Download Reflectionless eigenstates of the sech 2 potential

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
␣ = 共␯ + 1 + ika兲,
2
It is remarkable, even amazing, that the potential well
V共x兲 = −
ប2 ␯共␯ + 1兲
2ma2 cosh2共x/a兲
共1兲
is reflectionless, at any energy, if ␯ is a positive integer. The
potential 共1兲 共or the equivalent dielectric function profile兲
was first considered by Epstein1 共see also Eckart2兲, and is
treated in some quantum mechanics texts.3,4 The ␯ = 1 form
of Eq. 共1兲 appears as the simplest of a family of reflectionless
profiles.5
When ␯ is a positive integer, there is no reflection at any
energy, and thus no reflection of any wave packet formed by
superposition of positive energy eigenstates. A numerical
study has reported6 that Gaussian wave packets are made
narrower by passage over a reflectionless sech2 well, and
that they can travel faster than a free-space packet. In this
paper we shall study these effects analytically, by first reducing the positive energy eigenstates to elementary form.
For positive energies we write E = ប k / 2m, and the
Schrödinger equation with potential energy given by Eq. 共1兲
reads
2 2
冋
册
共2兲
The potential 共1兲 is even in x, so parity is a good quantum
number, and the two independent solutions of Eq. 共2兲 can be
taken to be the even and odd functions4
冉 冊 冉
冉 冊
冉
冊
␺␯e 共x兲 = cosh
␺␯o共x兲
x
a
x
= cosh
a
− sinh2
␯+1
␯+1
1
x
F ␣, ␤ ; ;− sinh2
2
a
冊
⬁
F共␣, ␤ ; ␥ ; ␨兲 =
⌫共␣ + n兲⌫共␤ + n兲 ␨n
⌫共␥兲
兺 ⌫共␥ + n兲 n!
⌫共␣兲⌫共␤兲 n=0
共5兲
within the unit circle 兩␨兩 = 1. Because F共␣ , ␤ ; ␥ ; ␨兲 is symmetric with respect to interchange of ␣ and ␤, and ␣ and ␤ are
complex conjugates, with ␥ and sinh2 x / a real, ␺␯e and ␺␯o are
real.
When ␯ = 0 there is no potential, so ␺e0 and ␺o0 are what
mathematicians refer to as trivial solutions. However, their
structure gives us a hint for the reduction of solutions for all
integer ␯. In Eq. 共15.1.18兲 of Ref. 7, namely
冉
冊
1
cos共2␣ − 1兲z
,
F ␣,1 − ␣ ; ;sin2 z =
2
cos z
冉
冊
3
sin共2␣ − 2兲z
,
F ␣,2 − ␣ ; ;sin2 z =
2
共␣ − 1兲sin 2z
共6兲
共7兲
we set ␣ = 1 + 共i / 2兲ka. In both equations we replace z by ix / a.
These identities then give us the expected cos kx and sin kx
eigenstates:
␺e0 = cos kx,
␺o0 =
sin kx
.
ka
共8兲
The integer ␯ solutions can be obtained by using Eqs. 共6兲
and 共7兲 and the differentiation formulas of Sec. 15.2 of Ref.
7. For example 共with z = ix / a as above, and ␯ = 1兲
冉
冊
i
␣ = 1 + ka.
2
共9兲
Using Eq. 共7兲 and 共15.2.4兲 with n = 1, namely
共3兲
d ␥−1
†␨ F共␣, ␤ ; ␥ ; ␨兲‡ = 共␥ − 1兲␨␥−2F共␣, ␤ ; ␥ − 1; ␨兲, 共10兲
d␨
we find, after some reduction, that
where
1151
F denotes the hypergeometric function,7 which can be represented by the Gauss hypergeometric series
1
␺e1 = cos2 zF ␣,2 − ␣ ; ;sin2 z ,
2
1
1 3
x
sinh F ␣ + , ␤ + ; ;
2
2 2
a
x
,
a
共4兲
we set ␣ = 21 共1 + ika兲. Likewise in Eq. 共15.1.16兲 of Ref. 7,
which reads
II. REFLECTIONLESS POSITIVE ENERGY
EIGENSTATES
␯共␯ + 1兲
d 2␺
2
␺ = 0.
2 + k + 2
dx
a cosh2共x/a兲
1
␤ = 共␯ + 1 − ika兲.
