Solutions to Chapter 1 exercises

Transcription

1 Appendx S Solutons to Chapter exercses Soluton to Exercse.. Usng the result of Ex. A.5, we wrte don t forget the complex conjugaton where approprate! ψ ψ N alve ψ dead ψ N 4alve alve + alve dead dead alve + dead dead. S. Now snce dead and alve are physcal states, ther norms equal. On the other hand, these states are ncompatble wth each other, so ther nner product vanshes. Hence we have ψ ψ N 4 + 5N and hence N / 5. Soluton to Exercse.. Although the moton s one-dmensonal, all poston states are ncompatble wth each other: x x 0 unless x x. Therefore there are nfntely many lnearly ndependent states, and the dmenson of the Hlbert space s nfnte. Soluton to Exercse.3. There are two vectors n each set. Because our Hlbert space s two-dmensonal, and accordng to Ex. A.9, t s enough to show that each set s orthonormal n order for t to be a bass. We calculate the nner products by expressng the vectors n the matrx form, n the canoncal bass accordng to Table.. a For the dagonal states, we fnd + + ; + 0; + 0;. b Smlarly, for the crcular states [we perform complex conjugaton accordng to Eq. A.5]: S

2 S S Solutons to Chapter exercses R R ; L R 0; R L 0; L L. Soluton to Exercse.4. For the dagonal bass, we fnd, usng Table.: + H ; 0 H ; + V 0 V 0 0 ;, and thus H + + / ; V + /. Smlarly for the crcular polarzaton bass, R H ; 0 L H ; 0 R V 0 ; L V 0, hence H R + L/ ; V R + L/. Soluton to Exercse.5. Usng Table., we express states a and b n the canoncal bass: 3 H + V a H V b We now follow the same approach as n the prevous exercse.

3 S Solutons to Chapter exercses + a a canoncal bass canoncal bass ; 3 3 S3 and thus, the decomposton of a n the dagonal polarzaton bass s, Smlarly, we obtan a For the crcular polarzaton bass, therefore and smlarly R a L a dagonal bass dagonal bass b canoncal bass canoncal bass + a a 3 +. S. 3 crcular bass a crcular bass b S ; To fnd the nner product n each of the three bases, we use Eq. A.5. a b canoncal bass 4 a b dagonal bass 8 a b crcular bass 8 S S and all three nner products are the same, confrmng the theory. Soluton to Exercse.6. Accordng to Eq. A.7, state ψ decomposes nto bass v accordng to

4 S4 S Solutons to Chapter exercses ψ v ψ v. S.6 Hence, ψ ψ v j ψ v ψ v j v S.7, j v j ψ v ψδ j S.8, j v ψ S.9.3 pr. S.0 Soluton to Exercse.7. Suppose state ψ, measured n bass { v }, gves probabltes pr v ψ. Then for state ψ e φ ψ, we have pr eφ v ψ e φ v ψ v ψ pr. Soluton to Exercse.8. a As we found n Ex. C.8, state 45 after propagatng through a waveplate at.5 wll become H and subsequently be transmtted through the PBS. State 45, on the other hand, wll become V and reflect from the PBS. Hence these two states wll generate clcks n two dfferent photon detectors and can therefore be dstngushed by ths apparatus. b As found n Ex. C.9, the two crcularly polarzed states, propagatng through a quarter-waveplate at 0, wll transform nto the dagonally polarzed states. The subsequent part of the apparatus s equvalent to that of part a, and hence can dstngush between these states. Soluton to Exercse.0. The apparatus would be smlar to that n Fg..b, but the waveplate s optc axs would need to be at the angle θ/ to horzontal. Such a waveplate wll convert state θ nto H and π + θ nto V. Soluton to Exercse.. One λ/4 waveplate wth ts optc axs orented at 45 from horzontal would do the trck. In the reference frame of that waveplate, states H and V appear dagonally polarzed, so the waveplate would nterconvert states between the canoncal and crcular bases accordng to H R V L H. Hence such a waveplate, followed by a polarzng beam spltter, wll send all rght crcularly polarzed photons to one detector and left to another. Soluton to Exercse.. a We use the theorem of total probablty Ex. B.6. Gven that the nput s ether H wth probablty / or V wth probablty /, and usng the result of Ex..9, we fnd

