|
|
|
|
@@ -269,6 +269,7 @@ The microscopic state of a system of quantum particles is given by a wavefunctio
|
|
|
|
|
\label{H}
|
|
|
|
|
\end{equation}
|
|
|
|
|
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.
|
|
|
|
|
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.
|
|
|
|
|
@@ -296,10 +297,9 @@ and the other is that exchanging two particles leads to the wavefunction changin
|
|
|
|
|
.
|
|
|
|
|
\end{equation}
|
|
|
|
|
Why are these the only two possibilities (in three dimensions)?
|
|
|
|
|
Mathematically, this comes from the fact that there are only two irreducible representations of the braid group in three dimensions.
|
|
|
|
|
More intuitively, if two particles are exchanged, the wavefunction should pick up a phase $e^{i\theta}$, but if I exchange the particles back, we should return to the original wavefunction, and thus pick up the phase $e^{-i\theta}$.
|
|
|
|
|
In three dimensions, there is no way of knowing which way is ``back'', so $e^{i\theta}=e^{-i\theta}$ and so $\theta$ is either $0$ or $\pi$.
|
|
|
|
|
(In two dimensions, one can know which way is ``back'', so there are other possible phases, but this is a discussion for another time).
|
|
|
|
|
Mathematically, this comes from the fact that there are only two irreducible representations of the permutation group in three dimensions.
|
|
|
|
|
More intuitively, if two particles are exchanged, the wavefunction should pick up a phase $e^{i\theta}$, but if I exchange the particles back, we should return to the original wavefunction, and thus $e^{2i\theta}=1$ and so $\theta$ is either $0$ or $\pi$.
|
|
|
|
|
(This is not the case in two dimensions, for reasons we will not elaborate here.)
|
|
|
|
|
\bigskip
|
|
|
|
|
|
|
|
|
|
\indent
|
|
|
|
|
@@ -636,6 +636,7 @@ Let us consider a more realistic Hamiltonian:
|
|
|
|
|
\label{Ham}
|
|
|
|
|
\end{equation}
|
|
|
|
|
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.
|
|
|
|
|
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.
|
|
|
|
|
@@ -689,9 +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.
|
|
|
|
|
The Schr\"odinger equation for this process is
|
|
|
|
|
\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}
|
|
|
|
|
(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
|
|
|
|
|
|
|
|
|
|
\indent
|
|
|
|
|
@@ -814,11 +816,13 @@ Alternatively, the scattering length can be computed as follows.
|
|
|
|
|
|
|
|
|
|
\theo{Lemma}\label{lemma:scattering}
|
|
|
|
|
If $v$ is spherically symmetric, compactly supported and integrable, then
|
|
|
|
|
\nopagebreakaftereq
|
|
|
|
|
\begin{equation}
|
|
|
|
|
a=\frac m{4\pi}\int dx\ v(x)\psi(x)
|
|
|
|
|
.
|
|
|
|
|
\end{equation}
|
|
|
|
|
\endtheo
|
|
|
|
|
\restorepagebreakaftereq
|
|
|
|
|
\bigskip
|
|
|
|
|
|
|
|
|
|
\indent\underline{Proof}:
|
|
|
|
|
@@ -881,7 +885,7 @@ We will be interested in the {\it thermodynamic limit}, in which
|
|
|
|
|
\frac NV=\rho
|
|
|
|
|
\end{equation}
|
|
|
|
|
with $\rho$ (the density) fixed.
|
|
|
|
|
In this case, the Hamiltonian\-~(\ref{Ham}) has pure-point spectrum (it is a compact operator).
|
|
|
|
|
In this case, the Hamiltonian\-~(\ref{Ham}) has pure-point spectrum (by Theorem\-~\ref{theo:compact_schrodinger}, $e^{-H_N}$ is compact, so it has discrete spectrum, and therefore so does $H_N$).
|
|
|
|
|
Its lowest eigenvalue is denoted by $E_0$, and is called the {\it ground-state energy}.
