Error Correction of Quantum Reference Frame Information

Transcription

1 Error Correction of Quantum Reference Frame Information Patrick Hayen, 1 Sepehr Nezami, 1 Sanu Popescu, an Grant Salton 1 1 Stanfor Institute for Theoretical Physics, Stanfor University, Stanfor, California 94305, USA H H Wills Physics Laboratory, University of Bristol, Tynall Avenue, Bristol, BS8 1TL, Unite Kingom In each of the examples escribe above, Bob s lack of knowlege about Alice s reference frame or time stanar is mathematically moele by the action of an unknown element of some group on Alice s state For irectional reference frames, Alice an Bob are relate by an unknown rotation ie, an element of SO3, whereas in the example of clock synchronization Alice an Bob s clocks are relate by an unknown time translation which can be thought of as an element of U1 In the spirit of [14] which generalizes reference frame information to general asymmetry information in the context of resource theory [14 19], we stuy error correction of physical information that transforms uner an arbitrary group G An important reuction following from the analysis of [14] is that the existence of encoing schemes for this type of informaarxiv: v1 [quant-ph] 13 Sep 7 The existence of quantum error correcting coes is one of the most counterintuitive an potentially technologically important iscoveries of quantum information theory However, stanar error correction refers to abstract quantum information, ie, information that is inepenent of the physical incarnation of the systems use for storing the information There are, however, other forms of information that are physical one of the most ubiquitous being reference frame information Here we analyze the problem of error correcting physical information The basic question we seek to answer is whether or not such error correction is possible an, if so, what limitations govern the process The main challenge is that the systems use for transmitting physical information, in aition to any actions applie to them, must necessarily obey these limitations Encoing an ecoing operations that obey a restrictive set of limitations nee not exist a priori We focus on the case of erasure errors, an we first show that the problem is equivalent to quantum error correction using group-covariant encoings We prove a no-go theorem showing that that no finite imensional, group-covariant quantum coes exist for Lie groups with an infinitesimal generator eg, U1, SU, an SO3 We then explain how one can circumvent this no-go theorem using infinite imensional coes, an we give an explicit example of a covariant quantum error correcting coe using continuous variables for the group U1 Finally, we emonstrate that all finite groups have finite imensional coes, giving both an explicit construction an a ranomize approximate construction with exponentially better parameters Introuction One of Shannon s original insights in the formulation of information theory was to focus on the transmission of sequences of symbols, such as strings of 0 s an 1 s, without regar to the semantic content of the message This approach makes it possible to encoe an enormous variety of messages, from phone numbers to photos, as long as the original information can be faithfully represente in terms of a sequence of symbols The same situation exists in the quantum worl: quantum information theorists are primarily concerne with information that can be store in a system of qubits or larger quantum systems, inepenent of the type of information Here we stuy a situation in which the information is physical an cannot be represente simply as abstract qubits Consier the following purely classical scenario [1] Alice wishes to transmit some irectional information to Bob, eg, the axis of rotation of a gyroscope inicate by the vector n, so that Bob can prepare a gyroscope rotating aroun the same axis as Alice s If Alice an Bob share a reference frame, Alice can measure ifferent components of n an escribe the result in wors to Bob, who then prepares his own gyroscope to match However, if Alice an Bob o not share a reference frame, ie, if Alice an Bob o not know the relative alignment of their coorinate systems, then this task is impossible Without a share reference frame, Alice has no way to communicate a set of symbols to Bob inicating the axis of rotation of her gyroscope Another simple example is clock synchronization, wherein two istant observers want to synchronize their clocks, but it is not possible to o so by sening purely symbolic messages [] Of course, the simple examples escribe above o not mean that sening physical information is impossible For example, in a classical worl, Alice can prepare an sen a physical copy of her gyroscope to Bob, thereby inicating her irection In this way, Alice an Bob can even establish a share reference frame Similarly, in the clock synchronization problem Alice can sen a copy of her clock to Bob [3] to establish a common time stanar ignoring relativistic effects Quantum mechanically, Alice can sen irection information by sening polarize spins, while timing information can be sent by quantum clocks such as two-level atoms As is common in the quantum worl, many interesting an counterintuitive effects occur For example, sening two anti-parallel spins polarize along the esire irection is a better irection inicator than sening two parallel spins [1, 4] The problem of aligning quantum reference frames has garnere significant attention in recent years [1, 3 13] In this paper we are intereste in quantum error correction of physical information Crucially, physical information can only be communicate using systems that themselves have the physical property of interest This places constraints on the actions that can be performe on the physical systems, since we can t, for example, estroy or change physical information arbitrarily In particular, there may be constraints on the set of possible encoing/ecoing schemes that one might have use to make the system more robust to errors, thereby limiting our ability to perform quantum error correction In this paper, we will characterize the constraints place on quantum error correction of physical information

