TheBlochSphere

Asinglespin-1/2state,or"qubit",isrepresentedasanormalizedstate ab

!

"#

$

%& where

|a|^2+|b|^2=1.Thephaseofthisisirrelevant,soyoucanalwaysmultiplyboth"a"and"b"byexp(ic)withoutchangingthestate.Notethatthisformalismcanbeusedforany2DHilbertspacestate;itdoesn't*have*tobeaspin-1/2particle.Foranysuchstate,youcanalwaysfindadirectionthatyoucanmeasurethespinandalwaysgetaresultof / 2 .Thisisthesamedirectionastheexpectationvalueofthevectorofspin-operators<S>.Thisdirectioncanberepresentedasaunitvector,pointingtoalocationonaunitsphere,orthe"Blochsphere".Forexample,spin-up(a=1,b=0)correspondstotheintersectionoftheunitspherewiththepositivez-axis.Spin-down(a=0,b=1)isthe-zaxis.

Foranarbitrarypointonthissphere,measuredinusualsphericalcoordinates θ,φ( ),thecorrespondingspin-1/2stateis(seeproblem4.30inGriffiths):Eqn[4.155]:

cosθ2

sinθ2eiφ

!

"

####

$

%

&&&&

orcosθ

2e−iφ /2

sinθ2eiφ /2

"

#

$$$$

%

&

''''

(seewhythesetwoformsarereallythesame?)

UsefulExercise:Checkthatthisworksfortheabovecasesinthediagram.

Noticethespecialnotationforspinup: 0 ,andforspindown: 1 .Thesearethe"0"'sand"1"'sofaquantumcomputer.Ofcourse,anysinglequbitstatecanbe

writtenintheseterms: ab

⎛

⎝⎜

⎞

⎠⎟= a 0 + b 1 .(Don'tforgetaandbarecomplex.)

SingleQubitMeasurements:Wealreadyknowhowtodosingle-qubitmeasurements,inprinciple,butthereisausefulshortcutwhenthinkingaboutstatesontheBlochsphere.Foranyqubit-statepointinginthef-directionontheBlochsphere,supposeyoumeasureitontheg-axis(foraspin-1/2particle,youcoulddothisbyputtingamagneticfieldintheg-directionandmeasuringtheenergy.)Itturnsoutthattheprobabilityoftheoutcomesonlydependsontheanglebetweenfandg:callthisangleθ .Itisnothardtoprove(doneinclass)thattheprobabilityofmeasuringtheeigenvaluecorrespondingtothestatepointinginthethe+gdirectionis cos2 θ 2( ) .Ifthiswasthemeasurementresult,wealreadyknowthatthequbitwould"collapse"intoastatewhoseBlochspherevectorpointedinthe+gdirectioninsteadofthefdirection.TheotherpossibilityisthatitmightcollapseintoastatewhoseBlochspherevectorpointedinthe-gdirection.Obviously,theprobabilityofthisothermeasurementoutcomewouldbe sin2 θ 2( ) .Therearenootherpossibleoutcomesforsuchameasurement.SingleQubitGates:RotationsontheBlochSphereAquantum"gate"isatransformationthatyoucanperformonaquantumstate.IMPORTANTNOTE:THISISNOTAMEASUREMENT.It'sjustalineartransformation,andcanberepresentedbyanoperator: Q ψ = ψ ' .Ingeneral,thegate(Q)takestheinputstate ψ ,andspitsouttheoutputstate ψ ' .Thereisnocollapse,justatransformation.(Technically,a"unitary"transformation.)Single-qubitgatesarebestenvisionedasrotationsontheBlochsphere.Youcanrotatearoundanyaxis,byanyangle--infact,wealreadyknowhowtodothistoaspin-1/2statewithanappropriatelyalignedmagneticfield.(ThestateprecessesaroundtheB-fielddirection.)Thesegates/rotationsarediscussedinanimportantprobleminGriffiths:Problem4.56.Theexponentialnotationintheearlierpartoftheproblemisnotneeded;theupshotofthisproblemisthelastequation[4.201](althoughthebook'sequation

dropstheIdentitymatrixinmyversion).ThistellsusthattheoperatorRcorrespondingtoarotationofanangleφ aroundanaxis n is:

