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\begin{flushright}
UWThPh-1995-10\\
\today
\end{flushright}
\vspace{2cm}
\begin{center}
{\Large \bf Interchannel Resonances at a Threshold}\\[50pt]
B. Baumgartner \\
Institut f\"ur Theoretische Physik \\
Universit\"at Wien
\vfill
{\bf Abstract} \\
\end{center}
Fermi's Golden Rule, the perturbation theoretic formula for calculating
the half width of a resonance, is not applicable to the case of a
transition from a bound state into an open channel, when the energy of
the bound state is exactly at the threshold of the continuum. We study
solvable models of this phenomenon. The exact results coincide in
leading order with the formulas found by modifying Fermi's Golden Rule.
\vfill
\end{titlepage}
\section{Introduction}
An interchannel resonance in quantum mechanical scattering is considered
as arising by a perturbation of a bound state in a closed channel, with
the bound state energy lying in a region where other channels are open
\cite{Newton}. In the case of a small perturbation the half--width of
the resonance is in general calculated with {\bf Fermi's Golden Rule},
FGR \cite{Sak,CoTa,Thi}. If the unpertubed Hamiltonian $H_0$ (which is the
kinetic energy plus interchannel potentials) with a bound state
$|\vp \rangle$ in the closed channel and continuum wave functions
$\psi_k$ in the open channel (we will consider only one open channel)
is perturbed by $\alpha V$, then FGR gives the imaginary part of the
resonance energy as a product of:
\begin{enumerate}
\item $\alpha^2$,
\item the absolute square of the transition matrix element
$\langle \vp|V \psi_k\rangle$,
\item the spectral density in the open channel, and
\item the constant $\pi$.
\end{enumerate}
A mathematical treatment, using complex scaling \cite{ABC}, confirms
Fermi's Golden Rule as the contribution to perturbation theory in
second order, with the exception of {\em the case, where the energy
of the bound state does coincide with the threshold of the continuous
spectrum in the open channel}. When trying to apply FGR to such a case
one observes the following facts:
\begin{enumerate}
\item[a)] The transition matrix element is zero in general; only in the
case of an open $s$--channel with a zero--energy inner--channel resonance
may it be nonvanishing at the threshold.
\item[b)] The spectral density at the threshold is infinite.
\end{enumerate}
So FGR gives either infinity or an undecided value of zero times infinity;
up to now we don't know for sure whether the bound state turns into a
resonance, and, in case it should do so, we don't have a
perturbation--theoretic formula for its lifetime.
Such cases actually do occur with atoms \cite{BB}. In principle they could
also appear with nuclei or in solid state physics, where the channels
are the bands of electron states, and where a bound state is located at
an impurity. For semiconductors just the resonances near a band edge
(= threshold) are of particular importance \cite{Ridley}.
In this paper we study simple solvable models: First a one--dimensional
lattice model with two bands. Then we make a comparison with a
simplified model, which keeps only the barest essentials. This serves as
an argument for the validity of the solvable continuum models with a
separable potential for coupling of the channels. The continuum models
are one--dimensional or the radial parts of a rotation--invariant
three--dimensional system.
The leading term in an expansion of the lifetime turns out to be
proportional to a half--integer--exponentiated coupling constant. We end
up with a conjecture how to modify FGR to get a generally valid
formula for the imaginary part of the resonance energy.
\section{A one--dimensional lattice model}
We consider a one--dimensional lattice {\bf Z}, carrying two channels,
which would be
translation invariant, if there were not an impurity at the point denoted
by $n = 0$. We consider only a local action of the impurity, so we can
restrict the Hilbert space for each channel
to the space of even functions. (The odd
functions are zero at $n = 0$ and are not affected by the impurity.)
So it is enough to consider the half space with points $n \geq 0$, and
the even wave functions can be represented by elements of the Hilbert
space $\ell^2({\bf N}) $ with the norm for
$\psi = (\psi_n) \in \ell^2({\bf N})$ given by
\beq
\| \psi\|^2 = |\psi_0|^2 + 2 \sum_{n=1}^\infty |\psi_n|^2.
