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This article is a contribution to the proceedings of the conference "Probability and Mathematical Physics", which was held at CRM, University of Montreal, June 27-July 1, 2005, in the honor of Stanislav Molchanov's 65th birthday.
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energy level statistics, point processes
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\begin{document}
\newtheorem{prop}{Proposition}
\newtheorem{thm}{Theorem}
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\newtheorem{cor}{Corollary}
\newtheorem{lemma}{Lemma}
\newtheorem{definition}{Definition}
\newtheorem{ex}{Example}
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\title{Theory of Point Processes and Some Basic Notions in \\
Energy Level Statistics 
\thanks{Dedicated to Professor S.A.~Molchanov on the occasion of 
his 65th birthday}} 
\author{Nariyuki MINAMI \\
Institute of Mathematics, Unversity of Tsukuba, \\
Tsukuba-shi, Ibaraki 305-8571, Japan. \\
e-mail: minami@sakura.cc.tsukuba.ac.jp}
\date{}
\maketitle

\begin{abstract}
Although it is not explicitly stated in physics literature, energy 
level statistics is based upon the hypothesis that the spectrum of a 
quantum Hamiltonian, after suitably normalized (unfolded),  and 
observed in thermodynamic or semiclassical limit, looks like a typical 
realization of a stationary point process. In this survey article, 
we present a purely phenomenological view of energy level statistics 
based on the theory of point processes.
\end{abstract}

%\keywords{level statistics; stationary point process; Palm measure; 
%Palm-Khinchin formula; unfolding; level clustering; quantum chaos} 
%\ams{81Q50}{82B44;60G55;60H25} 

\tableofcontents

\section{Introduction}
\label{sec;1}
Energy level statistics is a way to observe the discrete spectra of 
quantum Hamiltonians, in which one is interested in statistical fluctuation 
of energy levels (eigenvalues) looked at as a whole, rather than the 
location of individual levels nor the asymptotic or \lq\lq average\rq\rq 
distribution of levels. It originates in the study by nuclear physicists 
in early 60's (\cite{Po}), in which the excitation spectra of heavy nuclei 
were modelled by the spectra of random matrices. This led to an extensive 
study by Dyson, Mehta et al of random matrices. Their theory was also 
applied to fluctuation of spectra of small metallic particles (\cite{GE}). 
Later, Berry and Tabor (\cite{BT}) proposed to apply energy level statistics 
to semi-classical Hamiltonians in order to distinguish between regular 
and irregular spectra, a classification of discrete spectra of quantum 
Hamiltonians made by Percival (\cite{Pe}) according to the property of the 
underlying classical Hamiltonian dynamics describing the system being 
integrable or non-integrable. They argued that the spectrum of a quantum 
Hamiltonian which are obtained through quantization of a classical 
completely integrable Hamiltonian looks, after suitably normalized, like 
a typical realization of the Poisson point process. (This assertion is now 
called the \lq\lq Berry-Tabor conjecture\rq\rq \cite{Markl}.) They also 
conjectured that if a classical Hamiltonian dynamical system has chaotic 
behavior, then the spectrum of its quantization has fluctuation property 
similar to random matrices. This conjecture was later confirmed numerically 
(\cite{Bo}), but it is yet to be clarified mathematically. 
Fluctuation 
of spectra was investigated also for one-dimensional random Schr\"odinger 
operators (\cite{Mo}, \cite{MoR}, \cite{Rez1}, \cite{Rez2}), and for 
multi-dimensional Anderson tight binding models (\cite{Mi2}). In connection 
to Anderson transition, it is believed that the spectrum of random 
Schr\"odinger operator, considered on a large finite domain, looks like 
Poisson point process in the energy region where Anderson localization 
takes place, while in the energy region where delocalization of wave functions 
is expected, it looks like the spectrum of a random matrix. 

In order to give a meaning to the vague notion of \lq\lq fluctuation of 
levels\rq\rq, suppose we are given a (random) Hamiltonian $H$ with 
discrete spectrum
$$\cdots<E_n<E_{n+1}<\cdots<E_{n+m}<\cdots\ .$$
Among various statistics of the spectrum, one is often interested in spacings 
between consecutive levels, $S_i=E_{i+1}-E_i$, and wishes to know their 
statistical characteristics. (see e.g. \S1.4 of Mehta \cite{Me}.) But the 
following questions immediately arise:

\begin{enumerate}
	\item The \lq\lq density\rq\rq of levels is in general different in 
	different energy regions. For example, if $\{E_i\}$ are eigenvalues of the 
	Dirichlet Laplacian $-\Delta_D$ on a bounded domain 
	$\Omega\subset\mathbf{R}^d$, the cumulative distribution function 
	$$N(E):=\sum_i1_{\{E_i\leq E\}}$$
	has the asymptotic behavior $N(E)\sim\bar{N}(E):=CE^{d/2}$ 
	($E\to\infty$) with a positive constant $C$. Hence for $d>2$, levels 
	are distributed more densely for higher energy region, while for 
	$d=1$, they are distributed more sparsely. Thus in general, $S_i$'s 
	cannot have the same statistical nature.
	\item When one is observing the spectrum in the high energy region, 
	the exact numbering of levels is practically meaningless.
	\item Even if the density of levels is uniform over the entire energy 
	axis, it is not obvious if $S_i$'s have the same probability distribution 
	so that one can discuss the \lq\lq typical\rq\rq level spacing.
\end{enumerate}

In physics literature, the first obstacle is overcome by 
\lq\lq unfolding\rq\rq of the spectrum. In the case of Dirichlet Laplacian 
for example, this amounts to considering the sequence $\{\lambda_i\}_i$, 
where $\lambda_i:=\bar{N}(E_i)$, instead of $\{E_i\}$. By doing so, we 
obtain a sequence which has asymptotically uniform distribution. In fact, 
as $L\to\infty$,
$$\sum_i1_{\{\lambda_i\leq L\}}=\sum_i1_{\{\bar{N}(E_i)\leq L\}}
=N(\bar{N}^{-1}(L))\sim \bar{N}(\bar{N}^{-1}(L))=L\ .$$
In \S5 of this article, we shall discuss the notion of unfolding in a 
slightly generalized setting.

In order that level statistics makes sense, the unfolded levels 
$\{\lambda_i\}_i$ are expected to be distributed homogeneously over 
$\mathbf{R}^1$. Namely, the statistical nature of levels should not depend 
on the choice of the origin: for any choice of a real number $\lambda$, the 
sequence
$$\cdots,\lambda_n-\lambda,\ \lambda_{n+1}-\lambda,\ \ldots,
\lambda_{n+m}-\lambda,\ldots$$
as a whole should look like 
$$\cdots,\ \lambda_n,\ \lambda_{n+1},\ \ldots,\ \lambda_{n+m},\ldots\ .$$
Thus we should look at the levels $\{\lambda_i\}$ not one by one, but as a 
(random) configuration on $\mathbf{R}^1$. In particular, the exact numbering 
of levels is not essential, and we are thus led to formulate the basic 
assumption of energy level statistics, which is not clearly stated in 
physics literature:
\begin{quote}
After suitably unfolded, the discrete spectrum of a quantum Hamiltonian 
looks like a typical realization of a stationary point process on 
$\mathbf{R}^1$.
\end{quote}

The purpose of the present article is to present a phenomenological theory 
of basic notions in statistics of energy spectrum, viewed as a realization 
of a fictitious stationary point process on $\mathbf{R}$. Much of 
materials are taken from author's notes (\cite{Mi4}, \cite{Mi5}, \cite{Mi6}) 
published in preliminary forms, with additional new comments and improvements.

The outline of the article is as follows: In \S\ref{sec;2}, 
we summarize briefly 
some basic facts about stationary point processes on the real line, 
especially the Palm-Khinchin formula. Relations among various statistics 
such as $E(k,L)$, $F(k,L)$ or $P(S)$ (see e.g. \cite{Au}, \cite{Me}) 
are derived as 
direct corollaries of Palm-Khinchin formula. In \S\ref{sec;3}, 
we assume that our 
point process is ergodic. Then as a result of ergodic theorem, various 
statistics taken along the energy axis coincide with their counterparts 
taken on the ensemble (namely the probability space at the basis). It will 
be stressed that even if the unfolded spectrum $\{\lambda_i\}$ is realized 
as a sample of ergodic point process, the probability distribution of spacings 
$s_i:=\lambda_{i+1}-\lambda_i$ can differ from $i$ to $i$, and that 
{\em the} spacing should be observed not under the probability 
measure $\mathbf{P}$ given at the beginning, but under the Palm distribution 
$\mathbf{\hat{P}}$ which is naturally associated to 
$\mathbf{P}$ when $\{\lambda_i\}$ has 
unit mean density. In the next section \S\ref{sec;4}, we consider 
a deterministic 
family of sequences of real numbers, which we consider samples of a fictitious 
stationary point process. This time, there is no ensemble average, but one can 
consider various statistics taken along $\mathbf{R}$ and deduce relations 
among them. In particular, a deterministic analogue of Palm-Khinchin formula 
is obtained. The level statistics discussed in this section is what is usually 
done in quantum chaology. In \S\ref{sec;5}, 
we shall discuss two notions of unfolding. 
While the first one is rather well defined and applied in physics literature 
(see e.g. \cite{MT1}), the second idea, which is essentially due to Berry and 
Tabor (\cite{BT}), is not commonly used. Unfortunately, the present author 
is unable to clarify the general relationship of these two notions. 
For some examples, these two notions of unfolding give the same results, but 
there are also some cases where they lead to different mathematical problems. 
In the final section \S\ref{sec;6}, we give sufficient conditions for 
repulsion and clustering of levels in a wide sense.

\section{Stationary point processes on the real line; 
statistics over the ensemble}
\label{sec;2}
\subsection{Generalities and Palm-Khinchin formula}
\label{sec;2-1}
Let $M_p(\mathbf{R})$ be the totality of integer valued Radon measures 
on $\mathbf{R}$, and let ${\cal M}_p(\mathbf{R})$ be the $\sigma$-field 
on $M_p(\mathbf{R})$generated by mappings 
$$M_p(\mathbf{R})\ni N\mapsto N(I)\in\mathbf{Z}_+$$
for all bounded intervals $I\subset\mathbf{R}$. ${\cal M}_p(\mathbf{R})$ 
is equal to the Borel $\sigma$-field generated by the topology of vague 
convergence of Radon measures in $M_p(\mathbf{R})$ (see \cite{D}, 
\cite{N}). This topology is known to be metrizable.

\begin{definition}
A random variable $N_{\omega}$ defined on a probability space 
$(\Omega,{\cal F},\mathbf{P})$ and taking values in 
$(M_p(\mathbf{R}),{\cal M}_p(\mathbf{R}))$ is called a point process on 
$\mathbf{R}$. It is said to be {\em stationary}, if 
its probability law induced 
on the space $(M_p(\mathbf{R}),{\cal M}_p(\mathbf{R}))$ is invariant under 
the translation $N(\cdot)\mapsto N(\cdot+x)$ for all $x\in\mathbf{R}$.
\label{def;1}
\end{definition}

To each sample $N_{\omega}$ of a point process, we can associate sequences 
of random variables $\{x_n(\omega)\}_{n=-\infty}^{\infty}$ and 
$\{m_n(\omega)\}_{n=-\infty}^{\infty}$ so that 
$$N_{\omega}(dx)=\sum_{n\in\mathbf{Z}}m_n(\omega)\delta_{x_n(\omega)}(dx)\ ,$$
$\delta_{a}(dx)$ being the unit mass concentrated at $a$. Here the random 
variables $m_n(\omega)\geq1$ are integer valued, and $x_n(\omega)$'s satisfy 
\begin{equation}
\cdots<x_{-1}(\omega)<x_0(\omega)\leq0<x_1(\omega)<x_2(\omega)<\cdots\ .
\label{eqn;a1}
\end{equation}
$m_n(\omega)$ is called the multiplicity of the point $x_n(\omega)$. The 
point process $N_{\omega}$ is said to be {\em simple} if 
$m_n(\omega)=1$ almost surely for 
all $n\in\mathbf{Z}$. In the following, we shall assume that 
$m:=\mathbf{E}[N(0,1]]<\infty$, and call it the mean density of 
the point process $N_{\omega}$. (For notational simplicity, we have 
written $N(0,1]$ instead of $N((0,1])$ or $N_{\omega}((0,1])$. ) 

Now let us summarize some of the fundamental properties of stationary point 
processes which are derived directly from the definition. For detailed proofs, 
the reader is refered to the book by Daley and Vere-Jones \cite{D}. 

\begin{prop}
For any $t\in\mathbf{R}$ and any $x\geq0$, one has 
$\mathbf{E}[N(t,t+x]]=mx$.
\end{prop}

Indeed, by the stationarity we have $\mathbf{E}[N(t,t+x]]=\mathbf{E}[N(0,x]]$, 
which we shall denote by $M(x)$. Then for any $x,y\geq0$, 
$$M(x+y)=\mathbf{E}[N(0,x+y]]=\mathbf{E}[N(0,x]]+\mathbf{E}[N(x,x+y]]
=M(x)+M(y)\ .$$
Under the assumption $m=M(1)<\infty$, $M(x)$ is right continuous in $x$, 
hence we must have $M(x)=mx$ by a well known argument.

\begin{prop}
The limit 
$$\lambda:=\lim_{h\searrow0}\frac1h\mathbf{P}(N(0,h]>0)$$
exists and satisfies $\lambda\leq m$. $\lambda$ is called the {\em intensity} 
of the stationary point process $N$. 
\label{prop;2}
\end{prop}

For the proofs of limit theorems of this kind, the following lemma is 
repeatedly used:

\begin{lemma}
For any function $g(x)$ defined on $[0,\infty)$ which satisfies the 
conditions
\begin{enumerate}
	\item $\lim_{x\searrow0}g(x)=g(0)=0$ ;
	\item {\em (sub-additivity)} $g(x+y)\leq g(x)+g(y)$ $(x,y\geq0)$,
\end{enumerate}
the limit
$$\alpha:=\lim_{x\searrow0}\frac1x g(x)\leq\infty$$
exists and satisfies
$$\alpha=\sup_{x>0}\frac1x g(x)\ .$$
\label{lem;1}
\end{lemma}

For example, Proposition \ref{prop;2} is easily proved if you consider the 
function 
$$\phi(x):=\mathbf{P}(N(0,x]>0)\ ,$$ 
which obviously satisfies the 
conditions of the lemma and $\phi(x)\leq mx$. Hence the desired limit 
$\lambda$ exists and we have $\lambda\leq m$.

\begin{prop}
If the stationary point process $N_{\omega}$ is simple, then $\lambda=m$.
\label{prop;3}
\end{prop}

From a given stationary point process
$$N_{\omega}(dx)=\sum_{n\in\mathbf{Z}}m_n(\omega)\delta_{x_n(\omega)}(dx)\ ,$$
we obtain a new point process
$$N_{\omega}^{\ast}(dx):=\sum_{n\in\mathbf{Z}}\delta_{x_n(\omega)}(dx)\ ,$$
which is stationary and simple. Obviously $N$ and $N^{\ast}$ have the same 
intensity $\lambda$, and by Proposition \ref{prop;3}, $\lambda$ is equal 
to the mean density of $N^{\ast}$.

\begin{prop}
Under the assumption $m<\infty$, a stationary point process $N$ is simple 
if and only if it satisfies the condition 
$$\mathbf{P}(N(0,h]\geq2)=o(h)\quad(h\searrow0)\ .$$
\label{prop;4}
\end{prop}

As an important example of a point process, let us introduce the stationary 
Poisson point process:

\begin{definition}
A point process $N$ is called the stationary Poisson point process with 
parameter $c>0$, if it satisfies the following two conditions:
\begin{enumerate}
	\item For any finite family $A_1,\ldots,A_n$ of disjoint Borel subsets 
	of $\mathbf{R}$, $N(A_1),\ldots,N(A_n)$ are independent random variables;
	\item If $A$ is a bounded Borel subset of $\mathbf{R}$ with Lebesgue 
	measure $\vert A\vert$, then $N(A)$ obeys the Poisson distribution with 
	mean $c\vert A\vert$, namely
	$$\mathbf{P}(N(A)=k)=e^{-c\vert A\vert}\frac{(c\vert A\vert)^k}{k!}\ ,
	\quad k=0,1,2,\ldots\ .$$
\end{enumerate}
\label{def;2}
\end{definition}

It is easy to see that a 
stationary Poisson point process $N$ is \lq\lq stationary\rq\rq  in the 
sense of Definition \ref{def;1}. Moreover since
$$\mathbf{P}(N(0,h]\geq2)=1-e^{-ch}-ch\cdot e^{-ch}={\cal O}(h^2)$$
as $h\searrow0$, we see that $N$ is simple by Proposition \ref{prop;4}, 
which has been not quite obvious from Definition \ref{def;2}. Of course, we 
have $\lambda=m=c$. 

\begin{prop}
For $k=1,2,\ldots$, the limits
$$\pi_k:=\lim_{h\searrow0}\mathbf{P}(N(0,h]=k\vert\ N(0,h]>0)$$
exist, and satisfy $\sum_{k\geq1}\pi_k=1$.
\end{prop}

Intuitively, $\pi_k$ is the conditional probability that a point has 
multiplicity $k$, given that the origin is occupied by that point. When 
the stationary point process $N$ is simple, Proposition \ref{prop;2} and 
\ref{prop;4} show that $\pi_k=0$ for $k\geq2$.

\begin{prop}
For each $x>0$ and $k=0,1,2,\ldots$, the limit
$$Q_k(x):=\lim_{h\searrow0}\mathbf{P}(N(0,x]\leq k\vert\ 
N(-h,0]>0)$$
exists, and is right-continuous and non-decreasing as a function of $x$.
\label{prop;6}
\end{prop}

Now we can state the celebrated Palm-Khinchin formula:

\begin{thm}[the Palm-Khinchin formula]
If $N$ is a simple stationary point process (so that $\lambda=m$) such that 
\begin{equation}
\mathbf{P}(N(-\infty,0]=N(0,\infty)=\infty)=1\ ,
\label{eqn;a2}
\end{equation}
then one has
\begin{equation}
\mathbf{P}(N(0,x]\leq k)=\lambda\int_x^{\infty}q_k(u)du\ ,\ k=0,1,2,\ldots,
\label{eqn;a3}
\end{equation}
where we have set
$$q_k(x):=Q_k(x)-Q_{k-1}(x)=\lim_{h\searrow0}\mathbf{P}(N(0,x]=k\vert\ 
N(-h,0]>0)\ ,$$
with the convention $Q_{-1}(x)\equiv0$. Moreover for $k\geq1$,
\begin{equation}
R_k(x):=1-\sum_{j=0}^{k-1}q_j(x)=1-Q_{k-1}(x)
\label{eqn;a4}
\end{equation}
is a probability distribution function on $(0,\infty)$ with mean
\begin{equation}
\int_0^{\infty}xd_xR_k(x)=\frac{k}{\lambda}\ .
\label{eqn;a5}
\end{equation}
\label{thm;1}
\end{thm}

Note that if we let $x=0$ in (\ref{eqn;a3}), we get
\begin{equation}
\lambda\int_0^{\infty}q_k(u)du=1\ ,
\label{eqn;a6}
\end{equation}
and hence
\begin{equation}
\mathbf{P}(N(0,x]>k)=\lambda\int_0^{x}q_k(u)du\ .
\label{eqn;a7}
\end{equation}

We omit the proof of this theorem as well as of Propositions \ref{prop;3}
-- \ref{prop;6}, leaving them to \cite{D}. We only note that the fact 
$R_k(\infty)=1$ is a consequence of the condition (\ref{eqn;a2}). In fact, 
we can write intuitively
$$R_k(x)=1-\mathbf{P}(N(0,x]\leq k-1\vert\ N(\{0\})>0)
=\mathbf{P}(N(0,x]\geq k\vert\ N(\{0\})>0)\ .$$
If we let $x\to\infty$, we should have
$$R_k(\infty)=\mathbf{P}(N(0,\infty)\geq k\vert\ N(\{0\})>0)\ .$$
According to (\ref{eqn;a2}), this probability should be equal to $1$ for all 
$k$.

