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\title{An Algebraic Approach to Reflectionless Potentials in One Dimension}
\author{R.L.~Jaffe\\
\small Center for Theoretical Physics\\
\small MIT 6-311, 77 Massachusetts Ave.\\
\small Cambridge, MA 02139-4307\\[-0.25in]}
\date{}
%% You could use \date{\small\today\today} to keep track
%% of preliminary versions
\maketitle
\pagestyle{myheadings}
\markboth{R.L. Jaffe}{An Algebraic Approach to Reflectionless Potentials in
One Dimension}
\thispagestyle{empty}
\begin{abstract}
\noindent
We develop algebraic methods to find the eigenenergies and
eigenstates of reflectionless potentials in one dimension.
\end{abstract}
\section{Introduction}
A few interesting problems in wave mechanics have exact
solutions in terms of simple functions. The best known examples -- the
harmonic oscillator and the hydrogen atom -- teach us so much about the
structure of quantum systems that they are firmly established in the syllabi
of elementary courses. Another class of one-dimensional potential problems
also have exact solutions in terms of simple functions. The potentials are
inverse hyperbolic cosines,
%
\begin{equation}
v_\ell (x) = -\ell(\ell+1)\sech^2x
\label{1.1}
\end{equation}
%
where $\ell$ is any positive integer ($\ell = 0,1,2,\ldots$). Both the
bound states and scattering states can be found analytically for these
potentials in terms of elementary functions. In fact this is the only
example (other than step potentials and $\delta$-functions) I know of
where the scattering states can be found by elementary means.
These potentials have remarkable
properties including bound states at zero energy, and reflectionless
scattering. The latter means that a particle incident on the
potential is transmitted with unit probability, albeit with an
interaction-dependent phase. As a result they are known as
``reflectionless potentials''.
The Schr\"odinger equation for the harmonic oscillator and the Coulomb
potential can be \emph{either} by the more-or-less standard analysis
of differential equations, \emph{or} by algebraic methods. The
algebraic solution to the harmonic oscillator using raising and
lowering operators can be found in any textbook. The algebraic
solution to the hydrogen atom using the commutation relations of the
``Runge-Lenz'' vector and the angular momentum is treated in some
texts~\cite{ref:1.1}.
Eq.\,(\ref{1.1}) can also be solved by a direct attack on the differential
equation. The approach can be found in Ref.~\cite{ref:1.2}. The solutions
are expressed in terms of confluent hypergeometric functions that
reduce to elementary functions when the strength of the potential is
$\ell(\ell+1)$. On the other hand, the eigenstates of reflectionless
potentials can be found very easily using operator methods very
similar to those that are used to solve the harmonic oscillator in
elementary quantum mechanics texts. This does not appear to be very
widely known. In this short paper, I will develop the operator
solution to reflectionless potentials~\cite{ref:1.3}.
\section{General Formalism}\label{s2}
We begin with the one-dimensional Schr\"odinger equation,
%
\begin{equation}
\Bigl[-\frac{\hbar^2}{2m}\frac{\rd{^2}}{\rd{\xi^2}}-
V_0\sech^2(b\xi)\Bigr]\psi(\xi) = E\psi(\xi)\ .
\label{2.0}
\end{equation}
%
For convenience we scale out the dimensionful quantities by defining
$x=b\xi$, $v_0=2mV_0b^2/\hbar^2$, and
$k^2=2mb^2E/\hbar^2$, so for $v_0=\ell(\ell+1)$,
%
\begin{equation}
{\cal H}_\ell \;\psi(x) = \bigl[p^2 - \ell(\ell+1)\sech^2x\bigr]\psi(x) =
k^2\psi(x)
\label{2.1}
\end{equation}
%
where $p = -i \frac {\rd{}}{\rd{x}}$. $k^2\le 0$ corresponds to bound
states and
$k^2>0$ corresponds to scattering. Bound states should have normalizable
wavefunctions, $\int \rd x \left|\psi(x)\right|^2 < \infty$, and scattering states
should be defined in terms of incoming, transmitted, and reflected
waves. I will show that ${\cal H}_\ell $ has $\ell$ bound states and also
exhibit explicit wave functions for the bound and scattering states
of ${\cal H}_\ell $.
In an analogy to the harmonic oscillator we introduce operators
%
\begin{align}
a_\ell &= p-i\ell\tanh x\nonumber\\
a_\ell^\dagger &= p+i\ell\tanh x\ .
\label{2.2}
\end{align}
%
Using the canonical commutator between $p$ and $x$, $[x,p]=i$, it is
easy to show that
%
\begin{align}
{\cal A}_\ell \equiv a_\ell^\dagger a_\ell &=
p^2+\ell^2-\ell(\ell+1)\sech^2x
\nonumber\\
{\cal B}_\ell \equiv a_\ell a_\ell^\dagger &=
p^2+\ell^2-\ell(\ell-1)\sech^2x\ .
