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\begin{document}
\title[Resonances in One Dimension and Fredholm Determinants]{Resonances in One Dimension \\ 
and Fredholm Determinants}
\author[B. Simon]{Barry Simon}
\address{Division of Physics, Mathematics, and Astronomy 253-37 \\
California Institute of Technology \\
Pasadena, CA 91125-3700}
\email{bsimon@caltech.edu}

\date{July 31, 2000}



\thanks{This material is based upon work supported by the National 
Science Foundation under Grant No.~DMS-9707661. The Government has 
certain rights in this material.}

\begin{abstract} We discuss resonances for Schr\"odinger operators in 
whole- and half-line problems. One of our goals is to connect the Fredholm 
determinant approach of Froese to the Fourier transform approach of 
Zworski. Another is to prove a result on the number of antibound states 
--- namely, in a half-line problem there are an odd number of antibound 
states between any two bound states.
\end{abstract}

\maketitle

\section{Introduction} \lb{s1}

\medskip
In this paper, we want to discuss resonances and antibound states in one 
dimension for Schr\"odinger operators $H=H_0 + V$, where $H_0$ is one of the 
following:

\smallskip
{\bf Case 1:}  $-\f{d^2}{dx^2}$ on $L^2(\bbR)$

\smallskip
{\bf Case 2:} $-\f{d^2}{dx^2}$ with $u(0)=0$ boundary conditions on $L^2(0,\infty)$

\smallskip
{\bf Case 3:} $-\f{d^2}{dx^2}$ with $u'(0) + hu(0)=0$ boundary condition on 
$L^2 (0,\infty)$

\smallskip
{\bf Case 4:} $-\f{d^2}{dx^2} + \f{\ell(\ell + 1)}{x^2}$ on $L^2 (0,\infty)$; $\ell=1,2,
\dots$.

\smallskip
We will often consider Case 2 as the $\ell=0$ case of Case 4. We will normally suppose 
$V$ has compact support, although many of our results only require
\begin{equation} \lb{1.1}
\int e^{a|x|} |V(x)|\, dx < \infty
\end{equation}
for all $a >0$ (and some only that \eqref{1.1} hold for suitable $a>0$).

The operator $H_0$ has spectrum $[0,\infty)$ except for Case~3 with $h<0$ which has 
a single eigenvalue at energy $-h^2$. That means that $(H_0 + \kappa^2)^{-1}$ is a 
well-defined operator in the region $\Real \kappa >0$ (except for a pole at $\kappa 
=-h$ in Case~3 with $h<0$). Its integral kernel
\[
(H_0 + \kappa^2)^{-1} (x,y) \equiv G_0 (x,y; \kappa)
\]
has an explicit formula in terms of exponential functions in Cases~1--3 and Bessel 
functions in Case~4. For example
\begin{alignat}{2}
G_0 (x,y; \kappa) &= e^{-\kappa|x-y|}\big/ 2\kappa \quad && \text{(Case 1)} \lb{1.2} \\
G_0 (x,y; \kappa) &= e^{-\kappa x_>} \sinh (\kappa x_<) \big/ \kappa 
\quad &&\text{(Case 2)} \lb{1.3} 
\end{alignat}
where $x_< = \min(x,y)$ and $x_> = \max (x,y)$.

From these explicit formulae, $G_0 (x,y; \kappa)$ as a function of $\kappa$ has an analytic 
continuation to the entire $\kappa$ plane except for a simple pole at $\kappa =0$ in Case~1 
and at $\kappa =-h$, in Case~3 with $h\geq 0$.

By the explicit formula for $G_0$, it can be seen that when \eqref{1.1} holds for 
all $a>0$:
\begin{equation} \lb{1.4}
K(\kappa)(x,y) = V(x)^{1/2}  G_0(x,y; \kappa)|V(y)|^{1/2}
\end{equation}
is in $L^2 (\bbR\times\bbR, dx\,dy)$ for all $\kappa$ (except for the simple pole in 
Cases~1 and 3) and so defines an analytic Hilbert-Schmidt operator-valued function 
on $\bbC$ (or $\bbC\backslash \{0\}$, or $\bbC\backslash\{-h\}$). In \eqref{1.4}, 
$V(x)^{1/2}$ is short for 
\begin{alignat*}{2}
V(x)^{1/2} &= V(x)\big/ |V(x)|^{1/2} \quad && \text{if } V(x)\neq 0 \\
&= 0 \quad && \text{if } V(x) =0.
\end{alignat*}
$K(\kappa)$ is called the Birman-Schwinger kernel.

It is well known (see, e.g., \cite{Cyc,RS} that
\smallskip

\noindent{\bf Birman-Schwinger Principle.} $H_0 + V$ has $-\kappa^2$ as an eigenvalue 
(with $\Real \kappa>0$) if and only if $K(\kappa)$ has eigenvalue $-1$.

\smallskip
With this in mind, one defines

\smallskip
\noindent{\bf Definition.} $\kappa$ is called a resonance energy if $\Real \kappa<0$ 
and $K(\kappa)$ has eigenvalue $-1$. The multiplicity of the resonance is the algebraic 
multiplicity of the eigenvalue. If $\Ima \kappa=0$ as well, we call $\kappa$ an 
antibound state.

\smallskip
Of course, $-\kappa^2$ is really the energy, but since $\kappa$ is the natural 
parameter, we will abuse terminology. This definition is the one of Froese \cite{Fr1,Fr2}. 
Melrose and Zworski and their school in numerous papers (e.g., \cite{Zw1,Zw2}) have defined 
resonance in terms of poles of suitable elements of the analytically contained 
$S$-matrix. It follows from our results below (Proposition~\ref{p2.9} and \ref{2.10}) 
that the definitions agree and are equivalent to the solution of $-u'' + Vu = 
-\kappa^2 u$ with $u(x) = e^{-\kappa x}$ (a suitable Hankel function in Case~4) near 
$+\infty$ obeys the boundary condition at zero in Cases~2 and 3, is $e^{+\kappa x}$ 
near $-\infty$ in Case~1 and decreases as $x\downarrow 0$ in Case~4.

While one has obviously that $K(\kappa)$ is Hilbert-Schmidt, the following is true:

\begin{theorem}\lb{t1} $K(\kappa)$ is trace class for all $\kappa$ {\rm(}except for the 
poles in Cases~1 and 3{\rm)}.
\end{theorem}

This is a result of Froese \cite{Fr1} for Cases~1 and 2, but it seems worthwhile to give an 
alternate proof that also works in Case~4 (Case~3 is an easy consequence of Case~2). 
We do this in Section~\ref{s2}. Once one has Theorem~\ref{t1}, it is natural to define
\begin{equation} \lb{1.5}
d(\kappa) = \det(1+K(\kappa))
\end{equation}
in which case the resonances or bound states are precisely the zeros of $d(\kappa)$ 
(or $\kappa d(\kappa)$ in Case~1 or $(\kappa +h)\, d(\kappa)$ in Case~3). Determinants 
have also been used by Melrose \cite{Melr} and Zworski \cite{Zwo} in their work on 
resonances.

Our main goals in this paper are the following:
\begin{enumerate}
\item[(1)] Zworski \cite{Zw1} proved his theorem on counting resonances by realizing the 
inverse of a suitable $S$-matrix element as a Laplace transform. Froese's alternate 
proof directly analyzes the asymptotics of $d(\kappa)$. In this paper, we link the 
two methods by showing the Fredholm determinant can be realized as a Laplace transform. 
This provides an alternate to using Melin's theory \cite{Mel}.

\smallskip
\item[(2)] We want to present a new result on counting antibound states. During the 
final preparation of this manuscript, I received a preprint of Kargaev-Korotyaev 
\cite{KK} who independently found this result.

