Update to v0.1:
Added: references to [Nelson, 1964] and [Reed, Simon, 1975]. Fixed: sign of Laplacian in definition of \mathfrak K_e. Added: references to proofs in introduction. Fixed: miscellaneous typos.
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@ 114,7 +114,7 @@ Along with the computation of the low density energy of the simple equation in o


,\quad


\frac{2e}{\rho}=\int (1u(x))v(x)\ dx


\end{equation}


to be solved for an integrable function $u$ on $\R^3$ where $v$ is a given nonnegative radial function representing a repulsive interaction between particles with $(1+x^4)v\in L^1(\R^3)\cap L^2(\R^3)$, and where


to be solved for an integrable function $u$ on $\R^3$ where $v$ is a given nonnegative radial function representing a repulsive interaction between particles with $(1+x^4)v\in L^1(\R^3)\cap L^2(\R^3)$, and


where $e$ and $\rho$ are positive parameters representing, respectively, the energy perparticle and the density in the ground state of a Bose gas, and are related by the second equation in\~(\ref{simp}). As we explain below, the solution $u(x)$ specifies a pair correlation function for the Bose gas in terms of which many observable of physical interest can be computed.


This system was first introduced in \cite{Li63,LS64,LL64} and the equation on the left is referred to here as the {\em simple equation}; it results from applying some approximations to a more complicated equation derived in \cite{Li63}.


For the reader interested in the origins of this equation, we give a brief account of its derivation and motivation.


@ 123,15 +123,15 @@ Along with the computation of the low density energy of the simple equation in o


H_N:=\frac12\sum_{i=1}^N\Delta_i+\sum_{i<j}v(x_ix_j)


\end{equation}


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 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.


In \cite{Li63}, it is argued that, 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, the limiting twopoint correlation function


\begin{equation}


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}


The function $u(x)$ in \eqref{simp} is then defined as $u(x):=1 g(x)$. Note that since by definition $g(x) \geqslant 0$, $u(x) \leqslant 1$.


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 )}


\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 $g_{\infty}(x) \geqslant 0$, $u(x) \leqslant 1$.






Because the ground state expected values of many physical observables can be calculated in terms of $g$, any method for computing $g$ that bypasses direct solution of 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


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


understanding of how closely the solutions of\~(\ref{simp}) approximate the actual two point correlation function associated to the $N$body ground state $\psi_0$. However, we have conducted extensive numerical work, about\~(\ref{simp}) and other, more refined equations, and have found that these equations are surprisingly accurate.


Details on the numerical results will be published elsewhere\~\cite{CHe}.




@ 150,7 +150,7 @@ Along with the computation of the low density energy of the simple equation in o






If it is indeed true that the simple equation describes the manybody Bose gas in the thermodynamic limit with meaningful accuracy, then it seems important to understand this equation beyond simple numerics.


We have started this effort in a previous publication\~\cite{CJL20}, where we showed that, under the assumption that $v\geqslant 0$ and that $v\in L_1(\mathbb R^d)\cap L_{\frac32+\epsilon}(\mathbb R^d)$ but not necessarily radial, then in each dimension $d$, for each $e>0$, there is a unique value $\rho(e)$ for which \eqref{simp} has an integrable solution satisfying $u \leqslant 1$, and for each $e>0$, there is exactly one integrable solution $u$ with $u \leqslant 1$. (Recall that $u\leqslant 1$ is equivalent to $g\geqslant 0$, a necessary condition for the solution to be physically meaningful.) We also proved that all such solutions are necessarily positive, so that


We have started this effort in a previous publication\~\cite{CJL20}, where we showed that, under the assumption that $v\geqslant 0$ and that $v\in L^1(\mathbb R^d)\cap L^{\frac32+\epsilon}(\mathbb R^d)$ but not necessarily radial, then in each dimension $d$, for each $e>0$, there is a unique value $\rho(e)$ for which \eqref{simp} has an integrable solution satisfying $u \leqslant 1$, and for each $e>0$, there is exactly one integrable solution $u$ with $u \leqslant 1$. (Recall that $u\leqslant 1$ is equivalent to $g\geqslant 0$, a necessary condition for the solution to be physically meaningful.) We also proved that all such solutions are necessarily nonnegative, so that


\begin{equation}\label{uinf}


0\leqslant u(x)\leqslant 1


.


@ 165,7 +165,7 @@ Along with the computation of the low density energy of the simple equation in o








In addition, we showed that, under the further assumption that $v$ is of positive type (its Fourier transform is nonnegative), the quantity $e$ defined in\~(\ref{simp}) coincides with the ground stateenergy per particle of the manybody Bose gas, asymptotically both for low and high values of $\rho$.


In addition, we showed that, under the further assumption that $v$ is of positive type (its Fourier transform is nonnegative), the quantity $e$ defined in\~(\ref{simp}) coincides with the ground stateenergy per particle of the manybody Bose gas, asymptotically both for small and large values of $\rho$.


Finally, we showed that, if the potential $v$ is spherically symmetric and decays exponentially, then $u\simx^{4}$ for large $x$.




In the present paper we take this analysis further, and prove some of the conjectures in\~\cite{CJL20}, namely that the map $\rho\mapsto e(\rho)$ is strictly monotone increasing for small and for large $\rho$, as well as the fact that the map $\rho\mapsto \rho e(\rho)$ is convex for small values of $\rho$.


