Reflectionless eigenstates of the sech2 potential

John Leknera兲
School of Chemical and Physical Sciences, Victoria University of Wellington, PO Box 600, Wellington,
New Zealand
共Received 23 May 2007; accepted 28 August 2007兲
The one-dimensional potential well V共x兲 = −共ប2␯共␯ + 1兲 / 2ma2兲 sech2共x / a兲 does not reflect waves of
any energy when ␯ is a positive integer. We show that in this reflectionless case the solutions of
Schrödinger’s equation can be expressed in terms of elementary functions. Wave packets can be
constructed from these energy eigenstates, and the propagation of such wave packets through the
potential region can be studied analytically. We find that the group velocity of a particular packet can
substantially exceed the group velocity of a free-space Gaussian packet. The bound states of the
potential can also be expressed in terms of elementary functions when ␯ is an integer. The special
properties of the integer ␯ potentials are associated with critical binding. © 2007 American Association
of Physics Teachers.
关DOI: 10.1119/1.2787015兴

I. INTRODUCTION 1 1
␣ = 共␯ + 1 + ika兲, ␤ = 共␯ + 1 − ika兲. 共4兲
It is remarkable, even amazing, that the potential well 2 2
ប2 ␯共␯ + 1兲 F denotes the hypergeometric function,7 which can be repre-
V共x兲 = − 共1兲 sented by the Gauss hypergeometric series
2ma2 cosh2共x/a兲
⬁
⌫共␥兲 ⌫共␣ + n兲⌫共␤ + n兲 ␨n
is reflectionless, at any energy, if ␯ is a positive integer. The
potential 共1兲 共or the equivalent dielectric function profile兲
F共␣, ␤ ; ␥ ; ␨兲 = 兺 ⌫共␥ + n兲 n!
⌫共␣兲⌫共␤兲 n=0
共5兲
was first considered by Epstein1 共see also Eckart2兲, and is
treated in some quantum mechanics texts.3,4 The ␯ = 1 form within the unit circle 兩␨兩 = 1. Because F共␣ , ␤ ; ␥ ; ␨兲 is symmet-
of Eq. 共1兲 appears as the simplest of a family of reflectionless ric with respect to interchange of ␣ and ␤, and ␣ and ␤ are
profiles.5 complex conjugates, with ␥ and sinh2 x / a real, ␺␯e and ␺␯o are
When ␯ is a positive integer, there is no reflection at any real.
energy, and thus no reflection of any wave packet formed by When ␯ = 0 there is no potential, so ␺e0 and ␺o0 are what
superposition of positive energy eigenstates. A numerical mathematicians refer to as trivial solutions. However, their
study has reported6 that Gaussian wave packets are made structure gives us a hint for the reduction of solutions for all
narrower by passage over a reflectionless sech2 well, and integer ␯. In Eq. 共15.1.18兲 of Ref. 7, namely
that they can travel faster than a free-space packet. In this
paper we shall study these effects analytically, by first reduc-
ing the positive energy eigenstates to elementary form.
冉 1
F ␣,1 − ␣ ; ;sin2 z =
2
冊
cos共2␣ − 1兲z
cos z
, 共6兲

we set ␣ = 21 共1 + ika兲. Likewise in Eq. 共15.1.16兲 of Ref. 7,

II. REFLECTIONLESS POSITIVE ENERGY which reads
EIGENSTATES
For positive energies we write E = ប k / 2m, and the
2 2 冉 3
F ␣,2 − ␣ ; ;sin2 z =
2
冊
sin共2␣ − 2兲z
共␣ − 1兲sin 2z
, 共7兲
Schrödinger equation with potential energy given by Eq. 共1兲
reads we set ␣ = 1 + 共i / 2兲ka. In both equations we replace z by ix / a.

