Content-Type: multipart/mixed; boundary="-------------0408300356209" This is a multi-part message in MIME format. ---------------0408300356209 Content-Type: text/plain; name="04-266.comments" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="04-266.comments" AMS-Code: Primary 81V45, 35P20; Secondary 81Q20, 81R30 ---------------0408300356209 Content-Type: text/plain; name="04-266.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="04-266.keywords" Relativistic atoms, large-Z behaviour of energy, semi-classics ---------------0408300356209 Content-Type: application/x-tex; name="relTFmp-arc.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="relTFmp-arc.tex" \documentclass[reqno]{amsart} \NeedsTeXFormat{LaTeX2e}[1994/12/01] \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} %\usepackage{hyperref} \usepackage{enumerate} \usepackage{graphicx} \DeclareGraphicsExtensions{.ps,.eps,.eps.gz} \theoremstyle{plain} \newtheorem{thm}{Theorem}[section]{\bf}{\it} \newtheorem{prop}[thm]{Proposition}{\bf}{\it} \newtheorem{lemma}[thm]{Lemma}{\bf}{\it} \newtheorem{lem}[thm]{Lemma}{\bf}{\it} \newtheorem{conj}[thm]{Conjecture}{\bf}{\it} \newtheorem{cor}[thm]{Corollary}{\bf}{\it} \newtheorem{defn}[thm]{Definition}{\bf}{\rm} \newtheorem{rem}[thm]{Remark}{\it}{\rm} \newtheorem{remark}[thm]{Remark}{\it}{\rm} \newtheorem{obs}[thm]{Observation}{\it}{\rm} \newtheorem*{acknowledgement}{Acknowledgement} \newenvironment{pf}{\par\medskip\noindent\textit{Proof}:\,}{\hspace*{\fill}\qed\medskip\par\noindent} \newenvironment{pf*}[1]{\par\medskip\noindent\textit{#1}\,:}{\hspace*{\fill}\qed\medskip\par\noindent} \numberwithin{equation}{section} %\numberwithin{theorem}{section} \newcommand{\sphere}{{\mathbb S}} \newcommand{\supp}{\operatorname{supp}} \newcommand{\Span}{\operatorname{Span}} \newcommand{\dist}{{\operatorname{dist}}} \newcommand{\Vol}{{\operatorname{Vol}}} \newcommand{\Res}{{\operatorname{Res}}} \newcommand{\Ran}{{\operatorname{Ran}}} \newcommand{\Tr}{{\operatorname{Tr}}} \newcommand{\N}{{\mathbb N}} \newcommand{\R}{{\mathbb R}} \newcommand{\C}{{\mathbb C}} \begin{document} \title{The large - $Z$ behaviour of pseudo-relativistic atoms} \author{Thomas \O stergaard S\o rensen} \address{Mathematisches Institut, Universit\"at M\"unchen, Theresienstra\ss e 39, D-80333 {Munich}, Germany} \address[On leave from] {Department of Mathematical Sciences, Aalborg University, Fredrik Bajers Vej 7G, DK-9220 Aalborg East, Denmark} \address[]{} \email{sorensen@mathematik.uni-muenchen.de} \renewcommand{\datename}{} \date{\today} \thanks{\copyright\ 2004 by the author. This article may be reproduced in its entirety for non-commercial purposes.} \keywords{Relativistic atoms, large - $Z$ behaviour of energy, semi-classics} \subjclass{Primary 81V45, 35P20; Secondary 81Q20, 81R30} \renewcommand{\subjclassname}{\textup{2000} Mathematics Subject Classification} \begin{abstract} In this paper we study the large - $Z$ behaviour of the ground state energy of atoms with electrons having relativistic kinetic energy $\sqrt{p^{2}c^{2}+m^{2}c^{4}}-mc^{2}$. We prove that to leading order in $Z$ the energy is the same as in the non-relativistic case, given by (non-relativistic) Thomas-Fermi theory. For the problem to make sense, we