\subsection{Factorial moment measures of a point process}
\label{sec;2-2}
Given a (not necessarily stationary) point process $N_{\omega}(dx)$ on 
$\mathbf{R}$, we define its $k$-th factorial 
$N_{\omega}^{[k]}(dx_1,\ldots,dx_k)$ by
\begin{equation}
N_{\omega}^{[k]}(B):=(N_{\omega}\times\cdots\times N_{\omega})
(B\cap\mathbf{R}_k^{\ast})\ ,
\label{eqn;a8}
\end{equation}
where $B$ runs through ${\cal B}(\mathbf{R}^k)$, the totality of $k$-
dimensional Borel sets, and we have set 
$$\mathbf{R}_k^{\ast}:=\{(x_1,\ldots,x_k)\in\mathbf{R}^k\vert\ 
x_i\ne x_j\ (i\ne j)\}\ .$$
When the point process $N_{\omega}$ is simple and is written in the form 
\begin{equation}
N_{\omega}(dx)=\sum_{n\in\mathbf{Z}}\delta_{x_n(\omega)}(dx)\ ,
\label{eqn;a9}
\end{equation}
with $\{x_n(\omega)\}$ satisfying (\ref{eqn;a1}), then we have, 
for any intervals $I_1,\ldots,I_k$,
\begin{eqnarray}
& &N_{\omega}^{[k]}(I_1\times\cdots\times I_k) \nonumber\\
& &=\sharp\{(x_{i_1}(\omega),\ldots,x_{i_k}(\omega))\vert\ 
x_{i_j}(\omega)\ \mbox{are distinct and}\ x_{i_j}
(\omega)\in I_j\ (j=1,\ldots,k)\}\ .
\label{eqn;a10}
\end{eqnarray}
Also we define the $k$-th factorial moment measure 
$m^{[k]}(dx_1,\ldots,dx_k)$ as the expectation of $N^{[k]}$, namely
\begin{equation}
m^{[k]}(B):=\mathbf{E}[N^{[k]}(B)]\ ,\quad B\in{\cal B}(\mathbf{R}^k)\ .
\label{eqn;a11}
\end{equation}
Suppose $B\in{\cal B}(\mathbf{R}^k)$ is of the form 
\begin{equation}
B=B_1^{k_1}\times B_2^{k_2}\times\cdots\times B_{\ell}^{k_{\ell}}\ ,
\label{eqn;a12}
\end{equation}
where $B_1,\ldots,B_{\ell}\in{\cal B}(\mathbf{R})$ are disjoint and 
$k_1+\cdots+k_{\ell}=k$. Then it is easy to see that
\begin{equation}
N_{\omega}^{[k]}(B)=N_{\omega}(B_1)^{[k_1]}N_{\omega}(B_2)^{[k_2]}\cdots 
N_{\omega}(B_{\ell})^{[k_{\ell}]}\ ,
\label{eqn;a13}
\end{equation}
where for integers $n\geq0$ and $j\geq1$, we let
$$n^{[j]}=\left\{
\begin{array}{lr}
n(n-1)\cdots(n-j+1) & \mbox{for $n\geq1$} \\
0 & \mbox{for $n=0$} \ .
\end{array}\right.$$

\bigskip
As an important example, consider the case where $N_{\omega}$ is the Poisson 
point process with parameter $c$. When $B\in{\cal B}(\mathbf{R}^k)$ is of the 
form (\ref{eqn;a12}), then by (\ref{eqn;a13}) and by the definition of the 
Poisson point process, one has
\begin{equation}
\mathbf{E}[N^{[k]}(B_1^{k_1}\times\cdots\times B_{\ell}^{k_{\ell}})]=
\mathbf{E}[N(B_1)^{[k_1]}]\cdots\mathbf{E}[N(B_1)^{[k_{\ell}]}]
=\prod_{j=1}^{\ell}(c\vert B_j\vert)^{k_j}\ .
\label{eqn;a14}
\end{equation}
This can be rephrased as follows: when the one-dimensional Borel sets 
$B_1,B_2,\ldots,B_k$ are such that $B_i$ and $B_j$ ($i\ne j$) 
are either disjoint or identical, then 
\begin{equation}
\mathbf{E}[N^{[k]}(B_1\times\cdots\times B_k)]
=c^k\prod_{j=1}^k\vert B_j\vert\ .
\label{eqn;a15}
\end{equation}

This can be used to prove the following

\begin{prop}
If $N$ is the stationary Poisson point process with parameter $c$, then 
for any sequence $A_1,A_2,\ldots,A_k\in{\cal B}(\mathbf{R})$ 
of $k$ Borel sets, one has 
$$m^{[k]}(A_1\times\cdots\times A_k)=
\mathbf{E}[N^{[k]}(A_1\times\cdots\times A_k)]=
c^k\prod_{j=1}^k\vert A_j\vert\ .$$
\label{prop;7}
\end{prop}

\noindent{\it Proof.} 
Let $\{B_1,\ldots, B_m\}$ be the disjoint partition of the set 
$\cup_{i=1}^nA_i$ 
generated by $\{A_1,\ldots,A_n\}$, so that we can write
$$A_i=\bigcup_{p\in J_i}B_p$$
with a $J_i\subset\{1,2,\ldots,m\}$. Then we obtain
\begin{eqnarray*}
\mathbf{E}[N^{[k]}(A_1\times\cdots\times A_k)]&=&
\mathbf{E}\left[N^{[k]}\left(\bigcup
_{(p_1,\ldots,p_k)\in J_1\times\cdots\times J_k}
(B_{p_1}\times\cdots\times B_{p_k})\right)\right] \\
&=&\sum_{(p_1,\ldots,p_k)\in J_1\times\cdots\times J_k}
\mathbf{E}[N^{[k]}(B_{p_1}\times\cdots\times B_{p_k})] \\
&=&\sum_{(p_1,\ldots,p_k)\in J_1\times\cdots\times J_k}c^k\vert B_{p_1}\vert
\cdot\vert B_{p_2}\vert\cdots\vert B_{p_k}\vert 
=c^k\prod_{j=1}^k\vert A_j\vert
\end{eqnarray*}
as desired.


\subsection{Applications to level statistics}
\label{sec;2-3}
As we already noted, the unfolded spectrum of a quantum Hamiltonian is often 
implicitly assumed to be a realization $\{x_i\}$ of some stationary point 
process. In the context of quantum chaos, this identification is often 
something which might be called an intentional confusion. For example in 
\cite{Au}, 
the authors define $E(k,S)$, where $k=0,1,2,\ldots$ and $S>0$, to be 
\lq\lq the probability to find $k$ (unfolded) levels in a randomly chosen 
interval of length $L$\rq\rq. They even write it as $E(k,S,\hat{x})$ in 
order to indicate that this interval is of the form $(x,x+S]$ with $x$ 
randomly chosen from $(\hat{x},c\hat{x}]$ ($c>1$) according to the uniform 
distribution. This is actually a right way to consider the statistics 
of levels, because the quantal spectrum under consideration does not have the 
true stochasticity. (We shall return to the discussion of this situation 
in \S\ref{sec;4}.) But immediately, they omit $\hat{x}$ \lq\lq to keep 
the notation 
simple\rq\rq, and continue their discussion as if the spectrum is a truely 
random sequence of levels. In this subsection, we shall assume 
{\em explicitly} that the unfolded quantal spectrum under consideration 
is actually represented by a simple stationary point process $N_{\omega}$ 
defined on a probability space $(\Omega,{\cal F},\mathbf{P})$. Thus we 
consider $E(k,S)$ to be the probability to find $k$ levels in a {\em fixed} 
interval $(x,x+S]$ of length $S$. By the stationarity, this probability does 
not depend on the choice of $x$, so we define
\begin{equation}
E(k,S):=\mathbf{P}(N(0,S]=k)\ .
\label{eqn;a16}
\end{equation}
Following \cite{Au} (see also \cite{H}), we also consider $F(k,S)$, \lq\lq 
the probability to count $k$ levels in an interval of length $S$ which 
starts at an arbitrary quantal level $x_i$\rq\rq, and $P(k,S)$, \lq\lq the 
probability to obtain exactly $k$ levels in an interval 
of length $S$ beginning at an 
arbitrary quantal level $x_i$ and ending at the quantal level 
$x_{i+k+1}$\rq\rq. $F(k,S)$ should be interpreted as the conditional 
probability to count $k$ levels in the interval $(0,S]$, given that the 
origin $0$ is occupied by a level. Since the event $\{N(\{0\})=1\}$ has 
probability zero, we are led to define this conditional probability as
\begin{equation}
F(k,S):=\lim_{h\searrow0}\mathbf{P}(N(0,S]=k\vert\ N(-h,0]>0)\ ,
\label{eqn;a17}
\end{equation}
or $F(k,S):=q_k(S)$ in the notation of Theorem 1. To give a meaning 
to $P(k,S)$, we interpret $P(k,S)dS$ as the conditional probability, 
given that the origin is occupied by a level, say $x_{i}$, to find 
$x_{i+k+1}$ between $S$ and $S+dS$. This leads us to consider
\begin{eqnarray*}
\mathbf{P}(k,S)dS&=&
\lim_{h\searrow0}\mathbf{P}(N(0,S]\leq k\vert\ N(-h,0]>0) \\
& &-\lim_{h\searrow0}\mathbf{P}(N(0,S+dS]\leq k\vert\ N(-h,0]>0)\ .
\end{eqnarray*}
In view of this, we define
\begin{equation}
P(k,S)dS:=-dQ_k(S)\ .
\label{eqn;a18}
\end{equation}
When $k=0$, we shall write $E(S)$, $F(S)$, and $P(S)$ respectively for 
$E(0,S)$, $F(0,S)$, and $P(0,S)$. $E(S)$ is often called the \lq\lq gap 
probability\rq\rq. $P(S)$, when it exists, is the probability density 
of the spacing to the level which lies next to the one occupying the 
origin, because $F(S)$ is the probability that this spacing is greater 
than $S$. 

We now apply the Palm-Khinchin formula, to obtain
\begin{eqnarray*}
E(k,S)&=&\mathbf{P}(N(0,S]\leq k)-\mathbf{P}(N(0,S]\leq k-1) \\
&=&\lambda\int_S^{\infty}\{q_k(x)-q_{k-1}(x)\}dx \\
&=&\lambda\int_S^{\infty}\{F(k,x)-F(k-1,x)\}dx\ ,
\end{eqnarray*}
for $k\geq1$, and
$$E(S)=E(0,S)=\lambda\int_S^{\infty}F(0,x)dx$$
for $k=0$, where $\lambda$ is the intensity of the point process $N$. 
Summing over $k$, we get 
$$\lambda\int_S^{\infty}F(k,x)dx=\sum_{j=0}^kE(j,S)\ ,$$
or
\begin{equation}
F(k,S)=-\frac1{\lambda}\frac{d^+}{dS}\sum_{j=0}^kE(j,S)\ ,
\label{eqn;a19}
\end{equation}
which is formula (13) of \cite{Au}. Recall that $F(k,S)$ is right-continuous 
in $S$, so that differentiation from the left is possible. If both sides 
of (\ref{eqn;a19}) are again differentiable in $S$, we can write
\begin{eqnarray}
P(k,S)&=&-\frac{d}{d S}\sum_{i=0}^k F(i,S)
=\frac1{\lambda}\frac{d^2}{d S^2}\sum_{i=0}^k\sum_{j=0}^i
E(j,S) \nonumber\\
&=&\frac1{\lambda}\frac{d^2}{d S^2}
\sum_{i=0}^k(k-i+1)E(i,S)\ .
\label{eqn;a20}
\end{eqnarray}
In particular, letting $k=0$,
\begin{equation}
P(S)=\frac1{\lambda}\frac{d^2}{d S^2}E(S)\ .
\label{eqn;a21}
\end{equation}

The following relations are also valid:
\begin{equation}
\sum_{k=0}^{\infty}E(k,S)=1\ ;
\label{eqn;a22}
\end{equation}
\begin{equation}
\sum_{k=0}^{\infty}F(k,S)=1\ ;
\label{eqn;a23}
\end{equation}
\begin{equation}
\int_0^{\infty}P(k,S)dS=1\ ;
\label{eqn;a24}
\end{equation}
\begin{equation}
\int_0^{\infty}F(k,S)dS=\frac1{\lambda}\ ;
\label{eqn;a25}
\end{equation}
\begin{equation}
\int_0^{\infty}SP(k,S)dS=\frac{k+1}{\lambda}\ .
\label{eqn;a26}
\end{equation}
(\ref{eqn;a22}) and (\ref{eqn;a24}) are obvious from the definitions 
of $E(k,S)$ 
and $P(k,S)dS$. (\ref{eqn;a25}) and (\ref{eqn;a26}) follow from 
(\ref{eqn;a6}) and (\ref{eqn;a5}) respectively. 
(\ref{eqn;a23}) can be verified in the following way. Since 
$F(k,S)=Q_k(S)-Q_{k-1}(S)$ and $Q_{-1}(S)=0$, we have 
$$\sum_{k=0}^{\infty}F(k,S)=\lim_{k\to\infty}Q_k(S)=\sup_{k\geq0}Q_k(S)\ .$$
On the other hand, by the definition of $Q_k(S)$ (Propsition \ref{prop;6}) 
and $\mathbf{P}(N(-h,0]>0)\sim\lambda h$, we can write
$$Q_k(S)=\lim_{h\searrow0}\frac1{\lambda h}\mathbf{P}(N(0,S]\leq k,\ 
N(-h,0]>0)\ .$$
Now it is not difficult to show that $\mathbf{P}(N(0,S]\leq k,\ 
N(-h,0]>0)$, as a function of $h>0$, satisfies the condition of Lemma 
\ref{lem;1}. (The proof of Propsition \ref{prop;6} depends on this fact.) 
Hence
$$Q_k(S)=\sup_{h>0}\frac1{\lambda h}\mathbf{P}(N(0,S]\leq k,\ N(-h,0]>0)\ .$$
Interchanging $\sup_{k}$ and $\sup_{h>0}$, we see
\begin{eqnarray*}
\sum_{k=0}^{\infty}F(k,S)&=&\sup_{h>0}\sup_{k}\frac1{\lambda h}
\mathbf{P}(N(0,S]\leq k,\ N(-h,0]>0) \\
&=&\sup_{h>0}\frac1{\lambda h}\mathbf{P}(N(-h,0]>0)=1\ ,
\end{eqnarray*}
by Proposition \ref{prop;2}.

Let us now look for a representation of the {\em number variance} 
$\Sigma^2(S)$ in terms of $F(k,S)$ or $P(k,S)$. $\Sigma^2(S)$ is defined 
as the variance of the random variable $N(0,S]$, i.e. 
\begin{equation}
\Sigma^2(S):=\mathbf{E}\left[N(0,S]^2\right]-(\lambda S)^2\ .
\label{eqn;a27}
\end{equation}
We can compute, using Theorem 1 and $\lambda S=\mathbf{E}[N(0,S]]$,
\begin{eqnarray*}
\mathbf{E}\left[N(0,S]^2\right]-\lambda S&=&
\sum_{k=2}^{\infty}k(k-1)\mathbf{P}(N(0,S]=k) \\
&=&\sum_{k=2}^{\infty}k(k-1)\left\{\mathbf{P}(N(0,S]>k-1)
-\mathbf{P}(N(0,S]>k)\right\} \\
&=&2\sum_{k=1}^{\infty}k\mathbf{P}(N(0,S]>k) \\
&=&2\lambda\int_0^S\sum_{k=1}^{\infty}kq_k(u)du\ .
\end{eqnarray*}
Thus we obtain
$$\Sigma^2(S)=\lambda S-(\lambda S)^2+2\lambda\int_0^S
\left(\sum_{k=1}^{\infty}kF(k,u)\right)du\ ,$$
or
$$\Sigma^2(S)=\lambda S+2\lambda\int_0^S
\left(\sum_{k=1}^{\infty}kF(k,u)-\lambda u\right)du\ .$$
Formally, this can be rewritten as
\begin{equation}
\Sigma^2(S)=\lambda S+2\lambda\int_0^S
(S-u)\left(\sum_{k=0}^{\infty}P(k,u)-\lambda\right)du\ .
\label{eqn;a28}
\end{equation}
Here
\begin{equation}
\rho_2(u):=\sum_{k=0}^{\infty}P(k,u)
\label{eqn;a29}
\end{equation}
and
\begin{equation}
Y_2(S):=\rho_2(u)-\lambda\ ,
\label{eqn;a30}
\end{equation}
which are in general singular, 
are often called {\em two level correlation} and 
{\em two level cluster function} respectively.

\begin{ex}
Let $N$ be the stationary Poisson point process with parameter $c>0$. Then 
by the independence (see Definition \ref{def;2}), 
\begin{equation}
F(k,S)=q_k(S)=\mathbf{P}(N(0,S]=k)=E(k,S)=e^{-cS}\frac{(cS)^k}{k!}\ ;
\label{eqn;a31}
\end{equation}
\begin{equation}
P(k,S)=-\frac{d}{dS}\sum_{i=0}^kF(i,S)=ce^{-cS}\frac{(cS)^k}{k!}\ ,
\label{eqn;a32}
\end{equation}
and hence
\begin{equation}
\rho_2(u)=c\ ;\quad Y_2(u)=0\ ;\quad \Sigma_2(S)=cS\ .
\label{eqn;a33}
\end{equation}
\label{ex;1}
\end{ex}

\begin{ex}
For each $\omega$ in the probability space 
$(\Omega,{\cal B}[0,1),\mathbf{P})$, 
where $\Omega=[0,1)$, ${\cal B}[0,1)$ is the 
Borel $\sigma$-field on $[0,1)$, and $\mathbf{P}$ is 
the Lebesgue measure, define
\begin{equation}
N_{\omega}(dx):=\sum_{n=-\infty}^{\infty}\delta_{(n+\omega)/\lambda}(dx)\ ,
\label{eqn;a34}
\end{equation}
where $\lambda>0$. The resulting point process is obviously simple and 
stationary. It is just the random translation of the lattice 
$\lambda^{-1}\mathbf{Z}$. 

For any bounded measurable function $f(x)$ with compact support, one has
$$\mathbf{E}\left[\int_{-\infty}^{\infty}f(x)N_{\omega}(dx)\right]
=\int_0^1\sum_{n=-\infty}^{\infty}f(\lambda^{-1}(n+\omega))d\omega
=\lambda\int_{-\infty}^{\infty}f(x)dx\ .$$
In particular letting $f(x)=1_{(0,1]}(x)$,
$$\mathbf{E}[N(0,1]]=\lambda\ .$$

Now fix $k\geq0$ and $x>0$. Then for sufficiently small $h>0$, 
$N_{\omega}(-h,0]>0$ implies $N_{\omega}(0,x]\leq k$ when $x<(k+1)/\lambda$, 
and $N_{\omega}(0,x]>k$ when $x\geq(k+1)/\lambda$. On the other hand,
$\mathbf{P}(N(-h,0]>0)=\lambda h$ for small $h>0$. Hence 
\begin{equation}
Q_k(x)=\lim_{h\searrow0}\frac1{\lambda h}\mathbf{P}(N(0,x]\leq k,
N(-h,0]>0)=1_{[0,(k+1)/\lambda)}(x)\ .
\label{eqn;a35}
\end{equation}
Thus we get
\begin{equation}
F(k,S)=q_k(S)=1_{[k/\lambda,(k+1)/\lambda)}(S)\ ;\quad 
P(k,S)=\delta\left(S-\frac{k+1}{\lambda}\right)\ .
\label{eqn;a36}
\end{equation}
By direct computation, one can verify
\begin{equation}
\Sigma^2(S)=\{\lambda S\}(1-\{\lambda S\})\ ,
\label{eqn;a37}
\end{equation}
where $\{\lambda S\}$ is the fractional part of $\lambda S$.
\label{ex;2}
\end{ex}

\subsection{Superposition of independent point processes}
\label{sec;2-4}
In several situations, we can consider the spectrum of a quantum system 
as a superposition of spectra of independent or very weakly interacting 
subsystems (see e.g. \cite{BR}, \cite{MT1}, \cite{MT2}, 
and also \cite{Mo}, \cite{Mi2}). In our language, 
this amounts to consider a point process of the form
\begin{equation}
N(dx)=N_1(dx)+N_2(dx)+\cdots+N_p(dx)\ ,
\label{eqn;a38}
\end{equation}
where $N_1(dx),\ N_2(dx),\ \ldots,N_p(dx)$ are stationary point processes 
which are statistically independent from each other. Let $m_i$, $m$, 
$\lambda_i$, $\lambda$ be respectively the mean density and the intensity 
$N_i$ and $N$ respectively. Also let $\Sigma^2_i(S)$, $\Sigma^2(S)$, 
$Y_2^{(i)}(u)$, $Y_2(u)$ be the corresponding number variances and two 
level cluster functions, and let $E_i(S)$, $E(S)$, $F_i(S)$, $F(S)$ 
be defined similarly. 

