - the $3+16$ model (blue) and those of the complexity (yellow). Equilibrium - parameters are plotted as functions of the average energy $\langle - E\rangle=\partial_\beta(\beta F)$ and complexity parameters as functions of - fixed energy $E$. Solid lines show the result of a 2RSB ansatz, dashed - lines that of a 1RSB ansatz, and dotted lines that of a RS ansatz. All - paired parameters coincide at the ground state energy, as expected. - } \label{fig:2rsb.comparison} -\end{figure} - -\subsection{Full RSB complexity} - -\begin{figure} - \centering - \includegraphics{figs/24_complexity.pdf} - \caption{ - The complexity $\Sigma$ of the mixed $2+4$ spin model as a function of - distance $\Delta\epsilon=\epsilon-\epsilon_0$ of the ground state. The - solid blue line shows the complexity of dominant saddles given by the FRSB - ansatz, and the solid yellow line shows the complexity of marginal minima. - The dashed lines show the same for the annealed complexity. The inset shows - more detail around the ground state. - } \label{fig:frsb.complexity} -\end{figure} - -\begin{figure} - \centering - \includegraphics{figs/24_func.pdf} - \hspace{1em} - \includegraphics{figs/24_qmax.pdf} - - \caption{ - \textbf{Left:} The spectrum $\chi$ of the replica matrix in the complexity - of dominant saddles for the $2+4$ model at several energies. - \textbf{Right:} The cutoff $q_{\mathrm{max}}$ for the nonlinear part of the - spectrum as a function of energy $E$ for both dominant saddles and marginal - minima. The colored vertical lines show the energies that correspond to the - curves on the left. - } \label{fig:24.func} -\end{figure} - -\begin{figure} - \centering - \includegraphics{figs/24_comparison_b.pdf} - \hspace{1em} - \includegraphics{figs/24_comparison_Rd.pdf} - \raisebox{3em}{\includegraphics{figs/24_comparison_legend.pdf}} - - \caption{ - Comparisons between the saddle parameters of the equilibrium solution to - the $3+4$ model (black) and those of the complexity (blue and yellow). Equilibrium - parameters are plotted as functions of the average energy $\langle - E\rangle=\partial_\beta(\beta F)$ and complexity parameters as functions of - fixed energy $E$. Solid lines show the result of a FRSB ansatz and dashed - lines that of a RS ansatz. All paired parameters coincide at the ground - state energy, as expected. - } \label{fig:2rsb.comparison} -\end{figure} - -In the case where any FRSB is present, one must work with the functional form -of the complexity \eqref{eq:functional.action}, which must be extremized with -respect to $\chi$ under the conditions that $\chi$ is concave, monotonically -decreasing, and $\chi(1)=0$, $\chi'(1)=-1$. The annealed case is found by -taking $\chi(q)=1-q$, which satisfies all of these conditions. $k$-RSB is -produced by breaking $\chi$ into $k+1$ piecewise linear segments. - -Forget for the moment these tricky requirements. The function would then be -extremized by satisfying -\begin{equation} - 