From 3bd7865e4567b9e8c2b82d5a284e1e5e08993a99 Mon Sep 17 00:00:00 2001 From: Ian Jauslin Date: Wed, 26 Jul 2023 15:57:58 -0500 Subject: [PATCH] Update to v0.2 Added: Discussion of uniqueness of the solution of the Simple Equation Added: Exercise: prove \psi_0>=0 using |\psi_0| Added: Links in bibliography Removed: Commented blocks Fixed: Formatting --- Changelog | 12 +++ Jauslin_2023c.tex | 134 +++++++++++++++++++--------------- bibliography/bibliography.tex | 6 +- 3 files changed, 92 insertions(+), 60 deletions(-) diff --git a/Changelog b/Changelog index 49d79c2..091e29a 100644 --- a/Changelog +++ b/Changelog @@ -1,3 +1,15 @@ +v0.2: + * Added: Discussion of uniqueness of the solution of the Simple Equation + + * Added: Exercise: prove \psi_0>=0 using |\psi_0| + + * Added: Links in bibliography + + * Removed: Commented blocks + + * Fixed: Formatting + + v0.1: * Fixed: H_N is not compact, e^{-H_N} is. diff --git a/Jauslin_2023c.tex b/Jauslin_2023c.tex index 32beec0..c9fd15f 100644 --- a/Jauslin_2023c.tex +++ b/Jauslin_2023c.tex @@ -269,6 +269,7 @@ The microscopic state of a system of quantum particles is given by a wavefunctio \label{H} \end{equation} where $\Delta_i$ is the Laplacian with respect to $x_i$. +(Throughout this document, we choose units so that $\hbar=1$; we keep track of the mass because different authors use $m=1$ or $m=1/2$, so we keep it to avoid confusion.) This is the microscopic description of the system. Similarly to the classical case, to treat situations with a huge number of particles, we will approach the system statistically. In the classical case, we considered a probability distribution over all possible configurations, without fixing the number of particles. @@ -635,6 +636,7 @@ Let us consider a more realistic Hamiltonian: \label{Ham} \end{equation} where $v$ is the potential that accounts for the interaction between pairs of particles. +(Throughout this document, we choose units so that $\hbar=1$; we keep track of the mass because different authors use $m=1$ or $m=1/2$, so we keep it to avoid confusion.) This potential could be the Coulomb potential $1/|x|$ for charged particles, or the Lennard-Jones potential for interacting atoms, or something more realistic that accounts for the fine structure of atoms. Here, we will restrict our focus a bit and assume that $v$ is integrable ($v\in L_1(\mathbb R^3)$) and non-negative: $v(x)\geqslant 0$. In addition, we will only consider the zero-temperature state ($T=0$ that is $\beta=\infty$), in other words, we will be considering only the ground state of $H_N$, which is the eigenstate with lowest eigenvalue. @@ -688,9 +690,10 @@ The physical picture to have in mind is two particles coming closer together in By translation invariance, we can work in the center of mass reference frame, in which the scattering event can be seen as a single particle flying through a fixed potential. The Schr\"odinger equation for this process is \begin{equation} - -\frac1m\Delta\psi+v(x)\psi=i\hbar\partial_t\psi + -\frac1m\Delta\psi+v(x)\psi=i\partial_t\psi \end{equation} (note the absence of a $1/2$ factor in front of the Laplacian which comes from the fact that the effective mass in the center