keep the product $Z\alpha$ fixed (here $\alpha$ is Sommerfeld's fine structure constant), and smaller than, or equal to, $2/\pi$, which means that as $Z$ tends to infinity, $\alpha$ tends to zero. \end{abstract} \maketitle \section{Introduction and results} \label{intro} As a model for a relativistic atom with nuclear charge $Z$ and $N$ electrons, we consider the operator \begin{eqnarray*} H_{\text{\it{rel}}}= \sum_{i=1}^{N}\left\{\sqrt{ -\alpha^{-2}\Delta_{i}+\alpha^{-4}} - \alpha^{-2} - \frac{Z}{|x_{i}|}\right\} + \sum_{1\leq i0\) be fixed and let $H_{\text{\it{rel}}}$ and $E_{N=\lambda Z}(Z,\delta)$ be as above. Then \begin{equation} \label{eq:TF} E_{\lambda Z}(Z,\delta)= {}-C_{\text{TF}}(\lambda)Z^{7/3} + \text{o}(Z^{7/3})\quad, \quad Z\to\infty, \end{equation} where $-C_{\text{TF}}(\lambda)Z^{7/3}$ is the (non-relativistic) Thomas-Fermi energy of the atom. \end{thm} This shows that, to leading order, the ground-state energy of a relativistic atom is given by the (non-relativistic) semi-classical Thomas-Fermi energy approximation, as it is for the non-relativistic atom (note that the case $\delta=\frac{2}{\pi}$ is included). (In the non-relativistic case this was first proved by Lieb and Simon \cite{LiSi77}; see also Lieb \cite{Li81b}.) This expresses the fact that for large atoms the majority of the electrons are non-relativistic. The second term in the expansion~\eqref{eq:TF} will be studied in a forthcoming paper~\cite{RelScott}. The proof of Theorem~\ref{thm:TF} will be by finding upper and lower bounds on $E_{\lambda Z}(Z,\delta)$. Note that the relativistic kinetic energy is always lower than the non-relativistic one: \begin{equation} \sqrt{\alpha^{-2}q^{2}+\alpha^{-4}}-\alpha^{-2}= \alpha^{-2}\left(\sqrt{1+(\alpha q)^2}-1\right) \leq \frac{q^2}{2} . \end{equation} (Note: since we will later make Taylor expansions of the square root in the relativistic kinetic energy, we will have to insist on the non-relativistic kinetic energy being $-\Delta/2$). This means that all upper bounds derived earlier \cite{LiSi77,Li81b} for the non-relativistic operator \begin{equation*} H_{cl}= \sum_{i=1}^{N}\left\{ \frac{p_{i}^2}{2}- \frac{Z}{|x_{i}|}\right\} + \sum_{1\leq i0$ ($a$ to be chosen later), $\phi_{a}(x)=a^{-3}\phi(x/a)$, so that $\int\!\phi_{a}(x)\,d^{3}\!x=1$. Then for all $\rho: \R^{3}\to\R$ we have: \begin{align*} &\sum_{1\leq i 1+\beta$} \\ \end{array} \right. \qquad,\qquad \theta_{2}(\xi) = \left\{ \begin{array}{ll} 0 & \text{if $\xi < 1-\beta$} \\ 1 & \text{if $\xi > 1+\beta$} \\ \end{array} \right. , \end{split} \end{equation*} and such that $\theta_{1}(\xi)^{2}+\theta_{2}(\xi)^{2}=1$ for all $\xi\in\R_{+}$. Now define, with $8/90$, \begin{equation*} |x|> a l \ \text{ and } \ |y|\gamma|x| \ \text{ on }\ \supp\ \chi_{+}L_{j}^{(\alpha)}\chi_{-}. \end{equation*} Since both $|x|^{-2}$ and $K_{2}(\alpha^{-1}|x|)$ are decreasing in $|x|$ (the last is obvious from the definition of $K_{2}$), and since $(\chi_{j}(x)-\chi_{j}(y))^{2}\leq 1$, we get that pointwise, \begin{equation*} \chi_{+}(x)L_{j}^{(\alpha)}(x,y)\chi_{-}(y) \leq \chi_{+}(x)\frac{\alpha^{-2}}{4\pi^{2}} \frac{K_{2}(\alpha^{-1}\gamma|x|)}{(\gamma|x|)^{2}} \,\chi_{-}(y) \end{equation*} on $\supp\ \chi_{+}L_{j}^{(\alpha)}\chi_{-}$. Therefore \begin{align} \label{eq:loc1} |(f,\chi_{+}&L_{j}^{(\alpha)}\chi_{-}f)| \\ &\leq \left(\int |f(y)|\,\chi_{-}(y)\,d^{3}\!y\right) \left( \frac{(\alpha\gamma)^{-2}}{4\pi^{2}} \int |f(x)| \,\chi_{+}(x) \frac{K_{2}(\alpha^{-1}\gamma|x|)}{|x|^{2}}\,d^{3}\!x\right). \nonumber \end{align} We estimate both of these terms using the Cauchy-Schwartz inequality. For the first we get \begin{align} \label{eq:loc2} \int |f(y)|\,\chi_{-}(y)\,d^{3}\!y \leq \|f\|_{2}\,\|\chi_{-}\|_{2} =C l^{3/2}\|f\|_{2}, \end{align} and for the second \begin{align} \label{eq:loc3} \int |f(x)| \,\chi_{+}(x) \frac{K_{2}(\alpha^{-1}\gamma|x|)}{|x|^{2}}\,d^{3}\!x \leq \|f\|_{2}\, \left(\int \Big(\chi_{+}(x) \frac{K_{2}(\alpha^{-1}\gamma|x|)}{|x|^{2}}\Big)^{2} \,d^{3}\!x\right)^{1/2}. \end{align} Using the estimate~\eqref{eq:K_{2}-est} in Appendix~\ref{appendixA} on $K_{2}$, we get the estimate \begin{align*} \int \Big(&\chi_{+}(x) \frac{K_{2}(\alpha^{-1}\gamma|x|)}{|x|^{2}}\Big)^{2}\,d^{3}\!x \\ &\leq 4\pi\int_{2l}^{\infty} \frac{16}{|x|^{4}}\frac{\pi e^{-2\alpha^{-1}\gamma|x|}}{2\alpha^{-1} \gamma|x|} \Big(1+(2\alpha^{-1}\gamma|x|)^{-1} +(2\alpha^{-1}\gamma|x|)^{-2}\Big)^{2}|x|^{2}\, d|x| \\ &= 128\pi^{2}\alpha^{-1}\gamma\int_{4\gamma l\alpha^{-1}} ^{\infty}t^{-3}e^{-t}(1+\frac{1}{t}+\frac{1}{t^{2}})^{2}\, dt, \end{align*} where the last equality follows by the change of variables $t=2\gamma\alpha^{-1}|x|$. Dominating $e^{-t}$ in the integrand by $e^{-4\gamma l\alpha^{-1}}$ and working out the resulting integral, we arrive at (using~\eqref{eq:loc1}, \eqref{eq:loc2}, and \eqref{eq:loc3}; recall that $l=\alpha^{r}$) \begin{equation*} |(f, \chi_{+}L_{j}^{(\alpha)}\chi_{-}f)| \leq C\,\|f\|_{2}^{2}\, \alpha^{(3r-5)/2}e^{-2\gamma\alpha^{r-1}} \Big\{\dots\Big\}^{1/2} \end{equation*} where \begin{multline*} \Big\{\dots\Big\}^{1/2}= \Big\{\frac{1}{4}(4\gamma)^{-4}\alpha^{4(1-r)}+\frac{2}{5}(4\gamma)^{-5} \alpha^{5(1-r)} \\ +\frac{1}{2}(4\gamma)^{-6}\alpha^{6(1-r)}+ \frac{2}{7}(4\gamma)^{-7}\alpha^{7(1-r)} +\frac{1}{8}(4\gamma)^{-8}\alpha^{8(1-r)}\Big\}^{1/2}. \end{multline*} Now, since $8/92l \text{ and } |y|<(1+\beta)l \\ \chi_{+}(x)L_{3}^{(\alpha)}(x,y)\chi_{-}(y) &\neq0\ \text{ for } |x|>(1-\beta)\alpha^{t} \text{ and } |y|<2l \\ \chi_{+}(x)L_{2}^{(\alpha)}(x,y)\chi_{-}(y)&\neq0 \ \text{ for }|x|>(1-\beta)\alpha^{t} \text{ and } |y|<2l \\ &\!\!\!\!\text{ and }\ \text{ for } |x|\in [2l,(1-\beta)\alpha^{t}]\text{ and } |y|<(1+\beta)l . \end{align*} This is due to the fact that for $\alpha$ small enough, $\alpha^{t}>\alpha^{r}$, since $t<2/3<8/92l}|\nabla \chi_{j}|^{2} = c_{j}^{+}\alpha^{-2t}\quad , \quad j=2,3, \end{equation*} and since $t<2/3$ we get, using \eqref{eq:temp1}, that \begin{equation*} \sum_{i=1}^{N}(m_{i},\chi_{+}L_{3}^{(\alpha)}\chi_{+}m_{i}) \leq N\frac{3}{2}c_{j}^{+}\alpha^{1-2t} = o(\alpha^{-4/3}) \quad,\quad\alpha\to0, \end{equation*} as $N=\lambda Z=\lambda\delta\alpha^{-1}$ ($\lambda$ and $\delta$ fixed) and $\|m_{i}\|_{2}=1$. Similarly for $\chi_{+}L_{2}^{(\alpha)}\chi_{+}$. For the other two terms in \eqref{eq:last_terms}, note that \begin{align*} \|f\,\chi_{-}\|_{2}^{2} &=\int |f(x)|^{2}\,|\chi_{-}(x)|^{2}\,d^{3}\!x =\int |f(x)|^{2}\,\chi_{-}(x)\,d^{3}\!x =(f,\chi_{-}f)\\ & = (f,\chi_{-}(\chi_{1}^{2}+\chi_{2}^{2})f) =(\chi_{1}f,\chi_{-}\chi_{1}f)+(\chi_{2}f,\chi_{-}\chi_{2}f)\ , \end{align*} since $\chi_{-}^{2}=\chi_{-}$ and $\chi_{1}^{2}+\chi_{2}^{2}=1$ on $\supp\chi_{-}$. Using this and \eqref{eq:temp1}, we obtain (since $\chi_{-}\chi_{1}=\chi_{1}$): \begin{align} \label{eq:rest_error} \sum_{i=1}^{N}(m_{i},\chi_{-}(L_{1}^{(\alpha)}&+L_{2}^{(\alpha)}) \chi_{-}m_{i})\\ & \leq C\,\alpha^{1-2r} \Big(\sum_{i=1}^{N}(\chi_{1}m_{i},\chi_{1}m_{i})+ \sum_{i=1}^{N}(\chi_{2}m_{i},\chi_{-}\chi_{2}m_{i})\Big) \nonumber \end{align} where \begin{align*} C&=\frac{3}{2}(c_{1}+c_{2})\quad , \quad c_{j}\alpha^{-2r}= \sup_{|x|<2l}|\nabla \chi_{j}(x)|^{2}\ , \ j=1,2. \end{align*} The two terms in~(\ref{eq:rest_error}) will be estimated in the following two sections, the first one along with the energy at the nucleus, the second one with the energy in the intermediary zone. \section{The energy near the nucleus} \label{inner} In this section we estimate the energy at the nucleus, that is (see \eqref{eq:slater}), the term \begin{equation} \label{eq:inner} \sum_{i=1}^{N}(m_{i},\chi_{1}\big\{T(p)-V(x) +\alpha\,\rho*\phi_{a}*|x|^{-1}\big\}\chi_{1}m_{i}). \end{equation} Also, half of the remaining term~(\ref{eq:rest_error}) of the localisation error, treated in the previous section, will be estimated here. We start by noting that $\rho*\phi_{a}*|x|^{-1}$ is positive, so that we get a lower bound to~(\ref{eq:inner}) by dropping this term. The remaining expression will be treated by using the following result by Lieb and Yau~\cite[Theorem 11]{LiYau88} on the hyper-relativistic operator $|p|$: \begin{thm} \label{thm:LiYau} Let $C_{0}>0$ and $R>0$ and let \begin{equation*} H_{C_{0}R}=|p|-\frac{2}{\pi}|x|^{-1}-C_{0}/R \end{equation*} be defined on $L^2(\R^3)$ as a quadratic form. Let $0\leq\gamma\leq q$ be a density matrix (that is, any bounded operator on $L^2(\R^3)$ which satisfies the operator inequality $0\leq\gamma\leq q$ and for which $\Tr(\gamma)<\infty$) and let $\chi$ be any function with support in $B_{R}=\{x\, |\,|x|\leq R\}$. Then \begin{equation} \label{eq:LiYau} \Tr(\bar\chi\gamma\chi H_{C_{0}R}) \geq - 4.4827\,C_{0}^4R^{-1}q\{(3/4\pi R^3)\int |\chi(x)|^2\,d^{3}\!x\}. \end{equation} Note, that when $\chi\equiv1$ in $B_{R}$, then the factor in braces \{\} in \eqref{eq:LiYau} is 1. \end{thm} Here, $\Tr(\gamma h)$ is shorthand for $\sum_{k}(f_{k},hf_{k})\gamma_{k}$, where $(f_{k},\gamma_{k})$ are the eigenfunctions and eigenvalues of $\gamma$. For more details, see Lieb \cite{Li83}. In our situation, $q=2$. For our purpose, let $\Pi$ be the projection on span$\{m_{i}\,|\,i=1,\dots N\}$, then $\Pi$ is a density matrix as above, and \begin{equation*} \Tr(\chi_{1}\Pi\chi_{1} H_{C_{0}R}) = \sum_{i=1}^{N} (m_{i}, \chi_{1} H_{C_{0}R}\chi_{1} m_{i}). \end{equation*} Since $\supp\chi_{1}\subseteq B(0,(1+\beta)\alpha^{r})$ with $8/9-4/3$ (as \(8/9t$), \begin{equation*} \sum_{i=1}^{N}(m_{i},\chi_{3}\big\{\frac{\delta}{|x_{i}|}*\phi_{\alpha^{s}} -\frac{\delta}{|x_{i}|}\big\}\chi_{3}m_{i}) = 0. \end{equation*} This is one of the reasons for introducing the intermediary zone by the function $\chi_{2}$. Note also that $N\,o(\alpha^{-1/3}) = o(\alpha^{-4/3})$. Now, for $\alpha$ small