Solutions to Problems in Merzbacher, Quantum Mechanics, Third Edition. Chapter 7

Transcription

1 Solutions to Problems in Merzbcher, Quntum Mechnics, Third Edition Homer Reid April 5, 200 Chpter 7 Before strting on these problems I found it useful to review how the WKB pproimtion works in the first plce. The Schrödinger eqution is or 2 d 2 Ψ() + V ()Ψ() = EΨ() d2 d 2 d 2 Ψ() + k2 ()Ψ() = 0, We postulte for Ψ the functionl form Ψ() = Ae is()/ in which cse the Schrödinger eqution becomes k() [E V ()]. 2 i S () = [S ()] 2 2 k 2 (). () This eqution cn t be solved directly, but we obtin guidnce from the observtion tht, for constnt potentil, S() = ±k, so tht S vnishes. For nonconstnt but slowly vrying potentil we might imgine S () will be smll, nd we my tke S = 0 s the seed of series of successive pproimtions to the ect solution. To be specific, we will construct series of functions S 0 (), S (),, where S 0 is the solution of () with 0 on the left hnd side; S is solution with S 0 on the left hnd side; nd so on. In other words, t the nth step in the pproimtion sequence (by which point we hve computed S n ()), we compute S n() nd use tht s the source term on the LHS of () to clculte S n+ (). Then we compute the second derivtive of S n+ () nd use this s the source term for clculting S n+2, nd so on d infinitum. In

2 Homer Reid s Solutions to Merzbcher Problems: Chpter 7 2 symbols, Eqution (2) is clerly solved by tking 0 = [S 0 ()]2 2 k 2 () (2) i S 0 = [S ()] 2 2 k 2 () (3) i S = [S 2()] 2 2 k 2 () () S 0 () = ± k() S 0() = S 00 ± for ny constnt S 00. Then S 0 () = ± k (), so (3) is S () = ± k 2 () ± ik (). k( )d (5) With the two ± signs here, we pper to hve four possible choices for S. But let s think little bout the ± signs in this eqution. The ± sign under the rdicl comes from the two choices of sign in (5). But if we chose, sy, the plus sign in tht eqution, so tht S 0 > 0, we would lso epect tht S > 0. Indeed, if we choose the plus sign in (5) but the minus sign in (3), then S 0 nd S hve opposite sign, so S differs from S 0 by n mount t lest s lrge s S 0, in which cse our pproimtion sequence S 0, S, hs little hope of converging. So we choose either both plus signs or both minus signs in (3), whence our two choices re S = + k 2 () + ik () or S = k 2 () ik (). (6) If V () is constnt, k() is constnt, nd, s we observed before, the sequence of pproimtions termintes t 0th order with S 0 being n ect solution. By etension, if V () is not constnt but chnges little over one prticle wvelength, we hve k ()/k 2 (), so we my epnd the rdicls in (6): or Integrting, { } S k() + ik () 2k 2 () S () = S () ± { } or S k() ik () 2k 2 () S ± k() + i k () 2k(). (7) = S () ± k(u)du + i 2 k(u)du + i 2 ln k() k() k (u) k(u) d where is some point chosen such tht the pproimtion (7) is vlid in the full rnge < <. We could go on to compute S 2, S 3, etc., but in prctice it seems the pproimtion is lwys terminted t S.

3 Homer Reid s Solutions to Merzbcher Problems: Chpter 7 3 where The wvefunction t this order of pproimtion is Ψ() = ep(is ()/ ) = (e is()/ )( e ±i R k(u)du)( e ln k()/k()) /2 k() R = Ψ() k() e±i k(u)du = Ψ()G ± (; ) (8) k() R G ± (; ) k() e±i k(u)du. (9) We hve written it this wy to illustrte tht the function G(, ) is kind of like Green s function or propgtor for the wvefunction, in the sense tht, if you know wht Ψ is t some point, you cn just multiply it by G ± (; ) to find out wht Ψ is t. But this doesn t seem quite right: Schrödinger s eqution is second-order differentil eqution, but (8) seems to be sying tht we need only one initil condition the vlue of Ψ t = to find the vlue of Ψ t other points. To clrify this subtle point, let s investigte the equtions leding up to (8). If the pproimtion (7) mkes sense, then there re two solutions of Schrödinger s eqution t =, one whose phse increses with incresing (positive derivtive), nd one whose phse decreses. Eqution (8) seems to be sying tht we cn use either G + or G to get to Ψ() from Ψ(); but the requirement the dψ/d be continuous t = mens tht only one or the other will do. Indeed, in using (8) to continue Ψ from to we must choose the pproprite propgtor either G + or G, ccording to the derivtive of Ψ t = ; otherwise the overll wve function will hve discontinuity in its first derivtive t =. So to use (8) to obtin vlues for Ψ t point, we need to know both Ψ nd Ψ t nerby point =, s should be the cse for second-order differentil eqution. If we wnt to de-emphsize this nture of the solution with the propgtor we my write R Ψ() = C k() e±i k(u)du (0) where C = Ψ() k(). In regions where V () > E, k() is imginry, so it s useful to define κ() = ik() = [V () E] () 2 nd R Ψ() = C κ() e± κ(u)du. (2)

