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@@ -269,6 +269,7 @@ The microscopic state of a system of quantum particles is given by a wavefunctio
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\label{H}
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\label{H}
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\end{equation}
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\end{equation}
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where $\Delta_i$ is the Laplacian with respect to $x_i$.
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where $\Delta_i$ is the Laplacian with respect to $x_i$.
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(Throughout this document, we choose units so that $\hbar=1$; we keep track of the mass because different authors use $m=1$ or $m=1/2$, so we keep it to avoid confusion.)
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This is the microscopic description of the system.
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This is the microscopic description of the system.
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Similarly to the classical case, to treat situations with a huge number of particles, we will approach the system statistically.
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Similarly to the classical case, to treat situations with a huge number of particles, we will approach the system statistically.
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In the classical case, we considered a probability distribution over all possible configurations, without fixing the number of particles.
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In the classical case, we considered a probability distribution over all possible configurations, without fixing the number of particles.
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@@ -635,6 +636,7 @@ Let us consider a more realistic Hamiltonian:
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\label{Ham}
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\label{Ham}
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\end{equation}
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\end{equation}
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where $v$ is the potential that accounts for the interaction between pairs of particles.
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where $v$ is the potential that accounts for the interaction between pairs of particles.
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(Throughout this document, we choose units so that $\hbar=1$; we keep track of the mass because different authors use $m=1$ or $m=1/2$, so we keep it to avoid confusion.)
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This potential could be the Coulomb potential $1/|x|$ for charged particles, or the Lennard-Jones potential for interacting atoms, or something more realistic that accounts for the fine structure of atoms.
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This potential could be the Coulomb potential $1/|x|$ for charged particles, or the Lennard-Jones potential for interacting atoms, or something more realistic that accounts for the fine structure of atoms.
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Here, we will restrict our focus a bit and assume that $v$ is integrable ($v\in L_1(\mathbb R^3)$) and non-negative: $v(x)\geqslant 0$.
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Here, we will restrict our focus a bit and assume that $v$ is integrable ($v\in L_1(\mathbb R^3)$) and non-negative: $v(x)\geqslant 0$.
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In addition, we will only consider the zero-temperature state ($T=0$ that is $\beta=\infty$), in other words, we will be considering only the ground state of $H_N$, which is the eigenstate with lowest eigenvalue.
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In addition, we will only consider the zero-temperature state ($T=0$ that is $\beta=\infty$), in other words, we will be considering only the ground state of $H_N$, which is the eigenstate with lowest eigenvalue.
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@@ -688,9 +690,10 @@ The physical picture to have in mind is two particles coming closer together in
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By translation invariance, we can work in the center of mass reference frame, in which the scattering event can be seen as a single particle flying through a fixed potential.
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By translation invariance, we can work in the center of mass reference frame, in which the scattering event can be seen as a single particle flying through a fixed potential.
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The Schr\"odinger equation for this process is
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The Schr\"odinger equation for this process is
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\begin{equation}
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\begin{equation}
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-\frac1m\Delta\psi+v(x)\psi=i\hbar\partial_t\psi
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-\frac1m\Delta\psi+v(x)\psi=i\partial_t\psi
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\end{equation}
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\end{equation}
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(note the absence of a $1/2$ factor in front of the Laplacian which comes from the fact that the effective mass in the center of mass frame is $\frac{m_1m_2}{m_1+m_2}=m/2$.)
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(note the absence of a $1/2$ factor in front of the Laplacian which comes from the fact that the effective mass in the center of mass frame is $\frac{m_1m_2}{m_1+m_2}=m/2$.)
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(Recall that we chose units so that $\hbar=1$.)
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\bigskip
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\bigskip
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\indent
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\indent
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@@ -813,11 +816,13 @@ Alternatively, the scattering length can be computed as follows.
