Solutions to Problems in Merzbacher, Quantum Mechanics, Third Edition Homer Reid April 5, 2001 Chapter 7 Before starting on these problems I found it useful to review how the WKB approximation works in the first place. The Schrödinger equation is − ~2 d 2 Ψ(x) + V (x)Ψ(x) = EΨ(x) 2m dx2 or d2 Ψ(x) + k 2 (x)Ψ(x) = 0, dx2 We postulate for Ψ the functional form k(x) ≡ r 2m [E − V (x)]. ~2 Ψ(x) = AeiS(x)/~ in which case the Schrödinger equation becomes i~S 00 (x) = [S 0 (x)]2 − ~2 k 2 (x). (1) This equation can’t be solved directly, but we obtain guidance from the observation that, for a constant potential, S(x) = ±kx, so that S 00 vanishes. For a nonconstant but slowly varying potential we might imagine S 00 (x) will be small, and we may take S 00 = 0 as the seed of a series of successive approximations to the exact solution. To be specific, we will construct a series of functions S0 (x), S1 (x), · · · , where S0 is the solution of (1) with 0 on the left hand side; S1 is a solution with S000 on the left hand side; and so on. In other words, at the nth step in the approximation sequence (by which point we have computed Sn (x)), we compute Sn00 (x) and use that as the source term on the LHS of (1) to calculate Sn+1 (x). Then we compute the second derivative of Sn+1 (x) and use this as the source term for calculating Sn+2 , and so on ad infinitum. In 1 Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 2 symbols, 0 = [S00 (x)]2 − ~2 k 2 (x) i~S000 i~S100 = [S10 (x)]2 [S20 (x)]2 = ···· (2) 2 2 − ~ k (x) (3) 2 2 (4) − ~ k (x) Equation (2) is clearly solved by taking S00 (x) = ±~k(x) ⇒ S0 (x) = S00 ± ~ Z x k(x0 )dx0 (5) −∞ for any constant S00 . Then S000 (x) = ±~k 0 (x), so (3) is p S10 (x) = ±~ k 2 (x) ± ik 0 (x). With the two ± signs here, we appear to have four possible choices for S10 . But let’s think a little about the ± signs in this equation. The ± sign under the radical comes from the two choices of sign in (5). But if we chose, say, the plus sign in that equation, so that S00 > 0, we would also expect that S10 > 0. Indeed, if we choose the plus sign in (5) but the minus sign in (3), then S00 and S10 have opposite sign, so S10 differs from S00 by an amount at least as large as S00 , in which case our approximation sequence S0 , S1 , · · · has little hope of converging. So we choose either both plus signs or both minus signs in (3), whence our two choices are p p S10 = +~ k 2 (x) + ik 0 (x) or S10 = −~ k 2 (x) − ik 0 (x). (6) If V (x) is constant, k(x) is constant, and, as we observed before, the sequence of approximations terminates at 0th order with S0 being an exact solution. By extension, if V (x) is not constant but changes little over one particle wavelength, we have k 0 (x)/k 2 (x) 1, so we may expand the radicals in (6): ik 0 (x) ik 0 (x) S10 ≈ ~k(x) 1 + 2 or S10 ≈ −~k(x) 1 − 2 2k (x) 2k (x) or i~k 0 (x) S10 ≈ ±~k(x) + . (7) 2k(x) Integrating, Z i~ x k 0 (u) 0 dx 2 a k(u) Za x i~ k(x) = S1 (a) ± ~ k(u)du + ln 2 k(a) a S1 (x) = S1 (a) ± ~ Z x k(u)du + where a is some point chosen such that the approximation (7) is valid in the full range a < x0 < x. We could go on to compute S2 , S3 , etc., but in practice it seems the approximation is always terminated at S1 . Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 3 The wavefunction at this order of approximation is −1/2 Rx Ψ(x) = exp(iS1 (x)/~) = eiS1 (a)/~ e±i a k(u)du eln k(x)/k(a) s k(a) ±i R x k(u)du = Ψ(a) e a k(x) = Ψ(a)G± (x; a) where (8) G± (x; a) ≡ s k(a) ±i R x k(u)du e a . k(x) (9) We have written it this way to illustrate that the function G(x, a) is kind of like a Green’s function or propagator for the wavefunction, in the sense that, if you know what Ψ is at some point a, you can just multiply it by G± (x; a) to find out what Ψ is at x. But this doesn’t seem quite right: Schrödinger’s equation is a second-order differential equation, but (8) seems to be saying that we need only one initial condition—the value of Ψ at x = a—to find the value of Ψ at other points. To clarify this subtle point, let’s investigate the equations leading up to (8). If the approximation (7) makes sense, then there are two solutions of Schrödinger’s equation at x = a, one whose phase increases with increasing x (positive derivative), and one whose phase decreases. Equation (8) seems to be saying that we can use either G+ or G− to get to Ψ(x) from Ψ(a); but the requirement the dΨ/dx be continuous at x = a means that only one or the other will do. Indeed, in using (8) to continue Ψ from a to x we must choose the appropriate propagator—either G+ or G− , according to the derivative of Ψ at x = a; otherwise the overall wave function will have a discontinuity in its first derivative at x = a. So to use (8) to obtain values for Ψ at a point x, we need to know both Ψ and Ψ0 at a nearby point x = a, as should be the case for a second-order differential equation. If we want to de-emphasize this nature of the solution with the propagator we may write s 1 ±i R x k(u)du Ψ(x) = C e a (10) k(x) p where C = Ψ(a) k(a). In regions where V (x) > E, k(x) is imaginary, so it’s useful to define r 2m κ(x) = −ik(x) = [V (x) − E] (11) ~2 and s 1 ± R x κ(u)du e a . (12) Ψ(x) = C κ(x) Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 4 If we have a region of space in which the WKB approximation is valid, knowing the value of Ψ (and its derivative) at one point within the region is equivalent to knowing it everywhere, because we can use the propagator (9) to get from that one point to every other point within the region. The WKB method, however, gives us no way of determining the value of Ψ at that one starting point. Furthermore, even if we know Ψ at one point within a region of validity, we can’t use (8) to determine Ψ in other, nonadjacent regions, because we can’t carry the propagator across regions of invalidity. So basically what we need is a way of finding one starting value for Ψ(x) in every region of validity of the WKB approximation. How do we find such points? Well, one sure-fire way to get starting points in regions of validity is to identify regions of invalidity, of which there will be at least one adjacent to each region of validity, and then get values of Ψ at the boundaries of the regions of invalidity—which will also count as values in the regions of validity. So we need to identify the regions of invalidity of the WKB approximation and do a more accurate solution of the Schrödinger equation there. The WKB approximation breaks down when k 0 /k 2 1 ceases to hold, which is true when k ≈ 0 but k 0 6= 0, which happens near a classical turning point of the motion—i.e., a point x0 at which V (x0 ) = E. But near such a point we may expand V (x) − E in a Taylor series around the point x0 ; if we keep only the first (linear in x) term in the series, we arrive at a Schrödinger equation which we can solve exactly in the vicinity of x0 . To do this, suppose the point x0 is a classical turning point of the motion, so that V (x0 ) = E. In the neighborhood of x0 we may expand V (x): V (x) = E + (x − x0 )V 0 (x0 ) + · · · (13) Then the Schrödinger equation becomes d2 2m Ψ(x) − 2 V 0 (x0 )(x − x0 )Ψ(x) = 0. 