2 tion is equivalent to the existence of orinary, yet G-covariant, encoing schemes, which can correct the same errors In this paper, we first stuy the case in which the group G has at least one infinitesimal generator eg, the rotation group an time translation examples above In this first case, we fin a result strikingly ifferent from conventional, abstract quantum information: we prove a no-go theorem showing that it is impossible to encoe physical information in any number of finite imensional systems in such a way that the encoing allows for perfect correction of any erasure error We then show that both conitions of the no-go theorem are necessary by constructing coes that circumvent the theorem when either of the conitions is violate Specifically, we first emonstrate how one can encoe physical information to protect against erasure errors when one uses continuous-variable moes with infinite imensional Hilbert spaces Since continuous-variable moes are use, we expect this result to be of practical interest We then construct a perfect encoing scheme for any finite group G into finite imensional spaces, which is again robust to erasure errors Finally, we stuy a family of group covariant ranom coes an show that they can provie encoing schemes with better parameters than the perfect schemes for finite groups It is worth noting that the covariant channel formulation of the problem is closely relate to other results in the literature that have very ifferent motivations, incluing the Eastin- Knill Theorem [0] an recent stuies of invariant perfect tensors [1] We present a more etaile comparison in the iscussion Reference frame error correction We begin with a more formal escription of error correction in the familiar case of spatial reference frames, which correspons to G = SO3; the generalization to other groups is immeiate Suppose Alice an Bob share a possibly noisy quantum channel Alice wants to communicate some irectional quantum information a single spin, say to Bob, but Alice an Bob o not share a common reference frame Specifically, their reference frames are relate by an unknown rotation R SO3 Alice an Bob will claim success if Bob receives the spin in the same irection that it was sent by Alice ie, the irectional information is unchange a conition they coul check at a later stage If the task is successful, Bob can use the receive spin for various tasks, such as establishing a share reference frame The simplest metho of sening any quantum information is to sen the quantum state itself This is also true of irectional information, but since the quantum channel between Alice an Bob is noisy, we must account for the possibility of error We fix our error moel to be erasure of a single spin or moe, an our goal will be to esign an error correcting coe to protect the irectional information from this noise For simplicity of presentation, we will iscuss an encoing scheme that encoes one spin into three see fig 1, but the reaer is cautione that the choice of one-into-three is just for clarity; our result hols for an arbitrary one-to-many encoing We split the process into 6 steps: 1 Figure 1a Alice starts with an unknown input state ρ in, a ensity operator on H in, representing some irectional f Encoing Alice a b e Decoing Figure 1 Setup: Alice wants to sen a spin to Bob, but Alice an Bob o not share a reference frame a Alice encoes her spin into an error correcting coe b The environment erases one of the spins c Bob receives the encoe spins in his reference frame Bob then ecoes the remaining spins to reveal the original state e Bob sens the ecoe spin using a hypothetical perfect channel to Alice for verification f Alice confirms that the recovere state is the same as her original state information Alice encoes this initial state using an encoing channel E A We use the subscript A to inicate that E A is the encoing map in Alice s reference frame, an to istinguish it from the map as seen in Bob s frame: E B, to which we will return shortly Thus, the encoe state σ 13 on three spins is given by σ 13 = E A ρ in Figure 1b Spin j {1,, 3} is lost, which is known as an erasure error The erase spin coul be any one of the three, but it is assume that Bob can infer which 3 Figure 1c Prior to the erasure error, the encoe state as seen by Bob woul be U 1 U U 3 σ 13 U 1 U U 3, where U i = U i R is a unitary representation of the unknown rotation R mapping Alice s coorinate system Bob s Bob then receives the state tr j U 1 U U 3 σ 13 U 1 U U 3 4 Figure 1 Bob ecoes the state with an R-inepenent [ ecoing map D j to obtain D j tr j U 1 U U 3 σ 13 U 1 U U 3 ] This is the state recovere in Bob s reference frame If the protocol is successful, this state shoul be equal to ρ in = U in ρ in U in in orer to match Alice s original state, where U in = U in R is the representation of the rotation group acting on the initial state, an the tile signals that this is the input state as seen from Bob s rotate reference frame 5 Figure 1e Bob sens the ecoe state through a hypothetical perfect channel to Alice for verification 6 Figure 1f Success is claime if the receive state is the same as the initial state in Alice s frame Using ρ in = U in ρ in U in, the success conition becomes [ ρ in =D j trj U1 U U 3 E A U in ρ inu in U 1 U U ] 3, 1 for all R SO3, states ρ in H in, an j {1,, 3} c Bob

3 3 Covariant error correction Covariant quantum error correction is a seemingly ifferent problem in which the encoing map is require to commute with the action of the group Continuing the example of mapping a single spin into three, the covariance requirement is that the encoing map satisfy U 1 U U 3 EU in ρ in U in U 1 U U 3 = Eρ in for all R SO3 an initial states ρ in To be clear, in this problem, Alice an Bob are assume to share a single reference frame Imposing the simple constraint on the encoing map, however, efines an error correction problem equivalent to reference frame erasure correction Return now to the setting of reference frame error correction to see why Alice performs the encoing E A in her reference frame In Bob s reference frame, this operation is enote by E B,R E B,R is the quantum channel corresponing to the operation Alice performs as seen in Bob s reference frame For a fixe E A in Alice s reference frame, E B in Bob s frame is still parametrize by the unknown rotation R, ie, E B = E B,R Specifically, E B,R ρ in = U 1 U U 3 E A U in ρ in U in U 1 U U 3 The success conition simplifies to D j tr j E B,R ρ in = ρ in, 3 for all states ρ in an j {1,, 3} Now introuce the average channel E = E R [E B,R ], where the average is over all rotations R SO3 accoring to the Haar measure By the linearity of the partial trace an the ecoing channels, the error correction relation 3 hols for the average channel: D j tr j E ρ in = ρ in Moreover, the average channel is clearly covariant in the sense of, provie we substitute ρ in for ρ in in the equation So if reference frame error correction 1 is possible, we have foun a covariant erasure-correcting encoing Moreover, it is straightforwar to confirm that by choosing E to be E A, eqs 3 an lea to eq1 Therefore, reference frame error correction an covariant error correction are equivalent Results Let us now stuy a more general question Consier an encoing map E which encoes an initial state on H in into n encoe systems on H out = H 1 H n We o not impose any constraints on the output Hilbert spaces at this point ie, they can be the same or ifferent, finite or infinite imensional, etc Suppose there exists a group G, an representations U in, U 1,, U n acting on the ifferent Hilbert spaces Moreover, suppose that the channel is covariant uner the action of the group: Eρ in = U 1 U n EU in ρ inu in U 1 U n 4 Our goal is to answer the following question: is it possible to recover the original state after erasure of an arbitrary set of at most k subsystems which we henceforth refer to as moes? We will stuy this question in ifferent scenarios: 1 G is a Lie group an the coe is finite imensional We prove a no-go theorem: no perfect covariant error correcting scheme can be implemente in this case This applies to the example of sening spins, as in the original reference frame error correction task In fact, the no-go theorem applies to all groups with at least one infinitesimal generator, an it states that such generators can only act trivially on encoe states G is a Lie group an the coe is infinite imensional We show that G-covariant error correcting coes are possible when the encoing uses infinite imensional systems This illustrates the existence of interesting error correcting coes for a Lie group when the conitions of the no-go theorem above are not satisfie We provie an explicit coe for G = U1 in appenix A 3 G is a finite group an the coe is finite imensional For any finite group G, we fin examples of perfect covariant error correcting schemes This is again consistent with our no-go theorem since finite groups o not have infinitesimal generators We also provie a ranomize construction in appenix B to obtain approximate coes with better parameters Case 1: G is a Lie group an the coe is finite imensional In this case, suppose that the local Hilbert space imensions are finite, an that the group G is a Lie group [] Choose one infinitesimal generator of the Lie group, without loss of generality We enote this generator acting on the input moe by T in an on the ith output moe by T i Thus, the generator acting on the full set of output moes is T out = T 1 + T n Assume that T in is non-trivial; our goal will be to show that covariant quantum error correction is impossible with this assumption Consier an initial state ρ in an a slightly rotate state ρ in ɛ = e iɛtin ρ in e iɛtin These states are encoe as σ out = Eρ in an σ out ɛ = Eρ in ɛ Using the fact that Eρ in is invertible on its range, we can fin a set of orthogonal isometries {E i }, E i E j = δ ij I an probabilities p i such that Eρ in = i p i E i ρ in E i see, eg, [3], theorem 1 an its proof using H in as the coe space The inverse channel E 1 σ out can be escribe by the same set of isometries on the range of E E 1 ρ out = i E i ρ oute i + Π ρ out Π, where Π = I i E ie i A crucial but elementary property of E 1 is that if σ out = Eρ in an A is some arbitrary operator, then E 1 Aσ out = E Aρ in, where E A = i p ie i AE i Expaning the relation ρ in ρ in ɛ = E 1 σ out σ out ɛ to first orer in ɛ we obtain [T in, ρ in ] = E 1 [T out, σ out ] 5 = [E T out, ρ in ] Uner the assumption that error correction succees, we can then recover the original state from any of the n k subsets

