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\begin{document}
\newtheorem{df}{Definition}
\newtheorem{prop}{Proposition}
\title{On Level Clustering in Regular Spectra}
\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
\section{Introduction.}
Let $H(\hbar)$ be the quantum Hamiltonian describing a bounded, isolated
physical system of finite degrees of freedom. ( $\hbar$ is the Planck's
constant.) Then the spectrum of $H(\hbar)$ is purely discrete, consisting of
energy levels $\{E_n(\hbar)\}_{n\geq1}$ . Suppose further that $H(\hbar)$ ,
or its spectrum itself, is obtained by quantizing (in a certain way) the
classical Hamiltonian $H(p,q)$ defining the system. Now the classical
dynamical system corresponding to $H(p,q)$ may belong to one of two extreme
cases of being completely integrable or chaotic. It was Percival
\cite{percival} who proposed to distinguish the spectrum of $H(\hbar)$
according to the category to which the classical counterpart of $H(\hbar)$
belongs. He called the spectrum regular [resp. irregular] if $H(p,q)$ is
integrable [resp. chaotic], and discussed possible difference between these
two kinds of spectra. Later, Berry and Tabor \cite{berry} argued that
regular and irregular spectra are distinguished by looking at the probability
distribution of the spacing between adjacent energy levels. Namely they argued
that in the regular spectra, the level spacing obeys the exponential
distribution $e^{-t}dt$ , so that the spectrum looks like a typical
realization of the Poisson point process, which is the phenomenon they
called \lq\lq level clustering \rq\rq . On the other hand, they conjectured
that in the irregular spectra, the level spacing distribution $p(t)dt$
satisfies $p(t)\sim\mbox{const.}t^{\gamma}$ , $t\searrow0$ , with $\gamma>0$
(\lq\lq level repulsion \rq\rq). Hence irregular spectra should typically
look like the spectra of large random matrices. This conjecture is
supported by numerical studies performed later. (See e.g. \cite{bohigas}
for a review.)
However, neither the precise meaning of the \lq\lq probability distribution
\rq\rq of level spacing, nor a mathematical formulation of the statistics
for the spectrum of $H(\hbar)$ --the level statistics--in general seems
to have been explicitly given in physics litterature, although some
important ideas are stated in \cite{berry}. The present paper, which is an
elaboration of a part of the author's previous note \cite{minami3},
aims at giving a mathematical formulation of level statistics based
on the idea of Berry and Tabor, and applying it to regular spectra.
In \S2, we give the definition of strict (Definition 1) and wide
(Definition 2) sense level clustering, and prove some preliminary results
for later references. These results are formulated in analogy with
corresponding propositions in the theory of stationary point processes.
In \S3, we shall apply the level statistics thus formulated to regular
spectra, and discuss the closely related theorems by Sinai \cite{sinai2}
and Major \cite{major}. We argue that although it is probably very difficult
to apply theorems of Sinai and Major to prove the strict sense level
clustering in generic regular spectra, there is some hope in proving the
wide sense level clustering for some concrete Hamiltonian such as rectangular
billiard. Finally in \S4, we shall propose another formulation of level
statistics, and with an example, shall argue again that \lq\lq level
clustering \rq\rq should not always mean strict Poissonian property of
the spectrum.
\section{Strict and wide sense level clustering.}
\subsection{The unfolding of Berry and Tabor.}
When one speaks of the \lq\lq probability distribution \rq\rq of the spacing
of the adjacent energy levels of $H(\hbar)$ , the follwing two questions
immediately arise:
\begin{enumerate}
\item There is no stochasticity in $H(\hbar)$ , hence in
$\{E_n(\hbar)\}_{n\geq1}$ too. Therfore, it will be necessary to take
the semiclassical limit $\hbar\searrow0$ to have sufficiently many
levels in a fixed energy intervals, and we will have to take statistics
among these levels.
\item The mean level density is not uniform over the entire spectrum,
so that the statistical property of $E_{n+1}-E_n$ may not be uniform in
$n$ . Hence we will need to \lq\lq unfold \rq\rq the spectrum so that
it will look like a uniformly distributed sequence.
\end{enumerate}
Inspired by \cite{berry}, we make a normalization of the spectrum
$\{E_n(\hbar)\}_{n\geq1}$ of $H(\hbar)$ which meets the above two requirements
simultaneously. For this purpose, we make the following assumptions:
\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 $\hbar_j=\hbar_j(E)$ as the unique solution of the equation
$E_j(\hbar)=E$ . If we define
\begin{equation}
\lambda_j=\nu(E)\hbar_j(E)^{-d}\ ,
\end{equation}
then from (A3), it is easily seen that
\begin{equation}
n(L)\equiv\sharp\{j\geq1\ \vert\ \lambda_j\leq L\}\sim L\quad(L\to\infty)\ .