2 dx ~ The useful substitution here is 1/3 2m 0 u(x) = γ(x − x0 ) γ≡ V (x0 ) ~2 so x(u) = u + x0 . γ If we define Φ(u) = Ψ(x(u)) then dΦ dΨ dx 1 = = Ψ0 (x(u)) du dx du γ 1 00 d2 Φ = 2 Ψ (x(u)) du2 γ (14) Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 5 so (14) becomes γ2 d2 Φ(u) − γ 3 (x − x0 )Φ(u) = 0 du2 or d2 Φ(u) − uΦ(u) = 0. du2 The solution to this differential equation is Φ(u) = β1 Ai(u) + β2 Bi(u) (15) so the solution to the Schrödinger equation (14) is Ψ(x) = β1 Ai γ(x − x0 ) + β2 Bi γ(x − x0 ) . For γ(x − x0 ) 1 we have the asymptotic expression 3/2 2 −1/4 β1 − 23 |γ(x−x0 )|3/2 Ψ(x) ≈ π −1/2 [γ(x − x0 )] e + β2 e+ 3 |γ(x−x0 )| 2 (16) and for γ(x − x0 ) −1 we have Ψ(x) ≈ π −1/2 |γ(x − x0 )| −1/4 h 2 π 3/2 |γ(x − x0 )| − 3 4 2 π i 3/2 − β2 sin . |γ(x − x0 )| − 3 4 β1 cos (17) To simplify these, we need to consider two possible kinds of turning point. Case 1: V 0 (x0 ) > 0. In this case the potential is increasing through the turning point at x0 , which means that V (x) < E for x < x0 , and V (x) > E for x > x0 . Hence the region to the left of the turning point is the classically accessible region, while the right of the turning point is classically forbidden. Since V 0 (x0 ) > 0, γ > 0, so for x < x0 (??) holds. For points close to the turning point on the left side, r 1/2 2m 0 2m [E − V (x)] ≈ V k(x) = (x0 − x)1/2 = γ 3/2 (x0 − x)1/2 ~2 ~2 so |γ(x − x0 )| −1/4 and Z x0 k(u)du = γ x 3/2 Z x0 = r γ k(x) (x0 − x)1/2 du = x 2 = |γ(x − x0 )|3/2 . 3 (18) 2 3/2 γ (x0 − x)3/2 3 (19) On the other hand, for points close to the turning point on the left side we have x > x0 , so γ(x − x0 ) > 0. In this region, Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 κ(x) = r 1/2 2m 2m 0 = γ 3/2 (x − x0 )1/2 V (x0 )(x − x0 ) V (x) − E ≈ ~2 ~2 so, for x near x0 , Z x Z x 2 κ(u)du = γ 3/2 (u − x0 )1/2 du = γ 3/2 (x − x0 )3/2 3 x0 x0 and also |γ(x − x0 )| −1/4 = r γ . κ(x) 6 (20) (21) (22) Using (18) and (19) in (17), and (21) and (22) in (16), the solutions to the Schrödinger equation on either side of a classical turning point x0 at which V 0 (x0 ) > 0 are s Z x0 1 h π k(u)du − 2β1 cos Ψ(x) = k(x) 4 x Z x0 π i k(u)du − − β2 sin , x < x0 (23) 4 x s i R 1 h Rxx κ(u)du − x κ(u)du Ψ(x) = β1 e 0 + β 2 e x0 , x < x0 (24) κ(x) (we redefined the β constants slightly in going to this equation). Case 2: V 0 (x0 ) < 0. In this case the potential is decreasing through the turning point, so the classically accessible region is to the right of the turning point, and the forbidden region to the left. Since V 0 (x0 ) < 0, γ < 0. That means that the regions of applicability of (16) and (17) are on opposite sides of the turning points as they were in the previous case. The solutions to the Schrödinger equation on either side of the turning point are s i R 1 h R x0 κ(u)du − x κ(u)du β1 e x + β 2 e x0 , x < x0 (25) Ψ(x) = κ(x) s Z x π 1 h k(u)du − Ψ(x) = 2β1 cos k(x) 4 x0 Z x π i − β2 sin , x > x0 (26) k(u)du − 4 x0 (27) So, to apply the WKB approximation to a given potential V (x), the first step is to identify the classical turning points of the motion, and to divide space Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 7 up into regions bounded by turning points, within which regions the WKB approximation (7) is valid. Then, for each turning point, we write down (23) and (24) (or (25) and (26)) at nearby points on either side of the turning point, and then use (10) to evolve the