2
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
共filled shape兲.
␺e1 = cos kx − tanh
x sin kx
.
a ka
共11兲
Likewise 共again with z = ix / a兲
冉
冊
3
␺o1 = − i cos2 z sin zF ␣,3 − ␣ ; ;sin2 z ,
2
␣=
3 i
+ ka.
2 2
共12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
d␨
共and using the ␣ ↔ ␤ symmetry兲 we find
再
␺o1 = †1 + 共ka兲2‡−1 ka sin kx + tanh
= †1 + 共ka兲 ‡
2 −1
− 3ka tanh
再冋
冎
x
cos kx .
a
2x
1 + 共ka兲 − 3 tanh
2
冎
a
册
cos kx
x
+ 3ka tanh cos kx .
a
ika†4 + 共ka兲2‡ o
␺2 = †1 + 共ka兲2‡−1
1 + 共ka兲2
再
⫻ 1 + 共ka兲2 − 3 tanh2
Am. J. Phys., Vol. 75, No. 12, December 2007
冎
x
x ikx
+ 3ika tanh
e .
a
a
共19兲
共20兲
Note that ␺␯e 共and thus also ␺␯±兲 are normalized to unity at
x = 0. At large 兩x兩 the modulus of ␺␯± is greater than unity for
␯ ⬎ 0: for example,
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
The probability flux density
共16兲
册
x
sin kx
a
共17兲
冉 冊
d␺
ប
Im ␺*
m
dx
共22兲
is zero for the 共real兲 even and odd eigenstates, and is independent of x for the ␺± eigenstates, as it must be by conservation of particles:
J+0 =
The even and odd eigenfunctions for ␯ = 2 are shown in
Fig. 2.
1152
␺+2 = ␺e2 +
J=
1 + 共ka兲2 − 3 tanh2
册
i†1 + 共ka兲2‡ o
x ikx
i
tanh
␺1 = 1 +
e ,
ka
ka
a
共15兲
␣␤
d
F共␣, ␤ ; ␥ ; ␨兲 =
F共␣ + 1, ␤ + 1; ␥ + 1; ␨兲.
d␨
␥
再冋
冎
冋
共18兲
␺+1 = ␺e1 +
兩␺±1 兩2 →
The ␯ = 2 odd eigenstate can be found by differentiation of ␺e1
on using Eq. 共15.2.1兲, which reads
␺o2 = 共ka兲−1†4 + 共ka兲2‡−2
␺+0 = ␺e0 + ika␺o0 = eikx ,
共14兲
x
sin kx .
a
The result is
We see that the integer-␯ eigenstates rapidly become more
complicated as ␯ increases, but they remain elementary functions, expressible in terms of cos kx, sin kx and tanh x / a.
From the even and odd eigenstates we can construct by
superposition the reflectionless energy eigenstates propagating in either the +x or −x directions, for example,
共13兲
The even and odd eigenfunctions for ␯ = 1 are shown in
Fig. 1.
Similarly, using Eq. 共10兲 we can get ␺e2 by differentiating
o
␺ 1:
␺e2
Fig. 2. The ␯ = 2 even and odd positive energy eigenstates, ␺e2 and ␺o2 共Eqs.
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.
បk
,
m
J+1 =
បk 1 + 共ka兲2
,
m 共ka兲2
J+2 =
បk 4 + 共ka兲2
.
m 1 + 共ka兲2
共23兲
III. CONSTRUCTION OF NONREFLECTING WAVE
PACKETS
Kiriushcheva and Kuzmin6 have numerically integrated
the time-dependent Schrödinger equation H⌽ = iប⳵t⌽ to follow the passage of a wave packet through the ␯ = 1 potential.
John Lekner
1152
pendence of the wave packet 共25兲 is obtained from
⌽0共x,t兲 =
1
冑2␲
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
冊
x
i
tanh eikx,
ka
a
␺+1 共k,x兲 = 1 +
Ek =
ប 2k 2
.
2m
共29兲
The wave packets formed by superposition of these eigenstates,
⌽1共x,t兲 =
Fig. 3. The Gaussian wave packet of Eq. 共25兲, with x0 = −5b, k0b = 1 at times
t = 0 and t = 10b / v0, where v0 = បk0 / m is the group speed. At t = 0 only the
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.