\label{eqn:2-1-1}
\end{equation}
Thus we have obtained a sequence $\{\lambda\}_j$ which has asymptotically
uniform distribution. We will call the $\lambda_j$'s the unfolded levels,
and call the above procedure \lq\lq the unfolding of Berry and Tabor
\rq\rq. (It is also called \lq\lq quantization of Planck's constant \rq\rq
in \cite{berry}.)
\subsection{Statistics for asymptotically uniformly distributed sequences.}
In this subsection, we suppose that $\{\lambda_j\}_{j=0}^{\infty}$ is
simply a strictly increasing sequence of real numbers satisfying
(\ref{eqn:2-1-1}) and $\lambda_0=0$ , but still call $\lambda_j$ the \lq\lq
$j$-th level \rq\rq . We introduce the following notation:
\begin{equation}
N(t)=N(t;c)=\sharp\{j\geq1\ \vert\ \lambda_j\in(t,t+c]\}
=\sum_{j\geq1}1_{(t,t+c]}(\lambda_j)\ ;
\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)=\frac1L\int_0^L \left(\begin{array}{c} N(t) \\ k \end{array} \right)
\ dt=\sum_{j\geq k}\left(\begin{array}{c} j \\ k \end{array} \right)
\pi_j(c;L)\ ,
\end{equation}
where
\begin{equation}
\left(\begin{array}{c} j \\ k \end{array} \right)=
\frac1{k!}j(j-1)\cdots(j-k+1)\ ;
\end{equation}
and finally
\begin{equation}
\rho(c;L)=\frac{\sharp\{j\geq1\ \vert\ \lambda_j\leq L\ ,\ \lambda_{j+1}-
\lambda_j\leq c\}}
{\sharp\{j\geq1\ \vert\ \lambda_j\leq L\}}\ .
\end{equation}
Here $c>0$ and $k=0,1,2,\ldots$ .
These notions have the following obvious probabilistic meanings. $N(t;c)$
is nothing but the number of levels in the interval $(t,t+c]$ .
$\pi_k(c;L)$ is the probability that this interval catches exactly $k$ levels
when $t$ is randomly chosen from $(0,L]$ according to the uniform
distribution, and $\mu_k(c;L)$ is the $k$-th
factorial moment of the random varible $N(\cdot;c)$ . Finally, $\rho(c;L)$
is the relative frequency of those pairs of
levels below $L$ which have spacing not
exceeding $c$ .
We also write
\begin{equation}
\bar{\mu}_k(c)=\limsup_{L\to\infty}\mu_k(c;L)\ ;\
\underline{\mu}_k(c)=\liminf_{L\to\infty}\mu_k(c;L)\ ;
\end{equation}
\begin{equation}
\bar{\rho}(c)=\limsup_{L\to\infty}\rho(c;L)\ ;\
\underline{\rho}(c)=\liminf_{L\to\infty}\rho(c;L)\ ,
\end{equation}
and also
\begin{equation}
\pi_k(c)=\lim_{L\to\infty}\pi_k(c;L)\ ;\
\mu_k(c)=\lim_{L\to\infty}\mu_k(c;L)\ ;\
\rho(c)=\lim_{L\to\infty}\rho(c;L)
\end{equation}
whenever these limits exist. This $\rho(c)$ will be called the limiting level
spacing distribution function.
\begin{prop}
Under (\ref{eqn:2-1-1}), we have $\mu_1(c)=c$ for any $c>0$ .
\end{prop}
\noindent{\em Proof.} By the definition,
\begin{equation}
\mu_1(c;L)=\frac1L\sum_j\int_0^L1_{[\lambda_j-c,\lambda_j)}(t)dt\ .
\end{equation}
But if $L>c>0$ , one has
\[\int_0^L1_{[\lambda_j-c,\lambda_j)}(t)dt=
\left\{\begin{array}{rl}
c,&\quad\mbox{if $c\leq\lambda_j\leq L$}\\
0,&\quad\mbox{if $\lambda_j>L+c$}
\end{array}\right.\]
and
$$\int_0^L1_{[\lambda_j-c,\lambda_j)}(t)dt\leq c\ ,\quad\mbox{if}\ \lambda_j0$ , it is obvious that
%\begin{equation}
$$\mu_1(c;L)\sim c\frac{n(L)}L\quad\mbox{as}\ L\to\infty\ ,$$
%\end{equation}
so that
%\begin{equation}
$$\mu_1(c)=\lim_{L\to\infty}\mu_1(c;L)=c\lim_{L\to\infty}\frac{n(L)}{L}=c\ ,$$
%\end{equation}
which was to be proved.