wavefunction from those points to other points within the separate regions. We should probably quantify the meaning of “nearby” in that last sentence. Suppose x0 is a classical turning point of the motion, and we are looking for points x0 ± at which to make the “handoff” from approximations (16) and (17) to the WKB approximation These points must satisfy several conditions. First, the approximate Schrödinger equation (14) is only valid as long as we can neglect the quadratic and higher-order terms in the expansion (13), so we must have 0 V (x0 ) . (28) |V 00 (x0 )| |V 0 (x0 )| ⇒ 00 V (x0 ) But at the same time, γ must be sufficiently greater than 1 to justify the approximation (16) (or sufficiently less than -1 to justify (17)); the condition here is 1/3 −1/3 2m 0 2m 0 1 ⇒ . (29) V (x ) V (x ) 0 0 ~2 ~2 Finally, the points x± must be sufficiently far away from the turning points that the approximation (7) is valid for the derivative of the phase of the wavefunction; the condition for this to be the case was 0 2 0 k (x) ~ V (x ± ) 1 1⇒ (30) 1. k 2 (x) 2 2m [E − V (x ± )]3/2 If there are no points x0 ± satisfying all three conditions, the WKB approximation cannot be used. To apply all of this to the problem of bound states in a potential well, consider a potential like that shown in Figure 1, with two classical turning points at x = a and x = b. Although there are no discontinuities in the potential here, the problem may be analyzed in a manner similar to that used in the consideration of one-dimensional piecewise constant potentials, as in Chapter 6: we divide space into a number of distinct regions, obtain solutions of the Schrödinger equation in each region, and then match values and derivatives at the region boundaries. To divide space into distinct regions in this case, we begin by identifying narrow regions around the turning points a and b in which the linear approximation (13) is valid. In the narrow region around x = a, we may use (25) and (26); around x = b we may use (23) and (24). Let the narrow such region around a be a − 1 < x < a + 1, and that around b be b − 2 < x < b + 2. Then space divides naturally into five regions: (a) x < a − In this region we are far enough to the left of the turning point that the WKB approximation is valid, and the wavefunction takes the form (10). However, we 8 Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 PSfrag replacements V (x) E a b Figure 1: A potential V (x) with two classical turning points for an energy E. 9 Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 must throw out the term that grows exponentially as x → −∞, so we are left with s 1 − R (a−) κ(u)du e x , x < a − . (31) Ψ(x) = A κ(x) (b) a − < x < a + In this region we are close enough to the turning point that (13) is valid, so (25) and (26) may be used. s Ra 1 − R a κ(u)du Ψ(x) = β1 e x + β2 e+ x κ(u)du , x < (a − ). (32) κ(x) From (31) and (32) we see that continuity of both the value and first derivative of Ψ(x) at x = a − requires taking β1 = A, β2 = 0. With this choice of constants, we achive continuity not only of the value and first derivative of Ψ but also of all higher derivatives, as must be the case since there is no discontinuity in the potential. But now that we know the value of Ψ at x = a − , we also know it at x = a + , because of course the solution of the Schrödinger equation in the narrow strip around a (to which (32) is an asymptotic approximation for x < a) is valid throughout the strip; the same solution that’s valid at x = a − is valid