From (\ref{eqn;a38}), we obviously have
\begin{equation}
m=m_1+m_2+\cdots+m_p
\label{eqn;a39}
\end{equation}
and
\begin{equation}
\Sigma^2(S)=\Sigma^2_i(S)+\Sigma^2_2(S)+\cdots+\Sigma^2_p(S)
\label{eqn;a40}
\end{equation}
from the independence. 
For small $h>0$, we have from (\ref{eqn;a38}), the independence, and Propsition 
\ref{prop;2},
\begin{eqnarray*}
\mathbf{P}(N(0,h]>0)&=&1-\mathbf{P}(N(0,h]=0)
=1-\prod_{i=1}^p\mathbf{P}(N_i(0,h]=0) \\
&=&1-\prod_{i=1}^p(1-\lambda_ih+o(h))
=\left(\sum_{i=1}^p\lambda_i\right)h+o(h)\ ,
\end{eqnarray*}
whence
\begin{equation}
\lambda=\sum_{i=1}^p\lambda_i\ .
\label{eqn;a41}
\end{equation}
Now suppose $N_1,\ \ldots,N_p$ are all simple, so that 
$\mathbf{P}(N_i(0,h]\geq2)=o(h)$ as $h\searrow0$ for $i=1,\ldots,p$. 
Since
$$\{N(0,h]\geq2\}\subset\bigcup_{1\leq i\leq p}\{N_i(0,h]\geq2\}\cup
\bigcup_{1\leq i<j\leq p}\{N_i(0,h]\geq1,\ N_j(0,h]\geq1\}\ ,$$
we see that 
\begin{eqnarray*}
\mathbf{P}(N(0,h]\geq2)&\leq&
\sum_{i=1}^p\mathbf{P}(N_i(0,h]\geq2)+
\sum_{i<j}\mathbf{P}(N_i(0,h]\geq1)\mathbf{P}(N_j(0,h]\geq1) \\
&=&o(h)+{\cal O}(h^2)=o(h)\ ,
\end{eqnarray*}
so that by Proposition \ref{prop;4}, the superposition 
$N$ is also simple. In this case, 
if we 
compare (\ref{eqn;a40}) with (\ref{eqn;a28}) and (\ref{eqn;a29}), we obtain 
(as in \cite{MT2})
\begin{equation}
\lambda Y_2(S)=\sum_{i=1}^p\lambda_i Y_2^{(i)}(u)\ .
\label{eqn;a42}
\end{equation}
If we apply (\ref{eqn;a19}) to 
\begin{equation}
E(S)=\mathbf{P}(N(0,S]=0)=\prod_{i=1}^p\mathbf{P}(N_i(0,S]=0)
=\prod_{i=1}^pE_i(S)\ ,
\label{eqn;a43}
\end{equation}
we get
\begin{equation}
F(S)=\sum_{i=1}^p\frac{\lambda_i}{\lambda}F_i(S)\prod_{j\ne i}E_i(S)\ .
\label{eqn;a44}
\end{equation}
(Compare \cite{BR}, \cite{MT1}.)

\begin{rem}
It is natural to consider the case where the system is decomposed into 
infinitely many non-interacting small subsystems, so that the 
spectrum of the system is a superposition of a large number of 
subspectra. To model this situation, suppose for each $n=1,2,\ldots$, 
we are given mutually independent point processes $N^{(n)}_1(dx)$,$\ldots$,
$N^{(n)}_{p_n}(dx)$ such that $p_n\to\infty$ and that 
\begin{equation}
\lim_{n\to\infty}\max_{1\leq i\leq p_n}\mathbf{P}(N^{(n)}_{i}(I)\geq1)=0
\label{eqn;a45}
\end{equation}
holds for each bounded interval $I$. Such a triangular array of point 
processes is called {\em infinitesimal}. 
The class of all point processes which 
are obtained as a weak limit of the superposition
\begin{equation}
N^{(n)}(dx)=N^{(n)}_{1}(dx)+\cdots+N^{(n)}_{p_n}(dx)
\label{eqn;a46}
\end{equation}
from an infinitesimal triangular array coincides with the class of all 
infinitely divisible point processes (see \cite{MKM}), among which the Poisson 
point process is only a special example. That is, independent superposition 
of a large number of sparse point processes does not necessarily imply 
the limiting Poissonian nature, as is often naively supposed. In physics 
literature, the possibility of obtaining non-Poissonian limits was pointed 
out in a recent work of Makino and Tasaki (\cite{MT1}).

It is known (\cite{MKM}) that 
when each constituent $N_j^{(n)}(dx)$ is a simple 
stationary point process, the superposition (\ref{eqn;a46}) 
converges weakly to a Poisson 
point process as $n\to\infty$, if and only if one has 
\begin{equation}
\lim_{n\to\infty}\sum_{j=1}^{p_n}\lambda^{(n)}_j\{1-Q^{(n,j)}_0(u)\}=0
\label{eqn;a47}
\end{equation}
for any $u>0$, where $\lambda^{(n)}_j$ is the intensity of $N_j^{(n)}(dx)$ 
and 
$$Q^{(n,j)}_0(u):=\lim_{h\searrow0}\mathbf{P}(N_j^{(n)}(0,u]=0\vert\ 
N_j^{(n)}(-h,0]>0)\ .$$
The condition (\ref{eqn;a47}) is exactly the same as \lq\lq Case 1\rq\rq 
which is formulated in \cite{MT1}.
\end{rem}

\section{Ergodic point processes; statistics along the sequence}
\label{sec;3}
\subsection{Ergodic point processes on $\mathbf{R}$}
\label{sec;3-1}
In this section, we strengthen the assumption on the point process 
$N_{\omega}$, and suppose that our probability space 
$(\Omega,{\cal F},\mathbf{P})$ is equipped with a measure preserving 
{\em ergodic} flow 
$\{\theta_t;\ t\in\mathbf{R}\}$ such that
\begin{equation}
N_{\theta_t\omega}(A)=N_{\omega}(t+A)
\label{eqn;a48}
\end{equation}
holds for all $t\in\mathbf{R}$ and any Borel set $A\in{\cal B}(\mathbf{R})$. 
Here $t+A=\{t+x;\ x\in A\}$ is the translation of the set $A$. A point 
process satisfying the condition (\ref{eqn;a48}) is called a $\{\theta_t\}$-
ergodic point 
process on the probability space $(\Omega,{\cal F},\mathbf{P};\{\theta_t\})$. 
As before, we shall assume $m:=\mathbf{E}[N(0,1]]<\infty$.

\begin{ex}
Suppose $N$ is a stationary Poisson point process with parameter 
$c>0$, and let $P_N$ be its law 
induced on the measurable space $(M_p(\mathbf{R}),{\cal M}_p(\mathbf{R}))$. 
Let us then forget the original probability space on which $N=N_{\omega}$ 
was given, and take $(M_p(\mathbf{R}),{\cal M}_p(\mathbf{R}),P_N)$ as our 
new probability space, the mapping 
$M_p(\mathbf{R})\ni\omega\mapsto N_{\omega}$ being the identity, and the 
tranformation $\theta_t\omega$ being the spatial translation of 
$\omega\in M_p(\mathbf{R})$. Then the flow $\{\theta_t\}$ is ergodic, 
and the new point process satisfies (\ref{eqn;a48}) and the conditions of 
Definition \ref{def;1} with the same parameter $c>0$.
Thus a stationary Poisson point process can always be supposed to be 
ergodic.
\label{ex;3}
\end{ex}

\begin{ex}
Define the transformations $\{\theta_t;\ t\in\mathbf{R}\}$ on the unit circle 
$\Omega=[0,1)$ by $\theta_t\omega=\{t+\omega\}$. It is well known that 
$\{\theta_t;\ t\in\mathbf{R}\}$ is an ergodic flow on the probability space 
$(\Omega,{\cal B}[0,1),\mathbf{P})$ of Example \ref{ex;2}, 
and it is easy to see that 
the point process defined by (\ref{eqn;a34}) is a $\{\theta_t\}$-ergodic point 
process.
\label{ex;4}
\end{ex}

Let $N_{\omega}$ be an ergodic point process, which we consider as 
representing an unfolded quantal spectrum. The individual ergodic theorem 
then tells us that various statistics taken along the spectral sequence 
has non-random limits with probability one. In the following proposition, 
we formulate some of such limiting relations, for later comparison with 
level statistics in \S\ref{sec;4}:

\begin{prop}
(i) With probability one, we have
\begin{equation}
\lim_{L\to\infty}\frac1L N_{\omega}(0,L]=m\ .
\label{eqn;a49}
\end{equation}

\noindent
(ii) For any $n\geq1$, any bounded intervals $I_1,\ldots,I_n$, and any 
integers $k_1,\ldots,k_n\geq0$, define 
$$F_{\omega}:=\{t\geq0;\ N_{\omega}(t+I_j)=k_j\ ,\ j=1,\ldots,n\}\ .$$
Then we have with probability one
\begin{equation}
\lim_{L\to\infty}\frac1L\int_0^L1_{F_{\omega}}(t)dt=
\mathbf{P}(N(I_j)=k_j,\ j=1,\ldots,n)\ .
\label{eqn;a50}
\end{equation}
In particular, for any $k=0,1,\ldots$ and $c>0$, the limit
\begin{equation}
\pi_k(c):=\lim_{L\to\infty}\frac1L\int_0^L1_{\{N_{\omega}(t,t+c]=k\}}dt
\label{eqn;a51}
\end{equation}
exists almost surely, and is equal to $\mathbf{P}(N(0,c]=k)$.

\noindent
(ii) For any $k\geq1$ and any bounded intervals $I_1,\ldots,I_n$, we have 
\begin{eqnarray}
& &\lim_{L\to\infty}\frac1L\int_0^LN_{\omega}^{[k]}\left((t+I_1)\times
\cdots\times(t+I_k)\right)dt \nonumber \\
&=&\lim_{L\to\infty}\frac1L\int_0^LN_{\theta_t\omega}^{[k]}
(I_1\times\cdots\times I_k)dt 
=m^{[k]}(I_1\times\cdots\times I_k)\ .
\label{eqn;a52}
\end{eqnarray}
\end{prop}
In particular, for any $k=0,1,\ldots$ and $c>0$, the limit
\begin{equation}
\mu_k(c):=\lim_{L\to\infty}\frac1L\int_0^L\left(N_{\omega}(t,t+c]\right)^{[k]}
dt
\label{eqn;a53}
\end{equation}
exists almost surely, and is equal to 
$\mathbf{E}\left[\left(N_{\omega}(0,c]\right)^{[k]}\right]$, the $k$-th 
factorial moment of the random variable $N_{\omega}(0,c]$.

\subsection{The Palm measure}
\label{sec;3-2}
To each ergodic point process $N_{\omega}$ defined on a probability space 
$(\Omega,{\cal F},\mathbf{P};\{\theta_t\})$, there corresponds a measure 
$\mathbf{\hat{P}}$ on $(\Omega,{\cal F})$ called the Palm measure of the point 
process $N_{\omega}$. The Palm measure clarifies the basic structure of a 
given ergodic point process. Although $\mathbf{\hat{P}}$ and $\mathbf{P}$ are 
mutually singular, it will turn out that the level spacing should be observed 
under $\mathbf{\hat{P}}$, but not under $\mathbf{P}$. 

The basic materials of this section are taken from \cite{N}, to which the 
reader is referred for detailed proofs.

\begin{thm}[Existence and definition of the Palm measure]
If $N_{\omega}$ is a $\{\theta_t\}$-ergodic point process defined on the 
probability space $(\Omega,{\cal F},\mathbf{P};\{\theta_t\})$, then 
there is a unique measure $\mathbf{\hat{P}}$ on $(\Omega,{\cal F})$ 
with the following property: for any jointly measurable function 
$f(\omega,s)\geq0$ on $\Omega\times\mathbf{R}$,
\begin{equation}
\int_{\Omega}\mathbf{P}(d\omega)\int_{\mathbf{R}}N_{\omega}(ds)f(\theta_s\omega,s)=
\int_{\Omega}\mathbf{\hat{P}}(d\omega)\int_{\mathbf{R}}dsf(\omega,s)\ .
\label{eqn;a54}
\end{equation}
If we let $f(\omega,s)=g(\theta_{-s}\omega,s)$ with $g(\omega,s)\geq0$ jointly 
measurable, the defining condition (\ref{eqn;a54}) of the Palm measure 
is equivalent to
\begin{equation}
\int_{\Omega}\mathbf{P}(d\omega)\int_{\mathbf{R}}N_{\omega}(ds)g(\omega,s)=
\int_{\Omega}\mathbf{\hat{P}}(d\omega)\int_{\mathbf{R}}ds
g(\theta_{-s}\omega,s)\ .
\label{eqn;a55}
\end{equation}
\end{thm}

In particular, taking $g(\omega,s)=1_{A}(s)$, with $A\in{\cal B}(\mathbf{R})$, 
we get 
$$\mathbf{E}[N(A)]=\vert A\vert\mathbf{\hat{P}}(\Omega)\ .$$
Letting $A=(0,1]$, we obtain

\begin{prop}
$$\mathbf{\hat{P}}(\Omega)=\mathbf{E}[N(0,1]]=m\ .$$
\label{prop;9}
\end{prop}

Hence $\mathbf{\hat{P}}$ is a finite measure if $m<\infty$, 
but is a probability measure only if $m=1$.

\begin{prop}
The Palm measure $\mathbf{\hat{P}}$ is concentrated on the set 
$$\hat{\Omega}:=\{\omega\in\Omega\vert\ N_{\omega}(\{0\})>0\}\ .$$
\label{prop;10}
\end{prop}

To see this, take a strictly positive function $u$ with 
$\int_{\mathbf{R}}u(s)ds=1$. Then for $k=0,1,\ldots$,
\begin{eqnarray*}
\mathbf{\hat{P}}(N(\{0\})=k)
&=&\int_{\Omega}\mathbf{\hat{P}}(d\omega)\int_{\mathbf{R}}ds
1_{\{N_{\omega}(\{0\})=k\}}u(s) \\
&=&\int_{\Omega}\mathbf{P}(d\omega)\int_{\mathbf{R}}N_{\omega}(ds)
1_{\{N_{\theta_s\omega}(\{0\})=k\}}u(s) \\
&=&\int_{\Omega}\mathbf{P}(d\omega)\int_{\mathbf{R}}N_{\omega}(ds)
1_{\{N_{\omega}(\{s\})=k\}}u(s)\ .
\end{eqnarray*}
This shows $\mathbf{\hat{P}}(N(\{0\})=0)=0$, and when $N$ is simple, 
also $\mathbf{\hat{P}}(N(\{0\})\geq2)=0$.

\bigskip
For non-negative continuous functions $\varphi(x)$ on $\mathbf{R}$, 
we set
\begin{equation}
{\cal L}_N(\varphi):=\mathbf{E}[\exp(-N(\varphi))]\ ,
\label{eqn;a56}
\end{equation}
where
$$N_{\omega}(\varphi):=\int_{\mathbf{R}}\varphi(x)N_{\omega}(dx)\ ,$$
and call it the Laplace functional of the (not necessarily stationary) 
point process $N$. The probability law of a point process is 
uniquely determined by its Laplace functional. 

By definition, a sequence $\{N^{(n)}\}$ of point processes converges weakly to 
a point process $N$ if their law induced on the metric space $M_p(\mathbf{R})$ 
converge weakly to that of $N$. It is well known that this is equivalent 
to the convergence 
of ${\cal L}_{N^{(n)}}(\varphi)$ to ${\cal L}_{N}(\varphi)$ for all 
$\varphi$ in the class $C_0^+(\mathbf{R})$, 
the totality of non-negative continuous 
functions with a compact support. 
The following proposition then tells us that the conditional law 
$\mathbf{P}(N\in\bullet\vert\ N(-h,0]>0)$ of a simple ergodic point 
process $N$ converges weakly to the probability law 
$\mathbf{\hat{P}}(N\in\bullet)/\mathbf{\hat{P}}(\Omega)$ as $h\searrow0$. 

\begin{prop}
Let a point process $N_{\omega}$ be $\{\theta_t\}$-ergodic and 
simple, so that $\mathbf{\hat{P}}(\Omega)=m=\lambda<\infty$. 
Then for any $\varphi\in C_0^+(\mathbf{R})$, one has
\begin{equation}
\frac1{\hat{P}(\Omega)}\int_{\Omega}\exp(-N_{\omega}(\varphi))
\mathbf{\hat{P}}(d\omega)=
\lim_{h\searrow0}\int_{\Omega}\exp(-N_{\omega}(\varphi))
\mathbf{P}(d\omega\vert\ N(-h,0]>0)\ .
\label{eqn;a57}
\end{equation}
\label{prop;11}
\end{prop}

\noindent{\it Proof.} If we set $H(\omega):=\exp[-N_{\omega}(\varphi)]$, 
then $H(\theta_t\omega)$ is continuous in $t$. Hence letting 
$g(\omega,s)=H(\omega)1_{(-h,0]}(s)$ in (\ref{eqn;a55}), we get 
$$\frac1{\lambda h}\int_{\Omega}\mathbf{P}(d\omega)H(\omega)N_{\omega}(-h,0]
=\frac1{\lambda h}\int_{\Omega}\mathbf{\hat{P}}(d\omega)
\int_{-h}^0dsH(\theta_{-s}\omega)\ .$$
Since $N_{\omega}$ is simple, we have $m=\lambda$, 
$\mathbf{P}(N(-h,0]>0)\sim\lambda h=\mathbf{E}[N(-h,0]]$ as $h\searrow0$. 
Hence $N_{\omega}(-h,0]$ in the above equality can be replaced by 
$1_{\{N_{\omega}(-h,0]>0\}}$ in the limit $h\searrow0$, and the 
conclusion follows.

\begin{rem}
If we let 
$f(\omega,s)=1_{(-\infty,0]}(s)1_{\{N_{\omega}(0,x-s]=k\}}$ in 
(\ref{eqn;a54}), we obtain the equality
\begin{equation}
\mathbf{P}(N(0,x]\leq k)=\int_x^{\infty}\mathbf{\hat{P}}(N(0,s]=k)ds\ .
\label{eqn;a58}
\end{equation}
\label{rem;1}
\end{rem}
Since the integrand on the right hand side of (\ref{eqn;a58}) is easily 
seen to be right continuous, we can conclude from Palm-Khinchin formula 
(Theorem \ref{thm;1}) that
\begin{equation}
\mathbf{\hat{P}}(N(0,x]=k)=\lambda q_k(x):=
\lambda\lim_{h\searrow0}\mathbf{P}(N(0,x]=k\vert\ N(-h,0]>0)
\label{eqn;a59}
\end{equation}
holds for $k=0,1,2,\ldots$ and $x\geq0$. (Note that $q_k(x)$ is also 
right continuous by Proposition \ref{prop;6}.) 
This is not a direct consequence of the weak convergence of 
$\mathbf{P}(N\in\bullet\vert\ N(-h,0]>0)$ to 
$\mathbf{\hat{P}}(N\in\bullet)/\lambda$. In fact, weak convergence of 
a sequence of point process is in general equivalent to 
the convergence
$$\mathbf{P}(N^{(n)}(I_j)=k_j\ ,\ j=1,2,\ldots,p)
\longrightarrow \mathbf{P}(N(I_j)=k_j\ ,\ j=1,2,\ldots,p)$$
for all $p\geq1$, all bounded intervals 
$I_1,\ldots,I_p$ such that $\mathbf{P}(N(\partial I_j)>0)=0$, 
and for all integers $k_1,\ldots,k_p$. But $\mathbf{\hat{P}}$ always 
has a mass at the end point $0$ of $(0,x]$.

\begin{ex}
[Palm measure of the stationary Poisson point process]
Let $N_{\omega}$ be a stationary Poisson point 
process with parameter $c>0$ which is $\{\theta_t\}$-ergodic. 
(See Example 3.) Then 
Proposition \ref{prop;11} can be applied to obtain its Palm 
measure in the following way. If we define for $h>0$, 
$$N_{\omega}^{1,h}(\cdot):=N_{\omega}(\cdot\cap(-h,0])\ ;
\quad N_{\omega}^{2,h}(\cdot):=N_{\omega}(\cdot\setminus
(-h,0])\ ,$$
then $N_{\omega}^{1,h}$ and $N_{\omega}^{2,h}$ are independent. 
Hence for each $\varphi\in C_0^+(\mathbf{R})$,
\begin{eqnarray*}
& &\int_{\Omega}\exp[-N_{\omega}(\varphi)]
\mathbf{P}(d\omega\vert\ N(-h,0]>0) \\
&=&\int_{\Omega}\exp[-N_{\omega}^{1,h}(\varphi)]\exp[-N_{\omega}^{2,h}(\varphi)]\mathbf{P}(d\omega\vert\ N(-h,0]>0) \\
&=&\mathbf{E}[\exp[-N_{\omega}^{2,h}(\varphi)]
\mathbf{E}[1_{\{N(-h,0]>0\}}\exp[-N^{1,h}(\varphi)]]/
\mathbf{P}(N(-h,0]>0)\ .
\end{eqnarray*}
It is easy to see that this converges to 
$$\frac1c e^{-\varphi(0)}\mathbf{\hat{E}}[\exp[-N(\varphi)]]\ .$$
as $h\searrow0$. Hence, under the probability 
measure $\frac1c\mathbf{\hat{P}}$, 
$N$ has the distribution of the stationary Poisson point 
process with an extra non-random point placed at the origin.
\label{ex;5}
\end{ex}

\begin{ex}
For the point process treated in Example \ref{ex;2} and 
\ref{ex;4}, it is easy to see that $\mathbf{\hat{P}}(d\omega)
=\lambda\delta_{0}(d\omega)$. Under $\frac1{\lambda}\mathbf{\hat{P}}$, 
$N$ is the non-random lattice $\lambda\mathbf{Z}$.
\label{ex;6}
\end{ex}