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\theo{Lemma}\label{lemma:scattering}
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\theo{Lemma}\label{lemma:scattering}
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If $v$ is spherically symmetric, compactly supported and integrable, then
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If $v$ is spherically symmetric, compactly supported and integrable, then
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\nopagebreakaftereq
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\begin{equation}
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\begin{equation}
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a=\frac m{4\pi}\int dx\ v(x)\psi(x)
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a=\frac m{4\pi}\int dx\ v(x)\psi(x)
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.
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\end{equation}
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\end{equation}
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\endtheo
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\endtheo
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\restorepagebreakaftereq
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\bigskip
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\bigskip
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\indent\underline{Proof}:
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\indent\underline{Proof}:
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@@ -907,11 +912,13 @@ Lee, Huang and Yang predicted the following low-density expansion for the ground
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\label{lhy_e}
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\label{lhy_e}
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\end{equation}
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\end{equation}
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where $a$ is the scattering length of the potential, and
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where $a$ is the scattering length of the potential, and
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\nopagebreakaftereq
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\begin{equation}
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\begin{equation}
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\lim_{\rho\to 0}\frac{o(\sqrt{\rho})}{\sqrt{\rho}}=0
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\lim_{\rho\to 0}\frac{o(\sqrt{\rho})}{\sqrt{\rho}}=0
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\end{equation}
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\end{equation}
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\endtheo
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\endtheo
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\restorepagebreakaftereq
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\bigskip
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\bigskip
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The first two orders of this expansion thus only depend on the potential through its scattering length.
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The first two orders of this expansion thus only depend on the potential through its scattering length.
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@@ -1313,6 +1320,7 @@ We thus define an equation obtained from the Complete Equation in which we drop
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\label{bigeqL}
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\label{bigeqL}
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\end{equation}
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\end{equation}
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\endtheo
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\endtheo
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\restorepagebreakaftereq
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\bigskip
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\bigskip
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In practice, we have found that the predictions of the Big Equation are extremely close to those of the Complete Equation.
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In practice, we have found that the predictions of the Big Equation are extremely close to those of the Complete Equation.
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@@ -1710,7 +1718,7 @@ We are almost there: we have proved the existence of the solution of the modifie
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To prove Theorem\-~\ref{theo:existence}, we need to prove that $e\mapsto\rho(e)$ can be inverted locally.
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To prove Theorem\-~\ref{theo:existence}, we need to prove that $e\mapsto\rho(e)$ can be inverted locally.
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\bigskip
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\bigskip
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\theo{Lemma}
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\theo{Lemma}\label{lemma:surjective}
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The map $e\mapsto\rho(e)$ is continuous, and $\rho(0)=0$ and $\lim_{e\to\infty}\rho(e)=\infty$.
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The map $e\mapsto\rho(e)$ is continuous, and $\rho(0)=0$ and $\lim_{e\to\infty}\rho(e)=\infty$.
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Therefore $e\mapsto\rho(r)$ can be inverted locally.
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Therefore $e\mapsto\rho(r)$ can be inverted locally.
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\endtheo
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\endtheo
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@@ -1838,6 +1846,13 @@ But what of the uniqueness?
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\end{equation}
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\end{equation}
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\qed
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\qed
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\bigskip
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\indent
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Thus, taking the point of view in which the energy $e$ is fixed and $\rho$ is computed as a fuction of $e$, the solution is unique.
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However, this does not imply that the solution to the problem in which $\rho$ is fixed is unique: the mapping $e\mapsto\rho(e)$ may not be injective (it is surjective by Lemma\-~\ref{lemma:surjective}).
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To prove that it is injective, one could prove that $\rho$ is an increasing function of $e$ (physically, it should be: the higher the density is, the higher the energy should be because the potnetial is repulsive).
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This has been proved for small and large values of $e$\-~\cite{CJL21}, but, in general, it is still an open problem.