冋 册
These identities then give us the expected cos kx and sin kx
d 2␺ ␯共␯ + 1兲 eigenstates:
2
2 + k + 2 ␺ = 0. 共2兲
dx a cosh2共x/a兲 sin kx
The potential 共1兲 is even in x, so parity is a good quantum
␺e0 = cos kx, ␺o0 = . 共8兲
ka
number, and the two independent solutions of Eq. 共2兲 can be
taken to be the even and odd functions4 The integer ␯ solutions can be obtained by using Eqs. 共6兲

冉 冊 冉 冊
␯+1
and 共7兲 and the differentiation formulas of Sec. 15.2 of Ref.
x 1 x 7. For example 共with z = ix / a as above, and ␯ = 1兲
␺␯e 共x兲 = cosh F ␣, ␤ ; ;− sinh2
a 2 a
冉 1
␺e1 = cos2 zF ␣,2 − ␣ ; ;sin2 z , 冊 i
␣ = 1 + ka. 共9兲
␺␯o共x兲 冉 冊
= cosh 冉
x
a
␯+1
x 1 1 3
sinh F ␣ + , ␤ + ; ;
a 2 2 2
2
Using Eq. 共7兲 and 共15.2.4兲 with n = 1, namely
2

冊
− sinh2
x
a
, 共3兲
d ␥−1
d␨
†␨ F共␣, ␤ ; ␥ ; ␨兲‡ = 共␥ − 1兲␨␥−2F共␣, ␤ ; ␥ − 1; ␨兲, 共10兲

where we find, after some reduction, that

1151 Am. J. Phys. 75 共12兲, December 2007 http://aapt.org/ajp © 2007 American Association of Physics Teachers 1151
Fig. 1. The ␯ = 1 even and odd positive energy eigenstates, ␺e1 and ␺o1 共Eqs.
共11兲 and 共14兲兲, for ka = 1. The wave functions are raised from the x axis to
separate them from 2ma2 / ប2 times the ␯ = 1 potential, namely −2 sech2共x / a兲 Fig. 2. The ␯ = 2 even and odd positive energy eigenstates, ␺e2 and ␺o2 共Eqs.
共filled shape兲. 共15兲 and 共17兲兲, drawn for ka = 1. The filled shape is 2ma2 / ប2 times the ␯
= 2 potential, namely −6 sech2共x / a兲.

x sin kx We see that the integer-␯ eigenstates rapidly become more

␺e1 = cos kx − tanh . 共11兲
a ka complicated as ␯ increases, but they remain elementary func-
tions, expressible in terms of cos kx, sin kx and tanh x / a.
Likewise 共again with z = ix / a兲 From the even and odd eigenstates we can construct by

冉 3
␺o1 = − i cos2 z sin zF ␣,3 − ␣ ; ;sin2 z ,
2
冊 ␣=
3 i
+ ka.
2 2
superposition the reflectionless energy eigenstates propagat-
ing in either the +x or −x directions, for example,

共12兲 ␺+0 = ␺e0 + ika␺o0 = eikx , 共18兲

From Eq. 共7兲 and 共15.2.3兲 with n = 1, i.e.,

d ␣
†␨ F共␣, ␤ ; ␥ ; ␨兲‡ = ␣␨␣−1F共␣ + 1, ␤ ; ␥ ; ␨兲 共13兲
␺+1 = ␺e1 +
i†1 + 共ka兲2‡ o
ka
␺1 = 1 + 冋i
ka
tanh
x ikx
a
e , 册 共19兲

d␨
ika†4 + 共ka兲2‡ o
共and using the ␣ ↔ ␤ symmetry兲 we find ␺+2 = ␺e2 + ␺2 = †1 + 共ka兲2‡−1

再 冎
1 + 共ka兲2
␺o1 = †1 + 共ka兲2‡−1 ka sin kx + tanh
x
a
cos kx . 共14兲
再
⫻ 1 + 共ka兲2 − 3 tanh2
x
a
+ 3ika tanh
x ikx
a
e . 冎 共20兲
The even and odd eigenfunctions for ␯ = 1 are shown in
Note that ␺␯e 共and thus also ␺␯±兲 are normalized to unity at
Fig. 1.
x = 0. At large 兩x兩 the modulus of ␺␯± is greater than unity for
Similarly, using Eq. 共10兲 we can get ␺e2 by differentiating
␯ ⬎ 0: for example,
␺ 1:
o