\bigskip
Now we give a definition of $\{\lambda_j\}$'s looking like Poisson point
process in the following way:
\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
Note that if $\{\lambda_j\}_{j\geq1}$ actually is the realization of Poisson
point process on $[0,\infty)$ , then by the ergodic theorem, one has the
strict sense level clustering with probability one.
Obviously, the strict sense level clustering is equivalent to the weak
convergence of the family of probability distributions
$P(c;L)\equiv\{\pi_k(c;L)\}_{k=0}^{\infty}$ ($L>0$) , on $\mathbf{Z}_+$
to the Poisson
distribution $\{e^{-c}c^k/k!\}_{k=0}^{\infty}$ as $L\to\infty$ . As is well
known in elementary probability theory, a sufficient condition for this
weak convergence is that the factorial moments $\mu_k(c;L)$ of all
order of $P(c;L)$ converges to those corresponding to the Poisson
distribution. Namely we have
\bigskip
\begin{prop}
If
\begin{equation}
\mu_k(c)=\lim_{L\to\infty}\mu_k(c;L)=\frac{c^k}{k!}
\end{equation}
holds 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:
\bigskip
\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 is given by
\begin{equation}
\rho(c)=1+\frac d{dc}\pi_0(c)\ .
\end{equation}
\end{prop}
\bigskip
In fact, if the strict sense level clustering holds, then one has in
particular $\pi_0(c)=e^{-c}$ . Hence by the above proposition,
$\rho(c)=1-e^{-c}$ .
In \S1 of \cite{sinai1}, Sinai gave two definitions of $\{\lambda_j\}$'s
similarity
to the Poisson point process. One is the strict sense level clustering
as defined above, and the other is the existence of the limiting level
spacing distribution $\rho(c)$ and its equality to the exponential
distribution. The above proposition shows that these two definitions
are not independent.
\bigskip
\noindent
{\em Proof of Proposition 3.} If $\lambda_j\leq L$ and
$\lambda_{j+1}-\lambda_j>c$ , then for any $\delta\in(0,c)$ , we have the
implication
%\begin{equation}
$$
t\in[\lambda_j-\delta,\lambda_j)\ \Longrightarrow\ N(t;\delta)>0\ ,\
N(t+\delta;c-\delta)=0\ ,
$$
%\end{equation}
where the intervals $[\lambda_j-\delta,\lambda_j)$ are disjoint.
Hence if we set
%\begin{equation}
$$
n(c;L)=\sharp\{j\geq1\ \vert\ \lambda_j\leq L\ ,\ \lambda_{j+1}-\lambda_j>c\}
\ ,
%\end{equation}
$$
then we have
\begin{eqnarray*}
\delta n(c;L)
&=&\sum_{\lambda_j\leq L\ ;\ \lambda_{j+1}-\lambda_j>c}
\int_{-\delta}^L1_{[\lambda_j-\delta,\lambda_j)}(t)dt \\
&\leq&\delta+\int_0^L 1_{\{N(t;c)>0,N(t+\delta;c-\delta)=0\}}(t)dt \\
&=&\delta+\int_0^L 1_{\{N(t+\delta;c-\delta)=0 \}}(t)dt-
\int_0^L 1_{\{N(t;c)=0\}}(t)dt\ ,
\end{eqnarray*}
namely
$$\delta\frac{n(c;L)}{L}\leq\frac{\delta}{L}+\pi_0(c-\delta;L)-
\pi_0(c;L)\ .$$
Letting $L\to\infty$ and noting $n(L)\sim L$ , we get
$$\delta\limsup_{L\to\infty}\frac{n(c;L)}{n(L)}=
\delta(1-\underline{\rho}(c))\leq\pi_0(c-\delta)-\pi_0(c)\ ,$$
or
$$\underline{\rho}(c)\geq1+\frac{\pi_0(c)-\pi_0(c-\delta)}{\delta}\ .$$
Letting $\delta\searrow0$ , we arrive at
$$\underline{\rho}(c)\geq1+\frac{d}{dc}\pi_0(c)\ .$$
On the other hand, if $N(t;\delta)>0$ and $N(t+\delta;c)=0$ , then there is
a $\lambda_j\in(t,t+\delta]$ such that $\lambda_{j+1}-\lambda_j>c$ .