enough, $\alpha^{s-t}<1/4$, since $s>t$, so that if $|q|<\frac{1}{4}\alpha^{t}$, then \begin{equation*} |x-q|<\alpha^{s} \Rightarrow |x|<\frac{1}{2}\alpha^{t}, \end{equation*} and so $(m_{i}\chi_{3},g_{\alpha}^{p,q})=0$, since $\supp\ g_{\alpha}\subset B(0,\alpha^{s})$ and $\supp\ \chi_{3}\subset\R^{3}\setminus B(0,\frac{1}{2}\alpha^{t})$. That is, for $\alpha$ small enough \begin{equation*} \supp_{q} |(m_{i}\chi_{3},g_{\alpha}^{p,q})|^{2} \subseteq \R^{3}\setminus B(0,\tfrac{1}{4}\alpha^{t}), \end{equation*} so that for any $\mu\geq0$ we have, with $M(p,q)=\sum_{i=1}^{N}|(m_{i}\chi_{3},g_{\alpha}^{p,q})|^2$ and $\big[f\big]_{\pm}=\max\{\pm f,0\}$: \begin{align*} \frac{1}{(2\pi)^{3}}&\iint d^{3}\!p\,d^{3}\!q\,\big(\tilde T(p)- \tilde V(q)\big)\big( \sum_{i=1}^{N}|(m_{i}\chi_{3},g_{\alpha}^{p,q})|^2\big)\\ &=\frac{1}{(2\pi)^{3}}\iint\limits_{|q|>\frac{1}{4}\alpha^{t}} d^{3}\!p\,d^{3}\!q\, \big(\tilde T(p)-(\tilde V(q)-\alpha\mu)\big) \,M(p,q) -\alpha\mu\sum_{i=1}^{N}(\chi_{3}m_{i},\chi_{3}m_{i})\\ &\geq -\frac{1}{(2\pi)^{3}}\iint\limits_{|q|>\frac{1}{4}\alpha^{t}} d^{3}\!p\,d^{3}\!q\, \Big[ \tilde T(p)-(\tilde V(q)-\alpha\mu)\Big]_{-} -\alpha\mu N, \end{align*} since $0\leq M(p,q)\leq1$ and $(\chi_{3}m_{i},\chi_{3}m_{i})\leq\|m_{i}\|_{2}^{2}=1$. The first is seen by Bessel's inequality, since the $m_{i}$'s are orthonormal and $\|\chi_{3}g_{\alpha}^{p,q}\|_{2}\leq\|g_{\alpha}^{p,q}\|_{2}=1$. In this way we have shown that for $\mu\geq0$, $\rho:\R^{3}\to\R$ and $\psi\in{\mathcal H}_F=\bigwedge^N L^2(\R^3,\C^2)\,$: \begin{align} \label{eq:prelimi} \langle\psi,H\psi\rangle&\geq -\frac{1}{(2\pi)^{3}}\iint\limits_{|q|>\frac{1}{4}\alpha^{t}} d^{3}\!p\,d^{3}\!q\, \Big[\tilde T(p)-(\tilde V(q)-\alpha\mu)\Big]_{-} \notag\\ &\qquad-\frac{\alpha}{2}\iint \frac{\rho(x)\rho(y)}{|x-y|}\,d^{3}\!x\,d^{3}\!y -\alpha\mu N -o(\alpha^{-4/3}). \end{align} Choose now $\rho$ to be the Thomas-Fermi density $\rho_{TF}^{N,Z}$, that is, the function that minimises the Thomas-Fermi functional (here, $\gamma=(3\pi^{2})^{2/3}$): \begin{equation} \label{eq:TF-func} \mathcal{E}_{TF}(\rho)=\frac{3}{5}\gamma\int\rho(x)^{5/3}\,d^{3}\!x -\int\rho(x)\frac{Z}{|x|}\,d^{3}\!x +\frac{1}{2}\iint\frac{\rho(x)\rho(y)}{|x-y|}\,d^{3}\!x\,d^{3}\!y \end{equation} over the set \begin{equation*} \Big\{\rho\in L^{5/3}(\R^{3})\cap L^{1}(\R^{3})\, \big|\,\rho\geq0,\int\rho(x)\,d^{3}\!x\leq N\Big\}. \end{equation*} (For the Thomas-Fermi theory, see Lieb and Simon~\cite{LiSi77} and Lieb~\cite{Li81b}). Then $\rho_{TF}^{N,Z}$ satisfies the Thomas-Fermi equation: \begin{equation} \label{eq:TF-eq} \gamma\,\rho(x)^{2/3}=\Big[\frac{Z}{|x|}-\rho*|x|^{-1}-\mu\Big]_{+} \end{equation} for some unique $\mu=\mu(N)$. Furthermore, \begin{align*} &\text{for $N\leq Z$: }\quad \int\rho_{TF}^{N,Z}(x)\,d^{3}\!x=N \quad\text{and}\quad\mu(N)>0,\\ &\text{for $N>Z$: }\quad \int\rho_{TF}^{N,Z}(x)\,d^{3}\!x=Z \quad\text{and}\quad\mu(N)=0 \end{align*} (see Lieb and Simon~\cite[Theorems II.17, 18 and 20]{LiSi77}). In this way, $\int\rho_{TF}^{N,Z}(x)\,d^{3}\!xZ$, and therefore $\mu(N)=0$, so that we always have \begin{equation} \label{eq:mu-rho-N} \mu(N)\int\rho_{TF}^{N,Z}(x)\,d^{3}\!x=\mu(N)N. \end{equation} Let $\mathcal{E}_{TF}(N,Z)\equiv\mathcal{E}_{TF}(\rho_{TF}^{N,Z})$ and