Ian Jauslin
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authorIan Jauslin <ian@jauslin.org>2020-02-06 02:44:20 -0500
committerIan Jauslin <ian@jauslin.org>2020-02-06 02:44:20 -0500
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+\documentclass{ian-presentation}
+
+\usepackage[hidelinks]{hyperref}
+\usepackage{graphicx}
+\usepackage{array}
+
+\begin{document}
+\pagestyle{empty}
+\hbox{}\vfil
+\bf\Large
+\hfil A simplified approach to interacting Bose gases\par
+\vfil
+\large
+\hfil Ian Jauslin
+\normalsize
+\vfil
+\hfil\rm joint with {\bf Eric Carlen}, {\bf Elliott H. Lieb}\par
+\vfil
+arXiv:{\tt \href{https://arxiv.org/abs/1912.04987}{1912.04987}}
+\hfill{\tt \href{http://ian.jauslin.org}{http://ian.jauslin.org}}
+\eject
+
+\setcounter{page}1
+\pagestyle{plain}
+\title{Lieb's simple equation}
+\vskip-10pt
+\begin{itemize}
+ \item \href{https://doi.org/10.1103/PhysRev.130.2518}{[Lieb, 1963]}: $x\in\mathbb R^3$
+ $$
+ (-\Delta+v(x)+4e)u(x)=v(x)+2e\rho\ u\ast u(x)
+ $$
+ $$
+ e=\frac\rho2\int dx\ (1-u(x))v(x)
+ $$
+ \item with
+ $$
+ \rho>0
+ ,\quad
+ v(x)\geqslant 0
+ ,\quad
+ v\in L_1\cap L_2(\mathbb R^3)
+ $$
+ \item and
+ $$
+ u\in L_1(\mathbb R^3)
+ ,\quad
+ u\ast u(x):=\int dy\ u(x-y)u(y)
+ $$
+\end{itemize}
+\vfill
+\eject
+
+
+\title{Interacting Bose gas}
+\vskip-10pt
+\begin{itemize}
+ \item State: symmetric wave functions in a finite box of volume $V$ with periodic boundary conditions:
+ $$
+ \psi(x_1,\cdots,x_N)
+ ,\quad
+ x_i\in \Lambda_d:=V^{\frac1d}\mathbb T^d
+ $$
+ \item Probability distribution: $|\psi(x_1,\cdots,x_N)|^2$
+ \item $N$-particle Hamiltonian:
+ $$
+ H_N:=
+ -\frac12\sum_{i=1}^N\Delta_i
+ +\sum_{1\leqslant i<j\leqslant N}v(x_i-x_j)
+ $$
+ with $v(x-y)\geqslant 0$ and $v\in L_1\cap L_{\frac d2+\epsilon}(\mathbb R^d)$.
+\end{itemize}
+\vfill
+\eject
+
+\title{Interacting Bose gas}
+\begin{itemize}
+ \item Ground state:
+ $$
+ H_N\psi_0=E_0\psi_0
+ ,\quad
+ E_0=\min\mathrm{spec}(H_N)
+ $$
+ \item Compute the ground state-energy per particle in the thermodynamic limit:
+ $$
+ e_0:=\lim_{\displaystyle\mathop{\scriptstyle V,N\to\infty}_{\frac NV=\rho}}\frac{E_0}N
+ $$
+\end{itemize}
+\vfill
+\eject
+
+\title{Energy}
+\begin{itemize}
+ \item Integrate $H_N\psi_0=E_0\psi_0$:
+ $$
+ \frac{E_0}N=\frac{N-1}{2V}\int dx\ v(x)g_2(0,x)
+ $$
+ \item $g_n$: marginal of $\psi_0$
+ $$
+ g_n(x_1,\cdots,x_n):=\frac{V^n\int dx_{n+1}\cdots dx_N\ \psi_0(x_1,\cdots,x_N)}{\int dx_1\cdots dx_N\ \psi_0(x_1,\cdots,x_N)}
+ $$
+ \item $\psi_0\geqslant 0$, so it can be thought of as a probability distribution
+\end{itemize}
+\vfill
+\eject
+
+\title{Hierarchy}
+\vskip-10pt
+\begin{itemize}