\label{2.3}
\end{align}
%
First we look for the ground state -- a normalizable state
annihilated by $a_\ell $. We define the state $\ket0_\ell $ by
the equation
%
\begin{align}
a_\ell \ket0_\ell &= 0 \nonumber\\
\llap{\text{or}\qquad} %fussy, to move centering left a little!
\bigl(-i\frac{\rd{}}{\rd x}-i\ell\tanh x\bigr)\psi_{0\ell}(x) &= 0
\nonumber\\
\text{where}\qquad
\psi_{0\ell}(x) &= %\langle x\ket0_\ell
\braket{x}{0}_\ell
\label{2.5}\\
\intertext{which has the normalizable solution}
\psi_{0\ell}(x) &= N_\ell \sech^\ell (x).
\label{2.6}
\end{align}
%
Since $\psi_{\ell 0}$ is normalizable\footnote{$\ell=0$ is a special,
very simple, case that is treated in the next section.} it is a bound
state. Since it has no nodes, a standard theorem on the
one-dimensional Schr\"odinger equation guarantees it is the
ground state.
Now consider the relation between the operators ${\cal A}_\ell $, ${\cal
B}_\ell $, and ${\cal H}_\ell $. Comparing eqs.\,(\ref{2.1}) and (\ref{2.3}),
%
\begin{align}
{\cal A}_\ell &= {\cal H}_\ell + \ell^2\nonumber\\
{\cal B}_\ell &= {\cal H}_{\ell-1} + \ell^2\ .
\label{2.7}
\end{align}
%
Suppose $\psi$ is an eigenstate of ${\cal A}_\ell $,
%
\begin{equation}
{\cal A}_\ell \ket\psi = \alpha\ket\psi\ .
\label{2.7a}
\end{equation}
%
Then it is also an eigenstate of ${\cal B}_\ell $ \emph{with the same
eigenvalue}, $\alpha$, as shown by the following algebra:
%
\begin{align}
a_\ell \{{\cal A}_\ell \ket\psi\} &= \alpha a_\ell \ket\psi \nonumber\\
&= \{a_\ell a_\ell^\dagger \}a_\ell \ket\psi =
{\cal B}_\ell a_\ell \ket\psi\ .
\label{2.8}
\end{align}
%
The only exception to this is
the state $\ket0_\ell $, because $a_\ell \ket0_\ell =0$.
So $\ket0_\ell $ is an eigenstate of ${\cal A}_\ell $ with
eigenvalue $\alpha=0$, which has no corresponding
eigenstate of ${\cal B}_\ell $.
Eq.\,(\ref{2.7}) enables us to turn this into a statement about the
Hamiltonians, ${\cal H}_\ell $: ${\cal H}_{\ell-1}$ and ${\cal H}_\ell $
\emph{must share the same spectrum except for the single state
$\ket0_\ell $}. These simple results allow us to construct
the eigenstates and eigenenergies of all the Hamiltonians.
\section{Eigenstates and Eigenenergies}
\noindent
The easiest way to see how the information of the preceding section
enables us to solve for eigenstates and eigenenergies is to start
with $\ell=0$, then consider $\ell=1$, and so on until the pattern
becomes obvious.
\subsection{\protect\boldmath$\ell=0$}
%% If you really must use symbols in headings (better to use words),
%% use bold symbols to match the bold fonts.
\noindent
For $\ell=0$, ${\cal H}_0=p^2$. This is a free particle. We
know the eigenstates, $\ket k_0$. They are labeled by the wave
number
$k$, and the subscript, 0, which refers to $\ell=0$,
%
\begin{equation}
\psi_0(k,x) \equiv \braket xk_0 = \exp ikx\ .
\label{3.1}
\end{equation}
%
The corresponding eigenenergies are $E(k)=k^2$. According to our operator analysis, there should
be a state, $\ket0_0$, determined by $a_0\ket 0_0=0$, or
$\frac{\rd{}}{\rd x}\psi_0(0,x)=0$. The solution is simply a
constant, $\psi_0(0,x)={\rm const}$.
\subsection {\protect\boldmath$\ell=1$}
\noindent
For $\ell=1$ the results become nontrivial. The Hamiltonian is
%
\begin{equation}
{\cal H}_1 = p^2-2\sech^2x\ .
\label{3.2}
\end{equation}
%
According to our work in Section~\ref{s2}, the spectrum of ${\cal H}_1$ is
identical to that of ${\cal H}_0$ except for the state
$\ket0_1$. Thus we have established that ${\cal H}_1$ has a
continuum of eigenstates with $E=k^2$.
