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
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@ 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 nonnegative, as can be shown using the PerronFrobenius 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 nonnegative 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 twopoint correlation function associated to this probability measure:


\begin{equation}


g_N(x_1x_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_1x_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 twopoint 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 twopoint 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 ex}


\frac1\pi \left(\delta(x)+\frac{(\beta+1)e}\pi\frac{e^{2\sqrt ex}}{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}{\pix} \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 ex)e^{(2\sqrt e)x}}{x}


\label{g}


\end{equation}


Using


\begin{equation}


\frac{1}{xy^2} = \frac{1}{1+x^2} + \frac{1y^2 + 2x\cdot y}{ (1+x^2)xy^2}


\frac{1}{xy^2} = \frac{1}{x^2} + \frac{y^2 + 2x\cdot y}{x^2xy^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{(1y^2 + 2x\cdot y)^2}{ xy^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}{xy^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}{ xy^2}e^{(2\sqrt e)y} {\rm d} y


.


\end{equation}


We split the integral into two parts: $yx>x$ and $yx<x$.


We have, (recalling $x>\frac1{\sqrt e}$),


\begin{equation}


\int_{yx>x} \frac{(y^2 + 2x\cdot y)^2}{ xy^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_{yx<x} \frac{(y^2 + 2x\cdot y)^2}{ xy^2}e^{(2\sqrt e)y} {\rm d} y


\leqslant


e^{\sqrt ex}\int_{yx<x} \frac{(y^2 + 2xy)^2}{ xy^2} {\rm d} y


\leqslant


x^5e^{\sqrt ex}C


.


\end{equation}


Therefore, for all $x$,


\begin{equation}


f_1(x)\geqslant


\frac{1}{x^4}C(e^{\frac12}+e^2x^4e^{\sqrt ex})


.


\end{equation}


Finally, by use\~(\ref{boundf2}),


\begin{equation}


x^4\left(\delta(x)+\frac{(\beta+1)e}{\pi}\frac{e^{2\sqrt ex}}{x}\right)\ast f_1\ast f_2(x)\geqslant


Ce^{\frac12}


.


\end{equation}


All in all, by\~(\ref{U1}), (since $x^4e^{\frac32}e^{2\sqrt ex}<Ce^{\frac12}$)


\begin{equation}


x^4U_1(x)\leqslant Ce^{\frac12}


.


\label{boundU1}


\end{equation}


\bigskip




\point


We now show that $\Delta^2\widehat U_2$ is integrable and squareintegrable.



@ 716,6 +767,7 @@ This provides the control on $\u'\_q$ that we need to prove the theorem on co


\begin{equation}


16e^2\Delta^2\equiv\partial_\kappa^4+\frac4\kappa\partial_\kappa^3


.


\label{D2}


\end{equation}


We have, by the Leibniz rule,


\begin{equation}



@ 739,6 +791,7 @@ This provides the control on $\u'\_q$ that we need to prove the theorem on co


for some family of constants $c_{l_1,\cdots,l_p}^{(p,n)}$ which can easily be computed explicitly, but this is not needed.


Now, since $S\geqslant 0$, $\frac\rho{1e}\widehat S\leqslant 1$, so $\zeta_1\leqslant\frac12$ and $\zeta_1=\frac12$ if and only if $\kappa=0$.


Therefore, $\widehat U_2$ is bounded when $\kappa$ is away from 0, so it suffices to show that $\Delta^2\widehat U_2$ is integrable and square integrable at infinity and at 0.


\bigskip






\subpoint We first consider the behavior at infinity, and assume that $\kappa$ is sufficiently large.



@ 746,7 +799,7 @@ This provides the control on $\u'\_q$ that we need to prove the theorem on co


To prove the corresponding claim for $\zeta_1$, we use the fact that $x^4 v$ square integrable, which implies that $\widehat S$ is as well.


Therefore, by\~(\ref{zetas}) for $0\leqslant n\leqslant 4$, $\kappa^2\partial_\kappa^n\zeta_1$ is integrable at infinity, and, therefore, squareintegrable at infinity.


Furthermore, by\~(\ref{zetas}), $\zeta_1<\frac12\epsilon$ for large $\kappa$, and $\partial^n\zeta_1$ is bounded, so $\partial_\kappa^{ni}(\kappa^2+1)\partial_\kappa^i(1\sqrt{12\zeta_1})$ is integrable and square integrable.




\bigskip




\subpoint As $\kappa\to0$


\begin{equation}



@ 787,7 +840,7 @@ This provides the control on $\u'\_q$ that we need to prove the theorem on co


=O(\kappa^{1i})


.


\end{equation}


Thus, by\~(\ref{leibnitz}),


Thus, by\~(\ref{leibnitz}), as $\kappa\to0$,


\begin{equation}


\partial_\kappa^4\widehat U_2=O(\kappa^{1})


,\quad



@ 795,7 +848,15 @@ This provides the control on $\u'\_q$ that we need to prove the theorem on co


.


\end{equation}


Thus, $\Delta^2\widehat U_2$ is integrable and square integrable.


And since the $O(\cdot)$ hold uniformly in $e$ on all compact sets, $U_2x^4$ is bounded and square integrable uniformly in $e$ on all compact sets.


And since the $O(\cdot)$ hold uniformly in $e$ on all compact sets, by\~(\ref{D2}),


\begin{equation}


x^4U_2(x)


\leqslant


\frac{8e^{\frac32}}{16e^2}\int \left(\partial_{k}^4+\frac4{k}\partial_{k}^3\right)\hat U_2(k)\ dk


\leqslant\frac C{\sqrt e}


.


\end{equation}


This along with\~(\ref{U1decay}) and\~(\ref{boundU1}) implies\~(\ref{udecay}) and\~(\ref{fourdecay}).


\qed







@ 1849,7 +1910,7 @@ $$


$$


\end{proof}




It now follows that with $e_\star$ define as in Theorem~\ref{Mon}, on account of the bound on $\rho'$ proved there, and on account of Theorem~\ref{theo:pointwise} that there is a constant independent of $e$ such that for all $e\leqslant e_\star$,


It now follows that with $e_\star$ defined as in Theorem~\ref{Mon}, on account of the bound on $\rho'$ proved there, and on account of Theorem~\ref{theo:pointwise} that there is a constant independent of $e$ such that for all $e\leqslant e_\star$,


\begin{equation}\label{varphipointwise}


\varphi(x) \leqslant Ce^{3/2}(1 + x^4)^{1}\ .


\end{equation}





@ 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].




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