define the Thomas-Fermi potential by \begin{equation*} V_{TF}^{N,Z}(x)\equiv Z/|x|-\rho_{TF}^{N,Z}*|x|^{-1}-\mu(N), \end{equation*} then we have the following scaling (\cite[(2.24) p.608]{LiSi77}) (remember, that $\lambda=N/Z$ is fixed): \begin{align} \label{eq:scale_E} \mathcal{E}_{TF}(N,Z)&=Z^{7/3}\mathcal{E}_{TF}(\lambda,1) \equiv -C_{TF}(\lambda)Z^{7/3},\\ \label{eq:scale_V} V_{TF}^{N,Z}(x)&=Z^{4/3}V_{TF}^{\lambda,1}(Z^{1/3}x)\equiv Z^{4/3}V_{TF}(Z^{1/3}x). \end{align} The idea is now to estimate the difference between the integral in~(\ref{eq:prelimi}) (with $\rho=\rho_{TF}^{N,Z}$ and $\mu=\mu(N)$) and \begin{equation*} -\frac{\alpha}{(2\pi)^{3}}\iint\limits_{|q|>\frac{1}{4}\alpha^{t}} d^{3}\!p\,d^{3}\!q\, \Big[ \frac{p^{2}}{2}-\big(\frac{Z}{|q|}- \rho_{TF}^{N,Z}*|q|^{-1}-\mu(N)\big)\Big]_{-} . \end{equation*} This is done in two steps: first, we change the domain of the integration, then we change the integrand, each time estimating the error. First, \begin{eqnarray*} -\frac{1}{(2\pi)^{3}}\iint\limits_{|q|>\frac{1}{4}\alpha^{t}} d^{3}\!p\,d^{3}\!q\, \Big[ \tilde T(p)-\alpha V_{TF}^{N,Z}(q)\Big]_{-} =\frac{1}{(2\pi)^{3}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \iint\limits_{|q|>\frac{1}{4}\alpha^{t};\ \tilde T(p)<\alpha V_{TF}^{N,Z}(q)} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! d^{3}\!p\,d^{3}\!q\, \big(\tilde T(p)-\alpha V_{TF}^{N,Z}(q)\big)\\ =\frac{1}{(2\pi)^{3}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \iint\limits_{|q|>\frac{1}{4}\alpha^{t};\ \alpha\frac{p^{2}}{2}<\alpha V_{TF}^{N,Z}(q)} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! d^{3}\!p\,d^{3}\!q\, \big(\tilde T(p)-\alpha V_{TF}^{N,Z}(q)\big) + \frac{1}{(2\pi)^{3}} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \iint\limits_{|q|>\frac{1}{4}\alpha^{t};\ \tilde T(p)<\alpha V_{TF}^{N,Z}(q)<\alpha\frac{p^{2}}{2}} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! d^{3}\!p\,d^{3}\!q\, \big(\tilde T(p)-\alpha V_{TF}^{N,Z}(q)\big). \end{eqnarray*} Since $\tilde T(p)\geq0$, we get \begin{align*} \iint\limits_{|q|>\frac{1}{4}\alpha^{t};\ \tilde T(p)<\alpha V_{TF}^{N,Z}(q)<\alpha\frac{p^{2}}{2}} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! d^{3}\!p\,d^{3}\!q\, \big(\alpha V_{TF}^{N,Z}(q)&-\tilde T(p)\big) \leq \alpha\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \iint\limits_{|q|>\frac{1}{4}\alpha^{t};\ \tilde T(p)<\alpha V_{TF}^{N,Z}(q)<\alpha\frac{p^{2}}{2}} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! d^{3}\!p\,d^{3}\!q\, V_{TF}^{N,Z}(q). \end{align*} Using the scaling~(\ref{eq:scale_V}) and the change of variables $\omega=\delta^{1/3}\alpha^{-1/3}q$, the above is equal to \begin{align} \label{eq:area_1} \delta^{1/3}\alpha^{2/3} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! &\iint\limits_{\substack{|\omega|>\frac{1}{4}\delta^{1/3}\alpha^{t-1/3}\\ \tilde T(p)<\delta^{4/3}\alpha^{-1/3}V_{TF}(\omega)<\alpha\frac{p^{2}}{2}}} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! d^{3}\!p\,d^{3}\!