Quantum Magic

28 Pages·2015·19.05 MB·English

by  Jeff Bub

6 Quantum Magic
ThemathematicianAndrewGleasonusedtheterm‘intertwined’todescribetheintricatemeshing
ofcommutingandnoncommutingobservables.1 Inaseminalpaper,heprovedthatitfollowsfrom
thisfeaturethatthereisonlyonewaytodefineprobabilitiesonthestructureofquantumproperties.
I’llfollowmycolleagueAllenStairsandusetheterm‘entwined.’ Observablesareentwinedinthe
sense that a ‘yes–no’ observable representing a property of a system can belong to different sets
of commuting observables that don’t commute with each other. Entwinement shows up in the
‘contextuality’ of measurement outcomes, the dependence of an outcome of a measurement on
thecontextdefinedbywhatothercommutingobservablesaremeasured, wherethecontextinthis
sensecanbelocalornonlocal.
The Bananaworld chapter was all about nonlocal correlations in Bananaworld—correlations
between Alice-events and Bob-events that can’t be explained by either a direct causal influence
betweenAliceandBob,orbyacommoncause.
This chapter is about more quantum magic associated with weird correlations, specifically so-
called ‘pseudo-telepathic’ games and the feature of entwinement associated with local contextu-
ality. I’ll first consider a tripartite correlation between Alice, Bob, and Clio as an example of a
pseudo-telepathic game. Then I’ll look at an exotic pentagram correlation between bananas that
grow in bunches of ten that combines nonlocality and contextuality. The final section is about
contextualityandtheKochen–Speckertheorem. I’llshowhowtosimulatethesecorrelationswith
entangledquantumstatesintheMoresectionattheendofthechapter.
6.1 Greenberger–Horne–Zeilinger bananas
The Einstein–Podolsky–Rosen correlation can be simulated with local classical resources, specif-
ically with shared randomness: there is a common cause explanation of the correlation. The
Popescu–Rohrlichcorrelationcan’tbeperfectlysimulatedwithlocalclassicalresources,andcan’t
be perfectly simulated with nonlocal quantum resources either. Is there a correlation that can’t
be simulated with local resources, but can be perfectly simulated with measurements on shared
entangledquantumstates?
Infact,thereisaclassofsuchcorrelationsassociatedwithpseudo-telepathicgames. GillesBras-
sard,AnneBroadbent,andAlainTappintroducedtheterm‘pseudo-telepathy’ina2003paper:2
Wesayofaquantumprotocolthatitexhibitspseudo-telepathyifitisperfectprovided
thepartiessharepriorentanglement,whereasnoperfectclassicalprotocolcanexist.
Theydiscussseveralcorrelationsthatcan’tbesimulatedwithlocalresourcesbutcanbesimulated
perfectly if the players in the simulation game are allowed to share entangled quantum states be-
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fore the start of the game. From the perspective of players limited to classical or local resources,
the ability to win such games seems inexplicable without telepathic communication between the
players—hencethename. Ofcourse,nocheatingortelepathyisinvolved,onlyquantummagic.
