### Update to v0.2:

Changed: more precise statement of the bound on the large x decay of u (Theorem 1.2)

Added: more details in the proof of the large x decay of u (Theorem 1.2)

Fixed: miscellaneous typos.
master v0.2
parent f8a94b06fa
commit ed63f14f57
2 changed files with 93 additions and 23 deletions

#### 107 Carlen_Jauslin_Lieb_2020.tex View File

 @ -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|}

#### 9 Changelog View File

 @ -1,3 +1,12 @@ v0.2:    * Changed: more precise statement of the bound on the large x decay of u (Theorem 1.2)    * Added: more details in the proof of the large x decay of u (Theorem 1.2)    * Fixed: miscellaneous typos.     v0.1:    * Added: references to [Nelson, 1964] and [Reed, Simon, 1975].`