Update to v0.2

Added: Discussion of uniqueness of the solution of the Simple Equation

  Added: Exercise: prove \psi_0>=0 using |\psi_0|

  Added: Links in bibliography

  Removed: Commented blocks

  Fixed: Formatting
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Ian Jauslin 2023-07-26 15:57:58 -05:00
parent e76604e394
commit 3bd7865e45
3 changed files with 92 additions and 60 deletions

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@ -1,3 +1,15 @@
v0.2:
* Added: Discussion of uniqueness of the solution of the Simple Equation
* Added: Exercise: prove \psi_0>=0 using |\psi_0|
* Added: Links in bibliography
* Removed: Commented blocks
* Fixed: Formatting
v0.1: v0.1:
* Fixed: H_N is not compact, e^{-H_N} is. * Fixed: H_N is not compact, e^{-H_N} is.

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@ -269,6 +269,7 @@ The microscopic state of a system of quantum particles is given by a wavefunctio
\label{H} \label{H}
\end{equation} \end{equation}
where $\Delta_i$ is the Laplacian with respect to $x_i$. where $\Delta_i$ is the Laplacian with respect to $x_i$.
(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.)
This is the microscopic description of the system. This is the microscopic description of the system.
Similarly to the classical case, to treat situations with a huge number of particles, we will approach the system statistically. Similarly to the classical case, to treat situations with a huge number of particles, we will approach the system statistically.
In the classical case, we considered a probability distribution over all possible configurations, without fixing the number of particles. In the classical case, we considered a probability distribution over all possible configurations, without fixing the number of particles.
@ -635,6 +636,7 @@ Let us consider a more realistic Hamiltonian:
\label{Ham} \label{Ham}
\end{equation} \end{equation}
where $v$ is the potential that accounts for the interaction between pairs of particles. where $v$ is the potential that accounts for the interaction between pairs of particles.
(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.)
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. 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.
Here, we will restrict our focus a bit and assume that $v$ is integrable ($v\in L_1(\mathbb R^3)$) and non-negative: $v(x)\geqslant 0$. Here, we will restrict our focus a bit and assume that $v$ is integrable ($v\in L_1(\mathbb R^3)$) and non-negative: $v(x)\geqslant 0$.
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. 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.
@ -688,9 +690,10 @@ The physical picture to have in mind is two particles coming closer together in
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. 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.
The Schr\"odinger equation for this process is The Schr\"odinger equation for this process is
\begin{equation} \begin{equation}
-\frac1m\Delta\psi+v(x)\psi=i\hbar\partial_t\psi -\frac1m\Delta\psi+v(x)\psi=i\partial_t\psi
\end{equation} \end{equation}
(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 $\frac{m_1m_2}{m_1+m_2}=m/2$.) (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 $\frac{m_1m_2}{m_1+m_2}=m/2$.)
(Recall that we chose units so that $\hbar=1$.)
\bigskip \bigskip
\indent \indent
@ -813,11 +816,13 @@ Alternatively, the scattering length can be computed as follows.
\theo{Lemma}\label{lemma:scattering} \theo{Lemma}\label{lemma:scattering}
If $v$ is spherically symmetric, compactly supported and integrable, then If $v$ is spherically symmetric, compactly supported and integrable, then
\nopagebreakaftereq
\begin{equation} \begin{equation}
a=\frac m{4\pi}\int dx\ v(x)\psi(x) a=\frac m{4\pi}\int dx\ v(x)\psi(x)
. .
\end{equation} \end{equation}
\endtheo \endtheo
\restorepagebreakaftereq
\bigskip \bigskip
\indent\underline{Proof}: \indent\underline{Proof}:
@ -907,11 +912,13 @@ Lee, Huang and Yang predicted the following low-density expansion for the ground
\label{lhy_e} \label{lhy_e}
\end{equation} \end{equation}
where $a$ is the scattering length of the potential, and where $a$ is the scattering length of the potential, and
