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| -rw-r--r-- | stokes.tex | 60 |
1 files changed, 38 insertions, 22 deletions
@@ -680,11 +680,11 @@ These eigenvalues associated with the Takagi vectors can be further related to p complex symmetric matrix $\beta\operatorname{Hess}\mathcal S$. Suppose that $u\in\mathbb R^N$ satisfies the eigenvalue equation \begin{equation} - (\beta\operatorname{Hess} S)^\dagger(\beta\operatorname{Hess} S)u + (\beta\operatorname{Hess}\mathcal S)^\dagger(\beta\operatorname{Hess}\mathcal S)u =\sigma u \end{equation} for some positive real $\sigma$ (because $(\beta\operatorname{Hess} -S)^\dagger(\beta\operatorname{Hess} S)$ is self-adjoint). The square root of these +S)^\dagger(\beta\operatorname{Hess}\mathcal S)$ is self-adjoint). The square root of these numbers, $\sqrt{\sigma}$, are the definition of the \emph{singular values} of $\beta\operatorname{Hess}\mathcal S$. A direct relationship between these singular values and the eigenvalues of the real hessian immediately follows by taking a @@ -693,10 +693,10 @@ and writing \begin{equation} \eqalign{ \sigma v^\dagger u - &=v^\dagger(\beta\operatorname{Hess} S)^\dagger(\beta\operatorname{Hess} S)u - =(\beta\operatorname{Hess} Sv)^\dagger(\beta\operatorname{Hess} S)u\\ - &=(\lambda v^*)^\dagger(\beta\operatorname{Hess} S)u - =\lambda v^T(\beta\operatorname{Hess} S)u + &=v^\dagger(\beta\operatorname{Hess}\mathcal S)^\dagger(\beta\operatorname{Hess}\mathcal S)u + =(\beta\operatorname{Hess}\mathcal Sv)^\dagger(\beta\operatorname{Hess}\mathcal S)u\\ + &=(\lambda v^*)^\dagger(\beta\operatorname{Hess}\mathcal S)u + =\lambda v^T(\beta\operatorname{Hess}\mathcal S)u =\lambda^2 v^\dagger u } \end{equation} @@ -1080,8 +1080,8 @@ which is a sum of the `pure' actions \mathcal S_p(x)=\frac1{p!}\sum_{i_1\cdots i_p}J_{i_1\cdots i_p}x_{i_1}\cdots x_{i_p} \end{equation} The variables $x\in\mathbb R^N$ are constrained to lie on the sphere $x^2=N$, -making the model $D=N-1$ dimensional. The couplings $J$ form a totally symmetry -$p$-tensor whose components are normally distributed with zero mean and +making the model $D=N-1$ dimensional. The couplings $J$ form totally symmetric +$p$-tensors whose components are normally distributed with zero mean and variance $\overline{J^2}=p!/2N^{p-1}$. The `pure' $p$-spin models have $a_i=\delta_{ip}$, while the mixed have some more complicated coefficients $a$. @@ -1106,10 +1106,15 @@ The gradient of the constraint is simple with $\partial g=z$, and \eqref{eq:mult =\frac1Nz^T\partial\mathcal S =\sum_{p=2}^\infty a_pp\frac{\mathcal S_p(z)}N \end{equation} -which for the pure $p$-spin in particular implies that $\mu=p\epsilon$ for specific energy $\epsilon$. +which for the pure $p$-spin in particular implies that $\mu=p\epsilon$ for +specific energy $\epsilon$. \subsection{2-spin} +The pure 2-spin model is diagonalizable and therefore exactly solvable, and is +not complex in the sense of having a superextensive number of stationary points +in its action. However, it makes a good exercise of how the ideas of analytic +continuation will apply in the literally more complex case of the $p$-spin for $p>2$. The Hamiltonian of the pure $2$-spin model is defined by \begin{equation} \mathcal S_2(z)=\frac12z^TJz. @@ -1132,20 +1137,27 @@ zero. In the direction in question, \frac1N\frac12\lambda_iz_i^2=\epsilon=\frac12\lambda_i, \end{equation} whence $z_i=\pm\sqrt{N}$. Thus there are $2N$ stationary points, each -corresponding to $\pm$ the cardinal directions in the diagonalized basis. - -Suppose that two stationary points have the same imaginary energy; without loss -of generality, assume these are associated with the first