\end{enumerate}

\smallskip
To be more specific, in Section~\ref{s2}, we will prove expansion formulae for the 
Fredholm determinants $d(\kappa)$ in Cases~1 and 2 (this part also works for Cases~3 and 
4, but we do not state these results explicitly since we will only handle Cases~1 and 
2 in Section~\ref{s3}).

\begin{theorem} \lb{t2} In Case~2 {\rm(}half-line with $u(0)=0$ boundary conditions{\rm)}:
\begin{equation} \lb{1.6}
d(\kappa) = 1 + \sum_{n=1}^\infty d_n(\kappa)
\end{equation}
with
\begin{equation} \lb{1.7}
\begin{split}
&d_n(\kappa) = \\
&\quad \int_{x_0 \equiv 0<x_1 < \cdots < x_n} V(x_1) \dots V(x_n) 
\prod_{j=1}^n \f{\sinh(\kappa(x_j - x_{j-1}))}{\kappa}\, e^{-\kappa x_n}\, 
dx_1 \dots dx_n.
\end{split}
\end{equation}
\end{theorem}

\begin{theorem}\lb{t3} In Case~1 {\rm(}$H_0 \equiv -\f{d^2}{dx^2}$ on $L^2 (\bbR)${\rm)}:
\begin{equation} \lb{1.8}
d (\kappa) = 1 + \sum_{n=1}^\infty d_n(\kappa)
\end{equation}
with
\begin{equation} \lb{1.9}
\begin{split}
d_n(\kappa) = &\int_{x_1 <\cdots < x_n} V(x_1) \dots V(x_n) \\
&\qquad  \bigg[ \prod_{j=2}^n \f{\sinh (\kappa (x_j - x_{j-1}))}{\kappa}\bigg]\, 
\f{e^{-\kappa (x_n - x_1)}}{2\kappa}\, 
dx_1 \dots dx_n.
\end{split}
\end{equation}
\end{theorem}

\noindent{\it Notes.} 1. The product in \eqref{1.9} has $n-1$ terms and is empty in 
case $n=1$ so that
\begin{equation} \lb{1.9a}
d_1 (k) = \f{1}{2\kappa} \int V(x)\, dx.
\end{equation}

2. \eqref{1.9} implies that if we look at the problem of $\lambda V$ with a coupling 
constant $\lambda$ added
\[
\kappa d(\kappa) = \kappa + \lambda \int V(x)\, dx + \lambda^2 F(\lambda, \kappa)
\]
with $F$ analytic near $\lambda = \kappa =0$. It follows that if $\int V(x)\, dx < 0$, 
there is a bound state for small $\lambda$ with energy $(-\kappa^2)$ given by $-\lambda^2 
(\int V(x)\, dx)^2 + O(\lambda^3)$ and an antibound state at that energy if $\int V(x)\, dx >0$. 
This recovers old results of Landau-Lifshitz \cite[Section 45]{LL} (see also Simon \cite{Si1}).

Once we have these expansions, we can use 
\[
\f{\sinh \kappa y}{\kappa} = \int_0^y e^{-2\kappa \alpha}\, e^{\kappa y}\, d\alpha
\]
to obtain the following Laplace transform representations:

\begin{theorem}\lb{t4} In Case~2, let $a=\sup(\supp(V))$. Then
\[
d(\kappa) = 1 + \int_0^a t(\alpha) \, e^{-2\alpha\kappa}\, d\alpha
\]
for a suitable $L^1$-function $t$ on $[0,a]$ with $a\in\supp (t)$.
\end{theorem}

\begin{theorem} \lb{t5} In Case~1, let $[a,b]$ be the convex hull of the support of $V$. Then
\[
\kappa d(k) = \kappa + \tfrac12 \int_a^b V(x)\, dx + \tfrac12 \int_0^{b-a} t(\alpha) \, 
e^{-2\alpha\kappa}\, d\alpha,
\]
where $b-a\in\supp(t)$.
\end{theorem}

Following Zworski, these Laplace transform formulae allow one to use Titchmarsh's 
theorem \cite{Tit} to obtain the result of Regge \cite{Reg} and Zworski \cite{Zw1} on 
the density of resonances.

While we will directly write down these Laplace transforms, we could use a less direct 
result. As we will show in Section~\ref{s2}, the expansion in Theorem~2 lets us identify 
$d(\kappa)$ with a Jost function, the value of the Jost solution at $x=0$ (this is not a 
new result; it is due to Jost-Pais \cite{JP}). But Levin \cite{Lev} has a Laplace transform 
formula for the Jost solution as used extensively by Marchenko \cite{Mar} and that allows 
one to prove Theorem~4 from Theorem~2 (the estimate we use to show $a\in\supp(t)$ is in 
Marchenko's book).

\smallskip
Our final results are on a different subject
\begin{theorem} \lb{t6} In a half-line problem {\rm(}Cases~2, 3 or 4{\rm)}, suppose $h$ has 
$n$ bound states $0<\kappa_1 < \cdots < \kappa_n$. Then each interval $(-\kappa_{j+1}, 
-\kappa_j)$ has an odd number of antibound states and, in particular, at least one antibound 
state. In particular, there are at least $(n-1)$ antibound states.
\end{theorem}

We will prove Theorems~1, 2, and 3 in Section~\ref{s2}, Theorems~4 and 5 in 
Section~\ref{s3} and Theorem~6 in Section~\ref{s4}. An appendix has a result on 
finite determinants that we need in Section~\ref{s2}.

\medskip
I would like to thank F.~Gesteszy, M.~Hitrik, K.~Makarov, and A.~Pushnitski for 
useful discussions.



\bigskip
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Expansion of the Fredholm Determinant} \lb{s2}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\medskip
In this section, we want to first prove that $K(\kappa)$ is trace class for all 
$\kappa\in\bbC$ (or $\bbC\backslash\{0\}$), that is, prove Theorem~\ref{t1}. Then 
we present an expansion of the determinant, $\det(1+K(\kappa))$, in the $\ell=0$ case 
(Theorems~2 and 3).

We begin with the whole-line case, so
\[
K(\kappa)(x,y) \equiv V(x)^{1/2} \, \f{e^{-\kappa|x-y|}}{2\kappa}\, |V(y)|^{1/2}.
\]
Since $V$ has compact support, $K(\kappa)(x,y)$ is $L^2$ in $\bbR\times\bbR$ for all 
$\kappa\in\bbC\backslash\{0\}$ and the kernel is analytic. Thus,

\begin{proposition} \lb{p2.1} $K(\kappa)$ is Hilbert-Schmidt for all $\kappa\in\bbC
\backslash \{0\}$ and analytic in $\kappa$.
\end{proposition}

For $\Real \kappa >0$, we can write
\[
K(\kappa) = V^{1/2}(x) (p^2 + \kappa^2)^{-1} |V(x)|^{1/2}.
\]
It is a basic fact \cite[Theorem~4.1]{Si2} that if $f,g\in L^2(\bbR)$, then $f(p)g(x)$ is 
Hilbert-Schmidt with
\begin{equation} \lb{2.1}
\| f(p) g(x)\|_2 \leq (2\pi)^{-1/2} \|f\|_2 \|g\|_2.
\end{equation}
It follows immediately that

\begin{proposition} \lb{p2.2} If $g,h\in L^2 (\bbR)$ and $f\in L^1 (\bbR)$, then 
$g(x) f(p) h(x)$ is trace class and
\[
\| g(x) f(p) h(x)\|_1 \leq (2\pi)^{-1} \|g\|_2 \|h\|_2 \|f\|_1.
\]
\end{proposition}

Note next that if $|\Arg\kappa|\in (\f{\pi}{4}, \f{\pi}{2})$,
\[
\int \f{d\rho}{|p^2 + \kappa^2|} \leq \f{c}{|\Ima \kappa|} \, \big[ 1+\log 
\big(|\Ima \kappa |\big/ \Real\kappa\big)\big],
\]
so we have

\begin{proposition} \lb{p2.3} For $|\Arg\kappa | \in (\f{\pi}{4}, \f{\pi}{2})$, 
$K(\kappa)$ is trace class with
\begin{equation} \lb{2.2}
\| K(\kappa)\|_1 \leq c\|V\|_1 \bigg[ \f{1}{|\Ima\kappa|} \, \big[ 1 + \log 
\big( |\Ima\kappa| \big/ \Real\kappa\big)\big]\bigg].
\end{equation}
\end{proposition}