The packet is most compact at t = 0; at later 共and earlier兲 times its spread is
关b2 + 共បt / mb兲2兴1/2.
1
冑2␲
⌽0共x,0兲 = expˆik0共x − x0兲 − 共x − x0兲 /2b ‰
2
8,9
is known
再冉
to have the time development
冑
⌽0共x,t兲 =
1
exp ik0 x − x0 − v0t
2
iបt
b2 +
m
b
共24兲
2
冊
冎
− 共x − x0 − v0t兲2/2共b2 + iបt/m兲 ,
A0共k兲 =
冑2␲
冕
再
−⬁
冎
共25兲
⌽0共x,0兲 =
冑2␲
冕
共26兲
dxe A0共k兲.
共27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 de1153
⌽1共x,t兲 =
Am. J. Phys., Vol. 75, No. 12, December 2007
− ib
冑2␲
冉
冕
⬁
dke−ik
冎
共31兲
2បt/m
−⬁
⫻ ka + i tanh
冊
x ik共x−x 兲−共k − k 兲2b2/2
0
0
e
.
a
共32兲
The integrals may be evaluated in terms of the Gaussian
integral, since the exponent
ab
冑2␲
2 2
e−k0b /22
冕
⬁
dkk sin kze−k
0
=−
ab
冑2␲
共33兲
2 2
e−k0b /22⳵z
冕
2关b2+iបt/m兴/2
⬁
dk cos kze−k
2关b2+iបt/m兴/2
0
冉
冊
z2/2
ab
2 2
exp
−
k
b
/2
−
,
0
关b2 + iបt/m兴3/2
b2 + iបt/m
共34兲
The tanh x / a term gives the contribution
tanh
x b −k2b2/2
e 0 2
a 冑2␲
⬁
ikx
共30兲
再
z = x − x0 − ik0b2 .
The Fourier inverse is a superposition of free-space energy
eigenstates eikx:
1
␺+1 共k,x兲A共k兲,
−⬁
1
A共k兲 = − ikaA0共k兲 = − ikab exp − ikx0 − 共k − k0兲2b2 .
2
=
1
= b exp − ikx0 − 共k − k0兲2b2 .
2
2បt/2m
is quadratic in k. The ka term in the integrand gives the
contribution
⬁
dxe−ikx⌽0共x,0兲
dke−ik
− k20b2/2 − k2关b2 + iបt/m兴/2 + ik共x − x0 − ik0b2兲
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
the propagation and spreading of the Gaussian packet.
The Fourier transform of Eq. 共24兲 is
1
⬁
satisfy the time-dependent Schrödinger equation, for any
Fourier amplitude A共k兲. Because of the 共ka兲−1 term in ␺+1 , a
relatively simple wave packet is produced by taking
This gives
They reported that the packet accelerated and narrowed
slightly 共relative to the zero-potential case兲. Their wave
packet was taken to be Gaussian at time zero. In the absence
of a potential 共in our case for ␯ = 0兲 such a wave packet,
starting at t = 0 centered on x = x0,
冕
= tanh
冕
⬁
0
dke−k
2共b2+iបt/m兲
cos kz
冊
冉
z2/2
b
x
2 2
.
exp
−
k
b
/2
−
0
a 关b2 + iបt/m兴1/2
b2 + iបt/m
共35兲
Thus the wave packet built up from the nonreflecting energy
John Lekner
1153
If we take A共k兲 = A0共k兲 in Eq. 共30兲, a different but qualitatively similar wave packet results, namely
⌽1⬘共x,t兲 = ⌽0共x,t兲
+
冑
冉
冊
x − x0 − ik0b2
␲b
x 22
tanh e−k0b /2 erfc
冑2冑b2 + iបt/m ,
2a
a
共37兲
where the complementary error function is defined by10
erfc共␨兲 =
冑 冕
2
␲
⬁
␨
2
d ␰ e −␰ = 1 −
冑 冕
2
␲
␨
2
d␰e−␰ = 1 − erf共␨兲.