␺e2 = †1 + 共ka兲 ‡
2 −1
再冋 1 + 共ka兲 − 3 tanh
2 2x
册 cos kx
兩␺±1 兩2 →
1 + 共ka兲2
共ka兲2
, 兩␺±2 兩2 →
4 + 共ka兲2
1 + 共ka兲2
. 共21兲

冎
a
The probability flux density

冉 冊
x
− 3ka tanh sin kx . 共15兲 ប d␺
a J= Im ␺* 共22兲
m dx
The ␯ = 2 odd eigenstate can be found by differentiation of ␺e1
on using Eq. 共15.2.1兲, which reads is zero for the 共real兲 even and odd eigenstates, and is inde-
pendent of x for the ␺± eigenstates, as it must be by conser-
d ␣␤
F共␣, ␤ ; ␥ ; ␨兲 = F共␣ + 1, ␤ + 1; ␥ + 1; ␨兲. 共16兲 vation of particles:
d␨ ␥
បk បk 1 + 共ka兲2 បk 4 + 共ka兲2
J+0 = , J+1 = , J+2 = . 共23兲

再冋
The result is m 共ka兲2 m 1 + 共ka兲2

册
m
x
␺o2 = 共ka兲−1†4 + 共ka兲2‡−2 1 + 共ka兲2 − 3 tanh2 sin kx

冎
a
III. CONSTRUCTION OF NONREFLECTING WAVE
x PACKETS
+ 3ka tanh cos kx . 共17兲
a
Kiriushcheva and Kuzmin6 have numerically integrated
The even and odd eigenfunctions for ␯ = 2 are shown in the time-dependent Schrödinger equation H⌽ = iប⳵t⌽ to fol-
Fig. 2. low the passage of a wave packet through the ␯ = 1 potential.

1152 Am. J. Phys., Vol. 75, No. 12, December 2007 John Lekner 1152
 pendence of the wave packet 共25兲 is obtained from

⌽0共x,t兲 =
冑2␲
1
冕 −⬁
⬁
dkeikx−ik
2បt/2m
A0共k兲 共28兲

because each energy eigenstate, with energy Ek = ប2k2 / 2m,

develops in time according to the phase factor exp
共−iEkt / ប兲.
Our aim is to construct wave packets which solve the
time-dependent Schrödinger equation in the presence of the
sech2 potential. In particular, we shall use the reflectionless
energy eigenstates corresponding to ␯ = 1:

冉
␺+1 共k,x兲 = 1 +
i
ka
x
tanh eikx,
a
冊 Ek =
ប 2k 2
2m
. 共29兲

The wave packets formed by superposition of these eigen-

states,

⌽1共x,t兲 =
冑2␲
1
冕 −⬁
⬁
dke−ik
2បt/2m
␺+1 共k,x兲A共k兲, 共30兲

satisfy the time-dependent Schrödinger equation, for any

Fig. 3. The Gaussian wave packet of Eq. 共25兲, with x0 = −5b, k0b = 1 at times Fourier amplitude A共k兲. Because of the 共ka兲−1 term in ␺+1 , a
t = 0 and t = 10b / v0, where v0 = បk0 / m is the group speed. At t = 0 only the

再 冎
relatively simple wave packet is produced by taking
envelope ±兩⌽0兩 is shown. At t = 10b / v0 the packet is centered on x = 5b; the
envelope and the real 共solid兲 and imaginary 共dashed兲 parts of ⌽ are shown. 1
The packet is most compact at t = 0; at later 共and earlier兲 times its spread is A共k兲 = − ikaA0共k兲 = − ikab exp − ikx0 − 共k − k0兲2b2 .
关b2 + 共បt / mb兲2兴1/2. 2
共31兲
This gives

