From 12fe57bbf4180ab9d9214591a222c3261434d826 Mon Sep 17 00:00:00 2001 From: Jaron Kent-Dobias Date: Tue, 8 Feb 2022 14:08:46 +0100 Subject: Corrections and elaboration on evaluating thimbles section. --- stokes.tex | 64 +++++++++++++++++++++++++++++++++++++++++--------------------- 1 file changed, 42 insertions(+), 22 deletions(-) diff --git a/stokes.tex b/stokes.tex index 343714e..a03f901 100644 --- a/stokes.tex +++ b/stokes.tex @@ -723,43 +723,63 @@ value at and around the critical point, where the real part of the action is by construction at its minimum on the thimble and the integrand is therefore largest. -We will make a change of coordinates $u(s)\in\mathbb R^N$ such that $\beta\mathcal -S(s)=\beta\mathcal S(s_\sigma)+\frac{|\beta|}2 u(s)^Tu(s)$ \emph{and} the direction of each $\partial u/\partial -s$ is along the direction of the contour. This is possible because the thimble... - - -Then -\begin{equation} -Z_\sigma(\beta)=e^{-\beta\mathcal S(s_\sigma)}\int du\,\det\frac{ds}{du}e^{-\frac{|\beta|}2u^Tu} +We will make a change of coordinates $u(s)\in\mathbb R^D$ such that +\begin{equation} \label{eq:thimble.integration.def} + \beta\mathcal S(s)=\beta\mathcal S(s_\sigma)+\frac{|\beta|}2 u(s)^Tu(s) \end{equation} +\emph{and} the direction of each $\partial u/\partial s$ is along the direction +of the contour. This is possible because, in the absence of any Stokes points, +the eigenvectors of the hessian at the critical point associated with positive +eigenvalues provide a basis for the thimble. The coordinates $u$ can be real +because the imaginary part of the action is constant on the thimble, and +therefore stays with the value it holds at the stationary point, and the real +part is at its minimum. + +The coordinates $u$ can be constructed implicitly in the close vicinity of the stationary point, with \begin{equation} - Z_\sigma(\beta)\simeq e^{-\beta\mathcal S(s_\sigma)}\left.\det\frac{ds}{du}\right|_{s=s_\sigma}\int du\,e^{-\frac{|\beta|}2u^Tu} - =e^{-\beta\mathcal S(s_\sigma)}\left.\det\frac{ds}{du}\right|_{s=s_\sigma}\left(\frac{2\pi}{|\beta|}\right)^{D/2} -\end{equation} -We are left with evaluating the determinant of the coordinate transformation. The eigenvectors of the hessian corresponding to positive eigenvalues provide a basis for the thimble globally, and locally they provide the coordinates $u$ by -\begin{equation} - s(u)=s_\sigma+\sum_{i=1}^{D/2}\sqrt{\frac{|\beta|}{\lambda^{(i)}}}v^{(i)}u_i+O(u^2) + s(u)=s_\sigma+\sum_{i=1}^{D}\sqrt{\frac{|\beta|}{\lambda^{(i)}}}v^{(i)}u_i+O(u^2) \end{equation} +where the sum is over pairs $(\lambda, v)$ which satisfy +\eqref{eq:generalized.eigenproblem} and have $\lambda>0$. It is straightforward +to confirm that these coordinates satisfy \eqref{eq:thimble.integration.def} asymptotically close to the stationary point, as \begin{eqnarray} \beta\mathcal S(s(u)) &=\beta\mathcal S(s_\sigma) - +\frac12(s(u)-s_\sigma)^T(\beta\operatorname{Hess}\mathcal S)(s(u)-s_\sigma)+O((s(u)-s_\sigma)^3) \\ + +\frac12(s(u)-s_\sigma)^T(\beta\operatorname{Hess}\mathcal