4 4 of the encoe moes This means that upon tracing out all output moes except the ith moe, the remaining state ρ i is inepenent of the initial state since if it weren t the moe number i woul contain information about the input state Thus, for any state ρ in, we fin that trt i σ out = α i, where α i is inepenent of ρ in It is easy to see that α i = trt i σ out = trt i Eρ in = tre T i ρ in for all ρ in Hence E T i I, an consequently E T out I This implies that the last term in eq 5 is zero, which means that [T in, ρ in ] = 0 for all ρ in In orer for T in to commute with all ρ in it must be trivial, which is a contraiction of our assumption We conclue that perfect recoverability is impossible Case : G is a Lie group an the coe is infinite imensional If we allow Alice the ability to use infinite imensional Hilbert spaces violating one of the hypotheses of our no-go theorem, then even a naïve solution to the problem exists Intuitively, a simple way to achieve the task is for Alice to appen a classical gyroscope to the encoe state that she sens to Bob [4] Bob can then infer information about Alice s reference frame by measuring the state of the gyroscope, thereby establishing a common reference frame Inee, this is one strategy we will outline below Since the full state is sent through the noisy channel, Alice actually sens two gyroscopes in orer to safeguar against loss of one of the encoe shares Any reaer isappointe by the construction s use of effectively classical gyroscopes shoul be heartene to know that the 1-into-3 encoing escribe in appenix A achieves covariant error correction without them In the reference frame error correction paraigm, Alice chooses her favourite non-covariant erasure coe an appens two reunant ancilla the classical gyroscopes inicating her reference frame to the encoe state The ancilla must necessarily be states in infinite imensional Hilbert spaces so that the no-go theorem oes not apply an in this protocol this is also necessary so that Alice can specify her reference frame with perfect precision[5] If any shares of the erasure coe are lost, Bob can first measure the gyroscopes to learn Alice s reference frame, an then use the stanar ecoing on the remaining shares in the right frame Since Alice sent two ancilla, one can freely be lost without failure Let us now stuy this problem in the covariant quantum error correction paraigm Let H G = span{ g }, where g G The group acts via Ug h = gh [6] To encoe her state, Alice chooses her favourite, non-covariant erasure correcting coe enote by E 0, such as the C 3 C 3 3 qutrit coe for example wlog As before, we efine the rotate encoing map ie, the map in Bob s frame by E g Ψ = Ug 3 E 0 [ U gψug ] U 3 g 6 To complete the encoing, Alice appens two ancilla in the state e e where e G is the ientity element for a full encoe state E 0 Ψ e e as seen in her frame The two e e registers represent the classical gyroscopes above The encoing is mae covariant by averaging over the group G Thus, the full encoing is efine by symmetrizing the channel an ancilla together: EΨ = g G g E g Ψ g g, which can be easily seen to be covariant Our ecoing proceure is then fairly simple: one nee only measure any ancilla that are not lost, collapsing the state to one corresponing to the measure group element We can then recover the encoe state from the any qutrit shares they were not erase by the noisy channel The proceure escribe above is not the only metho one can use in this case In appenix A we escribe an explicit, group covariant, continuous-variable quantum erasure coe for the example of G = U1 An input continuous variable moe is mappe into three physical moes via the encoing E U1 = 3y, x + y, y + x 13 x in x,y Z We leave all relevant etails for appenix A Case 3: G is a finite group an the coe is finite imensional Consier a finite group G Here we show that it is possible to fin G-covariant channels that encoe the input Hilbert space into finite imensional Hilbert spaces while satisfying the erasure correction conitions Suppose the group G acts on some set A By efinition, the action of G permutes the elements of A Our goal is to construct an error correction scheme for which the action of the group commutes with the process of encoing, erasure, an ecoing To achieve our goal, we first start with a noncovariant erasure We then consier a tensor prouct of many copies of this non-covariant coe, one tensor factor for each element of A As it happens, this coe efine using many copies of a non-covariant coe is alreay a covariant coe! To see this, note that the encoing acts as a tensor prouct over the factors, while the group action simply permutes the factors Therefore, the encoing map an the group action commute, which implies that the encoing is G-covariant To be more precise, consier a channel E 0 : SH in SH out := H n where SH enotes the space of ensity matrices on the Hilbert space H Suppose that E 0 is an encoing map that allows for recovery after erasure of an arbitrary set of k of the n output moes However, we make no assumptions about the covariance of E 0 it is an arbitrary erasure correcting map We now introuce a new encoing E = E 0 = E A 0, E : SH A in SH A a A out, where we have use a A E 0 to inicate that the ifferent tensor copies are labele by elements of A For each g G the action of the representation on H A is efine by Ug φ a1 φ a φ a A = φ g 1 a 1 φ g 1 a φ g 1 a A Here a 1 a A is a list of the elements of A The covariance of E follows from the efinition, an the error correction properties of E are irectly inherite from those of E 0 Therefore,

