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1 files changed, 166 insertions, 196 deletions
diff --git a/2-point.tex b/2-point.tex index 2c61253..2fc35d4 100644 --- a/2-point.tex +++ b/2-point.tex @@ -429,18 +429,20 @@ It remains only to apply the doubled operators to $f$ and then evaluate the simp \subsection{Hubbard--Stratonovich} Having expanded this expression, we are left with an argument in the exponential which is a function of scalar products between the fields $\mathbf s$, $\hat{\mathbf s}$, $\pmb\sigma$, and $\hat{\pmb\sigma}$. We will change integration coordinates from these fields to matrix fields given by the scalar products, defined as -\begin{align} \label{eq:fields} - C^{00}_{ab}=\frac1N\pmb\sigma_a\cdot\pmb\sigma_b && - R^{00}_{ab}=-i\frac1N{\pmb\sigma}_a\cdot\hat{\pmb\sigma}_b && - D^{00}_{ab}=\frac1N\hat{\pmb\sigma}_a\cdot\hat{\pmb\sigma}_b \\ - C^{01}_{ab}=\frac1N\pmb\sigma_a\cdot\mathbf s_b && - R^{01}_{ab}=-i\frac1N{\pmb\sigma}_a\cdot\hat{\mathbf s}_b && - R^{10}_{ab}=-i\frac1N\hat{\pmb\sigma}_a\cdot{\mathbf s}_b && - D^{01}_{ab}=\frac1N\hat{\pmb\sigma}_a\cdot\hat{\mathbf s}_b \\ - C^{11}_{ab}=\frac1N\mathbf s_a\cdot\mathbf s_b && - R^{11}_{ab}=-i\frac1N{\mathbf s}_a\cdot\hat{\mathbf s}_b && - D^{11}_{ab}=\frac1N\hat{\mathbf s}_a\cdot\hat{\mathbf s}_b -\end{align} +\begin{equation} \label{eq:fields} + \begin{aligned} + C^{00}_{ab}=\frac1N\pmb\sigma_a\cdot\pmb\sigma_b && + R^{00}_{ab}=-i\frac1N{\pmb\sigma}_a\cdot\hat{\pmb\sigma}_b && + D^{00}_{ab}=\frac1N\hat{\pmb\sigma}_a\cdot\hat{\pmb\sigma}_b \\ + C^{01}_{ab}=\frac1N\pmb\sigma_a\cdot\mathbf s_b && + R^{01}_{ab}=-i\frac1N{\pmb\sigma}_a\cdot\hat{\mathbf s}_b && + R^{10}_{ab}=-i\frac1N\hat{\pmb\sigma}_a\cdot{\mathbf s}_b && + D^{01}_{ab}=\frac1N\hat{\pmb\sigma}_a\cdot\hat{\mathbf s}_b \\ + C^{11}_{ab}=\frac1N\mathbf s_a\cdot\mathbf s_b && + R^{11}_{ab}=-i\frac1N{\mathbf s}_a\cdot\hat{\mathbf s}_b && + D^{11}_{ab}=\frac1N\hat{\mathbf s}_a\cdot\hat{\mathbf s}_b + \end{aligned} +\end{equation} We insert into the integral the product of $\delta$ functions enforcing these definitions, integrated over the new matrix fields, which is equivalent to multiplying by one. Once this is done, the many scalar products appearing @@ -466,7 +468,7 @@ the resulting complexity is =\frac1N\lim_{n\to0}\lim_{m\to0}\frac\partial{\partial n}\int d\mathcal Q_{00}\,d\mathcal Q_{11}\,d\mathcal Q_{01}\,e^{Nm\mathcal S_0(\mathcal Q_{00})+Nn\mathcal S_1(\mathcal Q_{00},\mathcal Q_{11},\mathcal Q_{01})} \end{equation} where -\begin{equation} +\begin{equation} \label{eq:one-point.action} \begin{aligned} &\mathcal S_0(\mathcal Q_{00}) =\hat\beta_0E_0-r^{00}_\mathrm d\mu_0-\frac12\hat\mu_0(1-c^{00}_\mathrm d)+\mathcal D(\mu_0)\\ @@ -478,7 +480,7 @@ where \end{aligned} \end{equation} is the action for the ordinary, one-point complexity, and remainder is given by -\begin{equation} +\begin{equation} \label{eq:two-point.action} \begin{aligned} &\mathcal S(\mathcal Q_{00},\mathcal Q_{11},\mathcal Q_{01}) =\hat\beta_1E_1-r^{11}_\mathrm d\mu_1-\frac12\hat\mu_1(1-c^{11}_\mathrm d)+\mathcal D(\mu_1) \\ @@ -514,13 +516,13 @@ 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}\\ + &=-\frac{(E_0+\mu_0)f'(1)+E_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} \\ + &=\frac{\mu_0f(1)+E_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} + \frac{\mu_0f(1)+E_0f'(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2} \right)^2 \end{align} @@ -744,7 +746,7 @@ giving 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 \begin{equation} \begin{aligned} - F_H(\beta\mid\epsilon_1,\mu_1,\epsilon_2,\mu_2,q) + F_H(\beta\mid E_1,\mu_1,E_2,\mu_2,q) &=\int\frac{d\nu_H(\pmb\sigma,\varsigma\mid E_0,\mu_0)}{\int d\nu_H(\pmb\sigma',\varsigma'\mid E_0,\mu_0)}\,F_H(\beta\mid E_1,\mu_1,q,\pmb\sigma) \\ &=\lim_{m\to0}\int\left[\prod_{a=1}^m d\nu_H(\pmb\sigma_a,\varsigma_a\mid E_0,\mu_0)\right]\,F_H(\beta\mid E_1,\mu_1,q,\pmb\sigma_1) \end{aligned} @@ -752,7 +754,7 @@ again anticipating the use of replicas. Finally, the reference configuration $\p This formidable expression is now ready to be averaged over the disordered Hamiltonians $H$. Once averaged, the minimum eigenvalue of the conditioned Hessian is then given by twice the ground state energy, or \begin{equation} - \lambda_\text{min}=2\lim_{\beta\to\infty}\overline{F_H(\beta\mid\epsilon_1,\mu_1,\epsilon_2,\mu_2,q)} + \lambda_\text{min}=2\lim_{\beta\to\infty}\overline{F_H(\beta\mid E_1,\mu_1,E_2,\mu_2,q)} \end{equation} For this calculation, there are three different sets of replicated variables. Note that, as for the computation of the complexity, the $\pmb\sigma_1$ and @@ -813,11 +815,13 @@ which the Hessian is evaluated. \end{tikzpicture} \caption{ A sketch of the vectors involved in the calculation of the isolated - eigenvalue. All replicas $\mathbf x$ sit in an $N-2$ sphere corresponding - with the tangent plane (not to scale) of the first $\mathbf s$ replica. All of the - $\mathbf s$ replicas lie on the sphere, constrained to be at fixed overlap - $q$ with the first of the $\pmb\sigma$ replicas, the reference - configuration. All of the $\pmb\sigma$ replicas lie on the sphere. + eigenvalue. All replicas $\mathbf x$, which correspond with candidate + eigenvectors of the Hessian evaluated at $\mathbf s_1$, sit in an $N-2$ + sphere corresponding with the tangent plane (not to scale) of the first + $\mathbf s$ replica. All of the $\mathbf s$ replicas lie on the sphere, + constrained to be at fixed overlap $q$ with the first of the $\pmb\sigma$ + replicas, the reference configuration. All of the $\pmb\sigma$ replicas lie + on the sphere. } \end{figure} @@ -830,6 +834,10 @@ Using the same methodology as above, the disorder-dependant terms are captured i -\frac12 \delta(\mathbf t-\mathbf s_1)\beta\sum_c^\ell(\mathbf x_c\cdot\partial_{\mathbf t})^2 \end{equation} +that is applied to $H$ by integrating over $\mathbf t\in\mathbb R^N$. The +resulting expression for the integrand is produces dependencies only on the +scalar products in \eqref{eq:fields} and on the new scalar products involving +the tangent plane vectors $\mathbf x$, \begin{align} A_{ab}=\frac1N\mathbf x_a\cdot\mathbf x_b && @@ -841,23 +849,28 @@ Using the same methodology as above, the disorder-dependant terms are captured i && \hat X^1_{ab}=-i\frac1N\hat{\mathbf s}_a\cdot\mathbf x_b \end{align} - +Defining as before a block variable $\mathcal Q_x=(A,X^0,\hat X^0,X^1,\hat X^1)$ +and consolidating the previous block variables $\mathcal Q=(\mathcal Q_{00}, +\mathcal Q_{01},\mathcal Q_{11})$, we can write the minimum eigenvalue +schematically as \begin{equation} \lambda_\mathrm{min} =-2\lim_{\beta\to\infty} \lim_{\substack{\ell\to0\\m\to0\\n\to0}}\frac\partial{\partial\ell}\frac1{\beta N} - \int d\mathcal Q\,d\mathcal X\, + \int d\mathcal Q\,d\mathcal Q_x\, e^{N[ m\mathcal S_0(\mathcal Q_{00}) - +n\mathcal S_1(\mathcal Q_{11},\mathcal Q_{01}\mid\mathcal Q_{00}) - +\ell\mathcal S_x(\mathcal X\mid\mathcal Q_{00},\mathcal Q_{01},\mathcal Q_{11}) + +n\mathcal S(\mathcal Q_{11},\mathcal Q_{01}\mid\mathcal Q_{00}) + +\ell\mathcal S_x(\mathcal Q_x\mid\mathcal Q_{00},\mathcal Q_{01},\mathcal Q_{11}) ]} \end{equation} - -\begin{equation} +where $\mathcal S_0$ is given by \eqref{eq:one-point.action}, $\mathcal S$ is +given by \eqref{eq:two-point.action}, and the new action $\mathcal S_x$ is +given by +\begin{equation} \label{eq:action.eigenvalue} \begin{aligned} - \ell\mathcal S_x(\mathcal X\mid\mathcal Q) - =-\frac12\beta\mu+ + \ell\mathcal S_x(\mathcal Q_x\mid\mathcal Q) + =-\frac12\ell\beta\mu+ \frac12\beta\sum_b^\ell\bigg\{ \frac12\beta&f''(1)\sum_a^lA_{ab}^2\\ &+\sum_a^m\left[ @@ -886,7 +899,9 @@ Using the same methodology as above, the disorder-dependant terms are captured i \right) \end{aligned} \end{equation} +As usual in these quenched Franz--Parisi style computations, the saddle point expressions for the variables $\mathcal Q$ in the joint limits of $m$, $n$, and $\ell$ to zero are independent of $\mathcal Q_x$, and so these quantities take the same value they do for the two-point complexity that we computed above. The saddle point conditions for the variables $\mathcal Q_x$ are then fixed by extremizing with respect to the final action. +To evaluate this expression, we need a sensible ansatz for the variables $\mathcal Q_x$. The matrix $A$ we expect to be an ordinary hierarchical matrix, and since the model is a spherical 2-spin the finite but low temperature order will be {\oldstylenums1}\textsc{rsb}. The expected form of the $X$ matrices follows our reasoning for the 01 matrices of the previous section: namely, they should have constant rows and a column structure which matches that of the level of \textsc{rsb} order associated with the degrees of freedom that parameterize the columns. Since both the reference configurations and the constrained configurations have replica symmetric order, we expect \begin{align} X^0 = @@ -941,34 +956,16 @@ Using the same methodology as above, the disorder-dependant terms are captured i \hat x_1^1&\cdots&\hat x_1^1 \end{bmatrix} \end{align} -\begin{equation} - \begin{aligned} - \frac2\beta\lim_{\ell\to0}\mathcal S_x(\mathcal X\mid\mathcal Q) - &= - -\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 \\ - &- - \big(\hat\beta_1f''(q^{11}_0)+r^{11}_0f'''(q^{11}_0)\big)x_1^2 - -2f''(q^{11}_0)x_1\hat x_1^1 \\ - &+\lim_{\ell\to0}\frac1\ell\frac1\beta\log\det\left( - A- - \begin{bmatrix} - X^0\\\hat X^0\\X^1\\\hat X^1 - \end{bmatrix}^T - \begin{bmatrix} - C^{00}&iR^{00}&C^{01}&iR^{01}\\ - iR^{00}&D^{00}&iR^{10}&D^{01}\\ - (C^{01})^T&(iR^{10})^T&C^{11}&iR^{11}\\ - (iR^{01})^T&(D^{01})^T&iR^{11}&D^{11}\\ - \end{bmatrix}^{-1} - \begin{bmatrix} - X^0\\\hat X^0\\X^1\\\hat X^1 - \end{bmatrix} - \right) - \end{aligned} -\end{equation} - +Here, the lower block of the 0 matrices is zero, because these replicas have no +overlap with the reference or anything else. The first row of the $X^1$ matrix +needs to be zero because of the constraint that the tangent space vectors lie +in the tangent plane to the sphere, and therefore have $\mathbf x_a\cdot\mathbf +s_1=0$ for any $a$. This produces five parameters to deal with, which we +compile in the vector $\mathcal X=(x_0,\hat x_0,x_1\hat x_1^1,\hat x_1^0)$. + +Inserting this ansatz is straightforward in the first part of +\eqref{eq:action.eigenvalue}, but the term with $\log\det$ is more complicated. +We must invert the block matrix inside. Defining \begin{equation} \begin{bmatrix} C^{00}&iR^{00}&C^{01}&iR^{01}\\ @@ -978,12 +975,13 @@ Using the same methodology as above, the disorder-dependant terms are captured i \end{bmatrix}^{-1} = \begin{bmatrix} - A & B \\ - C & D + M_{11} & M_{12} \\ + M_{12}^T & M_{22} \end{bmatrix} \end{equation} -\begin{equation} - A= +where the blocks inside the inverse are given by +\begin{align} + M_{11}&= \left( \begin{bmatrix} C^{00}&iR^{00}\\iR^{00}&D^{00} @@ -1001,56 +999,9 @@ Using the same methodology as above, the disorder-dependant terms are captured i iR^{10}&D^{01} \end{bmatrix}^T \right)^{-1} -\end{equation} -\begin{align} - \hat c^{11}=\sum_{ij}^n[(C^{11}D^{11}+R^{11}R^{11})^{-1}D^{11}]_{ij}\\ - \hat r^{11}=\sum_{ij}^n[(C^{11}D^{11}+R^{11}R^{11})^{-1}R^{11}]_{ij}\\ - \hat d^{11}=\sum_{ij}^n[(C^{11}D^{11}+R^{11}R^{11})^{-1}D^{11}]_{ij} -\end{align} -Based on the structure of the 01 matrices established above, the second term inside the inverse is -\begin{equation} - \begin{aligned} - & \begin{bmatrix} - C^{01}&iR^{01}\\iR^{10}&D^{01} - \end{bmatrix} - \begin{bmatrix} - C^{11}&iR^{11}\\iR^{11}&D^{11} - \end{bmatrix}^{-1} - \begin{bmatrix} - C^{01}&iR^{01}\\iR^{10}&D^{01} - \end{bmatrix}^T \\ - &\qquad=\begin{bmatrix} - q^2\hat d^{11}+2qr_{10}\hat r^{11}-r_{10}^2\hat d^{11} - & - i\left[d_{01}(r_{10}\hat c^{11}-q\hat r^{11})+r_{01}(r_{10}\hat r^{11}+q\hat d^{11})\right] - \\ - i\left[d_{01}(r_{10}\hat c^{11}-q\hat r^{11})+r_{01}(r_{10}\hat r^{11}+q\hat d^{11})\right] - & - d_{01}^2\hat c^{11}+2r_{01}d_{01}\hat r^{11}-r_{01}^2\hat d^{11} - \end{bmatrix} - \end{aligned} -\end{equation} -where each block is proportional to the $m\times m$ matrix -\begin{equation} - \begin{bmatrix} - 1&0&\cdots&0\\ - 0&0&\cdots&0\\ - \vdots&\vdots&\ddots&\vdots\\ - 0&0&\cdots&0 - \end{bmatrix} -\end{equation} -which is \emph{not} a hierarchical matrix! But, all these new hat variables are proportional