Quantum teleportation with one classical bit | Scientific Reports

Quantum teleportation is a strikingly curious quantum phenomenon with myriads of applications ranging from secure quantum communication to ... Skiptomaincontent Thankyouforvisitingnature.com.YouareusingabrowserversionwithlimitedsupportforCSS.Toobtain thebestexperience,werecommendyouuseamoreuptodatebrowser(orturnoffcompatibilitymodein InternetExplorer).Inthemeantime,toensurecontinuedsupport,wearedisplayingthesitewithoutstyles andJavaScript. Advertisement nature scientificreports articles article Quantumteleportationwithoneclassicalbit DownloadPDF DownloadPDF Subjects ComputerscienceInformationtechnologyQuantumphysics AbstractQuantumteleportationallowsonetotransmitanarbitraryqubitfrompointAtopointBusingapairof(pre-shared)entangledqubitsandclassicalbitsofinformation.Theconventionalprotocolforteleportationusestwobitsofclassicalinformationandassumesthatthesenderhasaccesstoonlyonecopyofthearbitraryqubittobe sent.Here,weaskwhetherwecandobetterthantwobitsofclassicalinformationifthesenderhasaccesstomultiplecopiesofthequbittobeteleported.Weplacenorestrictionsonthequbitstates.Consequently,weproposeamodifiedquantumteleportationprotocolthatallowsAlicetoresetthestateoftheentangledpairtoitsinitialstateusingonlylocaloperations.Asaresult,theproposedteleportationprotocolrequiresthetransmissionofonlyoneclassicalbitwithaprobabilitygreaterthanone-half.Thishasimplicationsforefficientquantumcommunicationsandthe securityofquantumcryptographicprotocolsbasedonquantumentanglement. IntroductionQuantumteleportation1,2isastrikinglycuriousquantumphenomenonwithmyriadsofapplicationsrangingfromsecurequantumcommunicationtodistributedquantumcomputing3,4,5,6,7.FirstpresentedbyBennettetal.1,quantumteleportationallowsonetorecreateanarbitraryqubitfromjusttwobitsofclassicalinformation.Thepreconditionisthatthetwocommunicatingpartiesshareanentangledpairofqubits.Entanglement,essentially,transmitstherawvaluesofamplitudespresentinthearbitraryqubitandtheclassicalbitsprovideafinal“correction”tothesevalues.Quantumteleportationhasproventobeaninvaluabletoolinquantuminformationscience8,9,10.Quantumteleportationenablesthedevelopmentofquantumrepeaters3,11,apivotaltechnologyfortheestablishmentofquantumcommunicationnetworks12,andingeneraltheframeworkofquantumInternet.Quantumteleportationhasbeenimplementedinlabsettingusinganumberofdifferentresources13,14.Lastly,anumberofcryptographicprotocolsusesharedentangledqubitsandoperationsequivalenttoteleportationtoproducerandomnumbersandencryptionkeys15.Itisknownthat,givenapre-sharedBellpair,theteleportationofanunknownqubit,requiresthetransmissionoftwoclassicalbits1.Moregenerally,theteleportationofanN-staterequires2\(log_2N\)classicalbits16.This,however,assumesthatthesenderhasaccesstoonlyonecopyofthearbitraryunknownqubitthatsheisteleporting.Kakinvestigatedtheminimumnumberofclassicalbitsrequiredforteleportation17.Inturn,heproposedthreevariationsonBennett’sprotocolthatrequiredfewerthantwoqubits.Twoofthesevariations,however,useanon-traditionalsetupwheretheentangledqubitsarenotpre-sharedbetweenthecommunicatingpartiesbuttransmittedalongwiththeclassicalbitsofinformation.Thischangeinthesetupcanbelookeduponasanencryptionsystemwheretheclassicalbitofinformationactsastheencryptionkeyforthequbitandmultipathroutingcanbeusedfortheirtransmission.Fromtheviewpointofteleportation,nevertheless,ifthequbitsaretobetransmittedalongwithclassicalbits,thenonecouldtransmitthearbitraryqubitdirectlyatthattimewithouttheneedtoresortingtoteleportation.However,itdoespointouttothepossibilityofsome“non-standard”settingofquantumteleportationhavinglowerclassicalcostofcommunication.Onesuch“non-standard”settingwasinvestigatedbyPati18wheretheteleportedqubitarosefromaspecificportionoftheBlochsphereleadingtoaone-bitclassicalcommunicationcost.ThiswasfurtherinvestigatedbyLoasremotestatepreparation19.Inthispaper,wetooconsiderasomewhatnon-standardsettingbutintermsofresources.Likethetraditionalteleportationprotocol,weassumethatthecommunicatingpartiespre-shareentangledqubits.However,weconsiderthecasewhenthesenderhasmorethanonecopyofan(unknown)arbitraryqubit\(\left|{\phi}\right\rangle=a\left|{0}\right\rangle+b\left|{1}\right\rangle\),\(|a|^2+|b|^2=1\).Wedonotplaceanyotherrestrictionsonthequbits.Suchapossibilityhasnotbeeninvestigatedbefore.Ourresultshowsthatundersuchassetting,wecandobetterthantwoclassicalbitsofcommunication.Asendermayhavemultiplecopiesofthequbittobeteleportedinmanypracticalsituationswheretheunknownqubitiseithertheresultofaperiodicoron-demandnaturalprocessoroutputofaquantumalgorithm.Onemayneedtosendthis(unmeasured)outputtothereceiverforfurtherprocessing.Weshowthattheteleportationofsuchaqubitrequiresthetransmissionofonlyoneclassicalbitwithaprobabilitygreaterthanone-half.Inordertodoso,weproposearesetprocedurethatallowsfortherecreationoftheoriginalsharedBellstateusingonlylocaloperationsatthesender’sendandnoconsumptionofresourcesatreceiver’send.Intheproposedresetprocedure,aswithconventionalteleportationprotocol,webeginwithmaximallyentangledqubits.Atthefirststageoftheprotocol,ifthismaximallyentangledqubitentersintoadesiredstatewemoveontothesecondstageoftheprotocol.If,ontheotherhand,theentangledpartendsupinanundesiredstateweresetitbacktotheoriginalstateusingonlylocaloperationsatthesender’send.Thecostofdoingsoisonecopyofthearbitraryqubit,\(\left|{\phi}\right\rangle\).Furthermore,iftheamplitudesaandbareknownand/orthequbit,\(\left|{\phi}\right\rangle\),generatedondemand,thentheresetoperationmaybeattemptedmanytimesover.Theresultpresentedinthispaper,webelieve,representsafundamentaladvanceintheclassicalburdenofteleportation,sinceitwasproposedbyBennettetal.1,becausewedonotrestrictthestatesofthequbitandallowthepartiestopre-shareentangledpairsasinthetraditionalsetting.Thepresentedresultalsohaswidespreadconsequencesforquantumcryptographicprotocolsthatuseentangledpairsandassumeasymmetricpowerbetweencommunicatingparties.ResultItisusefultothinkoftheconventionalteleportationprotocolasatwostagesystem19.WestartwithasharedBellpair\(\frac{\left|{00}\right\rangle+\left|{11}\right\rangle}{\sqrt{2}}\)betweensender,Alice,andreceiver,Bob.Thefirststageoftheteleportationprotocolintroducestheamplitudesaandbintotheentangledpair.However,thelocationsoftheamplitudesmaybeflippedandiscorrectedwiththeapplicationofPauliXgatebyboththesenderandthereceiver.Thisrequiresthetransmissionof1-bitofclassicalinformationtoBob.Inthesecondstageoftheprotocol,AliceterminatestheentanglementbymeasuringherqubitoftheBellpairintheHadamardbasis.ThisintroducesanotherundesirableerrorintothestateofthequbitthatBobholdscorrespondingtothePauliZgate.ThisiscorrectedbythetransmissionofanotherclassicalbitofinformationtoBob.Tobemoreprecise,inBennett’sprotocol1attheendofthefirststagethetwopartiessharethestate\(a\left|{00}\right\rangle+b\left|{11}\right\rangle\)or\(b\left|{00}\right\rangle+a\left|{11}\right\rangle\).Thefirstofthesestatesisdesirableandthesecondoneisconvertedtothefirstonewiththeapplicationofabi-localunitarytransformation\(\left|{0}\right\rangle\rightarrow\left|{1}\right\rangle\).ThisrequiresonebittobesenttoBob.Inthesecondstageoftheprotocol,Bobendsupwith\(a\left|{0}\right\rangle+b\left|{1}\right\rangle\)or\(a\left|{0}\right\rangle-b\left|{1}\right\rangle\).ThecorrectionofthelaststaterequirestheotherbitofinformationtobesenttoBob.IntheendBobisleftwith\(a\left|{0}\right\rangle+b\left|{1}\right\rangle\),whichcorrespondstotheunknownstate\(\left|{\phi}\right\rangle\)thatAlicewantedtoteleport.Inthispaper,weaskasimplequestion:ifAliceendsupintheundesiredstateattheendofthefirststage,i.e.\(b\left|{00}\right\rangle+a\left|{11}\right\rangle\),cansheusingonlylocaltransformationsresetthestateofthesharedentanglementtotheoriginalstate\(\frac{\left|{00}\right\rangle+\left|{11}\right\rangle}{\sqrt{2}}\)?Ifso,shecanreattempttheteleportationprotocol.ResettingtheentangledstateInordertoresettheentangledstate,Aliceintroducesanancillaryqubit\(\left|{\theta}\right\rangle=c\left|{0}\right\rangle+d\left|{1}\right\rangle\)intothesystem.Itwill becomeapparentthroughthefollowingdiscussionthat\(\left|{\theta}\right\rangle=\left|{\phi}\right\rangle\).Theresultingsystemstateisgivenby\(\left|{\psi}\right\rangle\),$$\begin{aligned}\begin{aligned}\left|{\psi}\right\rangle&=\left|{\theta}\right\rangle\left(a\left|{1_A1_B}\right\rangle+b\left|{0_A0_B}\right\rangle\right)\\&=\left(c\left|{0_l}\right\rangle+d\left|{1_l}\right\rangle\right)\left(a\left|{1_A1_B}\right\rangle+b\left|{0_A0_B}\right\rangle\right)\\&=ca\left|{0_l1_A1_B}\right\rangle+cb\left|{0_l0_A0_B}\right\rangle+da\left|{1_l1_A1_B}\right\rangle+db\left|{1_l0_A0_B}\right\rangle\end{aligned}\end{aligned}$$ (1) QubitswithsubscriptslandAarewithAliceandthequbitdenotedbysubscriptBiswithBob.Now,AliceappliesCNOTwiththeancillaryqubit(thefirstqubit)asthecontrolqubitandsecondqubitasthetargetqubit.Thestatethenbecomes,$$\begin{aligned}\begin{aligned}\left|{\psi}\right\rangle&=ca\left|{0_l1_A1_B}\right\rangle+cb\left|{0_l0_A0_B}\right\rangle+da\left|{1_l0_A1_B}\right\rangle+db\left|{1_l1_A0_B}\right\rangle\\&=\left|{1_A}\right\rangle\left(ca\left|{0_l1_B}\right\rangle+db\left|{1_l0_B}\right\rangle\right)+\left|{0_A}\right\rangle\left(cb\left|{0_l0_B}\right\rangle+da\left|{1_l1_B}\right\rangle\right)\end{aligned}\end{aligned}$$ (2) InEq. (2)above,wehaveremovedthesecondqubitasthecommonqubitforsimplificationoftheexpression.Weseethat\(\left|{0_A}\right\rangle(cb\left|{0_l0_B}\right\rangle+da\left|{1_l1_B}\right\rangle)\)isclosesttothedesiredstate\(\frac{\left|{00}\right\rangle+\left|{11}\right\rangle}{\sqrt{2}}\).Ifweset\(c=a\)and\(d=b\)thenweget,$$\begin{aligned}\left|{\psi}\right\rangle=\left|{1_A}\right\rangle\left(aa\left|{0_l1_B}\right\rangle+bb\left|{1_l0_B}\right\rangle\right)+\left|{0_A}\right\rangle\left(ab\left|{0_l0_B}\right\rangle+ba\left|{1_l1_B}\right\rangle\right)\end{aligned}$$ (3) Alicenowmeasuresthesecondqubit.Shewillsee\(\left|{0}\right\rangle\)withprobability\(2|ab|^2\)andthesharedentangledqubitsendupinstate\(\frac{\left|{00}\right\rangle+\left|{11}\right\rangle}{\sqrt{2}}\).ThisconstitutesasuccessfulresetbyAlice.Notethat,setting\(c=a\)and\(d=b\)meansthatAliceissimplyusingacopyoftheoriginalqubit\(\left|{\phi}\right\rangle\)fortheresetoperationanddoesnotneedtocreateanynewspecialancillaryqubits.OncearesettotheoriginalBellstateisachieved,Alicereinitiatestheteleportationprotocolusingafreshcopyof\(\left|{\phi}\right\rangle\).Ontheotherhand,Alice’smeasurementresultsin\(\left|{1}\right\rangle\)withprobability\(|a^2|^2+|b^2|^2\).Thistakesthesystemto thestate\(\frac{a^2\left|{01}\right\rangle+b^2\left|{10}\right\rangle}{\sqrt{\left|{a^2}\right|^2+\left|{b^2}\right|^2}}\).AlicecanapplytheXgatetoherqubitandreattempttheresetoperationwith\(\frac{a^2}{\sqrt{\left|{a^2}\right|^2+\left|{b^2}\right|^2}}\)and\(\frac{b^2}{\sqrt{\left|{a^2}\right|^2+\left|{b^2}\right|^2}}\)asthenewaandbvalues,respectively.Atthispoint,Alicehasconsumedtwocopiesof\(\left|{\phi}\right\rangle\).Oneduringstageoneoftheteleportationprotocolandsecondastheancillaryqubitintroducedintothesystemduringtheresetoperation.Figure 1showsthebranchesthatateleportationattemptwiththeproposedresetprocedurecanfollow.Figure1DecisiontreeatAlice’sendforteleportationwiththeproposedresetoperation.Probabilitiesareshownalongthearmsbetweenthenodes.StatesintheredboxesresidewithBob.FullsizeimageDiscussionTheprobabilityofsuccessoftheresetattemptsvarywiththevaluesofamplitudesaandb,inthestate\(\left|{\phi}\right\rangle\),andisgivenby\(2|ab|^2\)forthefirstresetattempt.Since,\(|ab|^2=|a|^2|b|^2=|a|^2(1-|a|^2)=|a|^2-|a|^4\)and\(0\le|a|^2\le1\),weget\(|ab|^2\le\frac{1}{4}\)and\(2|ab|^2\le\frac{1}{2}\).Forexample,when\(a=\frac{i}{\sqrt{2}}\)and\(b=\frac{1}{2}+\frac{i}{2}\),theprobabilitythatthefirstresetattemptwillsucceedis\(\frac{1}{2}\).Afterasuccessfulreset,Alicemakesasecondattempttoteleportthequbit.Theprobabilitythatentangledqubitswillcollapseintothedesirablestate(\(a\left|{00}\right\rangle+b\left|{11}\right\rangle\))isagain\(\frac{1}{2}\).Therefore,withprobability\(\frac{1}{2}+\frac{1}{4}\)Alicewouldtransmitjustonebitforteleportationbutmusthaveatleastthreecopiesof\(\left|{\phi}\right\rangle\)atherdisposal.Remarkably,thequbit\(\left|{\phi}\right\rangle\)remainsunknownandarbitrary.Consequently,ingeneral,consideringonlythefirstresetattemptandanunknownqubit\(\left|{\phi}\right\rangle\)theprobabilityofendingupinthedesiredstate(\(a\left|{00}\right\rangle+b\left|{11}\right\rangle\)),thatisasuccessfulteleportationwithonlyone-classicalbit,is\(\frac{1}{2}+\frac{1}{2}\cdot2|ab|^2=\frac{1}{2}+|ab|^2\).Note,thisprobabilityisalwayssetto\(\frac{1}{2}\)inBennettetal.’squantumteleportation1protocolwhichalwaysrequirestwobitsofinformationtobetransmitted.Alice,however,doesrequirethevaluesofaandbfortheresetattemptsbeyondthefirstone;inthecaseofaresetfailure.Thislattercaseisusefulwhentheprecisionrequiredforclassicalrepresentationoftheaandbvaluesexceedstheavailablebandwidthorresourcesavailableontheclassicalchannel.Inotherwords,ifthevaluesofaandbareknown,Alicecanpursuetheresetattemptsuntilitsucceeds.Notethatoncearesetattemptsucceeds,theprobabilitythatwewillendupinthedesiredentangledstateisagain\(\frac{1}{2}\).Therefore,theexpectednumberofattempts,givenasuccessfulreset,is2.Theresetprocedurepresentedabove,whensuccessful,reducestheteleportationprotocoltoonestageprotocolunderthestandardsettingofpre-sharedBellstates.Furthermore,Aliceknowsateachstagewhethertheresethassucceededornot.Intheworstcase,ifthevaluesofaandbarenotknownandtheresetfails,shecanabandonthatparticularpairofentangledqubitsanduseanother.Thecomputationalburdenofteleportationwithresetcanthenbestatedas,$$\begin{aligned}H(T)=\left[\left(\frac{1}{2}+|ab|^2\right)(1)+\left(\frac{1}{2}-|ab|^2\right)(2)\right]\text{bits}\end{aligned}$$ (4) Itiseasytoseethat\(1.25\leH(T)\le1.5\)dependingonthevaluesofaandb.Forcomparison,oneoftheprotocolsbyKak17reducesthecomputationalburdento1.5bitswhentheBellstatesarepre-shared.Thecostofreducingthecomputationalburdento1.25bits,therefore,istheexpenditureofextracopiesoftheunknownqubit\(\left|{\phi}\right\rangle\).Inotherwords,theprobability\(\frac{1}{2}+|ab|^2\)onlyrepresentstheaveragecase.GiventhattheproposedprotocolisprobabilisticinnatureandifAlicehasseveralqubitstoteleport,shemayendupwithasituationwhereallherresetattemptsmaysucceedresultingina1bitperqubitforteleportationevenforacompletelyarbitraryunknownqubit!OnemayarguethateveninBennett’soriginalteleportationprotocol1,AliceandBobmayagreeonaschemesuchasthefollows:ifstageoneoftheprotocolsucceedsincreating\(a\left|{00}\right\rangle+b\left|{11}\right\rangle\)thentheywillproceedwithstagetwoandifstageoneresultsin\(a\left|{11}\right\rangle+b\left|{00}\right\rangle\),theywillabandontheentangledpair.Suchamodificationwouldalsobeprobabilisticandresultin1bitperqubitforteleportation.However,wepointoutthattheprobabilityof1bitperqubitinsuchamodifiedBennett’sprotocolwouldbestuckat\(\frac{1}{2}\).Whereas,ourresetprocedureallowsthisprobabilitytobe\(\frac{1}{2}+|ab|^2\)whichisgreaterthan\(\frac{1}{2}\)forallpracticalpurposes.Theprobabilityofsuccessfortheresetoperationateveryattemptchangeswiththeamplitudesaandbandisgivenbyarecursiverelationship.If\(R_0,R_1,R_2,\ldots\)representsuccessiveresetattemptsthentheprobabilityofsuccessofeachresetisgivenby,$$\begin{aligned}\begin{aligned}P(R_0)&=2|ab|^2\\P(R_1)&=2|a_1b_1|^2\text{where}a_1=\frac{a^2}{\sqrt{\left|{a^2}\right|^2+\left|{b^2}\right|^2}}\text{and}b_1=\frac{b^2}{\sqrt{\left|{a^2}\right|^2+\left|{b^2}\right|^2}}\\P(R_2)&=2|a_2b_2|^2\text{where}a_2=\frac{a_1^2}{\sqrt{\left|{{a_1^2}}\right|^2+\left|{{b_1^2}}\right|^2}}\text{and}b_2=\frac{b_1^2}{\sqrt{\left|{{a_1^2}}\right|^2+\left|{{b_1^2}}\right|^2}}\\&\quad\text{and}\text{so}\text{on.