0=\frac{\delta\Sigma}{\delta\lambda(q)}=\frac12\hat\beta^2f''(q)-\frac12\frac1{(\lambda(q)+R_d^2/D_d)^2} -\end{equation} -which implies the solution -\begin{equation} - \lambda^*(q)=\frac1{\hat\beta}f''(q)^{-1/2}-\frac{R_d^2}{D_d} -\end{equation} -If $f''(q)^{-1/2}$ is not concave anywhere, there is little use of this -solution. However, if it is concave everywhere it may constitute a portion of -the full solution. - -We suppose that solutions are given by -\begin{equation} - \lambda(q)=\begin{cases} - \lambda^*(q) & q<q_\textrm{max} \\ - 1-q & q\geq q_\textrm{max} - \end{cases} -\end{equation} -Continuity requires that $1-q_\textrm{max}=\lambda^*(q_\textrm{max})$. - -Here a picture of $\chi$ vs $C$ or $X$ vs $C$ showing limits $q_{max}$, $x_{max}$ -for different energies and typical vs minima. - -\section{Ultrametricity rediscovered} - -TENTATIVE BUT INTERESTING - -The frozen phase for a given index ${\cal{I}}$ is the one for values of $\hat \beta> \hat \beta_{freeze}^{\cal{I}}$. - -[Jaron: does $\hat \beta^I_{freeze}$ have a relation to the largest $x$ of the ansatz? If so, it would give an interesting interpretation for everything] - - The complexity of that index is zero, and we are looking at the lowest saddles -in the problem, a question that to the best of our knowledge has not been discussed -in the Kac--Rice context -- for good reason, since the complexity - the original motivation - is zero. -However, our ansatz tells us something of the actual organization of the lowest saddles of each index in phase space. - - - -\section{Conclusion} -We have constructed a replica solution for the general problem of finding saddles of random mean-field landscapes, including systems -with many steps of RSB. -The main results of this paper are the ansatz (\ref{ansatz}) and the check that the lowest energy is the correct one obtained with the usual Parisi ansatz. -For systems with full RSB, we find that minima are, at all energy densities above the ground state, exponentially subdominant with respect to saddles. -The solution contains valuable geometric information that has yet to be -extracted in all detail. - -\paragraph{Funding information} -J K-D and J K are supported by the Simons Foundation Grant No. 454943. - -\begin{appendix} - -\section{RSB for the Gibbs-Boltzmann measure} - -\begin{equation} - \beta F=-\frac12\lim_{n\to0}\frac1n\left(\beta^2\sum_{ab}f(Q_{ab})+\log\det Q\right)-\frac12\log S_\infty -\end{equation} -$\log S_\infty=1+\log2\pi$. -\begin{align*} - \beta F= - -\frac12\log S_\infty - -\frac12\lim_{n\to0}\frac1n\left(\beta^2nf(1)+\beta^2\sum_{i=0}^kn(x_i-x_{i+1})f(q_i) - +\log\left[ - \frac{ - 1+\sum_{i=0}^k(x_i-x_{i+1})q_i - }{ - 1+\sum_{i=1}^k(x_i-x_{i+1})q_i-x_1q_0 - } - \right]\right.\\ - +\frac n{x_1}\log\left[ - 1+\sum_{i=1}^k(x_i-x_{i+1})q_i-x_1q_0 - \right]\\ - \left.