5 5 Figure Permutation covariance for the group S 3 acting on S 3 ie, G = A = S 3 Each fork represents a coe that maps one quit into three, an can correct an erasure error on any one output quit π 1 G is the transposition that swaps systems 1 an Left The map EU inπ 1ρ inu inπ 1 The group action permutes the inputs to the channel Right The map U outπ 1Eρ inu outπ 1 As it is evient from the wiring of the forks, these two maps are equivalent we have succeee in fining a perfect G-covariant channel Figure shows an example in which G = S 3 the permutation group on 3 elements an A = {1,, 3} While our construction can be formally extene to infinite groups with their associate infinite imensional representations, we have not etermine which aitional conitions nee to be impose in orer for the argument to remain mathematically rigorous The construction presente in this section provies coes in which the Hilbert spaces can be exponentially large in G However, it is known that in many cases ranom coes give near optimal error correcting schemes with goo parameters [7 31] In appenix B, we show that choosing a ranom covariant isometry yiels approximate error correcting coes for which the imension of each moe is just G For these coes, the worst-case fielity of recovery, F worst, behaves well with high probability Specifically, PF worst < 1 ɛ ecays exponentially in G For example, we will show that: P F worst < 1 G 9 n 8 ] exp [ G G n log 30 G 7+n It is clear that for n 5 an G sufficiently large, the exponent on the right-han sie becomes arbitrarily negative, inicating that the worst-case fielity of recovery is very close to 1 with very high probability Discussion We showe that perfect error correction of physical information against erasure is a process that epens on the etails of the symmetry group an imensions of the coe For example, covariant error correction is impossible when the symmetry group is a Lie group an the coe is finite imensional This is connecte to the following no-go theorems in the literature: Eastin-Knill theorem [3] Eastin an Knill prove [0] that it is not possible to encoe information in an erroretecting coe in such a way that a set of universal gates can be implemente transversally We can reprouce the main thrust of the Eastin-Knill theorem [33] using an instance of our no-go theorem in which the input space is the set of N logical quits, the output consists of physical quits, an letting the group be G = UN Moreover, our continuous variable, infinite imensional coe construction provies a emonstration that the Eastin-Knill theorem can be circumvente in principle, although our examples o not appear to be useful for fault-tolerant quantum computation Invariant perfect tensors A quantum state on the tensor prouct of a number of Hilbert spaces is a perfect tensor if, for any bipartition of the Hilbert space into two collections of constituent factors, it forms an isometry from the smaller space to the larger [34] Motivate by the construction of physical states in the Hilbert space of loop quantum gravity, the authors in [1] efine the notion of invariant perfect tensors as those perfect tensors which are invariant with respect to the action of SU In [1], it was prove that there are no invariant perfect tensors with four tensor factors This can be seen as a irect consequence of our no-go theorem for G = SU, by consiering a four-partite invariant perfect tensor as a 1 moe to 3 moe isometry Such an invariant perfect tensor with 4 tensor factors woul efine an SU-covariant erasure correcting coe, which is prohibite by our no-go theorem Furthermore, our no-go theorem states that there are no invariant perfect tensors with higher numbers of tensor factors, thereby solving an open question in [1] One might hope to fin a more quantitative relation between some measure of the size of the group an the imension of the coe when error correction is possible For example, a conition of the form G imcoe ie, imension of the physical Hilbert space is consistent with our no-go theorem an the examples in Cases 1 an Another interesting avenue for future research relates to approximate error correction, in which one might like to fin a relationship between the error tolerance ɛ, group size G, an imension of the coe Acknowlegements We thank Dawei Ding, Iman Marvian, Michael Walter, an Beni Yoshia for helpful iscussion SN acknowleges support from Stanfor Grauate Fellowship GS acknowleges support from a NSERC postgrauate scholarship This work was supporte by the CIFAR an the Simons Founation [1] N Gisin an S Popescu, Physical Review Letters 83, [] If the transmission time for each message is preetermine, that provies a resource that coul itself be use for clock synchronization To avoi this loophole, symbolic messages shoul not be receive at preetermine times [3] J Preskill, arxiv preprint quant-ph/ [4] S Massar an S Popescu, Physical review letters 74, [5] E Bagan, M Baig, A Brey, R Munoz-Tapia, an R Tarrach,

