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\title{Level Statistics for Quantum Hamiltonians \\
--Some Preliminary Ideas toward Mathematical Justification of the
Theory of Berry and Tabor}
\author{Nariyuki MINAMI\thanks{e-mail: minami@sakura.cc.tsukuba.ac.jp} \\
Institute of Mathematics, University of Tsukuba \\
Tsukuba-shi, Ibaraki, 305-8571 Japan}
\date{}
\maketitle
\begin{abstract}
According to Berry and Tabor, the spectrum of a classically integrable
quantum Hamiltonian (the regular spectrum)
should generically have the same statistical property
of the Poisson point process. In particular, the spacings between energy
levels obey the exponential distribution, which is the phenomena called
\lq\lq level clustering \rq\rq . Strict justification of this assertion
seems to be a very difficult mathematical problem. In this note, the author
shall introduce a weakened notion of level clustering and give a sufficient
condition for this in terms of factorial moments of the number of levels
in intervals of fixed length. A preliminary result is also stated, which is
closely related to the justification of this sufficient condition in the case
of regular spectra. All proofs are omitted, and will be published elsewhere.
\end{abstract}
\section{An outline of Berry-Tabor's theory and the definition of clustering
and repulsion of energy levels.}
Consider a bounded physical system of $d-$degrees of freedom which is
described by a classical Hamiltonian $H(p,q)$ , $(p,q)\in\mathbf{R}^d\times
\mathbf{R}^d$ . Let $H(\hbar)$ be its quantization, and
$\{E_n(\hbar)\}_{n\geq1}$ be the energy levels (eigenvalues) of $H(\hbar)$ .
Here $\hbar$ is the Planck's constant, and we suppose that the spectrum
of $H(\hbar)$ is purely discrete.
Now the assertion of Berry and Tabor (\cite{berry})
is the following: Suppose the classical
dynamical system associated to $H(p,q)$ is completely integrable. Then the
spectrum $\{E_n(\hbar)\}_{n\geq1}$ of $H(\hbar)$ resembles, in the limit of
$\hbar\searrow0$ , the Poisson point process. In particular, the level
spacing distribution is $e^{-t}dt$ under suitable normalization. This
phenomena was called \lq\lq level clustering \rq\rq by Berry and Tabor
because the large part of
spacing distribution is concentrated near zero, which means
the statistical tendency of the appearance of closely lying levels.
On the other
hand, when the classical dynamical system is chaotic, the level spacing
distribution $p(t)dt$ should satisfy $p(t)\sim Ct^{\gamma}$ , $t\searrow0$
with $\gamma>0$ , which is the phenomena called level repulsion.
In short, Berry and Tabor conjectured that the regular or chaotic properties
of a classical dynamical system manifest themselves in the level statistics
of its quantization.
In order to give a precise meaning to the level spacing distribution, we need
to somehow normalize (unfold) the spectrum of $H(\hbar)$ , so that it looks
like a typical realization of a stationary point process. In doing this,
we follow the idea of Berry and Tabor, thereby accomplishing simultaneously
this normalization and the limiting procedure $\hbar\searrow0$ . Let us
assume that the specrum $\{E_n(\hbar)\}_{n\geq1}$ satisfies the following
conditions:
\begin{description}
\item[(A1)] All levels are non-degenerate and $E_n(\hbar)\geq0$ .
\item[(A2)] $E_n(\hbar)\searrow0$ monotonically as $\hbar\searrow0$ .
\item[(A3)] For each fixed $E>0$ , there is a constant $\nu(E)>0$ such
that
\begin{equation}
N_{\hbar}(E)\equiv\sharp\{n\geq1\ \vert\ E_n(\hbar)\leq E\}
\sim\nu(E)\hbar^{-d}\ (\hbar\searrow0)\ .