to $n$, and will vanish when the eventual limit is taken. So, the whole contribution is zero, and -\begin{equation} - A= - \begin{bmatrix} - C^{00}&iR^{00}\\iR^{00}&D^{00} - \end{bmatrix}^{-1} -\end{equation} -\begin{equation} - B=- - \begin{bmatrix} - C^{00}&iR^{00}\\iR^{00}&D^{00} - \end{bmatrix}^{-1} + \\ + M_{12}&=- + M_{11} \begin{bmatrix} C^{01}&iR^{01}\\ iR^{10}&D^{01} @@ -1058,22 +1009,8 @@ which is \emph{not} a hierarchical matrix! But, all these new hat variables are \begin{bmatrix} C^{11}&iR^{11}\\iR^{11}&D^{11} \end{bmatrix}^{-1} -\end{equation} -\begin{equation} - C=- - \begin{bmatrix} - C^{11}&iR^{11}\\iR^{11}&D^{11} - \end{bmatrix}^{-1} - \begin{bmatrix} - C^{01}&iR^{01}\\ - iR^{10}&D^{01} - \end{bmatrix}^T - \begin{bmatrix} - C^{00}&iR^{00}\\iR^{00}&D^{00} - \end{bmatrix}^{-1}=B^T -\end{equation} -\begin{equation} - D= + \\ + M_{22}&= \left( \begin{bmatrix} C^{11}&iR^{11}\\iR^{11}&D^{11} @@ -1091,42 +1028,50 @@ which is \emph{not} a hierarchical matrix! But, all these new hat variables are iR^{10}&D^{01} \end{bmatrix} \right)^{-1} -\end{equation} - -\begin{equation} +\end{align} +Here, $M_{22}$ is the inverse of the matrix already analyzed in as part of +\eqref{eq:two-point.action}. Following our discussion of the inverses of block +replica matrices above, and reasoning about their products with the rectangular +block constant matrices, things can be worked out from here. For instance, the +second term in $M_{11}$ contributes nothing once the appropriate limits are +taken, because each contribution is proportional to $n$. + +The contribution con be written as +\begin{equation} \label{eq:inverse.quadratic.form} \begin{bmatrix} X_0\\i\hat X_0 - \end{bmatrix}^TA + \end{bmatrix}^TM_{11} \begin{bmatrix} X_0\\i\hat X_0 \end{bmatrix} + 2\begin{bmatrix} X_0\\i\hat X_0 - \end{bmatrix}^TB + \end{bmatrix}^TM_{12} \begin{bmatrix} X_1\\i\hat X_1 \end{bmatrix} + \begin{bmatrix} X_1\\i\hat X_1 - \end{bmatrix}^TD + \end{bmatrix}^TM_{22} \begin{bmatrix} X_1\\i\hat X_1 \end{bmatrix} \end{equation} - - -\[ +and without too much reasoning one can see that the result is an $\ell\times\ell$ constant matrix. If $A$ is a replica matrix and $c$ is a constant, then +\begin{equation} \log\det(A-c)=\log\det A-\frac{c}{\sum_{i=0}^k(a_{i+1}-a_i)x_{i+1}} -\] +\end{equation} where $a_{k+1}=1$ and $x_{k+1}=1$. -So the basic form of the action is (for replica symmetric $A$) -\[ - -\mu+\frac14\beta^2f''(1)(1-a_0^2)+\frac12\log(1-a_0)+\frac12\frac{a_0}{1-a_0}+\frac12\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1^1\\\hat x_1^0\end{bmatrix}^T\left(\beta B-\frac1{1-a_0}C\right)\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1^1\\\hat x_1^0\end{bmatrix} -\] -for -\[ +The basic form of the action is (for replica symmetric $A$) +\begin{equation} + 2\mathcal S_x(\mathcal Q_x\mid\mathcal Q) + =-\beta\mu+\frac12\beta^2f''(1)(1-a_0^2)+\log(1-a_0)+\frac{a_0}{1-a_0}+\mathcal X^T\left(\beta B-\frac1{1-a_0}C\right)\mathcal X +\end{equation} +where the matrix $B$ comes from the $\mathcal X$-dependant parts of the first +lines of \eqref{eq:action.eigenvalue} and