+ \item Equation for $g_2$: integrate $H_N\psi_0=E_0\psi_0$ with respect to $x_3,\cdots,x_N$:
+ $$
+ \begin{array}{>\displaystyle l}
+ -\frac12(\Delta_x+\Delta_y) g_2(x,y)
+ +\frac{N-2}V\int dz\ (v(x-z)+v(y-z))g_3(x,y,z)
+ \\[0.5cm]\hfill
+ +v(x-y)g_2(x,y)
+ +\frac{(N-2)(N-3)}{2V^2}\int dzdt\ v(z-t)g_4(x,y,z,t)
+ =E_0g_2(x,y)
+ \end{array}
+ $$
+ \item Factorization assumption:
+ $$
+ g_3(x_1,x_2,x_3)=g_2(x_1,x_2)g_2(x_1,x_3)g_2(x_2,x_3)
+ $$
+ $$
+ g_4(x_1,x_2,x_3,x_4)=\prod_{i<j}(g_2(x_i,x_j)+O(V^{-1}))
+ $$
+\end{itemize}
+\vfill
+\eject
+
+\title{Lieb's simple equation}
+\begin{itemize}
+ \item In the thermodynamic limit, after making a few additional assumptions, \href{https://doi.org/10.1103/PhysRev.130.2518}{[Lieb, 1963]}:
+ $$
+ (-\Delta+v(x)+4e)u(x)=v(x)+2e\rho\ u\ast u(x)
+ $$
+ $$
+ e=\frac\rho2\int dx\ (1-u(x))v(x)
+ $$
+ \item with $\rho:=\frac NV$
+ $$
+ g_2(x,y)=1-u(x-y)
+ $$
+\end{itemize}
+\vfill
+\eject
+
+\title{One dimension}
+\hfil\includegraphics[height=6cm,angle=0.4]{1d.png}
+\vfill
+\eject
+
+\title{Numerical solution for $v(x)=e^{-|x|}$ in 3 dimensions}
+\hfil\includegraphics[height=6cm]{erho_bare.pdf}
+\vfill
+\eject
+
+\title{Numerical solution for $v(x)=e^{-|x|}$ in 3 dimensions}
+\hfil\includegraphics[height=6cm]{erho.pdf}
+\vfill
+\eject
+
+\title{Asymptotics for the Bose gas}
+\vskip-10pt
+\begin{itemize}
+ \item {\bf Theorem} \href{https://doi.org/10.1103/PhysRev.130.2518}{[Lieb, 1963]}: if $\hat v(k):=\int dx\ e^{ikx}v(x)\geqslant 0$, then
+ $$
+ \frac{e_0}{\rho}\mathop{\longrightarrow}_{\rho\to\infty}\frac12\int dx\ v(x)
+ $$
+ \item {\bf Theorem} \href{https://doi.org/10.1103/PhysRevLett.80.2504}{[Lieb, Yngvason, 1998]}: in 3 dimensions ($a$: scattering length)
+ $$
+ \frac{e_0}{\rho}\mathop{\longrightarrow}_{\rho\to0}2\pi a
+ $$
+ \href{https://doi.org/10.1103/PhysRev.106.1135}{[Lee, Huang, Yang, 1957]}, \href{https://doi.org/10.1007/s10955-009-9792-3}{[Yau, Yin, 2009]}, \href{https://arxiv.org/abs/1904.06164}{[Fournais, Solovej, 2019]}:
+ $$
+ e_0=2\pi\rho a\left(1+\frac{128}{15\sqrt\pi}\sqrt{\rho a^3}+o(\sqrt\rho)\right)
+ $$
+\end{itemize}
+\vfill
+\eject
+
+\title{Comparison with Bose gas (Monte Carlo)}
+\vskip-10pt
+\hfil\includegraphics[width=8cm]{erho_holzmann.pdf}
+
+\hfil{\footnotesize Monte Carlo computation courtesy of M. Holzmann}
+\vfill
+\eject
+
+\title{Main Theorem}
+\vskip-5pt
+\begin{itemize}
+ \item If $v(x)\geqslant 0$ and $v\in L_1\cap L_{\frac d2+\epsilon}(\mathbb R^d)$, then Lieb's simple equation
+ $$
+ (-\Delta+4e+v)u=v+2e\rho u\ast u
+ ,\quad
+ e=\frac\rho2\int dx\ (1-u(x))v(x)
+ $$
+ has an integrable solution (proved constructively), with $0\leqslant u\leqslant 1$.
