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An Algebraic Approach to Reﬂectionless Potentials in One Dimension

R.L. Jaﬀe

Center for Theoretical Physics, 77 Massachusetts Ave., Cambridge, MA 02139-4307

(Dated: January 31, 2009)

We develop algebraic methods to ﬁnd the eigenenergies and eigenstates of reﬂectionless potentials

in one dimension.

I. 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 hy-

drogen 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,

v`(x) = −`(`+ 1) sech2x(1)

where `is any positive integer (`= 0,1,2,...). Both the

bound states and scattering states can be found analyt-

ically for these potentials in terms of elementary func-

tions. In fact this is the only example (other than step

potentials and δ-functions) I know of where the scattering

states can be found by elementary means. These poten-

tials have remarkable properties including bound states

at zero energy, and reﬂectionless scattering. The latter

means that a particle incident on the potential is trans-

mitted with unit probability, albeit with an interaction-

dependent phase. As a result they are known as “reﬂec-

tionless potentials”.

The Schr¨odinger equation for the harmonic oscillator

and the Coulomb potential can be either by the more-

or-less standard analysis of diﬀerential equations, or by

algebraic methods. The algebraic solution to the har-

monic 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 [1].

Eq. (1) can also be solved by a direct attack on the dif-

ferential equation. The approach can be found in Ref. [2].

The solutions are expressed in terms of conﬂuent hyper-

geometric functions that reduce to elementary functions

when the strength of the potential is `(`+ 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 poten-

tials [3].

II. GENERAL FORMALISM

We begin with the one-dimensional Schr¨odinger equa-

tion,

h−¯h2

2m

d2

dξ2−V0sech2(bξ)iψ(ξ) = Eψ(ξ).(2)

For convenience we scale out the dimensionful quanti-

ties by deﬁning x=bξ,v0= 2mV0b2/¯h2, and k2=

2mb2E/¯h2, so for v0=`(`+ 1),

H`ψ(x) = p2−`(`+ 1) sech2xψ(x) = k2ψ(x) (3)

where p=−id

dx.k2≤0 corresponds to bound states and

k2>0 corresponds to scattering. Bound states should

have normalizable wavefunctions, Rdx|ψ(x)|2<∞, and

scattering states should be deﬁned in terms of incoming,

transmitted, and reﬂected waves. I will show that H`has

`bound states and also exhibit explicit wave functions

for the bound and scattering states of H`.

In an analogy to the harmonic oscillator we introduce

operators

a`=p−i` tanh x

a†

`=p+i` tanh x . (4)

Using the canonical commutator between pand x,

[x, p] = i, it is easy to show that

A`≡a†

`a`=p2+`2−`(`+ 1) sech2x

B`≡a`a†

`=p2+`2−`(`−1) sech2x . (5)

First we look for the ground state – a normalizable

state annihilated by a`. We deﬁne the state |0i`by the

equation

a`|0i`= 0

or −id

dx−i` tanh xψ0`(x) = 0

where ψ0`(x) = hx|0i`(6)

which has the normalizable solution

ψ0`(x) = N`sech`(x).(7)

Since ψ`0is normalizable[4] 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.

An Algebraic Approach to Reﬂectionless Potentials in One Dimension 2

Now consider the relation between the operators A`,

B`, and H`. Comparing eqs. (3) and (5),

A`=H`+`2

B`=H`−1+`2.(8)

Suppose ψis an eigenstate of A`,

A`|ψi=α|ψi.(9)

Then it is also an eigenstate of B`with the same eigen-

value,α, as shown by the following algebra:

a`{A`|ψi} =αa`|ψi

={a`a†

`}a`|ψi=B`a`|ψi.(10)

The only exception to this is the state |0i`, because

a`|0i`= 0. So |0i`is an eigenstate of A`with eigen-

value α= 0, which has no corresponding eigenstate of

B`.