Hence the set
$$\{t\in[0,L]\ \vert\ N(t;\delta)>0\ ,\ N(t+\delta;c)=0\}$$
is the union of finitely many disjoint intervals $I_{\ell}$ each with
length no greater than $\delta$ , and for each $I_{\ell}$ , there corresponds
a $\lambda_j\leq L+\delta$ satisfying $\lambda_{j+1}-\lambda_j>c$ .
Hence the number of these intervals does not exceed $1+n(c;L)$ , and so
\begin{eqnarray*}
\delta(1+n(c;L))&\geq&\int_0^L1_{\{N(t;\delta)>0,N(t+\delta;c)=0\}}(t)dt \\
&=&\int_0^L 1_{\{N(t+\delta;c)=0 \}}(t)dt-
\int_0^L 1_{\{N(t;c+\delta)=0 \}}(t)dt\ .
\end{eqnarray*}
Dividing by $L$ and letting $L\to\infty$ , we have
$$\delta\liminf_{L\to\infty}\frac{n(c;L)}{n(L)}=
\delta(1-\bar{\rho}(c))\geq\pi_0(c)-\pi_0(c+\delta)\ ,$$
or
$$\bar{\rho}(c)\leq 1+\frac{\pi_0(c+\delta)-\pi_0(c)}{\delta}\ .$$
Again letting $\delta\searrow0$ , we get
$$\bar{\rho}(c)\leq 1+\frac{d}{dc}\pi_0(c)\ ,$$
completing the proof.
\bigskip
As will be explained in \S3, it seems to be a difficult problem to prove
the strict sense level clustering for any concrete Hamiltonian. In fact,
no explicit example of regular spectrum is known which shows strict sense
level clustering. Moreover, as was shown in \cite{minami1} and will be
discussed in \S4,
there is an example of one-dimensional Hamiltonian for which the level
statistics, in a somewhat 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 situation suggests us that the
strict sense level clustering can only be proved in some ideal limit,
and generically level clustering should not mean the strict Poissonian
character of the unfolded spectrum. Thus we are led to define the notion
of level clustering in much weaker sense, nevertheless retaining some
physical significance. Taking the broadest statistical sense of the words
\lq\lq clustering \rq\rq and \lq\lq repulsion \rq\rq, we now make the
following definition.
\bigskip
\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.
\end{df}
\bigskip
We can give a criteria for the wide sense repulsion and clustering in
terms of $\mu_k(c)$ , $k\leq3$ . Recall that we always have $\mu_1(c)=c$ .
\begin{prop}
(i) If $\bar{\mu}_2(c)=o(c^2)$ as $c\searrow0$ , then one has the wide sense
level repulsion.
\medskip
(ii) If $\mu_2(c)=\frac12 c^2$ for any $c>0$ and $\bar{\mu}_3(c)=o(c^2)$
as $c\searrow0$ , then one has the wide sense level clustering
\end{prop}
\bigskip
\noindent{\em Proof.} Successively applying the equality
%\begin{equation}
$$\mu_k(c;L)=\sum_{j\geq k}\left(\begin{array}{c} j \\ k \end{array} \right)
\pi_j(c;L)\ ,\ k\geq1\ ,$$
%\end{equation}
we obtain
%\begin{equation}
$$1-\pi_0(c;L)=\sum_{j\geq1}\pi_j(c;L)=
\sum_{j=1}^k(-1)^{j-1}\mu_j(c;L)+\sum_{j=k+1}^{\infty}\sum_{i=0}^k
(-1)^i \left(\begin{array}{c} j \\ i \end{array} \right)\pi_j(c;L)$$
%\end{equation}
for all $k\geq1$ . But sinse
%\begin{equation}
$$\sum_{i=0}^k(-1)^i\left(\begin{array}{c} j \\ i \end{array} \right)=
(-1)^k\left(\begin{array}{c} j-1 \\ k \end{array} \right)\ ,\
j\geq k+1\ ,\ k\geq0\ ,$$
%\end{equation}
we have the inequality
\begin{equation}
\pi_0(c;L)\leq1-\mu_1(c;L)+\mu_2(c;L)-\cdots+(-1)^k\mu_k(c;L)
\label{eqn:even}
\end{equation}
when $k$ is even and
\begin{equation}
\pi_0(c;L)\geq1-\mu_1(c;L)+\mu_2(c;L)-\cdots+(-1)^k\mu_k(c;L)
\label{eqn:odd}
\end{equation}
when $k$ is odd.