at x = a + . With β1 = A and β2 = 0, (26) becomes Ψ(x) = 2A =A s s 1 cos k(x) Z x k(u)du − a π 4 i Rx 1 h +i(R x k(u)du−π/4) a + e−i( a k(u)du−π/4) , e k(x) x = (a + )− (33) (c) a + < x < b − In this region the WKB approximation (7) is valid, so we may use (8) to find the wavefunction at any point within the region. Using the expression (33) for the wavefunction at x = a + , integrating from a + to x in the propagator (9), and using G+ and G− , respectively, to propagate the first and second terms in Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 10 (33), we obtain for the wavefunction at a point x in this region s “R “R ” ” Rx Rx 1 −i a(a+) k(u)du+ (a+) −π/4 +i a(a+) k(u)du+ (a+) k(u)du−π/4 +e Ψ(x) = A e k(x) s i Rx 1 h +i(R x k(u)du−π/4) a =A e + e−i( a k(u)du−π/4) k(x) s Z x 1 π = 2A cos k(u)du − k(x) 4 a s ! Z b Z b 1 π , a+<x <b− cos = 2A k(u)du − k(u)du − k(x) 4 a x (34) Okay, I have now carried this analysis far enough to see for myself exactly where the Bohr-Sommerfeld quantization condition Z b 1 k(u)du = n + π, n = 1, 2, · · · (35) 2 a comes from, which was my original goal, so I am now going to stop this exercise and proceed directly to the problems. Problem 7.1 Apply the WKB method to a particle that falls with acceleration g in a uniform gravitational field directed along the z axis and that is reflected from a perfectly elastic plane surface at z = 0. Compare with the rigorous solutions of this problem. We’ll start with the exact solution to the problem. The requirement of perfect elastic reflection at z = 0 may be imposed by taking V (x) to jump suddenly to infinity at z = 0, i.e. ( mgz, z > 0 V (x) = ∞, z ≤ 0. For z > 0, the Schrödinger equation is d2 Ψ(x) + dx2 d2 = 2 Ψ(x) + dx d2 = 2 Ψ(x) − dx 0= 2m [E − mgz] Ψ(x) ~2 2m2 g E − z Ψ(x) ~2 mg 2m2 g [z − z0 ] Ψ(x) ~2 (36) 11 Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 where z0 = E/mg. With the substitution u = γ(z − z0 ) γ= 2m2 g ~2 1/3 and taking Φ(u) = Ψ(x(u)), we find that (36) is just the Airy equation for Φ(u), d2 Φ(u) − uΦ(u) = 0 du2 with solutions Φ(u) = β1 Ai(u) + β2 Bi(u). Since we require a solution that remains finite as z → ∞, we must take β2 = 0. The solution to (36) is then Ψ(x) = β1 Ai γ(z − z0 ) , (z > 0). (37) For z < 0, I wasn’t quite sure how to account for the infinite potential jump at z = 0, so instead I supposed the potential for z < 0 to be a constant, V (z) = V0 , where eventually I’ll take V0 → ∞. Then the Schrödinger equation for z < 0 is d2 2m Ψ(z) − 2 [V0 − E] Ψ(z) = 0 2 dz ~ with solution r 2m −kz [V0 − E]. (38) Ψ(z) = Ae , k= ~2 Matching values and derivatives of (37) and (38) at z = 0, we have β1 Ai(−γz0 ) = A γβ1 Ai0 (−γz0 ) = −kA Dividing, we obtain 1 Ai(−γz0 ) 1 =− γ Ai0 (−γz0 ) k Now taking V0 → ∞, we also have k → ∞, so the RHS of this goes to zero; thus the condition is that −γz0 be a zero of the Airy function, which means the energy eigenvalues En are given by 2m2 g ~2 1/3 En = xnm ⇒ En = mg where xn is the nth root of the equation Ai(−xn ) = 0. mg 2 ~2 2 1/3 xn (39) Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 12 So that’s the exact solution. In the WKB approximation, the spectrum of energy eigenvalues is determined by the condition (35). In this case the classical turning points are at z = 0 and z = z0 , so we have Z z0 1 n+ π= k(z) dz 2 0 r Z 2m z0 = [E − mgz]1/2 dz ~2 0 r Z 2m2 g z0 [z0 − z]1/2 du = ~2 0 r z0 2m2 g 2 3/2 − (z0 − z) = 2 ~ 3 0 r 3/2 2 2 2m g E = 3 ~2 mg so the nth eigenvalue is given by En = mg 2 ~2 2 1/3 2/3 1 3 n+ π . 2 2