In the remaining part of this section, we assume that our 
ergodic point process $N_{\omega}$ is simple and has the 
representation (\ref{eqn;a9}), with $x_n(\omega)$'s satisfying 
(\ref{eqn;a1}). Its Palm 
measure is concentrated on the set 
$$\hat{\Omega}:=\{\omega\in\Omega\vert\ x_0(\omega)=0\}\ .$$
In terms of $x_0(\omega)$ and $x_1(\omega)$, the defining 
condition (\ref{eqn;a54}) can be rephrased in the following way:

\begin{prop}
For any jointly measurable function $g(\omega,s)\geq0$ on 
$\Omega\times[0,\infty)$, 
\begin{equation}
\int_{\Omega} \mathbf{P}(d\omega)g(\theta_{x_0(\omega)}\omega,-x_0(\omega))=
\int_{\hat{\Omega}}\mathbf{\hat{P}}(d\omega)
\int_0^{x_1(\omega)}dsg(\omega,s)\ .
\label{eqn;a60}
\end{equation}
In particular, if we let $g(\omega,s)=f(\omega)$, we get
\begin{equation}
\int_{\Omega}\mathbf{P}(d\omega)f(\theta_{x_0(\omega)}\omega)=
\int_{\hat{\Omega}}\mathbf{\hat{P}}(d\omega)x_1(\omega)f(\omega)\ ,
\label{eqn;a61}
\end{equation}
and letting $g(\omega,s)=f(\theta_s\omega)$ and noting 
$g(\theta_{x_0(\omega)}\omega,-x_0(\omega))=f(\omega)$, we obtain
\begin{equation}
\int_{\Omega}\mathbf{P}(d\omega)f(\omega)=
\int_{\hat{\Omega}}\mathbf{\hat{P}}(d\omega)
\int_0^{x_1(\omega)}dsf(\theta_s\omega)\ .
\label{eqn;a62}
\end{equation}
\label{prop;12}
\end{prop}

\noindent{\it Proof.} The function 
$a_0(\omega,s)=1_{\{x_0(\omega)\leq s<x_1(\omega)\}}$ satisfies 
the relation 
$$a_0(\theta_s\omega,-s)=a_0(\omega,s)\ .$$
If we take 
$$f(\omega,s)=a_0(\omega,-s)g(\omega,-s)$$
in (\ref{eqn;a54}), we get
$$\int_{\Omega}\mathbf{P}(d\omega)\int_{\mathbf{R}}N_{\omega}
(ds)a_0(\theta_s\omega,-s)g(\theta_s\omega,-s)=
\int_{\hat{\Omega}}\mathbf{\hat{P}}(d\omega)
\int_{\mathbf{R}}dsa_0(\omega,-s)g(\omega,-s)\ .$$
But
\begin{eqnarray*}
\int_{\mathbf{R}}N_{\omega}(ds)a_0(\theta_s\omega,-s)g(\theta_s\omega,-s)
&=&
\int_{\mathbf{R}}N_{\omega}(ds)a_0(\omega,s)g(\theta_s\omega,-s) \\
&=&g(\theta_{x_0(\omega)}\omega,-x_0(\omega))\ ,
\end{eqnarray*}
and since $x_0(\omega)=0$ for $\omega\in\hat{\Omega}$ on which 
$\mathbf{\hat{P}}$ is concentrated, we have
$$\int_{\mathbf{R}}ds a_0(\omega,-s)g(\omega,-s)
=\int_{0}^{x_0(\omega)}ds g(\omega,s)\ .$$
Hence we obtain (\ref{eqn;a60}).

\bigskip
As corollaries, we obtain the following formulae 
for the probability distributions of $x_0$ and $x_1$ (\cite{N}):

\begin{prop}
If we define the measure $\hat{F}(dv)$ on $(0,\infty)$ by 
$\hat{F}(dv)=\mathbf{\hat{P}}(x_1\in dv)$, then
\begin{description}
	\item[(0)] \begin{equation}
	\int_0^{\infty}\ v\hat{F}(dv)=1\ ;
	\label{eqn;a63}
	\end{equation}
	\item[(i)] 
	\begin{equation}
	\mathbf{P}(-x_0\in ds,\ x_1-x_0\in dv)
	=1_{[0,v)}(ds)\hat{F}(dv)\ ;
	\label{eqn;a64}
	\end{equation}
	\item[(ii)] \begin{equation}
     \mathbf{P}(x_1-x_0\in dv)=v\hat{F}(dv)\ ;
     \label{eqn;a65}
     \end{equation}
	\item[(iii)] \begin{equation}
     \mathbf{P}(-x_0\in ds)=\mathbf{P}(x_1\in ds)=p(s)ds\ ,\ 
	\mbox{where}\ p(s)=\hat{F}((s,\infty))\ .
	\label{eqn;a66}
	\end{equation}
\end{description}
\label{prop;13}
\end{prop}

In particular, the distributions of $x_0$ and $x_1$ are 
absolutely continuous.

\bigskip
\noindent{\em Proof.} Applying (\ref{eqn;a60}) to the function of the 
form $g(\omega,s)=h(x_1(\omega)-x_0(\omega),s)$, with $h(v,s)$ an 
arbitrary non-negative Borel function on $\mathbf{R}_+^2$, 
we obtain, after noting 
$g(\theta_{x_0(\omega)}\omega,s)=g(\omega,s)$ and that $x_0(\omega)=0$ 
for $\omega\in\hat{\Omega}$,
\begin{equation}
\mathbf{E}[h(x_1(\omega)-x_0(\omega),-x_0(\omega))]=
\int_0^{\infty}ds\int_{\hat{\Omega}}\mathbf{\hat{P}}(d\omega)
1_{[0,x_1(\omega))}(s)h(x_1(\omega),s)\ ,
\label{eqn;a67}
\end{equation}
which proves the assertion (i). When $h\equiv1$, (\ref{eqn;a67}) 
reduces to the assertion (0), and 
when $h(v,s)=\psi(v)$, with $\psi\geq0$ Borel, 
then we have from (\ref{eqn;a67})
$$\mathbf{E}[\psi(x_1(\omega)-x_0(\omega))]
=\int_{\hat{\Omega}}\mathbf{\hat{P}}(d\omega)\ x_1(\omega)\psi(x_1(\omega))\ ,$$
proving (ii). On the other hand if $h$ is of the form $h(v,s)=\varphi(s)$, 
we get 
$$\mathbf{E}[\varphi(-x_0(\omega))]=\int_0^{\infty}ds\varphi(s)
\mathbf{\hat{P}}(x_1>s)\ ,$$
or
$$\mathbf{P}(-x_0\in ds)=\hat{F}((s,\infty))ds\ .$$
Finally, let $h(v,s)=1_{(u,\infty)}(v-s)$, where $u>0$ is arbitrarily 
fixed. Then we have
$$g(\theta_{x_0(\omega)}\omega,-x_0(\omega))=1_{(u,\infty)}(x_1(\omega))$$
for $\omega\in\Omega$, and 
$$g(\omega,s)=1_{(u+s,\infty)}(x_1(\omega))$$
for $\omega\in\hat{\Omega}$ and $s\geq0$. Thus (\ref{eqn;a67}) reduces 
to 
$$\mathbf{P}(x_1>u)=\int_u^{\infty}\mathbf{\hat{P}}(x_1>s)ds\ ,$$
which completes the proof of (iii).


\bigskip
Now we state a theorem which describes the fundamental structure of 
ergodic simple point processes. For its proof, we refer the reader to \cite{N} 
or \cite{Hn}. 

\begin{thm}
Let $N_{\omega}=\sum_{n\in\mathbf{Z}}\delta_{x_n(\omega)}$ be a simple, 
$\{\theta_t\}$-ergodic point process. Then for each $n\in\mathbf{Z}$, 
the mapping $\theta_{x_n}$, defined by $\theta_{x_n(\omega)}\omega$, maps 
$\Omega$ onto $\hat{\Omega}$. If we denote by $\hat{\theta}_n$ the 
restriction of $\theta_{x_n}$ to $\hat{\Omega}$, then 
$\{\hat{\theta}_n\vert\ n\in\mathbf{Z}\}$ is an ergodic 
group of measure preserving 
transformation on $(\hat{\Omega},\frac1{\lambda}\mathbf{\hat{P}})$.
\end{thm}

Let $\tau_n(\omega):=x_n(\omega)-x_{n-1}(\omega)$ ($n\in\mathbf{Z}$). Then 
obviously $\tau_n=\tau\circ\hat{\theta}_1^{n-1}$ holds on 
$\hat{\Omega}$, for all $n\in\mathbf{Z}$. By the theorem above, $\{\tau_n\}$, 
which is the sequence of spacings between points $\{x_n(\omega)\}$, is 
stationary and ergodic under the probability measure 
$\frac1{\lambda}\mathbf{\hat{P}}$.

\begin{ex}
When the random variables $\{\tau_n\}$ are independent, the point process 
$N_{\omega}$ is called a {\em renewal process}. The stationary Poisson point 
process (Example \ref{ex;3}) and the randomly shifted periodic lattice 
(Examples \ref{ex;2} and \ref{ex;4}) are both renewal processes. For the 
case of the Poisson point process, the distribution of $\tau_n$ under 
$\frac1{\lambda}\mathbf{\hat{P}}$ is the exponential distribution 
$\lambda e^{-\lambda t}dt$.
\label{ex;7}
\end{ex}

\begin{rem}
If $N_{\omega}$ is a renewal process, the distributions of $\tau_n$ for 
$n\ne1$ under $\mathbf{P}$ and 
$\frac1{\lambda}\mathbf{\hat{P}}$ coincide. But according 
to (ii) of Proposition \ref{prop;13}, $\tau_1=x_1-x_0$ and $\tau_n$ ($n\ne1$) 
have different distribution in general. When $\{x_n\}$ is Poissonian, 
this phenomena is called the \lq\lq waiting time paradox\rq\rq (\cite{D}). 
In fact, the only renewal process 
such that $\tau_1$ and $\tau_n$ ($n\ne1$) are identically distributed 
under $\mathbf{P}$ is the randomly shifted lattice $\lambda^{-1}\mathbf{Z}$ 
discussed in Example \ref{ex;2}, \ref{ex;4} and \ref{ex;6}. This is seen 
by solving the equation 
$$\frac1{\lambda}\hat{F}(dv)=v\hat{F}(dv)\ .$$
\label{rem;2}
\end{rem}

\subsection{The level spacing distribution}
\label{sec;3-3}
Let $N_{\omega}=\sum_{n\in\mathbf{Z}}\delta_{x_n(\omega)}$ be a simple, 
ergodic point process. In the previous section, we saw that the spacings 
$\tau_n(\omega):=x_n(\omega)-x_{n-1}(\omega)$ of points are identically 
distributed under the probability measure $\frac1{\lambda}\mathbf{\hat{P}}$. 
But under the original probability measure $\mathbf{P}$, the 
distribution of $\tau_n$ can be different for different $n$. To see this, 
first note that $\tau_n(\omega)=\tau_n(\theta_{x_0}\omega)$. By (\ref{eqn;a61}) 
of Proposition \ref{prop;12}, for any $t\geq0$,
\begin{equation}
P(\tau_n(\omega)>t)=P(\tau_n\circ\theta_{x_0}>t)=
\int_{\hat{\Omega}}\mathbf{\hat{P}}(d\omega)
x_1(\omega)1_{\{\tau_n(\omega)>t\}}\ .
\label{eqn;a68}
\end{equation}
If $N$ is not a renewal process, $x_1$ and $\tau_n$ ($n\ne1$) are correlated 
under $\mathbf{\hat{P}}$. Hence the right hand side of (\ref{eqn;a68}) can 
be different for different $n$.

This poses a question: Suppose we are given a sample 
$$\cdots<x_{-1}(\omega)<x_0(\omega)\leq0<x_1(\omega)<x_2(\omega)<\cdots$$
of an ergodic point process, which we regard as an unfolded quantal 
spectrum. If you wish to determine the level spacing distribution 
experimentally, then what you naturally do would be to collect a large 
set of data $\tau_k(\omega)=x_k(\omega)-x_{k-1}(\omega)$, $1\leq k\leq n$, 
of level spacings, and then draw a histogram of their frequency, 
expecting that 
it will approximately represent some smooth curve when the data size $n$ 
is large enough and the histogram is fine enough. Mathematically 
speaking, this amounts to considering the empirical distribution 
\begin{equation}
F_{\omega}^n(dt):=\frac1n\sum_{k=1}^n\delta_{\tau_k(\omega)}(dt)\ ,
\label{eqn;a69}
\end{equation}
and expecting this to converge, as $n\to\infty$, to some fixed distribution 
$F(dt)$ almost surely. But if the random variables $\tau_n$ are observed 
under the probability measure $\mathbf{P}$, they are not identically 
distributed. Hence mixing these to form empirical distributions seems 
meaningless. On the other hand, if the data $\{\tau_n\}$ are observed 
under $\frac1{\lambda}\mathbf{\hat{P}}$, they form a stationary ergodic 
sequence. In this case, the ergodic theorem works well, and $F_{\omega}^n(dt)$ 
converges to $\frac1{\lambda}\mathbf{\hat{P}}(\tau_1\in dt)$ as 
$n\to\infty$, $\mathbf{\hat{P}}$-almost surely. But $\mathbf{\hat{P}}$ 
and $\mathbf{P}$ are mutually singular. Which of them should be considered 
as the real law governing $\{\tau_n\}$ ? 

Fortunately, this puzzle is easily solved:

\begin{prop}
The empirical distribution $F_{\omega}^n(dt)$ defined by (\ref{eqn;a69}) 
converges weakly, as $n\to\infty$, to 
$\frac1{\lambda}\mathbf{\hat{P}}(\tau_1\in dt)$ for $\mathbf{P}$-almost 
all $\omega\in\Omega$.
\label{prop;14}
\end{prop} 

\noindent{\it Proof.} For each $t\geq0$, define 
$$A_t=\left\{\omega\in\Omega\left\vert\ \lim_{n\to\infty}\frac1n\sum_{k=1}^n
1_{\{\tau_k(\omega)>t\}}
=\frac1{\lambda}\mathbf{\hat{P}}(\tau_1>t)
\right.\right\}\ .$$
As was already pointed out, $\tau_k=\tau_1\circ\hat{\theta}_1^{k-1}$ holds 
on $(\hat{\Omega},\frac1{\lambda}\mathbf{\hat{P}})$, $\hat{\theta}_1$ 
being an ergodic automorphism. Hence by the individual ergodic theorem,
$$\lim_{n\to\infty}\frac1n\sum_{k=1}^n1_{\{\tau_k(\omega)>t\}}=
\lim_{n\to\infty}\frac1n\sum_{k=1}^n1_{\{\tau_1>t\}}\circ\hat{\theta}_1^{k-1}
(\omega)=\frac1{\lambda}\mathbf{\hat{P}}(\tau_1>t)$$
holds for $\frac1{\lambda}\mathbf{\hat{P}}$-almost all 
$\omega\in\hat{\Omega}$, namely
$$\frac1{\lambda}\mathbf{\hat{P}}(\hat{\Omega}\cap A_t)=1\ .$$
It remains to show $\mathbf{P}(A_t)=1$. But since $\omega\in A$ if and only if 
$\theta_{x_0}\omega\in A$, we have from Proposition 
\ref{prop;12},
\begin{eqnarray*}
\mathbf{P}(A)&=&\int_{\Omega}\mathbf{P}(d\omega)1_A(\omega)
=\int_{\Omega}\mathbf{P}(d\omega)1_A(\theta_{x_0}\omega) \\
&=&\int_{\hat{\Omega}}\mathbf{\hat{P}}(d\omega)x_1(\omega)
1_{A}(\omega)
=\int_{\hat{\Omega}}\mathbf{\hat{P}}(d\omega)x_1(\omega)=1\ ,
\end{eqnarray*}
the last equality following from (0) of Proposition \ref{prop;13}.

\begin{rem}
By the Palm-Khinchin formula, we have
$$\mathbf{P}(N(0,x]=0)=\int_x^{\infty}\mathbf{\hat{P}}(N(0,s]=0)ds
=\int_x^{\infty}\mathbf{\hat{P}}(x_1>t)dt\ .$$
Thus if we put $\pi_0(x):=\mathbf{P}(N(0,x]=0)$, and 
$\rho(t):=\mathbf{\hat{P}}(x_1>t)$, we get
\begin{equation}
\pi_0(x)=\int_x^{\infty}\rho(t)dt=1-\int_0^x\rho(t)dt\ ,
\label{eqn;a70}
\end{equation}
or
\begin{equation}
\rho(x)=-\frac{d^+}{dx}\pi_0(x)\ .
\label{eqn;a71}
\end{equation}
\label{rem;3}
\end{rem}

The consideration in the proof of Proposition \ref{prop;14} can be easily 
generalized to yield the following assertion:

\begin{prop}
For any non-negative measurable function $f$ on 
$(\Omega,{\cal F},\mathbf{P})$, 
one has
\begin{equation}
\lim_{L\to\infty}\frac1L\int_{(0,L]}N_{\omega}(dt)f(\theta_t\omega)=
\int_{\hat{\Omega}}\mathbf{\hat{P}}(d\omega)f(\omega)
\label{eqn;a72}
\end{equation}
for $\mathbf{P}$-almost all $\omega\in\Omega$.
\label{prop;15}
\end{prop}

\noindent{\it Proof.} If we let
$$A=\left\{\omega\in\Omega\left\vert\ \lim_{n\to\infty}\frac1n\sum_{k=1}^n
f(\theta_{x_k}\omega)
=\frac1{\lambda}\int_{\hat{\Omega}}\mathbf{\hat{P}}(d\omega)f(\omega)
\right.\right\}\ ,$$
then by the ergodic theorem applied to 
$(\hat{\Omega},\mathbf{\hat{P}},\hat{\theta}_1)$, we have 
$\mathbf{\hat{P}}(\hat{\Omega}\cap A)=\lambda$. But since $\omega\in A$ is 
equivalent to $\theta_{x_0}\omega\in A$, the argument in Proposition 
\ref{prop;14} gives $\mathbf{P}(A)=1$. On the other hand, we have 
$$\lim_{L\to\infty}\frac1L N_{\omega}(0,L]=\lambda$$
$\mathbf{P}$-almost surely. Hence for $\mathbf{P}$-almost all 
$\omega\in\Omega$, we have $N_{\omega}(0,L]\to\infty$ and
\begin{eqnarray*}
\lim_{L\to\infty}\frac1L\int_{(0,L]}N_{\omega}(dx)f(\theta_t\omega)
&=&\lim_{L\to\infty}\frac1L\sum_{j}1_{(0,L]}(x_j(\omega))
f(\theta_{x_j}\omega) \\
&=&\lim_{L\to\infty}\frac{N_{\omega}(0,L]}{L}
\cdot\lim_{n\to\infty}\frac1n\sum_{j=1}^n f(\theta_{x_j}\omega) \\
&=&\lambda\cdot\frac1{\lambda}\int_{\hat{\Omega}}
\mathbf{\hat{P}}(d\omega)f(\omega)
=\int_{\hat{\Omega}}\mathbf{\hat{P}}(d\omega)f(\omega)\ .
\end{eqnarray*}

\bigskip
Let us apply this proposition to the limiting $k$-level correlation, for 
later comparison with level statistics in \S\ref{sec;4} and \ref{sec;6}.

\begin{prop}
For realizations $\{x_n(\omega)\}$ of a simple ergodic point process, 
define the $k$-level correlation
\begin{eqnarray}
R_L^{\omega}(I_2,\ldots,I_k):=\frac1L
\sharp\{(x_{i_1}(\omega),\ldots,x_{i_k}(\omega))&\vert&
x_{i_1}(\omega)\leq L,\ i_1,\ldots,i_k\ \mbox{are distinct and} \nonumber\\
& &x_{i_j}(\omega)-x_{i_1}(\omega)\in I_j\ ,j=2,\ldots,k\}\ ,
\label{eqn;a73}
\end{eqnarray}
where $L>0$ and $I_2,\ldots,I_k$ are any bounded intervals. Then
\begin{equation}
\lim_{L\to\infty}R_L^{\omega}(I_2,\ldots,I_k)=
\int_{\hat{\Omega}}\mathbf{\hat{P}}(d\omega)N_{\omega}^{[k-1]}
(I^{\ast}_2\times\cdots\times I^{\ast}_k)
=:R_k(I_2,\ldots,I_k)
\label{eqn;a74}
\end{equation}
holds for $\mathbf{P}$-almost all $\omega\in\Omega$, where we have set 
$I^{\ast}_j:=I_j\setminus\{0\}$.
\label{prop;16}
\end{prop}