On the other hand, if $L>0$ , $N=n(L)+1$ , namely $\lambda_N\leq L<
\lambda_{N+1}$ , then
\begin{eqnarray*}
L\pi_0(c;L)&=&\sum_{n=0}^{N-1}(\lambda_{n+1}-\lambda_n-c)_++
\{(\lambda_{N+1}-\lambda_N-c)_+\wedge(L-\lambda_N)\} \\
&=&\sum_{n=0}^{N-1}(\lambda_{n+1}-\lambda_n-c)+
\sum_{n=0}^{N-1}\{c-(\lambda_{n+1}-\lambda_n)\}_+ \\
& &+\{(\lambda_{N+1}-\lambda_N-c)_+\wedge(L-\lambda_N)\} \\
&\leq&\lambda_N-cN+c\sharp\{n\leq N-1\vert\lambda_{n+1}-\lambda_n\leq c\}
+(L-\lambda_N) \\
&=&L-cN+cL\rho(c;L)\ .
\end{eqnarray*}
Dividing both sides by $L$ and letting $L\to\infty$ , we obtain from
(\ref{eqn:odd}),
\begin{eqnarray*}
1+c\underline{\rho}(c)-c&\geq&\underline{\pi}_0(c) \\
&\geq&1-\mu_1(c)+\mu_2(c)-\bar{\mu}_3(c) \\
&=&1-c+\frac12c^2+o(c^2)\ ,
\end{eqnarray*}
when the conditions of (ii) hold.
Hence we get
%\begin{equation}
$$\liminf_{c\searrow}\frac1c\underline{\rho}(c)\geq\frac12>0\ ,$$
%\end{equation}
namely the wide sense level clustering.
Now suppose the condition of (i) hold. For each $0<\delta<1$ , one has
\begin{eqnarray*}
L\pi_0(c;L)
&=&\sum_{n=0}^{N-1}(\lambda_{n+1}-\lambda_n-c)+
\sum_{n=0}^{N-1}\{c-(\lambda_{n+1}-\lambda_n)\}_+ \\
& &+\{(\lambda_{N+1}-\lambda_N-c)_+\wedge(L-\lambda_N)\} \\
&\geq&\lambda_N-cN+\sum_{\lambda_{n+1}-\lambda_n\leq\delta c}
\{c-(\lambda_{n+1}-\lambda_n)\}+(L-\lambda_N-c) \\
&\geq&(L-c)-cN+(1-\delta)c\rho(\delta c;L)\ .
\end{eqnarray*}
Again dividing by $L$ and letting $L\to\infty$ and noting (\ref{eqn:even}),
\begin{eqnarray*}
1+(1-\delta)c\bar{\rho}(\delta c)-c&\leq&\bar{\pi}_0(c) \\
&\leq&1-\mu_1(c)+\bar{\mu}_2(c) \\
&=&1-c+o(c^2)\ .
\end{eqnarray*}
Hence we have
%\begin{equation}
$$\limsup_{c\searrow}\frac1c\bar{\rho}(c)=
\limsup_{c\searrow}\frac1{\delta c}\bar{\rho}(\delta c)=0\ ,$$
%\end{equation}
namely the wide sense level repulsion.
\bigskip
At this point, let us discuss the relation of the factorial moment
$\mu_k(c)$ to the so called $k$-th correlation $R_k(c)$ defined by
\begin{equation}
R_k(c)=\lim_{L\to\infty}R_k(c;L)\ ;
\end{equation}
\begin{equation}
R_k(c;L)=\sharp\{(\lambda_{j_1},\ldots,\lambda_{j_k})\ \vert\
\lambda_{j_1}<\cdots<\lambda_{j_k}\leq L\ ,\ \lambda_{j_k}-\lambda_{j_1}\leq c\}
\ ,
\end{equation}
whenever the limit exists. In particular, $R_2(c)$ is called the pair
correlation (see e.g. \cite{sarnak}).
\bigskip
\begin{prop}
(i) If $\mu_k(c)$ exists for each $c>0$ and is differentiable with respect to
$c$ , then $R_k(c)$ also exists and
\begin{equation}
R_k(c)=\frac{d}{dc}\mu_k(c)\ .
\end{equation}
(ii) If $R_k(c)$ exists for each $c>0$ , then $\mu_k(c)$ also exists and
\begin{equation}
\mu_k(c)=\int_0^LR_k(c')dc'\ .