R = cos φ2⎛

⎝⎜⎞

⎠⎟I+ i n ⋅σ( )sin φ

2⎛

⎝⎜⎞

⎠⎟ [4.201]

HereIisthe2x2identitymatrix,andσ isthevectorofPaulimatrices.(Soifyouwantedtorotatearoundthez-axis,youwouldputin n ⋅σ( ) =σ z .ObviouslyRwouldalsobea2x2matrix,sothatitcanoperateonaqubit.(Note:ThesematricesarenotHermitian!Theyare"unitary".)Special/usefulsingle-qubitgatesinclude:TheNOTgate(alsoknownasthePauliX-gate);a180orotationaroundthex-axis.ThePauli-Zgate:a180orotationaroundthez-axis.ThePauli-Ygate:a180orotationaroundthey-axis.The NOT gate;a90orotationaroundthex-axis.Phaseshiftgates,R φ( ) ;aφ -anglerotationaroundthez-axis.Usefulexercise:Buildthese2x2matrices,andcheckthattheyworkasadvertised!BuildingTwoQubitStates:TensorProductsInQM,whenyouhavetwosingle-particleHilbertspaces,thetotalwavefunctionlivesinalargerHilbertspacethatisthe"tensorproduct"ofthosetwospaces.If

qubit1isinstate ab

!

"#

$

%& ,andqubit2isinstate c

d

!

"#

$

%& ,thenthetotalstateofthetwo

qubitsystemis:

€

ab⎛

⎝ ⎜ ⎞

⎠ ⎟ ⊗

cd⎛

⎝ ⎜ ⎞

⎠ ⎟ =

acadbcbd

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

,ina4DHilbertspace.

Now,youcanalsowritethissameequationintermsofthe 0 , 1 notation: a 0 + b 1⎡⎣ ⎤⎦⊗ c 0 + d 1⎡⎣ ⎤⎦= ac 00 + ad 01 + bc 10 + bd 11 Noticethenewnotation:Thestate 00 isjust 0 ⊗ 0 ,etc.

Youcanalsotensorproductsingle-qubitoperatorstogethertomaketwo-qubitoperators.Basically,youjustmultiply*each*elementofthefirstmatrixbythe*entire*secondmatrix!Thisclearlywillmakeabiggermatrix:

A BC D

⎛

⎝⎜

⎞

⎠⎟⊗ E F

G H

⎛

⎝⎜

⎞

⎠⎟=

AE AF BE BFAG AH BG BHCE CF DE DFCG CH DG DH

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

(seethepattern??)

Ifthe2x2matrixMisanoperatorononequbit,thisclearlycan'toperateona4DHilbertspace.ButtheoperatorM⊗ I (whereIisthe2x2identity)representsameasurementofMonqubit#1;it'sa4x4operator.Theoperator I⊗M iswhatyouwouldusetomeasureMonqubit#2.Two-Qubits:EntangledStatesNotallstatesin4DHilbertspacescanbeseparatedintotwodistinctsingle-particle

statesasintheaboveexample.Iftheyareintheform ab

!

"#

$

%&⊗ c

d

!

"#

$

%&=

acadbcbd

!

"

####

$

%

&&&&

,the

statesare"separable";otherwisetheyare"entangled".Entangledstatesaremathematicallypossiblebecauseyoucanaddupsuperpositionsofseparablestates,thatnolongerneatlysplitintotwoqubits.We'lltrytomakesenseofthemlater.Themostgeneraltwo-qubitstatecanbewritten:A 00 +B 01 +C 10 +D 11 Oneeasywaytoseeifsuchastateisseparableisifthequantity2|AD-BC|=0.Thisquantityiscalledthe"Concurrence",andisameasureofentanglement.(Themaximumpossiblevalueis2|AD-BC|=1;thisoccursfora"maximallyentangledstate".)Two-QubitGatesA"controllednot"or"CNOT"isanexampleofagatethatactsona2-qubitstate.Thisoperationcanturnseparablestatesintoentangledstates(andvice-versa)byswappingthebottomtwovaluesinthis4DHilbertspace:

efgh

!