5 S Solutons to Chapter exercses S5 pr H pr H H + pr H V ; pr V pr V H + pr V V. b Smlarly, pr + pr + H + pr + V ; pr pr H + pr V. c Smlarly, pr R pr R H + pr R V ; pr L pr L H + pr L V. Soluton to Exercse.3. We use the decompostons found n Ex..5. a b c pr H 30 H pr V 30 V Hence the measurement gves a 75% chance of beng n state H and a 5% chance of beng n state V. pr pr 30 pr R 30 R pr L 30 L

6 S6 S Solutons to Chapter exercses Soluton to Exercse.4. pr + + ψ H + V H + e ϕ V S.a 4 H H + eϕ H V + V H + e ϕ V V eϕ + cosϕ; pr ψ H V H + e ϕ V S.b 4 H H + eϕ H V V H e ϕ V V eϕ cosϕ. As expected, pr + + pr. Soluton to Exercse.5. For a general polarzaton state ψ ψ H H + ψ V V, let us express ψ H a H e φ H and ψ V a V e φ V, where the φ s are real, a s are both real and non-negatve. Recallng that a change n the overall phase of the system does not affect ts physcs, we may multply the state ψ by the phase factor e φ V, so that ψ r H H + r V e φ V where we have defned the new varable φ φ V φ H. Our task s to fnd three unknown varables, r H, r V and φ. We frst look at the measurement n the canoncal bass. The probablty to detect a horzontally polarzed photon s from whch we fnd pr H H ψ r 0 H r V e φ rh, r H pr H r V rh pr H. Here we used the facts that both r H and r V are real and postve, as well as the normalzaton condton ψ ψ r H + r V e φ r H + r V. It remans to determne φ. We wrte the probablty of detectng the +45 polarzed state as follows:

7 S Solutons to Chapter exercses pr + + ψ r H r V e φ / r H + r V e φ / [ r H + r V + r H r V e φ + e / + r H r V cosφ φ ] S7 and for the rght crcularly polarzed state pr R R ψ r H r V e φ / r H r V e φ / [ r H + r V r H r V e φ e / + r H r V snφ. Each of these equatons do not dentfy φ unambguously. For example, states R and L for whch φ ±π/, respectvely both have cos φ 0 and hence cannot be dstngushed through a canoncal and dagonal bass measurements alone. However, the two equatons taken together gve us both the sne and cosne, allowng us to unambguously determne the value of φ. Soluton to Exercse.6. a Assume for smplcty that the space we want to measure states n s two-dmensonal. An apparatus that s able to resolve one of the states, say a, wth perfect certanty, must nclude a projectve measurement assocated wth the bass { a, a }, where a s some state that s orthogonal to a. If ths apparatus s now appled to perform measurement on state b, t has a nonzero probablty a b to pont onto a. Therefore, wth some probablty, the projectve measurement wll yeld the same outcome for a and b. No matter what classcal processng the output of ths measurement s subjected to, ths nconclusveness wll persst. b Such a devce could be made wth two sub-devces, one measurng n the { a, a } bass, one measurng n the { b, b } bass, and a randomzer that sends states to one or the other sub-devce randomly for example, a non-polarzng beam spltter. If the second sub-devce detects a b state, one would know the nput state was certanly not b, so t must be a. Smlarly, f the frst sub-devce detects a a state, the nput state s certanly b. In the event of any other outcome, the nput state s uncertan. Soluton to Exercse.7. The photon has a 50% probablty to go ether nto the upper or lower path. If t goes nto the lower path, the bomb wll detonate. If t goes nto the upper path, t wll leave the nterferometer n the vertcal polarzaton state and wll wth equal probablty land on ether detector + or. The probablty of the event n each detector s thus 5%. In the event of a clck of detector, the bomb s detected. If detector + clcks, the result s nconclusve. Soluton to Exercse.8. Frst, note that only those events wll contrbute to the error rate n whch Alce and Bob use the same bass. In approxmately / events, Eve wll also be usng ths bass, then she wll not ntroduce an error. In the remanng / events, Eve wll ntercept and resend a photon n a wrong bass, whch wll be then randomly detected by one of Bob s detectors. Wth probablty /, t wll be the ncorrect detector, so φ ]