\end{equation}
\end{prop}
\bigskip
{\em Proof.} Since
\begin{eqnarray*}
\left(\begin{array}{c} N(t) \\ k \end{array} \right)
&=&\sum_{\lambda_{j_1}<\cdots<\lambda_{j_k}}1_{(t,t+c]}(\lambda_{j_1})
\cdots1_{(t,t+c]}(\lambda_{j_k}) \\
&=&\sum_{\lambda_{j_1}<\cdots<\lambda_{j_k}}
1_{[\lambda_{j_1}-c,\lambda_{j_1})}(t)\cdots
1_{[\lambda_{j_k}-c,\lambda_{j_k})}(t)\ ,
\end{eqnarray*}
we can compute
\begin{eqnarray*}
L\mu_k(c;L)
&=&\sum_{\lambda_{j_1}<\cdots<\lambda_{j_k}}
\vert[0,L]\cap\cap_{p=1}^k[\lambda_{j_p}-c,\lambda_{j_p})\vert \\
&=&\sum_{\lambda_{j_1}<\cdots<\lambda_{j_k}}\{(L\wedge\lambda_{j_1})
-(\lambda_{j_k}-c)_+\}_+ \\
&=&\sum_{\lambda_{j_1}<\cdots<\lambda_{j_k};\atop \lambda_{j_1}\leq L,
\lambda_{j_k}\geq c} (c-\lambda_{j_k}+\lambda_{j_1})_+ \\
& &+\sum_{L<\lambda_{j_1}<\cdots<\lambda_{j_k}}(c-\lambda_{j_k}+L)_+
+\sum_{\lambda_{j_1}<\cdots<\lambda_{j_k}L+c$ and
$1_{(t,t+c]}(\lambda_{j_k})=0$ for $tL+c$ and
$1_{(t,t+c]}(\lambda_{j_1})=0$ for $t0\}$ and turns out to be a probability measure when $m=1$ . In
this case, $\hat{P}(d\omega)$ has an intuitive meaning of the conditional
probability $P(d\omega\vert N(\{0\})>0)$ . (See \cite{neveu} for detail.)
Now we can apply the individual ergodic theorem and (\ref{eqn:palm}),
to prove that the following
relations hold with probability one:
\begin{description}
\item[(0)]
$$\lim_{L\to\infty}\frac1L\sharp\{j\geq1\vert\lambda_j(\omega)\leq L\}
=\lim_{L\to\infty}\frac1LN_{\omega}(0,L]=E[N_{\omega}(0,1]]=1\ ;$$
\item[(i)]
$$\pi_k(c)\equiv\lim_{L\to\infty}\frac1L\int_0^L
1_{\{N_{\omega}(t,t+c]=k\}}dt=\lim_{L\to\infty}
\frac1L\int_0^L 1_{\{N_{\theta_t\omega}(0,c]=k\}}dt=
P(N(0,c]=k)\ ;$$
\item[(ii)]
$$\rho(c)\equiv\lim_{L\to\infty}\frac1L
\sharp\{j\geq1\vert\lambda_j(\omega)\leq L\ ,\
\lambda_{j+1}(\omega)-\lambda_{j}(\omega)\leq c\}=
\hat{P}(N(0,c]>0)\ ;$$
\item[(iii)]
$$\mu_k(c)\equiv\lim_{L\to\infty}\frac1L\int_0^L
\left(\begin{array}{c} N_{\omega}(t;t+c] \\ k \end{array} \right)dt
=E\left[\left(\begin{array}{c} N(0;c] \\ k \end{array} \right)
\right]\ ;$$
\item[(iv)]
$$R_k(c)\equiv\lim_{L\to\infty}\frac1L\int_0^L
\sum_{\lambda_{j_1}(\omega)<\cdots<\lambda_{j_k}(\omega)\leq L}
1_{\{\lambda_{j_k}(\omega)-\lambda_{j_1}(\omega)\leq c\}}
=\hat{E}\left[\left(\begin{array}{c} N(0;c] \\ k-1 \end{array} \right)
\right]\ ,$$
\end{description}
where $E[\cdot]$ and $\hat{E}[\cdot]$ denote the integration with respect to
the measures $P$ and $\hat{P}$ respectively.
If we let
$$f(\omega,s)=1_{(-\infty,0]}(s)1_{\{N_{\omega}(0,x-s]=j\}}$$
in (\ref{eqn:palm}), we obtain the so called Palm-Khinchin formula:
\begin{equation}
P(N(0,x]\leq j)=\int_x^{\infty}\hat{P}(N(0,s]=j)ds\ .
\end{equation}
(See \cite{daley} for another approach.)
