diff --git a/Carlen_Jauslin_Lieb_2020.tex b/Carlen_Jauslin_Lieb_2020.tex index 1bbf607..b98df80 100644 --- a/Carlen_Jauslin_Lieb_2020.tex +++ b/Carlen_Jauslin_Lieb_2020.tex @@ -125,10 +125,10 @@ Along with the computation of the low density energy of the simple equation in o for $N$ particles in a cubic box of finite volume $V$ with periodic boundary conditions. The ground state eigenfunction $\psi_0$ is unique and non-negative, as can be shown using the Perron-Frobenius theorem, and thus we may normalize $\psi_0$ to obtain a probability measure. This is not the usual probability measure associated to a quantum state, which would be quadratic in the wave function, but since $\psi_0$ is non-negative and integrable ($\|\psi_0\|_1 \leqslant V^{1/2}\|\psi_0\|_2$), we may use it directly to define a probability measure, and this is the starting point of \cite{Li63}. Because particles interact pairwise, the ground state energy and other observables can be calculated in terms of the two-point correlation function associated to this probability measure: \begin{equation} - g_N(x_1-x_2) := \lim_{N,V\to\infty, N/V = \rho}\frac{V^2\int dx_3\cdots dx_N\ \psi_0(x_1,x_2,x_3,\dots,x_N)}{\int dy_1\cdots dy_N\ \psi_0(y_1,\dots,y_N )} + g(x_1-x_2) := \lim_{N,V\to\infty, N/V = \rho}\frac{V^2\int dx_3\cdots dx_N\ \psi_0(x_1,x_2,x_3,\dots,x_N)}{\int dy_1\cdots dy_N\ \psi_0(y_1,\dots,y_N )} \end{equation} - In \cite{Li63}, under a few physically motivated approximations, in the thermodynamic limit, in which the number of particles $N$ and the volume of the gas $V$ are taken to infinity, with $\rho:=\frac NV$ fixed, an equation for the limiting two-point correlation function $g_{\infty}$ is derived. - The function $u(x)$ in \eqref{simp} is then defined as $u(x):=1- g_{\infty}(x)$. Note that since by definition $g_{\infty}(x) \geqslant 0$, $u(x) \leqslant 1$. + In \cite{Li63}, under a few physically motivated approximations, in the thermodynamic limit, in which the number of particles $N$ and the volume of the gas $V$ are taken to infinity, with $\rho:=\frac NV$ fixed, an equation for the limiting two-point correlation function $g$ is derived. + The function $u(x)$ in \eqref{simp} is then defined as $u(x):=1- g(x)$. Note that since by definition $g(x) \geqslant 0$, $u(x) \leqslant 1$. Because the expected values in the ground state of many physical observables can be calculated in terms of $g$, any method for computing $g$ that bypasses directly solving the $N$-body Schr\"odinger equation for the Hamiltonian \eqref{ham0} provides an effective means for the computation of these values, and this motivates the study of the simple equation system\-~(\ref{simp}). Indeed, the ground state energy per particle is given in terms of $g$ by the second equation in \eqref{simp}. There is so far no rigorous derivation of \eqref{simp} from the $N$-body Schr\"odinger equation, and hence there is no mathematical @@ -325,7 +325,7 @@ We shall also need various $L^p$ bounds on $u'$, and for these we need a detaile If $(1+ |x|^4)v(x)\in L^1(\mathbb R^3)\cap L^2(\mathbb R^3)$, then \begin{equation} \rho