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\subsection{Energy of the Simple Equation}
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\subsection{Energy of the Simple Equation}
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\indent
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\indent
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@@ -2324,24 +2339,6 @@ In this appendix we gather a few useful definitions and results from functional
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\endtheo
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\endtheo
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\bigskip
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\bigskip
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%\theo{Lemma}
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% If $A$ and $B$ are positivity preserving, $A$ and $A^{-1}-B$ are invertible, and $\|BA\|<1$ or $\|AB\|<1$, then $(A^{-1}-B)^{-1}$ is positivity preserving.
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%\endtheo
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%\bigskip
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%
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%\indent\underline{Proof}:
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% We expand
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% \begin{equation}
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% (A^{-1}-B)^{-1}
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% =
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% \sum_{n=0}^\infty A(BA)^n
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% =
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% \sum_{n=0}^\infty (AB)^nA
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% .
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% \end{equation}
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%\qed
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%\bigskip
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\theo{Lemma}\label{lemma:add_inv}
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\theo{Lemma}\label{lemma:add_inv}
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If $\mathrm{spec}(A+B)>\epsilon>0$, and $e^{-tA}$ and $e^{-t B}$ are positivity preserving for all $t>0$, then for all $t>0$, $e^{-t(A+B)}$ and $(A+B)^{-1}$ are positivity preserving.
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If $\mathrm{spec}(A+B)>\epsilon>0$, and $e^{-tA}$ and $e^{-t B}$ are positivity preserving for all $t>0$, then for all $t>0$, $e^{-t(A+B)}$ and $(A+B)^{-1}$ are positivity preserving.
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\endtheo
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\endtheo
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@@ -2374,20 +2371,6 @@ In this appendix we gather a few useful definitions and results from functional
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\qed
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\qed
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\bigskip
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\bigskip
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%\theo{Lemma}\label{lemma:diff}
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% If $B$ is positivity preserving and $(A+\eta B)^{-1}$ is positivity preserving for all $\eta$, then for any $f\in\mathcal B_1$ such that $f\geqslant 0$, $\eta\mapsto(A+\eta B)^{-1}f$ is monotone decreasing pointwise in $x$.
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%\endtheo
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%\bigskip
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%
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%\indent\underline{Proof}:
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% We have
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% \begin{equation}
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% \partial_\eta(A+\eta B)^{-1}f
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% =-(A+\eta B)^{-1}B(A+\eta B)^{-1}f\leqslant 0
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% .
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% \end{equation}
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%\qed
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\theoname{Theorem}{Positivity of the heat kernel}\label{theo:heat}
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\theoname{Theorem}{Positivity of the heat kernel}\label{theo:heat}
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Given $t>0$, the operator $e^{t\Delta}$ from $L_2(\mathbb R^d)$ to $L_2(\mathbb R^d)$ is positivity preserving.
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Given $t>0$, the operator $e^{t\Delta}$ from $L_2(\mathbb R^d)$ to $L_2(\mathbb R^d)$ is positivity preserving.
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\endtheo
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\endtheo
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@@ -2452,30 +2435,6 @@ In this appendix we gather a few useful definitions and results from functional
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{\bf Remark}: In particular, taking the potential to be $\eta+v(x)$, this implies that $e^{-t(-\Delta+v(x)+\eta)}$ and $(-\Delta+v(x)+\eta)^{-1}$ are positivity preserving for any $\eta,v(x)\geqslant 0$.
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{\bf Remark}: In particular, taking the potential to be $\eta+v(x)$, this implies that $e^{-t(-\Delta+v(x)+\eta)}$ and $(-\Delta+v(x)+\eta)^{-1}$ are positivity preserving for any $\eta,v(x)\geqslant 0$.
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\bigskip
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\bigskip
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%\theo{Lemma}\label{lemma:yukawa}
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% Given $\epsilon>0$, the operator $(-\Delta+\epsilon)^{-1}$ from $L_p(\mathbb R^3)$ to $W_{2,p}(\mathbb R^3)$ is positivity preserving.