ThesimulationgamefortheGreenberger–Horne–Zeilingercorrelation,firstdiscussedbyDaniel
Greenberger, Michael Horne, and Anton Zeilinger,3 is the simplest pseudo-telepathic game if the
responses to the prompts are just 0 or 1 (as opposed to longer strings of 0’s and 1’s).4 In Banana-
world, Greenberger–Horne–Zeilinger bananas grow on trees with bunches of three bananas. The
tastes of a Greenberger–Horne–Zeilinger banana triple are correlated for certain combinations of
peelings (unlike a Popescu–Rohrlich correlation, which is defined for all possible peeling combi-
nations),andthecorrelationpersistsifthebananasareseparatedbyanydistance,asfollows:
ifonebananaispeeledS andtwobananasarepeeledT (soforpeelingsSTT,TST,TTS),
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anoddnumberofbananastasteintense(oneorthree)
ifallthreebananasarepeeledS (soforthepeelingSSS),anevennumberofbananastaste
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intense(zeroortwo)
if one banana is peeled S or T, there is an equal probability that the banana tastes ordinary
•
(0) or intense (1), irrespective of whether or not other bananas are peeled, or how they are
peeled(sotheno-signalingconstraintissatisfied)
ConsiderasimulationgameforthecorrelationofGreenberger–Horne–Zeilingerbananas,where
the players are restricted to local resources. The moderator sends separate prompts, S or T, ran-
domlytoAlice,Bob,andClio,whoareseparatedandcannotcommunicatewitheachotherduring
thegame,withthepromisethateachroundofpromptswillincludetwoT’sornoT’s. Thecondi-
tionsforwinningaroundare:
if the prompts are STT,TST,TTS, the responses should be 001,010,100 or 111 (an odd
•
numberof1’s)
if the prompts are SSS, the responses should be 000,011,101 or 110 (an even number of
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1’s)
theresponsesareotherwiserandom(themarginalprobabilitiesforAlice, Bob,andClioare
•
all1/2)
Call the responses of Alice, Bob, and Clio a,b,c. To win a round with local resources, Alice,
Bob,andCliomusteachseparatelyusesomeruleforchoosing0’sor1’s,basedonsharedlistsof
randombitsgeneratedbeforethestartofthesimulation,thatsatisfiestheconditions:
a +b +c = odd
S T T
a +b +c = odd
T S T
a +b +c = odd
T T S
a +b +c = even
S S S
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6.1 Greenberger–Horne–Zeilingerbananas
Figure6.1:Alice,Bob,andCliopeelingabunchofGreenberger–Horne–Zeilingerbananas.
wherethesubscriptsindicatetheprompt,S orT.
Thesumofthetermsonthelefthandsidesofthefourequationsisequaltothesumoftheterms
on the right hand sides. If you add the terms on the left hand sides, you get an even number, for
anyvalues,0or1,oftheterms,becauseeachtermoccurstwiceinthesum,asyoucancheck. But
ifyouaddthetermsontherighthandsides,yougetanoddnumber. Sothereisnoassignmentof
0’sand1’sbasedonsharedrandombitsthatsatisfiesthecorrelation,andalocalsimulationshared
randomnessisimpossible.
If Alice, Bob, and Clio can’t simulate the Greenberger–Horne–Zeilinger correlation with local
resources, neither can the bananas, which means that they can’t have pre-assigned tastes before
they are peeled, nor can their tastes for particular peelings be determined by the values of some
variable. As Einstein might put it, a banana in a Greenberger–Horne–Zeilinger triple can’t have a
‘being–thus’thatdeterminesitstasteforaparticularpeelingpriortothepeeling: thethreebananas
exhibitnonlocalcorrelationsforwhichthereisnocommoncauseexplanation.
The Greenberger–Horne–Zeilinger correlation can’t be simulated with local resources, but a
perfectsimulationofthecorrelationispossibleifAlice, Bob, andClioareallowedtosharemany
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copiesofthreequbitsinacertainentangledstatepreparedbeforethestartofthesimulation. They
each store one qubit of the three-particle state, and base their responses to the prompts on appro-
priatemeasurementsperformedonthestoredqubits. SeethesubsectionSimulatingGreenberger–
Horne–ZeilingerBananasintheMoresectionattheendofthechapterforthesimulationprotocol.
Thebottomline
A pseudo-telepathic game is a game (think of a simulation game for a correlation)
•
for which there is no winning strategy if the players are limited to classical (local)
resources,butwhichcanbewonbyplayerswithquantumresources.
Thethree-playersimulationgameforthecorrelationofGreenberger–Horne–Zeilinger
•
bananas, which grow on trees with bunches of three bananas, is the simplest pseudo-
telepathicgameforplayerswhorespondtopromptswith0or1(asopposedtolonger
stringsof0’sand1’s).
TastesandpeelingsforatripleofGreenberger–Horne–Zeilingerbananasarecorrelated
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asfollows:
(i) if one banana is peeled S and two bananas are peeled T (so for peelings
STT,TST,TTS),anoddnumberofbananastasteintense(oneorthree)
(ii) if all three bananas are peeled S (so for the peeling SSS), an even number of
bananastasteintense(zeroortwo)
(iii) if one banana is peeled S or T, there is an equal probability that the banana
tastesordinary(0)orintense(1),irrespectiveofwhetherornototherbananasare
peeled,orhowtheyarepeeled(sotheno-signalingconstraintissatisfied)
Thebananascan’thavepre-assignedtastesbeforetheyarepeelediftheyarecorrelated
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in this way, so Alice, Bob, and Clio can’t win the simulation game if they are limited
tolocalresources.
