Physics 74 Grdute Quntum Mechnics Solutions to Finl Exm, Fll 0 You my use () clss notes, () former homeworks nd solutions (vilble online), (3) online routines, such s Clebsch, provided by me, or (4) ny mth references, such s integrl tbles, Mple, etc Some possibly useful integrls cn be found t the end of the test If you cnnot do n integrl, or if you cnnot solve n eqution, try to go on, s if you knew the nswer Feel free to contct me with questions Work: 758-4994 Home: 74-008 Cell: 407-658 [5 points] A prticle of mss m lies in the one-dimensionl potentil 4 V x x Let the uncertinty in the position be x () [7] Find n inequlity for the uncertinty in the momentum Assuming this inequlity is sturted (tht mens it s n equlity), nd ssuming tht we cn replce x x nd p p, estimte the energy in terms of The uncertinty principle sttes tht xp Chnging this inequlity to n equlity, nd setting x, we hve p x We then substitute these expressions into the Hmiltonin, which is p p H V x x m m 8m 4 4 (b) [8] Estimte the ground stte energy by minimizing the function you found in prt () This function goes to infinity t either = 0 or The minimum, therefore, must lie between these two vlues To find this minimum, we tke the derivtive nd set it to zero to yield d H d 4m 3 0, 3 6 8 m, /3 /3 8m m We then substitute this bck into the expression to estimte the energy: /3 /3 4/3 /3 m 3 E /3 8m 4m 8m
[0 points] A prticle in one dimension hs wve function x x N xe x 0, 0 x 0 () [4] Wht is the normliztion constnt N? To find the normliztion constnt, we demnd tht the integrl of the squre of the wve function equl So we hve x x x x N 4 0 0 4 x dx N xe dx N e e, N (b) [6] Wht is the probbility tht if the prticle s position is mesured, it will give vlue x? We simply repet the previous integrl, chnging the limits ppropritely nd substituting our vlue for N P x x x dx N xe dx e e 4 4 e 3e 0406 4 4 x x x (c) [0] If you were to mesure the momentum of this prticle, wht is the probbility tht it will give vlue p? We must clculte the Fourier trnsform, which is given by 3/ dx ikx ikx x k xe xdxe ik ik Since the momentum is given by p k, demnding tht p is the sme s demnding tht k The probbility of this is given by dk P k k dk dk 4 ik ik 4 k 3 3 3 3 This eqution cn then be evluted by mking the trigonometric substitution k tn, nd we obtin
d tn sec d P k / / / 3/ 3 4 tn tn / sin 046 4 4 4 /4 4 cosd 3 [5 points] A prticle of mss m is in one-dimensionl hrmonic oscilltor potentil V x m x () [6] Work out the first two eigensttes, n = 0 nd n =, of the hrmonic oscilltor (eqution 53) Wht re the energies of these sttes? re Eqution (53) tell us tht the first two wve functions of the Hrmonic oscilltor 0 x x /4 m mx exp, /4 m m d mx x exp m dx /4 m m mx m mx x exp xexp m /4 m m mx x exp The energies re E n n 3, or E nd E 0 (b) [8] The wve function t t = 0 is given by mx mx mx xt, 0ABxexp Aexp Bxexp, where A nd B re rel constnts Wht is the wve function x, t t ll times? We see tht the two terms re proportionl to eigensttes of the Hmiltonin ie t n x e, so the A term will simply get multiplied by Eigensttes simply evolve s n i t e nd B by e 3i t So we hve it i t m x 3 x t Ae Bxe, exp