S)(s(u)-s_\sigma)+\cdots \\ &=\beta\mathcal S(s_\sigma) - +\frac{|\beta|}2\sum_i\sum_j\frac{v^{(i)}_k}{\sqrt{\lambda^{(i)}}}(\beta\partial_k\partial_\ell\mathcal S)\frac{v^{(j)}_\ell}{\sqrt{\lambda^{(j)}}}u_iu_j+\cdots \\ + +\frac{|\beta|}2\sum_{ij}\frac{v^{(i)}_k}{\sqrt{\lambda^{(i)}}}(\beta[\operatorname{Hess}\mathcal S]_{k\ell})\frac{v^{(j)}_\ell}{\sqrt{\lambda^{(j)}}}u_iu_j+\cdots \\ &=\beta\mathcal S(s_\sigma) - +\frac{|\beta|}2\sum_i\sum_j\frac{v^{(i)}_k}{\sqrt{\lambda^{(i)}}}\frac{\lambda^{(j)}(v^{(j)}_k)^*}{\sqrt{\lambda^{(j)}}}u_iu_j+\cdots \\ + +\frac{|\beta|}2\sum_{ij}\frac{v^{(i)}_k}{\sqrt{\lambda^{(i)}}}\frac{\lambda^{(j)}(v^{(j)}_k)^*}{\sqrt{\lambda^{(j)}}}u_iu_j+\cdots \\ &=\beta\mathcal S(s_\sigma) - +\frac{|\beta|}2\sum_i\sum_j\frac{\sqrt{\lambda^{(j)}}}{\sqrt{\lambda^{(i)}}}\delta_{ij}u_iu_j+\cdots \\ + +\frac{|\beta|}2\sum_{ij}\frac{\sqrt{\lambda^{(j)}}}{\sqrt{\lambda^{(i)}}}\delta_{ij}u_iu_j+\cdots \\ &=\beta\mathcal S(s_\sigma) - +\frac{|\beta|}2\sum_iu_i^2+\cdots \\ + +\frac{|\beta|}2\sum_iu_i^2+\cdots \end{eqnarray} +The Jacobian of this transformation is \begin{equation} - \frac{\partial s_i}{\partial u_j}=\lambda_0^{(j)}v^{(j)}_i + \frac{\partial s_i}{\partial u_j}=\sqrt{\frac{|\beta|}{\lambda^{(j)}}}v^{(j)}_i + =\sqrt{\frac1{\lambda_0^{(j)}}}v^{(j)}_i \end{equation} -This is the product of a diagonal matrix of positive eigenvalues with the unitary matrix of their associated eigenvectors. Therefore, its determinant is +where $\lambda_0=\lambda/|\beta|$ is the eigenvalue of the hessian for +$\beta=1$. This is a $N\times D$ matrix. Since the eigenvectors $v$ are +mutually complex orthogonal, $v_i^{(j)}$ is nearly a unitary matrix, and it can +be made one by including $v^{(N)}=\widehat{\partial g}$, the unit normal to the +constraint manifold. This lets us write $U_{ij}=v_i^{(j)}$ an $N\times N$ +unitary matrix, whose determinant will give the correct phase for the measure. + +We therefore have \begin{equation} - \det\frac{\partial s}{\partial u}=\det_{ij}v_i^{(j)}\prod_{\lambda>0}\lambda_0 +Z_\sigma(\beta)=e^{-\beta\mathcal S(s_\sigma)}\int du\,\det\frac{ds}{du}e^{-\frac{|\beta|}2u^Tu} \end{equation} +Now we take the saddle point approximation, assuming the integral is dominated +by its value at the stationary point such that the determinant can be +approximated by its value at the stationary point. This gives +\begin{eqnarray} + Z_\sigma(\beta) + &\simeq e^{-\beta\mathcal S(s_\sigma)}\left.\det\frac{ds}{du}\right|_{s=s_\sigma}\int du\,e^{-\frac{|\beta|}2u^Tu} \\ + &=e^{-\beta\mathcal S(s_\sigma)}\left(\prod_i^D\sqrt{\frac1{\lambda_0^{(i)}}}\right)\det U\left(\frac{2\pi}{|\beta|}\right)^{D/2} +\end{eqnarray} +We are left with evaluating the determinant of the unitary part of the coordinate transformation. In circumstances you may be used to, only the absolute value of the determinant from the coordinate transformation is relevant, and since the determinant of a unitary matrix is always magnitude one, it doesn't enter the computation. -- cgit v1.2.3-70-g09d2