6 6 Physical Review A 63, [6] R Jozsa, D S Abrams, J P Dowling, an C P Williams, Physical Review Letters 85, [7] C Souza, C Borges, A Khoury, J Huguenin, L Aolita, an S Walborn, Physical Review A 77, [8] S D Bartlett, T Ruolph, R W Spekkens, an P S Turner, New Journal of Physics 8, [9] S D Bartlett, T Ruolph, an R W Spekkens, Physical review letters 91, [10] A Peres an P F Scuo, Physical review letters 86, [11] S D Bartlett, T Ruolph, an R W Spekkens, Reviews of Moern Physics 79, [1] G Gour an R W Spekkens, New Journal of Physics 10, [13] I Marvian an R W Spekkens, Physical Review A 90, [14] I Marvian an R W Spekkens, New Journal of Physics 15, [15] I Marvian an R W Spekkens, Physical Review A 90, [16] I Marvian an R W Spekkens, Nature communications 5 4 [17] F G Branao, M Horoecki, J Oppenheim, J M Renes, an R W Spekkens, Physical review letters 111, [18] V Veitch, S H Mousavian, D Gottesman, an J Emerson, New Journal of Physics 16, [19] I Devetak, A W Harrow, an A J Winter, IEEE Transactions on Information Theory 54, [0] B Eastin an E Knill, Phys Rev Lett 10, [1] Y Li, M Han, M Grassl, an B Zeng, arxiv preprint arxiv: [] We exclue the case of 0-imensional Lie groups Also, G oes not actually nee to be a Lie group, but it must have at least one infinitesimal generator In that case, our proof shows that the infinitesimal generators of the group can only act trivially on the system [3] M A Nielsen an I Chuang, Quantum computation an quantum information, 00 [4] To be precise, each classical gyroscope etermines one axis In orer to sen a classical reference frame we nee at least two gyroscopes for the x an y axes By gyroscope we mean a complete inicator of the reference frame [5] Some reaers might take issue with calling this an erasure coe, since such coes are usually constructe such that the Hilbert spaces of each share are the same so that all Hilbert spaces in this case must then be infinite imensional If esire, one can simply embe finite imensional Hilbert spaces into the infinite imensional spaces such that the group acts on these subspaces accoring to the associate finite imensional representation an trivially on the rest [6] In fact, it is not necessary to assume that the basis is inexe by group elements they can be inexe by any set on which the group acts faithfully [7] P W Shor, in lecture notes, MSRI Workshop on Quantum Computation 00 [8] I Devetak, IEEE Transactions on Information Theory 51, [9] S Lloy, Physical Review A 55, [30] P Hayen, M Horoecki, A Winter, an J Yar, Open Systems & Information Dynamics 15, [31] M Hamaa, IEEE transactions on information theory 51, [3] We thank Beni Yoshia pointing out the connection to the Eastin- Knill Theorem [33] The Eastin-Knill theorem also iscusses the possibility of encoing information in the isconnecte components of the Lie group, a point that is absent in our work Furthermore, the full Eastin-Knill theorem makes reference to universal gates In orer to fully reprouce the theorem, we woul nee aitional arguments concerning continuity of the channel an error etection [34] F Pastawski, B Yoshia, D Harlow, an J Preskill, arxiv preprint arxiv: [35] P Hayen, D W Leung, an A Winter, Communications in Mathematical Physics 65, [36] A Harrow, P Hayen, an D Leung, Physical review letters 9, [37] P Hayen, D Leung, P W Shor, an A Winter, Communications in Mathematical Physics 50, [38] C H Bennett, P Hayen, D W Leung, P W Shor, an A Winter, IEEE Transactions on Information Theory 51,

7 7 Appenix A: G = U1 an the coe is continuous-variable Here we provie an explicit U1-covariant 1 3 encoing The construction presente in this section oes not violate the no-go theorem state in Case 1 above as the local systems are infinite imensional Since the symmetry group in question is U1, this coe coul be implemente in optical moes, an it is arguably more natural than the construction presente in Case We take the Hilbert space to be the space of functions on a circle using the position basis { φ } φ [0,π U1 acts on this space via the regular representation: if g = e iθ U1, then the action of the regular representation is efine by Ug α = α + θ It is convenient to work in the Fourier basis where the Hilbert space is escribe by the conjugate momentum basis { n } n Z an the group acts by Ug n = e inθ n We efine the isometry to be the following operator expresse in the conjugate momentum basis E U1 = 3y, x + y, y + x 13 x in x,y Z More explicitly, the isometry maps the state x φx x in to Ψ 13 = x,y φx 3y, x + y, y + x 13 It is easy to see that this isometry is U1-covariant: Ug 3 E U1 Ug = e i 3y x+y+y+x E U1 e ix = E U1 Here we show, step by step, that this mapping can correct an erasure error Consier the encoe ensity matrix Ψ 13 = x 1,y 1,x,y Z We will stuy the loss of moes 1, an 3, in turn 1 Loss of the first moe The resulting ensity matrix is Ψ 3 = φx 1 φx 3y 1, x 1 + y 1, y 1 + x 1 3y, x + y, y + x 13 x 1,x,y Z Decoing starts with the linear map a, b a, b a, yieling x 1,x,y Z φx 1 φx x 1 + y, y + x 1 x + y, y + x 3 φx 1 φx x 1 + y, 4x 1 x + y, 4x 3 We then use an isometry which maps the states of the form a, 4b to a, b x 1,x,y Z Finally by a, b a b, b, we obtain φx 1 φx x 1 + y, x 1 x + y, x 3 x 1,x,y Z Therefore, tracing out moe reveals the original state Loss of the secon moe The resulting ensity matrix is Ψ 13 = x 1,y,x Z or, equivalently by the change of variable y y + x 1, Ψ 13 = x 1,y,x Z φx 1 φx y, x 1 y, x 3 φx 1 φx 3y, y + x 1 3 x 1 + y + x, x 1 + y + x 13, φx 1 φx 3y + x 1, y + x 1 3y + x, y + x 13

8 8 We now use an isometry which maps states of the form 3a, b to a, b φx 1 φx y + x 1, y + x 1 y + x, y + x 13 x 1,y,x Z By a, b a, a + b, we have x 1,y,x Z We now use a, b a + b, b to obtain φx 1 φx y + x 1, y y + x, y 13 x 1,y,x Z Tracing out moe 3 reveals the original state φx 1 φx x 1, y x, y 13 3 Loss of the thir moe Again, the resulting ensity matrix is Ψ 1 = φx 1 φx 3y, x 1 + y 3y + x 1 x, x + y + x 1 1 x 1,y,x Z Using the change of variable y y + x 1 we have Ψ 1 = φx 1 φx 3y x 1, x 1 + y 3y x, x + y 1 x 1,y,x Z Applying an isometry that maps 3a, b to a, b yiels φx 1 φx y x 1, x 1 + y y x, x + y 1 x 1,y,x Z Using a, b a, a + b, x 1,y,x Z φx 1 φx y x 1, x 1 y x, x 1 Finally the isometry a, b a + b, a turns the state to φx 1 φx y, x 1 y, x 1 x 1,y,x Z Thus we can recover the state on moe Appenix B: G is a finite group an the coe is a ranom G-covariant isometry In the construction presente for Case 3, the local Hilbert space imension can grow exponentially with G In this section we present an alternative, approximate metho for error correction in which the local Hilbert space imemsions are equal to G Our goal will be to prove eq 7 Consier a 1 n encoing We will look for isometries that map H G H n G, where H G enotes the Hilbert space associate to the regular representation of G with the basis { g } g G Thus im H G = G = We represent the action of the regular representation of g G on H G by Ug To construct a ranom covariant map, we start with a ranom invariant state Ψ H n+1 G For our purposes: a ranom state is one that is chosen ranomly with respect to the unitary invariant measure; ranom unitaries are unitaries chosen ranomly with respect to the Haar measure; an a state is invariant if Ug n+1 Ψ = Ψ for all g G By projecting our chosen state onto an un-normalize, maximally entangle state φ + AB = i A i B we obtain a map E which is close to an isometry whp from H in H n, E in,1 n = φ + in,0 Ψ 0 n