Next, we note that
\begin{align*}
(K(\kappa) - K(-\kappa))(x,y) &= V^{1/2}(x) \, \f{2\cosh (\kappa|x-y|)}{\kappa}\, 
V(y)^{1/2} \\
&= V^{1/2}(x)\, \f{2\cosh (\kappa(x-y))}{\kappa}\, |V(y)|^{1/2}
\end{align*}
is a rank 2 operator. It follows that

\begin{proposition} \lb{p2.4} If $\Real(\kappa) > 0$, $K(\kappa) - K(-\kappa)$ is 
trace class with
\[
\| K(\kappa) - K(-\kappa)\|_1 \leq 2 \int |V(x)|\, e^{2(\Real\kappa) |x|}\, dx.
\]
\end{proposition}

As a result
\begin{proposition} \lb{p2.5} If $\Real(\kappa) < 0$, $K(\kappa)$ is trace class with 
\begin{equation} \lb{2.3}
\| K(\kappa)\|_1 \leq C (1 + \log (|\Ima\kappa|)),
\end{equation}
where $C$ is a constant bounded on each half-annulus $\{ \kappa \mid \Real\kappa < 0, \ 0 < 
A < |\kappa| < B\}$.
\end{proposition}

Now given $\kappa_0$ with $\kappa_0 = i\alpha$, $\alpha\in\bbR\backslash\{0\}$, write
\[
K(\kappa_0) = \int_{-\pi}^\pi K\bigg(\kappa_0 +\f{\alpha}{2}\, e^{i\theta}\bigg)\, 
\f{d\theta}{2\pi}\, .
\]
Since $\|K(\kappa_0 + \f{\alpha}{2}\, e^{i\theta}\|_1$ has only logarithmic singularities 
at $\theta = \pm \f{\pi}{2}$, we conclude

\begin{proposition} \lb{p2.6} $K(\kappa)$ is trace class on $\bbC\backslash\{0\}$. 
\end{proposition}
This is just Theorem~\ref{t1} in this case.

\smallskip
The proof of Theorem~\ref{t1} in the $\ell(\ell + 1)/r^2$ case depends on an eigenfunction 
expansion for $h^{(0)}_\ell$. Let
\[
u_\ell (k,x) = kx\, j_\ell(kr),
\]
where $j_\ell$ is a spherical Bessel function. Then \cite{Am},
\begin{equation} \lb{2.4}
(F_\ell \varphi)(k) = \sqrt{\tfrac{2}{\pi}} \int_0^\infty u_\ell (k,x) 
\varphi(x)\, dx
\end{equation}
is a unitary map of $L^2 (\bbR, dx)$ to $L^2 (\bbR, dk)$, that is,
\begin{equation} \lb{2.5}
\int_0^\infty |(F_\ell\varphi)(k)|^2 \, dk = \int_0^\infty |\varphi(x)|^2\, dx
\end{equation}
and
\begin{equation} \lb{2.6}
(F_\ell h^{(0)}_\ell \varphi)(k) = k^2 (F_\ell \varphi)(k).
\end{equation}

While the usual proof \cite{Si2} that $\|f(p) g(x)\|_2 = (2\pi)^{-1/2} \|f\|_2 \|g\|_2$ 
comes from writing $f(p)$ as a convolution operator, here is another proof. Let 
$V$ be the Fourier transform and let $M_f$ be multiplications by $f(x)$. Then
\[
f(p) g(x) = V^{-1} M_f VM_g
\]
so since $V^{-1}$ is unitary $\|f(p) g(x)\|_2 = \|M_f V M_g\|$ but $M_f V M_g$ has 
integral kernel:
\[
(M_f V M_g)(p,x) = f(p)\, \f{e^{-ipx}}{\sqrt{2\pi}}\, g(x)
\]
from which the Hilbert-Schmidt norm formula is immediate. Similarly,

\begin{proposition} \lb{p2.7} If $f,g \in L^2 (0,\infty)$, then $f(\sqrt{h^{(0)}_\ell}\,) 
g(x)$ is Hilbert-Schmidt and
\begin{equation} \lb{2.7}
\bigg\| f\bigg(\sqrt{h^{(0)}_\ell}\,\bigg) g(x)\bigg\|_2 \leq C \|f\|_2 \|g\|_2.
\end{equation}
\end{proposition}

\begin{proof} As above,
\[
f\bigg(\sqrt{h^0_\ell}\,\bigg) g(x) = F^{-1}_\ell M_f F_\ell M_g,
\]
so since $F_\ell$ is unitary \eqref{2.5}, we have
\[
\bigg\| f\bigg(\sqrt{h^{(0)}_\ell}\, \bigg) g(x)\bigg\|_2 = \|M_f F_\ell M_g\|
\]
and the latter has integral kernel
\[
f(k) \sqrt{\tfrac{2}{\pi}}\, u_\ell (k,x) g(x)
\]
so \eqref{2.7} follows from $|u(k,x)| \leq C_1$ (see \cite{Am}). 
\end{proof}

With \eqref{2.7} in hand, the proof of Theorem~\ref{t1} is essentially identical 
to the proof in the whole-line case. The $\ell=0$ situation is identical to the 
$u(0)=0$ boundary conditions for the half-line $-\f{d^2}{dx^2}$ case and then the 
general $h$ bound condition situation follows from the fact that the difference 
of the resolvents is rank 1 when $h$ changes. That completes the proof of 
Theorem~\ref{t1}.

Once we know $K(\kappa)$ is trace class, we can form $d(\kappa)=\det(1 + K(\kappa))$. 
We turn to the expansion of the Fredholm determinant. We will do this for a general 
trace class operator, $A$, with an integral kernel $A(x,y)$ on $\bbR$ (or $[0,\infty)$) 
of the form
\begin{equation} \lb{2.8}
A(x,y) = V(x)^{1/2} G_0(x,y) |V(y)|^{1/2},
\end{equation}
where $G_0$ has the form
\begin{equation} \lb{2.9}
G_0 (x,y) = f_- (x_<) f_+(x_>),
\end{equation}
with $x_< = \min(x,y)$, $x_> = \max(x,y)$. We will show that

\begin{proposition} \lb{p2.8} Let $A$ be a trace class operator of the form 
\eqref{2.8}/\eqref{2.9}. Then
\begin{equation} \lb{2.10}
\begin{split}
\det(1+A) = &1 + \sum_{n=1}^\infty \int_{x_1 < \cdots < x_n} V(x_1) \dots V(x_n) 
f_-(x_1) f_+(x_n) \\
&\qquad \prod_{j=1}^{n-1} [ f_+(x_{j+1}) f_-(x_j) - f_+ (x_j) f_-(x_{j+1})]\, 
dx_1 \dots dx_n.
\end{split}
\end{equation}
\end{proposition}

\smallskip
\noindent{\it Remarks.} 1. We are vague about convergence issues since in the 
applications we make, they are trivial. In general, it certainly suffices that 
$\int |V(x)|\, [|f_+(x)| + |f_-(x)| + 1]^2\, dx < \infty$.

\smallskip
2. Theorems~2 and 3 are immediate corollaries of Proposition~\ref{p2.8} given that 
\[
e^{-\kappa x}(2\kappa)^{-1}\, e^{+\kappa y} - e^{-\kappa y} (2\kappa)^{-1}\, 
e^{+\kappa x} = \kappa^{-1} \sinh (\kappa (y-x))
\]
and
\[
e^{-\kappa x} \kappa^{-1}\sinh(\kappa y) - e^{-\kappa y} \kappa^{-1}\sinh(\kappa x) 
= \kappa^{-1}\sinh (\kappa(y-x)).
\]

\smallskip
3. While we are interested in the cases given by Theorems~2 and 3 because of 
the application in the next section, the proposition applies directly to the 
$\ell(\ell+1)/x^2$ example; $f_-$ is a Bessel function (of imaginary argument) 
and $f_+$ is a suitable Hankel function.

\smallskip
4. Jost-Pais \cite{JP} have a related result.