@ 188,13 +188,13 @@ Along with the computation of the low density energy of the simple equation in o


While the results presented in this paper may seem disparate, for the most part they are obtained through the use of a common set of mathematical tools.


To see this, let us first consider the monotonicity result.


To prove that the map $e\mapsto\rho(e)$ is monotone increasing,


formally differentiate\~(\ref{simp}) with respect to $e$, and find that, denoting derivatives with respect to $e$ by primes, to find


formally differentiate\~(\ref{simp}) with respect to $e$, and find that, denoting derivatives with respect to $e$ by primes,


\begin{equation}\label{upform}


u'=\mathfrak K_e(4u+2\rho u\ast u+2\rho'u\ast u)


\end{equation}


with


\begin{equation}\label{bg6}


\mathfrak{K}_e = (\Delta + v + 4e(1  C_{\rho u}))^{1}


\mathfrak{K}_e = (\Delta + v + 4e(1  C_{\rho u}))^{1}


\end{equation}


in which $C_{\rho u}$ denotes the convolution by $\rho u$. Now, differentiating the second equation in \eqref{simp} in $e$ yields


\begin{equation}\label{bg6X}


@ 207,7 +207,7 @@ Along with the computation of the low density energy of the simple equation in o








Justifying these formal calculations and analyzing the resulting expression for $\rho'$, we can prove its strict positivity at all sufficiently low or high densities, and in some cases, depending on $v$, for all densities. It is easy to see that the same operator $\mathfrak{K}_e$ will again show up in the computations we do to prove convexity of $e\rho(e)$. It is probably less clear that it will again show up when we derive formulas for other observable such as the condensate fraction, and we now explain why this is the case.


Justifying these formal calculations and analyzing the resulting expression for $\rho'$, we will prove its strict positivity at all sufficiently low or high densities, and in some cases, depending on $v$, for all densities (see Theorem~\ref{Mon}). It is easy to see that the same operator $\mathfrak{K}_e$ will again show up in the computations we do to prove convexity of $e\rho(e)$. It is probably less clear that it will again show up when we derive formulas for other observable such as the condensate fraction, and we now explain why this is the case.






Let $A$ be a self adjoint operator on the $N$particle Hilbert space, representing some observable whose ground state expectation value $\langle \psi_0,A\psi_0\rangle$ we would like to evaluate.


@ 250,9 +250,9 @@ Along with the computation of the low density energy of the simple equation in o


.


\label{eta}


\end{equation}


Differentiating \eqref{simpleq_eta} leads once more to the operator $\mathfrak{K}_\mu$.


Note that, since approximations were made in computing the two particle correlation function, it is not immediately clear that the quantity $\eta$ defined in \eqref{etaf} satisfies $0 \leqslant \eta \leqslant 1$.


In the rest of this paper, we always use $\eta$ to mean the quantity defined in \eqref{eta}, and not the true condensate fraction, defined in \eqref{cond2}.


Differentiating \eqref{simpleq_eta} leads once more to the operator $\mathfrak{K}_{e_\mu}$.


Note that, since approximations were made in computing the twopoint correlation function, it is not immediately clear that the quantity $\eta$ defined in \eqref{etaf} satisfies $0 \leqslant \eta \leqslant 1$.


In the rest of this paper, we always use $\eta$ to mean the quantity defined in \eqref{eta}, and not the true uncondensed fraction, defined in \eqref{cond2}.


We shall see that at least for small $\rho$, the approximation is very good.




\indent


@ 272,7 +272,7 @@ A well known prediction\~\cite{CAL09} is that, for a deltafunction potential,


\label{tan}


\end{equation}


which is knwon as the {\it universal Tan relation}\~\cite{Ta08,Ta08b,Ta08c}.


We have found that the simple equation reproduces this prediction, even when the potential is finite, when the density is asymptotically small, see theorem\~\ref{theo:tan} below.


We have found that the simple equation reproduces this prediction, even when the potential is finite, when the density is asymptotically small (see Theorem\~\ref{theo:tan}).


To compute an approximation for $\mathfrak M(k)$, we follow the same procedure as above, which leads us to the following equation: for $k\neq0$,


\begin{equation}


(\Delta+4e_\mu) u_\mu=(1u_\mu)v+2\rho e_\mu u_\mu\ast u_\mu


@ 288,7 +288,7 @@ and


\label{gamma1}


\end{equation}




Differentiating \eqref{simpleq_momentum} leads once more to the operator $\mathfrak{K}_\mu$.


Differentiating \eqref{simpleq_momentum} leads once more to the operator $\mathfrak{K}_{e_\mu}$.




Therefore, a significant part of the analysis in this paper is aimed at understanding the operator $\mathfrak K_e$, as well as properties of solutions $u$ of the simple equation.


Consider for example the problem of showing that $\rho'(e) > 0$ using the formula in \eqref{rpratio}. We will need to have $L^p$ to $L^q$ mapping properties of $\mathfrak{K}_e$, among other things, but all $L^p$ bounds on solutions $u$ of the simple equation system.


@ 297,11 +297,11 @@ Integrating both sides of the simple equation, one sees that all solutions of th


\int u(x)\ dx=\frac1\rho \ .