\noindent{\it Proof.} Noting (\ref{eqn;a10}), we can write
\begin{eqnarray*}
LR_L^{\omega}(I_2,\ldots,I_k) 
&=&\sum_i1_{(0,L]}(x_i(\omega))
\sharp\{(x_{i_2}(\omega),\ldots,x_{i_k}(\omega))\vert\ 
i_2,\ldots,i_k\ \mbox{are distinct,} \\ 
& &\qquad x_{i_j}(\omega)\ne x_{i_1}(\omega),
\ x_{i_j}(\omega)\in I_j+x_{i_j}(\omega),\ j=2,\ldots,k\} \\
&=&\int N_{\omega}(ds)1_{(0,L]}(s)N_{\omega}^{[k-1]}
\left((I_2+s)\setminus\{s\})\times\cdots\times((I_k+s)\setminus\{s\})\right) \\
&=&\int N_{\omega}(ds)1_{(0,L]}(s)N_{\theta_s\omega}^{[k-1]}
(I^{\ast}_2\times\cdots\times I^{\ast}_k)\ ,
\end{eqnarray*}
from which the conclusion follows by Proposition \ref{prop;15}.

\begin{rem}
In case $I_2=\cdots=I_k=(0,c]$, let us write $R_k(I_2,\ldots,I_k)=R_k(c)$. 
By the Palm-Khinchin formula, we can compute
\begin{eqnarray*}
\frac1{k!}\mathbf{E}\left[(N(0,L])^{[k]}\right]
&=&\sum_{n\geq k}{n-1 \choose k-1}\mathbf{P}(N(0,c]\geq n) \\
&=&\sum_{n\geq k}{n-1 \choose k-1}\int_0^c\mathbf{\hat{P}}(N(0,s]=n-1)ds \\
&=&\frac1{(k-1)!}\int_0^c\int_{\hat{\Omega}}\mathbf{\hat{P}}(d\omega)
(N_{\omega}(0,s])^{[k-1]}\ ,
\end{eqnarray*}
namely
\begin{equation}
\mu_k(c)=k\int_0^cR_k(s)ds\ .
\label{eqn;a75}
\end{equation}
See \cite{CLS} for related formulas.
\label{rem;4}
\end{rem}


\begin{ex}
If $N_{\omega}$ is the stationary Poisson point process with parameter 
$\lambda>0$, then by Example \ref{ex;5} and Proposition \ref{prop;7}, 
\begin{equation}
R_k(I_2,\ldots,I_k)=\lambda\int_{\Omega}\mathbf{P}(d\omega)
N_{\omega}^{[k-1]}(I_2\times\cdots\times I_k)
=\lambda^k\prod_{j=2}^k\vert I_j\vert\ .
\label{eqn;a76}
\end{equation}
In particular, $R_k(c)=\lambda^kc^{k-1}$.
\label{ex;8}
\end{ex}

\begin{rem}
If $\{x_j(\omega)\}$ is a renewal 
process, then $R_2(c)=\lambda^2c$ implies that $N$ is Poissonian. 
Indeed, if we set 
$$F(c)=\frac1{\lambda}\hat{P}(N(0,c]>0)=
\frac1{\lambda}\hat{P}(x_{n+1}-x_n\leq c)\ ,$$
which does not depend on $n$ , then 
\begin{eqnarray*}
\frac1{\lambda}R_2(c)
&=&\frac1{\lambda}\int_{\hat{\Omega}}\mathbf{\hat{P}}(d\omega)N_{\omega}(0,c]
=\frac1{\lambda}\sum_{n=1}^{\infty}\mathbf{\hat{P}}(N(0,c]\geq n) \\
&=&\frac1{\lambda}\sum_{n=1}^{\infty}\mathbf{\hat{P}}
\left(\sum_{j=0}^{n-1}(x_{j+1}-x_j)\leq c\right)
=\sum_{n=1}^{\infty}(F^{n\ast})(c)\ ,
\end{eqnarray*}
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 $R_2(c)=\lambda^2c$ , we get 
$$\sum_{n=1}^{\infty}({\cal L}(\xi))^n=
\frac{{\cal L}(\xi)}{1-{\cal L}(\xi)}
=\frac1{\lambda}\int_0^{\infty}e^{-\xi c}dR_2(c)
=\frac{\lambda}{\xi}\ ,$$
namely ${\cal L}(\xi)=\lambda/(\lambda+\xi)$. 
Hence $dF(c)=\lambda e^{-\lambda c}dc$ . But the Poisson 
point process is the only renewal process for which $x_{n+1}-x_n$ 
obeys the exponential distribution.
\label{rem;5}
\end{rem}

\section{Statistics for asymptotically uniformly distributed sequences}
\label{sec;4}
Suppose we are given a finite or infinite sequence 
$$0\leq x_1(L)<x_2(L)<\cdots<x_n(L)<\cdots$$
depending on a parameter $L$, which we shall let tend to infinity. 
We think of this sequence $\{x_j(L)\}_j$ as the unfolded spectrum 
of a quantum Hamiltonian depending on a parameter, or more directly 
a realization of some fictitious simple ergodic point process with 
mean density (or intensity) $\lambda$, if it is asymptotically 
uniformly distributed in the following sense:

\begin{cond}
There is a positive constant $A\leq\infty$ and $\lambda<\infty$ such that 
for any $0\leq\alpha<A$,
\begin{equation}
\sharp\{j\geq1\vert\ x_j(L)\leq\alpha L\}\sim\lambda\alpha L
\quad(L\to\infty)\ .
\label{eqn;a77}
\end{equation}
\label{cond;1}
\end{cond}

In particular, when the sequence $\{x_j(L)\}_j$ does not depend on $L$ and 
satisfies
\begin{equation}
\sharp\{j\geq1\vert\ x_j\leq L\}\sim\lambda L \quad(L\to\infty)\ ,
\label{eqn;a78}
\end{equation}
then it satisfies Condition \ref{cond;1}. 

For each $c>0$ and for each interval $I=(a,b]$, we consider the following 
quantity (statistics): 
\begin{equation}
N_L(I):=\sharp\{j\geq1\vert\ x_j(L)\in I\}\ ;
\label{eqn;a79}
\end{equation}
\begin{equation}
\pi_k(L;I):=\frac1L\int_0^L1_{\{N_L(t+I)=k\}}dt\quad(k\geq0)\ ;
\label{eqn;a80}
\end{equation}
\begin{equation}
\mu_k(L;I):=\frac1L\int_0^L N_L(t+I)^{[k]}dt=\sum_{j\geq k}j^{[k]}\pi_j(L;I)
\quad(k\geq1)\ ;
\label{eqn;a81}
\end{equation}
\begin{equation}
\rho(L;c):=\frac1L\sharp\{j\geq1\vert\ x_j(L)\leq L,\ 
x_{j+1}(L)-x_j(L)>c\}\quad(c>0)\ ;
\label{eqn;a82}
\end{equation}
\begin{eqnarray}
& &R_k(L;c) \nonumber\\
& &:=\frac1L\sharp\{(x_{i_1}(L),\ldots,x_{i_k}(L))\vert\ 
i_1,\ldots,i_k\ \mbox{are distinct,}\ x_{i_1}(L)\leq L,
\ x_{i_j}(L)-x_{i_1}(L)\in(0,c]\} \nonumber\\
& &\quad(c>0,\ k\geq2)\ .
\label{eqn;a83}
\end{eqnarray}

As an immediate consequence of Condition \ref{cond;1}, we obtain the following

\begin{prop}
For any $c>0$, 
\begin{equation}
\mu_1(c):=\lim_{L\to\infty}\frac1L\int_0^L N_L(t,t+c]\ dt=
\lim_{L\to\infty}\mu_1(L;(0,c])=\lambda c\ .
\label{eqn;a84}
\end{equation}
\label{prop;17}
\end{prop}

\noindent{\em Proof.} 
By the definition, 
\begin{equation}
\mu_1(L;(0,c])=\frac1L\sum_{j}\int_0^L1_{(t,t+c]}(x_j(L))dt=
\frac1L\sum_j\int_0^L1_{[x_j(L)-c,x_j(L))}(t)dt\ .
\label{eqn;a85}
\end{equation}
But when $L>c>0$ , one has
$$\int_0^L1_{[x_j(L)-c,x_j(L))}(t)dt=
\left\{\begin{array}{rl}
c,&\quad\mbox{if $c\leq x_j(L)\leq L$}\\
0,&\quad\mbox{if $x_j(L)>L+c$}\ ,
\end{array}\right.$$
while
$$\int_0^L1_{[x_j(L)-c,x_j(L))}(t)dt\leq c\ ,\quad\mbox{if}\ x_j(L)<c
\quad
\mbox{or}\quad L<x_j(L)\leq L+c\ .$$
Hence for each fixed $c>0$ , it is obvious from Condition \ref{cond;1} that
%\begin{equation}
$$\mu_1(L;(0,c])\sim c\frac1L\sharp\{j\geq1\vert\ x_j(L)\leq L\}
\quad\mbox{as}\ L\to\infty\ ,$$
%\end{equation}
so that
%\begin{equation}
$$\mu_1(c)=\lim_{L\to\infty}\mu_1(L;(0,c])=\lambda c\ ,$$
%\end{equation}
which was to be proved.

\bigskip
Except for this trivial case, Condition \ref{cond;1} does not imply that 
any of the statistics (\ref{eqn;a80})--(\ref{eqn;a83}) converges to some 
limit as $L\to\infty$. But when some of these convergences occur, we can 
prove equalities between them, which may be called deterministic 
analogues of the Palm-Khinchin formula, as the following two propositions 
show. (Compare Remarks \ref{rem;3} and \ref{rem;4}.)

\begin{prop}
Let $k\geq2$ and assume that
\begin{equation}
\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L)<c}(c-x_{j_k}(L))
=o(L)\quad(L\to\infty)
\label{eqn;a86}
\end{equation}
holds for any $c>0$.

\noindent(i) 
If the limit
\begin{equation}
\mu_k(c):=\lim_{L\to\infty}\mu_k(L;(0,c])
\label{eqn;a87}
\end{equation}
exists for each $c>0$ and is differentiable with respect to $c$, then 
\begin{equation}
R_k(c):=\lim_{L\to\infty}R_k(L;c)
\label{eqn;a88}
\end{equation}
also exists and one has 
\begin{equation}
R_k(c)=\frac1k\frac{d}{dc}\mu_k(c)\ .
\label{eqn;a89}
\end{equation}

\noindent (ii) 
If on the other hand, $R_k(c)$ exists for each $c>0$, then 
$\mu_k(c)$ also exists and 
\begin{equation}
\mu_k(c)=k\int_0^LR_k(c')dc'\ .
\label{eqn;a90}
\end{equation}
\label{prop;18}
\end{prop}

\begin{rem}
The condition (\ref{eqn;a86}) is obviously satisfied when the sequence 
$\{x_j(L)\}_j$ does not depend on $L$.
\label{rem;6}
\end{rem}

\noindent{\it Proof of Propsition \ref{prop;18}.} 
Since
\begin{eqnarray*}
\frac1{k!}(N_L(t,t+c])^{[k]}
&=&\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L)}1_{(t,t+c]}(x_{j_1}(L))
\cdots1_{(t,t+c]}(x_{j_k}(L)) \\
&=&\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L)}
1_{[x_{j_1}(L)-c,x_{j_1}(L))}(t)\cdots
1_{[x_{j_k}(L)-c,x_{j_k}(L))}(t)\ ,
\end{eqnarray*}
we can compute
\begin{eqnarray*}
\frac{L}{k!}\mu_k(L;(0,c])
&=&\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L)}
\left\vert[0,L]\cap\bigcap_{p=1}^k[x_{j_p}(L)-c,x_{j_p}(L))\right\vert \\
&=&\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L)}\{(L\wedge x_{j_1}(L))
-(x_{j_k}(L)-c)_+\}_+ \\
&=&\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L);\atop x_{j_1}(L)\leq L,
x_{j_k}(L)\geq c} \{c-x_{j_k}(L)+x_{j_1}(L)\}_+ \\
& &+\sum_{L<x_{j_1}(L)<\cdots<x_{j_k}(L)}\{c-x_{j_k}(L)+L\}_+ 
+\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L)<c}x_{j_1}(L) \\
&=&\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L)\leq L}
\{c-x_{j_k}(L)+x_{j_1}(L)\}_+ \\
& &+\left[\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L);\atop x_{j_1}(L)\leq L
<x_{j_k}(L)} \{c-x_{j_k}(L)+x_{j_1}(L)\}_+
+\sum_{L<x_{j_1}(L)<\cdots<x_{j_k}(L)}\{c-x_{j_k}(L)+L\}_+\right] \\
& &+\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L)<c}[x_{j_1}(L)-
\{c-x_{j_k}(L)+x_{j_1}(L)\}_+]\ .
\end{eqnarray*}
The third term equals
$$-\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L)<c}(c-x_{j_k}(L))\ ,$$
and is $o(L)$ as $L\to\infty$ by the assumption.
 On the other hand, 
the second term, which is enclosed in the big bracket and 
which we denote by $M(L)$ , can be estimated in two ways. 
First of all, we have 
\begin{eqnarray*}
M(L)&\leq&\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L);\atop x_{j_1}(L)\leq L
<x_{j_k}(L)\ \mbox{{\em\scriptsize or}}\ 
L<x_{j_1}(L)<\cdots<x_{j_k}(L)\leq L+c}
 (c-x_{j_k}(L)+x_{j_1}(L))_+ \\
 &=&\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L);\atop x_{j_1}(L)\leq L
<x_{j_k}(L)\ \mbox{{\em\scriptsize or}}\ 
L<x_{j_1}(L)<\cdots<x_{j_k}(L)\leq L+c}
\int_{-\infty}^{\infty}1_{(t,t+c]}(x_{j_1}(L))\cdots
1_{(t,t+c]}(x_{j_k}(L))dt\ .
\end{eqnarray*}
Suppose $x_{j_1}(L)\leq L<x_{j_k}(L)$ . Then we have 
$1_{(t,t+c]}(x_{j_1}(L))=0$ for $t>L+c$ and 
$1_{(t,t+c]}(x_{j_k}(L))=0$ for $t<L-c$ . If, on the other hand, 
$L<x_{j_1}(L)<\cdots<x_{j_k}(L)\leq L+c$ , then we have 
$1_{(t,t+c]}(x_{j_k}(L))=0$ for $t>L+c$ and 
$1_{(t,t+c]}(x_{j_1}(L))=0$ for $t<L-c$ . Hence
\begin{eqnarray*}
M(L)&\leq&\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L)}
\int_{L-c}^{L+c}1_{(t,t+c]}(x_{j_1}(L))\cdots
1_{(t,t+c]}(x_{j_k}(L))dt \\
&=&\frac1{k!}\int_{L-c}^{L+c}(N_L(t,t+c])^{[k]}dt\ ,
\end{eqnarray*}
which is of $o(L)$ ($L\to\infty$) when $\mu_k(c)$ exists.

Secondly we also have the inequality
\begin{eqnarray*}
M(L)&\leq&c\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L);\atop x_{j_1}(L)\leq L
<x_{j_k}(L)\ \mbox{{\em\scriptsize or}}\ 
L<x_{j_1}(L)<\cdots<x_{j_k}(L)\leq L+c}
 1_{\{x_{j_k}(L)-x_{j_1}(L)\leq c\}} \\
 &\leq&c\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L)\leq L+c}
 1_{\{x_{j_k}(L)-x_{j_1}(L)\leq c\}} 
-c\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L)\leq L}
1_{\{x_{j_k}(L)-x_{j_1}(L)\leq c\}}\ .
\end{eqnarray*}
Obviously, the right hand side is of $o(L)$ as $L\to\infty$ when $R_k(c)$ 
exists. 
Therefore, under either of the conditions of (i) and (ii), one has
\begin{equation}
\frac1{k!}L\mu_k(L;(0,c])=
\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L)\leq L}
\{c-x_{j_k}(L)+x_{j_1}(L)\}_++o(L)\ .
\label{eqn;a91}
\end{equation}

Let us now prove the assertion (i). From the above remark and 
$$\{c+\delta-x_{j_k}(L)+x_{j_1}(L)\}_+ -\{c-x_{j_k}(L)+x_{j_1}(L)\}_+
\geq\delta1_{\{x_{j_k}(L)-x_{j_1}(L)\leq c\}}\ ,$$
we see
$$\frac1{k!}(\mu_k(c+\delta)-\mu_k(c))
\geq\delta\limsup_{L\to\infty}\frac1L
\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L)\leq L}
1_{\{x_{j_k}(L)-x_{j_1}(L)\leq c\}}\ .$$
Similarly, from the inequality
$$\{c-x_{j_k}(L)+x_{j_1}(L)\}_+ -\{c-\delta-x_{j_k}(L)+x_{j_1}(L)\}_+
\leq\delta1_{\{x_{j_k}(L)-x_{j_1}(L)\leq c\}}\ ,$$
it follows that
$$\frac1{k!}(\mu_k(c)-\mu_k(c-\delta))
\leq\delta\liminf_{L\to\infty}\frac1L
\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L)\leq L}
1_{\{x_{j_k}(L)-x_{j_1}(L)\leq c\}}\ .$$
Under the assumption of the differentiability of $\mu_k(c)$ , we can let 
$\delta\searrow0$ , to conclude
$$\lim_{L\to\infty}\frac1L
\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L)\leq L}
1_{\{x_{j_k}(L)-x_{j_1}(L)\leq c\}}=\frac1{k!}\frac{d}{dc}\mu_k(c)\ .$$
But the limit on the left hand side is easily seen to coincide with
$$\frac1{(k-1)!}\lim_{L\to\infty}R_k(L;c)\ $$
and hence
$$R_k(c)=\frac1k\frac{d}{dc}\mu_k(c)\ ,$$
as was to be proved.

Let us turn to the proof of (ii). 
Let $\ell\in\mathbf{N}$, and write $\delta=c/\ell$  and 
$\Delta=x_{j_k}(L)-x_{j_1}(L)$ . Then we have
$$\delta\sum_{m=1}^{\ell}1_{\{c\geq\Delta+m\delta\}}
\leq(c-\Delta)_+\leq\delta\sum_{m=0}^{\ell}1_{\{c\geq\Delta+m\delta\}}\ ,$$
and hence
\begin{eqnarray*}
& &\frac{c}{\ell}\sum_{m=1}^{\ell}\frac1L
\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L)\leq L}
1_{\{x_{j_k}(L)-x_{j_1}(L)\leq c-mc/\ell\}} \\
&\leq&\frac1L
\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L)\leq L}
\{c-x_{j_k}(L)+x_{j_1}(L)\}_+ \\
&\leq&\frac{c}{\ell}\sum_{m=0}^{\ell}\frac1L
\sum_{x_{j_1}(L)<\cdots<x_{j_k}(L)\leq L}
1_{\{x_{j_k}(L)-x_{j_1}(L)\leq c-mc/\ell\}} .
\end{eqnarray*}
Letting $L\to\infty$ and noting (\ref{eqn;a91}), we obtain
\begin{eqnarray}
\frac1{(k-1)!}\frac{c}{\ell}\sum_{m=1}^{\ell}R_k(c(1-\frac{m}{\ell}))
&\leq&\frac1{k!}\liminf_{L\to\infty}\mu_k(L;(0,c]) \nonumber\\
&\leq&\frac1{k!}\limsup_{L\to\infty}\mu_k(L;(0,c]) \nonumber\\
&\leq&\frac1{(k-1)!}\frac{c}{\ell}
\sum_{m=0}^{\ell}R_k(c(1-\frac{m}{\ell}))\ ,
\label{eqn;a92}
\end{eqnarray}
for all $\ell\in\mathbf{N}$. Finally letting $\ell\to\infty$ , we arrive at 
$$\frac1{k!}\mu_k(c)=\frac1{k!}\lim_{L\to\infty}\mu_k(L;(0,c])
=c\int_0^1R_k(c(1-x))dx=
\int_0^cR_k(c')dc'\ ,$$
or
$$\mu_k(c)=k\int_0^cR_k(c')dc'\ ,$$
completing the proof.

\begin{prop}
Suppose $\pi_0(c):=\lim_{L\to\infty}\pi_0(L;(0,c])$ 
exists for each $c>0$ and is differentiable with 
respect to $c$. Then $\rho(c)=\lim_{L\to\infty}\rho(L;c)$ also exists, 
and is given by
\begin{equation}
\rho(c)=-\frac d{dc}\pi_0(c)\ .
\label{eqn;a93}
\end{equation}
Conversely if $\rho(c)$ exists for each $c>0$, then $\pi_0(c)$ also 
exists, and is given by
\begin{equation}
\pi_0(c)=1-\int_0^c\rho(c')dc'\ .
\label{eqn;a94}
\end{equation}
\label{prop;19}
\end{prop}

