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diff --git a/frsb_kac_new.tex b/frsb_kac_new.tex index beeab96..abbbac6 100644 --- a/frsb_kac_new.tex +++ b/frsb_kac_new.tex @@ -117,7 +117,7 @@ minima are exponentially subdominant with respect to saddles, because a saddle i 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=-\frac12\lim_{n\to0}\frac1n\left(\beta^2\sum_{ab}^nf(Q_{ab})+\log\det Q\right)-1-\log2\pi + \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} @@ -125,41 +125,54 @@ which must be extremized over the matrix $Q$. When the solution is a Parisi matr \end{equation} Since it is the double integral of a probability distribution, $\chi$ must be convex, 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=-\frac12\int dq\,\left(\beta^2f''(q)\chi(q)+\frac1{\chi(q)}\right)-1-\log2\pi + \beta F=-1-\log2\pi-\frac12\int dq\,\left(\beta^2f''(q)\chi(q)+\frac1{\chi(q)}\right) \end{equation} -We are especially interested We insert a $k$ step RSB ansatz (the steps are standard, and are reviewed in Appendix A) , and obtain -$q_0=0$ -\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*} -The zero temperature limit is most easily obtained by putting $x_i=\tilde -x_ix_k$ and $x_k=y/\beta$, $q_k=1-z/\beta$, which ensures the $\tilde x_i$, -$y$, and $z$ have nontrivial limits. Inserting the ansatz and taking the limit -we get -\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 +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] - -\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 $\tilde x_k=1$. + \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} +$F$ 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} {\em We have lost one level of RSB because at zero temperature the states become points.} @@ -223,34 +236,34 @@ $\hat \beta$ is a parameter conjugate to the state energies, i.e. playing the ro \begin{equation} \prod_a^n\delta(N\epsilon-H(s_a))\delta(\partial H(s_a)+\mu s_a) - =\int \frac{d\hat\epsilon}{2\pi}\prod_a^n\frac{d\hat s_a}{2\pi} - e^{\hat\epsilon(N\epsilon-H(s_a))+i\hat s_a\cdot(\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} \begin{equation} \begin{aligned} \overline{ \exp\left\{ - \sum_a^n(i\hat s_a\cdot\partial_a-\hat\epsilon)H(s_a) + \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\epsilon) - (i\hat s_b\cdot\partial_b-\hat\epsilon) + (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\epsilon) - (i\hat s_b\cdot\partial_b-\hat\epsilon) + (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{-13em}\exp\left\{ \frac N2\sum_{ab}^n \left[ - \hat\epsilon^2f\left(\frac{s_a\cdot s_b}N\right) - -2i\hat\epsilon\frac{\hat s_a\cdot s_b}Nf'\left(\frac{s_a\cdot s_b}N\right) + \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] @@ -268,10 +281,10 @@ Introducing the parameters: \begin{equation} \begin{aligned} S - =\mathcal D(\mu)+\hat\epsilon\epsilon+\lim_{n\to0}\frac1n\left( + =\mathcal D(\mu)+\hat\beta\epsilon+\lim_{n\to0}\frac1n\left( -\mu\sum_a^nR_{aa} +\frac12\sum_{ab}\left[ - \hat\epsilon^2f(Q_{ab})+2\hat\epsilon R_{ab}f'(Q_{ab}) + \hat\beta^2f(Q_{ab})+2\hat\beta R_{ab}f'(Q_{ab}) -D_{ab}f'(Q_{ab})+R_{ab}^2f''(Q_{ab}) \right] \right. \\ \left. +\frac12\log\det\begin{bmatrix}Q&iR\\iR&D\end{bmatrix} @@ -298,10 +311,10 @@ Following the usual steps (Appendix B) we arrive at the replicated action: \begin{equation} \begin{aligned} S - =\mathcal D(\mu)+\hat\epsilon\epsilon+\lim_{n\to0}\frac1n\left( + =\mathcal D(\mu)+\hat\beta\epsilon+\lim_{n\to0}\frac1n\left( -\mu\sum_a^nR_{aa} +\frac12\sum_{ab}\left[ - \hat\epsilon^2f(Q_{ab})+2\hat\epsilon R_{ab}f'(Q_{ab}) + \hat\beta^2f(Q_{ab})+2\hat\beta R_{ab}f'(Q_{ab}) -D_{ab}f'(Q_{ab})+R_{ab}^2f''(Q_{ab}) \right] \right. \\ \left. +\frac12\log\det\begin{bmatrix}Q&iR\\iR&D\end{bmatrix} @@ -331,10 +344,10 @@ Insert the diagonal ansatz \cite{diagonal} one gets S =\mathcal D(\mu) + - \hat\epsilon\epsilon-\mu R_d - +\frac12(2\hat\epsilon R_d-D_d)f'(1)+\frac12R_d^2f''(1)+\frac12\log R_d^2 + \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\epsilon^2\sum_{ab}f(Q_{ab})+\log\det((D_d/R_d^2)Q+I)\right) + +\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 @@ -343,11 +356,11 @@ Using standard manipulations (Appendix B), one finds S =\mathcal D(\mu) + - \hat\epsilon\epsilon-\mu R_d - +\frac12(2\hat\epsilon R_d-D_d)f'(1)+\frac12R_d^2f''(1)+\frac12\log R_d^2 + \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\epsilon^2f''(q)\chi(q)+\frac1{\chi(q)+R_d^2/D_d} + \hat\beta^2f''(q)\chi(q)+\frac1{\chi(q)+R_d^2/D_d} \right) \end{aligned} \end{equation} @@ -399,35 +412,35 @@ Consider the extremum problem of \eqref{eq:diagonal.action} with respect to $R_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 S}{\partial R_d} - =-\mu+\hat\epsilon 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} + =-\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\epsilon-D_d) + 0=-R_d\mu+1+R_d^2f''(1)+f'(1)(R_d\hat\beta-D_d) \end{equation} There are two scenarios: one where the dominant stationary points in the vicinity of the ground state are minima, and one where they are saddles. In the case where the dominant stationary points are minima, we can use the optimal $\mu$ from \S\ref{sec:counting.minima}, which gives \begin{equation} - 0=f'(1)(R_d\hat\epsilon-D_d) + 0=f'(1)(R_d\hat\beta-D_d) \end{equation} -Therefore, in any situation where minima dominate, the optimal $\mu$ will have $R_d\hat\epsilon=D_d$. +Therefore, in any situation where minima dominate, the optimal $\mu$ will have $R_d\hat\beta=D_d$. When the dominant stationary points are saddles, we can use the $\mu$ from \S\ref{sec:counting.saddles}, which implies $R_d=\mu/2f''(1)$ and \begin{equation} - 0=1-\frac{\mu^2}{4f''(1)}+f'(1)(R_d\hat\epsilon-D_d) + 0=1-\frac{\mu^2}{4f''(1)}+f'(1)(R_d\hat\beta-D_d) \end{equation} -If saddles dominate all the way to the ground state, then they must become marginal minima at the ground state. Therefore at the ground state energy $\mu=\mu_m=\sqrt{4f''(1)}$, and once again $R_d\hat\epsilon-D_d=0$. +If saddles dominate all the way to the ground state, then they must become marginal minima at the ground state. Therefore at the ground state energy $\mu=\mu_m=\sqrt{4f''(1)}$, and once again $R_d\hat\beta-D_d=0$. -In any case, at the ground state $D_d=R_d\hat\epsilon$. Substuting this