There is an arbitrary overall complex phase that could be added to A, but since this makes no difference we set it to zero and choose A real.

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1 Midtrm #, Physics 37A, Spring 07. Writ your rsponss blow or on xtra pags. Show your work, and tak car to xplain what you ar doing; partial crdit will b givn for incomplt answrs that dmonstrat som concptual undrstanding. Cross out or ras parts of th problm you wish th gradr to ignor. Som quations of potntial rlvanc ar givn on th back pag. Problm : 5 pts Considr a spin / particl in th spin stat writtn in th basis of Ŝz ignstats χ = A whr A and δ ar ral constants a Dtrmin th constant A ndd to normaliz χ ANSWER: Th normalization rquirmnt is χ χ = a + b = whr a and b ar th two complx cofficints of th spinor. In this xampl A A + AA = A + A = A = / 3 Thr is an arbitrary ovrall complx phas that could b addd to A, but sinc this maks no diffrnc w st it to zro and choos A ral. b What is th probability that a masurmnt of Ŝz for th particl rturns spin up in th z-dirction? ANSWER: This and th nxt part of this problm ar a spcific cas of Problm of HW#6.Th probability of finding spin up in z-dirction is Whr th stat rprsntd by χ writtn in its abstract form is Th orthonormality of th basis vctors mans P z = z χ 4 χ = z + z 5 z χ = 6 P z = z χ = = 7 Sinc χ is th rprsntation in th S z basis, w could hav just writtn th answr down immdiatly as th absolut valu squard of th first componnt of χ c What is th probability that a masurmnt of Ŝx for th particl rturns spin up in th x-dirction? ANSWER: Th probability of finding spin up in x-dirction is Whr x can b writtn in trms of th S z basis vctors as P x = x χ 8 x = z + z χ x = 9

2 If this solution for th spin up ignstat of S x was not known, it can b dtrmind in th ordinary way by solving for th ignvctors of th S x matrix givn on th quation sht. Thn x χ = z + z z + z = + 0 And so So P x = x χ = + + = = + cos δ 4 4 P x = + cos δ as a sanity chck, w look at obvious th limits. For δ = 0 w gt P =, which maks sns sinc thn th stat x is just th spin-up in x ignstat. For δ = π w gt P = 0, which maks sns sinc thn th stat x is just th spin-down in x ignstat. An altrnativ but mathmatically quivalnt approach to solving this problm is to writ th spinor χ in th S x basis by applying a similarity transformation χ Sx = S χ whr S = Th similarity matrix S is just th on mad up of th ignvctors of S x writtn in th S z basis as don in HW#5. Th spinor in th S x basis is thn χ Sx = = + 4 And sinc th spinor is now rprsntd in th S z basis, th answr is just th absolut valu squard of th first componnt. d For a stat of th form of Eq., us th non-commutation of Ŝz and Ŝx to driv an uncrtainty rlation writtn in th form σ Sz σ Sx fδ 5 whr fδ is a function of δ and fundamntal constants. Th quantitis σ Sz and σ Sx hr ar th uncrtaintis i.., standard dviations in masurmnts of Ŝz and Ŝx thy r not th Pauli spin matrics! ANSWER: Th gnralizd uncrtainty rlation givn on th back quation sht is σ Aσ B 3 [Â, ˆB] 6 i In this cas, th oprators ar Ŝz and ˆ S x. W know th commutation rlations for th spin oprators ar [Ŝz, Ŝz] = i hŝy 7 or this rlation could b dirctly calculatd from th spin matrics on th quation sht. Th uncrtainty rlation is thn σszσ Sx i i hŝy 8 W nd th xpctation valu of Ŝ y for this stat, which can b calculatd using th spin matrix Ŝy = χ S y χ = h 0 i i 0 9

3 Ŝy = h 4 So plugging this in abov givs i i σ Szσ Sx = h i + i = h 4 4 sin δ = h sin δ 0 h h sin δ = h4 6 sin δ Taking th squar root σ Sz σ Sx h sin δ 4 3

4 Problm : pts Lt us dnot th normalizd nrgy ignstats of th infinit squar wll as,, 3,... n. Considr th oprator ˆQ = 3 a Show that th valu Q = 0 for any of th nrgy ignstats n. ANSWER: Th quantity rqustd is Q = n ˆQn = n n 4 By th orthogonality of th ignstats, n = 0 unlss n =, but in this cas n = = 0, so Q = 0 for any stat n. b Find th valu Q for th stat α = i + ANSWER: Th quantity rqustd Q = i + i + 5 whr w wr carful to complx conjugat th i whn taking th corrsponding bra of th kt. Th rsult is, using th orthogonality of th ignstats Q = i = i 6 Sinc ˆQ is not Hrmitian, thr is no rquirmnt that this valu b ral as alludd to in th footnot. c Dos ˆQ commut with th hamiltonian Ĥ? Basd on this rsult dtrmin whthr Q rmains constant ovr tim. ANSWER: W writ th commutator [Ĥ, ˆQ] = Ĥ Ĥ 7 This can b calculatd abstractly by using th fact that and ar ignstats of Ĥ with ignvalus E and E, rspctivly. W us th fact that Ĥ is hrmitian to mov it ovr onto th [Ĥ, ˆQ] Ĥ = Ĥ = Ĥ Ĥ = E E 8 hnc [Ĥ, ˆQ] = E E 0 9 If you ar uncomfortabl with th moving of Ĥ into th trm, you can instad apply th commutator to a gnral stat writtn as a sum of ignstats [Ĥ, ˆQ] α = Ĥ Ĥ c n n 30 n which is sn to b = E E n c n n = E E α 3 So th commutator again is not zro. Not that if ˆQ is not Hrmitian, its ignvalus ar not guarantd to b ral, and so th quantity Q = α ˆQα for som stat α may not actually rprsnt th avrag valu of masurmnts of som ral physical obsrvabl. 4