4 Homer Reid s Solutions to Merzbcher Problems: Chpter 7 If we hve region of spce in which the WKB pproimtion is vlid, knowing the vlue of Ψ (nd its derivtive) t one point within the region is equivlent to knowing it everywhere, becuse we cn use the propgtor (9) to get from tht one point to every other point within the region. The WKB method, however, gives us no wy of determining the vlue of Ψ t tht one strting point. Furthermore, even if we know Ψ t one point within region of vlidity, we cn t use (8) to determine Ψ in other, nondjcent regions, becuse we cn t crry the propgtor cross regions of invlidity. So bsiclly wht we need is wy of finding one strting vlue for Ψ() in every region of vlidity of the WKB pproimtion. How do we find such points? Well, one sure-fire wy to get strting points in regions of vlidity is to identify regions of invlidity, of which there will be t lest one djcent to ech region of vlidity, nd then get vlues of Ψ t the boundries of the regions of invlidity which will lso count s vlues in the regions of vlidity. So we need to identify the regions of invlidity of the WKB pproimtion nd do more ccurte solution of the Schrödinger eqution there. The WKB pproimtion breks down when k /k 2 ceses to hold, which is true when k 0 but k 0, which hppens ner clssicl turning point of the motion i.e., point 0 t which V ( 0 ) = E. But ner such point we my epnd V () E in Tylor series round the point 0 ; if we keep only the first (liner in ) term in the series, we rrive t Schrödinger eqution which we cn solve ectly in the vicinity of 0. To do this, suppose the point 0 is clssicl turning point of the motion, so tht V ( 0 ) = E. In the neighborhood of 0 we my epnd V (): Then the Schrödinger eqution becomes The useful substitution here is so V () = E + ( 0 )V ( 0 ) + (3) d 2 Ψ() d2 2 V ( 0 )( 0 )Ψ() = 0. () u() = γ( 0 ) γ (u) = u γ + 0. [ ] /3 2 V ( 0 ) If we define then Φ(u) = Ψ((u)) dφ du = dψ d d du = γ Ψ ((u)) d 2 Φ du 2 = γ 2 Ψ ((u))

5 Homer Reid s Solutions to Merzbcher Problems: Chpter 7 5 so () becomes or γ 2 d2 du 2 Φ(u) γ3 ( 0 )Φ(u) = 0 Φ(u) uφ(u) = 0. du2 The solution to this differentil eqution is d 2 Φ(u) = β Ai(u) + β 2 Bi(u) (5) so the solution to the Schrödinger eqution () is ( ) ( ) Ψ() = β Ai γ( 0 ) + β 2 Bi γ( 0 ). For γ( 0 ) we hve the symptotic epression ( Ψ() π /2 [γ( 0 )] / β 2 e 2 3 γ( 0) 3/2 + β 2 e + 2 γ( 0) 3/2) 3 (6) nd for γ( 0 ) we hve [ ( Ψ() π /2 γ( 0 ) / 2 β cos 3 γ( 0) 3/2 π ) ( 2 β 2 sin 3 γ( 0) 3/2 π )]. (7) To simplify these, we need to consider two possible kinds of turning point. Cse : V ( 0 ) > 0. In this cse the potentil is incresing through the turning point t 0, which mens tht V () < E for < 0, nd V () > E for > 0. Hence the region to the left of the turning point is the clssiclly ccessible region, while the right of the turning point is clssiclly forbidden. Since V ( 0 ) > 0, γ > 0, so for < 0 (??) holds. For points close to the turning point on the left side, so nd k() = 0 [E V ()] 2 [ ] /2 2 V ( 0 ) /2 = γ 3/2 ( 0 ) /2 γ( 0 ) / γ = k() 0 k(u)du = γ 3/2 ( 0 ) /2 du = 2 3 γ3/2 ( 0 ) 3/2 (8) = 2 3 γ( 0) 3/2. (9) On the other hnd, for points close to the turning point on the left side we hve > 0, so γ( 0 ) > 0. In this region,