8 S8 S Solutons to Chapter exercses Bob wll record a bt value that s dfferent from that sent by Alce. Therefore the overall probablty of error n the fnal key s / / /4. Soluton to Exercse.9. The loss wll not affect the securty because Alce and Bob, when generatng the secret key, do not use the data from those events n whch the photon has been lost. Soluton to Exercse.0. a A loss rate of 5% per km mples that nl n 0 e βl 0.95n 0 for L km. Accordngly, β ln0.95 km km. b For L 300 km, e βl e Soluton to Exercse.. Out of the n 0 photons sent by Alce every second, n 0 e βl wll reach Bob; each of them wll be detected wth probablty η. Of the detected photons, one-half wll be used to generate the secret key, so the quantum bt transfer rate s ηn 0 e βl /. Addtonally, the two Bob s detectors generate dark counts at a rate f d, but a half of that rate wll correspond to events n whch Alce and Bob chose dfferent bases. Of the remanng half, only one-half of the events wll be n the wrong detector, thereby generatng an error n the secret key. So the quantum bt error rate s f d /. The fracton of error n the produced secret key s, accordngly, f d / f d + ηn 0 e βl. When ths fracton becomes hgher than %, securty s not guaranteed. Ths happens at L 00 km for n s and 340 km for n s Fg..5. Soluton to Exercse.. Usng the result of Exercse A.45, we wrte: canoncal bass +. Usng the same approach for the { R, L} bass would requre obtanng the expressons for + and n that bass. As an alternatve technque, we use the expresson A. for convertng an operator from the Drac to matrx form: The actual secret key transfer rate s somewhat lower due to the overhead assocated wth the prvacy amplfcaton.

9 S Solutons to Chapter exercses S9 R + R R + R [ ][ ] R + L R + L [ ][ ] L + R L + R [ L + L L + L [ ][ ][ ] ] and hence the matrx s Soluton to Exercse.3. a Usng Eq. A.5, we wrte crcular bass + Â RH + HV H + V H + HV HH + V H + HV, S. whch, accordng to Eq. A.4, corresponds to the matrx / Â /. S.3 0 b Smlarly, Â R+ + H H + V H + V + HH V + HH + HV + V H + V V,

10 S0 so we obtan the matrx  + S Solutons to Chapter exercses. S.4 Soluton to Exercse.4. a From Eqs. A.5 and.4, we obtan  ϕ e ϕ αα + π π. + α + α b We know from Table. that α wrte cosα and π snα + α cos π + α sn π + α  ϕ e ϕ cosα snα cosα snα + snα cosα snα cosα e ϕ cos α + sn α e ϕ snα cosα e ϕ snα cosα e ϕ sn α + cos. α S.5 snα. Hence we can cosα c For a half-wave plate, ϕ π so e ϕ. For a quarter-wave plate, ϕ π/ so e ϕ. Substtutng ths nto  ϕ, we obtan Eq..5 for the half-wave plate, we also need to apply the trgonometrc denttes for the sne and cosne of a double argument. Soluton to Exercse.5. cosθ a Wrtng θ, we fnd snθ  HWP α θ.5a cosα snα cosθ snα cosα snθ cosα cosθ snα snθ snα cosθ + cosα snθ cosα θ snα θ α θ. b A quarter-wave plate wth ts optc axs orented horzontally has α 0 so Eq..5b takes the form 0  QWP 0. Applyng these to the dagonal and crcular polarzaton states, we fnd 0