(c) [] Find the probbility density x, t nd probbility current, j xt t ll times nd ll loctions, nd check tht probbility is loclly conserved The probbility density is given by * it i3t it i3t xt, xt, Ae Bxe Ae Bxe exp mx it it A B x ABxe ABxe exp m x A B x ABxcostexp mx The probbility current (normlly vector, but we re in one dimension) is given by m x * j Im x, t x, t it i3t mx it i3t mx Im Ae Bxe exp Ae Bxe exp m x it i3t mx i3t m it i3t mx Im Ae Bxe exp Be xae Bxe exp m it i3t i3t it it Ae Bxe Be Axme Bx me m x exp Im m m x it exp Im AB x m e AB x m e m mx AB mx exp ABsin t x m x m exp sin t m m Conservtion of probbility sys tht we should hve it j mx AB mx mx ABx sintexp exp sint0 t x m
4 [5 points] Consider the three opertors A X Y, B XY, L XP YP, z y x Where the third of these is simply the stndrd ngulr momentum opertor () [6] Work out the three commuttors of these opertors with ech other Note tht the commuttors of L z with vrious opertors were worked out in problem 34, prt () It is cler tht A nd B will commute with ech other The remining commuttors need to be worked out: z, z, z, z, z, z, ixy iyx iyxixy4ixy ib, L A L X Y X L X L X X Y L Y L Y Y L, B L, XY X L, Y L, X Y ixx iyy ia z z z z (b) [5] Bsed on the results of prt (), deduce two non-trivil inequlities relting the uncertinties of these opertors nd their expecttion vlues The product of the uncertinties of two opertors cn lwys be relted to the expecttion vlue of the commuttor This isn t helpful for the trivil commuttor, but for the other two we obtin the reltion L A i L, A B B, z z L B i L, B A A z z (c) [4] Show tht if you re in n eigenstte of L z then A B 0 As we proved in previous homework, if you re in n eigenstte of ny opertor, then the uncertinty in tht opertor is zero As quick review, if the eigenvlue of Lz is m, then L L L m m 0 z z z From our previous prt, this implies 0 B nd 0 A But n bsolute vlue cn only be non-positive if its rgument is zero So A B 0
5 [5 points] A prticle is bout to be mesured by one of the two opertors 0 0 0 0 A 0 0, B 0 0 0 0 0 0 () [8] For ech of these opertors, find the eigenvlues nd orthonorml eigenvectors For ech opertor, if mesurement of n rbitrry stte occurs, wht re the possible outcomes? I hve drwn in dshed lines which demonstrte tht ech of these opertors is block digonl This immeditely gives us one eigenstte with eigenvlue The remining mtrix is fmilir enough tht we cn lso know its eigenvlues nd eigenvectors In fct, this sub-mtrix is the sme sub-mtrix which we clled H on pge 5 of the notes, nd whose eigenvectors re given by (36) In summry, the three eigenvectors for A nd the three for B work out to 0 0 0,, 3, 0 0 b, b, b3 0 0 0 For A the first two hve eigenvlue + nd the third hs eigenvlue - For B, the first nd third hve eigenvlue nd the middle one hs eigenvlue - Whichever opertor you mesure, the only possible outcomes re (b) [5] Opertor A is mesured first, nd the resulting vlue is negtive Wht is the quntum stte immeditely therefter? After mesuring n opertor, the system will lwys be in n eigenstte with the corresponding eigenvlue Therefore it must be in the stte 3 (c) [5] Opertor B is then mesured Wht is the probbility for ech possible outcome? The probbilities re the squre of the overlp with the corresponding eigenvectors, so 0 0 3 P b 3 b3 3 0 0 0, 4 4
0 P b 3 0 4 (d) [7] Assuming the outcome in prt (c) ws positive, wht is the quntum stte fter the mesurement? If the result is positive, the stte vector is now supposed to be P b b 3 b3 b3 3 0 0 0 4 0 0 0 0 3 0 0 6 4 0 6 3 0 3 6 [5 points] A prticle of mss m is in one-dimensionl hrmonic oscilltor with ngulr frequency is in the quntum stte N 0 i () [3] Wht is the normliztion constnt N? We simply use the normliztion condition: N 0 i 0 i N 4 N, N