\end{equation}


Then since all physical solutions (those satisfying $u(x) \leqslant 1$) satisfy $0 \leqslant u(x) \leqslant 1$, it follows that $u\in L^p(\R^3)$ for all $1 \leqslant p \leqslant \infty$, and the obvious estimate that follows from this information is


$\u\_p \leqslant \rho^{1/p}$. However, one can do significantly better. We shall prove:


$\u\_p \leqslant \rho^{1/p}$. However, one can do significantly better. We shall prove the following lemma in section~\ref{UB}.




\begin{lemma}\label{ul2b} For $1\leqslant p < 3$, solutions $u$ of \eqref{simp} satisfy


\begin{equation}\label{sim6B}


\u\_p \leqslant C_p e^{(p3/2p}\qquad{\rm where}\qquad C_p := 2(4\pi)^{1/p 1} \Gamma^{1/p}(3p) (2p)^{(p3)/p}\v\_1 \ .


\u\_p \leqslant C_p e^{(p3)/2p}\qquad{\rm where}\qquad C_p := 2(4\pi)^{1/p 1} \Gamma^{1/p}(3p) (2p)^{(p3)/p}\v\_1 \ .


\end{equation}


In particular,


\begin{equation}\label{sim6}


@ 349,13 +349,13 @@ We shall also need various $L^p$ bounds on $u'$, and for these we need a detaile






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 proved them for small density $\rho$ (and, in the case of the monotonicity, also for large density).


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).




\begin{theorem}[Monotonicity]\label{Mon} Assume that $(1+ x^4)v(x)\in L^1(\mathbb R^3)\cap L^2(\mathbb R^3)$.


For


$$


e < e_\star:=\frac{\sqrt 2 \pi^{3}}{\v\_1^2}\qquad{\rm and\ for}\qquad e > \frac{2^3\v\_2^4}{\pi^4} $$


$\rho(e)$ is strictly monotone increasing in $e$, and in these intervals $\rho(e)$ is continuously differentiable. If $u(e,\cdot)$ denotes the solution as a function of $e$, $u(e,\cdot)$ is continuously differentiable in $L^2(\R^3)$. Moreover,


$\rho(e)$ is strictly monotone increasing in $e$, and in these intervals $\rho(e)$ is continuously differentiable. If $u(e,\cdot)$ denotes the solution of (\ref{simp}) as a function of $e$, $u(e,\cdot)$ is continuously differentiable in $L^2(\R^3)$. Moreover,


\begin{equation}\label{rhopb2}


\mathrm{for}\ e < e_\star\equiv\frac{\sqrt 2 \pi^{3}}{\v\_1^2}


\quad\mathrm{we\ have}\quad


@ 457,10 +457,10 @@ On account of \eqref{intu}, $\rho u$ is a probability density, and


\begin{equation}


0 \leqslant 4e(I  C_{\rho u}) \leqslant 4e


\end{equation}


so that $\mathfrak K_e$ is an unbounded operator. However, $e^{t4e(I  C_{\rho u})}$ is easily seen to be a positivity preserving contraction semigroup on $L^p$ for all $p$, as is $e^{t(\Delta + v)}$. Then by the Trotter Product formula, so is


$e^{t(\Delta + v + 4e(1  C_{\rho u})}$. Since


so that $\mathfrak K_e$, as an operator from $L^1(\mathbb R^3)$ to $L^1(\mathbb R^3)$, is unbounded. However, $e^{t4e(I  C_{\rho u})}$ is easily seen to be a positivity preserving contraction semigroup on $L^p$ for all $p$, as is $e^{t(\Delta + v)}$~\cite{Ne64,RS75b}. Then by the Trotter Product formula, so is


$e^{t(\Delta + v + 4e(1  C_{\rho u})}$. Since


\begin{equation}\label{intrep}


\mathfrak{K}_e = \int_0^\infty dt e^{t(\Delta + v + 4e(1  C_{\rho u})}\ ,


\mathfrak{K}_e = \int_0^\infty dt e^{t(\Delta + v + 4e(1  C_{\rho u})}\ ,


\end{equation}


$\mathfrak{K}_e$ has a positive kernel denoted $\mathfrak{K}_e(x,y)$. We also define the convolution operator


\begin{equation}\label{sim76}


@ 480,16 +480,16 @@ $$


$$


Fourier transforming the simple equation, one finds


\begin{equation}\label{uhat}


\rho\hat u(k)=\frac{k^2}{4e}+1\sqrt{\left(\frac{k^2}{4e}+1\right)^2\frac\rho{2e}\hat S(k)}


\rho\widehat u(k)=\frac{k^2}{4e}+1\sqrt{\left(\frac{k^2}{4e}+1\right)^2\frac\rho{2e}\widehat S(k)}


,\quad


\hat S(k):=\int dx\ e^{ikx}(1u(x))v(x)\ .


\widehat S(k):=\int dx\ e^{ikx}(1u(x))v(x)\ .


\end{equation}


By \eqref{intu}, $\rho \widehat{u}(0) =1$ and by the second equation in \eqref{simp}, $\frac{\rho}{2e}\widehat{S}(0) =1$,


and from here one obtains


\begin{equation}\label{facA}


\left( k^2 + 4e(1  \rho \widehat{u}(k)) \right)^{1} \leqslant k^{1}\left(k^2 + 2\sqrt{2e}\right)^{1/2}


\end{equation}


The right side is square integrable, and in this way we obtain a bound on $\mathfrak{Y}_e$, and hence on $\mathfrak{K}_e$, from $L^1(\R^3)$ to $L^2(\R^3)$. The following lemma summarizes information that we obtain on $\mathfrak{K}_e$ that suffices to prove Theorem~\ref{Mon} on monotonicity.


