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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 + ikao0 = 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⌽ = iបt⌽ 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⌽ = iបt⌽ 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 /22z 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 de− = 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] P. S. Epstein, “Reflection of waves in an inhomogeneous absorbing medium,” Proc. Natl. Acad. Sci. U.S.A. 16, 627–637 共1930兲. 2 C. Eckart, “The penetration of a potential barrier by electrons,” Phys. Rev. 35, 1303–1309 共1930兲. 3 L. D. Landau and E. M. Lifshitz, Quantum Mechanics 共Pergamon, Oxford, 1965兲, 2nd ed., Secs. 23 and 25. 4 S. Flügge, Practical Quantum Mechanics 共Springer-Verlag, Berlin, 1974兲, Problem 39. 5 I. Kay and H. E. Moses, “Reflectionless transmission through dielectrics and scattering potentials,” J. Appl. Phys. 27, 1503–1508 共1956兲. 1 1157 Am. J. Phys., Vol. 75, No. 12, December 2007 N. Kiriushcheva and S. Kuzmin, “Scattering of a Gaussian wave packet by a reflectionless potential,” Am. J. Phys. 66, 867–872 共1998兲. 7 F. Oberhettinger, “Hypergeometric functions,” in Handbook of Mathematical Functions, edited by M. Abramowitz and I. A. Stegun 共NBS Applied Mathematics Series No. 55, 1964兲, Chap. 15. 8 E. H. Kennard, “Zur quantenmechanik einfacher bewegungstypen,” Z. Phys. 44, 326–352 共1927兲. 9 C. G. Darwin, “Free motion in the wave mechanics,” Proc. R. Soc. London, Ser. A 117, 258–293 共1928兲. 10 W. Gautschi, “Error function and Fresnel integrals,” in Handbook of Mathematical Functions, edited by M. Abramowitz and I. A. Stegun 共NBS Applied Mathematics Series No. 55, 1964兲, Chap. 7. 11 J. Lekner, “Nonreflecting stratifications,” Can. J. Phys. 68, 738–742 共1990兲. 12 J. Lekner, Theory of Reflection of Electromagnetic and Particle Waves 共Nijhoff/Springer, Dordrech/Berlin, 1987兲. 13 J. Lekner, “Critical binding of diatomic molecules,” Mol. Phys. 23, 619– 625 共1972兲. 14 F. Cooper, A. Khare, and U. Sukhatme, Supersymmetry in Quantum Mechanics 共World Scientific, Singapore, 2001兲, pp. 46–47. John Lekner 1157