9 Note that the covariance of E efine by Ug n E = EUg, which follows from the invariance of Ψ From E we can efine the exact isometry T as T := EE E 1/ One can verify that T is also a covariant map, since [E E, Ug] for all g G Our encoing is then efine by Eρ in = T ρ in T With the covariant encoing in han, we now turn our attention to the ecoing Before iving in, let us first efine two notational conventions that will be use frequently henceforth Firstly, we will use trˆx to inicate tracing out all subsystems except the set x Seconly, if there are two isomorphic Hilbert spaces H α an H β with the same preferre basis, an if the operator X α acts on H α, then by X α β we mean the operator X α acting on H β in the sense that the matrix corresponing to X α is simply applie to H β One can think of X α β as overriing the Hilbert space inices When it is clear to o so, we use X β instea of X α β for brevity To ecoe after loss of one of the moes, say moe 1 without loss of generality, Bob first replaces the lost moe by a maximally mixe state τ 1 an then ecoes the state τ 1 tr 1 [Eρ in ] The ecoing map is given by σ out = D 1 ρ 1 n = [ trˆ U T 3 V 3 n ρ 1 n U3V T 3 n ] out, where U, an V 3 n are unitaries that transform Ψ 0 n into its Schmit form: U V n Ψ 0 n = i,j λij ij ij0 0 3 n, 9 an U 3 = U 3 is the same operator as U but acting on the Hilbert spaces inexe by an 3 In other wors, U = U 3 With the ecoing above, our task is now to prove eq 7 Our first step will be bouning the worst-case fielity of recovery F worst in terms of the istance between Ψ the reuce ensity matrix of the invariant state Ψ an the maximally mixe state: Lemma 1 For an 0 ɛ 1, if Ψ τ Proof We will prove in three steps: ɛ 3, then 1 ɛ F worst Step 1 We first simplify the expression for the recovere state an show that D 1 τ 1 Eρ in = tr 1 Ψ T 1/ Ψ T 0 1/ ρin 0 Ψ T 0 1/ Ψ T 1/ Step We then use joint concavity of the fielity, an properties of the Schatten norm to boun the worst-case fielity F worst min κ κ 0 tr 1 Ψ 1/ Ψ 1/ 0 κ 0 From the above equation, it is alreay clear that if Ψ 0 an Ψ are close to the maximally mixe state, then the worst-case fielity will be close to 1 We quantify this in the last step Step 3 We show that for 0 ɛ 1, if Ψ τ Step 1 We begin with the expression for the recovere state, D 1 τ 1 Eρ in = trˆ ɛ 3, then 1 ɛ F worst U3V T 3 n tr 1 T ρ in T V 3 n U 3 B1 Using the fact that E E = Ψ T 0 in, an the efinition ρ in = 1 Ψ T 1/ 0 ρ in Ψ T 1/ 0, we have that T ρ in T = E ρ in E in in From the efinition of E we can simplify the formula for the encoing map: Eρ in = E ρ in E = tr 0 Ψ Ψ 0 n ρ T 0 Therefore, D 1 τ 1 Eρ in = trˆ [ ] U3V T n Ψ Ψ 0 n V n U 3 ρ T 0 B

10 However, recall that U V n Ψ 0 n = i,j Thus we obtain Using eq B, we fin λij ij ij0 0 3 n, an that U Ψ 1/ U = ij λij ij ij V n Ψ 0 n = Ψ 1/ U φ + 0 φ n = U 3 Ψ T 1/ φ + 0 φ n 3 D 1 τ 1 Eρ in = trˆ = tr 3 Ψ 1/ T 3 Ψ 1/ T 3 φ + 0 φ + 13 φ + 0 φ + 13 ρ 0 Ψ 1/ T 3 = tr 1 Ψ T 1/ Ψ T 0 1/ ρ0 Ψ T 0 1/ Ψ T 1/ Ψ 1/ T ρ T 0 3 Therefore, we have achieve the goal of step 1 Step Our goal now is to lower boun the fielity of recovery Since the fielity is jointly concave, we know that the minimum fielity of recovery for the channel is achieve with a pure input state, say ρ 0 = κ κ T, where we have ae the transpose to simplify the expressions In this case, the recovere state takes the following form: T D 1 τ 1 Eρ in = tr 1 Ψ 1/ Ψ 0 1/ κ 0 κ 0 Ψ 1/ 0 Ψ 1/, so that the minimum fielity is F min = min κ tr κ 0 Ψ 1/ Ψ 1/ 0 κ 0 κ 0 Ψ 1/ 0 Ψ 1/ κ 0 κ 0 = min Ψ 1/ κ Ψ 1/ 0 κ 0 To procee, we use the following basic property of the Schatten norm: for 1 p + 1 q = 1, Y p trxy if X q = 1 Applying this inequality when X = I 1 / an p = q = we fin: κ Ψ 1/ Ψ 1/ 0 κ = max 1 tr { tr X 1 κ 0 Ψ 1/ Ψ 1/ 0 κ 0 X = 1 κ 0 Ψ 1/ Ψ 1/ 0 κ 0 = κ 0 tr 1 Ψ 1/ Ψ 1/ 0 κ 0 This conlues step Step 3 We woul ultimately like to lower boun the worst-case fielity using concentration of measure techniques for Ψ an Ψ 0 We start by upper bouning tr 1 1 Upper boun for tr 1 tr 1 Ψ 1/ Ψ 1/ Ψ 1/ I 0 : I 0 = tr 1 I 0 an max α Ψ 1/ I Ψ 1/ 0 I 0, assuming that Ψ 0 τ 0 1 α 0 g 1 tr 1 g G I, Ψ1/ = max α α 0 tr 1 Ψ 1/ I } Ψ 1/ I α 0 α 0 g 1 where g, g G form a basis for evaluating the trace, the first inequality is the triangle inequality, an the secon inequality comes from the fact that the infinite Schatten norm of a Hermitian operator is equal to its maximum eigenvalue Now, one 10 B3