The subsection Simulating Greenberger–Horne–Zeilinger Bananas in the More sec-
•
tionattheendofthechaptershowshowAlice, Bob, andCliocanwinthesimulation
gamewithquantumresources.
6.2 The Aravind–Mermin Magic Pentagram
Padmanabhan K. Aravind has proposed a variation of a correlation by David Mermin that is par-
ticularlyinformativeinrevealingstructuralfeaturesofquantummechanics.5Aravind’smagicpen-
tagram correlation can be associated with a pseudo-telepathic game and also has a counterpart in
Bananaworld.
Aravind–Mermin bananas grow in bunches of ten bananas. The bananas fan out from a central
stem in two layers: an outer layer of five large bananas whose tops are arranged like the five
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6.2 TheAravind–MerminMagicPentagram
vertices 1,2,3,4,5 of the pentagram in Figure 6.2, and an inner layer of five smaller bananas
whose tops are at the five vertices 6,7,8,9,10 of the inner pentagon of the pentagram. Although
it’snottechnicallycorrect,I’llusetheword‘edge’forconveniencetorefertoalinesegmentofthe
pentagramjoiningtwovertices,forexamplethehorizontallinesegmentlabelede joiningvertices
1
3and2. Eachofthefiveedgesinthissensecontainstwoendnodescorrespondingtotwovertices,
andtwoinnernodescorrespondingtotheintersectionsoftheedgewithtwootheredges. Theedge
e contains the inner nodes 9 and 8. The inner node 9 is the intersection of the edge e with the
1 1
edgee ,andtheinnernode8istheintersectionoftheedgee withtheedgee . Therearetennodes
2 1 3
in total, and each banana is associated with a node. Any two edges intersect in just one node, or
onebanana. Foranytwoedges,therearefourbananasoneachedge,withonebananaincommon,
sosevendistinctbananasonanytwoedges.
Figure6.2:Aravind–Merminbananas.
The magical thing about these bananas is that if you peel the seven bananas on any two edges
(stem end or top end, it doesn’t matter), the probability of a particular banana tasting ordinary
or intense is 1/2, but an odd number of bananas on each edge always tastes ordinary, either one
bananaorthreebananas,andsincetherearejustfourbananasonanedge,anoddnumberalsotastes
intense. Once you peel the bananas on two edges, the remaining bananas turn out to be inedible.
What’s remarkable about this is that the bananas can’t have definite tastes before they are peeled,
and this is an immediate consequence of the odd number property (the ‘parity,’ to use a technical
term)ofanAravind–Merminbunch. TheillustrationinFigure6.2showsthebananasontheedges
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e ande highlightedasabouttobepeeled,withbanana9commontobothedges.
1 2
Supposethebananasdoallhavedefinitetastesbeforetheyarepeeled,orsomefeaturethatleads
toadefinitetastewhenabananaispeeled,regardlessofwhatotherbananasarepeeledonthetwo
edgestowhichabananabelongs. Allthebananaswouldhavetohavethisfeatureandnotjustthe
bananasthathappentobepeeled,becausethechoiceofwhichbananastopeelisuptotheperson
peelingthebananas. If,asusual,anordinarytasteisdesignatedbythebit0andanintensetasteby
thebit1,thesumofthesebitsforeachedgemustbeanoddnumber. Soifyouaddupallthesesums
forthefiveedges,thetotalmustalsobeanoddnumber,becausethesumoffiveoddnumbersisan
odd number. But—and here’s the punch line—each banana occurs twice in this count, because it
belongstotwoedges,soeachtaste,0or1,iscountedtwice,whichmeansthatthetotalmustbean
evennumber. Sincetheassumptionthatthebananasallhavedefinitetastesbeforetheyarepeeled
leadstoacontradiction,thebananascan’thavedefinitetastesbeforetheyarepeeled.