Eq. (8) enables us to turn this into a statement about

the Hamiltonians, H`:H`−1and H`must share the same

spectrum except for the single state |0i`. These simple

results allow us to construct the eigenstates and eigenen-

ergies of all the Hamiltonians.

III. EIGENSTATES AND EIGENENERGIES

The easiest way to see how the information of the preced-

ing section enables us to solve for eigenstates and eigenen-

ergies is to start with `= 0, then consider `= 1, and so

on until the pattern becomes obvious.

A. `= 0

For `= 0, H0=p2. This is a free particle. We know the

eigenstates, |ki0. They are labeled by the wave number

k, and the subscript, 0, which refers to `= 0,

ψ0(k, x)≡ hx|ki0= exp ikx . (11)

The corresponding eigenenergies are E(k) = k2. Accord-

ing to our operator analysis, there should be a state, |0i0,

determined by a0|0i0= 0, or d

dxψ0(0, x) = 0. The solu-

tion is simply a constant, ψ0(0, x) = const.

B. `= 1

For `= 1 the results become nontrivial. The Hamiltonian

is

H1=p2−2 sech2x . (12)

According to our work in Section II, the spectrum of H1is

identical to that of H0except for the state |0i1. Thus we

have established that H1has a continuum of eigenstates

with E=k2.

The `= 1 ground state is determined by A1|0i1= 0.

Using eq. (8), H1=A1+ 1, we ﬁnd the ground-state

energy,

H1|0i1=− |0i1.(13)

So `= 1 has a bound state with E(0)

1=−1. The spec-

trum of H1is now complete: a bound state at E=−1

and a continuum E=k2. It is shown in Fig. 1 along

with other values of `.

`= 0

−1

−4

−9

−16

`= 1 `= 2 `= 3 `= 4

0E

FIG. 1: Energy levels in reﬂectionless potentials.

The wavefunctions of the eigenstates can be con-

structed using methods quite similar to those used for

the harmonic oscillator. The ground state is easy; from

eq. (7) we have

ψ(0)

1(x) = hx|0i1=N1sech x . (14)

To construct the continuum eigenstates, consider the

state obtained by acting with a†

1on the continuum eigen-

states of H0,

|ki1≡a†

1|ki0.(15)

The action of H1on these states can be related to the

`= 0 problem as follows:

H1|ki1= (A1−1)a†

1|ki0

=a†

1B1|ki0−a†

1|ki0

= (k2+ 1)a†

1|ki0−a†

1|ki0

=k2a†

1|ki0=k2|ki1.(16)

Thus |ki1is an eigenstate of H1with eigenvalue k2.

The continuum state wavefunctions are given by

ψ1(k, x) = hx|ki1=hx|a†

1|ki0

= (−id/dx+itanh x) exp ikx

= (k+itanh x) exp ikx . (17)

An Algebraic Approach to Reﬂectionless Potentials in One Dimension 3

To interpret the continuum states we have to relate

them to the usual parameterization of scattering in one

dimension,

lim

x→−∞ ψ(k, x) = eikx +R(k)e−ikx

lim

x→∞ ψ(k, x) = T(k)eikx .(18)

When we take the appropriate limits of eq. (17),

lim

x→−∞ ψ1(k, x) = (k−i)eikx

lim

x→∞ ψ1(k, x) = (k+i)eikx (19)

and compare with eq. (18) we ﬁnd

R(k) = 0

T(k) = k+i

k−i(20)

As promised, the reﬂection coeﬃcient vanishes, and the

transmission coeﬃcient is a pure phase,

T(k) = exp2itan−1(1/k).(21)

This completes the construction for `= 1.

C. `= 2

Armed with the methods developed for `= 1, we can

construct the solution for `= 2 more quickly. The Hamil-

tonian is

H2=p2−6 sech2x . (22)

According to our general result, the spectrum of H2co-

incides with that of H1except for the ground state, |0i2.