11 11 can check that for any λ 0, λ 1/ 1/ λ 1/ Taking {λ i } to be the set of eigenvalues of Ψ 1/, an using the aforementione inequality, we obtain Ψ1/ I = max i λ1/ i 1 max i λ i 1 Ψ τ B4 Thus tr 1 Ψ 1/ I 0 Ψ τ Upper boun for Ψ0 1/ I 0 : One can simply check that for any real number λ such that λ 1/ 1/, then λ 1/ / 1 λ 1/ In particular, since this inequality hols for all of the eigenvalues of Ψ 0, we can erive the following boun for the operator norm: Ψ 1/ 0 I 0 Ψ 0 τ 0 B5 To procee, we assume that Ψ τ κ 0 tr 1 Ψ 1/ Ψ 1/ 0 κ 0 ] = 1 + κ 0 [tr 1 Ψ 1/ I 0 [ 1 κ 0 tr 1 Ψ 1/ I 0 ] ɛ 3 for 0 ɛ 1 Combining this assumption with eq B3, we have [ ] Ψ 1/ [ κ 0 + κ 0 0 I 0 κ 0 + κ 0 tr 1 [ ] Ψ 1/ κ 0 κ 0 0 I 0 κ 0 Ψ 1/ κ 0 [tr 1 ] [ ] Ψ 1/ I 0 0 I 0 κ 0 Ψ 1/ ] [ ] Ψ 1/ I 0 0 I 0 κ 0, where the inequality in the last line is the triangle inequality Now, since κ X κ X for any matrix X, we have κ 0 tr 1 Ψ 1/ Ψ 1/ 0 κ 0 1 tr 1 Ψ 1/ Ψ 1/ ] I 0 0 I 0 [tr [ ] 1 Ψ 1/ Ψ 1/ I 0 0 I 0 1 tr 1 Ψ 1/ Ψ 1/ I 0 0 I 0 tr 1 Ψ 1/ Ψ 1/ I 0 0 I 0, where the secon inequality follows from the fact that XY X Y for any pair of matrices X an Y Using eqs B4 an B5 above, Ψ 1/ 0 κ 0 tr 1 Ψ 1/ κ 0 1 Ψ τ Ψ 0 τ 0 Ψ τ Ψ0 τ 0 Note that the conition Ψ 0 τ 0 1 is satisfie, since Ψ 0 τ 0 Ψ τ an Ψ τ ɛ 3 Finally, since Ψ 0 τ 0 Ψ τ, we have that κ 0 tr 1 Ψ 1/ Ψ 1/ 0 κ 0 1 Ψ τ Ψ τ, an we therefore conclue that F worst κ 0 tr 1 which proves the lemma Ψ 1/ Ψ 1/ 0 κ 0 1 Ψ τ Ψ τ 1 ɛ,

12 To complete the proof, it remains to be shown that our assumption is vali Specifically, in orer to show that the worst-case fielity is close to 1, it suffices to prove that the reuce ensity matrix of ranom invariant states, Ψ, is very close to the maximally mixe state in operator norm ie, Ψ τ is small with high probability Since Ψ τ = max σ tr [σ Ψ τ ], where the maximization is one over all possible ensity matrices σ, we can instea stuy the quantity on the right han sie To show that this is small, we will follow the techniques use in [35 38] Before stating the proof in its full glory, let us first gain an imprecise, high-level overview of the strategy We will first efine an ɛ-net on the set of ensity matrices on H 0 H 1, ie, a finite set of ensity matrices σ such that any other ensity matrix σ is close to one of the elements of the net in the trace norm If we can then show that tr [ σ Ψ τ ] is small for every σ in the net, then it must be small for all ensity matrices σ Using large eviation methos, we will then prove that for any fixe ensity matrix σ incluing the elements of the net, tr [σ Ψ τ ] is small with very high probability Since the number of elements in the net is finite with a known upper boun, we can then use a union boun to show that tr [σ Ψ τ ] is small for all elements in the net with high probability Therefore, we can boun tr [σ Ψ τ ], arriving at our esire conclusion We will now we give a etaile proof of eq 7 Let P δ,σ be the probability that, for a fixe σ, tr [σ Ψ τ ] δ/, an let P δ = max σ P δ,σ, where the maximum is over all ensity matrices on H 0 H 1 The following lemma relates P Ψ τ ɛ 3 to Pδ Lemma For 0 α ɛ, we have P Ψ τ ɛ [ P ɛ α 3 α Proof Consier an α 3 -trace istance net M of pure states in H 0 H 1, with α ɛ For every pure state σ, there exists a pure state σ such that σ σ 1 by efinition It is known that we can choose M such that M 15 α [37, Lemma II4] Now if tr σ [Ψ τ ] ɛ α 3, then from eq B6 it follows that trσ [Ψ τ ] tr σ [Ψ τ ] + trσ σ [Ψ τ ] Therefore, P Ψ τ ɛ α 3 ɛ α ɛ 3 α 3, + σ σ 1 Ψ τ 3 + σ σ 1 ɛ 3 We can simplify eq B7 using a union boun: P σ M : tr σ [Ψ τ ] ɛ α 3 This, along with eq B7, conclue the proof of the lemma ] = P σ : tr σ [Ψ τ ] ɛ 3 P σ M : tr σ [Ψ τ ] ɛ α 3 = 1 P σ M : tr σ [Ψ τ ] ɛ α 3 σ M P ɛ α P ɛ α M 3, σ 3 1 B6 B7