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%\endtheo
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%\bigskip
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%
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%\indent\underline{Proof}:
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% We apply lemma\-~\ref{lemma:add_inv} with $A^{-1}=-\Delta$
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% We have
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% \begin{equation}
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% (-\Delta+\epsilon)^{-1}f(x)=\frac1{4\pi}\int dx\ \frac{e^{-\sqrt\epsilon|x-y|}}{|x-y|}f(y)
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% .
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% \end{equation}
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%\qed
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%\bigskip
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%
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%{\bf Example}:
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%This implies that if $v(x)\geqslant 0$, $v\in L_p(\mathbb R^3)$ and $\|v\|_p<\epsilon$, then $(-\Delta+\epsilon-v)^{-1}$ is positivity preserving.
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%Indeed, take $A=(-\Delta+\epsilon)^{-1}$, and use the fact that
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%\begin{equation}
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% \|(-\Delta+\epsilon)^{-1}\|=\frac1\epsilon
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% .
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%\end{equation}
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%\bigskip
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\theo{Lemma}\label{lemma:conv}
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\theo{Lemma}\label{lemma:conv}
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If $A$ is positivity preserving, $f\in L_1(\mathbb R^d)$ such that $f\geqslant 0$ and $\mathrm{spec}(A-f\ast)>\epsilon>0$ and $e^{-tA}$ is positivity preserving, then $(A-f\ast)^{-1}$ is positivity preserving.
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If $A$ is positivity preserving, $f\in L_1(\mathbb R^d)$ such that $f\geqslant 0$ and $\mathrm{spec}(A-f\ast)>\epsilon>0$ and $e^{-tA}$ is positivity preserving, then $(A-f\ast)^{-1}$ is positivity preserving.
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\endtheo
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\endtheo
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@@ -2552,12 +2511,14 @@ We need to compute these up to order $V^{-2}$, because one of the terms in the e
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u_3(x-y):=u(x-y)+\frac{w_3(x-y)}V
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u_3(x-y):=u(x-y)+\frac{w_3(x-y)}V
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\label{u3}
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\label{u3}
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\end{equation}
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\end{equation}
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\nopagebreakaftereq
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\begin{equation}
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\begin{equation}
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w_3(x-y):=(1-u(x-y))\int dz\ u(x-z)u(y-z)
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w_3(x-y):=(1-u(x-y))\int dz\ u(x-z)u(y-z)
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.
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.
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\label{w3}
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\label{w3}
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\end{equation}
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\end{equation}
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\endtheo
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\endtheo
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\restorepagebreakaftereq
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\bigskip
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\bigskip
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\indent\underline{Proof}:
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\indent\underline{Proof}:
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@@ -2953,6 +2914,26 @@ Check that Theorem\-~\ref{theo:schrodinger} applies, so that $e^{-tH_N}$ is posi
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Use the Perron-Frobenius theorem (Theorem\-~\ref{theo:perron_frobenius}) to conclude.
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Use the Perron-Frobenius theorem (Theorem\-~\ref{theo:perron_frobenius}) to conclude.
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\bigskip
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\bigskip
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\problem\label{ex:nonneg} (solution on p.\-~\ref{sol:nonneg})\par
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\smallskip
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In this exercise, we will derive an alternate proof that $\psi_0\geqslant 0$ under the extra assumption that $v$ is continuous.
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To do so, consider the energy of $\psi_0$:
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\begin{equation}
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\mathcal E(\psi_0):=\left<\psi_0\right|H_N\left|\psi_0\right>
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=\int_{(\mathbb R/(L\mathbb Z))^{3N}} dx\
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\left(
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-\frac1{2m}\psi_0^*(x)\Delta\psi(x)
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+V(x)|\psi_0(x)|^2
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\right)
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\end{equation}
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where $\Delta$ is the Laplacian on $\mathbb R^{3N}$ and $V(x)\equiv\sum_{i<j}v(x_i-x_j)$.