\end{equation}
\end{description}
Let us take $E>0$ , which we shall fix throughout. By (A1) and (A2), we can
define $\hbar_n=\hbar_n(E)$ as the unique solution of the equation
$E_n(\hbar)=E$ . Let us define the unfolded level $\lambda_n$ by
$\lambda_n=\nu(E)\hbar_n(E)^{-d}$ . Then from (A3), it is easily seen that
\begin{equation}
\sharp\{n\geq1\ \vert\ \lambda_n\leq L\}\sim L\quad(L\to\infty)\ .
\end{equation}
Thus the unfolded levels $\{\lambda\}_n$ are asymptotically uniformly
distributed.
We introduce the following notation:
\begin{equation}
N(t)=N(t;c)=\sharp\{n\geq1\ \vert\ \lambda_n\in(t,t+c]\}\ ;
\end{equation}
\begin{equation}
\pi_k(c;L)=\frac1L\int_0^L1_{\{N(t;c)=k\}}dt\ ;
\end{equation}
\begin{equation}
\mu_k(c;L)=\frac1{k!}\frac1L\int_0^L N(t)\{N(t)-1\}\cdots\{N(t)-k+1\}dt\ ;
\end{equation}
\begin{equation}
\rho(c;L)=\frac{\sharp\{n\geq1\ \vert\ \lambda_n\leq L\ ,\ \lambda_{n+1}-
\lambda_n\leq c\}}
{\sharp\{n\geq1\ \vert\ \lambda_n\leq L\}}\ .
\end{equation}
Here $c>0$ and $k=0,1,2,\ldots$ .
\bigskip
\begin{df}
We shall say that one has the strict sense level clustering if
\begin{equation}
\pi_k(c)\equiv\lim_{L\to\infty}\pi_k(c;L)=e^{-c}\frac{c^k}{k!}
\end{equation}
holds for each $c>0$ and $k\geq0$ .
\end{df}
\bigskip
Obviously, the strict sense level clustering is equivalent to the weak
convergence as $L\to\infty$ of the probability distribution
$\{\pi_k(c;L)\}_{k=0}^{\infty}$
on $\mathbf{Z}_+$ to the Poisson distribution with mean $c$ . As is well
known in probability theory, a sufficient condition for this weak convergence
is that the moments of all orders of $\{\pi_k(c;L)\}_{k=0}^{\infty}$
converges to those of the Poisson distribution. Namely we have
\begin{prop}
If one has
\begin{equation}
\mu_k(c)\equiv\lim_{L\to\infty}\mu_k(c;L)=\frac{c^k}{k!}
\end{equation}
for each $c>0$ and $k=1,2,\ldots$ , then one has the strict sense
level clustering.
\end{prop}
\bigskip
If one has the strict sense level clustering, then the limiting level spacing
distribution exists and is the exponential distribution $e^{-c}dc$ . This
is a direct consequence of the following more general proposition:
\begin{prop}
If $\pi_0(c)=\lim_{L\to\infty}\pi_0(c;L)$ exists and is differentiable with
respect to $c$ , then $\rho(c)=\lim_{L\to\infty}\rho(c;L)$ also exists,
and one has
\begin{equation}
\rho(c)=1+\frac d{dc}\pi_0(c)\ .
\end{equation}
\end{prop}
\bigskip
It seems to be a difficult problem to prove the strict sense level clustering
for any concrete Hamiltonian. Moreover, as was shown in \cite{minami1} ,
there is an example of one-dimensional Hamiltonian for which the level
statistics, in a slightly different formulation, can be rigorously performed,
but the obtained level spacing distribution is different from the exponential
distribution. In that case, we have still $\rho'(0+)>0$ , so that we should
say that the level clustering is taking place, and if we take the high
disorder limit in the system, then the data $\rho(c)$ converges to the
distribution function of $e^{-c}dc$ . This circumstance has led the present
author to the opinion that we should difine the notion of level clustering
in much weaker sense, nevertheless retaining physical significance. Thus
taking the broadest statistical sense of the clustering and repulsion, we
make the following definition:
\begin{df}
We shall say that one has the wide sense level clustering [resp. repulsion] if
\begin{equation}
\liminf_{c\searrow0}\frac{\underline{\rho}(c)}{c}>0\quad
\left[\mbox{resp.}\ \limsup_{c\searrow0}\frac{\bar{\rho}(c)}{c}=0\right]
\end{equation}
holds. Here we have set
\begin{equation}
\underline{\rho}(c)=\liminf_{L\to\infty}\rho(c;L)\quad;\quad
\bar{\rho}(c)=\limsup_{L\to\infty}\rho(c;L)\ .