"

#####

$

%

&&&&&

⇒ (CNOT )⇒

efhg

!

"

#####

$

%

&&&&&

UsefulExercises:What4x4matrixwoulddothis?Findaseparablestatethatturnsintoamaximally-entangledstateundertheCNOToperation.Obviously,twoconsecutiveCNOTsgiveyoubacktheoriginalstate.Anothertwo-qubitgateistheSWAPgate,whicheffectivelyswapsthetwoqubits.efgh

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

⇒ (SWAP)⇒

egfh

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

UsefulExercise:Checkthatthisworksasadvertisedforaseparablestate;alsofigureoutwhatthe4x4matrixmightlooklike.TheSWAPgatecannotbeusedtoentangleseparablestates;itjustswapsthem.Buttheroot-SWAP,or SWAP gate,candothis:

.UsefulExercise:CheckthattwoofthesegivesyouaSWAP!TwoParticleMeasurements:Evenifyouhaveanentangledstate,youcanofcoursechoosetomeasureeachqubitseparately.Sayyouchooseameasurementdirectionforqubit1andameasurementdirectionforqubit2.Therearetwopossibleoutcomesforeachdirection,sothereare4possiblecombinedoutcomes.(YouwouldbuildsuchanoperatorusingtheTensorProduct:seeabove.)Butfindingtheeigenstatesofa4x4operatorishard;it'salmostalwayseasiertostartwiththetwo2x2operatorsseparately.

Thepossibleoutcomesforqubit1areorthogonaleigenstatesofthequbit-1measurementoperator,andsocanalwaysberepresentedby

ψ1+ = ab

!

"#

$

%& , ψ1− = b*

−a*"

#$$

%

&'' (seewhythesearealwaysorthogonal?)

Youcangeneratethefirstofthesefromtheanglesofthemeasurementsetting(seefirstpage,"TheBlochSphere"),wherethe"+"outcomehasBlochspherevectorthatisalignedwiththesettingandthe"-"outcomehasavectorthatisanti-alignedwiththesetting.Inthesameway,thepossibleoutcomesforqubit2aredeterminedbyaseparatesettingchoice;callthose:

ψ2+ = cd

!

"#

$

%& , ψ2− = d*

−c*"

#$$

%

&'' .

Sothe4measurementeigenstatesofthefull4x4operatorcanbegeneratedviaatensorproduct.Forexample,theoutcome"++"(bothoutcomesalignedwiththecorrespondingsetting)is

ψ1+ ⊗ ψ2+ = ab

"

#$

%

&'⊗ c

d

"

#$

%

&'=

acadbcbd

"

#

$$$$

%

&

''''

= ++ .

Giventheactualstateinthefull4DHilbertspace,theprobabilityoftheoutcomeofthis"++"statecanbefoundfromthegeneralizedBornrule: ++ ψ

2.

Ifthestatesarenormalized,the4probabilitieswillalwaysaddtoone.MarginalProbabilitiesWe'reofteninterestedintheprobabilitiesofameasurementofoneparticularparticle,independentofwhathappenstotheotherone.(Thequantumno-signallingtheoremsaysthatthenetprobabilitiesofoneparticlecan'tdependonthechoiceofmeasurementsettingattheotherparticle.)Sotofindtheprobabilityofa+outcomeofqubit#1,wecansetthequbit#2measurementsettingtoanythingwewant,saythez-axis.Thenc=1,d=0.Sothetwopossibleoutcomeswhereparticle1isfoundtobe"+"are:

++ =

a0b0

!

"

####

$

%

&&&&

,and +− =

0a0b

"

#

$$$$

%

&

''''

.