is given by +\begin{equation} B=\begin{bmatrix} \hat\beta_0f''(q)+r_{10}f'''(q)&f''(q)&0&0&0\\ f''(q)&0&0&0&0\\ @@ -1134,33 +1079,39 @@ for 0&0&-f''(q_0^{11})&0&0\\ 0&0&0&0&0 \end{bmatrix} -\] +\end{equation} +and where the matrix $C$ encodes the coefficients of the quadratic form +\eqref{eq:inverse.quadratic.form}, and is given element-wise by \begin{align} + \notag & C_{11}=d^{00}_\mathrm df'(1) - \quad + \qquad C_{12}=r^{00}_\mathrm df'(1) - \quad + \qquad C_{22}=-f'(1) \\ + \notag & C_{13} =\frac1{1-q_0}\left( (r^{11}_d-r^{11}_0)\left(r^{01}-q\frac{r^{11}_d-r^{11}_0}{1-q_0}\right)(f'(1)-f'(q_0))+qf'(1)d^{00}_d+r^{00}_d(r^{10}f'(1)+(r^{11}_d-r^{11}_0)f'(q)) \right) \\ + \notag & C_{15}=r^{00}_df'(q)+\left(r^{01}-q\frac{r^{11}_d-r^{11}_0}{1-q_0}\right)(f'(1)-f'(q_0)) - \quad + \qquad C_{14}=-C_{15} \\ & C_{23}=\frac1{1-q_0}\left((qr^{00}_d-r^{10})f'(1)-(r^{11}_d-r^{11}_0)f'(q)\right) - \quad + \qquad C_{24}=f'(q) - \quad + \qquad C_{25}=-C_{24} \\ + \notag & C_{33} = @@ -1171,72 +1122,91 @@ for \right)(f'(1)-f'(q_0)) -2\frac{qr^{00}-r^{10}}{1-q_0}f'(q) \right]\\ + \notag &\qquad-\frac{1-q^2}{(1-q_0)^2}-\frac{(r^{10}-qr^{00}_d)^2}{(1-q_0)^2}f'(1) \\ + \notag & C_{34} =-(qr^{01}-r^{11}_0)\frac{f'(1)-f'(q_0)}{1-q_0}-\frac{r^{11}_d-r^{11}_0}{1-q_0}\left( \frac{1-q^2}{1-q_0}(f'(1)-f'(q_0))-f'(q_0) \right)-f'(q)\frac{qr^{00}_d-r^{10}}{1-q_0} \\ + \notag & C_{35}=-C_{34}-\frac{r^{11}_d-r^{11}_0}{1-q_0}(f'(1)-f'(q_0)) - \quad + \qquad C_{44}=f'(1)-2f'(q_0) - \quad + \qquad C_{45}=f'(q_0) - \quad + \qquad C_{55}=-f'(1) + \notag +\end{align} +The saddle point conditions read +\begin{align} + 0=-\beta^2f''(1)a_0+\frac{a_0-\mathcal X^TC\mathcal X}{(1-a_0)^2} + && + 0=\bigg(\beta B-\frac1{1-a_0}C\bigg)\mathcal X \end{align} -Use $X$ for the big vector. Then -\[ - 0=-\beta^2f''(1)a_0+\frac{a_0-X^TCX}{(1-a_0)^2} -\] -\[ - 0=\bigg(\beta B-\frac1{1-a_0}C\bigg)X -\] -For a non-trivial solution, we require the following: change of basis in $X$ to diagonalize the matrix $\beta B-C/(1-a_0)$. Then, all of the new basis vectors of $X$ are zero except one, and the second equation is satisfied by tuning $a_0$ to make the coefficient zero. Finally, the first equation is satisfied by choice of the magnitude of this basis vector. -Suppose $a_0=1-1/(y\beta)$, $X\sim X+O(1/\beta)$. Then -\[ - 0=-f''(1)(1-(y\beta)^{-1})+y^2\left(1-(y\beta)^{-1}-X^TCX\right) -\] -For large $\beta$ -\[ - 0=-f''(1)+y^2(1-X^TCX) -\] -\[ - 0=(B-yC)X -\] -which gives -\[ - \mathcal S=-\frac12\beta\mu+\frac14\beta^2f''(1)\big(2(y\beta)^{-1}-(y\beta)^{-2}\big)-\frac12\log(y\beta)+\frac12y\beta(1-(y\beta)^{-1})+\frac12\beta X^TBX-\frac12y\beta X^TCX -\] -so -\[ - \lambda_\mathrm{min}=-2\lim_{\beta\to\infty}\frac{\partial\mathcal S}{\partial\beta} - =\mu-\left(y+\frac1yf''(1)\right)-X^T(B-yC)X +Note that the second of these conditions implies that the quadratic form in +$\mathcal X$ in the action vanishes at the saddle. + +We would like to take the limit of $\beta\to\infty$. As is usual in the +two-spin model, the appropriate limits of the order parameter is +$a_0=1-(y\beta)^{-1}$. Upon