Practice Questions for Final Exam - Introduction to Quantum Mechanics 1 | PHYS 5250, Exams of Quantum Mechanics

Download Practice Questions for Final Exam - Introduction to Quantum Mechanics 1 | PHYS 5250 and more Exams Quantum Mechanics in PDF only on Docsity! —Fival Exan— a T+ is easy +o see that we are dealin with @ system describect by the j= “ze representation of orgy ter womewtum, Ove notes this by observing Guar +he measuvceol valve of SFe Bur, ant Is We» which is ouly ime case 1€ AS Gets ECW = Hy? ond oh = 2% the j« Ye vepreseutotion is given by the Pav matrices, 1e o | _ 4 Oo -e ' oO a (| 3). 42 3). BAC *) col re Jz 1S subsegveutl wmeasuvedct, ond we wort to Know what values might be Lounck, oue Simply exeords the current stare in terms of Jy eigenstates. Because we Kuow ylat we ove in the thy state of Sy, it must be Mat ovr eiagenr* s sey ze 0) in the basis +tat cogoualites 37, 52. The erqeuf*s here ove lye-%e>= (fh) enet aee-Bad= (2) loot like Je 50 [Tx- hed * (ll) #G)* eG) t Te Vee? # SEV - Me) Since both Tye thy ort 3,>-, States ove present in ye supecpesition, we Kuow +he pessible values of Tz immecliotela ~~ dey ove Jy > ity, CO) ue probability of Lindine, Jas thy, is fount in the vseval way, ie PCs, %)- KK 5, */, | Tye 4) ° | K 32+ | (se |e + = lae-%,.9) o ee Crete) © Cary aor) - |) " [PCs,-%)+ 4 (it) Finally, one con Lind the expectation value of JZ im the 4% state of Jy by remembering £35.) Ctx %] 5, 5.7% (aedswesl 65-4) Ss (dete helo This is easy, since Se \te- FAS = ws [Tar te) , ete, 6° uN C5.y = a] He Getslaey + HaCnetil ter ke) ~ 4 C5e° We | Serb > ~ Hee (Sas We \Ta= >| wake a Chore in vaviables X(t)? X, C4) + yCe) where Xeelt) is he classical path, ork yey is re otevietion from it. Now 5 t . . dt. Lint & V tee) 2 aie Sue Ly \ (amy 4s ? ) e U Gyt;%)* qo from herve i+ is Straight for work 42 Compute ite free-perticie propagator t Here VEXI=O cwrak z 2 . Ss faci tm (42%) = EM (Ke -%) “ eo *¢ © SO ° t . 2 crt (K&e-%e) i § to tammy? Utxyerx = @ te Dyaye jae amy o Tu pvinetple, +he iut@aral is dificult to evaluate, Fertunatel this task fs made for less formidable by remem bering * Wak inv Me lait where t3o0 the propagqater wust reduce to a §-f2, Using te definition a ~ (x= x") . i SCx-x) = lim e “a aso Nol? rh is easy to See that here Qt: 2Kit eo ae YY, ——_-_—_—_: ™ 2 ‘ Ne om (x- x YY Ure (az) ° “ | d) Given Waat a herwmouie oseillater is imetvatla tA @ it is possible to compute what colereut State, evolutions happeus +o tunis State under +ime a Au easy way to do is is to vse Sue Fock -state basis where t jz> ~ e** lod Yor mall zation, remember at to fix dhe Leled = AMA Cole tet oy = 1 od sivce AB 8 a CA,e) ee reee Whar tHe comumutater is aA cCc-number one fruds iz? Lor ard Bat 2 Kole e “lo\ 24 t ¢zle> At Ae 2 z 12 ~ het \ le > A"Ae@ - 1 so Ace in otter Words, [z> - 121%, zat e 7e los - 18%, 2" = 1] el 2, Bain a * K sitte it is easy +0 evolle euergy Ciqeustates : . wv time, we car now hope te clo the same for the cohereut States ~e ot cw, + jearse= U, 12> =e A, lz. 2 e PASa tay se 2 2 . ~ 12), pt slant -12#l4 (zeit)? J2za@)> = e€ Qs oa é =e &, Ge [2¢4)> = fee*) é ) - . . CO. Voth the above observation, it is not tee hora +o frudk he expectation value of We operator ate at some time €. Since cohevent states ove eigenstates ef the aunihilatian Operator al@>-= 2125 “ a we ot -< 7 Alzcry= alee "*> = aatre® [2 oere*> ard accordingly, Lzces| a | Zee) = zie Met dee te zoo ey éa’> = gre tet (ce) One cov le, alternatively, arrive at is vesolt by usin Yue Hersenbvera pictvre in whieh dre epevators 4 Whewtselves evolve (mn ture ¢ Heve thas La, l= he, [a,ata| z= hye, ((ae]at +e (aar)) = Awa $o : ~tuet alt)+ Ae eer ond ates? = ate which means Jalaarle> = @* "del ale> the brackets look lke ef \ Nig { “4 . (ate Kee thm) X Wigs » = \ dad, dx, Ga ( | e ent @ x = OnE EKER ROL Clealu is amounts +0 completma N integrals of the form y, Vg z,to pt tt -2 Re ax WK, tH] € oF ey * (=a AK oes Ya - (=e aX, - ( z ae 2% Ke %, Xe + Be ° “+ xt Xo ° $0 Cwhes oa ZK he Bee meee (Ga + x"? N Poo [Ceres leeesrl = (22% ) Xe t+ KS? es o tt OK, (ty To fiunol he propabili ty of being iam whe fiest excited State of He new trap, ove west cam pute Mm wueh he sawme Wau as ), we wetite ‘he integra hile oo zt 2 2 z - (x2 tg tt Be) = Kr Kyte t MY < Mees | Mes) ~ NN dx,dx,.-A%, ix aR + x, ZK ton a x HORE REA A KEL Here re normaligations oe less important, ort the fore brackets signify a fully symme tei geal First excites state wave £2, The importut observation to make is that +.) omy aivew imdeqratouw over Ax,, OWE will evalvate on inteara | of ne farm oo x i \ ax xe **~ 20 -~o integrarcl is tue prodvet ef ar even carol obh £4 as Nhe : interval. Ace ordinal, each inteqrateot over of Wwe N integrals with ave O ma so O Syume tre Ip Fes c mae ) Now we look for the probab btu deusitu for Findinay a pectele with Velocity vw for a pecticle wnat sterts ovr in Me aground State of a hewmonte. oscillator after the trap is tvneck off Before dhe tvap is turned off, the wave £2 looks like 2 i Yq = ¥en Wo): (Fe) 'e ™ 6S While aftterwords i car be written as a linear super position of free-~pecticte states; ckx 0) = pee $0, Fouvier +vaxsformina, oe -eKx % Ye Bye We) = dx (ax) GF) é€ Ant ~ co this suteqral con be e@ualuated With tue hele of she i deutity 2 - z b \ b \ a e ores = (Ete (as $0 a 2 VM =F MY We) = (% ) e 2 v intevesteck iw tthe probability oteus). because We we neek 42 see hows VYuis state as a £2 of time, we in Hime 7 evolves For a free pevticie, the HKwe ependence looks Itke “ify Ent -i RRL e = e SO ¥. ene, ck Ke Va ERA HEL, WOK, t) = ( © \ e e em On one finds a probability Olenrsth like Ponty = | Wes eyed |odk Mdv h : (2= \" ma ee v h % or better yet, 2 YH e P,* + Why 4 x2 ( a°G , rots? ) - 44*%,& x zZm™ z2™ T me me drs iM a vusetel form, one must complete Onc write On effective potential +hadt Oo Usrmontec asceiWator s to write the S@vave resem bles TE one has or expression of tue form ax*4 bx = a(xr+ gx) written b.y*. af(x+ &)- gah (er) HR it Can be Tn +he present case, call gta? z 4 f= + mw, a= 4 z me a aA =e -b b= - %@kK,& ~* Za mo so dat dhe Olaniltoniag becomes 2 Bar z Bye f= Px + ¥kh + 5 (EE +m) («- 4) + wg" B zm zm ™ 2 (ot + mate mick me defining om effective frequency v, tL ( ors" + mdf = Ww, = ™ me allecss one to write z zat Ye Po + HKG + bmwe(x-d)* 4 hah B em au Zz wiwetc® onk we COA, ak loug last Get arouncad +06 Lorde ring down a soles Since Ws EW define Es E- HK’ (1+ gist zm * mt oe te, which Mearng wt wavst solve RE OPW) + Ems (x-d > WO) * Ee WO) em This eg" is, by now, oid hat, I+ resoltts Mia Spectoum ike EL es tue Cre t+ e SO fer our problem, we fina = Awe 4 Hk eat le we Cnt hy aby (1 £ se.) | To solve for the eiqenstates, it 18 useful to intraduce tre Aimenstonless variable Xo, where X,2 Vee ° +hen khe sol*s ove ime preadvets of He rmirte Polyvoulrals exes in xX, ond free particles atona y 3 ) (xa) Ye.) = Hcttje Fre icky b) From the above Spectrum, one notes contributions +o the eneray from the free perticle motion in the y Alrectou as well as from the Maqgnedric fielA's presence. The } dispersion relation is easy to stetch, i+ looks like Ey) nae net neo Liwer