Tofindtheprobabilityoftheoutcome"+"forqubit1,wethereforeneedtocalculate++ ψ

2+ +− ψ

2.Inotherwords,weadduptheprobabilitiesof(+on1,+on2)

and(+on1,-on2).Thesumisjustthetotalprobabilityofmeasuring+onqubit1.Weareabouttolearnaneasierwaytodothis(andaneasierwaytomakesenseoftheresults),usingsomethingcalleda"partialtrace"ofa"densitymatrix".DensityMatrices:PureStatesGiventhecompletequantumstate ψ ,it'spossibletoforma"densitymatrix"thatencodesthisstate,usingtheequation:ρ = ψ ψ Thiscrazy-lookingequationiscalledan"outerproduct";youcanfigureoutwhatitmeansjustbyplugginginthebra-andket-invectorform.Forasinglequbit,thisgives:

ρ = ab

⎛

⎝⎜

⎞

⎠⎟ a* b*( ) = aa* ab*

a*b bb*⎛

⎝⎜⎜

⎞

⎠⎟⎟ .

Noticethetraceofthismatrixis1.Andit'sautomaticallyindependentoftheglobalphaseontheoriginalstate.Also,noticethat ρ2 = ψ ψ ψ ψ = ρ .Forthiscase,when ρ2 = ρ ,wesaythisisa"purestate".(meaning,itisgeneratedfromasingle,completewavefunction).We'reabouttoencounterotherdensitymatricesforwhich ρ2 ≠ ρ ;thesewillbe"mixedstates".Everystateiseitherpureormixed.PureStateofSingleQubits

ψ =cosθ

2

sinθ2eiφ

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟correspondstothestateofapointontheBlochSphere(recall).

Buildingthedensitymatrixfromthisstateonefinds:

ρ =cos2(θ / 2) cosθ

2sinθ2e−iφ

cosθ2sinθ2e+iφ sin2(θ / 2)

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

Then,usingthehalf-angleformulas,thiscanbewrittenintermsofthePauliMatrices:

ρ =12I+ sinθ cosφσ x + sinθ cosφσ y + cosθσ z( )

ButthisisjusttheCartesiancoordinatesoftheoriginalvectorontheBlochSphere!(ThinkabouttheconversionfromsphericaltoCartesian;thosearethe3terms.)Soapurestatecanbewrittenas

ρ =12I+ n ⋅σ( )

Where"n"istheunitvectorontheBlochsphere.Laterwe'llseethatwecangeneralizethisevenwhen"n"isnotaunitvector!Thisisoneoftheeasiestwaystogetfromanarbitrarysingle-qubitwavefunctiontothevectorontheBlochsphere--especiallyifyoudon'tlikeusingsphericalcoordinates.IntroductiontoMixedStatesThenicethingaboutdensitymatricesisthattheycanbecombinedaccordingtoordinaryprobabilityrules.Supposeyouhadamachinethatmadethestate ψ1 30%ofthetime,butmadethedifferentstate ψ2 70%ofthetime.(Thisisnotasuperposition!Justclassicalignorance.)Inthiscase,foranygivenparticle,youwouldn'tknowforsurewhatthestatewas--andyet,yourjobistomakepredictionsallthesame.Thiscanbedonebysimplyweightingthepossibledensitymatrices(accordingtotheirprobability),andaddingthemupintoasingledensitymatrix:ρ = 0.3 ψ1 ψ1 + 0.7 ψ2 ψ2 .Seehowthatworks?Youjustweightthepossibledensitymatricesandaddthemup.Itturnsoutthatfromthistotaldensitymatrixyoucanmakealltherightpredictions.Also,thisisclearlyamixedstate;ρ2 ≠ ρ .(Butthetraceofthemixedstatedensitymatrixisstill1.)

Foranygivenmixedstatedensitymatrix,thereareusuallymanywaystomakeit:different"ensembles".Buteventhoughthosedifferentensemblesmaybemadeupoftotallydifferentquantumstates,itturnsoutthereisabsolutelynowaytoexperimentallydistinguishthedifferentensembles,iftheyhavethesamedensitymatrix!Ifyouknowthedensitymatrix,youknowalltheprobabilitiesthatyoucanmeasure.ExtractingProbabilitiesIt'spossibletousetherulesfromordinaryQMtoshowtheexpectationvalueofanyoperatorQissimply:<Q>=Trace(ρQ)Inotherwords,yousimplymultiplythedensitymatrixtimestheoperatormatrix,andtakethetrace.Thisalwaysworks,evenformixedstates!(There'salsoacollapserulewemaygettolater.)Inprinciple,knowingalltheexpectationvaluesgivesyoualltheprobabilities,butwe'llgettoexactprobabilitieslater.(maybe!)TimeEvolutionTheSchrodingerEquationalsolooksquitenicewhenwrittenintermsofdensitymatricesThesematricesevolvewithtime,ofcourse: ρ(t) .Itsolvestheequation:

i! dρdt

= H,ρ[ ]