}\end{aligned}\end{aligned}$$ (5) Therefore,theprobabilitythatAlicewouldneedthreeresetattemptsisgivenby\((1-P(R_0))(1-P(R_1))P(R_2)\).Forastatesuchas\(\frac{\left|{0}\right\rangle+\left|{1}\right\rangle}{\sqrt{2}}\)thiswouldbe\((1-\frac{1}{2})(1-\frac{1}{2})(\frac{1}{2})=\frac{1}{8}\).AnimportantimplicationofEq. (4)isthatwhen\(|ab|^2>0\)thereisanon-zeroprobabilityofasuccessfulreset.Theequality,\(|ab|^2=0\),holdsonlywhenoneoftheamplitudesaorb=0,acornercase.Asaresult,theaboveprotocolbreaksthesymmetrybetweenthefourcasesoftheconventionalteleportationprotocol1.Moreprecisely,intheconventionalteleportationprotocol,thereisanequalprobabilityforBobtoendupinanyofthefourstates:\(\left|{\phi_0}\right\rangle=a\left|{0}\right\rangle+b\left|{1}\right\rangle\),\(\left|{\phi_1}\right\rangle=a\left|{0}\right\rangle-b\left|{1}\right\rangle\),\(\left|{\phi_2}\right\rangle=b\left|{0}\right\rangle+a\left|{1}\right\rangle\),and\(\left|{\phi_3}\right\rangle=b\left|{0}\right\rangle-a\left|{1}\right\rangle\).Consequently,twoclassicalbitswereneededforteleportation.However,withtheproposedresetoperationtherelationshipbetweentheprobabilitiesofthestatesthatBobseesisgivenby,$$\begin{aligned}P\left(\left|{\phi_0}\right\rangle\right)+P\left(\left|{\phi_1}\right\rangle\right)>P\left(\left|{\phi_2}\right\rangle\right)+P\left(\left|{\phi_3}\right\rangle\right)\end{aligned}$$ (6) Foramomentonlyconsiderthefirststageoftheteleportationprotocolandtheresetoperation.AssumethatAlicehasnottransmittedanyinformationtoBob.Weseethat,theresetoperationalsoimpliesthatwhen\(a,b\ne0\),Alicecanunilaterallyforcethesharedentanglement,\(\frac{\left|{00}\right\rangle+\left|{11}\right\rangle}{\sqrt{2}}\)tobeconvertedto\(a\left|{00}\right\rangle+b\left|{11}\right\rangle\)withaprobabilityhigherthan\(\frac{1}{2}\)andnocommunicationwithBob.Ifthequbit\(\left|{\phi}\right\rangle\)isdeliberatelyconstructedbyAliceandtheinformationAlicewantstotransmitisencodedinthevaluesofaandbratherthanthephasedifferencebetweenthem,thenAlicehassuccessfullyinducedabiasinBob’squbit.Theabilityofonepartyto,unilaterally,induceabiasinasharedentangledpairgivesapartywantingtocheat,inacryptographicprotocol,anadvantage.Forexample,severalquantumkeyagreementprotocolshavebeenproposedinliterature20,21,22,23,24,25.Theseprotocolsaredesignedsuchthatboththecommunicatingpartiesmakeequalcontributiontothefinalagreedkey.ThisisincontrasttotraditionalquantumkeydistributionprotocolssuchasBB84whereonepartydeterminestheentirekeyandsecurelydistributesittotheotherparty.Ifthequantumkeyagreementprotocolisbasedonentanglementthenoneofthecheatingpartiescaninduceabiasinthefinalkeystreambyintroducinganappropriate\(\left|{\phi}\right\rangle\)inthesystem.ThesimplestexampleofaweakquantumkeyagreementprotocolistheE91protocolwherebothpartiesmakerandommeasurementsonmaximallyentangledpairsofqubits.IfthesemaximallyentangledpairsaredistributedbyacentralauthoritythenAlice(cheatingparty)canintroduce\(\left|{\phi}\right\rangle=a\left|{0}\right\rangle+b\left|{1}\right\rangle\)suchthat\(|a|^2\neq|b|^2\)resultinginabiasedmeasurementresult,atBob’send,ofherownchoosing.Therefore,theasymmetrycausedbytheresetoperationgivesAlicemorepowerthanBobandisdevastatingforcryptographicprotocolsthatuseentanglementandrelyonthepowersofthecommunicatingpartiesbeingsymmetricforsecurity.Inthispaper,wehaveonlyconsideredtheidealcasewherethesharedentangledpairismaximallyentangled.Investigationintohowtheresetprocedurewouldproceedinthecaseofnon-maximallyentangledstates,mixedstates,non-idealmeasurementsandchannelerrorsareleftasfuturework.Entanglementdistillation26couldbeusedtomitigatetheeffectsofthesenon-idealcases. 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