It follows that Alice and Bob can’t simulate the behavior of an Aravind–Mermin bunch of ba-
nanasiftheyareseparatedandrestrictedtolocalresources. Inthesimulationgame,AliceandBob
eachreceiveapentagramedgeasaprompt,e ,e ,e ,e ,ore . Towinaroundofthegame,Alice
1 2 3 4 5
andBobshouldeachrespondwithaseriesoffour0’sand1’sassignedtothebananasontheedge
correspondingtothepromptsothat
(i) theyassignthesamebittothecommonbanana(thebananaatthenodethatistheintersection
ofthetwoedges)
(ii) Alice’s four bits and Bob’s four bits should each have an odd number of 0’s (and so an odd
numberof1’s)
(iii) over many rounds of the game, the bit assigned to a banana should be 0 or 1 with equal
probability
I’llcallthefirstrulethe‘agreementrule,’thesecondrulethe‘parityrule,’becauseitsaysthatthe
parity of the sum of the bits assigned to an edge should be odd, and the third rule the ‘probability
rule,’becauseitrequiresthattheprobabilityofabananatastingordinaryorintenseis1/2.
Theycan’tpullthisoffiftheyarerestrictedtolocalresources. Theagreementrulecanonlybe
satisfied with local resources if Alice and Bob each respond to a prompt by consulting the same
sharedlistwith0’sand1’sassignedtothenodessoastosatisfytheparityconstraint,forallpossible
ways of doing this in some random order (to satisfy the equal probability rule). In other words,
AliceandBobwouldhavetosharearandomlistofpentagramswithbits, 0’sand1’s, assignedto
allthenodes,wherethesumofthefourbitsoneachedgeisodd. Butnosuchassignmentofbitsis
possible.
There’s no way Alice and Bob can win the simulation game with shared randomness, but there
isaquantumwinningstrategyifAliceandBobsharemanycopiesofacertainentangledstate, so
thegameisapseudo-telepathicgame. SeethesubsectionSimulatingAravind–MerminBananasin
theMoresectionattheendofthechapterforthesimulationprotocol.
The interesting thing about the Aravind–Mermin correlation is that it combines contextuality
andnonlocality,thestructuralfeaturesof‘entwinement’ofquantumobservablesthatImentioned
atthebeginningofthechapter.
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6.2 TheAravind–MerminMagicPentagram
Eachnodeofthemagicpentagramistheintersectionoftwoedges. Inaquantumsimulationof
Aravind–Merminbananas,eachnodeisassociatedwithanobservableofathree-qubitsystem,and
each edge is associated with a context defined by a set of compatible or commuting observables
that can all be measured together. The observables on any two different edges don’t all commute
witheachother,soanobservableassociatedwithanodebelongstotwoincompatiblecontexts. The
agreement rule forces the taste of a banana to be the same for the two different edges to which it
belongs. It’sthenimpossibleforthetastestosatisfytheparityruleifallthebananashavedefinite
tastesbeforetheyarepeeled. Sothereisno‘noncontextual’assignmentoftastestothebananasin
anAravind–Merminbunch—noassignmentinwhichabananagetsadefinitetastethatisthesame
forthetwocontextsdefinedbythetwoedgestowhichitbelongs.
Sinceit’spossibleforAliceandBobtoperfectlysimulatetheAravind–Mermincorrelationwith
quantumresources,quantummechanicsiscontextual—thetheorydoessomethingthatnononcon-
textualtheorycando. Thecorrelationdefinedbyaparticularquantumstateforthetenobservables
associatedwiththetennodesofamagicpentagramcan’tarisefrommeasurementsofobservables
withdefinitenoncontextualvaluesbeforetheyaremeasured,whichistosaythatnononcontextual
hiddenvariabletheorycanproducethecorrelation.
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Thebottomline
Aravind–Merminbananasgrowinbunchesoftenbananas,fivelargebananasandfive
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smallbananas,radiatingoutwardsfromacentralstem. Thetopsofthebananasdefine
theverticesofanouterpentagram,andthetopsofthesmallbananasdefinethevertices
ofaninterpentagon(seeFigure6.2).