冕
They reported that the packet accelerated and narrowed ⬁
− ib 2បt/m
slightly 共relative to the zero-potential case兲. Their wave ⌽1共x,t兲 = dke−ik
packet was taken to be Gaussian at time zero. In the absence 冑2␲ −⬁
of a potential 共in our case for ␯ = 0兲 such a wave packet,
starting at t = 0 centered on x = x0,
⌽0共x,0兲 = expˆik0共x − x0兲 − 共x − x0兲 /2b ‰ 2 2
共24兲
冉
⫻ ka + i tanh
a
e 冊
x ik共x−x 兲−共k − k 兲2b2/2
0 0 . 共32兲

8,9 The integrals may be evaluated in terms of the Gaussian

再冉
is known to have the time development integral, since the exponent
b 1
冊 − k20b2/2 − k2关b2 + iបt/m兴/2 + ik共x − x0 − ik0b2兲 共33兲
冑
⌽0共x,t兲 = exp ik0 x − x0 − v0t
iបt 2
b2 + is quadratic in k. The ka term in the integrand gives the

冎
m contribution

− 共x − x0 − v0t兲2/2共b2 + iបt/m兲 , 共25兲 ab

冑2␲
2 2
e−k0b /22 冕0
⬁
dkk sin kze−k
2关b2+iបt/m兴/2

冕
where v0 = បk0 / m is the group velocity of the packet, and b ⬁
gives the spatial extent of the packet at t = 0. Figure 3 shows ab 2 2 2关b2+iបt/m兴/2
=− e−k0b /22⳵z dk cos kze−k
the propagation and spreading of the Gaussian packet. 冑2␲ 0

冉 冊
The Fourier transform of Eq. 共24兲 is
z2/2
冕 ⬁
ab 2 2
1 = exp − k 0 b /2 − ,
A0共k兲 = dxe−ikx⌽0共x,0兲 关b2 + iបt/m兴3/2 b2 + iបt/m
冑2␲

再 冎
−⬁
z = x − x0 − ik0b2 . 共34兲
1
= b exp − ikx0 − 共k − k0兲2b2 . 共26兲 The tanh x / a term gives the contribution

冕
2 ⬁
x b −k2b2/2 2共b2+iបt/m兲
The Fourier inverse is a superposition of free-space energy tanh e 0 2 dke−k cos kz
eigenstates eikx: a 冑2␲ 0

⌽0共x,0兲 =
冑2␲
1
冕 ⬁

−⬁
dxe A0共k兲.
ikx
共27兲 = tanh
x b
a 关b2 + iបt/m兴1/2
exp − k 2 2
0 b /2 − 冉
z2/2
b2 + iបt/m
. 冊
共35兲
Since a formal solution of the time-dependent Schrödinger
equation H⌽ = iប⳵t⌽ is ⌽共x , t兲 = e−iHt/ប⌽共x , 0兲, the time de- Thus the wave packet built up from the nonreflecting energy

1153 Am. J. Phys., Vol. 75, No. 12, December 2007 John Lekner 1153
 If we take A共k兲 = A0共k兲 in Eq. 共30兲, a different but qualita-
tively similar wave packet results, namely
⌽1⬘共x,t兲 = ⌽0共x,t兲

+ 冑 ␲b
2a
x 22
tanh e−k0b /2 erfc
a
冉
x − x0 − ik0b2
冑2冑b2 + iបt/m , 冊
共37兲
where the complementary error function is defined by10

冑 冕 冑 冕
⬁ ␨
2 2 2 2
erfc共␨兲 = d ␰ e −␰ = 1 − d␰e−␰ = 1 − erf共␨兲.
␲ ␨ ␲ 0

共38兲
We can verify by direct differentiation that the wave pack-
ets given by Eqs. 共36兲 and 共37兲 satisfy the Schrödinger equa-
tion with the ␯ = 1 potential.