5 An vn simplr way to do th problm would b to just look at th commutator on th spcific stat. Thn w s [Ĥ, ˆQ] = Ĥ Ĥ = E E 0 3 This is a spcific countr-xampl to th claim that Ĥ and ˆQ commut. You drivd in HW#7 and also in class that th tim volution of an xpctation valu is rlatd to th xpctation valu of th commutator by t Q = ī h [Ĥ, ˆQ] which uss th fact that ˆQ dos not chang with tim. Sinc Ĥ and ˆQ don t commut, w s that th Q is not a constant with tim in gnral unlss it happns to b zro. d Is th oprator ˆQ unitary? Dmonstrat why or why not thr is mor than on way to do so. ANSWER: A unitary oprator prsrvs th lngths of vctors. But w hav from part a that 33 ˆQ n = whr is th null vctor 34 for any ignstat. Th lngth of ˆQ n is zro, whil th lngth of n is not its lngth is assuming th ignstats ar normalizd thrfor ˆQ is not unitary. Mor dirctly, th Hrmitian conjugat of ˆQ is ˆQ = 35 And so ˆQ ˆQ = = Î = n n n 36 Whr in th last stp w conclud that th oprator is not qual to th idntity oprator sinc it includs only on trm in th sum giving th rsolution of th idntity. Sinc ˆQ ˆQ w s ˆQ is not unitary. Not that all of th parts of this problm bcom rathr straightforward if you raliz that th oprator ˆQ can b writtn as a matrix in th basis of nrgy ignvctors as was don in HW#7 part 4c. Th action of ˆQ is to just turn stat into stat so th matrix must b ˆQ = It is thn dirctly clar that ˆQ ˆQ so th oprator is not unitary. On can can also calculat xpctation valus dirctly from th matrix multiplication, and can dirctly comput th commutation with th Ĥ as matrix multiplications in this rprsntation Ĥ is just diagonal with ignvalus along th diagonal. 5

6 Problm 3: 8 pts Scintists hav just discovrd an amazing nw quantum proprty of particls, calld animalnss. Whn masurd, a particl is obsrvd to b ithr a duck or a rabbit. Th animalnss is associatd with a Hrmitian oprator  with ignvctors D for duck and R for rabbit with ignvalus λ D = + and λ R = rspctivly. In th D, R basis th  oprator is rprsntd by 0  = 38 0 A particl is put into a dvic whr th Hamiltonian writtn in th D, R basis is Ĥ = whr is a constant 39 3a What ar th possibl valus of nrgy that can b obsrvd? ANSWER: W want to calculat th ignvalus of Ĥ, similar to th calculations w hav don for othr x matrics on th homwork. Th charactristic quation is λ dt Ĥ λî = dt = 0 40 λ λ λ = 0 λ + λ = 0 4 λλ = 0 λ = 0 or λ = 4 3b A particl with th abov Hamiltonian is initially a duck i.., in th D stat. What is th uncrtainty in its nrgy? ANSWER: In this basis th stat D is just rprsntd by so w hav 0 E = 0 0 = 0 = 43 and E = 0 So th uncrtainty in nrgy is hnc σ E =. 0 = 0 0 = 44 σ E = E E = = 45 3c W mak a masurmnt and dtrmin th nrgy of th particl. Calculat th probability that a subsqunt masurmnt of  will find th particl to b a duck. ANSWER: If w masur th nrgy, w collaps th stat to on of th nrgy ignvctors. W can dtrmin th rprsntation of th ignvctors in th D, R basis in th usual way a a = λ 46 b b which implis a + b = λa 47 6

7 for λ = 0 this implis a = b, whil for λ = w hav a = b. Normalizing th two ignvctors so that a + b = givs = = 48 Th probability of finding th particl to b a duck in th D, R basis is th absolut valu squard of th first componnt of th vctor. Th problm dos not spcify which nrgy is masurd and hnc which ignvctor w collaps to but w s that it dosn t mattr. For both ignvctors th probability of bing in th D stat is / 3d W masur  and find th particl to b a duck. W thn lt th systm volv for a tim t and masur  again. Calculat th probability that w find th particl to b a rabbit i.., in th R stat. ANSWER: This is similar to problms and 3 on HW#6. Th ignstats of th Hamiltonian ar thos that volv simply with tim by application of a complx phas. So w writ th stat D as a suprposition of nrgy ignstats D = And sinc w know how th nrgy ignvctors volv w hav th stat at som tim t as χt = iet/ h + iet/ h 50 so plugging in E = 0, E =, and th xprssion for th ignvctors givs χt = + it/ h it/ h = it/ h + it/ h 0 Th rabbit stat is rprsntd by and so th probability of bing obsrvd a rabbit is th absolut valu squard of th scond componnt of χ P R = it/ h it/ h = it/ h it/ h P R = 4 cost/ h = cost/ h 53 and if w car to, w s th probability of bing a duck is P D = + cost/ h 54 W sanity chck that at t = 0 th probability of bing a rabbit is zro, and th maximum valu of P R is. Th sum of bing a duck or a rabbit is on at all tims. 7