13 13 In appenix C, we will use large eviation techniques to show that P δ exp n δ /6 for 0 δ 1 B8 We will efer the proof to appenix C but use the result immeiately Combining lemma 1, lemma an eq B8 we have P F worst 1 ɛ P Ψ τ ɛ 3 min P 0αɛ ɛ α 3 min exp n ɛ α /54 + log 0αɛ [ 15 α 15 One convenient choice of ɛ an α is ɛ = 9 n 8 an α = ɛ/ With this choice we fin ] P F worst 1 ɛ exp [ n log 30 7+n 8, 16 which reuces to eq 7 after substituting G for α ] Appenix C: Proof of eq B8 The goal of this appenix is to prove eq B8 The iscussion is split into two parts: we first explain the ranom invariant state construction, an then we prove the esire boun Construction of ranom invariant states Consier the invariant subspace of H n+1 it is easy to see that the invariant subspace is spanne by states of the form 1 gh 1, gh,, gh n, g 0 n, We now introuce an isometry M from H n to the invariant subspace of H n+1, M = 1 g G g,h 1,,h n gh 1, gh,, gh n, g 0 n h 1, h,, h n 0 n 1 The projector onto the invariant subspace is efine as Π 0 n = MM Π 0 n has the important property that, upon tracing out any one of the subsystems, it becomes the ientity operator on the remaining subsystems That is tr i Π 0 n = I 0 î n C1 A ranom invariant state Ψ 0 n is constructe by choosing a ranom state φ 0 n 1 in H n from the unitary invariant measure, an then mapping φ to H n+1 using the isometry M, Ψ 0 n = M φ 0 n 1 Proof of eq B8 To begin, we will upper boun the moment generating function, E Ψ exp t tr [σ Ψ ], for an arbitrary ensity matrix σ, where Ψ = trˆ0ˆ1 [Ψ 0 n] an the average is over ranom invariant states Ψ 0 n Note that tr [σ Ψ ] = tr [σ Ψ 0 n ] = tr [ σ Mφ 0 n 1 M ] = tr [ M σ Mφ 0 n 1 ] One can easily check that M σ M = σ G I n 1, where σ G = 1 g,h 1,h,h 1,h h 1, h gh 1, gh σ gh 1, gh h 1, h One can also check that σ G is a ensity matrix, specifically a version of σ symmetrize by the group G Therefore, tr [σ Ψ 0 n ] = φ σ G φ, where φ = φ 0 n 1 is a state on H n chosen from the unitary invariant measure see the first subsection of this appenix We now choose a Gaussian state g 0 n 1 in which the coefficients of the wave function are chosen ii from a complex Gaussian istribution centere at zero with variance n Thus E g g = 1 Therefore, we have E g exp t g σ G g = E φ E g exp t g φ σ G φ

14 E φ exp t [ E g g ] φ σ G φ = E φ exp t φ σ G φ = E Ψ exp t tr [σ Ψ 0 n ], C where the inequality follows from the convexity of the exponential function Now suppose that the eigenvalues of σ G are p i0,i 1 Since the Gaussian states are unitarily invariant, we can evaluate E g exp t g σ G g in a basis in which σ G I n 1 is iagonal In that basis, E g exp t g σ G g = E g exp t = i 0 i n 1 p i0,i 1 g i0 i n 1 i 0 i n 1 E gi0 i n 1 exp t p i0,i 1 g i0 i n 1 However, the raial probability ensity for each coefficient is p g i0 i n 1 = n g i0 i n 1 exp n g i0 i n 1 Using the probability ensity formula, we fin Assuming t n, we have E gi0 i n 1 exp tp i0,i 1 g i0 i n 1 1 = 1 t for t n /p i0,i pi 0,i 1 1 n E g exp t g σ G g = 1 t p n n Ultimately, we will fix the value of t to prove the boun in eq B8, but we nee to istinguish the cases in which t is positive or negative to boun the fluctuations of tr [σ Ψ ] above or below 1/ Therefore, we iscuss these two ifferent ranges for t separately: 1 Positive t: We will use the assumption that t is positive to limit the fluctuations of tr [σ Ψ ] above 1/ Let 0 < s < 1 be a fixe number, an restrict t to 0 t s n Uner these conitions, we have, 1 t p 1 n t p i0,i 1 1 s n Therefore, Combining with eq C, we have 1 t p n n s t p i0,i 1 n E g exp t g σ G g E g exp t g σ G g = n 1 exp 1 s t p 14 1 p n t 1 n exp 1 s t C3 To boun the probabilities, we use Bernstein s trick: P tr [σ Ψ ] 1 + δ = P exp t tr [σ Ψ ] exp t 1 + δ [ E Ψ exp t tr [σ Ψ ] ] exp t 1 + δ exp t 1 + δ 1, 1 s where we use Markov s inequality for the exponentials an eq C3 To obtain the best result, we now set t = s n an s = δ 1/ With this substitution, P tr [σ Ψ ] 1 + δ exp n 1 + δ 1 exp n δ /6 where the last inequality is vali for 0 δ 1

15 Negative t: We now use the constraint on t to limit the fluctuations of tr [σ Ψ ] below 1/ Assuming that s > 0 an s n t 0, one can show that Therefore, E g exp t g σ G g = 1 tp n log1 + s n exp t p i0,i s 1 1 p n t n log1 + s log1 + s exp t p i0,i s 1 exp t s 15 Thus, P tr [σ Ψ ] 1 δ 1 = P t tr [σ Ψ ] t δ = P exp t tr [σ Ψ ] exp t 1 δ [ ] E Ψ exp t tr [σ Ψ ] exp t 1 δ exp t log1 + s 1 δ s We now fix t = s n an s = δ/1 δ to get P tr [σ Ψ ] 1 δ exp [ n δ + log1 δ ] exp n δ / exp n δ /6 This conclues the proof of eq B8