Yes,that'sacommutator!IfHisconstant(asusual,forus),thesolutionis:

ρ(t) = e−iHt/!ρ(t = 0)e+iHt/! Wewon'tusethissolutionmuch,sodon'tworryaboutittoomuch.ButseeGriffithsproblem4.56ifyou'rewantingtounderstandwhatitmeanstohaveanoperatorintheexponent.GateEvolutionFromthedefinitionofthedensitymatrix,aswellastheknownactionofagateoperatorRon ψ ,it'seasytoshowthatifyouputadensitymatrixintoagateR,theoutputdensitymatrixwillbesimply:ρoutput =RρR

† .Don'tforgetthatlastmatrixneedstobeHermitian-conjugated.

PartialTraces:Howtodescribeasmallerpieceofamulti-qubitsystem.Consideratwoqubitsystem, A 00 +B 01 +C 10 +D 11 .Youshouldbeabletoformthe4x4densitymatrixofthewholesystem;itwillbeapurestate.Butthen,asbefore,youmayaskaquestionabouttheprobabilitiesofameasurementonjustqubit#1.Ifyouwanttodescribethisqubitalone,independentlyoftheotherone,itwouldbenicetohavethatqubit's2x2densitymatrix.Thiscanbeextractedfromthefull4x4matrix;theprocedureistotakethe"partialtrace"ofthefull4x4matrix.Basically,ifyouwanttofindthe 0 0 entryofqubit#1'sdensitymatrix,youneedtoaddupthe 00 00 andthe 01 01 componentsofthefull4x4matrix.Inthiscase,you'llgetAA*+BB*.(Getit?Youdon'tcareaboutthesecondqubit,soyoutrybothoptions,whilekeepingthefirstqubitfixed.)Ifyouwanttofindthe 0 1

component,youaddupthe 0n 1nn∑ components,orAC*+BD*.

Itworkstheotherway,tooTofindthe 0 0 entryofqubit#2'sdensitymatrix,youaddupthe 00 00 andthe 10 10 componentsofthefull4x4matrix(keepingthesecondqubitfixed,summingoverallpossibleentriesforthefirstone.)Ifyouwanttofindthe a b componentforqubit#2,youaddupthe na nb

n∑ components.

PiecesofEntangledSystemsactlikeMixedStatesIftheoriginaltwo-qubitstatewasseparable,thispartialtraceprocedureendsupwiththepropertwostatesforthetwoqubits.Inthiscase,thesewouldbepurestates.Butiftheoriginaltwo-qubitstatewasentangled,thiscan'tpossiblywork,becausethestatedoesn'tfactorintotwopurestates.Sowhathappens?Youendupwithtwo*mixed*states!PiecesofentangledsystemsactjustlikeMixedStates!Whenyoufindthepartialtracecorrespondingtoasinglequbit,itturnsoutyoucanstillalwayswriteitintheearlierform:

ρ =12I+n ⋅σ( )

Onlynow"n"isnolongeraunitvector.(It'saunitvectorwhenit'sapurestate,justnotwhenit'samixedstate.)ButtheBlochspherepictureisstilluseful:themixed

stateisavector*inside*thesphere!!(Or,inthe"BlochBall".)Knowingwhichwayit'spointingisstilluseful,asitknowingitslength.Maximally-entangledstateswindupattheverycenteroftheBlochBall,withn=0.However,knowinghowthetwoindividualqubitsmightbemeasured,independently,isnotthewholestory.You'velostsomeinformationinthepartial-traceprocess.What'smissingisthe*correlations*betweenthetwoqubits,whichcanonlybegleanedfromthefull4x4densitymatrix.Andthisbringsustothetopicof"Entanglement".EntanglementForagenericpure2-qubitstate, ,can'talwaysbeseparatedintotwopuresinglequbits.Onewaytoseethisistogeneratethepartialtraceforeachqubit,andseeifitliesonthesurfaceoftheBlochsphereornot(seepreviouspage).Anotherwayistocomputethe"Concurrence",ameasureofentanglement:Concurrence=2|AD-BC|.Inthehomeworkyou'llshowthatforeachindividualqubit,themagnitudeofitsBlochBallvectornisrelatedtotheconcurrencevia:n2 + 2 AD−BC( )