6 Homer Reid s Solutions to Merzbcher Problems: Chpter 7 6 κ() = so, for ner 0, nd lso [ ] [ ] /2 V () E 2 2 V ( 0 )( 0 ) = γ 3/2 ( 0 ) /2 (20) 0 κ(u)du = γ 3/2 0 (u 0 ) /2 du = 2 3 γ3/2 ( 0 ) 3/2 (2) γ( 0 ) / γ = κ(). (22) Using (8) nd (9) in (7), nd (2) nd (22) in (6), the solutions to the Schrödinger eqution on either side of clssicl turning point 0 t which V ( 0 ) > 0 re [ ( 0 Ψ() = 2β cos k(u)du π ) k() ( 0 β 2 sin k(u)du π Ψ() = κ() )], < 0 (23) [ R β e κ(u)du 0 + β 2 e R ] κ(u)du 0, < 0 (2) (we redefined the β constnts slightly in going to this eqution). Cse 2: V ( 0 ) < 0. In this cse the potentil is decresing through the turning point, so the clssiclly ccessible region is to the right of the turning point, nd the forbidden region to the left. Since V ( 0 ) < 0, γ < 0. Tht mens tht the regions of pplicbility of (6) nd (7) re on opposite sides of the turning points s they were in the previous cse. The solutions to the Schrödinger eqution on either side of the turning point re [ Ψ() = κ() Ψ() = k() β e R 0 [ ( 2β cos κ(u)du + β 2 e R 0 κ(u)du ], < 0 (25) ) k(u)du π 0 ( β 2 sin k(u)du π 0 )], > 0 (26) (27) So, to pply the WKB pproimtion to given potentil V (), the first step is to identify the clssicl turning points of the motion, nd to divide spce

7 Homer Reid s Solutions to Merzbcher Problems: Chpter 7 7 up into regions bounded by turning points, within which regions the WKB pproimtion (7) is vlid. Then, for ech turning point, we write down (23) nd (2) (or (25) nd (26)) t nerby points on either side of the turning point, nd then use (0) to evolve the wvefunction from those points to other points within the seprte regions. We should probbly quntify the mening of nerby in tht lst sentence. Suppose 0 is clssicl turning point of the motion, nd we re looking for points 0 ± ɛ t which to mke the hndoff from pproimtions (6) nd (7) to the WKB pproimtion These points must stisfy severl conditions. First, the pproimte Schrödinger eqution () is only vlid s long s we cn neglect the qudrtic nd higher-order terms in the epnsion (3), so we must hve ɛ V ( 0 ) V ( 0 ) ɛ V ( 0 ) V ( 0 ). (28) But t the sme time, γɛ must be sufficiently greter thn to justify the pproimtion (6) (or sufficiently less thn - to justify (7)); the condition here is /3 /3 2 V ( 0 ) ɛ ɛ 2 V ( 0 ). (29) Finlly, the points ±ɛ must be sufficiently fr wy from the turning points tht the pproimtion (7) is vlid for the derivtive of the phse of the wvefunction; the condition for this to be the cse ws k () k 2 () ( ) 2 V ( ± ɛ) 2 [E V ( ± ɛ)] 3/2. (30) If there re no points 0 ± ɛ stisfying ll three conditions, the WKB pproimtion cnnot be used. To pply ll of this to the problem of bound sttes in potentil well, consider potentil like tht shown in Figure, with two clssicl turning points t = nd = b. Although there re no discontinuities in the potentil here, the problem my be nlyzed in mnner similr to tht used in the considertion of one-dimensionl piecewise constnt potentils, s in Chpter 6: we divide spce into number of distinct regions, obtin solutions of the Schrödinger eqution in ech region, nd then mtch vlues nd derivtives t the region boundries. To divide spce into distinct regions in this cse, we begin by identifying nrrow regions round the turning points nd b in which the liner pproimtion (3) is vlid. In the nrrow region round =, we my use (25) nd (26); round = b we my use (23) nd (2). Let the nrrow such region round be ɛ < < + ɛ, nd tht round b be b ɛ 2 < < b + ɛ 2. Then spce divides nturlly into five regions: () < ɛ In this region we re fr enough to the left of the turning point tht the WKB pproimtion is vlid, nd the wvefunction tkes the form (0). However, we

8 Homer Reid s Solutions to Merzbcher Problems: Chpter 7 8 PSfrg replcements V () E b Figure : A potentil V () with two clssicl turning points for n energy E.