u(x)= - \frac{\sqrt{2+\beta}}{2\pi^2\sqrt{e}}\frac1{1+|x|^{4}} + \frac{\sqrt{2+\beta}}{2\pi^2\sqrt{e}}\frac1{|x|^{4}} + R(x) \label{udecay} @@ -336,18 +336,13 @@ We shall also need various $L^p$ bounds on $u'$, and for these we need a detaile , \end{equation} and where $|x|^4R(x)$ is in $L^2(\mathbb R^3)\cap L^\infty(\R^3)$, uniformly in $e$ on all compact sets. - Moreover, there is a constant $C$ independent of $e$ and $\rho$ such that for all $x$ + Moreover, for every $\rho_0>0$, there is a constant $C$ that only depends on $\rho_0$ such that for all $x$, for all $\rho<\rho_0$, \begin{equation}\label{fourdecay} - u(x) \leqslant C \rho^{-1}e^{-1/2}\frac{1}{1 + |x|^4}\ . + u(x) \leqslant \min\left\{1,\frac C{\rho e^{\frac12}|x|^4}\right\}\ . \end{equation} \end{theorem} -\begin{remark} - Theorem\-~\ref{theo:pointwise} actually holds with the condition that $|x|^4 v\in L^1(\mathbb R^3)$, without necessarily being in $L^2(\mathbb R^3)$, together with the condition $v\in L^1(\R^3)\cap L^2(\R^3)$, but the proof of that is a bit more involved, and will not be presented here. -\end{remark} - - The next two theorems concern the monotonicity of $\rho\mapsto e(\rho)$ and convexity if $\rho\mapsto\rho e(\rho)$. These were conjectured in\-~\cite{CJL20}, and here, we prove them for small density $\rho$ (and, in the case of the monotonicity, also for large density). @@ -600,7 +595,7 @@ This provides the control on $\|u'\|_q$ that we need to prove the theorem on co \begin{remark} As stated at the very beginning of the paper, we assume that $v$ is spherically symmetric. This is, however, used very little in the proofs. - In fact, the only theorem which relies on the spherical symmetry is theorem\-~\ref{theo:pointwise}. + In fact, the only theorem that relies on the spherical symmetry is theorem\-~\ref{theo:pointwise}. We believe it should still hold (provided the decay constant in\-~(\ref{udecay}) is suitably adapted) without the spherical symmetry. In this case, the other theorems would not require the spherical symmetry. \end{remark} @@ -665,6 +660,7 @@ This provides the control on $\|u'\|_q$ that we need to prove the theorem on co U_1(x)=\frac{e^{\frac32}}\pi e^{-2\sqrt e|x|} -\frac1\pi \left(\delta(x)+\frac{(\beta+1)e}\pi\frac{e^{-2\sqrt e|x|}}{|x|}\right) \ast f_1\ast f_2 + \label{U1} \end{equation} where \begin{equation} @@ -676,11 +672,13 @@ This provides the control on $\|u'\|_q$ that we need to prove the theorem on co =\frac{e}{\pi|x|} \int_0^\infty e^{-\sqrt{2+\beta + t}(2\sqrt{e}|x|)} t^{-1/2} dt\ , \end{equation} now, for all $T>0$, - \begin{eqnarray*} - \int_0^\infty e^{-\sqrt{2+\beta + t}(2\sqrt{e}|x|)} t^{-1/2}\ dt &=& \int_0^{T} e^{-\sqrt{2+\beta + t}(2\sqrt{e}|x|)} t^{-1/2}+\int_{T}^\infty e^{-\sqrt{2+\beta + t}(2\sqrt{e}|x|)} t^{-1/2}\ dt\\ - &\leqslant& \int_0^{T} e^{-\sqrt{2+\beta}(2\sqrt{e}|x|)} t^{-1/2}+\int_{T}^\infty e^{-\sqrt{ t}(2\sqrt{e}|x|)} t^{-1/2}\ dt\\ - &=& 2 T^{1/2} e^{-\sqrt{2+\beta}(2\sqrt{e}|x|)} + \frac{1}{\sqrt{e}|x|}e^{-\sqrt{T}(2\sqrt{e}|x|)}\ . - \end{eqnarray*} + \begin{eqnarray} + \int_0^\infty e^{-\sqrt{2+\beta + t}(2\sqrt{e}|x|)} t^{-1/2}\ dt &=& \int_0^{T} e^{-\sqrt{2+\beta + t}(2\sqrt{e}|x|)} t^{-1/2}+\int_{T}^\infty e^{-\sqrt{2+\beta + t}(2\sqrt{e}|x|)} t^{-1/2}\ dt + \nonumber\\ + &\leqslant& + 2 T^{1/2} e^{-\sqrt{2+\beta}(2\sqrt{e}|x|)} + \frac{1}{\sqrt{e}|x|}e^{-\sqrt{T}(2\sqrt{e}|x|)}\ . + \label{boundf2} + \end{eqnarray} Choosing $T = 2+\beta$, we see that for large $(2\sqrt{e}|x|)$, $0 \leqslant f_2(x) \leqslant C e^{-\sqrt{2+\beta}(2\sqrt{e}|x|)}$. Furthermore, \begin{equation} @@ -689,14 +687,16 @@ This provides the control on $\|u'\|_q$ that we need to prove the theorem on co \frac{e^{\frac32}}{\pi^3}\frac1{|x|^2}\ast g ,\quad g(x)=\frac{(1-\sqrt e|x|)e^{-(2\sqrt e)|x|}}{|x|} + \label{g} \end{equation} Using \begin{equation} - \frac{1}{|x-y|^2} = \frac{1}{1+|x|^2} + \frac{1-|y|^2 + 2x\cdot y}{ (1+|x|^2)|x-y|^2} + \frac{1}{|x-y|^2} = \frac{1}{|x|^2} + \frac{-|y|^2 + 2x\cdot y}{|x|^2|x-y|^2} \end{equation} twice and the fact that $g(y)$ is even, integrates to zero, and $\int y g(y)\ dy=0$, \begin{equation} - f_1(x) = \frac{1}{(1+|x|^2)^2} \frac{e^{\frac32}}{\pi^3}\left( - \int_{\R^3} |y|^2 g(y){\rm d} y + \int_{\R^3} \frac{(1-|y|^2 + 2x\cdot y)^2}{ |x-y|^2}g(y) {\rm d} y\right) + f_1(x) = \frac{1}{|x|^4} \frac{e^{\frac32}}{\pi^3}\left( - \int_{\R^3} |y|^2 g(y){\rm d} y + \int_{\R^3} \frac{(-|y|^2 + 2x\cdot y)^2}{|x-y|^2}g(y) {\rm d} y\right) + \label{f1} \end{equation} We compute $ @@ -708,7 +708,58 @@ This provides the control on $\|u'\|_q$ that we need to prove the theorem on co Therefore, \begin{equation} \lim_{|x|\to \infty} |x|^4 f_1(x) = -\frac{1}{2\pi^2\sqrt e} \qquad{\rm and}\qquad \lim_{|x|\to \infty} |x|^4 U_1(x) = \frac{1}{2\pi^2\sqrt e}\sqrt{2+\beta}\ . + \label{U1decay} \end{equation} + \bigskip + + \indent + We now turn to an upper bound of $U_1$. + First of all, if $|x|\leqslant\frac1{\sqrt e}$, then by\-~(\ref{g}) and\-~(\ref{f1}), + \begin{equation} + f_1(x)\geqslant 0 + \end{equation} + and if $|x|>\frac1{\sqrt e}$, then + \begin{equation} + f_1(x)\geqslant + -\frac{1}{|x|^4} \frac{e^{2}}{\pi^3}\int_{\mathbb R^3} \frac{(-|y|^2 + 2x\cdot y)^2}{ |x-y|^2}e^{-(2\sqrt e)|y|} {\rm d} y + . + \end{equation} + We split the integral into two parts: $|y-x|>|x|$ and $|y-x|<|x|$. + We have, (recalling $|x|>\frac1{\sqrt e}$), + \begin{equation} + \int_{|y-x|>|x|} \frac{(-|y|^2 + 2x\cdot y)^2}{ |x-y|^2}e^{-(2\sqrt e)|y|} {\rm d} y + \leqslant + e^{-\frac52}C + \end{equation} + for some constant $C$ (we use a notation where the constant $C$ may change from one line to the next). + Now, + \begin{equation} + \int_{|y-x|<|x|} \frac{(-|y|^2 + 2x\cdot y)^2}{ |x-y|^2}e^{-(2\sqrt e)|y|} {\rm d} y + \leqslant + e^{-\sqrt e|x|}\int_{|y-x|<|x|} \frac{(|y|^2 + 2|x||y|)^2}{ |x-y|^2} {\rm d} y + \leqslant + |x|^5e^{-\sqrt e|x|}C + . + \end{equation} + Therefore, for all $x$, + \begin{equation} + f_1(x)\geqslant + -\frac{1}{|x|^4}C(e^{-\frac12}+e^2|x|^4e^{-\sqrt e|x|}) + . + \end{equation} + Finally, by use\-~(\ref{boundf2}), + \begin{equation} + |x|^4\left(\delta(x)+\frac{(\beta+1)e}{\pi}\frac{e^{-2\sqrt e|x|}}{|x|}\right)\ast f_1\ast f_2(x)\geqslant + -Ce^{-\frac12} + . + \end{equation} + All in all, by\-~(\ref{U1}), (since $|x|^4e^{\frac32}e^{-2\sqrt e|x|}