IV. GROUP VELOCITY AND WIDTH OF THE

NONREFLECTING PACKET ⌽1„x , t…
Fig. 4. The ␯ = 1 nonreflecting wave packet of Eq. 共36兲, drawn for a = b and Kiriushcheva and Kuzmin6 made a numerical study of
otherwise using the same parameters and line types as in Fig. 3, namely wave packet propagation in the presence of the ␯ = 1 sech2
x0 = −5a, k0a = 1, at times t = 0 共envelope only兲 and t = 10a / v0. The constric- potential. They found that a wave packet, constructed to have
tion near x = 0 is due the potential −共ប2 / ma2兲sech2共x / a兲 共not shown兲. The the form 共24兲 at time zero, propagated through the potential
center of the packet is slightly to the left of x = −5a at t = 0, and at about x
= 10a at t = 10a / v0, consistent with the group speed estimate given in
region faster than at the group speed v0 = បk0 / m of the free-
Eq. 共47兲. space Gaussian solution, and also was narrower after passing
through the potential than the free-space Gaussian packet.
Here we shall examine the speed and width of the ⌽1共x , t兲
wave packet, given in Eq. 共36兲.
eigenstates ␺+1 共k , x兲 with Fourier amplitude given by Eq. 共31兲 The envelope of this packet is 兩⌽1共x , t兲兩, where

冤
is

⌽1共x,t兲 = ⌽0共x,t兲 冋 a共x − x0 − ik0b2兲

b2 + iបt/m
+ tanh
x
a
, 册 共36兲 兩⌽1共x,t兲兩2 = 兩⌽0共x,t兲兩2 tanh2
x
a

冥
where ⌽0 is the free-space Gaussian wave packet given in x
Eq. 共25兲. The propagation of the packet through the potential 2ab2共x − x0 − v0t兲tanh + a2†共x − x0兲2 + k20b4‡
a
well region is illustrated in Figs. 4 and 5. +
b4 + 共បt/m兲2
共39兲

再 冎
with
b − 共x − x0 − v0t兲2
兩⌽0共x,t兲兩2 = exp .
冑b2 + 共បt/mb兲2 关b2 + 共បt/mb兲2兴
共40兲
The Gaussian free-space wave packet has, by inspection
of Eq. 共40兲, group speed v0 = បk0 / m and width ␴0共t兲
= [b2 + 共បt / m兲2]1/2. The expectation values of the energy and
momentum are, respectively,

具H典 =
ប2 2
2m
冉 1
k0 + 2 ,
2b
冊 具p典 = បk0 . 共41兲

To find the group speed and width associated with the

wave packet ⌽1共x , t兲 we limit ourselves to regions outside
the potential 共so that tanh x / a takes the values ±1兲. We dif-
ferentiate 兩⌽1兩2 with respect to x, and set x = x0 + d. The zeros
Fig. 5. Another view of the ␯ = 1 nonreflecting wave packet, drawn as a
of this derivative give the stationary points of 兩⌽1兩2. There
spiral curve at t = 10a / v0, with parameters as in Fig. 4. The x axis is the result two cubic equations for d, corresponding to tanh x / a
direction of propagation, the transverse axes are real and imaginary parts of → 1 and −1. There are three intrinsic speeds in the problem,
⌽, also shown in Fig. 4 as solid and dashed curves, respectively. namely

1154 Am. J. Phys., Vol. 75, No. 12, December 2007 John Lekner 1154
 បk0 ប ប
v0 = , ua = and ub = . 共42兲
m ma mb
At times t such that v0t, uat and ubt are all large compared to
the lengths a and b 共the range of the potential and the initial
width of the free-space packet, respectively兲, both cubics re-
duce to
u2av0t3 + 共u2b − u2a兲t2d + v0td2 − d3 = 0. 共43兲
Equation 共43兲 determines the position x = x0 + d of the maxi-
mum of 兩⌽1兩2 at time t. The group speed v is found by setting
d = vt:
u2av0 + 共u2b − u2a兲v + v0v2 − v3 = 0. 共44兲
When v0 is large compared to ua and ub 共that is, at high
energy expectation values兲, v ⬇ v0, i.e., the reflectionless
packet travels at about the same speed as the free-space wave
packet ⌽0. More precisely, the high-energy asymptotic form