\eeq
In each channel there is a kinetic energy $K$:
$$
(K \psi)_n = \left\{ \ba{ll}
2 \psi_n - \psi_{n-1} - \psi_{n+1} & \qquad n \geq 1 \\
2 \psi_0 - 2 \psi_1 & \qquad n = 0. \ea \right.
$$
The complete system with two bands is modelled in the Hilbert space
$\ell^2({\bf N}) \oplus \ell^2({\bf N})$.
One channel gives a band shifted by a positive energy $b$, so, without
impurity, the Hamiltonian would be $(K + b {\bf 1}) \oplus K$. The
impurity is represented by a negative local inner--channel potential
$(- w \delta_{n,0}) \oplus (-v \delta_{n,0})$, where $w$ is positive.
We are interested in the bound state in the upper band. It has the wave
function
\beq
\vp_n = c e^{-n \gamma_0}, \qquad c = ( \coth \gamma_0)^{-1/2}
\eeq
where
\beq
2 \sinh \gamma_0 = w
\eeq
and a binding energy which we consider as exactly equal to the relative
shift of the bands:
\beq
b = 2 \cosh \gamma_0 - 2.
\eeq
Now we collect all these contributions as the unperturbed Hamiltonian
$H_0$. In matrix notation
\beq
H_0 = \left( \ba{cc} b {\bf 1} + K - w \delta_{n,0} & 0 \\
0 & K - v \delta_{n,0} \ea \right)
\eeq
it has the bound state $\left( \ba{c} \vp_n \\ 0 \ea \right)$ with
energy zero, which equals the lower band edge of the lower channel.
Next we add a perturbation $\alpha V$ by an impurity at the point
$n = 0$, which introduces a coupling of the channels. For explanations
see the discussion in part 3. We define
\beq
H_\alpha = H_0 + \alpha V, \qquad
V = \delta_{n,0} \left( \ba{cc} {\bf 1} & t^* \\ t & u \ea \right),
\qquad t \in {\bf C}, \qquad u \in {\bf R}.
\eeq
In the open channel the resonance state should behave like a
Gamow function \cite{G}, so, to solve the equation, we make
the ansatz
\beq
\psi_n = \left( \ba{c} e^{-n \gamma} \\ a \cdot e^{ikn} \ea \right)
\eeq
\beq
(H_\alpha - E)\psi = 0
\eeq
and search for solutions with Im~$(k) < 0$, $E \ra 0$, $a \ra 0$ and
$\gamma \ra \gamma_0$ as $\alpha \ra 0$.
Inserting (7) into (8) gives for the components
$((H_\alpha - E)\psi)_n$ for $n \geq 1$ the equations
\beq
E = 2 - 2 \cosh \gamma + b = 2 - 2 \cos k,
\eeq
\beqa
b + 2 - 2e^{-\gamma} - w - E + \alpha(1 + t^* a) &=& 0, \no \\
(2 - 2e^{ik} - v - E)a + \alpha(ua + t) &=& 0.
\eeqa
Expressing $E$ by the appropriate parts of (9) we transform (10) to
\beq
a = \frac{-\alpha t}{\alpha u - v - 2i \sin k} =
\frac{\alpha - w + 2 \sinh \gamma}{- t^* \alpha}
\eeq
and
\beq
\alpha^2(t^*t - u) + \alpha(f + ug) - fg = 0
\eeq
where we have introduced
\beqa
f &:=& v + 2 i \sin k \no \\
g &:=& w - 2 \sinh \gamma .
\eeqa
Considering, by way of equ. (9), $E$, $\gamma$, $f$, $g$ as analytic
functions of the uniformizing complex variable $k$:
\beq
\gamma = \gamma(k) = \mbox{ar cosh } (b/2 + \cos k)
\eeq
with
$$
\lim_{k \ra 0} \gamma(k) = \gamma_0 > 0,
$$
we will be mainly interested in the behaviour of these functions of
$k$ near zero. The expansions are:
\beqa
E &=& k^2 - \frac{k^4}{12} + O(k^6) \no \\
f &=& v + 2ik - \frac{i}{3} k^3 + O(k^5) \\
g &=& k^2 \coth \gamma_0 + \left( \frac{1}{w} + \frac{2}{w^3} -
\frac{\sqrt{w^2 + 4}}{12w}\right) k^4 + O(k^6) =:
c^{-2} k^2 + d k^4 + O(k^6). \no
\eeqa
The goal is to get $E$ as a function of real $\alpha$ by way of $k$.