The right side is square integrable, and in this way we obtain a bound on $\mathfrak{Y}_e$ from $L^1(\R^3)$ to $L^2(\R^3)$. The following lemma (proved in section~\ref{UB}) summarizes information that we obtain on $\mathfrak{K}_e$ that suffices to prove Theorem~\ref{Mon} on monotonicity.








@ 524,9 +524,9 @@ e\mapsto \int_{\R^3} \phi(x) ( \mathfrak{K}_e \psi)(x)\ dx


\rho^{2}\int_{\R^3} (\mathfrak{K}_e v) u'*u'\ dx


\end{equation}


which shows that for small $\rho$, this is negligible compared to $\rho^{2}$.


By Young's inequality for convolutions, if we have bounds on $\\mathfrak{K}_e v\_p$ and $\u'\_q$ with $1/p + 2/q = 2$, we can bound the integral in \eqref{toughnut}. As we shall see below, since $v\geqslant 0$, $\mathfrak{K}_e v$ can decay at infinity no faster than $x^{2}$, and hence cannot belong to $L^p$ for $p \leqslant 3/2$. Therefore, we will need to have a bound on $\u'\_q$ for fairly small $q$. We shall see that $\u'\_q < \infty$ for all $q>1$, and we shall obtain a bound on $\u'\_{4/3}$ that can be combined with our bound on $\\mathfrak{K}_e v\_2$ to obtain the necessary control on the integral in \eqref{toughnut}.


By Young's inequality for convolutions, if we have bounds on $\\mathfrak{K}_e v\_p$ and $\u'\_q$ with $1/p + 2/q = 2$, we can bound the integral in \eqref{toughnut}. As we shall see below, since $v\geqslant 0$, $\mathfrak{K}_e v$ can decay at infinity no faster than $x^{2}$, and hence cannot belong to $L^p$ for $p \leqslant 3/2$. Therefore, we will need to have a bound on $\u'\_q$ for fairly small $q$. We shall see that $\u'\_q < \infty$ for all $q>1$ (see Theorem~\ref{Lpu'}), and we shall obtain a bound on $\u'\_{4/3}$ that can be combined with our bound on $\\mathfrak{K}_e v\_2$ to obtain the necessary control on the integral in \eqref{toughnut}.




To do this, we need something more incisive than the bound \eqref{facA}. We shall show that


To do this, we need something more incisive than the bound \eqref{facA}. We shall show (see section\~\ref{sec:uprime}) that


$\mathfrak{Y}_e$, factors as the product of three commuting operators operators


\begin{equation}\label{3fac}


\mathfrak{Y}_e = (\Delta)^{1/2} (\Delta+ 8e)^{1/2} [I + \mathfrak{H}_e]


@ 614,53 +614,53 @@ This provides the control on $\u'\_q$ that we need to prove the theorem on co


\end{equation}


in terms of which (\ref{uhat}) becomes


\begin{equation}


\rho\hat u=(\kappa^2+1)\left(1\sqrt{1\frac{\frac\rho{2e}\hat S}{(\kappa^2+1)^2}}\right)


\rho\widehat u=(\kappa^2+1)\left(1\sqrt{1\frac{\frac\rho{2e}\widehat S}{(\kappa^2+1)^2}}\right)


.


\label{uhat_factorized}


\end{equation}


For small $\kappa$, since $x^4v$ is integrable, $\hat S$ is $\mathcal C^4$


For small $\kappa$, since $x^4v$ is integrable, $\widehat S$ is $\mathcal C^4$


\begin{equation}


\frac\rho{2e}\hat S=1\beta \kappa^2+O(e^2\kappa^4)


\frac\rho{2e}\widehat S=1\beta \kappa^2+O(e^2\kappa^4)


\label{expS}


\end{equation}


and $\beta$ is defined in\~(\ref{betadef}):


\begin{equation}\label{betabound}


\beta = \frac\rho{4e}\partial_\kappa^2 \hat S \leqslant \rho\x^2v\_1


\beta = \frac\rho{4e}\partial_\kappa^2 \widehat S \leqslant \rho\x^2v\_1


.


\end{equation}


Therefore, defining


\begin{equation}


\hat U_1:=(\kappa^2+1)^{2}\left(1\sqrt{1\frac{(1\beta \kappa^2)}{(\kappa^2+1)^2}}\right)


\widehat U_1:=(\kappa^2+1)^{2}\left(1\sqrt{1\frac{(1\beta \kappa^2)}{(\kappa^2+1)^2}}\right)


\end{equation}


$\hat U_1$ coincides with $\hat u$ asymptotically as $\kappa\to0$


and we chose the prefactor $(\kappa^2+1)^{2}$ in such a way that $\hat U_1$ is integrable.


$\widehat U_1$ coincides with $\widehat u$ asymptotically as $\kappa\to0$


and we chose the prefactor $(\kappa^2+1)^{2}$ in such a way that $\widehat U_1$ is integrable.