This can be used to relate the right hand sides of (i) and (ii), (iii) and
(iv). Indeed, letting $j=0$ in Palm-Khinchin formula above, one obtains
%\begin{equation}
$$P(N(0,c]=0)=\int_c^{\infty}\hat{P}(N(0,s]=0)ds\ ,$$
%\end{equation}
namely
%\begin{equation}
$$\pi_0(c)=\int_c^{\infty}(1-\rho(s))ds\ ,$$
%\end{equation}
or
%\begin{equation}
$$\rho(c)=1+\frac{d}{dc}\pi_0(c)\ ,$$
%\end{equation}
whenever $\rho(\cdot)$ is continuous at $c$ . On the other hand,
\begin{eqnarray*}
E\left[\left(\begin{array}{c} N(0;c] \\ k \end{array} \right)\right]&=&
\sum_{n\geq k}\left(\begin{array}{c} n-1 \\ k-1 \end{array} \right)
P(N(0,c]\geq n) \\
&=&\sum_{n\geq k}\left(\begin{array}{c} n-1 \\ k-1 \end{array} \right)
\int_0^c\hat{P}(N(0,s]=n-1)ds \\
&=&\int_0^c\hat{E}\left[
\left(\begin{array}{c} N(0;c] \\ k-1 \end{array} \right)\right]ds\ ,
\end{eqnarray*}
namely
%\begin{equation}
$$\mu_k(c)=\int_0^cR_k(s)ds\ .$$
%\end{equation}
(Compare \cite{cramer}.)
Hence when the sequence $\{\lambda_j\}$ is the typical realization of a
stationary point process, then $\pi_0(c)$ , $\rho(c)$ , $\mu_k(c)$ and
$R_k(c)$ all exist with probability one, and are expressed as appropriate
expectation values which are related with each other through Palm-
Khinchin formula. By this observation, we are inclined to call the
considerations developed in \S2 \lq\lq deterministic point process theory
\rq\rq. After all, the energy level statistics is based upon the hypothesis,
often tacitly supposed, that the spectrum of a quantum Hamiltonian looks,
after a suitable normalization, like a typical realization of a stationary
point process. Hence the phenomenological side of the theory of energy level
statistics should be developed in analogy of the theory of point processes.
\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)\in
\mathbf{R}^d\times\mathbf{R}^d$ 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)$ ,
the quantizationof $H(p,q)$ ,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 unfolding of Berry and Tabor as formulated in \S2-1
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\leq
(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)$ .
Thus the number $N(t)=N(t;c)$ of unfolded levels in $(t,t+c]$
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}
As $t$ gets large, then the domain $\Pi_c(t)$ expands in the space
$\mathbf{R}_+^d$ , at the same time getting thinner and thinner to keep its
volume constant. Hence provided the boundary of $\Pi_c(t)$ is not too
degenerate, e.g. flat, then each lattice point in $\mathbf{Z}_+^d$ would
randomly belong to $\Pi_c(t)$ if $t$ were chosen at random from a long
interval $[0,L]$ , so that the total number of lattice points in $\Pi_c(t)$ ,
namely $N(t)$ , would obey the Poisson distribution. This is conceivable
if one recalls how Poisson's law of small numbers was proved in elementary
probability theory. But it must be a very difficult peoblem to justify this
intuitive idea for concretely specified $\Pi_c(t)$ . In fact no example of
Hamiltonian is known for which this program is rigorously performed, to
prove the strict sense level clustering.
At this point, it is appropriate to discuss the results of Sinai \cite{sinai2}
and Major \cite{major} (see also \cite{minami2}). They 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)$ .
Although the randomness they assume is so strong that the curve $r=f(\varphi)$
consisting the boundary of $\Pi_c(t)$ cannot be smooth, violating the
natural connection between level statistics and the lattice points counting,
their proof suggests us that it would be possible to
prove $\mu_k(c)=c^k/k!$ for $k=1,\ldots,K$ with a finite $K$
if the boundary of $\Pi_c(t)$
has finite dimensional randomness as we are going to argue now.
The conditions for the random curve $r=f(\varphi)$ , $0\leq\varphi\leq\pi/2$
assumed by Sinai and
Major (see \cite{major} or \cite{minami2}) are the following:
\begin{description}
\item[(a)] $b_1\leq f(\varphi)\leq b_2$ , $\vert f(\varphi_2)-f(\varphi_1)
\vert\leq b_3\vert\varphi_2-\varphi_1\vert$
for some positive constants $b_j$ , $j=1,2,3$
\item[(b)] For any $k\geq1$ , $0\leq\varphi_1<\cdots<\varphi_k\leq2\pi$ ,
the joint probability distribution of $f(\varphi_j)$ , $j=1,\ldots,k$ has
a $C^1$ density
$$p_k(y_1,\ldots,y_k\vert\varphi_1,\ldots,\varphi_k)\ .$$
\item[(c)] For some $\tau\in(1,2)$ , one has
$$p_k(y_1,\ldots,y_k\vert\varphi_1,\ldots,\varphi_k)\leq
\mbox{const.}\prod_{j=2}^k(\varphi_{j}-\varphi_{j-1})^{-\tau}$$
\item[(d)] Similar conditions for derivatives of $p_k$ .