Using equation (12) it seems easier first to consider $\alpha$ as a
complex valued function of $k$, and then to invert this to a function
$k(\alpha)$. Four different cases have to be treated separately:
\begin{enumerate}
\item[a)] If $t^*t - u = 0$:
$$
\alpha = \frac{f \cdot g}{f + ug} \left\{ \ba{ll}
(c^{-2} k^2 + dk^4) \left( 1 - \dfrac{t^*t}{c^2 v} k^2 +
i \dfrac{2 t^*t}{c^2v} k^3 \right) + O(k^6) & \mbox{ if } v \neq 0 \\[10pt]
(c^{-2} k^2 + dk^4) \left( 1 + i \dfrac{t^*t}{2c^2} \right) k +
O(k^5) & \mbox{ if } v = 0.
\ea \right.
$$
\item[b)] If $t^*t - u \neq 0$, the solution of the quadratic equation
with $\alpha \ra 0$ as $k \ra 0$ has to be chosen:
\beqa
\alpha &=& \frac{f + ug}{2(t^*t - u)} \left( \sqrt{1 +
\frac{4fg(t^*t - u)}{(f + ug)^2}} - 1 \right) = \no \\
&=& \left\{ \ba{ll}
(c^{-2} k^2 + dk^4) \left( 1 - \dfrac{t^*t}{c^2v} k^2 + 2it^*t
\dfrac{c^{-2}}{v^2} k^3 \right) + O(k^6) & \mbox{ if } v \neq 0 \\[10pt]
(c^{-2} k^2 + dk^4) \left( 1 + i \dfrac{t^*t}{2c^2} k \right) + O(k^4)
& \mbox{ if } v = 0.
\ea \right. \no \\
\eeqa
\end{enumerate}
The leading contributions to the power series for the case b are
independent of $u$ and identical to the power series in case a, where
we may replace $u$ by $t^*t$. But the distinction between absence and
presence of an impurity well, $v = 0$ and $v \neq 0$, is important:
\noindent for $v \neq 0$:
$$
\alpha = c^{-2} k^2 - \left( \frac{t^*t}{v c^4} - d \right) + i
\frac{2t^*t}{v c^4} k^5 + O(k^6)
$$
for $v = 0$:
\beq
\alpha = c^{-2} k^2 + i \frac{t^*t}{2c^4} k^3 + O(k^4).
\eeq
In all the cases $\alpha(k)$ is an analytic function of $k$ in some
neighbourhood of $k = 0$, which is a zero point of second order, so the
function can be inverted to give $k$ as an analytic function of the
square root of $\alpha$ \cite{Osgood,Smirnov}. Its expansion is:
\noindent for $v \neq 0$:
\beqa
k &=& c \alpha^{1/2} + \frac{1}{2} \left( t^*t \frac{c}{v} - c^5 d\right)
\alpha^{3/2} - i t^*t \frac{c^2}{v^2} \alpha^2 + O(\alpha^{5/2}) \no \\
E &=& c^2 \alpha + \left( t^*t \frac{c^2}{v} - c^6 d - \frac{c^4}{12}\right)
\alpha^2 - i t^*t \frac{2c^3}{v^2} \alpha^{5/2} + O(\alpha^3)
\eeqa
for $v = 0$:
\beqa
k &=& c \alpha^{1/2} - i \frac{t^*t}{4} \alpha + O(\alpha^{3/2}) \no \\
E &=& c^2 \alpha - i t^*t \frac{c}{2} \alpha^{3/2} + O(\alpha^2).
\eeqa
\section{Discussion}
The model has been chosen to investigate the basic mathematical structure,
with best possible ease in computation. The impurity's potential and
the perturbation are therefore assumed to act locally at one point only.
>From there on everything is as general as possible. For the closed
channel the inner--channel part of the perturbing potential has been set
equal to one in equation (6). The reason is the following: If it were zero,
the bound state would under perturbation
not turn into a resonance, but would remain a bound
state. (This can be checked with the ansatz of a wave function
$\Phi_n = \left( \ba{c} e^{-\gamma n} \\ a \cdot e^{-\kappa n} \ea
\right)$, and proceeding in a similar way as for the resonance.)