Define the remainder term


\begin{equation}


\hat U_2:=\rho\hat u\hat U_1


\widehat U_2:=\rho\widehat u\widehat U_1


=(\kappa^2+1)\left(1\sqrt{12\zeta_1}\right)(\kappa^2+1)^{2}\left(1\sqrt{12\zeta_2}\right)


\end{equation}


with


\begin{equation}


\zeta_1:=\frac{\frac\rho{4e}\hat S}{(\kappa^2+1)^2}


\zeta_1:=\frac{\frac\rho{4e}\widehat S}{(\kappa^2+1)^2}


,\quad


\zeta_2:=\frac{1\beta \kappa^2}{2(\kappa^2+1)^2}


.


\label{zetas}


\end{equation}


The rest of the proof proceeds as follows: we show that the Fourier transform of $\hat U_1$ decays like $x^{4}$ by direct analysis, then we show that $\Delta^2\hat U_2$ is integrable and square integrable, which implies that it is subdominant as $x\to\infty$.


The rest of the proof proceeds as follows: we show that the Fourier transform of $\widehat U_1$ decays like $x^{4}$ by direct analysis, then we show that $\Delta^2\widehat U_2$ is integrable and square integrable, which implies that it is subdominant as $x\to\infty$.






\point We compute $U_1(x):=\int\frac{dk}{(2\pi)^3}e^{ikx}\hat U_1(k)$.


\point We compute $U_1(x):=\int\frac{dk}{(2\pi)^3}e^{ikx}\widehat U_1(k)$.


We write


\begin{equation}


\sqrt{1  \frac{1\beta \kappa^2}{(1+\kappa^2)^2}} = \frac{\kappa}{1+\kappa^2} \sqrt{2 + \beta + \kappa^2} = \frac{1}{\pi} \frac{\kappa(2 + \beta + \kappa^2)}{1+\kappa^2} \int_0^\infty \frac{1}{2+\beta + t + \kappa^2} t^{1/2} dt \ \ .


\end{equation}


Therefore,


\begin{equation}


\hat{U}_1 := (\kappa^2+1)^{2}  \frac{\kappa}{\pi} (\kappa^2+1)^{2} \left(1 + (\beta+1)\frac{1}{1+\kappa^2}\right) \int_0^\infty \frac{1}{2+\beta + t + \kappa^2} t^{1/2} dt


\widehat{U}_1 := (\kappa^2+1)^{2}  \frac{\kappa}{\pi} (\kappa^2+1)^{2} \left(1 + (\beta+1)\frac{1}{1+\kappa^2}\right) \int_0^\infty \frac{1}{2+\beta + t + \kappa^2} t^{1/2} dt


.


\end{equation}


We take the inverse Fourier transform of $\hat U_1$, recalling the definition of $\kappa$\~(\ref{kappa})


We take the inverse Fourier transform of $\widehat U_1$, recalling the definition of $\kappa$\~(\ref{kappa})


\begin{equation}


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)


@ 711,7 +711,7 @@ This provides the control on $\u'\_q$ that we need to prove the theorem on co


\end{equation}




\point


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


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


We use the fact that


\begin{equation}


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


@ 719,7 +719,7 @@ This provides the control on $\u'\_q$ that we need to prove the theorem on co


\end{equation}


We have, by the Leibniz rule,


\begin{equation}


\partial_\kappa^n\hat U_2=


\partial_\kappa^n\widehat U_2=


\sum_{i=0}^n{n\choose i}\left(


\partial_\kappa^{ni}(\kappa^2+1)\partial_\kappa^i(1\sqrt{12\zeta_1})





@ 737,13 +737,13 @@ This provides the control on $\u'\_q$ that we need to prove the theorem on co


\label{dsqrt}


\end{equation}


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


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


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.






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


The fact that $\partial_\kappa^{ni}(\kappa^2+1)^{2}\partial_\kappa^i(1\sqrt{12\zeta_2})$ is integrable and square integrable at infinity follows immediately from\~(\ref{zetas}).


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


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.




@ 789,12 +789,12 @@ This provides the control on $\u'\_q$ that we need to prove the theorem on co


\end{equation}


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


\begin{equation}


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


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


,\quad


\frac4\kappa\partial_\kappa^3\hat U_2=O(\kappa^{1})


\frac4\kappa\partial_\kappa^3\widehat U_2=O(\kappa^{1})


.


\end{equation}


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


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.


\qed




@ 1127,21 +1127,21 @@ in which $\mathfrak Y_e$ is defined in \eqref{sim76}.


uniformly in $x$.


We work in Fourier space: by \eqref{uhat},


\begin{equation}


\hat\xi(2\sqrt{e}k)=


\widehat\xi(2\sqrt{e}k)=


\frac1{4e}


\left(\frac{k^2+1}{\sqrt{(k^2+1)^2\frac\rho{2e}\hat S(2\sqrt{e}k)}}1\right)


\left(\frac{k^2+1}{\sqrt{(k^2+1)^2\frac\rho{2e}\widehat S(2\sqrt{e}k)}}1\right)


.


\label{x1rescale}


\end{equation}


Since $S(x)\geqslant 0$, $\hat S(k)\leqslant\hat S(0)=\frac{2e}\rho$, and since $S$ is symmetric, $\hat S$ is real, so


Since $S(x)\geqslant 0$, $\widehat S(k)\leqslant\widehat S(0)=\frac{2e}\rho$, and since $S$ is symmetric, $\widehat S$ is real, so


\begin{equation}


\hat\xi(2\sqrt{e}k)\leqslant


\widehat\xi(2\sqrt{e}k)\leqslant


\frac1{4e}


\left(\frac{k^2+1}{\sqrt{(k^2+1)^21}}\frac{k^2+1}{\sqrt{(k^2+1)^2+1}}\right)


\label{dominatedxi1}


\end{equation}


which is integrable.