\noindent
{\em Proof.} If $x_j(L)\leq L$ and 
$x_{j+1}(L)-x_j(L)>c$ , then for any $\delta\in(0,c)$ , we have the 
implication
%\begin{equation}
$$
t\in[x_j(L)-\delta,x_j(L))\ \Longrightarrow\ N_L(t;\delta)>0\ ,\ 
N_L(t+\delta;c-\delta)=0\ ,
$$
%\end{equation}
the intervals $[x_j(L)-\delta,x_j(L))$ corresponding to these $j$'s 
being disjoint. 
Hence if we set 
%\begin{equation}
$$
n(c;L)=\sharp\{j\geq1\ \vert\ x_j(L)\leq L\ ,\ x_{j+1}(L)-x_j(L)>c\}
\ ,
%\end{equation}
$$
then it holds that
\begin{eqnarray*}
\delta n(L;c)
&=&\sum_{x_j(L)\leq L\ ;\ x_{j+1}(L)-x_j(L)>c}
\int_{-\delta}^L1_{[x_j(L)-\delta,x_j(L))}(t)dt \\
&\leq&\int_{-\delta}^L 1_{\{N_L(t,t+\delta]>0,\ N_L(t+\delta,t+c]=0\}}dt \\
&=&\int_{-\delta}^L\left[1_{\{N_L(t+\delta,t+c]=0\}}
-1_{\{N_L(t,t+c]=0\}}\right]dt \\
&=&\int_0^{L+\delta}1_{\{N_L(t,t+c-\delta]=0\}}dt
-\int_{-\delta}^L1_{\{N_L(t,t+c]=0\}}dt \\
&\leq&\delta+\int_0^L 1_{\{N_L(t,t+c-\delta]=0 \}}dt-
\int_0^L 1_{\{N_L(t,t+c]=0\}}dt\ ,
\end{eqnarray*}
namely
$$\delta\rho(L;c)\leq\frac{\delta}{L}+\pi_0(L;(0,c-\delta])-
\pi_0(L;(0,c])\ .$$
Letting $L\to\infty$, we get 
$$\delta\limsup_{L\to\infty}\rho(L;c)=
\leq\pi_0(c-\delta)-\pi_0(c)\ .$$
Since $\delta>0$ is arbitrary, this shows
$$\limsup_{L\to\infty}\rho(L;c)\leq-\frac{d}{dc}\pi_0(c)\ .$$

On the other hand, if $N_L(t,t+\delta]>0$ and 
$N_L(t+\delta,t+c+\delta)=0$ , then there is 
an $x_j(L)\in(t,t+\delta]$ such that $x_{j+1}(L)-x_j(L)>c$ . 
Hence the set
$$\{t\in[0,L-\delta]\ \vert\ N_L(t,t+\delta]>0\ ,
\ N_L(t+\delta,t+c+\delta]=0\}$$
is a union of finitely many disjoint intervals $I_{\ell}$ each of 
length no greater than $\delta$ , and for each $I_{\ell}$ , there corresponds 
an $x_j(L)\leq L$ satisfying $x_{j+1}(L)-x_j(L)>c$. 
Hence the number of these intervals does not exceed $n(c;L)$ , and so
\begin{eqnarray*}
\delta n(c;L)&\geq&\sum_{\ell}\vert I_{\ell}\vert \\
&=&\int_0^{L-\delta}1_{\{N_L(t,t+\delta]>0,\ N_L(t+\delta,t+c+\delta]=0\}}dt \\
&=&\int_0^{L-\delta}\left[1_{\{N_L(t+\delta,t+c+\delta]=0\}}
-1_{\{N_L(t,t+c+\delta]=0\}}\right]dt \\
&=&\int_{\delta}^L 1_{\{N_L(t,t+c]=0 \}}dt-
\int_0^{L-\delta} 1_{\{N_L(t,t+c+\delta]=0 \}}dt \\
&\geq&-\delta+L\pi_0(L;(0,c])-L\pi_0(L;(0,c+\delta])\ .
\end{eqnarray*}
Dividing both sides by $L$ and letting $L\to\infty$ , we get 
$$\delta\liminf_{L\to\infty}\rho(L;c)
\geq\pi_0(c)-\pi_0(c+\delta)\ ,$$
and since $\delta>0$ is arbitrary, we obtain
$$\liminf_{L\to\infty}\rho(L;c)\geq-\frac{d}{dc}\pi_0(c)\ ,$$
which completes the proof of the first assertion.

Conversely suppose that the limit $\rho(c)$ exists for each $c>0$. As we 
have already seen, the following two inequalities are valid for any 
$L>0$ and $\delta\in(0,c)$:
\begin{equation}
\delta\rho(L;c)\leq\frac{\delta}{L}+\pi_0(L;(0,c-\delta])
-\pi_0(L;(0,c])\ ;
\label{eqn;a95}
\end{equation}
\begin{equation}
\delta\rho(L;c)\geq-\frac{\delta}{L}+\pi_0(L;(0,c])
-\pi_0(L;(0,c+\delta])\ .
\label{eqn;a96}
\end{equation}

Now let $0<c_1<c_2$ be given and let $\ell\in\mathbf{N}$ be large enough 
so that $\delta:=(c_2-c_1)/\ell<c_1$. If in (\ref{eqn;a95}) we take 
$c=c_1+k\delta$, $k=1,2,\ldots,\ell$, then 
$$\delta\rho(L;c_1+k\delta)\leq\frac{\delta}{L}+
\pi_0(L;(0,c_1+(k-1)\delta])-\pi_0(L;(0,c_1+k\delta])\ .$$
Summing over $k$, one obtains 
$$\delta\sum_{k=1}^{\ell}\rho(L;c_1+k\delta)
\leq\frac{c_2-c_1}{L}+\pi_0(L;(0,c_1])-\pi_0(L;(0,c_2])\ .$$
Letting first $L\to\infty$ and then $\ell\to\infty$, we arrive at
\begin{equation}
\limsup_{L\to\infty}\pi_0(L;(0,c_2])+
\int_{c_1}^{c_2}\rho(c')dc'\leq1\ .
\label{eqn;a97}
\end{equation}
In the same way, taking $c=c_1+k\delta$, $k=0,1,\ldots,\ell-1$, in 
(\ref{eqn;a96}), and summing over $k$, we obtain
\begin{equation}
\liminf_{L\to\infty}\pi_0(L;(0,c_2])+
\int_{c_1}^{c_2}\rho(c')dc'\geq\liminf_{L\to\infty}\pi_0(L;(0,c_1])\ .
\label{eqn;a98}
\end{equation}

Denote $\underline{\pi}_0(c):=\liminf_{L\to\infty}\pi_0(L;(0,c])$. 
We shall now verify that $\underline{\pi}_0(c)\nearrow1$ as $c\searrow0$. 
Suppose, on the contrary, that
$$\alpha:=\lim_{c\searrow0}\underline{\pi}_0(c)<1\ .$$
Then for any $\beta\in(\alpha,1)$ and $c>0$, we must have
$$\limsup_{L\to\infty}\frac1L\int_0^L1_{\{N_L(t,t+c]>0\}}dt>1-\beta\ .$$
Now the set 
$$\{t\in(0,L]\vert\ N_L(t,t+c]>0\}$$
is a disjoint union of intervals $I_1,\ldots,I_m$, and by the definition 
of this set, we see
$$\sharp\{j\geq1\vert\ x_j(L)\leq L\}\geq\frac1c\sum_{j=1}^m\vert I_j\vert\ ,$$
and hence
$$\lambda
=\lim_{L\to\infty}\frac1L\sharp\{j\geq1\vert\ x_j(L)\leq L\}
\geq\frac1c\limsup_{L\to\infty}\frac1L\sum_{j=1}^m\vert I_j\vert
\geq\frac{1-\beta}{c}\ .$$
Letting $c\searrow0$, we arrive at a contradiction.

Having shown $\lim_{c\searrow0}\underline{\pi}_0(c)=1$, we can let 
$c_1\searrow0$ in (\ref{eqn;a97}) and (\ref{eqn;a98}), to obtain
\begin{equation}
\limsup_{L\to\infty}\pi_0(L;(0,c])+\int_0^c\rho(c')dc'
\leq1\leq
\liminf_{L\to\infty}\pi_0(L;(0,c])+\int_0^c\rho(c')dc'\ ,
\end{equation}
which show the existence of $\pi_0(c)$ and the relation
$$\pi_0(c)=1-\int_0^c\rho(c')dc'\ .$$

\begin{rem}
Proposition \ref{prop;19} above shows the close relationship between the 
\lq\lq two possible ways of describing the randomness of the spectrum\rq\rq, 
proposed by Sinai in the Introduction of \cite{Si2}.
\end{rem}

\section{Two notions of \lq\lq unfolding\rq\rq}
\label{sec;5}
\subsection{\lq\lq Unfoldable\rq\rq family of sequences}
\label{sec;5-1}
In general, energy levels of a quantum Hamiltonian have a different local 
density in different parts of the energy axis. Hence, as we noted in the 
introduction, in order to discuss the fluctuating properties of energy 
levels, it is necessary to \lq\lq unfold\rq\rq the quantal spectrum 
under consideration, so that the transformed sequence has uniform 
asymptotic distribution. To formulate this notion, let us first define a 
class of spectra which can be unfolded.

Suppose we are given a finite or infinite increasing sequence 
$\{E_j(h);j\geq1\}$ of real numbers depending on a parameter $h>0$. 
We shall say that this family is {\em unfoldable} in the limit $h\searrow0$ 
if the following condition is satisfied:

\begin{cond}
There is an $\alpha>0$ and 
a non-decreasing function $\nu(E)\geq0$ on $\mathbf{R}$ with 
$\lim_{E\to-\infty}\nu(E)=0$ such that for any $E\in\mathbf{R}$, one has
\begin{equation}
N_h(E):=\sharp\{j\geq1;\ E_j(h)\leq E\}\sim\nu(E)h^{-\alpha}
\end{equation}
as $h\searrow0$.
\label{cond;2}
\end{cond}

Below we give some examples of quantum Hamiltonians whose spectrum is 
unfoldable in the sense defined above.

\begin{ex}
Let $H(\hbar):=-\hbar^2\Delta_D$, where $\Delta_D$ is the Laplacian on a 
connected, bounded domain $D\subset\mathbf{R}^d$ with smooth boundary, 
considered under the Dirichlet boundary condition. If $0<e_1<e_2<\cdots$ 
are the eigenvalues of $H(1)=-\Delta_D$, then the spectrum $\{E_j(\hbar)\}_j$ 
of $H(\hbar)$ is given by $E_j(\hbar)=\hbar^2e_j$. As is well known (Weyl's 
asymptotic), there is a universal constant $c_d>0$, depending only on the 
dimension of the space, such that 
\begin{equation}
N_1(E):=\sharp\{j\geq1;\ e_j\leq E\}\sim c_d\vert D\vert E^{d/2}\ ,
\quad E\to\infty\ .
\end{equation}
Hence for any fixed $E>0$, we have 
\begin{equation}
N_{\hbar}(E):=\sharp\{j\geq1;\ E_j(\hbar)\leq E\}
=N_1(E/\hbar^2)\sim (c_d\vert D\vert E^{d/2})\hbar^{-d}
\end{equation}
as $E\to\infty$. Thus $\{E_j(\hbar)\}_j$ is unfoldable in the (semiclassical) 
limit $\hbar\searrow0$ with $\alpha=d$ and 
$\nu(E)=c_d\vert D\vert(E\vee0)^{d/2}$.
\label{ex;9}
\end{ex}

\begin{ex}[Regular spectra (\cite{BT}, \cite{Pe})]
Suppose that the classical Hamiltonian system associated to $H(p,q)$ is 
completely integrable. Then one can transform the momentum-coordinate 
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{Arn}).
Following Percival (\cite{Pe}), Berry and Tabor (\cite{BT}), 
we shall say that $H(\hbar)$, the quantization of $H(p,q)$, 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_{\mathbf{n}}(\hbar)=
H\left(\hbar(\mathbf{n}+\frac14\mathbf{\alpha})\right)\quad;\quad 
\mathbf{n}=(n_1,\ldots,n_d)\in
\mathbf{Z}_+^d\ ,
\end{equation}
where $\mathbf{\alpha}\in\mathbf{Z}_+^d$ is called the Maslov index. 
(In the following discussion, we put $\mathbf{\alpha}=\mathbf{0}$ as in 
\cite{BT}.) 
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 harmonic 
oscillators or free motion on flat tori (\cite{Sa}, \cite{Va}). 

Suppose that $H(I)$, as a function on $\mathbf{R}_+^d$ satisfies the 
following conditions:
\begin{enumerate}
	\item $H(r\mathbf{x})$ is increasing in $r\geq0$ for any 
	$\mathbf{x}\in\mathbf{R}_+^d\setminus\{\mathbf{0}\}$;
	\item $H(\mathbf{0})=0$;
	\item $\{\mathbf{x};\ H(\mathbf{x})\leq E\}$ is compact 
	for any $E>0$.
\end{enumerate}
Then $\{E_{\mathbf{n}}(\hbar);\ \mathbf{n}\in\mathbf{Z}_+^d\}$ is 
unfoldable in the limit $\hbar\searrow0$ with $\alpha=d$ and
\begin{equation}
\nu(E)=\int\cdots\int_{\mathbf{R}_+^d}1_{\{H(I)\leq E\}}dI_1\cdots dI_d\ .
\end{equation}
\label{ex;10}
\end{ex}

\begin{ex}[Ergodic Schr\"odinger operators]
Let $\{V_{\omega}(x);x\in\mathbf{R}^d\ \mbox{or}\ \mathbf{Z}^d\}$ be an 
ergodic random field defined on some probability space 
$(\Omega,{\cal F},\mathbf{P})$. Consider the random Schr\"odinger operator 
$H_{\omega}:=-\Delta+V_{\omega}(x)$ and its restriction $H_{\omega}^L$ 
to the cube $[-L/2,L/2]^d$ under Dirichlet boundary condition. Here $\Delta$ 
denotes the continuum or discrete Laplacian. Let $\{E_j^{\omega}(L)\}_j$ 
be the spectrum of $H_{\omega}^L$. Then under mild conditions on the 
random potential $\{V_{\omega}(x)\}_x$, there exists a non-decreasing 
function ${\cal N}(E)\geq0$ (the integrated density of states) 
such that ${\cal N}(-\infty)=0$ and that for $\mathbf{P}$-almost all 
$\omega\in\Omega$, one has
\begin{equation}
\lim_{L\to\infty}\frac1{\vert\Lambda\vert}\sharp
\{j\geq1;\ E_j^{\omega}(L)\leq E\}={\cal N}(E)
\end{equation}
for all $E\in\mathbf{R}$. (See \cite{Ve} for a review of recent results.) 
This shows that for $\mathbf{P}$-almost all 
$\omega\in\Omega$, the spectrum $\{E_j^{\omega}(L)\}_j$ is unfoldable 
in the (thermodynamic) limit $h:=1/L\searrow0$, with $\alpha=d$ and 
$\nu(E)={\cal N}(E)$.
\label{ex;11}
\end{ex}

\subsection{The unfolding of the first kind}\label{sec;5-2}
\label{sec;5-2}
In general, the purpose of unfolding is to transform a given family of 
unfoldable sequences of levels into another family of sequences $\{x_j(L)\}_j$ 
which is asymptotically uniformly distributed so as to satisfy 
Condition \ref{cond;1}. 
In this subsection, we assume that the function $\nu(E)$ appearing in 
Condition \ref{cond;2} is continuous in $E$.

For each $t\in(0,\nu(+\infty))$, define
\begin{equation}
\nu^{-1}(t):=\inf\{E;\ \nu(E)>t\}\ .
\end{equation}
Then we have $\nu(\nu^{-1}(t))=t$ and the equivalence
$$\nu(E)\leq t\ \Leftrightarrow\ E\leq\nu^{-1}(t)\ .$$
If we let $e_j(h):=\nu(E_j(h))$, then for each $t>0$,
\begin{eqnarray*}
\sharp\{j\geq1;\ e_j(h)\leq t\}
&=&\sharp\{j\geq1;\ E_j(h)\leq\nu^{-1}(t)\} \\
&=&N_h(\nu^{-1}(t)) \\
&\sim&\nu(\nu^{-1}(t))h^{-\alpha}=th^{-\alpha}
\end{eqnarray*}
as $h\searrow0$ by Condition \ref{cond;2}. Hence if we let $L=h^{-\alpha}$, 
and $x_j(L)=Le_j(L^{-1/\alpha})$, then we have for any $\alpha>0$,
\begin{eqnarray*}
\sharp\{j\geq1;\ x_j(L)\leq\alpha L\}&=&
\sharp\{j\geq1;\ e_j(L^{-1/\alpha})\leq\alpha\} \\
&\sim&\alpha(L^{-1/\alpha})^{-\alpha}=\alpha L
\end{eqnarray*}
as $L\to\infty$, so that the parametrized sequence $\{x_j(L)\}_j$ 
satisfies Condition \ref{cond;1} with $\lambda=1$.

\subsection{The unfolding of the second kind}
\label{sec;5-3}
The unfolding formulated above is what is usually practiced in quantum 
chaos (see e.g. \cite{MT1}), and looks like a quite natural idea in that it 
leads one to directly observe energy levels, although under a non-linear 
transformation $e_j=\nu(E_j)$. But with regard to this, we notice a seemingly 
strange remark made by Berry and Tabor in their celebrated paper \cite{BT}. 
They argue that the energy $E$ itself
\begin{quote}
is an unsuitable parameter to describe the levels, for two reasons. 
The first is that the {\em mean level density} $\bar{n}(E)$ depends on 
energy.$\cdots$ The second reason is that the statistics of the levels 
might themselves depend on the region of energy being studied.
\end{quote}
As a solution to this problem, Berry and Tabor propose to observe the 
quantum system at a fixed energy value $E$, and to \lq\lq quantize the 
Planck's constant $\hbar$\rq\rq. They apply this procedure only to 
regular spectra discussed in Example \ref{ex;10}, but in a general setting, 
it would be formulated in the following way: In addition to Condition 
\ref{cond;2}, let us assume that $E_j(h)$ is a non-decreasing function 
of $h>0$ for each $j$. Let us fix $E$ such that $E_0<E<E_1$, where
\begin{equation}
E_0:=\inf\{E;\ \nu(E)>0\}\geq-\infty
\end{equation}
and
\begin{equation}
E_1:=\sup\{E;\ \nu(E)<\nu(\infty)\}\leq\infty\ .
\end{equation}
Now define
\begin{equation}
h_j(E):=\inf\{h>0;\ E_j(h)>E\}
\end{equation}
for each $j\geq1$, where we let $h_j(E)=0$ if $E_j(h)>E$ for all $h>0$, and 
$h_j(E)=+\infty$ if $E_j(h)\leq E$ for all $h>0$. Actually, the last two 
possibilities are ruled out by the following argument: Since $\nu(E)>0$, 
Condition \ref{cond;2} implies in particular that
\begin{equation}
\lim_{h\searrow0}\sharp\{j\geq1;\ E_j(h)\leq E\}=\infty\ .
\label{eqn;a110}
\end{equation}
If $h_j(E)=0$ for some $i$, then $E_j(h)>E$ for all $j\geq i$ and 
$h>0$. Hence 
$$\sharp\{j\geq1;\ E_j(h)\leq E\}\leq i-1\ ,$$ 
contradicting (\ref{eqn;a110}), and so $h_j(E)>0$ must hold for all $j\geq1$. 
On the other hand, since $E<E_1$, there is an $E'>E$ such that 
$\nu(E)<\nu(E')<\nu(\infty)$. Again by Condition \ref{cond;2}, we have 
\begin{equation}
\lim_{h\searrow0}\sharp\{j\geq1;\ E<E_j(h)\leq E'\}=\infty\ .
\label{eqn;a111}
\end{equation}
This shows that for only finitely many $j$'s, one can have 
$$\sup_{h>0}E_j(h)\leq E\ .$$ 
Thus there is a $j_0\geq1$ such that $h_j(E)<\infty$ for all $j\geq j_0$. 
We now define
\begin{equation}
x_j=x_j(E):=\nu(E)h_j(E)^{-\alpha}
\end{equation}
for $j\geq j_0$. In the sequel we assume $j_0=1$. By the definition, we have 
the following equivalence and implication for each $L>0$:
\begin{eqnarray*}
x_j<L&\Leftrightarrow&h_j(E)>\left(\nu(E)/L\right)^{1/\alpha} \\
&\Rightarrow&E_j\left((\nu(E)/L)^{1/\alpha}\right)\leq E \\
&\Rightarrow&h_j(E)\geq\left(\nu(E)/L\right)^{1/\alpha}
\Leftrightarrow x_j\leq L\ .
\end{eqnarray*}
This shows 
$$\sharp\{j\geq1;\ x_j<L\}
\leq
\sharp\left\{j\geq1;\ E_j\left(\left(\nu(E)/L\right)^{1/\alpha}\right)
\leq E\right\}
\leq\sharp\{j\geq1;\ x_j\leq L\}\ .$$
But by Condition \ref{cond;2}, we have
$$\lim_{L\to\infty}\frac1L
\sharp\left\{j\geq1;\ E_j\left(\left(\nu(E)/L\right)^{1/\alpha}\right)
\leq E\right\}
=\lim_{L\to\infty}\frac1L\nu(E)\left\{\left(\frac{\nu(E)}{L}\right)
^{1/\alpha}\right\}^{-\alpha}=1\ ,$$
so that
$$\limsup_{L\to\infty}\frac1L\sharp\{j\geq1;\ x_j<L\}
\leq1\leq\liminf_{L\to\infty}\sharp\{j\geq1;\ x_j\leq L\}\ ,$$
namely
$$\sharp\{j\geq1;\ x_j\leq L\}\sim L\ ,\quad L\to\infty\ .$$
Since $\{x_j\}$ itself does not depend on $L$, we obtain a sequence 
satisfying Condition \ref{cond;1} with $\lambda=1$.