So we are only interested in the case where this inner channel
perturbation is strictly positive, and we may use the freedom of shifting
the normalization between the coupling constant and the potential $V$,
to set this matrix element equal to one.
The threshold behaviour of the open channel, connected with the absence
or presence of an impurity well, determines the leading term for the
imaginary part: If $v \neq 0$, there is a nonvanishing effective range of
the impurity, the continuum wave functions at low energy tend to stay
away from it. The transition probability from the bound state to the
continuum increases with the coupling constant $\alpha$ more slowly
than in other cases. If $v = 0$, there is no effective range, the
transitions are not hindered, and the high spectral density of the
continuum effects an increase of Im~$(E(\alpha))$ faster than
$\alpha^2$. These effects will be discussed in closer detail for the
continuum models.
\section{The simplified lattice model}
To pave the way for simple continuum models, we simplify this
two--band model by ignoring any inner--channel effects in the closed channel,
except for the change in energy. So we replace $H_\alpha$ by $P H_\alpha P$,
with the projection operator
\beq
P = \left( \ba{cc} |\vp\rangle \langle \vp| & 0 \\ 0 & 1 \ea \right)
\eeq
where $|\vp\rangle$ is the unperturbed bound state (2).
So we work in the restricted Hilbert space $P[\ell^2({\bf N}) \oplus
\ell^2 ({\bf N})] \cong {\bf C} \oplus \ell^2({\bf N})$, where the
closed channel is represented by only one state. Moreover, we feel free
to replace the inner--channel part $u$ of the perturbation $V$ by
$t^*t$, because the leading terms in the expansion of 16b are identical
to those of 16a.
The Hilbert space is $\Ha = {\bf C} \oplus \ell^2({\bf N})$, the
Hamiltonian $H_\alpha = H_0 + \alpha V$,
\beq
H_0 = \left( \ba{cc} 0 & 0 \\ 0 & K - v \delta_{n0} \ea \right),
\qquad
V = \left( \ba{cc} c^2 & c t^* \delta_{n0} \\
c t \delta_{n0} & t^* t \delta_{n0} \ea \right).
\eeq
With the ansatz for the unnormalized resonance wave function
\beq
\Psi = \left( \ba{c} 1 \\ a \cdot e^{ikn} \ea \right)
\eeq
inserted into $(H_\alpha - E)\Psi = 0$, we get
\beqa
E &=& 2(1 - \cos k) \no \\
a &=& (E/\alpha c - c)/t^* \\
\alpha &=& \frac{c^{-2} (1 - \cos k)(v + 2i \sin k)}
{v/2 + i \sin k + t^*t c^{-2}(1 - \cos k)}. \no
\eeqa
This $\alpha$ is an analytic function of $k$, with the power series in a
neighbourhood of the origin
\beq
\alpha = \left\{ \ba{ll}
c^{-2} k^2 + \left( \frac{t^*t}{v c^4} - \frac{1}{12 c^2} \right) k^4
- i k^5 + O(k^6) & \mbox{ for } v \neq 0 \\[10pt]
c^{-2} k^2 + i \frac{t^*t}{2c^4} k^3 + O(k^4) & \mbox{ for } v = 0.
\ea \right.
\eeq
These functions can be inverted to a function of $\sqrt{\alpha}$,
analytic in a neighbourhood of the origin:
\noindent For $v \neq 0$:
\beqa
k &=& c \alpha^{1/2} + \left( \frac{t^*t}{2vc^2} - \frac{1}{24} \right)
c^3 \alpha^{3/2} - i \frac{t^*tc^2}{v^2} \alpha^2 + O(\alpha^{5/2})
\no \\
E &=& c^2 \alpha + \frac{t^*tc^2}{v} \alpha^2 - i \frac{2t^*tc^3}{v^2}
\alpha^{5/2} + O(\alpha^3).
\eeqa
For $v = 0$:
\beqa
k &=& c \alpha^{1/2} - i \left( \frac{t^*t}{4} \right) \alpha +
O(\alpha^{3/2}) \no \\
E &=& c^2 \alpha - i \frac{t^*tc}{2} \alpha^{3/2} + O(\alpha^2).
\eeqa
The leading contributions both to the real part and to the imaginary
part of the energy in (25) and (26) are identical to those in the
formulas (18) and (19).