+\sum_{j=1}^kn(x_{j+1}^{-1}-x_j^{-1})\log\left[ - 1+\sum_{i=j+1}^k(x_i-x_{i+1})q_i-x_{j+1}q_j - \right] -\right) -\end{align*} - -\begin{align*} - \lim_{n\to0}\frac1n - \log\left[ - \frac{ - 1+\sum_{i=0}^k(x_i-x_{i+1})q_i - }{ - 1+\sum_{i=1}^k(x_i-x_{i+1})q_i-x_1q_0 - } - \right] - &= - \lim_{n\to0}\frac1n - \log\left[ - \frac{ - 1+\sum_{i=0}^k(x_i-x_{i+1})q_i - }{ - 1+\sum_{i=0}^k(x_i-x_{i+1})q_i-nq_0 - } - \right] \\ - &=q_0\left(1+\sum_{i=0}^k(x_i-x_{i+1})q_i\right)^{-1} -\end{align*} - - -\begin{align*} - \beta F= - -\frac12\log S_\infty - -\frac12\left(\beta^2f(1)+\beta^2\sum_{i=0}^k(x_i-x_{i+1})f(q_i) - +q_0\left(1+\sum_{i=0}^k(x_i-x_{i+1})q_i\right)^{-1}\right. \\ - +\frac1{x_1}\log\left[ - 1+\sum_{i=1}^{k}(x_i-x_{i+1})q_i-x_1q_0 - \right]\\ - \left.+\sum_{j=1}^k(x_{j+1}^{-1}-x_j^{-1})\log\left[ - 1+\sum_{i=j+1}^{k}(x_i-x_{i+1})q_i-x_{j+1}q_j - \right] -\right) -\end{align*} -$q_0=0$ -\begin{align*} - \beta F= - -\frac12\log S_\infty - -\frac12\left(\beta^2f(1)+\beta^2\sum_{i=0}^k(x_i-x_{i+1})f(q_i) - +\frac1{x_1}\log\left[ - 1+\sum_{i=1}^{k}(x_i-x_{i+1})q_i - \right]\right.\\ - \left.+\sum_{j=1}^k(x_{j+1}^{-1}-x_j^{-1})\log\left[ - 1+\sum_{i=j+1}^{k}(x_i-x_{i+1})q_i-x_{j+1}q_j - \right] -\right) -\end{align*} -$x_i=\tilde x_ix_k$, $x_k=y/\beta$, $q_k=1-z/\beta$ -\begin{align*} - \beta F= - -\frac12\log S_\infty- - \frac12\left(\beta^2f(1)+\beta^2(y\beta^{-1}-1)f(1-z\beta^{-1})+y\beta\sum_{i=0}^{k-1}(\tilde x_i-\tilde x_{i+1})f(q_i)\right. \\ - +\frac\beta{\tilde x_1 y}\log\left[ - y\sum_{i=1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z-yz/\beta - \right]\\ - +\sum_{j=1}^{k-1}\frac\beta y(\tilde x_{j+1}^{-1}-\tilde x_j^{-1})\log\left[ - y\sum_{i=j+1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z-yz/\beta-y\tilde x_{j+1}q_j -\right]\\ - \left.-\frac\beta{\tilde x_1 y}\log\beta-\sum_{j=1}^{k-1}\frac\beta y(\tilde x_{j+1}^{-1}-\tilde x_j^{-1})\log\beta+(1-\beta y^{-1})\log\left[ - z/\beta - \right] -\right) -\end{align*} -\begin{align*} - \lim_{\beta\to\infty}F= - -\frac12\left(yf(1)+zf'(1)+y\sum_{i=0}^{k-1}(\tilde x_i-\tilde x_{i+1})f(q_i) - +\frac1{\tilde x_1 y}\log\left[ - y\sum_{i=1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z - \right]\right.\\ - \left.+\sum_{j=1}^{k-1}\frac1 y(\tilde x_{j+1}^{-1}-\tilde x_j^{-1})\log\left[ - y\sum_{i=j+1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z-y\tilde x_{j+1}q_j - \right] - -\frac1y\log z -\right) -\end{align*} -$F$ is a $k-1$ RSB ansatz with all eigenvalues scaled by $y$ and shifted by $z$. $\tilde x_0=0$ and $\tilde x_k=1$. - - - -\section{ RSB for the Kac--Rice integral} - - -\subsection{Solution} - - - -\begin{align*} - \lim_{n\to0}\frac1n\log\det(\hat\beta R_d^{-1} Q+I) - =x_1^{-1}\log\left(\hat\beta R_d^{-1}(1-\bar q_k)+1\right)+\int_{q_0^+}^{q_{k-1}}dq\,\mu(q)\log\left[\hat\beta R_d^{-1}\lambda(q)+1\right] -\end{align*} -where -\[ - \mu(q)=\frac{\partial x^{-1}(q)}{\partial q} -\] -Integrating by parts, -\begin{align*} - \lim_{n\to0}\frac1n\log\det(\hat\beta R_d^{-1} Q+I) - &=x_1^{-1}\log\left(\hat\beta R_d^{-1}(1-\bar q_k)+1\right)+\left[x^{-1}(q)\log[\hat\beta R_d^{-1}\lambda(q)+1]\right]_{q=q_0^+}^{q=q_{k-1}}-\frac{\hat\beta}{R_d}\int_{q_0^+}^{q_{k-1}}dq\,\frac{\lambda'(q)}{x(q)}\frac1{\hat\beta