0
共38兲
We can verify by direct differentiation that the wave packets given by Eqs. 共36兲 and 共37兲 satisfy the Schrödinger equation 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
otherwise using the same parameters and line types as in Fig. 3, namely
x0 = −5a, k0a = 1, at times t = 0 共envelope only兲 and t = 10a / v0. The constriction near x = 0 is due the potential −共ប2 / ma2兲sech2共x / a兲 共not shown兲. The
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
Eq. 共47兲.
eigenstates ␺+1 共k , x兲 with Fourier amplitude given by Eq. 共31兲
is
⌽1共x,t兲 = ⌽0共x,t兲
冋
册
a共x − x0 − ik0b2兲
x
+ tanh
,
b2 + iបt/m
a
共36兲
Kiriushcheva and Kuzmin6 made a numerical study of
wave packet propagation in the presence of the ␯ = 1 sech2
potential. They found that a wave packet, constructed to have
the form 共24兲 at time zero, propagated through the potential
region faster than at the group speed v0 = បk0 / m of the freespace 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.
The envelope of this packet is 兩⌽1共x , t兲兩, where
冤
兩⌽1共x,t兲兩2 = 兩⌽0共x,t兲兩2 tanh2
where ⌽0 is the free-space Gaussian wave packet given in
Eq. 共25兲. The propagation of the packet through the potential
well region is illustrated in Figs. 4 and 5.
x
a
冥
x
+ a2†共x − x0兲2 + k20b4‡
a
b4 + 共បt/m兲2
共39兲
2ab2共x − x0 − v0t兲tanh
+
with
兩⌽0共x,t兲兩2 =
b
冑b2 + 共បt/mb兲2
再
exp
冎
− 共x − x0 − v0t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 =
Fig. 5. Another view of the ␯ = 1 nonreflecting wave packet, drawn as a
spiral curve at t = 10a / v0, with parameters as in Fig. 4. The x axis is the
direction of propagation, the transverse axes are real and imaginary parts of
⌽, also shown in Fig. 4 as solid and dashed curves, respectively.
1154
Am. J. Phys., Vol. 75, No. 12, December 2007
冉
冊
1
ប2 2
k0 + 2 ,
2m
2b
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 differentiate 兩⌽1兩2 with respect to x, and set x = x0 + d. The zeros
of this derivative give the stationary points of 兩⌽1兩2. There
result two cubic equations for d, corresponding to tanh x / a
→ 1 and −1. There are three intrinsic speeds in the problem,
namely
John Lekner
1154
v0 =
បk0
,
m
ua =
ប
ma
and ub =
ប
.
mb
共42兲
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 reduce 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 maximum 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
of the group speed v is
再
v = v0 1 +
冎
1
a2 + b2
−
+ O共k−6
0 兲 .
共k0b兲2 a2b4k40
共45兲
At low energies the situation is more complicated, and the
group speed can be much greater than that of the free-space
packet. For example, when a = b 共and thus ua = ub兲,
v = v0
再
冎
1 1
1
+ 共k0a兲2/3 + O共k0a兲4/3 .
2/3 +
3 9
共k0a兲
共46兲
At k0a = 1 = k0b we find from Eq. 共44兲, with ␳ = 29+ 3冑93,
that
v0
v = 关共␳/2兲1/3 + 共2/␳兲1/3 + 1兴 ⬇ 1.46557v0 .
3
共47兲
The width of the free-space Gaussian wave packet, ␴0共t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],
and F is the content of the square bracket in Eq. 共39兲. Outside the range of the potential the maximum of 兩⌽1兩2 occurs
at x = x0 + vt where the group speed v is determined by the
cubic Eq. 共44兲. The location of the points x at which X
− ln F = 1 is determined by a transcendental equation, which
we shall simplify by neglecting the slowly varying ln F term.
The breadth of the wave packet, centered on xm = x0 + vt, is
then approximately given by the points x1 at which X = 1:
共x1 − x0 − v0t兲2 = b2 + 共បt/mb兲2 .
共48兲
␴共t兲 ⬇ ␴0共t兲 − 共v − v0兲t,
= b + 共បt/mb兲 .
2
2
共49兲
Thus when the group speed v exceeds the group speed v0 of
the free-space Gaussian packet, we expect a narrower packet
than the Gaussian 共at the same time of propagation兲. The
expression given in Eq. 共49兲 is a rough estimate only, and
cannot be valid when v − v0 ⬎ ub = ប / mb, since then ␴共t兲
would become negative at large times.
Our results for the group speed and wave packet width are
in qualitative agreement with the numerical example given
by Kiriushcheva and Kuzmin.6 The propagation of a nonre1155
Am. J. Phys., Vol. 75, No. 12, December 2007
flecting packet and of a free-space Gaussian packet is illustrated in Fig. 6 for the parameter choice leading to the velocity ratio of Eq. 共47兲.