+
+ \item In 3 dimensions,
+ $$
+ e=2\pi\rho a\left(1+\frac{128}{15\sqrt\pi}\sqrt{\rho a^3}+o(\sqrt\rho)\right)
+ ,\quad
+ \frac{e}\rho\mathop{\longrightarrow}_{\rho\to\infty}\frac12\int dx\ v(x)
+ $$
+
+ \item If $v(x)\equiv v(|x|)$ is radially symmetric and decays exponentially,
+ $$
+ u(|x|)\mathop\sim_{|x|\to\infty}\frac\alpha{|x|^4}
+ $$
+\end{itemize}
+\vfill
+\eject
+
+\title{Existence of a solution (sketch)}
+\begin{itemize}
+ \item Change the point of view: fix $e>0$, and compute $\rho$ and $u$.
+
+ \item Iteration: $u_0=0$,
+ $$
+ (-\Delta+4e+v)u_n=v+2e\rho_{n-1}u_{n-1}\ast u_{n-1}
+ ,\quad
+ \rho_n:=\frac{2e}{\int dx\ (1-u_n(x))v(x)}
+ .
+ $$
+
+ \item Prove by induction that $u_n(x)$ is an increasing sequence, and is bounded $u_n(x)\leqslant 1$. It therefore converges to a function $u$, which is the unique integrable solution of the equation with $e$ fixed.
+
+ \item In addition, we prove that $e\mapsto\rho(e)$ is continuous, and $\rho(0)=0$ and $\rho(\infty)=\infty$, which allows us to compute solutions for the problem at fixed $\rho$. This does not imply the uniqueness of the solution.
+\end{itemize}
+\vfill
+\eject
+
+\title{Asymptotics (sketch)}
+\vskip-10pt
+\begin{itemize}
+ \item When $\rho$ is small, $e$ is small as well, so the solution $u$ is {\it not too far from} the solution of the scattering equation
+ $$
+ (-\Delta+v)\varphi=v
+ .
+ $$
+
+ \item The energy of $\varphi$ is
+ $$
+ \frac\rho 2\int dx\ (1-\varphi(x))v(x)=2\pi\rho a
+ $$
+ which yields the first term in the expansion.
+
+ \item The second term comes from approximating
+ $$
+ (1-u(x))v(x)\approx\frac{2e}\rho\delta(x)
+ $$
+ and solving the equation in Fourier space.
+\end{itemize}
+\vfill
+\eject
+
+\title{Decay (sketch)}
+$$
+ (-\Delta+4e+v)u=v+2e\rho u\ast u
+ ,\quad
+ e=\frac\rho2\int dx\ (1-u(x))v(x)
+$$
+\begin{itemize}
+ \item $u$ and $u\ast u$ have to decay at the same rate. This is a property of algebraically decaying functions.
+
+ \item (Remark: if $f(x)\geqslant f\ast f(x)$ and $\int f=\frac12$, then (morally) $f\sim|x|^{d+1}$.)
+
+ \item (Remark: $u_n(x)$ decays exponentially).
+
+ \item Proof is based on the Fourier transform and complex analysis.
+
+ \item Remark: The truncated two-point correlation function of the Bose gas is also conjectured to decay like $|x|^{-4}$.
+\end{itemize}
+\vfill
+\eject
+
+\title{Conclusion}
+\vfill
+\begin{itemize}
+ \item Simple equation: correct asymptotics for the ground state energy at both high and low densities.
+
+ \item Good approximation for intermediate densities (relative error of 5\%).
+
+ \item Intriguing non-linear PDE.
+
+ \item Proved existence, asymptotics, and decay rate.
+\end{itemize}
+\vfill
+\eject
+
+\title{Open problems and conjectures}
+\begin{itemize}
+ \item Monotonicity of $e\mapsto\rho(e)$, and concavity of $e\mapsto\frac1{\rho(e)}$ (would imply uniqueness).
+
+ \item Other observables? Condensate fraction? (in progress)
+
+ \item Crystallization?
+
+ \item {\it Lieb's simple equation} is actually a simplified version of a more complicated one: {\it Lieb's full equation}. Can it improve on the simple one? (in progress)
+\end{itemize}
+
+\end{document}