\end{description}
Define
\begin{equation}
\Pi_c(t;f)=\{x\in\mathbf{R}^2\vert 0\leq\varphi(x)\leq\Theta\ ,\
\sqrt{t}f(\varphi(x))<\vert x\vert<\sqrt{t+c}f(\varphi(x))\}\ .
\end{equation}
Then the area of $\Pi_c(t;f)$ equals $\lambda(f)=\frac{c}{2}\int_0^{\Theta}
f(\varphi)^2d\varphi$ .
Let $\xi(t;f)=\sharp(\Pi_c(t;f)\cap\mathbf{Z}^2)$ be the number of lattice
points caught in $\Pi_c(t;f)$ .
Then the following proposition holds (\cite{sinai2}, \cite{major} and
\cite{minami2}):
\begin{prop}
(i) Under the conditions {\bf (a)} to {\bf (c)} above, we have
\begin{equation}
\lim_{t\to\infty}E_P\left[\left(
\begin{array}{c} \xi(t;f) \\ k \end{array} \right)
\right]=\frac1{k!}E_P[\lambda(f)^k]\ ,\ k\geq1\ .
\end{equation}
(ii) Under the conditions {\bf (a)} to {\bf (d)} above, we have
with probability one,
\begin{equation}
\mu_k(c)=\lim_{L\to\infty}\frac1{(a_2-a_1)L}
\int_{a_1L}^{a_2L}\left(\begin{array}{c}\xi(t;f)\\ k\end{array}\right)dt
=\frac1{k!}\lambda(f)^k\ ,\ k\geq1\ ,
\end{equation}
where $00)=\hat{P}(\lambda_{n+1}-\lambda_n\leq c)\ ,$$
which does not depend on $n$ , then
$$R_2(c)=\hat{E}[N(0,c]]=\sum_{n=1}^{\infty}\hat{P}(N(0,c]\geq n)
=\sum_{n=1}^{\infty}\hat{P}(\sum_{j=0}^{n-1}(\lambda_{j+1}-\lambda_j\leq c)
=\sum_{n=1}^{\infty}(F^{n\ast})(c)\ ,$$
where $F^{n\ast}$ is the $n$-fold convolution of $F$ . Hence if
${\cal L}(\xi)=\int_0^{\infty}e^{-\xi c}dF(c)$ is the Laplace transform
of $dF(c)$ , then from $R_2(c)=c$ , we get
$$\sum_{n=1}^{\infty}({\cal L}(\xi))^n=
\frac{{\cal L}(\xi)}{1-{\cal L}(\xi)}=\int_0^{\infty}e^{-\xi c}dc
=\frac1{\xi}\ ,$$
namely ${\cal L}(\xi)=1/(1+\xi)$ . Hence $dF(c)=e^{-c}dc$ . But the Poisson
point process is the only renewal process for which $\lambda_{n+1}-\lambda_n$
obeys the exponential distribution.
\section{An example of level statistics, in which non-Poissonian
level clustering is observed.}
Consider the one-dimensional Schr\"odinger operator
\begin{equation}
H_v(\hbar)=-\hbar^2\frac{d^2}{dx^2}+v\sum_{j=1}^n\delta(x-x_j)\quad
0\leq x\leq1
\label{eqn:sch}
\end{equation}
with Dirichlet boundary condition at $x=0,1$ . Here $v>0$ and
$$0=x_00$ such that $cy_j<1$ , $j=0,1,\ldots,n$. Hence by
Proposition 3 in \S2,
\begin{eqnarray*}
\rho(c)&=&1-(\sum_{j=0}^n\frac{y_j}{1-cy_j})\prod_{j=0}^n(1-cy_j) \\
&\sim&c(1-\sum_{j=0}^ny_j^2)\ ,\ c\searrow0
\end{eqnarray*}
so that $\rho'(0+)=1-\sum_{j=0}^ny_j^2>0$ . Hence we have level clustering,
but the level spacing distribution $d\rho(c)$ is different from the
exponential distribution.
\bigskip
Now let $X_1,X_2,\ldots$ be i.i.d random variables uniformly distributed
in $(0,1)$ , and let
$X_1^{(n)}<\cdots