\smallskip
\begin{proof} As shown in \cite{Si2} (and essentially due to Fredholm)
\begin{equation} \lb{2.11}
\det(1+A) = 1 + \sum_{n=1}^\infty \f{1}{n!} \int A\binom{x_1\dots x_n}{x_1\dots x_n}\, 
dx_1 \dots dx_n,
\end{equation}
where
\[
A\binom{x_1 \dots x_n}{y_1\dots y_n} = \det ([A(x_i, y_j)]_{i,j=1,\dots, n}).
\]
Since $A\binom{x_1\dots x_n}{x_1\dots x_n}$ is symmetric in the $x$'s, we can remove 
the $n!$ and integrate over $x_1 < \cdots <x_n$. In that case,
\[
A\binom{x_1 \dots x_n}{x_1 \dots x_n} = V(x_1) \dots V(x_n) \det (f_- (x_{\min(i,j)}) 
f_+ (x_{\max(i,j)})).
\]
Determinants of this form are discussed in the appendix where it is shown that
\begin{equation*}
\begin{split}
\det (f_- (x_{\min(i,j)}) &f_+(x_{\max(i,j)})) \\
&= f_+ (x_n) f_-(x_j) \prod_{j=1}^{n-1} [f_+ (x_{j+1}) f_- (x_j) - f_+ (x_j) 
f_- (x_{j+1})],
\end{split}
\end{equation*}
which proves the proposition. 
\end{proof}

We close this section by using the expansions in Theorem~2 and 3 to identify 
$d(\kappa)$ with quantities related to the Jost function. Consider first the 
half-line case. Let $d(\kappa, x_0)$ be the $d(\kappa)$-function for the 
potential $\{V(x+x_0)\}_{x\geq 0}$. Then, with
\begin{equation} \lb{2.12}
F(x,y; \kappa) = \f{\sinh (\kappa(y-x))}{\kappa}\, ,
\end{equation}
we have that
\begin{align*}
d(\kappa, & x_o) \\
& = 1 + \sum_{n=1}^\infty \, \int_{0<x_1 < \cdots < x_n} 
F(0, x_1; \kappa) V(x_0 + x_1) F(x_1, x_i; \kappa) \dots V(x_0 + x_n) \\
&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad e^{-\kappa x_n}\, dx_1 \dots dx_n \\
&= 1 + e^{\kappa x_0} \sum_{n=1}^\infty \, \int_{x_0 < x_1 < \cdots < x_n} 
F(x_0, x_1; \kappa) V(x_1) F(x_1, x_n) \dots V(x_n) \\
&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad e^{-\kappa x_n} \, dx_1 \dots dx_n
\end{align*}
by a change of variables. That means if we define
\begin{equation} \lb{2.13}
f(x,\kappa) = e^{-\kappa x}\, d(\kappa, x),
\end{equation}
then $f$ obeys
\begin{equation} \lb{2.14}
\begin{split}
f(x,\kappa) = & e^{-\kappa x} + \sum_{n=1}^\infty \, \int_{x < x_1 < \cdots <x_n} 
F(x, x_1; \kappa) V(x_1) F(x_1, x_2; \kappa) \dots V(x_n) \\
&\qquad \qquad \qquad \qquad \qquad \qquad e^{-\kappa x_n}\, 
dx_1 \dots dx_n,
\end{split}
\end{equation}
so that $f$ obeys the integral equation
\begin{equation} \lb{2.15}
f(x, \kappa) = e^{-\kappa x} + \int_x^\infty F(x,y; \kappa) V(y) f(y,\kappa)\, dy,
\end{equation}
which implies that $f$ obeys
\begin{equation} \lb{2.16}
-f'' + Vf = -\kappa^2 f
\end{equation}
with the boundary condition
\begin{equation} \lb{2.17}
f(x,\kappa) = e^{-\kappa x} \qquad\text{if } x \geq a = \sup[\supp V].
\end{equation}
Thus,

\begin{proposition} \lb{p2.9} $f(x,\kappa)$ is the Jost solution, that is, solution of 
\eqref{2.16} which obeys \eqref{2.17}. In particular, $d(\kappa)$ is the Jost function, 
that is, $f(x=0,\kappa)$.
\end{proposition}

\noindent{\it Remark.} It is a known result that the Jost function is a Fredholm 
determinant. See Jost-Pais \cite{JP} and Simon \cite{Si2}. Our proof is related to that 
of Jost-Pais. Similarly, Proposition~\ref{p2.10} below is known; it follows by 
combining formulae in Newton \cite{Ne}.

\smallskip
There is a related expression for the whole-line case. First, we need some notation. Let 
$[a,b]$ be the convex hull of the support of $V$. The Jost solutions $f_\pm (x,\kappa)$ 
are the solutions of \eqref{2.16} that obey the boundary conditions
\begin{equation} \lb{2.17a}
f_\pm (x,\kappa) = e^{\mp\kappa x} \qquad\text{for } \pm x \geq |a| + |b|.
\end{equation}
Given two $C^1$ functions $f,g$, we define their Wronskian by
\begin{equation} \lb{2.17b}
W(f,g)(x) = f' (x) g(x) - f(x) g'(x).
\end{equation}
As usual, if $f,g$ obey the same second-order differential equation, $W$ is constant and 
we denote its value as $W(f,g)$. Finally, we define the free Jost solutions
\[
f^{(0)}_\pm (x,\kappa) = e^{\mp\kappa x}.
\]

\begin{proposition} \lb{p2.10} In the whole-line case
\begin{equation} \lb{2.18}
\kappa d(\kappa) = \tfrac12\, W (f_- (\dott, \kappa), f_+ (\dott, \kappa)).
\end{equation}
\end{proposition}

\begin{proof} By looking at the expansion \eqref{2.14} of the Jost solution $f_+(x,\kappa)$, 
and the expansion of $d(\lambda)$ given by Theorem~3, we see that
\begin{equation} \lb{2.19}
d(\kappa) = 1 + \f{1}{2\kappa} \int_a^b e^{\kappa x} V(x) f_+(x,\kappa)\, dx
\end{equation}
so
\begin{equation} \lb{2.20}
\kappa d(\kappa) = \kappa + \tfrac12 \int_a^b f^{(0)}_- (x,\kappa) V(x) f_+ (x,\kappa)\, dx.
\end{equation}
But since $f_+ = f^{(0)}_+$ for $x > b$,
\[
\kappa = \tfrac12\, W(f^{(0)}_-, f_+)(b)
\]
and
\[
f^{(0)}_- (x,\kappa) V(x) f_+ (x,\kappa) = -\f{d}{dx}\, W(f^{(0)}_-, f_+),
\]
so \eqref{2.20} shows that
\[
\kappa d(\kappa) = \tfrac12\, W(f^{(0)}_-, f_+) (a).
\]
But for $x\leq a$, $f^{(0)}_- = f_-$, so
\[
\kappa d(\kappa) = \tfrac12\, W(f_-, f_+) (a) = \tfrac12 \, W(f_-, f_+)(x)
\]
for all $x$. 
\end{proof}

\smallskip
\noindent{\it Remarks.} 1. Proposition~\ref{p2.9} can be rephrased in a form close to 
\eqref{2.18}. Namely, if $u(x,\kappa)$ is the solution of \eqref{2.16} obeying 
$u(0,\kappa) = 0$, $u'(0,\kappa) = 1$, then $d(\kappa) = W(u,f)$.

\smallskip
2. Similarly, in the $\ell (\ell +1)/x^2$ case, if $f_\ell (x,\kappa)$ is the solution 
of the equation
\[
-u'' + \f{\ell(\ell+1)}{x^2}\, u + Vu =\kappa^2 u,
\]
which is given by $kxh^{(1)} (\kappa x)$ if $x\geq a = \sup (\supp(V))$ and $u_\ell$ is 
the solution that obeys $\lim_{x\downarrow 0} \f{u(x,r)}{kxj_\ell (kx)} = 1$, then
\[
d(\kappa) = W(u,f).
\]

In the next section, we will need an additional function and a relation between $d$ and 
this new function. Let $c(\kappa)$ be defined by
\begin{equation} \lb{2.21}
\kappa c(\kappa) \equiv \tfrac12\, W(f_- (\dott, \kappa), f_+(\dott, -\kappa)).
\end{equation}

\begin{proposition} \lb{p2.11} 
\begin{alignat}{2} 
&\text{\rm{(i)}} \qquad && f_- (\dott,\kappa) = -c(\kappa) f_+ (\dott, \kappa) + 
d(\kappa) f_+ (\dott, -\kappa) \lb{2.22} \\
&\text{\rm{(ii)}} \qquad && d(\kappa) d(-\kappa) = 1 + c(\kappa) c(-\kappa) \lb{2.23}
\end{alignat}
\end{proposition}

\begin{proof} \eqref{2.21} is a direct consequence of \eqref{2.18}, \eqref{2.21} and 
\[
W(f_+(\dott, \kappa), f_+ (\dott, -\kappa)) = -2\kappa
\]
since $f_+ (x,\kappa) = e^{-\kappa x}$ for $x$ near $+\infty$.