Next we consider the question “what is so special about
the integer values of ␯ in Eq. 共1兲?”
V. CRITICAL BINDING
The phenomenon of zero reflection of waves is common in
optics, acoustics and quantum mechanics.11 For example, antireflection coatings can make reflection zero at one wavelength, or very small over a range of wavelengths. What is
rare is zero reflection at any wavelength. In optics and acoustics there is zero reflection by a sharp interface at the Brewster and Green angles 共see, for example, Secs. 1-2 and 1-4 of
Ref. 12兲
tan2 ␪B =
The distance ␴共t兲 between x1 and xm is then roughly
␴20共t兲
Fig. 6. Comparison of the propagation of a free-space Gaussian packet and
of the ␯ = 1 nonreflecting packet, Eqs. 共25兲 and 共36兲, respectively. The parameters are a = b, k0a = 1, t = 0 共top兲, 1, 2, 3, 4 and 5a / v0 共bottom兲. The
potential well is also shown 共shaded兲. The nonreflecting packet has a group
speed roughly 1.5 times greater than the Gaussian. The curves show 兩⌽0兩2
and 兩⌽1兩2, with ⌽1 having the larger amplitude. 兩⌽1兩2 shows an indentation
near x = 0 as it passes over the potential; 兩⌽0兩2 共dashed curves兲 does not,
because it is a free-space solution of Schrödinger’s equation.
␧2
,
␧1
tan2 ␪G =
共 ␳ 2v 2兲 2 − 共 ␳ 1v 1兲 2
␳21共v21 − v22兲
,
共50兲
where ␧, ␳, v are dielectric constants, densities and sound
speeds, respectively. The reflection of the electromagnetic p
wave and of the acoustic wave is, however, zero only in the
limit when the step from ␧1 to ␧2 or 共␳1 , v1兲 to 共␳2 , v2兲 is very
rapid on the scale of the wavelength.
Here we have an example of a potential 共or dielectric function profile兲 with a characteristic length a, and zero reflection
for any values of a at any energy, provided ␯ is an integer.
Why?
The potential 共1兲 has, for given ␯, the bound states3
John Lekner
1155
En = −
ប2
共␯ − n兲2,
2ma2
n = 0,1,2 . . . ,关␯兴,
共51兲
where 关␯兴 is the integer part of ␯. The Schrödinger equation
for bound states, and the even and odd energy eigenstates,
can be obtained from Eqs. 共2兲 and 共3兲, respectively, by replacing k by q = ik. The energy becomes −ប2q2 / 2m. For example, for ␯ = 1 we have the bound state
1
q= ,
a
E共1兲
0 =
− ប2
,
2ma2
␺共1兲
0 = sech
x
a
共52兲
E共1兲
1 = 0,
x
␺共1兲
1 = tanh .
a
共53兲
For ␯ = 2 we have two bound states and one zero-energy
state:
2
q= ,
a
E共2兲
0
2ប2
=−
,
ma2
1
q= ,
a
E共2兲
1
ប2
=−
,
2ma2
q = 0,
E共2兲
2 = 0,
␺共2兲
0
x
= sech ,
a
␺共2兲
1
2
共54兲
共55兲
冉 冊
ប2
2m兩E兩
x
x
− 2 tanh2 .
a
a
共56兲
1/2
共57兲
共the positive sign is to be taken for bound states, E ⬍ 0, and
the negative sign for virtual states, E ⬎ 0兲. Thus systems near
critical binding can be thought of as having a reach far beyond the range of the binding potential, in our case the
length a. This is one reason for the nonreflecting integer-␯
sech2 potentials. Another is the V共−x兲 = V共x兲 symmetry of the
potential: as shown in Ref. 11, reflection amplitudes of symmetric profiles automatically have coincident zeros of their
real and imaginary parts, whereas general profiles do not.
The above association between critical binding strengths
of the sech2 potentials and zero reflection adds a physical
heuristic to the mathematical explanation of supersymmetric
quantum mechanics. In the latter the potentials 共1兲 and, with
integer n,
− ប2
x
共␯ − n兲共␯ − n + 1兲sech2
2ma2
a
共58兲
are shown to be partners in supersymmetric algebra.14 If one
of the partners has zero reflection amplitude they all do 共Ref.