\section{A simple continuum model}
The studies of the lattice model encourages us to use two simple building
blocks:
To keep only a single bound state, representing the closed channel,
and to consider a rank one operator as the perturbation $V$, coupling the
bound state to the open channel. In the open channel we consider the
particle to be free. Everything is reflection or rotation invariant in
one or three dimensions.
The Hilbert space is now
$$
\Ha = {\bf C} \oplus \cL^2({\bf R}_+) = \left\{ \left( \ba{c}
p \\ |\psi\rangle \ea \right), p \in {\bf C}, |\psi\rangle = \psi(x)
\in \cL^2({\bf R}_+) \right\}.
$$
As Hamiltonian we consider
\beq
H_\alpha = H_0 + \alpha V, \qquad
H_0 = \left( \ba{cc} 0 & 0 \\ 0 & K_\nu \ea \right), \qquad
V = \left( \ba{cc} c^2 & c \langle v| \\
c |v\rangle & |v\rangle \langle v| \ea \right)
\eeq
with $c > 0$, $v(x) = O(e^{-\kappa x})$ for some $\kappa > 0$, as
$x \ra \infty$. The kinetic energy, plus the angular momentum barrier,
is either
\beq
K_\nu = - \frac{d^2}{dx^2} + \frac{\ell(\ell + 1)}{x^2}, \qquad
\ell = \nu - \frac{1}{2} \in {\bf N}
\eeq
with Dirichlet--boundary conditions at $x = 0$, or $K_\nu = - d^2/dx^2$
with Neumann boundary conditions at $x = 0$ and $\nu = - 1/2$.
The models in one dimension are either of parity $-1$ or $+1$, concerning the
reflection at $x = 0$ in $\cL^2({\bf R})$. They can thus be reduced to
models on the half line ${\bf R}_+$, with $\nu = \pm 1/2$, the sign of
$\nu$ corresponding to the sign of the parity.
The index $\nu$ is chosen in such a way that it coincides with the
standard indication of the Bessel functions, which, multiplied by
$\sqrt{\pi k x/2}$, give the eigenfunctions of $K_\nu$.
The {\bf formal} ansatz for the resonance wave function at positive
$\alpha$ is
\beq
\Psi_\alpha = \left( \ba{c} 1 \\ \psi_\alpha(x) \ea \right),
\eeq
supposed to be a solution of the differential equation
\beq
(H_\alpha - E(\alpha)) \Psi_\alpha = 0,
\eeq
with the asymptotic behaviour
\beq
\psi_\alpha(x) \sim e^{ikx} \quad \mbox{at $x \ra \infty$, with
Im~$(k) < 0$, Re~$(k) > 0$.}
\eeq
A discussion of this asymptotic behaviour would involve mathematical
subtleties, especially if $v(x)$ is not of compact support. So we
will instead analyze these formulas first for negative coupling
constants $\alpha$. There we get unnormalized bound states
$\Psi_\alpha$ with negative energies $E(\alpha)$, the asymptotic
behaviour as in (31), but with $k = k(\alpha) = i \sqrt{- E(\alpha)}$.
Then we may consider the analytic continuation of $E(\alpha)$ and
$\psi_\alpha(x)$ to positive $\alpha$. This is again achieved by
inverting the function $\alpha(k)$.
Inserting (27) and (29) into (30) gives the pair of equations
\beqa
\alpha c^2 + \alpha c \; \langle v|\psi_\alpha \rangle - E &=& 0 \no \\
(K_\nu - E)|\psi_\alpha \rangle + \alpha(c + \langle v|\psi_\alpha
\rangle) |v\rangle &=& 0 .
\eeqa
The second equation is transformed to
\beq
|\psi_\alpha\rangle = - \alpha(c + \langle v|\psi_\alpha \rangle)
(K_\nu - E)^{-1} |v\rangle
= - \frac{E}{c} (K_\nu - E)^{-1} |v\rangle .
\eeq
Multiplying (33) by $\langle v|$, and eliminating $\langle v|\psi_\alpha
\rangle$ from the pair of equations, we get
\beq
\alpha = \frac{E}{c^2 - E \; \langle v|(K_\nu - E)^{-1}|v\rangle}.