\nopagebreakaftereq
\begin{equation} \begin{equation}
\lim_{\rho\to 0}\frac{o(\sqrt{\rho})}{\sqrt{\rho}}=0 \lim_{\rho\to 0}\frac{o(\sqrt{\rho})}{\sqrt{\rho}}=0
. .
\end{equation} \end{equation}
\endtheo \endtheo
\restorepagebreakaftereq
\bigskip \bigskip
The first two orders of this expansion thus only depend on the potential through its scattering length. The first two orders of this expansion thus only depend on the potential through its scattering length.
@ -1313,6 +1320,7 @@ We thus define an equation obtained from the Complete Equation in which we drop
\label{bigeqL} \label{bigeqL}
\end{equation} \end{equation}
\endtheo \endtheo
\restorepagebreakaftereq
\bigskip \bigskip
In practice, we have found that the predictions of the Big Equation are extremely close to those of the Complete Equation. In practice, we have found that the predictions of the Big Equation are extremely close to those of the Complete Equation.
@ -1710,7 +1718,7 @@ We are almost there: we have proved the existence of the solution of the modifie
To prove Theorem\-~\ref{theo:existence}, we need to prove that $e\mapsto\rho(e)$ can be inverted locally. To prove Theorem\-~\ref{theo:existence}, we need to prove that $e\mapsto\rho(e)$ can be inverted locally.
\bigskip \bigskip
\theo{Lemma} \theo{Lemma}\label{lemma:surjective}
The map $e\mapsto\rho(e)$ is continuous, and $\rho(0)=0$ and $\lim_{e\to\infty}\rho(e)=\infty$. The map $e\mapsto\rho(e)$ is continuous, and $\rho(0)=0$ and $\lim_{e\to\infty}\rho(e)=\infty$.
Therefore $e\mapsto\rho(r)$ can be inverted locally. Therefore $e\mapsto\rho(r)$ can be inverted locally.
\endtheo \endtheo
@ -1838,6 +1846,13 @@ But what of the uniqueness?
. .
\end{equation} \end{equation}
\qed \qed
\bigskip
\indent
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.
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}).
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).
This has been proved for small and large values of $e$\-~\cite{CJL21}, but, in general, it is still an open problem.
\subsection{Energy of the Simple Equation} \subsection{Energy of the Simple Equation}
\indent \indent
@ -2324,24 +2339,6 @@ In this appendix we gather a few useful definitions and results from functional
\endtheo \endtheo
\bigskip \bigskip
%\theo{Lemma}
% 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.
%\endtheo
%\bigskip
%
%\indent\underline{Proof}:
% We expand
% \begin{equation}
% (A^{-1}-B)^{-1}
% =
% \sum_{n=0}^\infty A(BA)^n
% =
% \sum_{n=0}^\infty (AB)^nA
% .
% \end{equation}
%\qed
%\bigskip
\theo{Lemma}\label{lemma:add_inv} \theo{Lemma}\label{lemma:add_inv}
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. 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.
\endtheo \endtheo
@ -2374,20 +2371,6 @@ In this appendix we gather a few useful definitions and results from functional
\qed \qed
\bigskip \bigskip
%\theo{Lemma}\label{lemma:diff}
% 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 $x$.
%\endtheo
%\bigskip
%
%\indent\underline{Proof}:
% We have
% \begin{equation}
% \partial_\eta(A+\eta B)^{-1}f
% =-(A+\eta B)^{-1}B(A+\eta B)^{-1}f\leqslant 0
% .
% \end{equation}
%\qed
\theoname{Theorem}{Positivity of the heat kernel}\label{theo:heat} \theoname{Theorem}{Positivity of the heat kernel}\label{theo:heat}
Given $t>0$, the operator $e^{t\Delta}$ from $L_2(\mathbb R^d)$ to $L_2(\mathbb R^d)$ is positivity preserving. Given $t>0$, the operator $e^{t\Delta}$ from $L_2(\mathbb R^d)$ to $L_2(\mathbb R^d)$ is positivity preserving.
\endtheo \endtheo
@ -2452,30 +2435,6 @@ In this appendix we gather a few useful definitions and results from functional
{\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 $\eta,v(x)\geqslant 0$. {\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 $\eta,v(x)\geqslant 0$.
\bigskip \bigskip
%\theo{Lemma}\label{lemma:yukawa}
% Given $\epsilon>0$, the operator $(-\Delta+\epsilon)^{-1}$ from $L_p(\mathbb R^3)$ to $W_{2,p}(\mathbb R^3)$ is positivity preserving.
%\endtheo
%\bigskip
%
%\indent\underline{Proof}:
% We apply lemma\-~\ref{lemma:add_inv} with $A^{-1}=-\Delta$
% We have
% \begin{equation}
% (-\Delta+\epsilon)^{-1}f(x)=\frac1{4\pi}\int dx\ \frac{e^{-\sqrt\epsilon|x-y|}}{|x-y|}f(y)