of mass frame is $\frac{m_1m_2}{m_1+m_2}=m/2$.) +(Recall that we chose units so that $\hbar=1$.) \bigskip \indent @@ -813,11 +816,13 @@ Alternatively, the scattering length can be computed as follows. \theo{Lemma}\label{lemma:scattering} If $v$ is spherically symmetric, compactly supported and integrable, then + \nopagebreakaftereq \begin{equation} a=\frac m{4\pi}\int dx\ v(x)\psi(x) . \end{equation} \endtheo +\restorepagebreakaftereq \bigskip \indent\underline{Proof}: @@ -907,11 +912,13 @@ Lee, Huang and Yang predicted the following low-density expansion for the ground \label{lhy_e} \end{equation} where $a$ is the scattering length of the potential, and + \nopagebreakaftereq \begin{equation} \lim_{\rho\to 0}\frac{o(\sqrt{\rho})}{\sqrt{\rho}}=0 . \end{equation} \endtheo +\restorepagebreakaftereq \bigskip The first two orders of this expansion thus only depend on the potential through its scattering length. @@ -1313,6 +1320,7 @@ We thus define an equation obtained from the Complete Equation in which we drop \label{bigeqL} \end{equation} \endtheo +\restorepagebreakaftereq \bigskip In practice, we have found that the predictions of the Big Equation are extremely close to those of the Complete Equation. @@ -1710,7 +1718,7 @@ We are almost there: we have proved the existence of the solution of the modifie To prove Theorem\-~\ref{theo:existence}, we need to prove that $e\mapsto\rho(e)$ can be inverted locally. \bigskip -\theo{Lemma} +\theo{Lemma}\label{lemma:surjective} The map $e\mapsto\rho(e)$ is continuous, and $\rho(0)=0$ and $\lim_{e\to\infty}\rho(e)=\infty$. Therefore $e\mapsto\rho(r)$ can be inverted locally. \endtheo @@ -1838,6 +1846,13 @@ But what of the uniqueness? . \end{equation} \qed +\bigskip + +\indent +Thus, taking the point of view in which the energy $e$ is fixed and $\rho$ is computed as a fuction of $e$, the solution is unique. +However, this does not imply that the solution to the problem in which $\rho$ is fixed is unique: the mapping $e\mapsto\rho(e)$ may not be injective (it is surjective by Lemma\-~\ref{lemma:surjective}). +To prove that it is injective, one could prove that $\rho$ is an increasing function of $e$ (physically, it should be: the higher the density is, the higher the energy should be because the potnetial is repulsive). +This has been proved for small and large values of $e$\-~\cite{CJL21}, but, in general, it is still an open problem. \subsection{Energy of the Simple Equation} \indent @@ -2324,24 +2339,6 @@ In this appendix we gather a few useful definitions and results from functional \endtheo \bigskip -%\theo{Lemma} -% If $A$ and $B$ are positivity preserving, $A$ and $A^{-1}-B$ are invertible, and $\|BA\|<1$ or $\|AB\|<1$, then $(A^{-1}-B)^{-1}$ is positivity preserving. -%\endtheo -%\bigskip -% -%\indent\underline{Proof}: -% We expand -% \begin{equation} -% (A^{-1}-B)^{-1} -% = -% \sum_{n=0}^\infty A(BA)^n -% = -% \sum_{n=0}^\infty (AB)^nA -% . -% \end{equation} -%\qed -%\bigskip - \theo{Lemma}\label{lemma:add_inv} If $\mathrm{spec}(A+B)>\epsilon>0$, and $e^{-tA}$ and $e^{-t B}$ are positivity preserving for all $t>0$, then for all $t>0$, $e^{-t(A+B)}$ and $(A+B)^{-1}$ are positivity preserving. \endtheo @@ -2374,20 +2371,6 @@ In this appendix we gather a few useful definitions and results from functional \qed \bigskip -%\theo{Lemma}\label{lemma:diff} -% If $B$ is positivity preserving and $(A+\eta B)^{-1}$ is positivity preserving for all $\eta$, then for any $f\in\mathcal B_1$ such that $f\geqslant 0$, $\eta\mapsto(A+\eta B)^{-1}f$ is monotone decreasing pointwise in $x$. -%\endtheo -%\bigskip -% -%\indent\underline{Proof}: -% We have -% \begin{equation} -% \partial_\eta(A+\eta B)^{-1}f -% =-(A+\eta B)^{-1}B(A+\eta B)^{-1}f\leqslant 0 -% . -% \end{equation} -%\qed - \theoname{Theorem}{Positivity of the heat kernel}\label{theo:heat} Given $t>0$, the operator $e^{t\Delta}$ from $L_2(\mathbb R^d)$ to $L_2(\mathbb R^d)$ is positivity preserving. \endtheo @@ -2452,30 +2435,6 @@ In this appendix we gather a few useful definitions and results from functional {\bf Remark}: In particular, taking the potential to be $\eta+v(x)$, this implies that $e^{-t(-\Delta+v(x)+\eta)}$ and $(-\Delta+v(x)+\eta)^{-1}$ are positivity preserving for any $\eta,v(x)\geqslant 0$. \bigskip -%\theo{Lemma}\label{lemma:yukawa} -% Given $\epsilon>0$, the operator $(-\Delta+\epsilon)^{-1}$ from $L_p(\mathbb R^3)$ to $W_{2,p}(\mathbb R^3)$ is positivity preserving. -%\endtheo -%\bigskip -% -%\indent\underline{Proof}: -% We apply lemma\-~\ref{lemma:add_inv} with $A^{-1}=-\Delta$ -% We have -% \begin{equation} -% (-\Delta+\epsilon)^{-1}f(x)=\frac1{4\pi}\int dx\ \frac{e^{-\sqrt\epsilon|x-y|}}{|x-y|}f(y) -% . -% \end{equation} -%\qed -%\bigskip -% -%{\bf Example}: -%This implies that if $v(x)\geqslant 0$, $v\in L_p(\mathbb R^3)$ and $\|v\|_p<\epsilon$, then $(-\Delta+\epsilon-v)^{-1}$ is positivity preserving. -%Indeed, take $A=(-\Delta+\epsilon)^{-1}$, and use the fact that -%\begin{equation} -% \|(-\Delta+\epsilon)^{-1}\|=\frac1\epsilon -% . -%\end{equation} -%\bigskip - \theo{Lemma}\label{lemma:conv} If $A$ is positivity preserving, $f\in L_1(\mathbb R^d)$ such that $f\geqslant 0$ and $\mathrm{spec}(A-f\ast)>\epsilon>0$ and $e^{-tA}$ is positivity preserving, then $(A-f\ast)^{-1}$ is positivity preserving. \endtheo @@ -2552,12 +2511,14 @@ We need to compute these up to order $V^{-2}$, because one of the terms in the e u_3(x-y):=u(x-y)+\frac{w_3(x-y)}V \label{u3} \end{equation} + \nopagebreakaftereq \begin{equation} w_3(x-y):=(1-u(x-y))\int dz\ u(x-z)u(y-z) . \label{w3} \end{equation} \endtheo +\restorepagebreakaftereq \bigskip \indent\underline{Proof}: @@ -2953,6 +2914,26 @@ Check that Theorem\-~\ref{theo:schrodinger} applies, so that $e^{-tH_N}$ is posi Use the Perron-Frobenius theorem (Theorem\-~\ref{theo:perron_frobenius}) to conclude. \bigskip +\problem\label{ex:nonneg} (solution on p.\-~\ref{sol:nonneg})\par +\smallskip +In this exercise, we will derive an alternate proof that $\psi_0\geqslant 0$ under the extra assumption that $v$ is continuous. +To do so, consider the energy of $\psi_0$: +\begin{equation} + \mathcal E(\psi_0):=\left<\psi_0\right|H_N\left|\psi_0\right> + =\int_{(\mathbb R/(L\mathbb Z))^{3N}} dx\ + \left( + -\frac1{2m}\psi_0^*(x)\Delta\psi(x) + +V(x)|\psi_0(x)|^2 + \right) +\end{equation} +where $\Delta$ is the Laplacian on $\mathbb R^{3N}$ and $V(x)\equiv\sum_{i