\eeq
For negative real energies $E$, (34) gives negative real $\alpha$. In
the uniformizing variable $k$, this function $\alpha(k)$ can be
analytically continued into the region Im~$(k) > - \kappa$ (Appendix A).
In the appendix B we prove that the energies $E = k^2$ in the unphysical
sheet, giving positive real $\alpha(k)$, are actually poles of the
resolvent of $H_\alpha$. In appendix A it is shown that
\beq
\langle v|(K_\nu - k^2)^{-1}|v\rangle = F(k) - i k^{2\nu} G(k),
\eeq
with $F(k)$ and $G(k)$ analytic functions of $k$, real and even for
$k \in {\bf R}$. $G(0)$ is the absolute square of a scaled
transition matrix element from $|v\rangle$ into the continuum of
$K_\nu$ at energy zero: $|t|^2 = G(0)$.
The analytic function $\alpha(k)$ can be split into a real and an
imaginary part, for $k \in {\bf R}$, and expanded in powers of $k$:
\beq
\alpha(k) = \frac{k^2}{c^2 - k^2 F(k) - ik^{2\nu +2} G(k)} =
\frac{k^2}{c^2} + O(k^4) + i \left[ \frac{|t|^2}{c^4} k^{2\nu +4} +
O(k^{2\nu + 6})\right].
\eeq
To invert it, we have to form the square root:
\beq
\alpha^{1/2}(k) = \frac{k}{c} + O(k^3) + i \left[ \frac{|t|^2}{2c^3}
k^{2\nu +3} + O(k^{2\nu +5})\right].
\eeq
In both equations the remaining contributions, which we indicate by
the order of their leading terms, have real--valued expansion
coefficients. The inversion of this function near $k = 0$ gives the
searched for expansions of $k$ and of $E$ in the square root of the
coupling constant:
\beqa
k(\alpha^{1/2}) &=& c \alpha^{1/2} + O(\alpha^{3/2}) - i \left[
\frac{c^{2\nu +1}}{2} |t|^2 \alpha^{\nu + 3/2} +
O(\alpha^{\nu + 5/2}) \right] \\
E(\alpha^{1/2}) &=& c^2 \alpha + O(\alpha^2) - i \left[ c^{2\nu +2}
|t|^2 \alpha^{\nu + 2} + O(\alpha^{\nu +3)}\right] .
\eeqa
\section{Discussion}
The leading term in the expansion of the real part of the resonance
energy is precisely as in ordinary first order perturbation theory.
The leading term in the expansion of the negative imaginary part
increases as $\alpha^{3/2}$ for $\nu = -1/2$, where standard FGR would
give infinity times $\alpha^2$. In the other cases, for $\nu \geq 1/2$,
the increase is slower than $\alpha^2$. Now it turns out that the
leading term in the expansion can be calculated with a
\subsection*{modified FGR:}
As a {\em first step\/} one has to take into account the shift
in energy of the bound state to $E_\alpha = c^2 \alpha$. \\
Then as a {\em
second step\/} one has to proceed with standard FGR, applied to the bound
state with the shifted energy. \\
In the present case this procedure gives the transition
matrix element to the wave function $b_\nu(kx)$ in the open channel
(see appendix A)
\beq
t_k := \sqrt{\frac{2}{\pi}} \; c \int_0^\infty v^*(x) b_\nu(kx) dx
\st{k \ra 0}{\sim} c k^{\nu + 1/2} \sqrt{\frac{2}{\pi}} \; t.
\eeq
The spectral density is $1/2k$. Evaluating all the contributions at
$k = \sqrt{E_\alpha}$, one gets
\beq
\frac{\Gamma}{2} = |\mbox{Im } E(\alpha)| = \frac{\pi}{2k} |t_k|^2
\alpha^2 = c^{2\nu +2} |t|^2 \alpha^{\nu +2},
\eeq
precisely the leading term in (39).
Concerning the relevance of the model for physics, one has to observe,
that the wave functions in the continuum of an open channel with a
short range potential behave as in the open channel without a potential
and with Dirichlet b.c., {\em unless there is a zero energy bound state
or innerchannel--resonance}. In an open channel with $\ell = 0$ and
a zero--energy resonance, the wave functions behave as in the open
channel with Neumann b.c. and the transition matrix element into the
continuum stays finite at $E = 0$.