再 冎
of the group speed v is
1 a2 + b2
0 兲 . 共45兲
v = v0 1 + − + O共k−6 Fig. 6. Comparison of the propagation of a free-space Gaussian packet and
共k0b兲2 a2b4k40 of the ␯ = 1 nonreflecting packet, Eqs. 共25兲 and 共36兲, respectively. The pa-
rameters are a = b, k0a = 1, t = 0 共top兲, 1, 2, 3, 4 and 5a / v0 共bottom兲. The
At low energies the situation is more complicated, and the potential well is also shown 共shaded兲. The nonreflecting packet has a group
group speed can be much greater than that of the free-space speed roughly 1.5 times greater than the Gaussian. The curves show 兩⌽0兩2
packet. For example, when a = b 共and thus ua = ub兲, and 兩⌽1兩2, with ⌽1 having the larger amplitude. 兩⌽1兩2 shows an indentation

再 冎
near x = 0 as it passes over the potential; 兩⌽0兩2 共dashed curves兲 does not,
1 1 1 because it is a free-space solution of Schrödinger’s equation.
v = v0 2/3 + + 共k0a兲2/3 + O共k0a兲4/3 . 共46兲
共k0a兲 3 9
At k0a = 1 = k0b we find from Eq. 共44兲, with ␳ = 29+ 3冑93,
that
flecting packet and of a free-space Gaussian packet is illus-
v0 trated in Fig. 6 for the parameter choice leading to the ve-
v = 关共␳/2兲1/3 + 共2/␳兲1/3 + 1兴 ⬇ 1.46557v0 . 共47兲 locity ratio of Eq. 共47兲.
3
Next we consider the question “what is so special about
The width of the free-space Gaussian wave packet, ␴0共t兲 the integer values of ␯ in Eq. 共1兲?”
= [b2 + 共បt / m兲2]1/2, was defined by the value of 兩x − x0 − vt兩 at
which 兩⌽0兩2 fell to 1 / e of its maximum value, that is, when
the exponent in Eq. 共40兲 decreased from 0 to −1. For the ⌽1
wave packet we shall use the same definition. The exponent
of 兩⌽1兩2 is −X + ln F, where X = 共x − x0 − v0t兲2 / [b2 + 共បt / mb兲2], V. CRITICAL BINDING
and F is the content of the square bracket in Eq. 共39兲. Out-
side the range of the potential the maximum of 兩⌽1兩2 occurs The phenomenon of zero reflection of waves is common in
at x = x0 + vt where the group speed v is determined by the optics, acoustics and quantum mechanics.11 For example, an-
cubic Eq. 共44兲. The location of the points x at which X tireflection coatings can make reflection zero at one wave-
− ln F = 1 is determined by a transcendental equation, which length, or very small over a range of wavelengths. What is
we shall simplify by neglecting the slowly varying ln F term. rare is zero reflection at any wavelength. In optics and acous-
The breadth of the wave packet, centered on xm = x0 + vt, is tics there is zero reflection by a sharp interface at the Brew-
then approximately given by the points x1 at which X = 1: ster and Green angles 共see, for example, Secs. 1-2 and 1-4 of
Ref. 12兲
共x1 − x0 − v0t兲2 = b2 + 共បt/mb兲2 . 共48兲
␧2 共 ␳ 2v 2兲 2 − 共 ␳ 1v 1兲 2
The distance ␴共t兲 between x1 and xm is then roughly tan2 ␪B = , tan2 ␪G = , 共50兲
␧1 ␳21共v21 − v22兲
␴共t兲 ⬇ ␴0共t兲 − 共v − v0兲t, ␴20共t兲 = b + 共បt/mb兲 .
2 2
共49兲
where ␧, ␳, v are dielectric constants, densities and sound
Thus when the group speed v exceeds the group speed v0 of speeds, respectively. The reflection of the electromagnetic p
the free-space Gaussian packet, we expect a narrower packet wave and of the acoustic wave is, however, zero only in the
than the Gaussian 共at the same time of propagation兲. The limit when the step from ␧1 to ␧2 or 共␳1 , v1兲 to 共␳2 , v2兲 is very
expression given in Eq. 共49兲 is a rough estimate only, and rapid on the scale of the wavelength.
cannot be valid when v − v0 ⬎ ub = ប / mb, since then ␴共t兲 Here we have an example of a potential 共or dielectric func-
would become negative at large times. tion profile兲 with a characteristic length a, and zero reflection
Our results for the group speed and wave packet width are for any values of a at any energy, provided ␯ is an integer.
in qualitative agreement with the numerical example given Why?
by Kiriushcheva and Kuzmin.6 The propagation of a nonre- The potential 共1兲 has, for given ␯, the bound states3