14, Eq. 共3.32兲兲. When ␯ is an integer one of the potentials
will be zero, and a null potential does not reflect, so all the
integer-␯ potentials are nonreflecting.
1156
Am. J. Phys., Vol. 75, No. 12, December 2007
冊
共59兲
A wavelength can be defined by ␾共x + ␭ , t兲 − ␾共x , t兲 = 2␲, or
␾共x + ␭ / n , t兲 − ␾共x , t兲 = 2␲ / n. In the limit of large n we obtain
the local wavelength
␭共x,t兲 =
2␲
.
⳵x␾共x,t兲
␭0共x,t兲 =
The special property of the integer-␯ potentials 共−ប2 /
2ma2兲␯共␯ + 1兲sech2 x / a is that they support a critically bound
state: one that has zero energy and a wave function of infinite range.13 Systems which are near critical binding have
special properties associated with the long range of the
wavefunction. In three dimensions, the zero-energy scattering length s goes to infinity as the binding energy E tends to
zero:3,13
s= ±
冉
共60兲
Verify that for the Gaussian packet Eq. 共60兲 gives
x
x
= sech tanh ,
a
a
2
␺共2兲
2 = sech
1. Is it true in general that symmetric potentials at critical
binding strength do not reflect? 关Hint: one counterexample is sufficient to disprove a conjecture. Try a square
well.兴
2. The Gaussian wave packet ⌽0 and the ␯ = 1 wave packet
⌽1 共Eqs. 共25兲 and 共36兲兲 both have shorter wavelengths at
the front: see Figs. 3–5. For the Gaussian pulse the phase
␾, defined by ⌽ = 兩⌽兩exp共i␾兲, is
1
共x − x0 − v0t兲2បt/2m
␾共x,t兲 = k0 x − x0 − v0t +
.
2
b4 + 共បt/m兲2
and the just-bound 共or just-unbound兲 state
q = 0,
VI. SUGGESTED PROBLEMS
2␲
.
2共x − x0 − v0t兲បt/m
k0 +
b4 + 共បt/m兲2
共61兲
Find the corresponding result for ⌽1 and compare with plots
of ⌽0 and ⌽1.
3. The phase ␾共x , t兲 of a wave packet is stationary when
d ␾ = ⳵x␾共x , t兲dx + ⳵t␾共x , t兲dt is zero. The phase velocity is
the value of dx / dt when d␾ is zero, v p = −⳵t␾共x , t兲 /
⳵x␾共x , t兲. Find v p for the Gaussian free-space packet, and
show that the phase speed is half the group speed
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.
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
coordinate system. A snapshot of the space curve of ⌽1 is
shown in Fig. 5. It is predominantly right handed, but
there is a left-handed portion in the tail. The curve is right
handed if the phase is increasing with x; the change in
handedness occurs at zeros of ⳵x␾共x , t兲. Verify that for the
Gaussian packet the change occurs at
x = x 0 + v 0t −
mko 4
†b + 共បt/m兲2‡,
2បt
共62兲
and find the corresponding result for ⌽1. Compare with plots
of the space curves.
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,
b, k0, and m. We saw in Sec. IV that the ⌽1 wave packet
generally has group speed greater than the group speed
v0 = បk0 / m of the Gaussian packet ⌽0. Is the same true of
⌽1⬘? Is ⌽1 or ⌽1⬘ faster for the same parameter set?
6. There is a one-to-one correspondence between the reflection of s-polarized electromagnetic waves and the reflection of particle waves 共see, for example, Ref. 12, Sec.
1-3兲:
John Lekner
1156
␧共x兲
2m
␻2
2 ↔ 2 关E − V共x兲兴.
c
ប
6
共63兲
Here ␧共x兲 is the dielectric function, ␻ the angular frequency,
and c is the speed of light. Write ␧共x兲 = 1 + ⌬␧, and find the
functional form of ⌬␧ corresponding to the V共x兲 given in Eq.
共1兲. Is ⌬␧ frequency dependent?
a兲
Electronic address: [email protected]
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2
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4
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5
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1
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by a reflectionless potential,” Am. J. Phys. 66, 867–872 共1998兲.
7
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8
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9
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11
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共1990兲.
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John Lekner
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