The exponential fall off of the perturbing potential in the model
corresponds to the exponential decrease of the bound state wave function
to be incorporated in the model: Consider the case of two particles with
the bound state $\Phi$ in the closed channel,
$$
\Phi (\vec x,\vec y) = \vp_1(\vec x) \vp_2(\vec y)
$$
perturbed by an interaction $\alpha W(\vec x - \vec y)$. It enables
transitions into an open channel, where the second particle with
$y$--coordinate is still bounded with the wave function
$\vp_3(\vec y)$, but the first
particle is free. In this open channel the general wave function is
$$
\Psi(x,y) = \psi(\vec x) \vp_3(\vec y),
$$
the transition matrix element is
$$
\langle \Psi|\alpha W|\Phi \rangle = \langle \psi| \alpha v \rangle
$$
with
$$
v(\vec x) = \vp_1(\vec x) \int d^3y \; W(\vec x - \vec y)
\vp_3^*(\vec y) \vp_2(\vec y).
$$
The decay of $v(\vec x)$ is thus determined by the decay of the bound
state wave function $\vp_1(\vec x)$.
There remains the problem of the behaviour of continuum wave function
in the open channel in the presence of a long range potential.
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\section*{Appendix A: Analytic continuation}
The solutions to the differential equation
$$
((K_\nu - k^2)f) (x) = 0
$$
are given by the transformed Bessel and Hankel functions $b_\nu$ and
$e_\nu$:
\beqa
b_\nu(kx) &:=& (\pi kx/2)^{1/2} J_\nu(kx) \no \\
e_\nu(kx) &:=& (\pi kx/2)^{1/2} H_\nu^{(1)} (kx).
\eeqa
The asymptotic behaviour is \cite{AbSt}
\beqa
b_\nu(kx) &\st{kx \ra 0}{\sim}& \frac{\sqrt{\pi}}{\Gamma(\nu +1)}
\left( \frac{kx}{2} \right)^{\nu + 1/2} \\
e_\nu(kx) &\st{kx \ra \infty}{\sim}& e^{i(kx - \pi(\nu + 1/2)/2)}.
\eeqa
The Wronskian
\beq
b_\nu \frac{d}{dx} e_\nu - e_\nu \frac{d}{dx} b_\nu = ik.
\eeq
The Green's function $G_{\nu,k}(x,y)$, the kernel of the resolvent
$(K_\nu - k^2)^{-1}$ for Im~$k > 0$, is thus
\beq
G_{\nu,k}(x,y) = \frac{i}{k} \left[ \Theta(y-x) b_\nu(kx) e_\nu(ky) +
\Theta(x-y) b_\nu(ky) e_\nu(kx)\right].
\eeq
For half--integer valued $\nu$, it is an entire function of $k$.
Since in the limit $x \ra \infty$ $v(x) = O(e^{-\kappa x})$, the
integration
$$
\int_0^\infty dx \int_0^\infty dy \; v^*(x) G_{\nu,k}(x,y) v(y)
$$
converges for Im~$k > - \kappa$. So does the integration with $G_{\nu,k}$
replaced by the kernel $\frac{d}{dk}(k G_{\nu,k})$, and the existence
of a continuation of the analytic function
$$
kQ(x) := k \langle v|
(K_\nu - k^2)^{-1}|v\rangle
$$
is established for Im~$k > - \kappa$.
In order to distinguish between real and imaginary coefficients of the
expansions, we use the transformed Neumann functions
$$
n_\nu(kx) := (\pi kx/2)^{1/2} N_\nu(kx)
$$
and split
\beq
e_\nu = b_\nu + i n_\nu.
\eeq
The Green's function is now written as
\beq
G_{\nu,k}(x) = \frac{i}{k} b_\nu(ky) b_\nu(kx) +
\frac{1}{k} \left[ \Theta(y-x) n_\nu(ky) b_\nu(kx) +
\Theta(x-y) b_\nu(ky) n_\nu(kx)\right].
\eeq
The Bessel and Neumann functions take real values at positive real
arguments. So, {\em for positive real $k$}, we may split $Q(k)$ into
real and imaginary parts:
\beq
Q(k) = F(k) + i k^{2\nu} G(k)
\eeq
\beq
F(k) = - \frac{1}{k} \int_0^\infty dx \int_0^\infty dy \; \Theta(y-x)
n_\nu(ky) b_\nu(kx) \left[ v^*(y) v(x) + v(y) v^*(x)\right].