Next, note that $\frac\rho{2e}\hat S(2\sqrt e k)\to1$ and


Next, note that $\frac\rho{2e}\widehat S(2\sqrt e k)\to1$ and


\begin{equation}\label{idint}


\int\left(\frac{k^2+1}{\sqrt{(k^2+1)^21}}1\right)\ \frac{dk}{8\pi^3}=\frac1{3\pi^2\sqrt2}


\end{equation}


@ 1153,13 +1153,13 @@ Next, by\~(\ref{x1rescale}) and\~(\ref{idint}),


=&


\frac{\sqrt{e}}{4\pi^3} \int (e^{i2\sqrt{e}kx}1)


\left(


\frac{k^2+1}{\sqrt{(k^2+1)^2\frac \rho{2e}\hat S(2\sqrt{e}k)}}1


\frac{k^2+1}{\sqrt{(k^2+1)^2\frac \rho{2e}\widehat S(2\sqrt{e}k)}}1


\right)\ dk


\\[0.5cm]


&+


\frac{\sqrt{e}}{4\pi^3}\int


\left(


\frac{k^2+1}{\sqrt{(k^2+1)^2\frac \rho{2e}\hat S(2\sqrt{e}k)}}


\frac{k^2+1}{\sqrt{(k^2+1)^2\frac \rho{2e}\widehat S(2\sqrt{e}k)}}


\frac{k^2+1}{\sqrt{(k^2+1)^21}}


\right)\ dk


.


@ 1169,25 +1169,25 @@ By\~(\ref{dominatedxi1}), the first integrand is absolutely integrable, so


\begin{equation}


\frac{\sqrt{e}}{4\pi^3} \int (e^{i2\sqrt{e}kx}1)


\left(


\frac{k^2+1}{\sqrt{(k^2+1)^2\frac \rho{2e}\hat S(2\sqrt{e}k)}}1


\frac{k^2+1}{\sqrt{(k^2+1)^2\frac \rho{2e}\widehat S(2\sqrt{e}k)}}1


\right)\ dk


=o(\sqrt e)


\end{equation}


uniformly in $x$.


Furthermore, since $\frac\rho{2e}\hat S\leqslant 1$ and $1(1+\epsilon)^{\frac12}\leqslant\frac\epsilon 2$ for all $\epsilon\geqslant 0$,


Furthermore, since $\frac\rho{2e}\widehat S\leqslant 1$ and $1(1+\epsilon)^{\frac12}\leqslant\frac\epsilon 2$ for all $\epsilon\geqslant 0$,


\begin{equation}


\left


\frac{k^2+1}{\sqrt{(k^2+1)^2\frac \rho{2e}\hat S(2\sqrt{e}k)}}


\frac{k^2+1}{\sqrt{(k^2+1)^2\frac \rho{2e}\widehat S(2\sqrt{e}k)}}


\frac{k^2+1}{\sqrt{(k^2+1)^21}}


\right


\leqslant


\frac{k^2+1}{((k^2+1)^21)^{\frac32}}\frac{1\frac\rho{2e}\hat S(2\sqrt{e}k)}2


\frac{k^2+1}{((k^2+1)^21)^{\frac32}}\frac{1\frac\rho{2e}\widehat S(2\sqrt{e}k)}2


\end{equation}


and since $\hat S$ is the Fourier transform of $(1u)v$ which is absolutely integrable, $\hat S$ is uniformly continuous, so


and since $\widehat S$ is the Fourier transform of $(1u)v$ which is absolutely integrable, $\widehat S$ is uniformly continuous, so


\begin{equation}


\frac{\sqrt{e}}{4\pi^3}\int dk\


\left


\frac{k^2+1}{\sqrt{(k^2+1)^2\frac \rho{2e}\hat S(2\sqrt{e}k)}}


\frac{k^2+1}{\sqrt{(k^2+1)^2\frac \rho{2e}\widehat S(2\sqrt{e}k)}}


\frac{k^2+1}{\sqrt{(k^2+1)^21}}


\right


=o(\sqrt{e})


@ 1261,7 +1261,7 @@ We now turn to the proof of\~(\ref{M_asym}).


First of all, by\~(\ref{neglect_denom}),


\begin{equation}


\mathfrak M(k)=


\rho\hat u_0(k)(1+O(\rho))\int v(x)\mathfrak K_e\cos(k\cdot x)\ dx


\rho\widehat u_0(k)(1+O(\rho))\int v(x)\mathfrak K_e\cos(k\cdot x)\ dx


\end{equation}


Proceeding as in section\~\ref{sec:condensate_fraction}, we use the resolvent identity to rewrite


\begin{equation}


@ 1275,7 +1275,7 @@ in which $\mathfrak Y_e$ is defined in \eqref{sim76}, so


\begin{equation}


\int v(x)\mathfrak K_e\cos(kx)\ dx


=


\frac{\hat v(k)\int e^{ikx}v\mathfrak K_e v\ dx}{k^2+4e(1\rho \hat u(k))}


\frac{\widehat v(k)\int e^{ikx}v\mathfrak K_e v\ dx}{k^2+4e(1\rho \widehat u(k))}


.