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Prove that
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\begin{equation}
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\mathcal E(|\psi_0|)=\mathcal E(\psi_0)
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\end{equation}
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and use that to prove that $\psi_0\geqslant 0$.
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\bigskip
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\problem\label{ex:feynman_hellman} (solution on p.\-~\ref{sol:feynman_hellman})\par
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\problem\label{ex:feynman_hellman} (solution on p.\-~\ref{sol:feynman_hellman})\par
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\smallskip
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\smallskip
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In this exercise, we will show how to compute the condensate fraction in terms of the ground state energy of an effective Hamiltonian.
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In this exercise, we will show how to compute the condensate fraction in terms of the ground state energy of an effective Hamiltonian.
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@@ -3135,6 +3116,43 @@ In addition, $e^{-tH_N}$ is compact by Theorem\-~\ref{theo:compact_schrodinger},
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Finally, since $\mathrm{spec}(H_N)\geqslant 0$ (because $-\Delta\geqslant 0$ and $v\geqslant 0$, we can apply the Perron-Frobenius theorem, which implies that $\psi_0$ is unique and $\geqslant 0$.
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Finally, since $\mathrm{spec}(H_N)\geqslant 0$ (because $-\Delta\geqslant 0$ and $v\geqslant 0$, we can apply the Perron-Frobenius theorem, which implies that $\psi_0$ is unique and $\geqslant 0$.
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\bigskip
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\bigskip
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\solution{nonneg}
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The potential term $\int dx\ V(x)|\psi_0|^2$ is obviously the same for $\psi_0$ and $|\psi_0|$, so we only need to worry about the kinetic term.
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Let
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\begin{equation}
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P:=\{x\in\mathbb R^{3N}:\ \psi_0(x)\neq0\}
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.
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\end{equation}
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Now, $\psi_0$ is twice contiuously differentiable, since it is an eigenfunction and $v$ is continous (see\-~\cite[Theorem 11.7(vi)]{LL01}), so $|\psi_0|$ is twice continuously differentiable on $P$, and so
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\begin{equation}
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-\int dx\ |\psi_0|\Delta|\psi_0|
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\equiv
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-\int_P dx\ |\psi_0|\Delta|\psi_0|
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=
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\int_P dx\ (\nabla|\psi_0|)^2
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+\int_{\partial P} dx\ |\psi_0(x)|(n(x)\cdot\nabla|\psi_0|)
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\end{equation}
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where $n(x)$ is the normal vector at $x$ ($\partial P$ is differentiable because $\psi_0$ is as well), but since $\psi_0(x)=0$ on $\partial P$,
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\begin{equation}
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-\int dx\ |\psi_0|\Delta|\psi_0|
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=
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\int_P dx\ (\nabla|\psi_0|)^2
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=
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\int_P dx\ \left(\nabla\psi_0\frac{\psi_0}{|\psi_0|}\right)^2
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=
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\int_P dx\ (\nabla\psi_0)^2
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=
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-\int_P dx\ \psi_0\Delta\psi_0
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.
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\end{equation}
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Thus, $\mathcal E(|\psi_0|)=\mathcal E(\psi_0)$.
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Since $\psi_0$ is the minimizer of $\mathcal E(\psi_0)$, so is $|\psi_0|$, but since the ground state is unique,
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\begin{equation}
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\psi_0=|\psi_0|
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\end{equation}
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so $\psi_0\geqslant 0$.
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\bigskip
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\solution{feynman_hellman}
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\solution{feynman_hellman}
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Let $\tilde\psi_0(\epsilon)$ denote the ground state of $\tilde H_N(\epsilon)$ with $\|\tilde\psi_0(\epsilon)\|_2=1$.
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Let $\tilde\psi_0(\epsilon)$ denote the ground state of $\tilde H_N(\epsilon)$ with $\|\tilde\psi_0(\epsilon)\|_2=1$.
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We have
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We have
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