\subsection{Examples of unfolding}
\label{sec;5-4}
Unfortunately, the present author is unable to clarify the general 
relationship between unfoldings of first and second kinds. 
For the time being, let us content ourselves by discussing several examples of 
quantal spectra to which both kinds of unfoldings can be applied. 

\subsubsection{The quantum billiard}
\label{sec;5-4-1}
For the Laplacian examined in Example \ref{ex;9}, we have 
$E_j(\hbar)=\hbar^2 e_j$, $\alpha=d$, and $\nu(E)=c_d\vert D\vert E^{d/2}$. 
Since $\nu(E)$ is continuous and $E_j(\hbar)$ is increasing in $\hbar$, 
both kinds of unfoldings can be applied. To perform the unfolding of the 
first kind, we put as in \ref{sec;5-2}, 
\begin{equation}
e_j(\hbar):=\nu(E_j(\hbar))=c_d\vert D\vert\hbar^d e_j^{d/2}\ ;
\end{equation}
and
\begin{equation}
x_j(L):=Le_j(L^{-1/d})=c_d\vert D\vert e_j^{d/2}\ .
\end{equation}
The resulting sequence $\{x_j(L)\}$ does not depend on the parameter $L$. 

To perform the unfolding of the second kind, let us fix an $E>0$. Then we 
put
\begin{equation}
\hbar_j(E):=\inf\{\hbar;\ \hbar^2 e_j=E\}=(E/e_j)^{1/2}
\end{equation}
and
\begin{equation}
x_j(E):=\nu(E)\hbar_j(E)^{-d}=c_d\vert D\vert E^{d/2}(E/e_j)^{-d/2}
=c_d\vert D\vert e_j^{d/2}\ .
\end{equation}
Thus both kinds of unfoldings yield the same sequence.

\subsubsection{A one-dimensional Schr\"odinger operator with 
$\delta$-potentials in the semi-classical limit}
\label{sec;5-4-2}
Consider the one-dimensional Schr\"odinger operator
\begin{equation}
H_v(\hbar)=-\hbar^2\frac{d^2}{dx^2}\ +\ v\sum_{s=1}^n\delta(x-x_s)\ ,\quad 
0\leq x\leq1
\end{equation}
under Dirichlet boundary conditions at $x=0,1$. Here $v>0$, and
$$0\equiv x_0<x_1<\cdots<x_n<x_{n+1}\equiv1\ .$$
Let $\{E_j(\hbar)\}_j$ be the spectrum of $H_v(\hbar)$. Then we have for 
each $E>0$
\begin{equation}
N_{\hbar}(E):=\sharp\{j\geq1;\ E_j(\hbar)\leq E\}\sim
\frac1{\pi}\sqrt{E}\hbar^{-1}\ ,\quad \hbar\searrow0\ .
\end{equation}
Thus Condition \ref{cond;2} is satisfied with $\alpha=1$ and 
$\nu(E)=\frac1{\pi}\sqrt{E}$, which is continuous. Moreover, we can verify 
that each energy level $E_j(\hbar)$ is non-decreasing in $\hbar>0$ by the 
min-max principle. Hence both kinds of unfoldings are applicable to our 
spectrum $\{E_j(\hbar)\}$: For each $L>0$, the unfolded levels of the 
first kind are given by
\begin{equation}
x_j(L):=L\nu(E_j(L^{-1}))=\frac{L}{\pi}\sqrt{E_j(L^{-1})}\ ,
\end{equation}
while for each fixed $E>0$, the unfolded levels of the second kind are 
given by
\begin{equation}
x_j(E):=\nu(E)\hbar_j(E)^{-1}=\frac{\sqrt{E}}{\pi}\hbar_j(E)^{-1}\ ,
\end{equation}
where $\hbar_j(E)$ is defined by the equation $E_j(\hbar)=E$. 

It was shown in \cite{Mi1} that $E$ is an energy level of $H_v(\hbar)$, 
if and only if it is a zero of the function
\begin{equation}
\Gamma_{\hbar}(E):=\sum_{p=0}^n\left(\frac{\hbar\sqrt{E}}{v}\right)^{n-p}
\sum_{1\leq j_1<\cdots<j_p\leq n}\prod_{s=1}^{p+1}
\sin\left\{(x_{j_s}-x_{j_{s-1}})\frac{\sqrt{E}}{\hbar}\right\}\ .
\end{equation}
This shows on one hand that for a fixed $L>0$, $\xi>0$ is an unfolded 
level of the first kind if and only if it is a zero of the function 
$G_L(\xi):=\Gamma_{L^{-1}}\left((\pi\xi/L)^2\right)$, while on the 
other hand $\zeta>0$ is an unfolded level of the second kind if and 
only if it is a zero of the function 
$\gamma_E(\zeta):=\Gamma_{\sqrt{E}/\pi\zeta}(E)$. If we define 
\begin{equation}
f(z):=\prod_{s=1}^{n+1}\sin\left\{\pi(x_s-x_{s-1})z\right\}\ ,
\end{equation}
then we can write
\begin{equation}
G_L(\xi)=f(\xi)+g_L(\xi)\ \mbox{and}\quad 
\gamma_E(\zeta)=f(\zeta)+\delta_E(\zeta)\ ,
\end{equation}
where $g_L(\xi)$ is small for large $L$ and for $\xi={\cal O}(L)$, while 
$\delta_E(\zeta)$ is small for large values of $\zeta$. Hence by the argument 
in \cite{Mi1} using Rouch\'e's theorem, the statistics for the zeros of the 
functions $G_L(\xi)$ and $\gamma_E(\zeta)$ are both reduced to the statistics 
for the zeros of the function $f(z)$, which are
\begin{equation}
\left\{\frac{m}{x_s-x_{s-1}}\ ;\ m\geq1\ ,\ s=0,1,\ldots,n\right\}\ .
\end{equation}
Thus two kinds of unfoldings applied to the spectrum $\{E_j(\hbar)\}$ 
lead to the same level statistics. 

To proceed, consider as in \S \ref{sec;4} the counting function
\begin{equation}
N(t,t+c]:=\sum_{s=0}^n\sum_{m=1}^{\infty}1_{(t,t+c]}\left(\frac{m}{x_s-x_{s-1}}
\right)\ ,
\end{equation}
and the probability
\begin{equation}
\pi_k(L;(0,c]):=\frac1L\int_0^L1_{\{N(t,t+c]=k\}}dt
\end{equation}
for $k=0,1,\ldots$, $c>0$, and $L>0$. Then we have
\begin{equation}
M_c\leq N(t,t+c]\leq M_c+n+1\ ,
\end{equation}
where
$$M_c:=\sum_{s=1}^{n+1}[c(x_s-x_{s-1})]\ .$$
Moreover if $x_s-x_{s-1}$, $s=1,2,\ldots,n+1$, are rationally independent, 
one can 
show (\cite{Mi1}) that
\begin{eqnarray}
\pi_k(c)&:=&\lim_{L\to\infty}\pi_k(L;(0,c]) \nonumber\\
&=&\sum_{A\in{\cal F}(k-M_c)}
\left(\prod_{s\in A}\{c(x_s-x_{s-1})\}\right)
\left(\prod_{s\in A^c}(1-\{c(x_s-x_{s-1})\})\right)
\end{eqnarray}
for each $k=M_c,M_c+1,\ldots,M_c+n+1$. Here $\{a\}$ is the fractional part 
of $a$, and ${\cal F}(j)$ is the totality of subsets of $\{0,1,\ldots,n\}$ 
having cardinality $j$. In particular, if $c>0$ is small, then $M_c=0$ and 
\begin{equation}
\pi_0(c)=\prod_{s=1}^{n+1}\left(1-c(x_s-x_{s-1})\right)\ .
\end{equation}
We can now apply Proposition \ref{prop;19}, to obtain
\begin{equation}
\rho(c)=-\frac{d}{dc}\pi_0(c)=
\left(\sum_{s=1}^{n+1}\frac{x_s-x_{s-1}}{1-c(x_s-x_{s-1})}\right)
\prod_{s=1}^n\left(1-c(x_s-x_{s-1})\right)\ .
\end{equation}
For large $n$ and for typical configuration $\{x_s\}$ of the 
$\delta$-potentials, $\pi_k(c)$, $k=0,1,\ldots$, and $\rho(c)$ are different 
from, but very close to being Poissonian. In fact, if $X_1,X_2,\ldots$ are 
independent random variables which are uniformly distributed on $(0,1)$, 
and if $0<x_1^{(n)}<\cdots<x_n^{(n)}<1$ is a rearrangement of 
$X_1,\ldots,X_n$ according 
to their magnitude, then one can show (\cite{Mi1}) that with probability one,
\begin{equation}
\lim_{n\to\infty}\pi_k(c;x_1^{(n)},\ldots,x_n^{(n)})
=e^{-c}\frac{c^k}{k!}\ ,\quad 
k=0,1,\ldots\ ,
\end{equation}
and
\begin{equation}
\lim_{n\to\infty}\rho(c;x_1^{(n)},\ldots,x_n^{(n)})=e^{-c}\ .
\end{equation}

\subsubsection{Regular spectra}
\label{sec;5-4-3}
Under the assumptions on the classical Hamiltonian of Example \ref{ex;10}, 
$\nu(E)$ is a continuous function of $E$, and 
$E_{\mathbf{n}}(\hbar)=H(\hbar\mathbf{n})$ is increasing in $\hbar$ for 
each $\mathbf{n}$. Hence both kinds of unfolding are applicable to our 
regular spectrum $\{E_{\mathbf{n}}(\hbar);\mathbf{n}\in\mathbf{Z}_+^d\}$. 
To apply the unfolding of the {\em second} kind, we first note that for each 
$\mathbf{x}\in\mathbf{R}_+^d\setminus\{\mathbf{0}\}$ and $E>0$, the equation 
$H(\hbar\mathbf{x})=E$ can be solved for $\hbar$. If we denote this 
solution by $\hbar_E(\mathbf{x})$, then the unfolded energy level is 
given by $x_{\mathbf{n}}(E)=U_E(\mathbf{n})$, where we have defined 
\begin{equation}
U_E(\mathbf{x})=\nu(E)\hbar_E(\mathbf{x})^{-d}\ ,\quad 
\mathbf{x}\in\mathbf{R}_+^d\ .
\end{equation}
Since $\beta\hbar_E(\beta\mathbf{x})=\hbar_E(\mathbf{x})$ for any $\beta>0$, 
we see that $U_E(\mathbf{x})$ is a homogeneous function of degree $d$, 
namely
\begin{equation}
U_E(\beta\mathbf{x})=\beta^dU_E(\mathbf{x})\ ,
\end{equation}
for any $\beta>0$. (See \cite{BT}.) 

To perform the level statistics of \S\ref{sec;4} for this sequence 
$\{U_E(\mathbf{n})\}_{\mathbf{n}}$, which is asymptotically uniformly 
distributed, we need to consider for $I=(a,b]$
\begin{equation}
N_L(t+I)=\sum_{\mathbf{n}\in\mathbf{Z}_+^d}1_{(t+a,t+b]}(U_E(\mathbf{n}))\ .
\end{equation}
If we define 
\begin{equation}
\theta(\mathbf{x})=\frac{\mathbf{x}}{\vert\mathbf{x}\vert}\ ,\quad 
\mathbf{x}\in\mathbf{R}_+^d\setminus\{\mathbf{0}\}
\end{equation}
and
\begin{equation}
\Pi_I(t):=\{\mathbf{x}\in\mathbf{R}_+^d\vert\ 
(t+a)^{1/d}U_E(\theta(\mathbf{x}))^{-1/d}<\vert\mathbf{x}\vert\leq
(t+b)^{1/d}U_E(\theta(\mathbf{x}))^{-1/d}\}\ ,
\end{equation}
then 
\begin{equation}
N_L(t+I)=\sum_{\mathbf{n}\in\mathbf{Z}_+^d}1_{\Pi_I(t)}(\mathbf{n})\ ,
\end{equation}
which is the number of lattice points contained in the $d$-dimensional 
domain $\Pi_I(t)$. It is then easy to see 
$\vert\Pi_I(t)\vert=b-a=\vert I\vert$, so that as $t$ gets large, the 
domain $\Pi_I(t)$ expands in the space $\mathbf{R}_+^d$, but at the 
same time gets thinner and thinner to keep its volume constant. Hence 
provided the boundary of $\Pi_I(t)$ is not too regular, e.g. flat or 
spherical, then each point in $\mathbf{Z}_+^d$ would randomly belong 
to $\Pi_I(t)$ if $t$ were chosen at random from a long interval 
$[0,L]$, so that the total number of lattice points in $\Pi_I(t)$, namely 
$N_L(t+I)$, would obey Poisson distribution with mean $\vert I\vert$. 
Namely we expect that
\begin{equation}
\pi_k(I):=\lim_{L\to\infty}\pi_k(L;I)=e^{-\vert I\vert}
\frac{\vert I\vert^k}{k!}\ ,\quad k=0,1,2,\ldots
\label{eqn;a139}
\end{equation}
holds. This is conceivable if we recall how Poisson's law of small numbers 
is proved in elementary probability theory. Actually, it is a difficult 
problem to rigorously justify this intuition. In fact, no example of 
classically integrable Hamiltonian is known for which (\ref{eqn;a139}) 
is verified.

Let us now apply the unfolding of the {\em first} kind to our regular 
spectrum. This gives 
$$x_{\mathbf{n}}(L):=L\cdot \nu(E_{\mathbf{n}}(L^{-1/d}))
=L\cdot(\nu\circ H)(L^{-1/d}\mathbf{n})\ ,\quad L>0,\ 
\mathbf{n}\in\mathbf{Z}_+^d\ ,$$
as unfolded levels. To proceed further, we assume that our Hamiltonian 
$H(\mathbf{x})$ satisfies the following
\begin{cond}
For any $\mathbf{x},\mathbf{z}\in\mathbf{R}_+^d$ and $\beta>0$, 
$H(\beta\mathbf{z})\leq H(\beta\mathbf{x})$ implies 
$H(\mathbf{z})\leq H(\mathbf{x})$.
\label{cond;3}
\end{cond}

Under this condition, the function $G:=\nu\circ H$ is homogeneous with 
degree $d$. In fact for any $\beta>0$,
$$G(\beta\mathbf{x})
=\int\cdots\int_{\mathbf{R}_+^d}1_{\{H(I)\leq H(\beta\mathbf{x})\}}
dI_1\cdots dI_d
=\beta^d\int\cdots\int_{\mathbf{R}_+^d}
1_{\{H(\beta\mathbf{z})\leq H(\beta\mathbf{x})\}}dz_1\cdots dz_d
=\beta^d G(\mathbf{x})\ .$$
Thus we have $x_{\mathbf{n}}(L)=G(\mathbf{n})$, and hence replacing 
$U_E$ by $G=\nu\circ H$, we are lead to the same lattice point problem 
as before. 

If in particular, the Hamiltonian $H(\mathbf{x})$ itself is a homogeneous 
function with degree $\gamma>0$, say, then Condition \ref{cond;3} is 
automatically satisfied and for each $E>0$, we have $\nu(E)=c_HE^{d/\gamma}$ 
with 
$$c_H=\int\cdots\int_{\mathbf{R}_+^d}1_{\{H(\mathbf{x})\leq 1\}}
dx_1\cdots dx_d\ ,$$
and $\hbar_E(\mathbf{x})=E^{1/\gamma}H(\mathbf{x})^{-1/\gamma}$. 
Hence 
$$U_E(\mathbf{x})\nu(E)\hbar_E(\mathbf{x})^{-d}=c_HH(\mathbf{x})^{d/\gamma}
=\nu(H(\mathbf{x}))=G(\mathbf{x})\ .$$
Thus the unfolded levels $x_{\mathbf{n}}(E)=U_E(\mathbf{n})$ of the 
second kind does not depend on $E$, and coincide with the unfolded levels 
$x_{\mathbf{n}}(L)=G(\mathbf{n})$ of the first kind.


At this point, it is appropriate to discuss the limit theorem of Sinai 
\cite{Si2} and Major \cite{Maj} (see also \cite{Mi3}). They consider the 
case $d=2$ in which the curve
$$r=f(\theta)\ ,\quad 0\leq\theta\leq\pi/2$$
defining the boundary of the domain 
$$\Pi_I(t):=\{\mathbf{x}\in\mathbf{R}_+^2\vert\ 
\sqrt{t+a}f(\theta)<r\leq\sqrt{t+b}f(\theta)\}$$
is very random. It was then proved by Major (\cite{Maj}) that (\ref{eqn;a139}) 
actually holds for almost all realization of this random curve, and by 
the present author (\cite{Mi3}) that 
\begin{equation}
\mu_k(I):=\lim_{L\to\infty}\mu_k(L;I)=\vert I\vert^k\ ,\quad k=1,2,\ldots,
\end{equation}
also holds almost surely, which is a slightly stronger assertion than 
(\ref{eqn;a139}). Since the randomness assumed in these 
works is so strong that the typical realizations of the curve is non-smooth, 
the direct connection of these limit theorems to the question of level 
statistics is lost. But the analysis in \cite{Si2}, \cite{Maj}, and 
\cite{Mi3} suggests us that it would be possible to prove that 
$\mu_k(I)=\vert I\vert^k$, $k=1,2,\ldots,K$, should hold almost surely 
with some finite $K$ if the curve has only a finite dimensional randomness. 
This was actually done by VanderKam (\cite{Va}), the significance of 
which we shall discuss again in the next section.

\subsubsection{Schr\"odinger operator with white noise potential in the 
thermodynamic limit}
\label{sec;5-4-4}
Let $\{B_{\omega}(t);\ t\geq0\}$ be the standard Brownian motion defined 
on a probability space $(\Omega,{\cal F},\mathbf{P})$, and consider for 
each $L>0$ the Schr\"odinger operator
\begin{equation}
H_{\omega}^L:=-\frac{d^2}{dt^2}\ +\ B_{\omega}'(t)\ ,\quad 0\leq t\leq L
\end{equation}
under Dirichlet boundary conditions at $x=0,L$. Here $B_{\omega}'(t)$ 
is the formal derivative of the sample function $B_{\omega}(t)$. Although 
$B'_{\omega}(t)$ is singular, $H_{\omega}^L$ is well defined 
as a 
random self-adjoint operator acting in $L^2([0,L];dt)$, which has a purely 
discrete spectrum. Compared to previous definitions 
(e.g. \cite{FN}, \cite{Mi0}) of the operators of the form 
\begin{equation}
H=-\frac{d^2}{dt^2}\ +\ Q'(t)
\end{equation}
with a non-smooth $Q(t)$, the recent method by Savchuk and Shkalikov 
(\cite{SS}) seems to be the most ingeneous in that it yields Sturm's 
oscillation theorem through almost identical argument as in the classical 
theory (\cite{W}, \cite{Na}). 

