Update to v0.1:

Fixed: H_N is not compact, e^{-H_N} is.

  Fixed: Removed confusing discussion of Anyons

  Added: Discussion of compactness.
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Ian Jauslin 2023-07-25 23:33:57 -05:00
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commit e76604e394
3 changed files with 77 additions and 12 deletions

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v0.1:
* Fixed: H_N is not compact, e^{-H_N} is.
* Fixed: Removed confusing discussion of Anyons
* Added: Discussion of compactness.

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@ -296,10 +296,9 @@ and the other is that exchanging two particles leads to the wavefunction changin
. .
\end{equation} \end{equation}
Why are these the only two possibilities (in three dimensions)? 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. 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 pick up the phase $e^{-i\theta}$. 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 $\pi$.
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 $\pi$. (This is not the case in two dimensions, for reasons we will not elaborate here.)
(In two dimensions, one can know which way is ``back'', so there are other possible phases, but this is a discussion for another time).
\bigskip \bigskip
\indent \indent
@ -881,7 +880,7 @@ We will be interested in the {\it thermodynamic limit}, in which
\frac NV=\rho \frac NV=\rho
\end{equation} \end{equation}
with $\rho$ (the density) fixed. 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}. Its lowest eigenvalue is denoted by $E_0$, and is called the {\it ground-state energy}.
The corresponding eigenvector is denoted by $\psi_0$: The corresponding eigenvector is denoted by $\psi_0$:
\begin{equation} \begin{equation}
@ -2201,7 +2200,7 @@ In this appendix we gather a few useful definitions and results from functional
\endtheo \endtheo
\bigskip \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 If $f\in L_p(\Omega)$ and $g\in L_q(\Omega)$, then $fg\in L_r(\Omega)$ with
\begin{equation} \begin{equation}
\frac1r=\frac1p+\frac 1q \frac1r=\frac1p+\frac 1q
@ -2253,18 +2252,72 @@ In this appendix we gather a few useful definitions and results from functional
\bigskip \bigskip
\subsection{The Trotter product formula} \subsection{The Trotter product formula}
\theoname{Theorem}{Trotter product formula}\label{theo:trotter} \theoname{Theorem}{Trotter product formula \cite[Theorem 5.11]{Te14}}\label{theo:trotter}
Given two operators $A$ and $B$ on a Banach space, 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 \nopagebreakaftereq
\begin{equation} \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} \end{equation}
\endtheo \endtheo
\restorepagebreakaftereq \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} \subsection{Positivity preserving operators}\label{app:positivity_preserving}
\theoname{Definition}{Positivity preserving operators}\label{def: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 $f(x)\geqslant0$, $Af(x)\geqslant 0$. 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 $f(x)\geqslant0$, $Af(x)\geqslant 0$.
@ -2475,7 +2528,7 @@ In this appendix, we state some useful results from harmonic analysis.
\restorepagebreakaftereq \restorepagebreakaftereq
\bigskip \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 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 $A$. A function that is superharmonic on $A$ achieves its minimum on the boundary of $A$.
\endtheo \endtheo
@ -3078,7 +3131,7 @@ This implies\-~(\ref{sol_softcore}).
\solution{perron_frobenius} \solution{perron_frobenius}
By Theorem\-~\ref{theo:schrodinger}, $e^{-tH_N}$ is positivity preserving. 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 $\geqslant 0$. 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

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@ -49,6 +49,8 @@ doi:{\tt\color{blue}\href{http://dx.doi.org/10.1103/PhysRev.130.2518}{10.1103/Ph
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\bibitem[LL01]{LL01}E.H. Lieb, M. Loss - {\it Analysis}, Second edition, Graduate studies in mathematics, Americal Mathematical Society, 2001.\par\medskip
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@ -66,6 +68,8 @@ doi:{\tt\color{blue}\href{http://dx.doi.org/10.4171/90-2/40}{10.4171/90-2/40}},
\bibitem[Se11]{Se11}R. Seiringer - {\it The Excitation Spectrum for Weakly Interacting Bosons}, Communications in Mathematical Physics, volume\-~306, issue\-~2, pages\-~565-578, 2011,\par\penalty10000 \bibitem[Se11]{Se11}R. Seiringer - {\it The Excitation Spectrum for Weakly Interacting Bosons}, Communications in Mathematical Physics, volume\-~306, issue\-~2, pages\-~565-578, 2011,\par\penalty10000
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