I use the word ‘edge’ for convenience to refer to a line segment of the pentagram
•
joiningtwovertices,forexamplethehorizontallinesegmentlabelede joiningvertices
1
3and2. Ifyoupeelthesevenbananasonanytwoedges(stemendortopend,itdoesn’t
matter), the probability of a particular banana tasting ordinary or intense is 1/2, but
an odd number of bananas on each edge always tastes ordinary, either one banana or
threebananas(andsincetherearejustfourbananasonanedge,anoddnumbertastes
intense). Once you peel the bananas on two edges, the remaining bananas turn out to
beinedible.
Inthesimulationgame,AliceandBobareeachreceiveapentagramedgeasaprompt
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and respond with a series of four0’s and 1’s assigned to the bananas on the edge in a
waythatshouldagreewiththecorrelation. It’sanimmediateconsequenceoftheodd
number property that the bananas can’t have definite tastes before they are peeled, so
AliceandBobcan’twinthesimulationgamewithlocalresources.
The interesting thing about the Aravind–Mermin correlation is that it combines con-
•
textualityandnonlocality. Eachnodeofthepentagramistheintersectionoftwoedges.
InaquantumsimulationofAravind–Merminbananas,eachnodeisassociatedwithan
observableofathree-qubitsystem,andeachedgeisassociatedwithacontextdefined
byasetofcompatibleorcommutingobservablesthatcanallbemeasuredtogether.
The impossibility of winning the simulation game with local resources means that
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there is no ‘noncontextual’ assignment of tastes to the bananas—no assignment in
which a banana gets a definite taste that is the same for the two contexts defined by
the two edges to which it belongs. Since it’s possible for Alice and Bob to perfectly
simulatetheAravind–Mermincorrelationwithquantumresources(seethesubsection
Simulating Aravind–Mermin Bananas in the More section at the end of the chapter),
quantum mechanics is contextual: the theory does something that no noncontextual
theorycando.
6.3 The Kochen–Specker Theorem and Klyachko Bananas
Intheprevioussection,Ishowedthatthere’snononcontextualassignmentoftastestothebananas
inanAravind–Merminbunch,butit’spossibletoperfectlysimulatetheAravind–Mermincorrela-
tionwithquantumresources. SimonKochenandErnstPaulSpecker,6 andseparatelyJohnBell,7
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6.3 TheKochen–SpeckerTheoremandKlyachkoBananas
proved a general version of this ‘no go’ result, now known as the Bell–Kochen–Specker theorem,
orjusttheKochen–Speckertheorem. Thetheoremsaysthattheobservablesofaquantumsystem
can’t have definite, noncontextual, pre-existing values before they are measured, not even for cer-
tain finite sets of observables, and it’s true for a qutrit, or any quantum system with observables
that have three or more possible values. The proof is state-independent: it doesn’t depend on any
particular quantum state. The way certain finite sets of observables are entwined is inconsistent
with the observables all having definite values noncontextually. So the question of whether you
couldrecovertheprobabilitiesdefinedbythequantumstatesofasystemiftheobservablesallhave
definitenoncontextualvaluesdoesn’tarise,becausenosuchassignmentofvaluesisevenpossible.
The theorem doesn’t apply to a qubit because there’s no entwinement for a qubit—you need
observables with at least three possible values. Think of a photon in a state of linear polarization,
say vertical polarization in the z direction, denoted by the state 1 . There’s a unique state of
| i
orthogonal polarization, horizontal polarization in the z direction, denoted by the state 0 . The
| i
twopolarizationstates 0 and 1 defineacoordinatesystemorbasisinthetwo-dimensionalstate
| i | i
space associated with the observable ‘polarization in the z direction.’ For a qutrit, the state space
is three-dimensional. Three orthogonal states in the qutrit state space associated with the three
possiblevaluesofaqutritobservabledefineabasis. Inthiscase, ifyoupickabasisstate, suchas
thestate 1 ontherightinFigure6.3,thereisn’tauniquepairoforthogonalstatesthatcompletes
| i
the basis. Rather, there are an infinite number of orthogonal pairs—Figure 6.3 shows two. Any
orthogonalpairwilldo,andeachchoiceforthesestatesisassociatedwithadifferentthree-valued
observablethatdoesn’tcommutewithanobservableassociatedwithanyotherchoice. Forthequtrit
in Figure 6.3, the basis 1 , 2 , 3 is associated with the three possible values of an observable
| i | i | i
O, and the basis 1 , 2 , 3 is associated with the three possible values of an observable O .