11 S Solutons to Chapter exercses  QWP L; 0  QWP 0 L 0 ; 0  QWP 0 0 R; 0  QWP 0 R S Soluton to Exercse.6. By referrng to Eq..5a we fnd that the the matrx representaton n the canoncal bass of a λ/ waveplate wth ts optc axs orented vertcally s the ˆσ z operator. Ths waveplate s all that s necessary to mplement the ˆσ z operator. Smlarly [Ex..4b], a λ/ waveplate wth ts optc axs orented at a 35 degree angle from horzontal s suffcent to mplement the ˆσ x operator. If we have a sequence of optcal elements beng appled to the photon, the operator for ths sequence can be found by multplyng the operators of the ndvdual elements together n the reverse order,.e. the operator correspondng to the frst optcal element s placed last n the expresson for the product. Because ˆσ z ˆσ x ˆσ y, the Paul operator ˆσ y may be mplemented up to an overall phase factor usng a λ/ waveplate wth ts optc axs orented at 35 followed by a λ/ waveplate wth ts optc axs orented vertcally. Soluton to Exercse.7. a Referrng to Eq. A.4, we fnd HH + HV + V H V V. S.6 b Ĥ H 0 Ĥ V S.7 c Matrx.5a takes the form of the Hadamard matrx for α 5π/4. The Hadamard operaton can be therefore be mplemented by a λ/ waveplate wth ts optc axs orented at 5π/8.5. Soluton to Exercse.8.

12 S ˆΠ S Solutons to Chapter exercses Soluton to Exercse.9. We begn by wrtng the observable operator n the Drac notaton accordng to the defnton.: HH + V V. S.8 Ths s equvalent to the Paul operator ˆσ z [see Eq. refpauldefndracc]. Smlarly, usng Table., we fnd for the dagonal bass measurement: ++ and for the crcular bass RR LL 0 0 ˆσ x S.9 0 ˆσ 0 y. S.0 Soluton to Exercse.30. a The observable operator ˆV s gven by Eq... Because the egenvalues of an observable are real.e. v v, the adjont operator s the same: b Ths follows from the spectral theorem Ex. A.60. ˆV v v v ˆV. Soluton to Exercse.3. We wll begn wth the Paul matrx 0 ˆσ x. 0 We are lookng for the egenvalues and egenvectors of ths matrx see e.g. the soluton to Ex. A.64 for more detals on ths procedure. The characterstc equaton takes the form ˆσ x vˆ v v v 0 By solvng for v, we fnd that our egenvalues are v, ±. α Now we fnd the egenvector v assocated wth each of these egenvalues by solvng the equaton β ˆσ x vˆ v 0. Ths equaton becomes

13 S Solutons to Chapter exercses v v α β 0, 0 S3 from whch, for v, we fnd α β. Also, we apply the normalzaton condton α + β to determne a normalzed egenvector v +. S. Usng the same procedure for the v we obtan v. S. Now, we follow the same procedure to calculate the egenvectors and egenbass for the other two Paul 0 matrces. For ˆσ y, we get v 0, ± and v R; v L. 0 The matrx ˆσ z s already dagonal, so v 0, ± and v H; v 0 0 V. S.3 These results are n agreement wth the alternatve defnton of the Paul matrces found n Ex..9. Note that n all three cases the matrx representatons for the Paul operators n ther own egenbases consst of the egenvalues placed along the dagonal: 0 ˆσ S.4 0 Soluton to Exercse.3. a From Eq. B. we may wrte that the expectaton value s gven by V N pr v S.5 where v s the value returned by the measurement and pr s the probablty to detect ψ n the state v. The latter equals pr v ψ ψ v v ψ S.6 and hence