and second -cardinal directions. Since the gradient is proportional to $z$, any components that are -zero at some time will be zero at all times. The dynamics for the components of -interest assuming all others are zero are +corresponding to $\pm$ the cardinal directions in the diagonalized basis. The +energy at each stationary point is real if the couplings are real, and +therefore there are no complex stationary points in the ordinary 2-spin model. + +Imagine for a moment that the coupling are allowed to be complex, giving the +stationary points of the 2-spin complex energies and therefore potentially +interesting thimble structure. Generically, the eigenvalues of the coupling +matrix will have distinct imaginary parts, and there will be no Stokes lines. +Suppose that two stationary points are brought to the same imaginary energy by +some continuation; without loss of generality, assume these are associated with +the first and second cardinal directions. Since the gradient is proportional to +$z$, any components that are zero at some time will be zero at all times. The +upward flow dynamics for the components of interest assuming all others are +zero are \begin{equation} \dot z_1 - =-z_1^*\left(d_1^*-\frac{d_1^*z_1^*z_1+d_2^*z_2^*z_2}{|z_1|^2+|z_2|^2}\right) - =-(d_1-d_2)^*z_1^*\frac{|z_2|^2}{|z_1|^2+|z_2|^2} + =-z_1^*\left(\lambda_1^*-\frac{\lambda_1^*z_1^*z_1+\lambda_2^*z_2^*z_2}{|z_1|^2+|z_2|^2}\right) + =-(\lambda_1-\lambda_2)^*z_1^*\frac{|z_2|^2}{|z_1|^2+|z_2|^2} \end{equation} -and the same for $z_2$ with all indices swapped. Since $\Delta=d_1-d_2$ is -real, if $z_1$ begins real it remains real, with the same for $z_2$. Since the +and the same for $z_2$ with all indices swapped. Since $\Delta=\lambda_1-\lambda_2$ is +real when the energies and therefore eigenvalues have the same imaginary part, if $z_1$ begins real it remains real, with the same for $z_2$. Since the stationary points are at real $z$, we make this restriction, and find \begin{equation} \frac d{dt}(z_1^2+z_2^2)=0 \qquad @@ -1174,12 +1186,16 @@ These trajectories are plotted in Fig.~\ref{fig:two-spin}. } \label{fig:two-spin} \end{figure} -Since they sit at the corners of a simplex, the stationary points of the 2-spin +Since they sit at the corners of a simplex, the distinct stationary points of the 2-spin model are all adjacent: no stationary point is separated from another by the separatrix of a third. This means that when the imaginary energies of two stationary points are brought to the same value, their surfaces of constant -imaginary energy join. +imaginary energy join. However, this is not true for stationary points related +by the symmetry $z\to-z$, as seen in Fig.~\ref{fig:3d.thimbles}. +Since the 2-spin model with real couplings does not have any stationary points +in the complex plane, analytic continuation can be made without any fear of +running into Stokes points. Starting from real, large $\beta$, making an infinitesimal phase rotation into the complex plane results in a decomposition into thimbles where that of each stationary point is necessary, because all stationary points are real. The curvature of the action at the stationary points lying at $z_i=\delta_{ik}$ in the $j$th direction is given by $\lambda_k-\lambda_j=2(\epsilon_k-\epsilon_k)$. Therefore the generic case of $N$ distinct eigenvalues of the coupling matrix leads to $2N$ stationary points with $N$ distinct energies, two at each index from $0$ to $N-1$. Starting with the expression \eqref{eq:real.thimble.partition.function} valid for the partition function contribution from the thimble of a real stationary point, we have \begin{equation} \eqalign{ Z(\beta) |