The $\ell=1$ ground state is determined by ${\cal
A}_1\ket0_1=0$. Using eq.\,(\ref{2.7}), ${\cal H}_1={\cal
A}_1+1$, we find the ground-state energy,
%
\begin{equation}
{\cal H}_1\ket0_1=-\ket0_1\ .
\label{3.5}
\end{equation}
%
So $\ell=1$ has a bound state with $E^{(0)}_1=-1$. The spectrum of
${\cal H }_1$ is now complete: a bound state at $E=-1$ and a
continuum $E=k^2$. It is shown in Fig.~\ref{samplefig1} along with other
values of~$\ell$.
\begin{figure}[ht]
$$\BoxedEPSF{EnergyLevels.eps scaled 900}$$
\caption{Energy levels in reflectionless potentials.}
\label{samplefig1}
\end{figure}
%% The double-$ signs center the figure, which is scaled to 90 percent.
The wavefunctions of the eigenstates can be constructed using methods
quite similar to those used for the harmonic oscillator. The ground
state is easy; from eq.\,(\ref{2.6}) we have
%
\begin{equation}
\psi_1^{(0)}(x) = \braket x0_1 = N_1\sech x\ .
\label{3.6}
\end{equation}
%
To construct the continuum eigenstates, consider the state obtained
by acting with $a_1^\dagger$ on the continuum eigenstates of
${\cal H}_0$,
%
\begin{equation}
\ket k_1 \equiv a_1^\dagger \ket k_0\ .
\label{3.3}
\end{equation}
%
The action of ${\cal H}_1$ on these states can be related to the
$\ell=0$ problem as follows:
%
\begin{align}
{\cal H}_1\ket k_1 &= ({\cal
A}_1-1)a_1^\dagger \ket k_0\nonumber\\
&= a_1^\dagger {\cal
B}_1\ket k_0-a_1^\dagger \ket k_0\nonumber\\
&= (k^2+1)a_1^\dagger \ket k_0-a_1^\dagger \ket k_0\nonumber\\
&= k^2a_1^\dagger \ket k_0 = k^2\ket k_1 .
\label{3.4}
\end{align}
%
Thus $\ket k_1$ is an eigenstate of ${\cal H}_1$ with
eigenvalue $k^2$.
The continuum state wavefunctions are given by
%
\begin{align}
\psi_1(k,x) &= \braket xk_1 = \bra x a_1^\dagger \ket k_0\nonumber\\
&= (-i\,\rd{\null} /\rd x + i\tanh x)\exp ikx\nonumber\\
&= (k+i\tanh x) \exp ikx\ .
\label{3.7}
\end{align}
%
To interpret the continuum states we have to relate them to
the usual parameterization of scattering in one dimension,
%
\begin{align}
\lim_{x\to -\infty}\psi(k,x) &= e^{ikx} + R(k)e^{-ikx}\nonumber\\
\lim_{x\to \infty}\psi(k,x) &= T(k) e^{ikx} .
\label{3.8}
\end{align}
%
When we take the appropriate limits of eq.\,(\ref{3.7}),
%
\begin{align}
\lim_{x\to -\infty}\psi_1(k,x) &= (k-i)e^{ikx}\nonumber\\
\lim_{x\to \infty}\psi_1(k,x) &= (k+i)e^{ikx}
\label{3.9}
\end{align}
%
and compare with eq.\,(\ref{3.8}) we find
%
\begin{align}
R(k)&= 0\nonumber\\
T(k)&= \frac{k+i}{k-i}
\label{3.10}
\end{align}
As promised, the reflection coefficient vanishes, and the transmission
coefficient is a pure phase,
%
\begin{equation}
T(k) = \exp \bigl(2i\tan^{-1}(1/k)\bigr).
\label{3.11}
\end{equation}
%
This completes the construction for $\ell=1$.
\subsection{\protect\boldmath$\ell=2$}
\noindent
Armed with the methods developed for $\ell=1$, we can construct the
solution for $\ell=2$ more quickly. The Hamiltonian is
%
\begin{equation}
{\cal H }_2 = p^2 - 6\sech^2x\ .
\label{3.12}
\end{equation}
%
According to our general result, the spectrum of ${\cal H }_2$
coincides with that of ${\cal H }_1$ except for the ground state,
$\ket0_2$. So there must be \emph{two} bound states. One
with energy $E=-1$ is obtained by acting with $a_2^\dagger $ on
$\ket0_1$, with energy $E=-1$, and wavefunction
%
\begin{align}
\psi_2^{(1)}(x) &= \bra x
a_2^\dagger \ket0_1\nonumber\\
&\propto (p+2i\tanh x)\sech x\nonumber\\
&\propto \tanh x/\cosh x\ .