\end{equation}
\end{df}
\bigskip
We can prove the following
\begin{prop}
(i) Suppose one has $\mu_1(c)=c$ for any $c>0$ and
\begin{equation}
\bar{\mu}_2(c)\equiv\limsup_{L\to\infty}\mu_2(c;L)=o(c^2)\quad(c\searrow0)\ .
\end{equation}
Then one has the wide sense level repulsion.
\noindent
(ii) Suppose one has $\mu_1(c)=c$ , $\mu_2(c)=\frac12 c^2$ for any $c>0$ and
\begin{equation}
\bar{\mu}_3(c)\equiv\limsup_{L\to\infty}\mu_3(c;L)=o(c^2)\quad(c\searrow0)\ .
\end{equation}
Then one has the wide sense level clustering.
\end{prop}
\section{Level statistics for regular spectra}
Suppose that the classical Hamiltonian system associated to $H(p,q)$ is
completely integrable. Then one can transform the variable $(p,q)$ into
the action--angle variable $(I,\varphi)=(I_1,\ldots,I_d;\varphi_1,\ldots,
\varphi_d)$ , and $H(p,q)=H(I)$ depends only on the action variable (see
\cite{arnold}).
Now following Percival (\cite{percival})
, Berry and Tabor (\cite{berry}), we shall say that $H(\hbar)$ has
regular spectrum if its energy levels are (approximately) given by
quantizing the action variables $I_1,\ldots,I_d$ appearing in $H(I)$ .
Namely, we suppose that the energy levels of $H(\hbar)$ are
\begin{equation}
E_n(\hbar)=H(\hbar(n+\frac14\alpha))\quad;\quad n=(n_1,\ldots,n_d)\in
\mathbf{Z}_+^d\ ,
\end{equation}
where $\alpha\in\mathbf{Z}_+^d$ is the Maslov index. This procedure is called
the EBK quantization after the names of Einstein, Brillouin and Keller, and
it gives the exact spectrum in such concrete examples as the rectangular
billiards and the harmonic oscillators.
Under mild conditions on $H(I)$ (e.g. $H(I)$ being convex, positive with
$H(0)=0$ ) , regular spectra satisfy the conditions (A1)--(A3) stated in \S1
with
\begin{equation}
\nu(E)=\int\cdots\int_{\mathbf{R}_+^d}1_{\{H(I)\leq E\}}dI_1\cdots dI_d\ .
\end{equation}
Let us apply the level statistics formulated in \S1 to our regular spectrum.
Note that the suffix $n$ distinguishing the levels is now $d$-dimensional.
For $x\in\mathbf{R}_+^d\setminus\{0\}$ define
$\hbar(x)=\hbar_E(x)\geq0$ by the equation
$H(\hbar(x)\cdot x)=E$ , where $E>0$ will be fixed throughout, and let
$\lambda(x)=\nu(E)\hbar_E(x)^{-d}$ . Our unfolded levels are then given by
$\lambda(n+\frac14\alpha)$ , $n\in\mathbf{Z}_+^d$ . Since
$\lambda(\beta x)=\beta^d\lambda(x)$ for $\beta>0$ ,
we have the equivalence
\begin{equation}
\lambda(n+\frac14\alpha)\in(t,t+c]\ \Leftrightarrow n+\frac14\alpha\in
\Pi_c(t)\ ,
\end{equation}
where we have defined
\begin{equation}
\Pi_c(t)=\{x\in\mathbf{R}_+^d\ \vert\
t^{1/d}\lambda(\varphi(x))^{-1/d}<\vert x\vert<
(t+c)^{1/d}\lambda(\varphi(x))^{-1/d}\}\ ,
\end{equation}
with
\begin{equation}
\varphi(x)=\frac{x}{\vert x\vert}\ .
