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Diffstat (limited to '2-point.tex')
| -rw-r--r-- | 2-point.tex | 63 |
1 files changed, 32 insertions, 31 deletions
diff --git a/2-point.tex b/2-point.tex index 3ab4b07..b6f13c4 100644 --- a/2-point.tex +++ b/2-point.tex @@ -140,7 +140,12 @@ where the function $f$ is defined from the coefficients $a_p$ by \end{equation} In this paper, we will focus on models with a replica symmetric complexity, but many of the intermediate formulae are valid for arbitrary replica symmetry -breakings. +breakings. At most {\oldstylenums1}\textsc{rsb} in the equilibrium is guaranteed if the function +$\chi(q)=f''(q)^{-1/2}$ is convex. The complexity at the ground state must +reflect the structure of equilibrium, and therefore be replica symmetric. We +are not aware of any result guaranteeing this for the complexity away from the +ground state, but we check that our replica-symmetric solutions satisfy the +saddle point equations at 1RSB. To enforce the spherical constraint at stationary points, we make use of a Lagrange multiplier $\omega$. This results in the extremal problem \begin{equation} @@ -408,7 +413,20 @@ is the action for the ordinary, one-point complexity, and remainder is given by Because of the structure of this problem in the twin limits of $m$ and $n$ to zero, the parameters $\mathcal Q_{00}$ can be evaluated at a saddle point of $\mathcal S_0$ alone. This means that these parameters will take the same value -they take when the ordinary, 1-point complexity is calculated. +they take when the ordinary, 1-point complexity is calculated. For a replica +symmetric complexity of the reference point, this results in +\begin{align} + \hat\beta_0 + &=-\frac{(\epsilon_0+\mu_0)f'(1)+\epsilon_0f''(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2}\\ + r_\mathrm d^{00} + &=\frac{\mu_0f(1)+\epsilon_0f'(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2} \\ + d_\mathrm d^{00} + &=\frac1{f'(1)} + -\left( + \frac{\mu_0f(1)+\epsilon_0f'(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2} + \right)^2 +\end{align} + In general, we except the $m\times n$ matrices $C^{01}$, $R^{01}$, $R^{10}$, and $D^{01}$ to have constant \emph{rows} of length $n$, with blocks of rows @@ -500,23 +518,6 @@ then the result is \end{equation} where each block is a constant $n\times n$ matrix. -We focus now on models whose equilibrium has at most one level of replica -symmetry breaking, which corresponds to a replica symmetric complexity. -For these models, the saddle point parameters for the reference stationary -point are well known, and take the values. -\begin{align} - \hat\beta_0 - &=-\frac{(\epsilon_0+\mu_0)f'(1)+\epsilon_0f''(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2}\\ - r_\mathrm d^{00} - &=\frac{\mu_0f(1)+\epsilon_0f'(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2} \\ - d_\mathrm d^{00} - &=\frac1{f'(1)} - -\left( - \frac{\mu_0f(1)+\epsilon_0f'(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2} - \right)^2 -\end{align} - -$(r^{00}_\mathrm d)^2+d^{00}_\mathrm d=1/f'(1)$ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case, the diagonals of the inverse block matrix from above are simple expressions: \begin{align} @@ -669,12 +670,12 @@ spherical constraint. The free energy of this model given a point $\mathbf s$ and a specific realization of the disordered Hamiltonian is \begin{equation} \begin{aligned} - \beta F_H(\beta\mid\mathbf s) + \beta F_H(\beta\mid\mathbf s,\omega) &=-\frac1N\log\left(\int d\mathbf x\,\delta(\mathbf x\cdot\mathbf s)\delta(\|\mathbf x\|^2-N)\exp\left\{ - -\beta\frac12\mathbf x^T\partial\partial H(\mathbf s)\mathbf x + -\beta\frac12\mathbf x^T\operatorname{Hess}H(\mathbf s,\omega)\mathbf x \right\}\right) \\ &=-\lim_{\ell\to0}\frac1N\frac\partial{\partial\ell}\int\left[\prod_{\alpha=1}^\ell d\mathbf x_\alpha\,\delta(\mathbf x_\alpha^T\mathbf s)\delta(N-\mathbf