Writing \eqref{2.22} for $\kappa$ and $-\kappa$ and
\[
W(f_- (\dott, \kappa), f_- (\dott, -\kappa) = 2\kappa
\]
(since $f_- (x,\kappa) = e^{\kappa x}$ for $x$ near $-\infty$), we have that 
\[
2\kappa = -2\kappa c(\kappa) c(-\kappa) + 2\kappa d(\kappa) d(-\kappa).
\]
which implies \eqref{2.23}.
\end{proof}

\bigskip
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Determinants as Laplace Transforms} \lb{s3}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\medskip

Our goal in this section is to prove Theorems~4 and 5. Given the expansion of 
Theorems~2 and 3, the argument is similar to part of the construction of the 
$A$-function in \cite{Si3}. We will use
\begin{align}
\f{\sinh\kappa x}{\kappa} &= \int_{-x/2}^{x/2} e^{-2\ell\kappa}\, d\ell \lb{3.1} \\
&= \int_0^x e^{x\kappa} e^{-2\ell\kappa}\, d\ell. \lb{3.2}
\end{align}

We start with the half-line case. Using \eqref{1.7} and \eqref{3.2}, we have that
\begin{align}
d(\kappa) &= 1 + \sum_{n=1}^\infty d_n (\kappa) \lb{3.3} \\
d_n(\kappa) &= \int_{ 0<x_1 < \cdots <x_n; \, 
0 < \ell_1 < x_1, \, 0 < \ell_2 < x_2 - x_1, \dots, 0 < \ell_n < x_n - x_{n -1}}  
V(x_1) \dots V(x_n) \, \notag \\
&\qquad\qquad\qquad\qquad e^{-2\sum_{j=1}^n \ell_j \kappa} \, dx_1 \dots dx_n\, 
d\ell_1 \dots d\ell_n \lb{3.4} \\
&\equiv \int t_n (\alpha)\, e^{-2\alpha\kappa}\, d\alpha \lb{3.5},
\end{align}
where
\begin{equation} \lb{3.6}
t_n (\alpha) = \int_{0<x_1 < \dots < x_n} V(x_1)\dots V(x_n) R_n(x_1, \dots, x_n; 
\alpha) \, dx_1 \dots dx_n
\end{equation}
with
\begin{equation} \lb{3.7}
\begin{split}
& R_n (x_1, \dots, x_n; \alpha) \\
&\quad = \int_{0 < \ell_1 < x_1, \, 0 < \ell_2 < x_2 - x_1, \dots, 0 < \ell_n < 
(x_n - x_{n-1})} \delta \bigg(\sum_{j=1}^n \ell_j - \alpha\bigg) \, d\ell_1 
\dots d\ell_n.
\end{split}
\end{equation}

\begin{lemma} \lb{l3.1}
\begin{enumerate}
\item[{\rm(i)}] $R_n = 0$ if $\alpha > x_n$
\item[{\rm(ii)}] $|R_n| \leq \f{\alpha^{n-1}}{(n-1)!}$
\item[{\rm(iii)}] For $n\geq 2$, $R_n$ is $C^1$ and $|\f{\partial R_n}{\partial\alpha}| 
\leq 2\, \f{\alpha^{n-2}}{(n-2)!}\,$.
\end{enumerate}
\end{lemma}

\begin{proof} (i) By the inequalitites on $\ell_j$, $\sum\ell_j \leq (x_1 + (x_2 - x_1) 
+ \cdots + (x_n - x_{n-1}))= x_n$.

\smallskip
(ii) Clearly, $|R_n|\leq \int_{0<\ell_1, \, 0 < \ell_2, \dots, 0 < \ell_n} 
\delta (\sum_{j=1}^n \ell_j - \alpha) \, d\ell_1 \dots d\ell_n = \f{\alpha^{n-1}}{(n-1)!}$ 
by induction.

\smallskip
(iii) 
\begin{align*}
\f{\partial R_n}{\partial \alpha} &= \int_{\dots} \delta' \bigg(\sum_{j=1}^n \ell_j - 
\alpha \bigg) \, d\ell_1 \dots d\ell_n  \\
&= \int_{\dots} \delta \bigg( \sum_{j=1}^{n-1} \ell_j - \alpha\bigg) \, d\ell_1 \dots 
d\ell_{n-1}  \\
&\qquad \qquad - \int_{\dots} \delta \bigg(\sum_{j=1}^{n-1} \ell_j - \alpha - 
(x_n - x_{n-1})\bigg)\, d\ell_1 \dots d\ell_{n-1}.
\end{align*}
Now use the estimation method of (ii). 
\end{proof}

\begin{proposition} \lb{p3.2} Let $a = \sup(\supp(V))$.
\begin{enumerate}
\item[{\rm(i)}]
\begin{equation} \lb{3.8}
d (\kappa) = 1 + \int_0^a t(\alpha)\, e^{-2\alpha\kappa} \, d\alpha
\end{equation}
with $t(\alpha)$ a continuous function.
\item[{\rm(ii)}] $t$ is an absolutely continuous function with
\begin{equation} \lb{3.9}
|t'(\alpha) + V(\alpha)| \leq C \int_\alpha^a |V(y)|\, dy.
\end{equation}
\item[{\rm(iii)}] The convex hull of the support of $\delta(\alpha)+ t(\alpha)$ is
$[0,a]$.
\end{enumerate}
\end{proposition}

\begin{proof} By \eqref{3.6} and (ii) of Lemma~\ref{l3.1}, we have that
\[
|t_n (\alpha)| \leq \f{\alpha^{n-1}}{(n-1)!}\, \bigg[ \int_\alpha^a |V(y)| \, dy 
\bigg] \f{1}{(n-1)!} \bigg( \int_0^a |V(y)|\, dy \bigg)^{n-1}
\]
so
\[
t(\alpha) = \sum_{n=1}^\infty t_n (\alpha)
\]
converges and we can justify the interchange of sum and integral to obtain \eqref{3.8}.

Similarly, by (iii) of Lemma~\ref{l3.1}, $\sum_{n=2}^\infty t_n (\alpha)$ is $C^1$ 
with derivative $g(\alpha)$ that obeys
\begin{align*}
|g(\alpha)| &\leq \sum_{n=2}^\infty (2\alpha^{n-2}) \big/ (n-2)!\, \bigg[ \int_\alpha^a V(y)\, 
dy\bigg] \f{1}{(n-1)!}\, \bigg(\int_0^a |V(y)|\, dy\bigg) \\
&\leq C \int_\alpha^a |V(y)|\, dy.
\end{align*}
By a direct calculation.
\[
t_1(\alpha) = \int_\alpha^a V(y)\, dy
\]
so that \eqref{3.9} is proven.

Clearly, $t(\alpha)$ is supported on $[0,a]$ and $\delta  + t$ has $0$ in its 
support. That means we need only show $t$ is non-zero on each interval $[a - \veps, 
a]$. Suppose $t$ vanishes identically on such an interval. Then by \eqref{3.9}, 
\[
|V(\alpha)| \leq C \int_\alpha^a |V(y)|\, dy
\]
for $\alpha > a-\veps$. So for $x\in [a-\veps, a]$,
\[
\int_x^a |V(y)|\, dy \leq |a-x| C \int_x^a |V(y)|\, dy
\]
so $V$ vanishes on $[a-\veps', a)$ with $\veps' = \min(\veps, \f{1}{C})$. Since 
$a=\sup(\supp(V))$ by hypothesis, this is a contradiction which shows that 
$a\in\supp(t)$.
\end{proof}

\smallskip
This implies Theorem~4. Proposition~\ref{p3.2} also makes it easy to construct 
potentials with an infinity of antibound states. The basic idea is close to those of 
Titchmarsh \cite{Tit} on zeros of Laplace transforms and a similar analysis (but using 
Melin's theory and on the whole line) has been made by Zworski \cite{Zw1}. Define intervals 
$I_1, I_2, \ldots \subset [0,1]$ by $I_1 = [0,\f12], I_2 = [\f12, \f34], \dots, I_n = 
[1 - \f{1}{2^{n-1}}, 1 - \f{1}{2^n}]$. Let $a_n = 2^n\, e^{-2^{n+1}}$ and
\[
V(x) = (-1)^n a_n \qquad\text{if } x\in I_n.
\]
It is easy to modify $V$ to be $C^\infty$. The infinity of oscillations of $V$ is 
critical since if $V$ has a definite sign near $x=a$, it is easy to see that $t(\alpha)$ 
has a definite sign, and so $f(-\kappa)$ does for $\kappa$ near infinity and thus 
there are only finitely many antibound states.