\end{equation}
It is easy to see $\vert\Pi_c(t)\vert=c$ , where
$\vert\Pi_c(t)\vert$ is the volume of $\Pi_c(t)$ .
Hence the number $N(t)=N(t;c)$ of unfolded levels is equal to the number of
lattice points (which are shifted by $\frac14\alpha$) in the domain
$\Pi_c(t)$ , namely we have
\begin{equation}
N(t)=\sharp\{\Pi_c(t)\cap(\mathbf{Z}_+^d+\frac14\alpha)\}\ .
\end{equation}
>From this geometric consideration and $\vert\Pi_c(t)\vert=c$ , it is obvious
that we always have $\mu_1(c)=c$ . It is, however, a non-trivial question to
prove or disprove $\mu_k(c)=c^k/k!$ for $k\geq2$ . Sarnak (\cite{sarnak})
proved $\mu_2(c)=c^2/2$ by a number theoretic method for the quantized
uniform motion on a two dimensional flat torus. (Actually he proved that the
pair correlation function is proportional to $c$ , which turns out to be
equivalent to $\mu_2(c)=c^2/2$ after a suitable normalization.)
On the other hand, Sinai \cite{sinai} and Major \cite{major} (see also
\cite{minami2}) considered the case in which $d=2$ , $\alpha=0$ and the curve
(written in polar coordinate)
$r=f(\varphi)=\lambda(\varphi)^{-1/d}$ which defines the boundary of
$\Pi_c(t)$
is very random. Especially, Major proved that
$\mu_k(c)=c^k/k!$ , $k\geq1$ holds
for almost all realization of the random curve $r=f(\varphi)$ .
But the randomness they assume is so strong that the curve $r=f(\varphi)$
consisting the boundary of $\Pi_c(t)$ cannot be smooth, which violates the
natural connection between level statistics and the lattice points counting.
On the other hand, their proof suggests us that it would be possible to
prove $\mu_k(c)=c^k/k!$ for $k=1,\ldots,K$ and for any finite $K$ ,
if the boundary of $\Pi_c(t)$
has \lq\lq $n$-dimensional randomness \rq\rq with $n$ sufficiently large.
(Presumably $n\geq2K$ would suffice.)
For example, consider the $d$-dimensional billiard of a particle of
mass $1$ in the rectangle
$\Lambda=\Pi_{j=1}^d[0,a_j]$\ . Then its classical Hamiltonian is
$H(I)=\frac{\pi^2}{2}\sum_{j=1}^d(I_j/a_j)^2$ , if expressed in terms
of action variables, and its quantization is $H(\hbar)=-\frac{\hbar^2}{2}
\Delta$ with the Dirichlet boundary condition on $\partial\Lambda$ .
Its energy levels (eigenvalues) are exactly given by
\begin{equation}
E_n(\hbar)=H(\hbar n)=\frac{\pi^2\hbar^2}{2}\sum_{j=1}^d(\frac{n_j}{a_j})^2
\quad(n=(n_1,\ldots,n_d)\ ,\ n_j\geq1)\ .
\end{equation}
We can then conjecture that if $(a_1\ldots,a_d)$ is a $d$-dimensional random
variable with smooth density, then for sufficiently large $d$ , one has
$\mu_k(c)=c^k/k!$ for $k=2$ and $3$ for almost all $(a_1\ldots,a_d)$ ,
yielding the wide sense level clustering according to Proposition 3.
\section{A preliminary result}
Let $d=2$ in the rectangular billiard of \S2. In this case we have
\begin{equation}
\Pi_c(t)=\{x\in\mathbf{R}_+^2\ \vert\ t<\frac{\pi}{4}a_1a_2
\sum_{j=1}^2(\frac{x_j}{a_j})^2