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-rw-r--r--2-point.bib19
-rw-r--r--2-point.tex362
2 files changed, 182 insertions, 199 deletions
diff --git a/2-point.bib b/2-point.bib
index 4c08cca..64adf31 100644
--- a/2-point.bib
+++ b/2-point.bib
@@ -115,13 +115,12 @@
author = {Folena, Giampaolo and Zamponi, Francesco},
title = {On weak ergodicity breaking in mean-field spin glasses},
year = {2023},
- month = {feb},
+ month = {2},
url = {http://arxiv.org/abs/2303.00026v2},
date = {2023-02-28T19:02:47Z},
eprint = {2303.00026v2},
eprintclass = {cond-mat.dis-nn},
- eprinttype = {arxiv},
- urldate = {2023-05-20T17:09:49.352227Z}
+ eprinttype = {arxiv}
}
@article{Franz_1995_Recipes,
@@ -138,6 +137,20 @@
doi = {10.1051/jp1:1995201}
}
+@article{Franz_1998_EffectivePotential,
+ author = {Franz, Silvio and Parisi, Giorgio},
+ title = {Effective potential in glassy systems: theory and simulations},
+ journal = {Physica A: Statistical Mechanics and its Applications},
+ publisher = {Elsevier BV},
+ year = {1998},
+ month = {12},
+ number = {3-4},
+ volume = {261},
+ pages = {317--339},
+ url = {https://doi.org/10.1016%2Fs0378-4371%2898%2900315-x},
+ doi = {10.1016/s0378-4371(98)00315-x}
+}
+
@article{Ikeda_2023_Bose-Einstein-like,
author = {Ikeda, Harukuni},
title = {{Bose}--{Einstein}-like condensation of deformed random matrix: a replica approach},
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)