Since $a_n$ is decreasing, $|t'(\alpha) + V(\alpha)| \leq C 2^{-n} a_n$ by 
(ii) of Proposition~\ref{p3.2}. Thus for $n$ large, $t'(\alpha)$ is very close 
to $(-1)^{n+1} a_n$ on $I_n$. Thus
\begin{align*}
\int_0^1 t'(\alpha) \, e^{2\alpha\kappa}\, d\alpha &\sim \sum_{n=1}^\infty 
(-1)^{n+1} a_n 2^{-n}\, e^{2(1-2^{-n})\kappa} \\
&\sim e^{2\kappa} \sum_{n=1}^\infty (-1)^{n+1} \exp(-2^{-n-1} \kappa -2^{n+1}).
\end{align*}
For $\kappa_m = 2^{2(m+1)}$, it is easy to see this sum is dominated by the 
term with $n=m$. Since $\int_0^1 t'(\alpha) \, e^{2\alpha\kappa}\,d\alpha \sim 
-2\kappa \int_0^1 t(\alpha)\, e^{2\alpha\kappa}\, d\alpha$, we concluded that
\[
(-1)^m f(-2^{2(m+1)}) >0
\]
for $m$ large, and so $f(-\kappa)$ has infinitely many zeros as $\kappa\to\infty$.

Once one has a half-line potential with an infinity of antibound states, the same 
is true of whole-line problems with suitable even potentials.

\smallskip
We now turn to the whole-line case and Theorem~5. We will use \eqref{2.19}
\begin{equation} \lb{3.10}
2\kappa d(\kappa) = 2\kappa + \int_a^b e^{\kappa x} V(x) f_+ (x,\kappa)\, dx,
\end{equation}
as well as the following formula proven in a similar way:
\begin{align}
2\kappa c(\kappa) &= \int_a^b e^{\kappa x} V(x) f_+ (x, -\kappa)\, d\lambda \lb{3.11} \\
&= \int_a^b f_- (x, \kappa) V(x) \, e^{\kappa x}\, dx. \lb{3.12}
\end{align}

\begin{proposition} \lb{p3.3} In the whole-line case:
\begin{enumerate}
\item[{\rm(i)}] 
\begin{equation} \lb{3.13}
\kappa d(\kappa) = \kappa + \tfrac12 \int_a^b V(x)\, dx + \tfrac12 \int_0^{b-a} 
t(\alpha) e^{-2\alpha\kappa}\, d\alpha
\end{equation}
for an $L^1$-function $t$ on $[0, b-a]$.
\item[{\rm(ii)}]
\begin{equation} \lb{3.14}
\kappa c(\kappa) = \tfrac12 \int_a^b e^{2\kappa\alpha} s(\alpha)\, d\alpha,
\end{equation}
where $s(\alpha)$ is a function in $L^1 (a,b)$.
\item[{\rm(iii)}]
\begin{align*}
|s(\alpha) - V(\alpha)| &\leq C\int_a^\alpha |V(y)|\, dy \\
&\leq C\int_\alpha^b |V(y)|\, dy.
\end{align*}
\item[{\rm(iv)}] $a,b \in\supp (s)$.
\end{enumerate}
\end{proposition}

\begin{proof} (i) We begin with the extension of \eqref{3.8} which implies that
\begin{equation} \lb{3.15}
f_+ (x,\kappa) = e^{-\kappa x} + \int_x^b t_+ (\alpha, x)\, e^{-2\kappa\alpha} \, 
e^{+\kappa x}.
\end{equation}
Plugging this into \eqref{3.10},
\begin{align*}
\kappa d(\kappa) &= \kappa + \tfrac12 \int_a^b V(x)\, dx + \tfrac12 \int_a^b dx \int_x^b 
t_+ (\alpha, x)\, e^{-2\kappa(\alpha-x)}\, d\alpha \\
&= \kappa + \tfrac12 \int_a^b V(x)\, dx + \tfrac12 \int_0^{b-a} \bigg( \int_a^{b-\beta} 
t_+ (\beta + x, x)\,dx \bigg)\, e^{-2\beta\kappa}\, d\beta,
\end{align*}
which is \eqref{3.13} if $t(\alpha) = \int_a^{b-\alpha} t_+ (\alpha + x, x)\, dx$.

\smallskip
(ii) Plugging \eqref{3.15} into \eqref{3.11} yields
\[
\kappa c(\kappa) = \tfrac12 \int_a^b e^{2\kappa x} V(x)\, dx + \tfrac 12 \int_a^b 
V(x)\, dx \bigg( \int_x^b t_+ (\alpha, x)\, e^{2\kappa\alpha}\, d\alpha\bigg)
\]
which is \eqref{3.14} if
\begin{equation} \lb{3.16}
s(\alpha) = V(\alpha) + \int_a^\alpha t_+ (\alpha, x) V(x)\, dx.
\end{equation}
Using \eqref{3.12} instead of \eqref{3.11} yields
\begin{equation} \lb{3.17}
s(\alpha) = V(\alpha) + \int_\alpha^b t_- (\alpha, x) V(x) \, dx.
\end{equation}

\smallskip
(iii) This follows from \eqref{3.16}/\eqref{3.17} with 
\[
c = \sup_{\alpha,x} |t_\pm (\alpha, x)|.
\]

\smallskip
(iv) This follows as in the proof of (iii) of Proposition~\ref{p3.2}. 
\end{proof}

\smallskip
\eqref{3.13} is Theorem~5 if we prove that $b-a\in\supp(t)$. We give a proof 
of this fact that is part of Zworski's proof \cite{Zw1} translated to our language; 
Froese \cite{Fr1} has a different proof.

\begin{proposition} \lb{p3.4} $b-a \in \supp(t)$.
\end{proposition}

\begin{proof} Write $\kappa d(\kappa) = \int m(\alpha) \, e^{-2\alpha\kappa}\, d\alpha$ 
and $\kappa c(\kappa) = \int n(\alpha)\, e^{-2\alpha\kappa}\, d\alpha$ in distributional 
sense where
\[
m(\alpha) = \tfrac12 \delta' (\alpha) + \bigg(\tfrac12 \int_a^b V(x) \, dx\bigg) 
\delta(\alpha) + \tfrac12\, t(\alpha)
\]
and
\[
n(\alpha) = \tfrac12 \, s(-\alpha).
\]
By \eqref{2.23} and the uniqueness of inverse Laplace transforms
\begin{equation} \lb{3.18}
(m * \tilde m)(\alpha) = \tfrac14 \, \delta'' (\alpha) + (n * \tilde n)(\alpha),
\end{equation}
where $\tilde m(\alpha) = m(-\alpha)$ and $*$ is convolution. Since $a,b\in\supp(s)$, 
we have that $a-b$ and $b-a$ lie in $\supp [n * \tilde n]$. If $\supp [m] \subset 
[0,c]$, then $\supp [m * \tilde m] \subset [-c, c]$. Hence by \eqref{3.18}, $b-a\in
\supp [m]$.
\end{proof}


\bigskip
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Antibound States} \lb{s4}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\medskip
Our goal in this section is to prove Theorem~6. We will provide three rather 
different proofs, two that I found and one supplied to me by G.M.~Graf \cite{Gr}. We will 
present them first in the case when $u(0)=0$ boundary conditions with $V$ of compact 
support, and then make some remarks about the other cases.