2=1

Fromthisitshouldbeclearthatwhentheconcurrenceiszerothestateisseparable(notentangledatall).Whentheconcurrenceismaximum,1,thestateis"maximallyentangled",andeachqubit'sBlochBallvectoriszero(atthecenteroftheBlochBall).Inthiscase,youhavenoinformationabouteachindividualqubit;itcouldbemeasuredinanydirection,withequalprobability.Buteveninthiscase,notalltheinformationhasbeenlost!Ifyouknowthefullpurestateofthe2-qubitsystem, ,it'strueyouknownothingabouteachindividualqubit,butyouknoweverythingaboutthe*relationship*betweenthetwoqubits.Theclearestexampleisthatofasingletstate,1201 − 1

210 .(We'veseenthisinchapter4ofGriffiths;it'sthespin-0statefora

2spin-1/2system.)Forthissystem,weknowthatameasurementofanyspindirectionofqubit#1should"collapse"qubit#2intotheoppositedirection.Note:thisisnottosaytheresultsoftwomeasurementsonthetwoqubitswillalwaysbeopposite,becausetheexperimentersmightchoosetomeasuredifferentorientations.Iftheyare

A 00 +B 01 +C 10 +D 11

A 00 +B 01 +C 10 +D 11

measuredinthesameorientation,theywillalwaysbeopposite.Butiftheyaremeasured(say),onewithaz-measurementandonewithanx-measurement,thentheoutcomeswillberandomlycorrelated.(Seewhy?Az-measurementof#1collapses#2intoaz-state,butthenanx-measurementonanyz-stategivesarandomresult.)No-SignalingThediscussioninthepreviousparagraphmadeitsoundlikethatqubit#2isbeingaffectedbyameasurementonqubit#1,inthat#2is"collapsing"intoaparticularstatebasedontheoutcomeofthe#1measurement.That'scertainlywhatthemathematicsofQMseemstobeimplying:there'sno"local"descriptionoftheseparatetwoqubits,evenatthefundamentallevelthatQMsupposedlyprovides.Thisraisestwoquestions:mighttherebealower-leveldescription,belowQM,thathasalocaldescription?Andifnot,ifnatureisreally"non-local",doesthatmeandistantexperimenterscansignaltoeachother,usingentangledparticles?We'lltacklethelastquestionfirst,wheretheanswerisaclear"no".IfAliceistheexperimenteratqubit#1,andBobistheexperimenteratqubit#2,theyeachhaveachoiceofwhatmeasurementtomake.ButnomatterwhatAlicechooses,itdoesn'tchangethepartialtraceofBob'squbit,andvice-versa.Sure,the*outcome*ofAlice'smeasurementseemstochangeBob'squbit,butnotthe*choice*;andsinceAlicecan'tcontrolheroutcome,shecan'tsignal.Forexample,consideraseparatedsingletstate.IfAlicemeasuresthespincomponentonanyaxis,50%ofthetimeshe'llgetonedirection,and50%ofthetimeshe'llgettheother.ThissupposedlycollapsesBob'sstatetojusttheoppositeoutcome;50%eachway,wherethisisjustclassicaluncertaintyforBobatthispoint:ordinaryignorance.Butnomatter*what*directionthesestatesarepointing,buildingBob'smixedstatefromsucha50/50(classical!)uncertaintymixalwaysyieldsthesameresult:

50% 0 0 + 50% 1 1 = 12I (CenteroftheBlochBall)

Inotherwords,Alicehasn'tdoneanythingtochangeBob'sdensitymatrix:beforeshemeasureditwasastateofmaximumuncertainty(inthemiddleoftheBlochBall)andaftershemeasuredit'sstillastateofmaximumuncertainty.ALICEmayknowBob'sstate,butBobdoesn'tknowituntilhehearstheresultsofAlice'smeasurement.Moreimportantly,hecan'ttellbymeasuringhisownstate,becausealltheprobabilitiesareentirelydeterminedbyhisowndensitymatrix,whichhasn'tchanged.Thereexistsageneral"no-signalingtheorem":AlicecanneversignaltoBobusingentanglement.