|
|
|
|
|
The corresponding eigenvector is denoted by $\psi_0$:
|
|
|
|
|
\begin{equation}
|
|
|
|
|
@@ -908,11 +912,13 @@ Lee, Huang and Yang predicted the following low-density expansion for the ground
|
|
|
|
|
\label{lhy_e}
|
|
|
|
|
\end{equation}
|
|
|
|
|
where $a$ is the scattering length of the potential, and
|
|
|
|
|
\nopagebreakaftereq
|
|
|
|
|
\begin{equation}
|
|
|
|
|
\lim_{\rho\to 0}\frac{o(\sqrt{\rho})}{\sqrt{\rho}}=0
|
|
|
|
|
.
|
|
|
|
|
\end{equation}
|
|
|
|
|
\endtheo
|
|
|
|
|
\restorepagebreakaftereq
|
|
|
|
|
\bigskip
|
|
|
|
|
|
|
|
|
|
The first two orders of this expansion thus only depend on the potential through its scattering length.
|
|
|
|
|
@@ -1314,6 +1320,7 @@ We thus define an equation obtained from the Complete Equation in which we drop
|
|
|
|
|
\label{bigeqL}
|
|
|
|
|
\end{equation}
|
|
|
|
|
\endtheo
|
|
|
|
|
\restorepagebreakaftereq
|
|
|
|
|
\bigskip
|
|
|
|
|
|
|
|
|
|
In practice, we have found that the predictions of the Big Equation are extremely close to those of the Complete Equation.
|
|
|
|
|
@@ -1711,7 +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.
|
|
|
|
|
\bigskip
|
|
|
|
|
|
|
|
|
|
\theo{Lemma}
|
|
|
|
|
\theo{Lemma}\label{lemma:surjective}
|
|
|
|
|
The map $e\mapsto\rho(e)$ is continuous, and $\rho(0)=0$ and $\lim_{e\to\infty}\rho(e)=\infty$.
|
|
|
|
|
Therefore $e\mapsto\rho(r)$ can be inverted locally.
|
|
|
|
|
\endtheo
|
|
|
|
|
@@ -1839,6 +1846,13 @@ But what of the uniqueness?
|
|
|
|
|
.
|
|
|
|
|
\end{equation}
|
|
|
|
|
\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}
|
|
|
|
|
\indent
|
|
|
|
|
@@ -2201,7 +2215,7 @@ In this appendix we gather a few useful definitions and results from functional
|
|
|
|
|
\endtheo
|
|
|
|
|
\bigskip
|
|
|
|
|
|
|
|
|
|
\theoname{Theorem}{H\"older's inequality}\label{theo:holder}
|
|
|
|
|
\theoname{Theorem}{H\"older's inequality \cite[Problem 0.26]{Te14}}\label{theo:holder}
|
|
|
|
|
If $f\in L_p(\Omega)$ and $g\in L_q(\Omega)$, then $fg\in L_r(\Omega)$ with
|
|
|
|
|
\begin{equation}
|
|
|
|
|
\frac1r=\frac1p+\frac 1q
|
|
|
|
|
@@ -2253,42 +2267,78 @@ In this appendix we gather a few useful definitions and results from functional
|
|
|
|
|
\bigskip
|
|
|
|
|
|
|
|
|
|
\subsection{The Trotter product formula}
|
|
|
|
|
\theoname{Theorem}{Trotter product formula}\label{theo:trotter}
|
|
|
|
|
Given two operators $A$ and $B$ on a Banach space,
|
|
|
|
|
\theoname{Theorem}{Trotter product formula \cite[Theorem 5.11]{Te14}}\label{theo:trotter}
|
|
|
|
|
Given two operators $A$ and $B$ on a Hilbert space such that $A$, $B$ and $A+B$ are self-adjoint, and bounded from below, then for $t\geqslant 0$,
|
|
|
|
|
\nopagebreakaftereq
|
|
|
|
|
\begin{equation}
|
|
|
|
|
e^{A+B}
|
|
|
|
|
e^{-t(A+B)}
|
|
|
|
|
=
|
|
|
|
|
\lim_{N\to\infty}\left(e^{\frac 1N A}e^{\frac1N B}\right)^N
|
|
|
|
|
\lim_{N\to\infty}\left(e^{-t\frac 1N A}e^{-t\frac1N B}\right)^N
|
|
|
|
|
.