0 0 0
| i | i | i
The observables O and O are entwined, in the sense that the state 1 belongs to both bases or
0
| i
measurementcontexts.
1 1
| i | i
qubit qutrit
|30i
0 3
| i | i
|2i |20i
Figure6.3:Aqubitcontext,andtwoqutritcontextssharingthestatevector 1 .
| i
Suppose you have a filter that transmits a qutrit in the state 1 and blocks a qutrit in a state
| i
orthogonal to 1 . The filter is a measuring instrument for a ‘yes–no’ qutrit observable P that has
| i
the value ‘yes’ or 1 if the filter transmits the qutrit, and ‘no’ or 0 if the filter blocks the qutrit.
The question of contextuality arises for P with respect to the observables O and O . Although
0
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O and O are associated with different bases and don’t commute, P commutes with both O and
0
O . (Why? Here’s the technical explanation. The operator P is just the projection operator onto
0
the state 1 . The operator O can be represented as a sum of projection operators onto the state
| i
1 and the orthogonal states 2 , 3 , with the eigenvalues or possible values of O as coefficients.
| i | i | i
Similarly,O canberepresentedasasumofprojectionoperatorsonto 1 andtheorthogonalstates
0
| i
2 , 3 ,withtheeigenvaluesofO ascoefficients. IshowthisforthePaulioperatorsinthesection
0 0 0
| i | i
The Pauli Operators in the Supplement, Some Mathematical Machinery, at the end of the book.
Since P commutes with itself and with any orthogonal projection operators, P commutes with
both O and O .) So the observable P, which can be measured with a P-filter with two possible
0
outcomes, can also be measured via a measurement of O with three possible outcomes, 1,2,3, or
via a measurement of O with three possible outcomes, 1,2,3. If you measure O or O and get
0 0 0 0
theoutcome1,thequtritendsupinthestate 1 ,whichisalsothestateofthequtritafteritpasses
| i
the P-filter. So P can be assigned a value ‘in the context of O’ via an O-measurement, or ‘in the
context of O ’ via a measurement of O . In the 2-dimensional state space of a qubit, there are no
0 0
observablesthatareentwinedlikeO,O andP,sothere’snoissueaboutcontextuality.
0
The Born rule for quantum probabilities guarantees that the outcome 1 for O, or for O , or for
0
P, has the same probability in any quantum state , because the probability is the square of
| i
the absolute value of the length of the projection of onto 1 . So quantum probabilities are
| i | i
noncontextual. (SeethesectionTheBornRuleintheSupplement,SomeMathematicalMachinery,
at the end of the book.) If quantum observables all had definite noncontextual pre-measurement
values and measurement merely revealed these values, that would explain the noncontextuality
of quantum probabilities. But since different contexts contain noncommuting observables, they
can’t be measured simultaneously. So there’s no reason to think that an observable like P, which
belongstodifferentcontextsdefinedbynoncommutingobservablesOandO ,wouldbemeasured
0
ashavingthesamevalue,whetherP ismeasuredtogetherwithOorwithO . Puttingitdifferently,
0
ifyoumeasureP viaameasurementofOandgettheoutcome1,there’snojustificationforsaying,
counterfactually, that you would have found the corresponding outcome if you had measured O
0
instead. Forallyouknow,youmighthavefoundtheoutcome2 or3 forO ,correspondingtothe
0 0 0
value0forP. There’snowaytocheck,becauseyoucan’tmeasureO andO atthesametime.
0
SodoesanobservablelikeP havethesamevalueinthecontextofO asinthecontextofO ,or
0
arethevaluescontext-dependentandpossiblydifferentinthetwocontexts? TheKochen–Specker
theorem answers the question: a noncontextual assignment of values to the observables of any
quantum system at least as complex as a qutrit is impossible. The proof of the Kochen–Specker
theoremforaqutritisequivalenttoshowingthatthere’snowaytoassign0’sand1’stoaparticular
finite set of points on the surface of a unit sphere, in such a way that every context defined by
an orthogonal triple of points gets one 1 and two 0’s, where orthogonality is defined for the lines
connectingthecenterofthespheretothepointsonthesurface.
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