into the action, and also substituting the optimal $\mu$ for saddles or minima, gives +In any case, at the ground state $D_d=R_d\hat\beta$. Substuting this into the action, and also substituting the optimal $\mu$ for saddles or minima, and taking $\Sigma(\epsilon_0,\mu^*)=0$, gives \begin{equation} - \epsilon_0 - =-\lim_{n\to0}\frac1n\frac12\left(nR_df'(1)+\hat\epsilon\sum_{ab} - f(Q_{ab}) - +\frac1{\hat\epsilon}\log\det(\hat\epsilon R_d^{-1} Q+I) + \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 $R_d=z$ and $\hat\epsilon=y$. +which is precisely \eqref{eq:ground.state.free.energy} with $R_d=z$ and $\hat\beta=\tilde\beta$. {\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 } @@ -563,10 +576,6 @@ $x_i=\tilde x_ix_k$, $x_k=y/\beta$, $q_k=1-z/\beta$ \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 $\tilde x_k=1$. -\begin{equation} \label{eq:ground.state.free.energy} - \lim_{\beta\to\infty}F=-\lim_{n\to0}\frac1n\frac12\left(nzf'(1)+y\sum_{ab}f(\tilde Q_{ab})+\frac1y\log\det(yz^{-1}\tilde Q+I) - \right) -\end{equation} @@ -578,8 +587,8 @@ $F$ is a $k-1$ RSB ansatz with all eigenvalues scaled by $y$ and shifted by $z$. \begin{align*} - \lim_{n\to0}\frac1n\log\det(\hat\epsilon R_d^{-1} Q+I) - =x_1^{-1}\log\left(\hat\epsilon R_d^{-1}(1-\bar q_k)+1\right)+\int_{q_0^+}^{q_{k-1}}dq\,\mu(q)\log\left[\hat\epsilon R_d^{-1}\lambda(q)+1\right] + \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 \[ @@ -587,26 +596,26 @@ where \] Integrating by parts, \begin{align*} - \lim_{n\to0}\frac1n\log\det(\hat\epsilon R_d^{-1} Q+I) - &=x_1^{-1}\log\left(\hat\epsilon R_d^{-1}(1-\bar q_k)+1\right)+\left[x^{-1}(q)\log[\hat\epsilon R_d^{-1}\lambda(q)+1]\right]_{q=q_0^+}^{q=q_{k-1}}-\frac{\hat\epsilon}{R_d}\int_{q_0^+}^{q_{k-1}}dq\,\frac{\lambda'(q)}{x(q)}\frac1{\hat\epsilon R_d^{-1}\lambda(q)+1}\\ - &=\log[\hat\epsilon R_d^{-1}\lambda(q_{k-1})+1]+\frac{\hat\epsilon}{R_d}\int_{q_0^+}^{q_{k-1}}dq\,\frac1{\hat\epsilon R_d^{-1}\lambda(q)+1} + \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*} \begin{align*} \Sigma - =-\epsilon\hat\epsilon+ - \frac12\hat\epsilon R_df'(1) + =-\epsilon\hat\beta+ + \frac12\hat\beta R_df'(1) +\frac12\int_0^1dq\,\left[ - \hat\epsilon^2\lambda(q)f''(q) - +\frac1{\lambda(q)+R_d/\hat\epsilon} + \hat\beta^2\lambda(q)f''(q) + +\frac1{\lambda(q)+R_d/\hat\beta} \right] \end{align*} for $\lambda$ concave, monotonic, $\lambda(1)=0$, and $\lambda'(1)=-1$ \[ - 0=\frac{\delta\Sigma}{\delta\lambda(q)}=\frac12\hat\epsilon^2f''(q)-\frac12\frac1{(\lambda(q)+R_d/\hat\epsilon)^2} + 0=\frac{\delta\Sigma}{\delta\lambda(q)}=\frac12\hat\beta^2f''(q)-\frac12\frac1{(\lambda(q)+R_d/\hat\beta)^2} \] \[ - \lambda^*(q)=\frac1{\hat\epsilon}\left[f''(q)^{-1/2}-R_d\right] + \lambda^*(q)=\frac1{\hat\beta}\left[f''(q)^{-1/2}-R_d\right] \] We suppose that solutions are given by @@ -621,25 +630,25 @@ where $1-q$ guarantees the boundary conditions at $q=1$, and corresponds to the Inserting this into the complexity, we find \begin{align*} \Sigma - &=-\epsilon\hat\epsilon+\frac12\hat\epsilon R_df'(1) + &=-\epsilon\hat\beta+\frac12\hat\beta R_df'(1) +\frac12\int_0^{q^*}dq\left[ - \hat\epsilon(f''(q)^{-1/2}-R_d)f''(q)+\hat\epsilon f''(q)^{1/2} + \hat\beta(f''(q)^{-1/2}-R_d)f''(q)+\hat\beta f''(q)^{1/2} \right] +\frac12\int_{q^*}^1dq\left[ - \hat\epsilon^2(1-q)f''(q)+\frac1{q-1+R_d/\hat\epsilon} + \hat\beta^2(1-q)f''(q)+\frac1{q-1+R_d/\hat\beta} \right] \\ - &=-\epsilon\hat\epsilon+\frac12\hat\epsilon R_d\left[f'(1)-f'(q^*)\right] - +\hat\epsilon\int_0^{q^*}dq\,f''(q)^{1/2} - +\frac12\hat\epsilon^2\int_{q^*}^1dq\, + &=-\epsilon\hat\beta+\frac12\hat\beta R_d\left[f'(1)-f'(q^*)\right] + +\hat\beta\int_0^{q^*}dq\,f''(q)^{1/2} + +\frac12\hat\beta^2\int_{q^*}^1dq\, (1-q)f''(q) - -\log\left[1-(1-q^*)\hat\epsilon/R_d\right] + -\log\left[1-(1-q^*)\hat\beta/R_d\right] \end{align*} $R_d$ can be extremized now, with \[ R_d=\frac12\left( - (1-q^*)\hat\epsilon\pm\sqrt{ + (1-q^*)\hat\beta\pm\sqrt{ (1-q^*)\left( - (1-q^*)\hat\epsilon^2+8/[f'(1)-f'(q^*)] + (1-q^*)\hat\beta^2+8/[f'(1)-f'(q^*)] \right) } \right) @@ -652,11 +661,11 @@ This all is for $\mu=\mu^*$, which counts the dominant saddles. We can also coun and \begin{align*} \Sigma - =-\epsilon\hat\epsilon+ - \frac12\hat\epsilon R_df'(1) + =-\epsilon\hat\beta+ + \frac12\hat\beta R_df'(1) +\frac12\int_0^1dq\,\left[ - \hat\epsilon^2\lambda(q)f''(q) - +\frac1{\lambda(q)+R_d/\hat\epsilon} + \hat\beta^2\lambda(q)f''(q) + +\frac1{\lambda(q)+R_d/\hat\beta} \right]-\mu R_d+\frac12R_d^2f''(1)+\log R_d\\ +\operatorname{Re}\left\{\frac12\left(1+\frac\mu{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right)-\log\left(\frac1{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right)\right\} \end{align*} @@ -669,13 +678,13 @@ and \begin{aligned} \overline{\Sigma(\epsilon,\mu)} =\mathcal D(\mu) - +\operatorname*{extremum}_{\substack{R_d,D_d,\hat\epsilon\in\mathbb R\\\chi\in\Lambda}} + +\operatorname*{extremum}_{\substack{R_d,D_d,\hat\beta\in\mathbb R\\\chi\in\Lambda}} \left\{ - \hat\epsilon\epsilon+\mu R_d - +\frac12(2\hat\epsilon R_d-D_d)f'(1)+\frac12R_d^2f''(1) + \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 \right.\\\left. +\frac12\int_0^1dq\,\left( - \hat\epsilon^2f''(q)\chi(q)+\frac1{\chi(q)+R_d^2/D_d} + \hat\beta^2f''(q)\chi(q)+\frac1{\chi(q)+R_d^2/D_d} \right) \right\} \end{aligned} @@ -705,7 +714,7 @@ $x_1,\ldots,x_k$ and $q_1,\ldots,q_k$, then the parameters $\tilde x_1,\ldots,\tilde x_{k-1}$ and $\tilde q_1,\ldots,\tilde q_{k-1}$ for the complexity in the ground state are \begin{align} - \hat\epsilon=\lim_{\beta\to\infty}\beta x_k + \hat\beta=\lim_{\beta\to\infty}\beta x_k && \tilde x_i=\lim_{\beta\to\infty}\frac{x_i}{x_k} && @@ -713,7 +722,7 @@ complexity in the ground state are && R_d=\lim_{\beta\to\infty}\beta(1-q_k) && - D_d=R_d\hat\epsilon + D_d=R_d\hat\beta \end{align} \section{ A motivation for the ansatz} @@ -736,9 +745,9 @@ Q_{ab} -i\left[\bar\theta_1\theta_1+\bar\theta_2\theta_2\right] R_{ab} \end{equation} \begin{equation} \overline{\Sigma(\epsilon,\mu)} - =\hat\epsilon\epsilon\lim_{n\to0}\frac1n\left[ + =\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\epsilon\bar\theta_1\theta_1)f(\mathbb Q_{ab}(1,2))(1+\hat\epsilon\bar\theta_2\theta_2) + +\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} |