|
|
|
|
|
\end{equation}
|
|
|
|
|
\endtheo
|
|
|
|
|
\restorepagebreakaftereq
|
|
|
|
|
|
|
|
|
|
\subsection{Compact and trace-class operators}\label{app:compact}
|
|
|
|
|
\theoname{Definition}{Trace}
|
|
|
|
|
Given a separable Hilbert space with an orthonormal basis $\{\varphi_i\}_{i\in\mathbb N}$, the trace of an operator $A$ is formally defined as
|
|
|
|
|
\begin{equation}
|
|
|
|
|
\mathrm{Tr}(A):=\sum_{i=0}^\infty \left<\varphi_i\right|A\left|\varphi_i\right>
|
|
|
|
|
.
|
|
|
|
|
\end{equation}
|
|
|
|
|
If this expression is finite, then $A$ is said to be trace-class.
|
|
|
|
|
\endtheo
|
|
|
|
|
\bigskip
|
|
|
|
|
|
|
|
|
|
\theoname{Definition}{Compact operators}
|
|
|
|
|
The set of compact operators on a Hilbert space is the closure of the set of finite-rank operators.
|
|
|
|
|
\endtheo
|
|
|
|
|
\bigskip
|
|
|
|
|
|
|
|
|
|
\theoname{Theorem}{\cite[Theorem 6.6]{Te14}}
|
|
|
|
|
If $A$ is a compact self-adjoint operator on a Hilbert space, then its spectrum consists of an at most countable set of eigenvalues.
|
|
|
|
|
\endtheo
|
|
|
|
|
\bigskip
|
|
|
|
|
|
|
|
|
|
\theo{Theorem}\label{theo:trace_compact}
|
|
|
|
|
Trace class operators are compact.
|
|
|
|
|
\endtheo
|
|
|
|
|
\bigskip
|
|
|
|
|
|
|
|
|
|
\theoname{Theorem}{\cite[Lemma 5.5]{Te14}}\label{theo:compact_ideal}
|
|
|
|
|
If $A$ is compact and $B$ is bounded, then $KA$ and $AK$ are compact.
|
|
|
|
|
\endtheo
|
|
|
|
|
\bigskip
|
|
|
|
|
|
|
|
|
|
\theo{Theorem}\label{theo:compact_schrodinger}
|
|
|
|
|
Let $\Delta$ be the Laplacian on $L_2(\mathbb R^d/(L\mathbb Z)^d)$ (the same result holds for Dirichlet and Neumann boundary conditions as well).
|
|
|
|
|
For any function $v$ on $[0,L]^d$ that is bounded below, $e^{t(-\Delta+v)}$ is compact for any $t\geqslant 0$.
|
|
|
|
|
\endtheo
|
|
|
|
|
\bigskip
|
|
|
|
|
|
|
|
|
|
\indent\underline{Proof}:
|
|
|
|
|
First of all, $e^{-t\Delta}$ is trace class: using the basis $e^{ikx}$ with $k\in(\frac{2\pi}L\mathbb Z)^d$
|
|
|
|
|
\begin{equation}
|
|
|
|
|
\mathrm{Tr}e^{-t\Delta}
|
|
|
|
|
=\sum_{k\in(\frac{2\pi}L\mathbb Z)^d}e^{-tk^2}<\infty
|
|
|
|
|
.
|
|
|
|
|
\end{equation}
|
|
|
|
|
By Theorem\-~\ref{theo:trace_compact}, $e^{-t\Delta}$ is compact.
|
|
|
|
|
By the Trotter product formula, Theorem\-~\ref{theo:trotter},
|
|
|
|
|
\begin{equation}
|
|
|
|
|
e^{-t\Delta+tv}
|
|
|
|
|
=\lim_{N\to\infty}(e^{-\frac tN\Delta}e^{\frac tNv})^N
|
|
|
|
|
\end{equation}
|
|
|
|
|
and by Theorem\-~\ref{theo:compact_ideal}, $(e^{-\frac tN\Delta}e^{\frac tNv})^N$ is compact.
|
|
|
|
|
Since the set of compact operators is closed, so is $e^{t(-\Delta+v)}$.