9 Homer Reid s Solutions to Merzbcher Problems: Chpter 7 9 must throw out the term tht grows eponentilly s, so we re left with R ( ɛ) Ψ() = A κ() e κ(u)du, < ɛ. (3) (b) ɛ < < + ɛ In this region we re close enough to the turning point tht (3) is vlid, so (25) nd (26) my be used. Ψ() = (β e R κ(u)du + β 2 e + R κ(u)du), < ( ɛ). (32) κ() From (3) nd (32) we see tht continuity of both the vlue nd first derivtive of Ψ() t = ɛ requires tking β = A, β 2 = 0. With this choice of constnts, we chive continuity not only of the vlue nd first derivtive of Ψ but lso of ll higher derivtives, s must be the cse since there is no discontinuity in the potentil. But now tht we know the vlue of Ψ t = ɛ, we lso know it t = + ɛ, becuse of course the solution of the Schrödinger eqution in the nrrow strip round (to which (32) is n symptotic pproimtion for < ) is vlid throughout the strip; the sme solution tht s vlid t = ɛ is vlid t = + ɛ. With β = A nd β 2 = 0, (26) becomes ( Ψ() = 2A k() cos k(u)du π ) = A [e +i(r k(u)du π/) + e i(r k(u)du π/)], k() = ( + ɛ) (33) (c) + ɛ < < b ɛ In this region the WKB pproimtion (7) is vlid, so we my use (8) to find the wvefunction t ny point within the region. Using the epression (33) for the wvefunction t = + ɛ, integrting from + ɛ to in the propgtor (9), nd using G + nd G, respectively, to propgte the first nd second terms in

10 Homer Reid s Solutions to Merzbcher Problems: Chpter 7 0 (33), we obtin for the wvefunction t point in this region [ R Ψ() = A e +i (+ɛ) k(u)du+ R R (+ɛ) k(u)du π/ + e i (+ɛ) k(u)du+ R ] (+ɛ) π/ k() = A [e +i(r k(u)du π/) + e i(r k(u)du π/)] k() ( = 2A k() cos k(u)du π ) ( b ) b = 2A k() cos k(u)du k(u)du π, + ɛ < < b ɛ (3) Oky, I hve now crried this nlysis fr enough to see for myself ectly where the Bohr-Sommerfeld quntiztion condition b ( k(u)du = n + ) π, n =, 2, (35) 2 comes from, which ws my originl gol, so I m now going to stop this eercise nd proceed directly to the problems. Problem 7. Apply the WKB method to prticle tht flls with ccelertion g in uniform grvittionl field directed long the z is nd tht is reflected from perfectly elstic plne surfce t z = 0. Compre with the rigorous solutions of this problem. We ll strt with the ect solution to the problem. The requirement of perfect elstic reflection t z = 0 my be imposed by tking V () to jump suddenly to infinity t z = 0, i.e. { mgz, z > 0 V () =, z 0. For z > 0, the Schrödinger eqution is 0 = d2 Ψ() + d2 2 = d2 d 2 Ψ() + 2 g 2 [E mgz] Ψ() [ ] E mg z Ψ() = d2 d 2 Ψ() 2 g 2 [z z 0 ] Ψ() (36)

11 Homer Reid s Solutions to Merzbcher Problems: Chpter 7 where z 0 = E/mg. With the substitution ( 2 g u = γ(z z 0 ) γ = 2 ) /3 nd tking Φ(u) = Ψ((u)), we find tht (36) is just the Airy eqution for Φ(u), with solutions d 2 Φ(u) uφ(u) = 0 du2 Φ(u) = β Ai(u) + β 2 Bi(u). Since we require solution tht remins finite s z, we must tke β 2 = 0. The solution to (36) is then Ψ() = β Ai ( γ(z z 0 ) ), (z > 0). (37) For z < 0, I wsn t quite sure how to ccount for the infinite potentil jump t z = 0, so insted I supposed the potentil for z < 0 to be constnt, V (z) = V 0, where eventully I ll tke V 0. Then the Schrödinger eqution for z < 0 is d 2 Ψ(z) dz2 2 [V 0 E] Ψ(z) = 0 with solution Ψ(z) = Ae kz, k = 2 [V 0 E]. (38) Mtching vlues nd derivtives of (37) nd (38) t z = 0, we hve Dividing, we obtin β Ai( γz 0 ) = A γβ Ai ( γz 0 ) = ka Ai( γz 0 ) γ Ai ( γz 0 ) = k Now tking V 0, we lso hve k, so the RHS of this goes to zero; thus the condition is tht γz 0 be zero of the Airy function, which mens the energy eigenvlues E n re given by ( 2 g 2 ) /3 En mg = nm E n = where n is the nth root of the eqution Ai( n ) = 0. ( mg 2 2 ) /3 n (39) 2

12 Homer Reid s Solutions to Merzbcher Problems: Chpter 7 2 So tht s the ect solution. In the WKB pproimtion, the spectrum of energy eigenvlues is determined by the condition (35). In this cse the clssicl turning points re t z = 0 nd z = z 0, so we hve ( n + ) z0 π = k(z) dz 2 0 z0 = 2 [E mgz] /2 dz 0 2 g z0 = 2 [z 0 z] /2 du 0 2 g z0 = 2 3 (z 0 z) 3/2 = 2 3 so the nth eigenvlue is given by 2 2 g 2 ( ) 3/2 E mg 0 ( mg 2 2 ) /3 [ ( 3 E n = n + ) 2/3 π]