\end{equation}


Since $v\mathfrak K_e \leqslant v(\Delta+v)^{1}v$ which is integrable: by\~(\ref{intscat})


@ 1288,7 +1288,7 @@ we have, by dominated convergence, in the limit $e\to0$ and $k\to0$,


\begin{equation}


\int v(x)\mathfrak K_e\cos(kx)\ dx


\sim


\frac{4\pi a_0}{k^2+4e(1\rho\hat u(k))}


\frac{4\pi a_0}{k^2+4e(1\rho\widehat u(k))}


\end{equation}


so, as $\kappa\to\infty$,


\begin{equation}


@ 1378,21 +1378,21 @@ Let $\mathfrak{Y}_e$ be defined as in \eqref{sim76}. Then the positive kernels o


$$ 0 \leqslant \mathfrak{K}_e \psi \leqslant \mathfrak{Y}_e\psi \qquad{\rm and\ hence}\qquad \\mathfrak{K}_e \psi\_2 \leqslant \\mathfrak{Y}_e \psi\_2 \ .$$


Then by \eqref{sim76}, for nonnegative $\psi\in L^1$, $\mathfrak{Y}_e\psi\in L^2(\R^3)$ with


\begin{equation}\label{sim30}\


\\mathfrak{Y}_e\psi \_2^2 \leqslant \frac{\\psi\_1^2}{(2\pi)^3} \int dk [k^2 +4e (1 \rho\hat u(k))]^{2} \ .


\\mathfrak{Y}_e\psi \_2^2 \leqslant \frac{\\psi\_1^2}{(2\pi)^3} \int dk [k^2 +4e (1 \rho\widehat u(k))]^{2} \ .


\end{equation}








Recall from \eqref{uhat} that


$$\rho \hat{u}(k) = 1 + \frac{k^2}{4e}  \sqrt{ \left(1 + \frac{k^2}{4e}\right)^2 \frac{\rho}{2e}\hat{S}(k)}\quad{\rm where}\quad \hat{S}(k) = \int v(1u)e^{ikx} dx\ ,$$


$$\rho \widehat{u}(k) = 1 + \frac{k^2}{4e}  \sqrt{ \left(1 + \frac{k^2}{4e}\right)^2 \frac{\rho}{2e}\widehat{S}(k)}\quad{\rm where}\quad \widehat{S}(k) = \int v(1u)e^{ikx} dx\ ,$$


and hence


$$(1 \rho \hat{u}(k)) \geqslant \sqrt{ \left(1 + \frac{k^2}{4e}\right)^2 \frac{\rho}{2e}\hat{S}(k)}\ .$$


$$(1 \rho \widehat{u}(k)) \geqslant \sqrt{ \left(1 + \frac{k^2}{4e}\right)^2 \frac{\rho}{2e}\widehat{S}(k)}\ .$$


By \eqref{simp} $\frac{\rho}{2e}S(k) \leqslant 1$ and hence


${\displaystyle


4e(1 \rho \hat{u}(k)) \geqslant \sqrt{k^4 + 8e k^2} \geqslant \sqrt{8e}k}$.


4e(1 \rho \widehat{u}(k)) \geqslant \sqrt{k^4 + 8e k^2} \geqslant \sqrt{8e}k}$.


Therefore


$$


\int [k^2 +4e (1 \rho\hat u(k))]^{2}dk \leqslant \int dk [k^2 + \sqrt{8e}k]^{2} = 4\pi \int_0^\infty dr \frac{r^2}{[r^2 + \sqrt{8e} r]^2} =\frac{2\pi}{\sqrt{2e}}\ .


\int [k^2 +4e (1 \rho\widehat u(k))]^{2}dk \leqslant \int dk [k^2 + \sqrt{8e}k]^{2} = 4\pi \int_0^\infty dr \frac{r^2}{[r^2 + \sqrt{8e} r]^2} =\frac{2\pi}{\sqrt{2e}}\ .


$$


For the final part, we apply the bound just proved to the positive and negative parts of $\psi$ separately.


\end{proof}


@ 1496,7 +1496,7 @@ Then $L_{p,q}$ consists of the measurable functions $f$ such that $\f\_{p,q} <


\begin{equation}\label{isias72}


\{x\ :\ f(x) > \lambda\} \leqslant C\lambda^{p}\ .


\end{equation}


That is $L_{p,\infty}$ is weak $L^p$ and hence $L_{p.\infty} \subset L^p$.


That is $L_{p,\infty}$ is weak $L^p$ and hence $L_{p,\infty} \subset L^p$.


and in this case,


$$


\f\_{p,\infty} \leqslant C^{1/p}


@ 1700,7 +1700,7 @@ $$


\end{proof}






\section{Bounds on $u'$  Proof of Theorem~\ref{Lpu'} } Recall from \eqref{upform} that


\section{Bounds on $u'$  Proof of Theorem~\ref{Lpu'} }\label{sec:uprime} Recall from \eqref{upform} that


\begin{equation}\label{bamb34}


u' = \mathfrak{K}_e \varphi \qquad{\rm where }\quad \varphi = 2\left( 1 + \rho'\frac{e}{\rho}\right)\rho u*u 4u\ ,


\end{equation}


@ 1753,23 +1753,23 @@ It then follows that for $3/2 \leqslant q \leqslant \infty$,




The operator $\mathfrak{Y}_e$ is also a convolution operator, but somewhat more complicated. It has a useful factorization that we now describe.




