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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.
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Ian Jauslin
2021-05-22 18:38:30 -04:00
parent f8a94b06fa
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107
Carlen_Jauslin_Lieb_2020.tex
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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 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 $|x|^4e^{\frac32}e^{-2\sqrt e|x|}<Ce^{-\frac12}$)
\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 square-integrable.
@@ -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, square-integrable at infinity.
Furthermore, by\-~(\ref{zetas}), $\zeta_1<\frac12-\epsilon$ for large $\kappa$, and $\partial^n\zeta_1$ is bounded, so $\partial_\kappa^{n-i}(\kappa^2+1)\partial_\kappa^i(1-\sqrt{1-2\zeta_1})$ 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^{1-i})
.
\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, $U_2|x|^4$ 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}

9
Changelog
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@@ -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].