\medskip
\noindent {\it First Proof of Theorem~6.} We begin by noting some properties of 
the Fredholm determinant $d(\kappa)$:
\begin{enumerate}
\item[(i)] If $\kappa\neq 0$, either $d(\kappa) \neq 0$ or $d(-\kappa)\neq 0$ (or both). 
This is true because if $d(\pm \kappa) =0$, then the Jost solution $f(x, \pm \kappa)$ 
vanishes at $x=0$, so the function solving $-u'' + qu = -\kappa^2 u$ with $u(0)=0$, 
$u'(0) =1$ is $\alpha e^{\mp \kappa x}$ for $x>a \equiv \max \supp (V)$. This cannot 
happen for both $\kappa$ and $-\kappa$ since $u\neq 0 $ for $x$ large.

\smallskip
\item[(2)] If $d(\kappa=0)=0$, then $d'(\kappa=0) \neq 0$. This well-known result holds 
because $\left. \f{\partial}{\partial \kappa} f(x,\kappa)\right|_{\kappa=0}$ and $\left. 
f(x,\kappa)\right|_{\kappa=0}$ both solve $-u'' + qu =0$ and are equal to $x$ and $1$ for 
$x>a$. As with the proof of (i), it cannot happen that both $d(\kappa = 0) = 0$ (in which 
case the solution with $u(0) = 0$, $u'(0)=1$ is a constant for $x>a$) and $\f{\partial d} 
{\partial \kappa} (\kappa=0)=0$ (in which case $u$ is equal to $\alpha x$ for $x > a$).

\smallskip
\item[(iii)] If $\kappa_n (\lambda)$ are the bound state ``energies" of $-\f{d^2}{dx^2} + 
\lambda q$, then $\kappa_n (\lambda)$ is increasing in $\lambda$ for $\lambda$ in 
$(L_n; \infty)$ where $L_n$ is the infimum over those $\lambda$ with $n$ or more 
bound states. This follows from first-order perturbation theory (Feynman-Hellman theorem), 
which implies that $\f{\partial (- \kappa^2_n)}{\partial\lambda} = \langle \varphi, q\varphi
\rangle \leq -\kappa^2_n / \lambda < 0$.

\smallskip
\item[(iv)] There are no zeros of $f$ in the quadrants $\Real \kappa > 0$, $\Ima \kappa 
\neq 0$ since they would correspond to imaginary eigenvalues.

\smallskip
\item[(v)] Resonances (zeros with $\Real \kappa < 0$, $\Ima \kappa\neq 0$) occur in complex 
conjugate pairs.
\end{enumerate}

This means as $\lambda$ increase and we look at eigenvalues of $-\f{d^2}{dx^2} + \lambda q$, 
the only way bound states can change is by an antibound state turning into a bound state. 
However, antibound states can change due to a complex pair of resonances turning into 
a pair of antibound states or vice-versa.

Now imagine $\lambda$ increasing past $L_1$. A single antibound state turns into a bound 
state $\kappa_1(\lambda)$. For $\lambda$ near $L_1$, there is exactly one bound 
state/antibound state near $0$ since $d' (L_1)\neq 0$ by (ii). In particular, there are 
no antibound states in $(-\kappa_1(\lambda), 0)$. Since $-\kappa_1 (\lambda)$ cannot be 
an antibound state, no antibound states can pass along the real axis into or out of 
$(-\kappa_1 (\lambda), 0)$ for $\lambda \in (L_1, L_2)$. Only pairs of resonances can 
produce pairs of antibound states. Thus, for $\lambda\in (L_1, L_2)$, there are an even 
number of antibound states in $(-\kappa_1 (\lambda), 0)$. At $\lambda =L_2$, a single 
antibound state turns into a bound state leaving an odd number in $(-\kappa_1 (\lambda), 
-\kappa_2(\lambda))$. As $\lambda$ increases, antibound states only get added or 
subtracted in pairs, so the number stays odd. The argument is simple for each similar 
interval $(-\kappa_n (\lambda), -\kappa_{n+1}(\lambda))$ for $\lambda> L_{n+1}$. \qed

\medskip

\noindent{\it Second Proof of Theorem~6.} This proof has some connection with an old 
paper of Ciafaloni-Menotti \cite{CM}, who discuss a related result concerning alternation 
as coupling constant is changed rather as $\kappa$ is varied. We define
\begin{equation} \lb{4.1}
\f{d(-\kappa)}{d(\kappa)} = e(\kappa). 
\end{equation}
If $\kappa_j$ is the $j$th bound state $0 < \kappa_n < \kappa_{n-1} < \cdots < \kappa_1$,  
then $d(\kappa_j) = 0$ and $d(-\kappa_j)\neq 0$ (as in (i) in the last proof). We will show 
that 
\begin{equation} \lb{4.2}
\lim_{\kappa\to \kappa_j} (\kappa - \kappa_j) e(\kappa) < 0.
\end{equation}
It follows that $e(\kappa_j + \veps) < 0$ and $e(\kappa_{j-1} - \veps) > 0$ for $\veps$ 
small and so, by continuity, $e(\kappa)$ has an odd number of zeros (counting multiplicity) 
in $(\kappa_j, \kappa_{j-1})$. But zeros of $d(-\kappa)$ are the same as zeros of $e(\kappa)$.

We will prove \eqref{4.2} by using a representation for $e(\kappa)$ due to Froese 
\cite{Fr1,Fr2}. In this case 
\begin{gather*} 
d(\kappa) = \det (1+ K(\kappa)) \\
K(\kappa)(x,y) = V(x)^{1/2}\, \f{\sinh (\kappa x_<)}{\kappa}\, e^{-\kappa x_>} |V(y)|^{1/2}
\end{gather*}
so
\[
[K(\kappa) - K(-\kappa)] (x,y) = 2\kappa^{-1} V(x)^{1/2} \sinh (\kappa x) 
\sinh (\kappa y) |V(y)|^{1/2}
\]
is rank $1$; call it $A(\kappa)$. Then
\begin{align*}
d(-\kappa) &= d(\kappa) \det (1 - (1 + K(\kappa))^{-1} A(\kappa)) \\
&= d(\kappa) [1 - \Tr (( 1 + K(\kappa))^{-1} A(\kappa))]
\end{align*}
so \cite{Fr1,Fr2}:
\begin{equation} \lb{4.3}
e(\kappa) = 1 - 2\kappa^{-1} \langle |V|^{1/2} \varphi_0, (1 + K(\kappa))^{-1} V^{1/2} 
\varphi_0\rangle , 
\end{equation}
where $\varphi_0 (x) = \sinh (\kappa x)$.

Let $L(\kappa) = V^{1/2} (h_0 + V+ \kappa^2)^{-1} |V|^{1/2}$. Then, by the resolvent formula:
\[
L(\kappa) = K(\kappa) - K(\kappa) L(\kappa)
\]
or
\[
(1 + K(\kappa))^{-1} = 1 - L(\kappa).
\]
Then by \eqref{4.3}
\[
e(\kappa) = 1 -2\kappa^{-1} \langle \varphi_0, V\varphi_0\rangle + 2\kappa^{-1} 
\langle V\varphi_0, (h_0 + V + \kappa^2)^{-1} V\varphi_0 \rangle.
\]
It follows from this that
\[
\lim_{\kappa \to \kappa_j} (\kappa - \kappa_j) e(\kappa_j) = -2\kappa^{-1}_j (2\kappa_j)^{-1} 
|\langle V\varphi_0, \eta_j \rangle |^2 ,
\]
where $\eta_j$ is the normalized eigenvector for $H_0 + V$ at energy $-\kappa^2_j$. Thus, 
the limit is non-positive. Since $e$ has a pole, the limit is non-zero, and so 
\eqref{4.2} is proven. \qed

\medskip
\noindent {\it Third Proof of Theorem~6} (Graf \cite{Gr}). Let $u(x,\kappa)$ solve the 
equation $-u'' + qu = -\kappa^2 u$ with $u(0)=0$, $u'(0) =1$. As is well known, $u(x,\kappa_j)$ 
has $j-1$ zeros, so since $u(x,\kappa) > 0$ for $x$ small, $(-1)^{j-1} u(x, \kappa_j) >0$ for 
$x$ near infinity. Since $q$ has compact support, say $[0,a]$,
\[
u(x, \kappa) = \alpha (\kappa) \, e^{-\kappa x} + \beta(\kappa) \, e^{\kappa x}, \quad 
x \geq a
\]
and $\alpha$ and $\beta$ are continuous since $u(x,\kappa)$, $u'(x,\kappa)$ are, and 
$\alpha,\beta$ can be expressed in terms of suitable $u,u'$ data. By the sign condition 
on $u$ near infinity 
\begin{equation} \lb{4.4}
(-1)^{j-1} \alpha (\kappa_j) > 0, 
\end{equation}
where one has strict positivity since $u$ cannot be identically zero on $(a,\infty)$.