14 S4 S Solutons to Chapter exercses V N ψ v ψ v v ψ N v v v ψ S.7. ψ ˆV ψ. S.8 b Smlarly to part a, V pr v V. Transformng the operator n the rght-hand sde of Eq..5, we have ˆV V ˆ { S.9 [v V v v ]} S.30 [ v V v j V v ] v v j v j S.3, j [ v V v v ]. S.3 The quantum mean value of ths operator s then ψ ˆV ˆV ˆ ψ ψ v v ψv V pr v V, whch s the same as the rght-hand sde of Eq. S.9. To prove Eq..6, we utlze the result of Ex. B. to argue that V pr v V. S.33 The frst term n the expresson above s the expectaton value of operator ˆV. Soluton to Exercse.34. The experment n queston s equvalent to measurng observable ˆσ z N tmes and takng the sum of all the results. The statstcs of such summaton has been calculated n Ex. B.5. Applyng the result of Ex..33, we fnd that the expectaton value s N σ z 0 and the uncertanty s N σ z N. Soluton to Exercse.35. If ψ s an egenstate of the operator ˆV, we have ˆV ψ v ψ and ˆV ψ v ψ. Therefore ˆV ψ ˆV ψ ψ ˆV ψ v ψ ψ vψ ψ v v 0. Conversely, suppose that the uncertanty of measurng the observable ˆV n the state ψ vanshes. Denotng ˆV ψ φ, we wrte ˆV ψ ˆV ψ ψ ˆV ψ φ φ ψ φ,

15 S Solutons to Chapter exercses S5 where n the last equalty we have utlzed the fact that ˆV, as an observable, s Hermtan, so ψ ˆV ψ ψ ˆV ˆV ψ φ φ. By assumpton, ˆV 0, so we have φ φ ψ φ. S.34 Because the state ψ s normalzed, we can rewrte Eq. S.34 as ψ ψφ φ ψ φ. We now notce that the above equaton saturates the Cauchy-Schwartz nequalty A.0. As determned n Ex. A.6, ths can happen f and only f states ψ and φ are collnear,.e. φ ˆV ψ v ψ. Soluton to Exercse.36. If both operators are smultaneously dagonalzable, we can present them n the form  A v v and ˆB B v v. Then:  ˆB j A B j v v v j }{{} δ j v j A B v v ˆBÂ. Let us now prove the converse statement. Consder one of  s egenvectors v : S.35  v v v Multply both sdes by ˆB on the left: ˆB v v ˆB v Commutng the operators n the left-hand sde above, we have  ˆB v v ˆB v, S.36 S.37 S.38 so ˆB v must be an egenstate of  wth the egenvalue A. If the egenvalue v s non-degenerate, ths s only possble f ˆB v s proportonal to v Ex. A.66, whch mens that v s an egenstate of ˆB. Now let us consder the case of v beng degenerate. As we know from Ex. A.70, egenstates of  wth egenvalue v form a subspace whch we shall call V. Moreover, Eq. S.38 tells us that operator ˆB maps any state n V onto another state n V. Because ˆB s a Hermtan operator n V, t dagonalzes n ths subspace. That s, there exsts an orthonormal bass n V consstng of egenvectors of ˆB. But because V only contans egenvectors of Â, each element of ths bass s a smultaneous egenvector of both operators. The same procedure can be appled to each of the subspaces assocated wth the egenvalues of operator Â. Soluton to Exercse.37.