(b) [7] Find the expecttion vlues X, X, P, nd P This is most esily done by using rising nd lowering opertors We hve X m i i 0 i 63 i 0 m 8m i 63 X X iii 0 i i 0 i 6 3 0, 8m 8m 6 i 0 i 6 3 i 0 i 6 3 8m 8m 5, 4m i m m P 0 i i 8 i 6 3 0 i, P m 0 i 8 0 3i i 6 3 m m 33, 8 8 m P 0 3i 8 i 6 3 0 3i i 6 3 m 9m 96 8 4 0 (c) [5] Find the uncertinties uncertinty reltion We hve X nd P nd check tht they stisfy the 5 X X X, 4m P P P m m m, 9 9 9 4 8 8 5 9m 45 XP 86 4m 8 3 The uncertinty reltion sys the product must exceed 05, which it certinly does
7 [5 points] A prticle of mss m in three dimensions hs Hmiltonin H P P P X Y X Y kz m 4 4 x y z cos () [4] Show the Hmiltonin is invrint under rottions round the z-xis by R zˆ, for n pproprite ngle 90 degrees, The kinetic term is lwys invrint under n rbitrry rottion of ny ngle The V X, Y, Z V X, Y, Z, potentil terms must be checked fter rottion by seeing if where X, Y, Z Y, X, Z We esily see tht 4 4 V Y, X, Z Y X Y X sin kz V X, Y, Z (b) [6] Show tht the Hmiltonin is invrint under trnsltions long the z-xis T z ˆ Wht is the smllest such trnsltion possible? by n mount, We similrly wish to hve V X, Y, Z V X, Y, Z kz k sin kz Clerly, this demnds sin Since sine is -periodic, k must be multiple of, nd the smllest vlue of for which this will work is k (c) [5] Suppose sttes re found tht re eigensttes of these two opertors, so R zˆ Tzˆ,,, nd,, R T R R T R T T R T Wht restrictions, if ny, re there on the eigenvlues R nd T? Symmetry opertors re lwys unitry, nd therefore must hve mgnitude However, if we perform the rottion four times, we end up bck where we strted, so we cn conclude tht R ˆ, 4 z It follows tht 4 4 zˆ R, T R, R, T R R, T, 4 R, R,,, i or i R * For the other eigenvlue, we only demnd tht T T
8 [0 points] The mesons re pir of spin (s = ) prticles with chrge e nd identicl mss m 775 MeV/ c Although it wouldn t work in prctice, one could imgine binding them together to mke hydrogen-like system Assume only electric interctions re relevnt for the resulting system () [5] Wht would be the binding energy of this stte in the n = stte? We first need to clculte the reduced mss, which is given by The energy is then given by, m m m m m m 775 MeV/ c 3875 MeV/ c E n c n 37 3875 MeV 0003 MeV 03 kev (b) [7] If one were to mesure the totl spin squred S for this system in the n = stte, wht re the possible outcomes? For ech of these possible outcomes, wht re the possible outcomes for S z? Becuse it is in n =, nd l is lwys less thn n, there is no orbitl ngulr momentum Hence the totl ngulr momentum of this tom will come just from the spins of the two mesons Since these ech hve spin s s, the totl spin must be in the rnge from ss 0 to s s, so the totl spin cn S Sz be either s = 0, or The corresponding eigenvlue of S is 0 0 given by s s, which is either 0, or 6 The,0, eigenvlues of Sz re then given by ms, where m s runs from s to s A complete tble of possibilities is given t right 6,,0,, (c) [8] The vlue of S nd S z re ctully mesured S yields the mximum possible vlue, nd S z yields 0 Write the spin stte of the two mesons in form sm, s nd explicitly in terms of the individul spin sttes ms, m s You my use the online Clebsch routine If the spin S z of just the were mesured, wht would be the possible outcomes, nd wht re their probbilities? The lrgest vlues of S corresponds to s =, nd obviously Sz = 0 corresponds to m = 0 So the stte is sm, s,0 We now wnt to write this in terms of the underlying spins, tht is, we wnt to write