BellInequalitiesJustbecauseQMgivesusanon-localaccountofthecorrelationsbetweenAliceandBobdoesn'tmeanthatQMmustbecorrect.Perhapsthere*is*alocaldescription,andQMisjustastatisticaltheorygoverningthatdeeperlevelofreality.ThisiscertainlywhatEinsteinthought,whenpresentedwithQM.Infact,thereseemstobeanobviouswaytoaccountforhavingcorrelationsbetweenthetwoqubits:acommoncause,orcommon"HiddenInstructions"sentalongtoAliceandBobthroughtheparticles.(Liketwooppositeglovessentthroughthemail,atrandom:it'snotsurprisingthatwhenAliceandBobreceivetheirpackage,theyalwayshaveoppositegloves!)Now,QMdoesnottellusanythingaboutwhatthese"hiddeninstructions"mightlooklike,buttheymightbethereallthesame,hidingintheparticles(justastheactualstateofthegloveswerehidinginthepackages).Orsopeoplethought,beforeJohnBellshowedthatthereisalimitoncorrelationsthatcanbeachievedinthismanner.AsdetailedinclassandChapter3oftheaccompanyingtext,onesuchlimitissomethingcalledtheCHSH-inequality.SayAlicecanmeasureeitherpropertya1ora2wherethevaluesofthesepropertiesarealwayseither+1or-1;numbersassignedtothe2-outcomesofameasurementlikeaspinorientation.(Forexample,removingthe! / 2 fromthevalueofaspin-componentmeasurement,leavingonlythesignoftheoutcome.)Bobcanmeasureeitherpropertyb1orb2.(Also+1or-1ineachcase.)Iftheoriginaldistributionofthesevariablescomesfromsomeinitialsourcewithanarbitrary"jointprobabilitydistribution"P(a1,a2,b1,b2),wherethe16possibleinitialdistributionsareeachassignedsome"generationprobability"bythesourceoftheentangledparticles,thesehiddeninstructionscouldthenbesenttoAliceandBobalongwiththeparticlesthemselves.Buteventhen,it'spossibletoprovetheCHSHinequality:a1b1 + a1b2 + a2b1 − a2b2 = S ≤ 2 (CHSHInequality)

Herethe<>'sareexpectationvalues(weightedaverages),overmanyidenticalmeasurements.Ofcourse,sinceAliceandBobeachhavetochoosewhichpropertytomeasure,andsodoesBob,theycan'tmeasureall4oftheseinasinglerun.Anygivenrunonlygivesapieceofinformationtowardsoneofthese4terms.(Aliceisn'tallowedtomeasurebotha1anda2atthesametime,anymorethanshecouldmeasuretwospinorientationsatthesametime.)Butthisinequalityis*violated*byQuantumExperiments!Inparticular,ifyoustartwithanentangledsingletstate,andAlicechoosesbetweenmeasuringspinonthex-axisorz-axis,whileBobchoosesbetweenmeasuringspinonthe x ± z( ) / 2 axes,