16 S6 S Solutons to Chapter exercses [Â, ˆB ] ψ  ˆB ψ ψ ˆB ψ ψ  ˆB ψ ψ ˆB  ψ [because  and ˆB are Hermtan] ψ  ˆB ψ ψ  ˆB ψ [by Eq. A.37]  ˆB  ˆB Im  ˆB [because z z Imz for any complex z] Smlarly: Fnally, {Â, ˆB } ψ  ˆB ψ + ψ  ˆB ψ  ˆB +  ˆB Re  ˆB [Â, ˆB] 4 Im  ˆB 4 Im  ˆB + 4 Re  ˆB 4  ˆB. S.39 Soluton to Exercse.38. The left-hand sde of the Cauchy-Schwarz nequalty a ab b a b S.40 for a  ψ and b ˆB ψ, where  and ˆB are Hermtan operators, takes the form a ab b ψ   ψ ψ ˆB ˆB ψ ψ  ψ ψ ˆB ψ. S.4 Smlarly, the rght-hand sde of Eq. S.40 becomes a b ψ  ˆB ψ, S.4 so nequalty S.40 takes the form of Eq..0. Soluton to Exercse.39. Because A B 0, we have  A and ˆB B, so the uncertanty prncple. takes the form: ψ  ψ ψ ˆB ψ ψ [Â, ˆB] ψ 4 Ths result obtans mmedately from Eqs..9 and.0. S.43 Soluton to Exercse.40. Let us defne operators    and ˆB ˆB ˆB. The expectaton values of these observables vansh, so we can use the smplfed uncertanty prncple from the prevous exercse to wrte At the same tme, we have  ˆB [Â, ˆB ]. 4 S.44

17 S Solutons to Chapter exercses S7 A A A A ; S.45 B B B B and [Â, ˆB ]  ˆB ˆB    ˆB ˆB ˆB ˆB    ˆB  ˆB  ˆB +  ˆB ˆB + ˆB  + ˆB  ˆB   ˆB ˆB [ Â, ˆB ]. S.46 Substtutng Eqs. S.45 and S.46 nto Eq. S.44, we obtan  ˆB [ Â, ˆB ]. 4 S.47 The uncertanty prncple would not reman vald f the commutator of  and ˆB were replaced by the antcommutator or product of these operators because n ths case Eq. S.46 would no longer apply. Soluton to Exercse.4.  ˆB [ Â, ˆB ] ε ˆ ε Soluton to Exercse.4. a ˆσ x H ˆσ x H 0 0 0; S ˆσy H ˆσ y H σˆ x H σˆ x H H σˆ x H ; ˆσ y H ˆσ y H H ˆσ y H ; S

18 S8 S Solutons to Chapter exercses On the other hand, we know from Ex. A.78 that [ ˆσ x, ˆσ y ] σ z so [ σx ˆ, σˆ y ] H ˆσ z H b The uncertanty prncple takes the form σˆ x σˆ y 4 [ σˆ x, σˆ y ]. Both sdes of the nequalty equal. c The uncertanty product can vansh for any state n whch the expectaton value of ˆσ z s zero. For example, f ψ +, observable ˆσ x has a certan value of +, and hence the uncertanty product s zero. Soluton to Exercse.43. Accordng to Eq..5, ψt e E t/ h E + e E t/ h E S.50 e E t/ h E + e E E t/ h E. S.5 Neglectng the overall phase factor, state ψt becomes physcally equvalent to E E / when e E E t/ h, or E E t/ h π. Soluton to Exercse.44. a Let { E k } be the energy egenbass. From Eq..5 we know that Û E k e E kt/ h E k. The matrx elements of the evoluton operator are therefore Hence U j E j Û Ek e E k t/ h E j Ek e E k t/ h δ jk. b Ths matrx can be rewrtten n the Drac notaton usng Eq. A.4 as e E t/ h 0 Û S e E Nt/ h Û e Ekt/ h E k E k. k S.53 Comparng ths wth the equaton.6 for the Hamltonan and the defnton A.49 of operator functons, we fnd that Ût e Ĥ h t. The Hamltonan operator Ĥ corresponds to a physcal observable, energy, and s thus Hermtan. The Schrödnger evoluton operator, e Ĥ h t must then be untary accordng to Ex. A.9. Soluton to Exercse.46. Accordng to the result of Ex. A.96:

19 S Solutons to Chapter exercses S9 d dt e ī h Ĥt ψ0 ī hĥe ī h Ĥt ψ0 ī hĥ ψt, whch s consstent wth the Schrödnger equaton.3. Soluton to Exercse.47. a Method I. The egenstates of the operator ˆσ z are H and V, wth the correspondng egenvalues ± Ex..9, whch means that the energy egenvalues are E H hω and E V hω. The ntal state ψ0 H s an egenstate of the Hamltonan and hence a statonary state, and evolves accordng to ψt.8 e h ī EHt H e ωt H. The ntal state ψ0 + H + V, and hence ψt.8 e ī h E Ht H + e ī h E V t V e ωt H + e ωt V. Method II. Snce ˆσ z HH V V, the evoluton operator s cf. Ex. A.94 Ût e ω ˆσ zt e ωt HH + e ωt V V. S.54 By applyng Eq..9 we have for a photon ntally n ψ0 H: For the ntal state ψ0 +: ψt e ωt HH + e ωt V V H e ωt H. ψt e ωt HH + e ωt V V H + V e ωt H + e ωt V. Method III. Let ψt ψh t ψ V t n the canoncal bass. The matrx of the Hamltonan s 0 Ĥ hω 0 S.55 and the Schrödnger equaton becomes or d dt ψh t ψ V t 0 ω 0 ψh t, ψ V t

20 S0 d dt ψh t ω ψ V t S Solutons to Chapter exercses. ψh t ψ V t Ths expresson means that the dfferental equaton must hold for each row of the matrces n the left- and rght-hand sde, so we can rewrte t as a system of ordnary dfferental equatons: { ψh t ωψ H t ψ V t ωψ V t whch has the followng soluton: { ψh t Ae ωt ψ V t Be ωt The coeffcents A and B are obtaned from the ntal condtons. If the ntal state s H have A,B 0 and thus e ωt ψt e ωt H. 0 If the ntal state s +, then we have A B and thus ψt e ωt e ωt,, then we 0 n agreement wth the answer obtaned usng the other two methods. b Method I. The egenstates of the Hamltonan are now + and, wth the correspondng egenvalues E ± ± hω. The ntal state H decomposes accordng to ψ0 + +, and evolves accordng to ψt e ī h E +t + + e ī h E t e ωt + + e ωt [ e ωt H + V + e ωt H V ] cosωt H snωt V. The ntal state + s an egenstate of the Hamltonan: Method II. The evoluton operator s now ψt.8 e ī h E +t + e ωt + e ωt H + V. Ût e ω ˆσ x e ωt ++ + e ωt. S.56 The tme evoluton for a photon ntally n ψ0 H s

21 S Solutons to Chapter exercses S For the ntal state ψ0 +: ψt Ût H [ e ωt ++ + e ωt ] + + e ωt + + e ωt cosωt H snωt V. ψt e ωt ++ + e ωt +e ωt +. Method III. In order to apply the Schrödnger equaton technque, we agan decompose ψt accordng to Eq. S.55. The matrx of the Hamltonan takes the form 0 Ĥ hω 0 and the Schrödnger equaton takes the form d dt ψh t ψ V t ω ψv t. ψ H t The system of equatons for the state components s { ψh t ωψ V t ψ V t ωψ H t. In order to solve ths system we can, for example, take the dervatve of both sdes of the frst equaton and substtute ψ V t from the second one: Ths equaton has soluton and, accordngly, ψ H t ω ψ V t ω ψ H t. ψ H t Ae ωt + Be ωt ψ V t ψ H t/ ω Ae ωt + Be ωt. For the ntal state H, we have ψ H 0, ψ V 0 0, hence A B / and thus ψt e ωt + e ωt cosωt e ωt + e ωt. snωt For the ntal state +, we have ψ H 0 ψ V 0 /, hence A 0, B and thus ψt e ωt e ωt e ωt +.

22 S S Solutons to Chapter exercses Soluton to Exercse.48. Polarzaton state transformatons by half-waveplates at 0 and 45 are gven by operators HH + V V and ++ +, respectvely see Ex..4. Comparng these wth the evoluton operators S.54 and S.56, respectvely, we see that they become dentcal, up to an overall phase factor, when ωt HWP π/ n both cases. The quarter-wave plate corresponds to the evoluton for a tme perod that s one-half of that for a half-wave plate,.e. ωt QWP π/4.