\eeq
The product $n_\nu(ky) b_\nu(kx)$ lifts the singularity of $1/k$. For
half--integer--valued $\nu$, $F(k)$ has an analytic expansion in
powers of $k^2$.
\beq
G(k) = k^{-(2\nu +1)} \left| \int_0^\infty dx \; v^*(x) b_\nu(kx)
\right|^2 \st{k \ra 0}{\longrightarrow} |t|^2,
\eeq
with the ``scaled transition matrix element''
\beq
t := \lim_{k \ra 0} k^{-(\nu + 1/2)} \int_0^\infty dx \; v^*(x)
b_\nu(kx) =
\frac{\sqrt{\pi}}{2^{\nu + 1/2} \Gamma(\nu + 1)}
\int_0^\infty dx \; v^*(x) x^{\nu + 1/2}.
\eeq
At $k \sim 0$, $G(k)$ has an analytic expansion in powers of $k^2$.
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\section*{Appendix B: Poles of the resolvent}
We split the Hamiltonian $H_\alpha$ as
\beq
H_\alpha = L + W, \qquad
L := \left( \ba{cc} \alpha c^2 & 0 \\ 0 & K + \alpha|v\rangle\langle v|
\ea \right), \qquad
W := \left( \ba{cc} 0 & \alpha c \langle v| \\ \alpha c|v\rangle & 0
\ea \right)
\eeq
(we simply write $K$ instead of $K_\nu$) and use the iterated resolvent
equation in the Hilbert space ${\bf C} \oplus \cL^2({\bf R})$
\beqa
(H_\alpha - E)^{-1} &=& (L-E+W)^{-1} = \frac{1}{L-E} - \frac{1}{L-E}
W \frac{1}{L - E + W} = \no \\
&=& \frac{1}{L-E} - \frac{1}{L-E} W \frac{1}{L-E} +
\frac{1}{L-E} W \frac{1}{L-E} W \frac{1}{L-E+W}
\eeqa
to calculate the matrix element with $\left( \ba{c} 1 \\ |0\rangle \ea
\right)$:
\beq
r := (1,\langle 0|) \frac{1}{L+W-E} \left( \ba{c} 1 \\ |0\rangle \ea
\right).
\eeq
We get
\beq
r = \frac{1}{\alpha c^2-E} + \frac{\alpha^2 c^2}{\alpha c^2 -E} \;
\left\langle v \left| \frac{1}{K + \alpha|v\rangle \langle v| -E}
\right| v \right\rangle r.
\eeq
Again we use the resolvent equation, now in the smaller Hilbert space
$\cL^2({\bf R}_+)$:
\beq
\frac{1}{K-E+\alpha|v\rangle \langle v|} =
\frac{1}{K-E} - \frac{\alpha}{K-E} \; |v\rangle \langle v|
\frac{1}{K-E+\alpha|v\rangle \langle v|}.
\eeq
Taking the expectation value of (B.5) with $|v \rangle$, we get an
equation, to be transformed to
\beq
\left\langle v \left| \frac{1}{K-E+\alpha |v\rangle \langle v|}
\right| v\right\rangle =
\frac{\langle v | \frac{1}{K-E} | v \rangle}
{1 + \alpha \langle v | \frac{1}{K-E} |v\rangle} .
\eeq
Inserting (B.6) in (B.4) and solving for $r$ results in
\beq
r = \frac{1 + \alpha \langle v|(K-E)|v\rangle}
{\alpha c^2-E-\alpha E \langle v|(K-E)^{-1}|v\rangle}.
\eeq
Due to the asymptotic bound at large $x$
\beq
v(x) =O(e^{-\kappa x}),
\eeq
the matrix element $\langle v|(K-k^2)^{-1}|v\rangle$ can be continued
to an analytic function of $k$, for Im~$k > - \kappa$. The poles of the
matrix element $r = r(k)$ in this region are the zeroes of the
denominator of (B.7). Finding these zeroes amounts to solving equ.
(34).
Other matrix elements of the resolvent involve the same denominator.
\newpage
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\end{document}