onefindsthatScanbeawhopping2 2 !(That'sthemaximumpossiblevalueofS,eveninQM;somethingcalled"Tsirelson'sBound".)Thismeansthatyoucan'taccountforsuchexperimentsusingcommon-causesandhiddeninstructions,nomatterwhatdetailsyoutrytodreamup.Ifthereisalocalaccountofentanglementexperiments,iteitherhastobe:1)Basedonthingsthatdon'tliveinordinaryspacetime(Inherentlynon-local)2)UsingFaster-than-Lightinfluences(Effectivelynon-local)3)RestrictiveonwhatAliceandBobareallowedtomeasureinanygivenrun(Superdeterministic)4)LocalhiddenvariablesthatdependonAliceandBob'sfuturesettings(Retrocausal)Forthislastcase,the"hiddenvariables"wouldnotbe"instructions";theywouldinfacthavedifferentinitialjointprobabilitydistributionsP(a1,a2,b1,b2),dependingonwhatAliceandBobwilleventuallychoosetomeasure.ItisakeyassumptionofallBell-inequalitiesthatsuchaprobabilitydistributionshould*not*dependonfutureevents.Ifyourelaxthisassumption,itistrivialtoviolatetheCHSHinequalityusinglocalparameters.ManipulatingEntangledStates:SingleQubitOperationsGivenatwo-qubitstate,theoperationsyoucanperformonitdependonwhetheryouhaveaccesstobothqubits(inthesameplace).Ifyouonlyhaveonequbitofanentangledpair(say,qubit#1),youcanputthatqubitthroughasinglequbitgate/operator:some2x2matrixR.(Thisisnotameasurement;justsomeunitarytransformation.)Asdescribedaboveinthisdocument,thatmeanstheoperationonthefullstatewouldbe:R⊗ I Here"I"isthe2x2identityforqubit#2.Obviously,ifyouwereoperatingonqubit#2,youwouldswaptheRandIintheaboveexpression.Suchasingle-qubitoperationislimitedinseveralways:1)Youcan'tchangethepartialtraceofthequbityouaren'toperatingon.(That'stheno-signalingtheorem!)2)Youcan'tchangetheconcurrenceofthefullstate.(Whichfollowsfrom#1,sinceknowledgeofngivesyouknowledgeoftheconcurrence.)

Butyou*can*changetherelationshipbetweenthetwostates,whichhopefullymakessense.Sayyouwereinasingletstate,whereyouknewthetwospinswouldbemeasuredinanoppositeorientation.Thenyoutakeonequbitandrotateitbya180orotationaroundthez-axis.Thatobviouslychangestherelationshipbetweenthetwoqubits!Nowameasurementinthex-ory-direction(onbothqubits)wouldyieldthe*same*result,notoppositeresults!Inthehomeworkyouwillshowthatsuchlocaltransformationscantakeyoufromonemaximally-entangledstatetoanother.ManipulatingEntangledStates:Two-QubitOperationsIfbothqubitsaretogether,youcanputthemthrougha2-qubitgate,asdescribedabove,suchasaCNOToraSWAP.Suchgatescaningeneralchangetheconcurrence,andcanbeusedtoentanglestatesthatare(initially)separable.Suchoperationscanalsobeusedtoimplement"jointmeasurements"onapairofqubits.Supposeyouwanttomeasurewhetherapairofspin-1/2particlesisinaspin-0(singlet)state,butyouonlywanttomeasureoneparticleatatime.ThesolutionwouldbetoputthetwoqubitsintoaCNOTgate,andthenmeasurethetwoparticlesseparately(inthecorrectbasis:anx-measurementonQubit#1andaz-measurementonQubit#2,asitturnsout).That'sbecause:

CNOT|singlet>=

0120

−12

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟

= 12

1−1

⎛

⎝⎜

⎞

⎠⎟⊗ 0

1

⎛

⎝⎜

⎞

⎠⎟ .

Inotherwords,theCNOToperationprojectsanysingletcomponentoftheoriginalstateintotwoseparablequbitswhosefirstqubitisinthe-xdirection(ontheBlochsphere)andwhosesecondqubitisinthe-zdirection.Asimplepairofsingle-qubitmeasurements(afteraCNOT)iseffectivelythesameasatotal-spinmeasurement(beforetheCNOT).Ofcourse,youmightfindanyof4resultsfromthismeasurement(2optionsforqubit#1,2optionsforqubit#2).Theother3casescorrespondtootherinitialcomponentsoftheentangledstate.Specifically,the4possiblemeasurementoutcomeseachcorrespondtoaninitialmaximally-entangledstateintheso-called"Bell-Basis".Therewillbeahomeworkquestionaboutthisbasis.