inserting this scaling and taking the limit, we +finally find +\begin{equation} + \lambda_\mathrm{min}=-2\lim_{\beta\to\infty}\frac1\beta\mathcal S_x =\mu-\left(y+\frac1yf''(1)\right) -\] -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 -\[ +\end{equation} +with associated saddle point conditions +\begin{align} + 0=-f''(1)+y^2(1-\mathcal X^TC\mathcal X) + && + 0=(B-yC)\mathcal X +\end{align} +The trivial solution, which gives the bottom of the semicircle, is for +$\mathcal X=0$. When this is satisfied, the first equation gives $y^2=f''(1)$, +and +\begin{equation} \lambda_\mathrm{min}=\mu-\sqrt{4f''(1)}=\mu-\mu_\mathrm m -\] -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. In this case, if the associated normalized eigenvector is $\hat X$, its magnitude is set by +\end{equation} +as expected. The nontrivial solutions have nonzero $\mathcal X$. The only way +to satisfy this with the saddle conditions is for $y$ such that one of the +eigenvalues of $B-yC$ is zero. In this case, if the normalized eigenvector +associated with the zero eigenvector is $\hat{\mathcal X}_0$, $\mathcal +X=\|\mathcal X_0\|\hat{\mathcal X}_0$ is a solution. The magnitude of the solution +is set by the other saddle point condition, namely \begin{equation} - \|X\|^2=\frac1{\hat X^TC\hat X}\left(1-\frac{f''(1)}{y^2}\right) + \|\mathcal X_0\|^2=\frac1{\hat{\mathcal X}_0^TC\hat{\mathcal X}_0}\left(1-\frac{f''(1)}{y^2}\right) +\end{equation} +In practice, we find that $\hat{\mathcal X}_0^TC\hat{\mathcal X}_0$ is positive +at the saddle point. Therefore, for the solution to make sense we must have +$y^2\geq f''(1)$. In practice, there is at most \emph{one} $y$ which produces a +zero eigenvalue of $B-yC$ and satisfies this inequality, so the solution seems +to be unique. + +In this solution, we simultaneously find the smallest eigenvalue and information +about the orientation of its associated eigenvector: namely, its overlap with +the tangent vector that points directly toward the reference spin. This is +directly related to $x_0$. This tangent vector is $\mathbf x_{0\leftarrow +1}=\frac1{1-q}\big(\pmb\sigma_0-q\mathbf s_a\big)$, which is normalized and +lies strictly in the tangent plane of $\mathbf s_a$. Then +\begin{equation} + \frac{\mathbf x_{0\leftarrow 1}\cdot\mathbf x_\mathrm{min}}N + =\frac{x_0}{1-q} \end{equation} -In practice, we find that $\hat X^TC\hat X$ is positive. Therefore, for the -solution to make sense we must have $y^2>f''(1)$. In practice, there is at most -\emph{one} $y$ which produces a zero eigenvalue of $B-yC$ and satisfies this -inequality, so the solution seems to be unique. \section{Franz--Parisi potential} \label{sec:franz-parisi} -\cite{Franz_1995_Recipes} +Here, we compute the Franz--Parisi potential for this model at zero +temperature, with respect to a reference configuration fixed to be a stationary +point of energy $E_0$ and stability $\mu_0$ as before. + +The Franz--Parisi +\cite{Franz_1995_Recipes, Franz_1998_EffectivePotential} \begin{equation} \label{eq:franz-parisi.definition} \beta V_\beta(q\mid E_0,\mu_0) |