1155 Am. J. Phys., Vol. 75, No. 12, December 2007 John Lekner 1155
 ប2 VI. SUGGESTED PROBLEMS
En = − 共␯ − n兲2, n = 0,1,2 . . . ,关␯兴, 共51兲
2ma2 1. Is it true in general that symmetric potentials at critical
binding strength do not reflect? 关Hint: one counterexam-
where 关␯兴 is the integer part of ␯. The Schrödinger equation
ple is sufficient to disprove a conjecture. Try a square
for bound states, and the even and odd energy eigenstates, well.兴
can be obtained from Eqs. 共2兲 and 共3兲, respectively, by re- 2. The Gaussian wave packet ⌽0 and the ␯ = 1 wave packet
placing k by q = ik. The energy becomes −ប2q2 / 2m. For ex- ⌽1 共Eqs. 共25兲 and 共36兲兲 both have shorter wavelengths at
ample, for ␯ = 1 we have the bound state the front: see Figs. 3–5. For the Gaussian pulse the phase
1 − ប2 x ␾, defined by ⌽ = 兩⌽兩exp共i␾兲, is
q= , E共1兲
0 = , ␺共1兲
0 = sech 共52兲

冉 冊
a 2ma2 a
1 共x − x0 − v0t兲2បt/2m
and the just-bound 共or just-unbound兲 state ␾共x,t兲 = k0 x − x0 − v0t + . 共59兲
2 b4 + 共បt/m兲2
x
q = 0, E共1兲
1 = 0, ␺共1兲
1 = tanh . 共53兲 A wavelength can be defined by ␾共x + ␭ , t兲 − ␾共x , t兲 = 2␲, or
a ␾共x + ␭ / n , t兲 − ␾共x , t兲 = 2␲ / n. In the limit of large n we obtain
For ␯ = 2 we have two bound states and one zero-energy the local wavelength
state:
2␲
2 2ប2 x ␭共x,t兲 = . 共60兲
q= , E共2兲
0 =− , ␺共2兲
0
2
= sech , 共54兲 ⳵x␾共x,t兲
a ma2 a
Verify that for the Gaussian packet Eq. 共60兲 gives
1 ប2 x x
q= , E共2兲 =− , ␺共2兲 = sech tanh , 共55兲
a 1
2ma2 1
a a 2␲
␭0共x,t兲 = . 共61兲
2共x − x0 − v0t兲បt/m
x x k0 +
q = 0, E共2兲 ␺共2兲 2
− 2 tanh2 . 共56兲 b4 + 共បt/m兲2
2 = 0, 2 = sech
a a
Find the corresponding result for ⌽1 and compare with plots
The special property of the integer-␯ potentials 共−ប2 / of ⌽0 and ⌽1.
2ma2兲␯共␯ + 1兲sech2 x / a is that they support a critically bound 3. The phase ␾共x , t兲 of a wave packet is stationary when
state: one that has zero energy and a wave function of infi- d ␾ = ⳵x␾共x , t兲dx + ⳵t␾共x , t兲dt is zero. The phase velocity is
nite range.13 Systems which are near critical binding have the value of dx / dt when d␾ is zero, v p = −⳵t␾共x , t兲 /
special properties associated with the long range of the
wavefunction. In three dimensions, the zero-energy scatter-
⳵x␾共x , t兲. Find v p for the Gaussian free-space packet, and
ing length s goes to infinity as the binding energy E tends to show that the phase speed is half the group speed
zero:3,13 共v p = v0 / 2兲 at the center of the packet, where x = x0 + v0t.
Explore the behavior of v p for ⌽0 and for ⌽1.
s= ± 冉 冊
ប2
2m兩E兩
1/2
共57兲
4. The locus of 关Re ⌽ , Im ⌽ , x兴 is, at a given time, a curve
in space. For ⌽ = eik共x−vt兲 the curve is a right-handed helix
of pitch 2␲ / k, if Re ⌽, Im ⌽, and x form a right-handed
共the positive sign is to be taken for bound states, E ⬍ 0, and coordinate system. A snapshot of the space curve of ⌽1 is
the negative sign for virtual states, E ⬎ 0兲. Thus systems near shown in Fig. 5. It is predominantly right handed, but
critical binding can be thought of as having a reach far be- there is a left-handed portion in the tail. The curve is right