R_d^{-1}\lambda(q)+1}\\ - &=\log[\hat\beta R_d^{-1}\lambda(q_{k-1})+1]+\frac{\hat\beta}{R_d}\int_{q_0^+}^{q_{k-1}}dq\,\frac1{\hat\beta R_d^{-1}\lambda(q)+1} -\end{align*} - -\section{ A motivation for the ansatz} - - We may encode the original variables in a superspace variable: -\begin{equation} - \phi_a(1)= s_a + \bar\eta_a\theta_1+\bar\theta_1\eta_a + \hat s_a \bar \theta_1 \theta_1 -\end{equation}Here $\theta_a$, $\bar \theta_a$ are Grassmann variables, and we denote the full set of coordinates -in a compact form as $1= \theta_1 \overline\theta_1$, $d1= d\theta_1 d\overline\theta_1$, etc. -The correlations are encoded in -\begin{equation} -\begin{aligned} - \mathbb Q_{a,b}(1,2)&=\frac 1 N \phi_a(1)\cdot\phi_b (2) = -Q_{ab} -i\left[\bar\theta_1\theta_1+\bar\theta_2\theta_2\right] R_{ab} - +(\bar\theta_1\theta_2+\theta_1\bar\theta_2)F_{ab} - + \bar\theta_1\theta_1 \bar \theta_2 \theta_2 D_{ab} \\ -&+ \text{odd terms in the $\bar \theta,\theta$}~. -\end{aligned} -\label{Q12} -\end{equation} -\begin{equation} - \overline{\Sigma(\epsilon,\mu)} - =\hat\beta\epsilon\lim_{n\to0}\frac1n\left[ - \mu\int d1\sum_a^n\mathbb Q_{aa}(1,1) - +\int d2\,d1\,\frac12\sum_{ab}^n(1+\hat\beta\bar\theta_1\theta_1)f(\mathbb Q_{ab}(1,2))(1+\hat\beta\bar\theta_2\theta_2) - +\frac12\operatorname{sdet}\mathbb Q - \right] -\end{equation} -The odd and even fermion numbers decouple, so we can neglect all odd terms in $\theta,\bar{\theta}$. - -\cite{Annibale_2004_Coexistence} - -This encoding also works for dynamics, where the coordinates then read -$1= (\bar \theta, \theta, t)$, etc. The variables $\bar \theta \theta$ and $\bar \theta ' \theta'$ play -the role of `times' in a superspace treatment. We have a long experience of -making an ansatz for replicated quantum problems, which naturally involve a (Matsubara) time. The dependence on this time only holds for diagonal replica elements, a consequence of ultrametricity. The analogy strongly -suggests that only the diagonal ${\bf Q}_{aa}$ depend on the $\theta$'s. This boils down the ansatz \ref{ansatz}. -Not surprisingly, and for the same reason as in the quantum case, this ansatz closes, as we shall see.For example, consider the convolution: - -\begin{equation} - \begin{aligned} - \int d3\,\mathbb Q_1(1,3)\mathbb Q_2(3,2) - =\int d3\,( - Q_1 -i(\bar\theta_1\theta_1+\bar\theta_3\theta_3) R_1 - +(\bar\theta_1\theta_3+\theta_1\bar\theta_3)F_1 - + \bar\theta_1\theta_1 \bar \theta_3 \theta_3 D_1 - ) \\ ( - Q_2 -i(\bar\theta_3\theta_3+\bar\theta_2\theta_2) R_2 - +(\bar\theta_3\theta_2+\theta_3\bar\theta_2)F_2 - + \bar\theta_3\theta_3 \bar \theta_2 \theta_2 D_2 - ) \\ - =-i(Q_1R_2+R_1Q_2) - +Q_1D_2\bar\theta_2\theta_2+D_1Q_2\bar\theta_1\theta_1 - -i\bar\theta_1\theta_1\bar\theta_2\theta_2R_1D_2 - -i\bar\theta_1\theta_1\bar\theta_2\theta_2D_1R_2 - \end{aligned} -\end{equation} -\end{appendix} - - -\bibliographystyle{plain} -\bibliography{frsb_kac-rice} - - - - -\end{document} |