8 Problm 4: 5 pts Th lowring laddr oprator of th harmonic oscillator â is not Hrmitian and so dos not corrspond to any obsrvabl but w can still find ignstats of th oprator that oby â α = α α 55 whr α is a normalizd ignstat of â and th associatd ignvalu α is som complx numbr. Th stats α turn out to b quit physically intrsting. 4a Find th xpctation valu of ˆx for an ignstat α in trms of α and othr variabls. ANSWER: Rcalling th us of laddr oprators in HW#7 problms 3 and 4, w writ th oprator ˆx in trms of laddr oprators h ˆx = x 0 â + + â whr x 0 = 56 mω 0 so th xpctation valu of ˆx is x α ˆxα = x 0 α â + + â α = x 0 α â + α + x 0 α â α 57 Th scond trm can b valuatd sinc α is an ignstat of aˆ x 0 α â α = x 0 α α α = x 0 α 58 which uss th fact hat α is normalizd. To dal with th first trm involving th raising oprator w not that th Hrmitian conjugat of â + is â + = â. Thus th trm is x 0 α â + α = x 0 â +α α = x 0 â α α = x 0 α α α = x 0 α 59 Not that sinc α is not ncssarily ral, w must complx conjugat th constant. This givs finally x = x 0 α + α 60 4b Dtrmin th uncrtainty in position σ x for α. ANSWER: To gt th uncrtainty w nd to find x. Th oprator is ˆx = x 0 â + + â x 0 â + + â = x 0â + + â + â + â â + + â 6 Th xpctation valu has four trms now. Thr of thm ar α â α = α 6 α â + α = â α α = α 63 α â + â α = â α â α = α α 64 For th last trm, w will nd to us th commutation rlation [â, â + ] = which allows us to writ â â + = + â + â α â â + α = α α + α â + â α = + α α 65 Putting it all togthr x = x 0α + α + α α + 66 so th uncrtainty squard is and so σ x = x 0 σ x = x x = x 0α + α + α α + x 0α + α + α α = x

9 For th sak of tim, I won t ask you to calculat th uncrtainty in momntum σ p, but if you did you would s that th α ar spcial stats that hav th minimum possibl uncrtainty σ x σ p = h/. W can xprss ths stats as w can any stat as a suprposition of th nrgy ignstats of th harmonic oscillator, n α = c n n 68 n=0 4c Driv an xprssion for th cofficints c n in trms of th cofficint c 0 of th ground stat th cofficint c 0 itslf can b dtrmind using th normalization condition, which you ndn t bothr to do. ANSWER: W want to us th fact that α is an ignstat of th lowring oprator, which mans c n n = α c n n 69 â n=0 It may b asist to s this by writing out trms xplicitly n=0 α = c c + c + c applying th lowring oprator its action is givn on th back quation sht w find thn th first quation is â α = c 0 + c + c c 0 + c + c = αc0 0 + c + c + c Now to pick out th cofficints, w just apply th appropriat bra. For xampl, applying 0 to ach sid givs c = αc 0 73 and applying to ach sid givs and applying to ach sid givs c = αc c = α c = α c 0 74 c 3 3 = αc c 3 = α 3 c = α3 3 c 0 75 and w s th trnd c n = αn n! c 0 76 To do this mor gnrally, w can writ th action of th lowring oprator â α = c n â n = c n n n 77 n=0 n=0 and sinc this is an ignstat, â α = α α so α c n n = c n n n 78 n= Applying n to both sids to pick off th cofficints givs so which is th pattrn abov. n= αc n = c n+ n + 79 c n+ = α n + c n 80 9

10 Equations and Extra Work Spac i h ψ = Ĥ ψ tim dpndnt Schrodingr quation 8 t Ĥ = ˆp m + mω 0 ˆx Harmonic oscillator Hamiltonian 8 Harmonic oscillator laddr oprators â + = â = m hω0 [mω 0ˆx iˆp] â + n = n + n + 83 m hω0 [mω 0ˆx + iˆp] â n = n n 84 Commutation rlations [ˆx, ˆp] = i h [â, â + ] = 85 Spin / oprator matrics writtn in Ŝz basis Sˆ x = h 0 Ŝ 0 y = h 0 i i 0 Ŝ z = h Expctation valu and uncrtainty Q = ψ ˆQψ σ Q = Q Q σ Aσ B [Â, ˆB] 87 i 0