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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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