So there must be two bound states. One with energy

E=−1 is obtained by acting with a†

2on |0i1, with en-

ergy E=−1, and wavefunction

ψ(1)

2(x) = hx|a†

2|0i1

∝(p+ 2itanh x) sech x

∝tanh x/ cosh x . (23)

Note that this wavefunction is antisymmetric in x→

−xas we would expect for the ﬁrst excited state in a

one-dimensional potential. The ground-state energy is

determined to be E(0)

2=−4 by following an argument

analogous to eq. (17). Its wavefunction is given by eq. (7),

ψ(0)

2(x) = hx|0i2=N2sech2x . (24)

Finally, the continuum state wavefunctions are con-

structed by following a procedure analogous to the `= 1

case. In short,

ψ2(k, x) = hx|a†

2|ki1

= (p+ 2itanh x)(k+itanh x)eikx

= (1 + k2+ 3ik tanh x−3 tanh2x)eikx .(25)

Comparison with the deﬁnition of transmission and re-

ﬂection coeﬃcients gives

R2(k) = 0

T2(k) = (k+i)(k+ 2i)

(k−i)(k−2i)

= exp 2itan−1(1/k) + tan−1(2/k).(26)

Clearly we have outlined a method that can be ex-

tended to arbitrary `. The explicit expressions for the

wavefunctions are not as interesting as the spectrum and

the transmission coeﬃcients.

•A sequence of bound states beginning at E(0)

`=

−`2and continuing with E(j)

`=−(`−j)2until

j=`and E(`)

`= 0.

•The scattering is reﬂectionless, and the transmis-

sion coeﬃcient is given by

T`(k) = exp2i

`

X

j=1

tan−1(j/`).(27)

IV. DISCUSSION

Many interesting features of scattering theory are nicely

illustrated by the bound states and transmission coeﬃ-

cients of reﬂectionless potentials. A full discussion would

lead us far aﬁeld, so we simply quote some of the most

important results:

1. The transmission coeﬃcient, T`(k), has a pole

at every value of kat which the potential

`(`+ 1) sech2xhas a bound state. For `= 1 we

see a pole at k=i. For `= 2 we see poles at k=i

and k= 2i.

2. In addition to the bound states at k=i, 2i, 3i, . . .,

the potential `(`+ 1) sech2xhas a bound state

at zero energy. The solution to the Schr¨odinger

equation at k2= 0 must become asymptotic to a

straight line, ψ`(0, x)→A+Bx as x→ ±∞. When

the slope (B) of the straight line vanishes, the sys-

tem 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 poten-

tials.

3. If we parameterize T`(k) in terms of a phase shift,

T`(k) = exp2iδ`(k), then it is easy to show that

the diﬀerence between the phase shift at k= 0 and

k→∞counts πtimes the number of bound states,

with the bound state at zero energy counting as 1

2.

This result, known as Levinson’s theorem, holds for

arbitrary potentials in three dimensions as well as

one dimension.

An Algebraic Approach to Reﬂectionless Potentials in One Dimension 4

In summary, reﬂectionless potentials form a simple and

versatile laboratory for studying the properties of bound

states and scattering.

Acknowledgments

The author is grateful to Jeﬀrey Goldstone for con-

versations on reﬂectionless potentials. This work is sup-

ported in part by funds provided by the U.S. Department

of Energy (D.O.E.) under cooperative research agreement

#DF-FC02-94ER40818.

[1] See, for example, R.L. Liboﬀ, Introductory Quantum Me-

chanics, 3rd Ed. (Addison-Wesley, Reading, MA, 1998)

problem 10.58, page 481.

[2] P. Morse and H. Feshbach, Methods of Mathematical

Physics (McGraw-Hill, New York, 1953), page 1650.

[3] I learned these methods in conversation with Jeﬀrey Gold-

stone, who claims they are well known.

[4] `= 0 is a special, very simple, case that is treated in the

next section.