Note that the Fourier transform renders $\mathfrak{Y}_e$ as the operation of multiplication by $[k^2 + 4e(1 \rho \hat{u}(k))]^{1}$.


Note that the Fourier transform renders $\mathfrak{Y}_e$ as the operation of multiplication by $[k^2 + 4e(1 \rho \widehat{u}(k))]^{1}$.


Recall that


$$\rho \hat{u}(k) = 1 + \frac{k^2}{4e}  \sqrt{ \left(1 + \frac{k^2}{4e}\right)^2 \frac{\rho}{2e}\hat{S}(k)}\ ,$$


$$\rho \widehat{u}(k) = 1 + \frac{k^2}{4e}  \sqrt{ \left(1 + \frac{k^2}{4e}\right)^2 \frac{\rho}{2e}\widehat{S}(k)}\ ,$$


where


$\hat{S}(k) = \int v(1u)e^{ikx} dx$, $\frac{\rho}{2e}S(k) \leqslant 1$ and hence


$\widehat{S}(k) = \int v(1u)e^{ikx} dx$, $\frac{\rho}{2e}S(k) \leqslant 1$ and hence


\begin{equation}\label{sim40}


[k^2 + 4e(1 \rho \hat{u}(k))]^{1}


[k^2 + 4e(1 \rho \widehat{u}(k))]^{1}


= \frac{1+\widehat H(k)}{k(k^2 + 8e)^{1/2}}


\end{equation}


with


$$\widehat{H}(k) := \left(1 + \frac{16e^2(1\frac{\rho}{2e}\hat{S}(k))}{ k^4 + 8ek^2 } \right)^{1/2} 1.$$


Since $\lim_{k\to\infty}\hat{S}(k) = 0$, $\widehat{H}$ is integrable.


$$\widehat{H}(k) := \left(1 + \frac{16e^2(1\frac{\rho}{2e}\widehat{S}(k))}{ k^4 + 8ek^2 } \right)^{1/2} 1.$$


Since $\lim_{k\to\infty}\widehat{S}(k) = 0$, $\widehat{H}$ is integrable.




It is more work to see that its inverse Fourier transform, $H(x)$ is also integrable, but we show this below. It turns out that $H(x)$ is not nonnegative. Had this been the case we would have that


$\H\_1 =\widehat{H}(0)$. To compute this, we expand


\begin{equation}


\frac{\rho}{2e}\hat{S}(k) = 1 + \beta k^2 + O(k^4) .


\frac{\rho}{2e}\widehat{S}(k) = 1 + \beta k^2 + O(k^4) .


\end{equation}


Therefore,


$$


@ 1783,7 +1783,7 @@ and in particular, for a different $C$ still independent of $e$, $\ H\_1 \leq


\end{lemma}


\begin{proof}[Proof of Lemma~\ref{Hlem}]


Recall that


$$\widehat{H}(k) := \left(1 + G(k) \right)^{1/2} 1 \qquad{\rm where}\qquad G(k) := \frac{16e^2(1\frac{\rho}{2e}\hat{S}(k))}{ k^4 + 8ek^2}\ .$$


$$\widehat{H}(k) := \left(1 + G(k) \right)^{1/2} 1 \qquad{\rm where}\qquad G(k) := \frac{16e^2(1\frac{\rho}{2e}\widehat{S}(k))}{ k^4 + 8ek^2}\ .$$


The proof is very similar to that of Theorem~\ref{theo:pointwise}, except simpler: One shows that $\widehat{H}$ and $\Delta^2 \widehat{H}$ are integrable, with $\\widehat{H}\_1 + \\Delta^2 \widehat{H}\_1 \leqslant Ce^{1/2}$. The claim now follows from the RiemannLebesgue lemma.


\end{proof}




@ 1961,7 +1961,7 @@ By direct computations, we find that


\rho=\frac1{\int dx\ u(x)}=\frac{b^3}{c\pi^2}


\label{rho}


,\quad


\hat u(k)=\frac{\pi^2 c}{b^3}e^{\frac{k}b}


\widehat u(k)=\frac{\pi^2 c}{b^3}e^{\frac{k}b}


,\quad


u\ast u=\frac{2\pi^2c^2}{b^3(4+b^2 x^2)^2}


\end{equation}



9
Changelog
Normal file
9
Changelog
Normal file
@ 0,0 +1,9 @@


v0.1:




* Added: references to [Nelson, 1964] and [Reed, Simon, 1975].




* Fixed: sign of Laplacian in definition of \mathfrak K_e.




* Added: references to proofs in introduction.




* Fixed: miscellaneous typos.

@ 1,15 +1,16 @@


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2Fs1095500997180}, year = 2009, month = {apr}, publisher = {Springer Science and Business Media {LLC}}, volume = {135}, number = {56}, pages = {915934}, author = {Alessandro Giuliani and Robert Seiringer}, title = {The Ground State Energy of the Weakly Interacting Bose Gas at High Density}, journal = {Journal of Statistical Physics} }


@article{KLS88, doi = {10.1007/bf01023854}, url = {https://doi.org/10.1007


@ 30,6 +31,8 @@


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2F13616633


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@article{Ne64, doi = {10.1063/1.1704124}, url = {https://doi.org/10.1063


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@book{RS75b, title={Methods of Modern Mathematical Physics II: Fourier Analysis, SelfAdjointness}, author={Reed, Michael and Simon, Barry}, edition={2}, year={1975}, publisher={Academic Press, New York}}



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