yond the range of the binding potential, in our case the handed if the phase is increasing with x; the change in
length a. This is one reason for the nonreflecting integer-␯ handedness occurs at zeros of ⳵x␾共x , t兲. Verify that for the
sech2 potentials. Another is the V共−x兲 = V共x兲 symmetry of the Gaussian packet the change occurs at
potential: as shown in Ref. 11, reflection amplitudes of sym-
metric profiles automatically have coincident zeros of their mko 4
real and imaginary parts, whereas general profiles do not. x = x 0 + v 0t − †b + 共បt/m兲2‡, 共62兲
The above association between critical binding strengths 2បt
of the sech2 potentials and zero reflection adds a physical
heuristic to the mathematical explanation of supersymmetric and find the corresponding result for ⌽1. Compare with plots
quantum mechanics. In the latter the potentials 共1兲 and, with of the space curves.
integer n, 5. The reflectionless wave packets ⌽1 in Eq. 共36兲 and ⌽⬘1 in
Eq. 共37兲 are both characterized by the same parameters a,
− ប2 x b, k0, and m. We saw in Sec. IV that the ⌽1 wave packet
共␯ − n兲共␯ − n + 1兲sech2 共58兲 generally has group speed greater than the group speed
2ma2 a
v0 = បk0 / m of the Gaussian packet ⌽0. Is the same true of
are shown to be partners in supersymmetric algebra.14 If one ⌽1⬘? Is ⌽1 or ⌽1⬘ faster for the same parameter set?
of the partners has zero reflection amplitude they all do 共Ref. 6. There is a one-to-one correspondence between the reflec-
14, Eq. 共3.32兲兲. When ␯ is an integer one of the potentials tion of s-polarized electromagnetic waves and the reflec-
will be zero, and a null potential does not reflect, so all the tion of particle waves 共see, for example, Ref. 12, Sec.
integer-␯ potentials are nonreflecting. 1-3兲:

1156 Am. J. Phys., Vol. 75, No. 12, December 2007 John Lekner 1156
 6
␻2 2m N. Kiriushcheva and S. Kuzmin, “Scattering of a Gaussian wave packet
␧共x兲 2 ↔ 2 关E − V共x兲兴. 共63兲 by a reflectionless potential,” Am. J. Phys. 66, 867–872 共1998兲.
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F. Oberhettinger, “Hypergeometric functions,” in Handbook of Math-
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Here ␧共x兲 is the dielectric function, ␻ the angular frequency,
Applied Mathematics Series No. 55, 1964兲, Chap. 15.
and c is the speed of light. Write ␧共x兲 = 1 + ⌬␧, and find the 8
E. H. Kennard, “Zur quantenmechanik einfacher bewegungstypen,” Z.
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C. G. Darwin, “Free motion in the wave mechanics,” Proc. R. Soc. Lon-
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a兲
W. Gautschi, “Error function and Fresnel integrals,” in Handbook of
Electronic address: john.lekner@vuw.ac.nz Mathematical Functions, edited by M. Abramowitz and I. A. Stegun
1
P. S. Epstein, “Reflection of waves in an inhomogeneous absorbing me- 共NBS Applied Mathematics Series No. 55, 1964兲, Chap. 7.
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1157 Am. J. Phys., Vol. 75, No. 12, December 2007 John Lekner 1157