% .
% \end{equation}
%\qed
%\bigskip
%
%{\bf Example}:
%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.
%Indeed, take $A=(-\Delta+\epsilon)^{-1}$, and use the fact that
%\begin{equation}
% \|(-\Delta+\epsilon)^{-1}\|=\frac1\epsilon
% .
%\end{equation}
%\bigskip
\theo{Lemma}\label{lemma:conv} \theo{Lemma}\label{lemma:conv}
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. 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.
\endtheo \endtheo
@ -2552,12 +2511,14 @@ We need to compute these up to order $V^{-2}$, because one of the terms in the e
u_3(x-y):=u(x-y)+\frac{w_3(x-y)}V u_3(x-y):=u(x-y)+\frac{w_3(x-y)}V
\label{u3} \label{u3}
\end{equation} \end{equation}
\nopagebreakaftereq
\begin{equation} \begin{equation}
w_3(x-y):=(1-u(x-y))\int dz\ u(x-z)u(y-z) w_3(x-y):=(1-u(x-y))\int dz\ u(x-z)u(y-z)
. .
\label{w3} \label{w3}
\end{equation} \end{equation}
\endtheo \endtheo
\restorepagebreakaftereq
\bigskip \bigskip
\indent\underline{Proof}: \indent\underline{Proof}:
@ -2953,6 +2914,26 @@ Check that Theorem\-~\ref{theo:schrodinger} applies, so that $e^{-tH_N}$ is posi
Use the Perron-Frobenius theorem (Theorem\-~\ref{theo:perron_frobenius}) to conclude. Use the Perron-Frobenius theorem (Theorem\-~\ref{theo:perron_frobenius}) to conclude.
\bigskip \bigskip
\problem\label{ex:nonneg} (solution on p.\-~\ref{sol:nonneg})\par
\smallskip
In this exercise, we will derive an alternate proof that $\psi_0\geqslant 0$ under the extra assumption that $v$ is continuous.
To do so, consider the energy of $\psi_0$:
\begin{equation}
\mathcal E(\psi_0):=\left<\psi_0\right|H_N\left|\psi_0\right>
=\int_{(\mathbb R/(L\mathbb Z))^{3N}} dx\
\left(
-\frac1{2m}\psi_0^*(x)\Delta\psi(x)
+V(x)|\psi_0(x)|^2
\right)
\end{equation}
where $\Delta$ is the Laplacian on $\mathbb R^{3N}$ and $V(x)\equiv\sum_{i<j}v(x_i-x_j)$.
Prove that
\begin{equation}
\mathcal E(|\psi_0|)=\mathcal E(\psi_0)
\end{equation}
and use that to prove that $\psi_0\geqslant 0$.
\bigskip
\problem\label{ex:feynman_hellman} (solution on p.\-~\ref{sol:feynman_hellman})\par \problem\label{ex:feynman_hellman} (solution on p.\-~\ref{sol:feynman_hellman})\par
\smallskip \smallskip
In this exercise, we will show how to compute the condensate fraction in terms of the ground state energy of an effective Hamiltonian. In this exercise, we will show how to compute the condensate fraction in terms of the ground state energy of an effective Hamiltonian.
@ -3135,6 +3116,43 @@ In addition, $e^{-tH_N}$ is compact by Theorem\-~\ref{theo:compact_schrodinger},
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 $\geqslant 0$. 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 $\geqslant 0$.
\bigskip \bigskip
\solution{nonneg}
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.
Let
\begin{equation}
P:=\{x\in\mathbb R^{3N}:\ \psi_0(x)\neq0\}
.
\end{equation}
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
\begin{equation}
-\int dx\ |\psi_0|\Delta|\psi_0|
\equiv
-\int_P dx\ |\psi_0|\Delta|\psi_0|
=
\int_P dx\ (\nabla|\psi_0|)^2
+\int_{\partial P} dx\ |\psi_0(x)|(n(x)\cdot\nabla|\psi_0|)
\end{equation}
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 $\partial P$,
\begin{equation}
-\int dx\ |\psi_0|\Delta|\psi_0|
=
\int_P dx\ (\nabla|\psi_0|)^2
=
\int_P dx\ \left(\nabla\psi_0\frac{\psi_0}{|\psi_0|}\right)^2
=
\int_P dx\ (\nabla\psi_0)^2
=
-\int_P dx\ \psi_0\Delta\psi_0
.
\end{equation}
Thus, $\mathcal E(|\psi_0|)=\mathcal E(\psi_0)$.
Since $\psi_0$ is the minimizer of $\mathcal E(\psi_0)$, so is $|\psi_0|$, but since the ground state is unique,
\begin{equation}
\psi_0=|\psi_0|
\end{equation}
so $\psi_0\geqslant 0$.
\bigskip
\solution{feynman_hellman} \solution{feynman_hellman}
Let $\tilde\psi_0(\epsilon)$ denote the ground state of $\tilde H_N(\epsilon)$ with $\|\tilde\psi_0(\epsilon)\|_2=1$. Let $\tilde\psi_0(\epsilon)$ denote the ground state of $\tilde H_N(\epsilon)$ with $\|\tilde\psi_0(\epsilon)\|_2=1$.
We have We have

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@ -29,7 +29,8 @@ doi:{\tt\color{blue}\href{http://dx.doi.org/10.4007/annals.2020.192.3.5}{10.4007
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