\label{3.13}
\end{align}
%
Note that this wavefunction is \emph{antisymmetric} in
$x\rightarrow -x$ as we would expect for the first excited state in a
one-dimensional potential. The ground-state energy is determined to
be $E_2^{(0)} = -4$ by following an argument analogous to eq.\,(\ref{3.7}).
Its wavefunction is given by eq.\,(\ref{2.6}),
%
\begin{equation}
\psi_2^{(0)}(x) = \braket x0_2 = N_2 \sech^2x\ .
\label{3.14}
\end{equation}
Finally, the continuum state wavefunctions are constructed by
following a procedure analogous to the $\ell=1$ case. In short,
%
\begin{align}
\psi_2(k,x) &= \bra x a_2^\dagger \ket k_1\nonumber\\
&= (p+2i\tanh x)(k+i\tanh x)\exp ikx\nonumber\\
&= (1+k^2 +3ik\tanh x -3\tanh^2x)\exp ikx\ .
\label{3.15}
\end{align}
%
Comparison with the definition of transmission and reflection
coefficients gives
%
\begin{align}
R_2(k) &= 0\nonumber\\
T_2(k) &= \frac{(k+i)(k+2i)}{(k-i)(k-2i)}\nonumber\\
&= \exp 2i \bigl(\tan^{-1}(1/k) + \tan^{-1}(2/k)\bigr)\ .
\label{3.16}
\end{align}
%
Clearly we have outlined a method that can be extended to arbitrary
$\ell$. The explicit expressions for the wavefunctions
are not as interesting as the spectrum and the transmission
coefficients.
\begin{itemize}
\item
A sequence of bound states beginning at $E_\ell ^{(0)} =
-\ell^2$ and continuing with $E_\ell^{(j)}=-(\ell - j)^2$
until $j=\ell$ and $E_\ell^{(\ell)}=0$.
\item
The scattering is reflectionless, and the transmission coefficient
is given by
%
\begin{equation}
T_\ell (k) = \exp\Bigl(2i \sum_{j=1}^\ell \tan^{-1}(j/\ell)\Bigr).
\label{3.17}
\end{equation}
%
\end{itemize}
\section{Discussion}
\noindent
Many interesting features of scattering theory are nicely illustrated
by the bound states and transmission coefficients of reflectionless
potentials. A full discussion would lead us far afield, so we simply
quote some of the most important results:
%
\begin{enumerate}
\item
The transmission coefficient, $T_\ell (k)$, has a pole at every
value of $k$ at which the potential $\ell(\ell+1)\sech^2x$ has a bound state. For $\ell=1$ we see a pole at
$k=i$. For $\ell=2$ we see poles at $k=i$ and $k=2i$.
\item
In addition to the bound states at $k=i,2i,3i,\ldots$, the
potential $\ell(\ell+1)\sech^2x$ has a bound state at
zero energy. The solution to the Schr\"odinger equation at $k^2=0$
must become asymptotic to a straight line, $\psi_\ell (0,x)
\rightarrow A+Bx$ as $x\rightarrow\pm\infty$. When the slope ($B$) of
the straight line vanishes, the system is said to possess a bound
state at zero energy. The name is justified by the fact that
making the potential infinitesimally deeper (and the problem no
longer exactly solvable) gives a state bound by an infinitesimal
binding energy. Bound states at zero energy are very special to
reflectionless potentials.
\item
If we parameterize $T_\ell (k)$ in terms of a phase shift,
$T_\ell (k)=\exp\bigl(2i\delta_\ell (k)\bigr)$, then it is easy to show
that the difference between the phase shift at $k=0$ and
$k\rightarrow\infty$ counts $\pi$ times the number of bound
states, with the bound state at zero energy counting as $\frac{1}{2}$.
This result, known as Levinson's theorem, holds for arbitrary
potentials in three dimensions as well as one dimension.
\end{enumerate}
In summary, reflectionless potentials form a simple and versatile
laboratory for studying the properties of bound states and
scattering.
\subsection*{Acknowledgments}
The author is grateful to Jeffrey Goldstone for conversations on
reflectionless potentials. This work is supported in part by funds
provided by the U.S. Department of Energy (D.O.E.) under cooperative
research agreement \#DF-FC02-94ER40818.
\begin{thebibliography} {9}
\bibitem{ref:1.1} See, for example, R.L.~Liboff, {\sl Introductory
Quantum Mechanics, 3rd Ed.} (Addison-Wesley, Reading, MA, 1998)
problem 10.58, page 481.
\bibitem{ref:1.2}P.~Morse and H.~Feshbach, {\sl Methods of Mathematical
Physics\/} (McGraw-Hill, New York, 1953), page 1650.
\bibitem{ref:1.3} I learned these methods in conversation with Jeffrey
Goldstone, who claims they are well known.
\end{thebibliography}
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%\section{Supplementary \LaTeX\ Notes}
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%\end{appendix}
\end{document}