|
|
|
|
|
\qed
|
|
|
|
|
|
|
|
|
|
\subsection{Positivity preserving operators}\label{app:positivity_preserving}
|
|
|
|
|
\theoname{Definition}{Positivity preserving operators}\label{def:positivity_preserving}
|
|
|
|
|
An operator $A$ from a Banach space of real-valued functions $\mathcal B_1$ to another $\mathcal B_2$ is said to be {\it positivity preserving} if, for any $f\in\mathcal B_1$ such that $f(x)\geqslant0$, $Af(x)\geqslant 0$.
|
|
|
|
|
\endtheo
|
|
|
|
|
\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}
|
|
|
|
|
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
|
|
|
|
|
@@ -2321,20 +2371,6 @@ In this appendix we gather a few useful definitions and results from functional
|
|
|
|
|
\qed
|
|
|
|
|
\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}
|
|
|
|
|
Given $t>0$, the operator $e^{t\Delta}$ from $L_2(\mathbb R^d)$ to $L_2(\mathbb R^d)$ is positivity preserving.
|
|
|
|
|
\endtheo
|
|
|
|
|
@@ -2399,30 +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 $\eta,v(x)\geqslant 0$.
|
|
|
|
|
\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}
|
|
|
|
|
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
|
|
|
|
|
@@ -2475,7 +2487,7 @@ In this appendix, we state some useful results from harmonic analysis.
|
|
|
|
|
\restorepagebreakaftereq
|
|
|
|
|
\bigskip
|
|
|
|
|
|
|
|
|
|
\theo{Theorem}\label{theo:harmonic}
|
|
|
|
|
\theoname{Theorem}{\cite[Theorem 9.4]{LL01}}\label{theo:harmonic}
|
|
|
|
|
A function that is subharmonic on $A$ achieves its maximum on the boundary of $A$.
|
|
|
|
|
A function that is superharmonic on $A$ achieves its minimum on the boundary of $A$.
|
|
|
|
|
\endtheo
|
|
|
|
|
@@ -2499,12 +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
|
|
|
|
|
\label{u3}
|
|
|
|
|
\end{equation}
|
|
|
|
|
\nopagebreakaftereq
|
|
|
|
|
\begin{equation}
|
|
|
|
|
w_3(x-y):=(1-u(x-y))\int dz\ u(x-z)u(y-z)
|
|
|
|
|
.
|
|
|
|
|
\label{w3}
|
|
|
|
|
\end{equation}
|
|
|
|
|
\endtheo
|
|
|
|
|
\restorepagebreakaftereq
|
|
|
|
|
\bigskip
|
|
|
|
|
|
|
|
|
|
\indent\underline{Proof}:
|
|
|
|
|
@@ -2900,6 +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.
|
|
|
|
|
\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
|
|
|
|
|
\smallskip
|
|
|
|
|
In this exercise, we will show how to compute the condensate fraction in terms of the ground state energy of an effective Hamiltonian.
|
|
|
|
|
@@ -3078,10 +3112,47 @@ This implies\-~(\ref{sol_softcore}).
|
|
|
|
|
|
|
|
|
|
\solution{perron_frobenius}
|
|
|
|
|
By Theorem\-~\ref{theo:schrodinger}, $e^{-tH_N}$ is positivity preserving.
|
|
|
|
|
In addition, $e^{-tH_N}$ is compact because $H_N$ is, and the spectrum of $e^{-tH_N}$ is $e^{-t\mathrm{spec}(H_N)}$, and the largest eigenvector of $e^{-tH_N}$ with the largest eigenvalue is the ground-state of $H_N$.
|
|
|
|
|
In addition, $e^{-tH_N}$ is compact by Theorem\-~\ref{theo:compact_schrodinger}, and the spectrum of $e^{-tH_N}$ is $e^{-t\ \mathrm{spec}(H_N)}$, and the largest eigenvector of $e^{-tH_N}$ with the largest eigenvalue is the ground-state of $H_N$.
|
|
|
|
|
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
|
|
|
|
|
|
|
|
|
|
\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}
|
|
|
|
|
Let $\tilde\psi_0(\epsilon)$ denote the ground state of $\tilde H_N(\epsilon)$ with $\|\tilde\psi_0(\epsilon)\|_2=1$.
|
|
|
|
|
We have
|
|
|
|
|
|