By the association of $f$ and the Jost function, $f(-\kappa) = 0$ if and only if the 
solution which is $e^{\kappa x}$ near infinity vanishes at $x=0$, that is, if and only 
if $\alpha (\kappa) =0$. By \eqref{4.4}, $\alpha$ has an odd number of zeros in 
$(\kappa_j, \kappa_{j-1})$. 
\qed

\medskip
These proofs were stated for the case where $H$ has $u(0)=0$ boundary conditions, but 
each proof can be modified to handle the other half-line cases. For example, the third proof 
accommodates $u'(0) + hu(0)=0$ boundary conditions by looking at the solution 
obeying that boundary condition and $u(0)=1$ normalization. And it handles an 
$\f{\ell (\ell + 1)}{x^2}$ term by replacing exponentials by suitable Bessel functions.

The proofs also accommodate potentials with superexponential decay with minor 
modification. For example, the second proof applies verbatim to such potentials.

An illuminating example is the Bargmann potential \cite{CS} with Jost function (suggested 
by Newton \cite{New})
\[
f(\kappa) = \f{(\kappa - k_1)(\kappa - k_2)}{(\kappa+ k_3 + ik_4)(\kappa +k_3 - ik_4)}\, ,
\]
where $k_j > 0$ for $j=1,2,3,4$. Such a potential has no antibound states and seemingly 
violates Theorem~6. The point, of course, is that $q$ is not superexponential in 
this case, but only decays as $e^{-\alpha x}$ with $\alpha = \min (k_1, k_2, k_3)$. 
In general, Theorem~6 does hold for potentials with exponential decay, say, 
bounded by $e^{-\alpha x}$, but only for intervals $(-\kappa_j, -\kappa_{j-1})$ 
with $\kappa_j < \alpha$.



\bigskip


%%%%%%%%%%%%%%%%%%%%
%Appendix
%%%%%%%%%%%%%%%%%%%%
\appendix
\section{The Determinant of a Green's Matrix} \lb{App}
\renewcommand{\theequation}{A.\arabic{equation}}
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\setcounter{theorem}{0}
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Let $\{a_i\}^N_{i=1}$, $\{b_i\}^N_{i=1}$ be two sequences and let $C$ be the 
$N\times N$ matrix
\begin{equation} \lb{A.1}
c_{ij} = a_{\min(i,j)} b_{\max(i,j)}; \qquad 1\leq i,j\leq N.  
\end{equation}
Our main goal in this appendix is to give a simple proof of the following theorem 
of Barrett-Feinsilver \cite{BF}, which we used in Section~\ref{s2}:

\begin{theorem} \lb{tA.1} Let $C$ be a matrix given by \eqref{A.1}. Then
\begin{equation} \lb{A.2}
 \det(C) = b_N (a_N b_{N-1} - a_{N-1} b_N)(a_{N-1} b_{N_2} - a_{N-2} b_{N-1})  
\dots (a_2 b_1 - a_1 b_2) a_1.
\end{equation}
\end{theorem}

\medskip
\noindent{\bf Remarks.} 1. Barrett-Feinsilver \cite{BF} actually state the theorem for matrices 
closely related to ones given by \ref{A.1} and express the result in terms of the $c_{ij}$, 
viz:
\[
\det(C) = \prod_{i=1}^{N-1} (c_{i,i} c_{i+1, i-1} - c_{i, i+1} c_{i + 1, i}) 
\bigg/ \prod_{i+1}^N c_{ii}.
\]

\smallskip
2. The proof in \cite{BF} is combinatoric but has the advantage of generalizations in a variety 
of directions \cite{Barr,Kl,BJ}.

\smallskip
3. An idea similar to our proof in a related context appears in Jost-Pais \cite{JP}.

\smallskip
4. Matrices of type \eqref{A.1} were dubbed Green's matrices by Karlin \cite{Kar} 
since they are a discrete analog of one-dimensional Green's functions. Indeed, they occur 
in Section~\ref{s2} in connection with free Green's functions. And they arise as the 
inverses of symmetric tridiagonal matrices
\begin{equation} \lb{A.3}
 J = 
\left(
\begin{matrix}
x_1      & y_1 \dots     & 0 \\
y_1      & \dots         & 0 \\
\vdots   & \ddots        & \vdots \\
0        & \dots y_{N-1} & x_N 
\end{matrix}
\right)
\end{equation} 

\smallskip
\noindent in that there is a one-one correspondence realized by the inverse between 
invertible matrices of the form \eqref{A.1} and invertible matrices of the form 
\eqref{A.3}. This is a theorem of Gantmacher-Krein \cite{GK} (see also Barrett \cite{Barr}). 
Indeed, given an invertible \eqref{A.3}, $a$ can be determined by
\begin{gather} 
a_1 = 1; \quad x_1 a_1 + y_1 a_2 = 0; \quad y_{n-1} a_{n-1} + x_n a_n + y_n a_{n+1} = 0, 
\quad n=2, \dots N-1 \lb{A.4} \\
y_{N-1} b_{N-1} + x_N b_n=0; \quad y_{n-1} b_{n-1} + x_n b_n + y_n b_{n+1} = 0, 
\quad n=2, \dots, N-1, \lb{A.5}
\end{gather}
where $b$ is normalized by
\begin{equation} \lb{A.6}
y_n (a_{n+1} b_1 - b_{n+1} a_n) = 1, \qquad n=1, \dots, N-1.
\end{equation}
[{\it Note:} \eqref{A.4}/\eqref{A.5} imply the left side of \eqref{A.6} is independent of 
$n$ and it is non-zero if $J$ is invertible.]  Conversely, given a matrix \eqref{A.1}, 
one defines $y_n$ by \eqref{A.6}. [{\it Note:} $\det(C)\neq 0$ means $(a_{n-1} b_n - 
b_{n-1} a_n)\neq 0$ by Theorem~\ref{tA.1}.] And then one defines $x_n$ by \eqref{A.4} or 
\eqref{A.5}.

\medskip
\noindent{\it Proof of Theorem~\ref{tA.1}.} We prove the result by induction in $N$. 
The result for $N=1$ is obvious. If we can prove the result when $b_{N-1}\neq 0$, it 
follows by continuity for all $b_{N-1}$, so suppose $b_{N-1}\neq 0$. Consider the last 
row and column of $C$. It has $a_N b_N$ in the corner and otherwise every element is 
$a_j b_N$ for some $j\in \{ 1,\dots, N-1\}$. It follows that
\[
\det(C) = a_N b_N \det (C_{N-1}) + b^2_N F(a_1, \dots, a_{n-1}; b_1, \dots b_{N-1})
\]
where $C_{N-1}$ is the $(N-1)\times (N-1)$ Green's matrix with the last row and column 
removed and $F$ is some function of $\{a_i\}^{N-1}_{i=1}$ and $\{b_i\}^{N-1}_{i=1}$.

Fix these values of $a$ and $b$ and think of $a_N$ and $b_N$ as variable and $\det (C)$ 
as a function of them. When $a_N = a_{N-1}$ and $b_N = b_{N-1}$, then $C$ has the identical 
rows, so $\det (C) = 0$, that is,
\[
a_{N-1} b_{N-1} \det (C_{N-1}) + b^2_{N-1} F = 0
\]
so, since $b_{N-1} \neq 0$, $F=-a_{N-1} \det (C_{N-1}) / b_{N-1}$ and thus
\begin{equation} \lb{A.7}
\det(C) = b_N (a_N b_{N-1} - a_{N-1} b_N) \det(C_{N-1})/b_{N-1}. 
\end{equation}
But by induction, we can assume \eqref{A.2} for $\det(C_{N-1})$ and thus 
\eqref{A.7} implies \eqref{A.2} for $\det(C)$. \qed

\bigskip

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\end{document}

\begin{equation} \lb{3.}
\end{equation}