Now let 
$$E_1^{\omega}(L)<E_2^{\omega}(L)<\cdots<E_j^{\omega}(L)<\cdots$$
be the eigenvalues of $H_{\omega}^L$. As a special case of an ergodic 
Schr\"odinger operator (see Example \ref{ex;11}), there exists a non-
decreasing function ${\cal N}(E)$ such that for almost all $\omega\in\Omega$, 
$$\sharp\{j\geq1;\ E_j^{\omega}(L)\leq E\}\sim{\cal N}(E)L\ ,\quad L\to\infty$$
holds for every $E\in\mathbf{R}$. Actually, it is known (\cite{FN}) that
\begin{equation}
{\cal N}(E)=\left\{\sqrt{2\pi}\int_0^{\infty}\exp\left(-\frac16u^3
-2Eu\right)\frac{du}{\sqrt{u}}\right\}^{-1}\ .
\end{equation}
Hence for almost all $\omega\in\Omega$, the spectrum of $H_{\omega}^L$ is 
unfoldable with $h=L^{-1}$, $\alpha=1$, and $\nu(E)={\cal N}(E)$, the latter 
being continuous in $E$. The first kind unfolding is carried through by 
letting
\begin{equation}
x_j^{\omega}(L):=L{\cal N}(E_j^{\omega}(L))\ .
\end{equation}
If we consider the statistics $\pi_k(L;I)$ defined by (\ref{eqn;a80}) for this 
family of sequences $\{x_j^{\omega}(L)\}_j$, then in view of the results 
by Molchanov (\cite{Mo}) and by McKean (\cite{Mc}) 
(see the remark below), we would expect that 
$$\lim_{L\to\infty}\pi_k(L;(0,c])=e^{-c}\frac{c^k}{k!}$$
holds almost surely for each $c>0$, or at least that 
$\{x_j^{\omega}(L);j\geq1\}$ converges in law to the Poisson point process 
on $(0,\infty)$ with mean density 1. But for the moment, this is only a 
conjecture.

\begin{rem}
McKean actually proved that as $L\to\infty$, the distribution of 
$x_1^{\omega}(L)=L{\cal N}(E_1^{\omega}(L))$ converges weakly 
to the exponential distribution 
$1_{(0,\infty)}(x)e^{-x}dx$. See \cite{GMS} for related topics.
\label{ex;7}
\end{rem}

Let $\psi_E^{\omega}(t)$ be the solution of the equation 
$H_{\omega}\psi=E\psi$ satisfying the initial condition $\psi(0)=0$, 
$\psi'(0)=1$. Then by the Sturm's oscillation theorem, $E=E_j^{\omega}(L)$ 
if and only if $t=L$ is the $j$-th zero of the function $\psi_E^{\omega}(t)$. 
Moreover $E_j^{\omega}(L)$ is increasing in $h:=L^{-1}$, and hence we can 
apply the unfolding of the second kind. For this purpose, fix an energy 
value $E\in\mathbf{R}$ and let $L_j^{\omega}(E)$ be defined by the 
equation $E_j^{\omega}(L)=E$. Then $L_j^{\omega}(E)$ has the following 
meaning. If the stochastic processes $\{r_E(t)\}$ and $\{\theta_E(t)\}$ 
are introduced through the Pr\"ufer's transformation
\begin{equation}
\psi_E(t)=r_E(t)\sin\theta_E(t)\ ;\quad
\psi'_E(t)=r_E(t)\cos\theta_E(t)\ ,
\end{equation}
then $\{\theta_E(t)\}_t$ is a diffusion process on the circle 
$\mathbf{R}/\pi\mathbf{Z}$ (\cite{Mc}, \cite{FN}) with $\theta_E(0)=0$ 
(mod $\pi$), 
and $L_j(E)$ ($j\geq1$) are the $j$-th hitting times to the origin $0$ 
of this process. Namely letting $L_0(E):=0$, we have
\begin{equation}
L_j(E):=\inf\{t>L_{j-1}(E);\ \theta_E(t)=0\ \mbox{(mod $\pi$)}\}
\end{equation}
for $j\geq1$. 
By the strong Markov property of $\{\theta_E(t)\}_t$, 
$\tau_j(E):=L_j(E)-L_{j-1}(E)$, $j=1,2,\ldots$, are independent 
and identically distributed 
random variables, and $L_j^{\omega}(E)=\sum_{k=1}^j\tau_k(E)$. Moreover one 
has $\mathbf{E}[\tau_1(E)]={\cal N}(E)^{-1}$ (\cite{FN}). Hence the 
unfolded levels of the second kind, given by
\begin{equation}
x_j=x_j(E):={\cal N}(E)L_j^{\omega}(E)\ ,\quad j=1,2,\ldots
\end{equation}
constitute a renewal process on $(0,\infty)$ with mean density 1, 
so that the corresponding level statistics is rather trivial. 
For example, one has
\begin{equation}
\lim_{L\to\infty}\rho(L;c)=\lim_{L\to\infty}\frac1L\sharp
\{j\geq1;\ x_j\leq L,\ x_{j+1}-x_j>c\}
=\mathbf{P}\left(\frac{\tau_1(E)}{\mathbf{E}[\tau_1(E)]}>c\right)
\end{equation}
with probability one. In this case, the result of unfolding of the second kind 
does depend on the fixed energy value $E$, the distribution of 
$\tau_1(E)/\mathbf{E}[\tau_1(E)]$ being dependent on $E$. As we let 
$E\to-\infty$, this distribution tends to the exponential distribution 
$1_{(0,\infty)}(x)e^{-x}dx$, as was proved by McKean (\cite{Mc}). 
Accordingly the renewal process $\{x_j(E)\}_j$ converges weakly to the 
Poisson point process on $(0,\infty)$ with mean density 1.

\section{Clustering and repulsion of levels}
\label{sec;6}
As we noted in \S\ref{sec;5-4-3}, no concrete example of regular spectrum 
is known for which Berry-Tabor conjecture is literally true in the sense 
that
\begin{equation}
\rho(c):=\lim_{L\to\infty}\rho(L;c)=e^{-c}\ \quad c>0
\label{eqn;138}
\end{equation}
holds for the unfolded spectrum. The assertion (\ref{eqn;138}) on the 
level spacing distribution function $\rho(c)$ is what Berry and Tabor 
called the \lq\lq level clustering\rq\rq. We have also seen in 
\S\ref{sec;5-4-2} that there is even an example of quantum Hamiltonian 
for which level statistics is carried out in full details, yet yielding 
a level spacing distribution function $\rho(c)$ which is close to, but 
different from $e^{-c}$. This situation suggests us that we should give 
some wider, or looser definition to the term \lq\lq level clustering\rq\rq 
than the strict assertion like (\ref{eqn;138}). To formulate this problem, 
suppose we are given a sequence $\{x_j(L)\}_j$ depending on a parameter 
$L>0$ which satisfies Condition \ref{cond;1} with $\lambda=1$, and which 
we regard as an unfolded quantal spectrum. For this sequence $\{x_j(L)\}_j$, 
let us consider statistics defined by (\ref{eqn;a80})--(\ref{eqn;a83}) with 
$I=(0,c]$. Except that we always have 
$\mu_1(c):=\lim_{L\to\infty}\mu_1(L;(0,c])=c$, none of these is guaranteed 
to converge to some limit as $L\to\infty$. Let us then define
\begin{equation}
\underline{\pi}_k(c):=\liminf_{L\to\infty}\pi_k(L;(0,c])\ ,\quad 
\overline{\pi}_k(c):=\limsup_{L\to\infty}\pi_k(L;(0,c])\ ;
\end{equation}
\begin{equation}
\underline{\mu}_k(c):=\liminf_{L\to\infty}\mu_k(L;(0,c])\ ,\quad 
\overline{\mu}_k(c):=\limsup_{L\to\infty}\mu_k(L;(0,c])\ ;
\end{equation}
\begin{equation}
\underline{\rho}(c):=\liminf_{L\to\infty}\rho(L;c)\ ,\quad 
\overline{\rho}(c):=\limsup_{L\to\infty}\rho(L;c)\ ;
\end{equation}
and
\begin{equation}
\underline{R}_k(c):=\liminf_{L\to\infty}R_k(L;c)\ ,\quad 
\overline{R}_k(c):=\limsup_{L\to\infty}R_k(L;c)\ ;
\end{equation}
When the corresponding upper and lower limits coincide, we shall denote 
their common value by $\pi_k(c)$, $\mu_k(c)$, $\rho(c)$, or $R_k(c)$ as 
before. 

Now one of the characteristic of the exponential distribution is that its 
density is strictly positive at the origin $c=0$. Hence if (\ref{eqn;138}) 
were true, we would have relatively many closely adjacent pairs of levels. 
This situation would be well described in the following way. 

\begin{definition}
We shall say that the unfolded spectrum $\{x_j(L)\}_j$ exhibits level 
clustering [resp. repulsion] in the wide sense if the condition 
\begin{equation}
\liminf_{c\searrow0}\frac{1-\overline{\rho}(c)}{c}>0\quad
\left[\mbox{resp.}\ \limsup_{c\searrow0}\frac{1-\underline{\rho}(c)}{c}=0
\right]
\label{eqn;143}
\end{equation}
holds.
\label{def;3}
\end{definition}

We can give sufficient conditions for repulsion or clustering in the wide 
sense in terms of $\mu_k(L;(0,c])$, $k=2,3$, as follows.

\begin{prop}
(i) If $\overline{\mu}_2(c)=o(c^2)$ as $c\searrow0$, then the spectrum 
exhibits level repulsion in the wide sense.

\noindent(ii) If for some $a>0$, 
$\underline{\mu}_2(c)\geq ac^2$ for all $c>0$ and if 
$\overline{\mu}_3(c)=o(c^2)$ as $c\searrow0$, then the spectrum exhibits 
level clustering in the wide sense.
\label{prop;20}
\end{prop}
\noindent{\em Proof.} Applying Lemma 2.2 of \cite{CLS} to the probability 
distribution $\{\pi_k(L;(0,c])\}_{k\geq0}$, we obtain the following equality 
which is valid for $n\geq1$ and $k\geq0$:
\begin{equation}
P_n(L;c):=\sum_{j\geq n}\pi_j(L;(0,c])=S_{n,k}(c)+(-1)^{k+1}T\ ,
\label{eqn;a155}
\end{equation}
where 
\begin{equation}
S_{n,k}(c)=\sum_{j=n}^{n+k}(-1)^{j-n}{j-1 \choose n-1}
\frac1{j!}\mu_j(L;(0,c])\ ,
\end{equation}
and $T=T_{n,k}\geq0$ is given by
\begin{equation}
T_{n,k}=\sum_{j\geq n+k+1}{j-1 \choose n-1}{j-n-1 \choose k}P_j(L;c)\ ,
\end{equation}
though this explicit expression for $T$ will not be used in the following 
argument. When $n=1$, formula (\ref{eqn;a155}) reduces to
\begin{equation}
1-\pi_0(L;(0,c])=\sum_{j=1}^{k+1}(-1)^{j-1}\frac1{j!}\mu_j(L;(0,c])
+(-1)^{k+1}T\ ,
\end{equation}
and hence
\begin{equation}
\pi_0(L;(0,c])\leq1-\mu_1(L;(0,c])+\frac1{2!}\mu_2(L;(0,c])+\cdots+
(-1)^k\frac1{k!}\mu_k(L;(0,c])
\label{eqn;a159}
\end{equation}
when $k$ is even, and
\begin{equation}
\pi_0(L;(0,c])\geq1-\mu_1(L;(0,c])+\frac1{2!}\mu_2(L;(0,c])+\cdots+
(-1)^k\frac1{k!}\mu_k(L;(0,c])
\label{eqn;a160}
\end{equation}
when $k$ is odd. (These are special cases of Theorem 2.3 of \cite{CLS}.) 

Given $L>0$, define $N$ by $x_N(L)\leq L<x_{N+1}(L)$, namely by 
\begin{equation}
N:=n(L)=\sharp\{j\geq1;x_j(L)\leq L\}\ ,
\end{equation} 
and put $x_0(L)=0$. Then we have 
\begin{eqnarray}
L\pi_0(L;(0,c])&=&\sum_{j=0}^{N-1}(x_{j+1}(L)-x_j(L)-c)_+
+\{(x_{N+1}(L)-x_N(L)-c)_+\wedge(L-x_N(L))\} \nonumber\\
&=&\sum_{j=0}^{N-1}(x_{j+1}(L)-x_j(L)-c)+
\sum_{j=0}^{N-1}(c-(x_{j+1}(L)-x_j(L)))_+ \nonumber\\
& &+\{(x_{N+1}(L)-x_N(L)-c)_+\wedge(L-x_N(L))\} \nonumber\\
& & \nonumber\\
&\leq&x_N(L)-cN+c\cdot\sharp\{j\geq0;\ x_{j+1}(L)\leq L,\ 
x_{j+1}(L)-x_j(L)\leq c\} \nonumber\\
& &+(L-x_N(L)) \nonumber\\
&=&L-c\cdot\sharp\{j\geq0;\ x_{j+1}(L)\leq L,\ 
x_{j+1}(L)-x_j(L)>c\}\ .
\label{eqn;a162}
\end{eqnarray}

Suppose that the condition of (ii) is satisfied. Then dividing bothe sides 
of (\ref{eqn;a162}) by $L$ and letting $L\to\infty$, we obtain from 
(\ref{eqn;a160})
\begin{eqnarray}
1-c\overline{\rho}(c)&\geq&\underline{\pi}_0(c)
\geq1-\mu_1(c)+\frac12\underline{\mu}_2(c)
-\frac1{3!}\overline{\mu}_3(c) \nonumber\\
&\geq&1-c+\frac a2c^2+o(c^2)\ ,
\end{eqnarray}
or
$$c(1-\overline{\rho}(c))\geq\frac{ac^2}2+o(c^2)\ .$$
Hence we get
\begin{equation}
\liminf_{c\searrow0}\frac{1-\overline{\rho}(c)}{c}\geq\frac a2>0\ .
\end{equation}

Next suppose that the condition of (i) is satisfied. For each $0<\delta<1$, 
we have 
\begin{eqnarray}
L\pi_0(L;(0,c])&\geq&x_N(L)-cN+(L-x_N(L)-c)  \nonumber\\
& &+\sum_{j=0}^{N-1}
1_{(0,\delta c]}(x_{j+1}(L)-x_j(L))\{c-(x_{j+1}(L)-x_j(L))\} \nonumber\\
&\geq&(L-c)-cN \nonumber\\
&+&(1-\delta)c
\left[N-\sharp\{j\geq0;\ x_{j+1}(L)\leq L,\ x_{j+1}(L)-x_j(L)>\delta c\}
\right]\ .
\end{eqnarray}
Again dividing by $L$ and noting $N/L\to1$ as $L\to\infty$, we obtain from 
(\ref{eqn;a159})
\begin{eqnarray}
1-c+(1-\delta)c[1-\underline{\rho}(\delta c)]&\leq&
\overline{\pi}_0(c) \nonumber\\
&\leq&1-\mu_1(c)+\frac12\overline{\mu}_2(c) \nonumber\\
&=&1-c+o(c^2)\ ,
\end{eqnarray}
and hence
\begin{equation}
\limsup_{c\searrow0}\frac{1-\underline{\rho}(c)}{c}
=\limsup_{c\searrow0}\frac{1-\underline{\rho}(\delta c)}{\delta c}=0\ ,
\end{equation}
as desired.

\begin{cor}
(i) If $\overline{R}_2(c)=o(c)$ as $c\searrow0$, then the spectrum 
exhibits level repulsion in the wide sense.

\noindent(ii) If for some $a>0$, 
$\underline{R}_2(c)\geq ac$ for all $c>0$ and if 
$\overline{R}_3(c)=o(c)$ as $c\searrow0$, then the spectrum exhibits 
level clustering in the wide sense.
\label{cor;1}
\end{cor}

\noindent{\em Proof.} By the inequality (\ref{eqn;a92}) in the proof of 
Proposition \ref{prop;18}, we see that
\begin{equation}
kc\frac1{\ell}\sum_{\ell=0}^m\underline{R}_k
\left(c\left(1-\frac{m}{\ell}\right)\right)
\leq \underline{\mu}_k(c)
\leq
\overline{\mu}_k(c)
\leq kc\frac1{\ell}\sum_{\ell=0}^m\overline{R}_k
\left(c\left(1-\frac{m}{\ell}\right)\right)
\end{equation}
holds for all integer $\ell>1$. Letting $\ell\to\infty$, we get 
\begin{equation}
k\int_0^c\underline{R}_k(c')dc
\leq \underline{\mu}_k(c)
\leq\overline{\mu}_k(c)\leq k\int_0^c\overline{R}_k(c')dc'\ .
\end{equation}
It is now obvious that the conditions of the corollary imply those 
of Proposition \ref{prop;20}.

\bigskip
If we impose conditions on $R_k(c)$, we get more precise estimates on 
$\overline{\rho}(c)$ and $\underline{\rho}(c)$.

\begin{prop}
If the limits $R_j(c)$ exist for $j=2,\ldots,k$, 
$k\geq2$, then we have
\begin{equation}
\overline{\rho}(c)\leq1+\sum_{j=1}^k(-1)^j\frac1{j!}R_{j+1}(c)
\end{equation}
when $k$ is even, and 
\begin{equation}
\underline{\rho}(c)\geq1+\sum_{j=1}^k(-1)^j\frac1{j!}R_{j+1}(c)
\end{equation}
when $k$ is odd.
\end{prop}

\noindent{\em Proof.} Let $n(L):=\sum_{j}1_{(0,L]}(x_j(L))$ and define the 
probability distribution $\{\rho_k(L;c)\}_{k\geq0}$ by
\begin{equation}
\rho_k(L;c):=\frac1{n(L)}\sum_j1_{(0,L]}(x_j(L))
1_{\{N_L(x_j(L),x_j(L)+c]=k\}}\ .
\end{equation}
Then 
\begin{equation}
\rho_0(L;c)=\frac{n(L)}{L}\rho(L;c)\ .
\end{equation}
Note that $n(L)\sim L$, and hence the upper and lower limits of 
$\rho_0(L;c)$ as $L\to\infty$ coincide with 
$\overline{\rho}(c)$ and $\underline{\rho}(c)$ respectively. 
If we let 
\begin{equation}
r_k(L;c):=\frac{L}{n(L)}R_k(L;c)
\end{equation}
for $k\geq2$, then it is not difficult to see
\begin{equation}
r_k(L;c)=\frac1{n(L)}\sum_j1_{(0,L]}(x_j(L))
\left(N_L(x_j(L),x_j(L)+c]\right)^{[k-1]}
=\sum_{i\geq k-1}i^{[k-1]}\rho_i(L;c)\ .
\end{equation}
Namely $r_k(L;c)$ is the $(k-1)$-th factorial moment of the probability 
distribution $\{\rho_i(L;c)\}_{i\geq0}$. Thus again by Lemma 2.2 of 
\cite{CLS}, we can write
\begin{equation}
Q_n(L;c):=\sum_{j\geq n}\rho_i(L;c)=\sigma_{n,k}(c)+(-1)^{k+1}\tau\ ,
\end{equation}
where $\tau\geq0$ and 
\begin{equation}
\sigma_{n,k}(c)=\sum_{j=n}^{n+k}(-1)^{j-n}{j-1 \choose n-1}\frac1{j!}
r_{j+1}(L;c)\ .
\end{equation}
When $n=1$, this reduces to 
\begin{equation}
1-\rho_0(L;c)=\sum_{j=1}^k(-1)^{j-1}\frac1{j!}r_{j+1}(L;c)+(-1)^k\tau\ ,
\end{equation}
which is valid for $k\geq1$. The assertion of the proposition immediately 
follows from this equality.

\begin{ex}
In \cite{Va}, VanderKam considered the level statistics for the Laplacian 
on a flat $d$-dimensional torus, which is the quantization of the uniform 
motion on that torus. The corresponding Hamiltonian, expressed in the 
action variables, is a positive definite quadratic form
\begin{equation}
P(x_1,\ldots,x_d)=\ ^t\mathbf{x}A\mathbf{x}\ ,
\end{equation}
where $A$ is a $d\times d$ real symmetric, positive definite matrix, and 
the eigenvalues of the Laplacian are values taken by $P$ on the lattice 
$\hbar\mathbf{Z}^d$. To eliminate the obvious degeneracy, $\mathbf{x}$ 
and $-\mathbf{x}$ are identified. Let us denote this equivalence 
relation by $\sim$. Then the spectrum 
$\{P(\hbar\mathbf{n});\mathbf{n}\in\mathbf{Z}^d/\sim\}$ is unfoldable with 
$\alpha=d$ and $\nu(E)=c_PE^{d/2}$, where $c_P$ is the volume of the 
domain $\{\mathbf{x}\in\mathbf{R}^d;P(\mathbf{x})\leq1\}/\sim$. 
Since $P(\mathbf{x})$ is a homogeneous function of degree 2, both kinds 
of unfoldings discussed in \S \ref{sec;5} give the same unfolded sequence 
$\{\xi_{\mathbf{n}};\mathbf{n}\in\mathbf{Z}^d\}$, where
\begin{equation}
\xi_{\mathbf{n}}:=c_P P(\mathbf{n})^{d/2}\ .
\label{eqn;a180}
\end{equation}
Then VanderKam proves that for almost all realization of the coefficient 
matrix $A$ of $P(\mathbf{x})$, one has $R_k(c)=c^{k-1}$ for 
$k=2,\ldots,[d/2]$, for the correlation functions of the unfolded spectrum 
(\ref{eqn;a180}). Hence if $d\geq6$, Corollary \ref{cor;1} tells us that 
$\{\xi_{\mathbf{n}}\}$ exhibits level clustering in the wide sense. Moreover, 
by Proposition \ref{prop;20}, we can assert that when $d$ is large and 
$c>0$ is small, $\overline{\rho}(c)$ and $\underline{\rho}(c)$ are 
very close to each other, and are approximated by $e^{-c}$.
\end{ex}

\noindent
{\em Acknowledgements.} The author is grateful to Professor V. Jaksic for 
giving him the opportunity to make this contribution, and to Dr. E. Giere 
for careful reading of the manuscript and useful comments.

\begin{thebibliography}{3}

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