\omega\, V_{TF}(\omega). \end{align} The limits in the integral means that \begin{equation*} 2\delta^{4/3}\alpha^{-4/3}V_{TF}(\omega)\leq p^{2} \leq 2\delta^{4/3}\alpha^{-4/3}V_{TF}(\omega) \big(1+\frac{1}{2}\delta^{4/3}\alpha^{2/3}V_{TF}(\omega)\big) \end{equation*} so that with \begin{align*} X= 2\delta^{4/3}\alpha^{-4/3}V_{TF}(\omega)\ ,\ Y=\frac{1}{2}\delta^{4/3}\alpha^{2/3}V_{TF}(\omega)\ ,\ Z=|p|^{2}\ ,\ W=\frac{1}{4}\delta^{1/3}\alpha^{t-1/3}, \end{align*} we have \begin{align*} \text{(\ref{eq:area_1})}&=(4\pi)^{2}\delta^{1/3}\alpha^{2/3} \int_{W}^{\infty}d|\omega|\,|\omega|^{2}\,V_{TF}(\omega) \Big(\int_{X}^{X(1+Y)}\frac{\sqrt{Z}}{2}\,dZ\Big)\\ &=(4\pi)^{2}\delta^{1/3}\alpha^{2/3} \int_{W}^{\infty}d|\omega|\,|\omega|^{2}\,V_{TF}(\omega) \frac{X^{3/2}}{3}\big((1+Y)^{3/2}-1\big). \end{align*} By the Taylor-expansion~(\ref{eq:Taylor}), we have \( (1+Y)^{3/2}\leq1+\frac{3}{2}Y+\frac{3}{8}Y^{2}\), and so \begin{align*} \text{(\ref{eq:area_1})}&\leq C\delta^{7/3}\alpha^{-4/3} \int_{W}^{\infty}|\omega|^{2}\,V_{TF}(\omega)^{5/2}\\ &\qquad\qquad\qquad\qquad\times \big(\frac{3}{4}\delta^{4/3}\alpha^{2/3}V_{TF}(\omega) +\frac{3}{32}\delta^{8/3}\alpha^{4/3}V_{TF}(\omega)^{2}\big) \,d|\omega|. \end{align*} Using that $V_{TF}^{N,Z}(x)\leq Z/|x|$, since $\mu(N)\geq0$ and $\rho_{TF}^{N,Z}\geq0$ (remember that $V_{TF}\equiv V_{TF}^{\lambda,1}$), we arrive at \begin{align*} \text{(\ref{eq:area_1})}&\leq C\delta^{11/3}\alpha^{-2/3}\int_{W}^{\infty}d|\omega|\, |\omega|^{-3/2}+\sqrt{2}\pi^{2}\delta^{5}\int_{W}^{\infty}d|\omega|\, |\omega|^{-5/2}\\ &\sim\alpha^{-2/3}W^{-1/2}+W^{-3/2}\sim \alpha^{-2/3}\alpha^{1/6-t/2}+\alpha^{(1-3t)/2}\\ &= o(\alpha^{-5/6})+ o(\alpha^{-1/2}) \end{align*} since $t<2/3$. This means, that \begin{align*} -\frac{1}{(2\pi)^{3}}&\iint\limits_{|q|>\frac{1}{4}\alpha^{t}} d^{3}\!p\,d^{3}\!q\, \Big[ \tilde T(p)-\alpha V_{TF}^{N,Z}(q)\Big]_{-} \\ &\geq \frac{1}{(2\pi)^{3}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \iint\limits_{|q|>\frac{1}{4}\alpha^{t};\ \alpha\frac{p^{2}}{2}<\alpha V_{TF}^{N,Z}(q)} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! d^{3}\!p\,d^{3}\!q\, \big( \tilde T(p)-\alpha V_{TF}^{N,Z}(q)\big) - o(\alpha^{-4/3}). \end{align*} Next note that since $|q|>\frac{1}{4}\alpha^{t}$ and $\alpha V_{TF}^{N,Z}(q)\leq\delta/|q|$ in the area of integration, we here have that \begin{equation*} \tilde T(p)=\sqrt{p^{2}+\alpha^{-2}} -\alpha^{-1} \geq \alpha\frac{p^{2}}{2}-\alpha^{3}\frac{p^{4}}{8}. \end{equation*} In this way, we get \begin{align} \label{eq:clas_E} \frac{1}{(2\pi)^{3}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! &\iint\limits_{|q|>\frac{1}{4}\alpha^{t};\ \alpha\frac{p^{2}}{2}<\alpha V_{TF}^{N,Z}(q)} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! d^{3}\!p\,d^{3}\!q\, \big(\tilde T(p)-\alpha V_{TF}^{N,Z}(q)\big) \notag\\ &\geq\frac{1}{(2\pi)^{3}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \iint\limits_{|q|>\frac{1}{4}\alpha^{t}; \ \alpha\frac{p^{2}}{2}<\alpha V_{TF}^{N,Z}(q)} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! d^{3}\!p\,d^{3}\!q\, \big(\alpha\frac{p^{2}}{2} -\alpha^{3}\frac{p^{4}}{8}-\alpha V_{TF}^{N,Z}(q)\big)\notag\\ &=\frac{1}{(2\pi)^{3}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \iint\limits_{|q|>\frac{1}{4}\alpha^{t}; \ \alpha\frac{p^{2}}{2}<\alpha V_{TF}^{N,Z}(q)} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! d^{3}\!p\,d^{3}\!q\, \big(\alpha\frac{p^{2}}{2}-\alpha