The Bloch Sphere A single spin-1/2 state, or "qubit", is represented as a normalized state a b ! " # $ % & where |a|^2+|b|^2=1. The phase of this is irrelevant, so you can always multiply both "a" and "b" by exp(ic) without changing the state. Note that this formalism can be used for any 2D Hilbert space state; it doesn't *have* to be a spin-1/2 particle. For any such state, you can always find a direction that you can measure the spin and always get a result of /2 . This is the same direction as the expectation value of the vector of spin-operators < S >. This direction can be represented as a unit vector, pointing to a location on a unit sphere, or the "Bloch sphere". For example, spin-up (a=1,b=0) corresponds to the intersection of the unit sphere with the positive z-axis. Spin-down (a=0,b=1) is the -z axis. For an arbitrary point on this sphere, measured in usual spherical coordinates θ, φ ( ) , the corresponding spin-1/2 state is (see problem 4.30 in Griffiths): Eqn [4.155]: cos θ 2 sin θ 2 e i φ ! " # # # # $ % & & & & or cos θ 2 e −i φ /2 sin θ 2 e i φ /2 " # $ $ $ $ % & ' ' ' ' (see why these two forms are really the same?) Useful Exercise: Check that this works for the above cases in the diagram.
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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: Q̂ 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 n̂ is:
R = cos φ2⎛
⎝⎜⎞
⎠⎟I+ i n̂ ⋅σ( )sin φ
2⎛
⎝⎜⎞
⎠⎟ [4.201]
HereIisthe2x2identitymatrix,andσ isthevectorofPaulimatrices.(Soifyouwantedtorotatearoundthez-axis,youwouldputin n̂ ⋅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+ n̂ ⋅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 00 10 andthe 01 11 components,orAC*+BD*.Andsoon.Itworkstheotherway,tooTofindthe 0 0 entryofqubit#2'sdensitymatrix,youaddupthe 00 00 andthe 10 10 componentsofthefull4x4matrix(keepingthesecondqubitfixed,summingoverallpossibleentriesforthefirstone.)Ifyouwanttofindthe 0 1 componentforqubit#2,youaddupthe 00 01 andthe 10 11 components.PiecesofEntangledSystemsactlikeMixedStatesIftheoriginaltwo-qubitstatewasseparable,thispartialtraceprocedureendsupwiththepropertwostatesforthetwoqubits.Inthiscase,thesewouldbepurestates.Butiftheoriginaltwo-qubitstatewasentangled,thiscan'tpossiblywork,becausethestatedoesn'tfactorintotwopurestates.Sowhathappens?Youendupwithtwo*mixed*states!PiecesofentangledsystemsactjustlikeMixedStates!Whenyoufindthepartialtracecorrespondingtoasinglequbit,itturnsoutyoucanstillalwayswriteitintheearlierform:
ρ =12I+ n̂ ⋅σ( ) .
Onlynow"n"isnolongeraunitvector.(It'saunitvectorwhenit'sapurestate,justnotwhenit'samixedstate.)ButtheBlochspherepictureisstilluseful:themixed
stateisavector*inside*thesphere!!Knowingwhichwayit'spointingisstilluseful,asitknowingitslength.Maximally-entangledstateswindupattheverycenteroftheBlochsphere,withn=0.However,knowinghowthetwoindividualqubitsmightbemeasured,independently,isnotthewholestory.You'velostsomeinformationinthepartial-traceprocess.What'smissingisthe*correlations*betweenthetwoqubits,whichcanonlybegleanedfromthefull4x4densitymatrix.Andthisbringsustothetopicof"Entanglement".
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