x_\alpha^T\mathbf x_\alpha)\exp\left\{ - -\beta\frac12\mathbf x^T_\alpha\partial\partial H(\mathbf s)\mathbf x_\alpha + -\beta\frac12\mathbf x^T_\alpha\big(\partial\partial H(\mathbf s)+\omega I\big)\mathbf x_\alpha \right\}\right] \end{aligned} \end{equation} @@ -688,8 +689,8 @@ such points, giving \begin{equation} \begin{aligned} F_H(\beta\mid E_1,\mu_1,q,\pmb\sigma) - &=\int\frac{d\nu_H(\mathbf s,\omega\mid E_1,\mu_1)\delta(Nq-\pmb\sigma\cdot\mathbf s)}{\int d\nu_H(\mathbf s',\omega'\mid E_1,\mu_1)\delta(Nq-\pmb\sigma\cdot\mathbf s')}F_H(\beta\mid\mathbf s) \\ - &=\lim_{n\to0}\int\left[\prod_{a=1}^nd\nu_H(\mathbf s_a,\omega_a\mid E_1,\mu_1)\,\delta(Nq-\pmb\sigma\cdot\mathbf s_a)\right]F_H(\beta\mid\mathbf s_1) + &=\int\frac{d\nu_H(\mathbf s,\omega\mid E_1,\mu_1)\delta(Nq-\pmb\sigma\cdot\mathbf s)}{\int d\nu_H(\mathbf s',\omega'\mid E_1,\mu_1)\delta(Nq-\pmb\sigma\cdot\mathbf s')}F_H(\beta\mid\mathbf s,\omega) \\ + &=\lim_{n\to0}\int\left[\prod_{a=1}^nd\nu_H(\mathbf s_a,\omega_a\mid E_1,\mu_1)\,\delta(Nq-\pmb\sigma\cdot\mathbf s_a)\right]F_H(\beta\mid\mathbf s_1,\omega_1) \end{aligned} \end{equation} again anticipating the use of replicas. Finally, the reference configuration $\pmb\sigma$ should itself be a stationary point of $H$ with its own energy density and stability. Averaging over these conditions gives @@ -808,7 +809,7 @@ Using the same methodology as above, the disorder-dependant terms are captured i \begin{equation} \begin{aligned} \ell\mathcal S_x(\mathcal X\mid\mathcal Q) - = + =-\frac12\beta\mu+ \frac12\beta\sum_b^\ell\bigg\{ \frac12\beta&f''(1)\sum_a^lA_{ab}^2\\ &+\sum_a^m\left[ @@ -896,7 +897,7 @@ Using the same methodology as above, the disorder-dependant terms are captured i \begin{aligned} \frac2\beta\lim_{\ell\to0}\mathcal S_x(\mathcal X\mid\mathcal Q) &= - \frac12\beta f''(1)(1-a_0^2) + -\mu+\frac12\beta f''(1)(1-a_0^2) +\big(\hat\beta_0f''(q)+r_{10}f'''(q)\big)x_0^2 +2f''(q)x_0\hat x_0 \\ &- @@ -1074,7 +1075,7 @@ which is \emph{not} a hierarchical matrix! But, all these new hat variables are where $a_{k+1}=1$ and $x_{k+1}=1$. So the basic form of the action is (for replica symmetric $A$) \[ - \frac14\beta^2f''(1)(1-a_0^2)+\frac12\log(1-a_0)+\frac12\frac{a_0}{1-a_0}+\frac12\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1^1\\\hat x_1^0\end{bmatrix}^T\left(\beta B-\frac1{1-a_0}C\right)\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1^1\\\hat x_1^0\end{bmatrix} + -\mu+\frac14\beta^2f''(1)(1-a_0^2)+\frac12\log(1-a_0)+\frac12\frac{a_0}{1-a_0}+\frac12\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1^1\\\hat x_1^0\end{bmatrix}^T\left(\beta B-\frac1{1-a_0}C\right)\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1^1\\\hat x_1^0\end{bmatrix} \] for \[ @@ -1160,17 +1161,17 @@ For large $\beta$ \] which gives \[ - \mathcal S=\frac14\beta^2f''(1)\big(2(y\beta)^{-1}-(y\beta)^{-2}\big)-\frac12\log(y\beta)+\frac12y\beta(1-(y\beta)^{-1})+\frac12\beta X^TBX-\frac12y\beta X^TCX + \mathcal S=-\frac12\beta\mu+\frac14\beta^2f''(1)\big(2(y\beta)^{-1}-(y\beta)^{-2}\big)-\frac12\log(y\beta)+\frac12y\beta(1-(y\beta)^{-1})+\frac12\beta X^TBX-\frac12y\beta X^TCX \] so \[ \lambda_\mathrm{min}=-2\lim_{\beta\to\infty}\frac{\partial\mathcal S}{\partial\beta} - =-\left(y+\frac1yf''(1)\right)-X^T(B-yC)X - =-\left(y+\frac1yf''(1)\right) + =\mu-\left(y+\frac1yf''(1)\right)-X^T(B-yC)X + =\mu-\left(y+\frac1yf''(1)\right) \] assuming the last equation is satisfied. The trivial solution, which gives the bottom of the semicircle, is for $X=0$, so the first equation is $y^2=f''(1)$, and \[ - \lambda_\mathrm{min}=-\sqrt{4f''(1)} + \lambda_\mathrm{min}=\mu-\sqrt{4f''(1)}=\mu-\mu_\mathrm m \] as expected. We need to first the nontrivial solutions with nonzero $X$, but because the coefficients are so nasty this will be a numeric problem... Specifically, we are good for $y$ where one of the eigenvalues of $B-yC$ is zero. |