V_{TF}^{N,Z}(q)\big) -\alpha^{3}\frac{1}{(2\pi)^{3}} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \iint\limits_{|q|>\frac{1}{4}\alpha^{t}; \ \alpha\frac{p^{2}}{2}<\alpha V_{TF}^{N,Z}(q)} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \frac{p^{4}}{8}\,d^{3}\!p\,d^{3}\!q. \end{align} Note that \begin{align*} \frac{1}{(2\pi)^{3}} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! &\iint\limits_{|q|>\frac{1}{4}\alpha^{t}; \ \alpha\frac{p^{2}}{2}<\alpha V_{TF}^{N,Z}(q)} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\! d^{3}\!p\,d^{3}\!q\, \big(\alpha\frac{p^{2}}{2}-\alpha V_{TF}^{N,Z}(q)\big)\\ &=-\frac{\alpha}{(2\pi)^{3}} \iint\limits_{|q|>\frac{1}{4}\alpha^{t}} d^{3}\!p\,d^{3}\!q\, \Big[\frac{p^{2}}{2}-\big(\frac{Z}{|q|} -\rho_{TF}^{N,Z}*|q|^{-1}-\mu(N)\big)\Big]_{-} \\ &\geq-\frac{\alpha}{(2\pi)^{3}} \iint d^{3}\!p\,d^{3}\!q\, \Big[\frac{p^{2}}{2}-\big(\frac{Z}{|q|} -\rho_{TF}^{N,Z}*|q|^{-1}-\mu(N)\big)\Big]_{-}. \end{align*} Let us now look at the last term in~(\ref{eq:clas_E}). Again using that $ V_{TF}^{N,Z}(x)\leq Z/|x|$, we have that \begin{align*} \iint\limits_{|q|>\frac{1}{4}\alpha^{t}; \ \alpha\frac{p^{2}}{2}<\alpha V_{TF}^{N,Z}(q)} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \frac{p^{4}}{8}\,&d^{3}\!p\,d^{3}\!q\, \leq \!\!\!\!\!\! \iint\limits_{|q|>\frac{1}{4}\alpha^{t}; \ \alpha\frac{p^{2}}{2}<\delta/|q|} \!\!\!\!\!\!\!\!\!\!\!\!\! \frac{p^{4}}{8}\,d^{3}\!p\,d^{3}\!q\,\\ &= (4\pi)^{2}\int_{\frac{1}{4}\alpha^{t}}^{\infty}d|q|\,\left(|q|^{2} \int_{0}^{\sqrt{2Z/|q|}}\frac{|p|^{4}}{8}|p|^{2}\,d|p|\right)\\ &=2\pi^{2}\int_{\frac{1}{4}\alpha^{t}}^{\infty}d|q|\, \left(|q|^{2}\Big[ t^{7}\!/7\Big]_{0}^{\sqrt{2Z/|q|}}\right)\\ &=\frac{2\pi^{2}(2Z)^{7/2}}{7}\int_{\frac{1}{4}\alpha^{t}}^{\infty} |q|^{-3/2}\,d|q| =\frac{8\pi^{2}(2Z)^{7/2}}{7}\alpha^{-t/2}. \end{align*} Using this, we then get the following \begin{align*} &\frac{1}{(2\pi)^{3}} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \iint\limits_{|q|>\frac{1}{4}\alpha^{t}; \ \alpha\frac{p^{2}}{2}<\alpha V_{TF}^{N,Z}(q)} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! d^{3}\!p\,d^{3}\!q\, \big(\tilde T(p)-\alpha V_{TF}^{N,Z}(q)\big)\\ &\geq -\frac{\alpha}{(2\pi)^{3}}\iint d^{3}\!p\,d^{3}\!q\, \Big[ \frac{p^{2}}{2}-\big(\frac{Z}{|q|} -\rho_{TF}^{N,Z}*|q|^{-1}-\mu(N)\Big]_{-} \\ &- \alpha^{(6-t)/2}\,\,\frac{(2Z)^{7/2}}{7\pi}. \end{align*} Hence, since $\delta=Z\alpha$ is fixed and $t<2/3$, we have \begin{equation*} \alpha^{(6-t)/2}\,\,\frac{(2Z)^{7/2}}{7\pi} =\frac{8\sqrt{2}}{7\pi}\alpha^{-(1+t)/2}\delta^{7/2} =o(\alpha^{-4/3})\quad,\quad\alpha\to0. \end{equation*} Summing up, we have now proved that for $\psi\in{\mathcal H}_F=\bigwedge^N L^2(\R^3,\C^2)\,$: \begin{align} \label{eq:before_TF} \langle\psi,H\psi\rangle&\geq -\frac{\alpha}{(2\pi)^{3}}\iint d^{3}\!p\,d^{3}\!q\, \Big[\frac{p^{2}}{2} -(\frac{Z}{|q|}-\rho_{TF}^{N,Z}*|q|^{-1}-\mu(N))\Big]_{-} \notag\\ &\quad-\frac{\alpha}{2}\iint \frac{\rho_{TF}^{N,Z}(x)\rho_{TF}^{N,Z}(y)}{|x-y|}\,d^{3}\!xd^{3}\!y -\alpha\mu(N)N - o(\alpha^{-4/3}). \end{align} Integrating firstly in $p$ in the first integral in~\eqref{eq:before_TF}, we get, for each $q$ fixed: \begin{align} \label{eq:5/2-int} \int &d^{3}\!p \Big[\frac{p^{2}}{2} -(\frac{Z}{|q|}-\rho_{TF}^{N,Z}*|q|^{-1}-\mu(N))\Big]_{-}\\ &=\!\!\!\! \int\limits_{\frac{p^{2}}{2}