,0 ms, ms ms, ms,0 ms, ms This is the purpose of Clebsch-Gordn coefficients The coefficients we wnt re,; ms, m s, 0 This cn only be non-vnishing if ms ms 0, so there re only three coefficients tht need to be clculted We will use the online routine > for m from - to do clebsch(,,m,-m,,0);end do; We find: 6 sm,,0, 0,0, s We now cn strightforwrdly clculte the probbility tht the spin of the first spin is in ech of these sttes We find: 6 6 6 3 6 6 z z z P S, P S 0, P S 9 [0 points] An electron of mss lies in region with electric nd mgnetic fields 0 E xxˆ yyˆ, B Bz ˆ e U r Demonstrte tht the vector () [5] Find suitble sclr potentil potentil Ar B xyˆ yx ˆ is one wy to describe this mgnetic field Since the vector potentil hs no time-dependence, it does not contributes only to the mgnetic field, for which we hve A Ay z Ax Az Ay A x BA xˆ yˆ zˆ 00 zˆ B BBz ˆ y z z x x y For the electric field, we wnt EU r, which suggests we need U 0 U 0 x nd y x e y e These equtions re esy to integrte to yield U x y e 0
(b) [6] Write the Hmiltonin explicitly The Hmiltonin is given by ge H PeA eur SB ge P eby P ebx P 0 X Y BS x y z z ge P P P ebxp YP 4e B X Y 0 X Y BS x y z y x z eb eb 0 4 Px Py Pz X Y Lz gsz (c) [9] Demonstrte tht this Hmiltonin commutes with one of the three momentum opertors P, one of the three ngulr momentum opertors L, nd one of the three spin opertors S Which one, in ech cse? Do they commute with ech other? Cll the three eigenvlues under the opertors k, m, nd m s respectively Are there ny restrictions on these eigenvlues? Every term mnifestly commutes with Pz nd Sz Also, it is esy to see tht Lz commutes with the combintion X Y, since this is just the distnce squred from the z-xis, nd invrint under rottion bout this xis So the three opertors tht commute with the Hmiltonin re Pz, Lz, nd Sz Since they lso commute with ech other, they cn ll be simultneously digonlized, nd we cn ssume our sttes re eigenvectors under these three opertors We will lbel these eigenvlues s pproprite, so P k, L m, S m z z z s As lwys, for spin one-hlf prticle, ms, nd the ngulr momentum quntum number m must be n integer, but there re no restrictions on k 0 [0 points] Two non-intercting spinless prticles re in the n = nd n = stte of n infinite squre well with llowed region 0 x () [8] Write explicitly the wve function x, x if the prticles re (i) nonidenticl, with the first prticle in n =, (ii) identicl bosons, nd (iii) identicl fermions The wve function for non-intercting single prticle in n infinite squre well is n x sin nx For two prticles, we simply tke the product, nd hve just x x, sin sin x x
However, boson sttes must be symmetrized, nd fermion sttes nti-symmetrized Tking this into ccount, we hve B F x x, sin sin sin sin, x x x x x x x x, sin sin sin sin x x (b) [] Find the probbility tht if the position of the two prticles is mesured, they will both lie in the region 0 x, if the prticles re (i) nonidenticl, (ii) bosons (iii) fermions The probbility in ech cse is simply given by For non-identicl prticles, we hve 0 0 P dx dx x, x 4 x x sin sin 0 0 P dx dx x 4x x x 0 0 4 4 sin sin 8 6 4 4 4 For the fermions nd bosons, the computtion is much more complicted, but it is clerly wise to do them together to sve time We hve x x x x P dx dx sin sin sin sin 0 0 x x x x sin sin sin sin dx dx 0 0 x x x x sin sin sin sin 4 x x x x sin sin dx sin sin dx 4 4 0 0 4 3 x x 4 6 sin sin 4 6 4 0 3 4 9 The probbilities work out to PB 0430 nd PF 00699 We see the strong tendency for bosons to tend to be ner ech other, while fermions void ech other