In how many ways can quantum information be split ?
Sreraman Muralidharan, Siddharth Karumanchi, Srinatha Narayanaswamy, R. Srikanth, Prasanta K. Panigrahi
aa r X i v : . [ qu a n t - ph ] M a r In how many ways can quantum information be split ?
S. Muralidharan, S. Karumanchi, S. Narayanaswamy, R. Srikanth,
3, 4 and P. K. Panigrahi ∗ School of EPS, Heriot-Watt University, Edinburgh, EH144AS, United Kingdom Birla Institute of Technology and Science, Pilani, Rajasthan- 333031, India Poornaprajna Institute of Scientific Research, Sadashivnagar, Bangalore 560 080, India Raman Research Institute, Sadashiva Nagar, Bangalore 560012, India Indian Institute of Science Education and Research - Kolkata, Mohanpur - 741252, India
We establish a theoretical understanding of the entanglement properties of a physical systemthat mediates a quantum information splitting protocol. We quantify the different ways in whichan arbitrary n qubit state can be split among a set of k participants using a N qubit entangledchannel, such that the original information can be completely reconstructed only if all the partic-ipants cooperate. Based on this quantification, we show how to design a quantum protocol withminimal resources and define the splitting efficiency of a quantum channel which provides a way ofcharacterizing entangled states based on their usefulness for such quantum networking protocols. PACS numbers: 03.67.Hk, 03.65.UdKeywords: Quantum information splitting, Teleportation
Introduction-
Splitting and sharing of secret informa-tion among a group of parties such that none of them cancompletely reconstruct the secret information by them-selves is a common requirement in financial and defencesectors [1]. Recently, it has been found that certain as-pects of quantum mechanics such as entanglement [2]can be effectively used for the splitting and sharing ofsecret information which can be either ‘classical’ (bits)or ‘quantum’ (qubits) [3]. Sharing of quantum informa-tion among a group of parties such that none of themcan reconstruct the unknown information completely byoperating on their own share is usually referred to asQuantum Information Splitting (QIS). Starting from theseminal work of Hillery et al. [4] who used a N -qubitGreenberger-Horne-Zeilinger (GHZ) state for the split-ting up of an unknown single qubit among many parties,several entangled states were found to be useful for thispurpose [5–10]. Among these, ‘graph-states’ are of par-ticular interest as they have been experimentally realizedupto six-qubits [11, 12]. A schematic view of a standardQIS protocol is presented in Fig. 1. FIG. 1:
Alice, Bobs and Charlie share an entangled chan-nel. Alice performs a joint-measurement on the secret andher share of the entangled channel. Each party performs ameasurement on his/her qubits and subsequently, the secretis split among other parties.
In a realistic situation, for the splitting of an arbitrarynumber of qubits, many parties need to be in an entan-gled quantum network. Keeping in mind the complex-ity of the multipartite entangled system, there will arisemore than one way of splitting and sharing of secret in-formation, given a fixed number of parties. For instance,if a dealer want to split an unknown single qubit state | ψ i = α | i + β | i ( α, β ∈ C, | α | + | β | = 1) amongtwo parties, using a symmetric four-qubit entangled GHZstate, | φ i as a resource [4], this could be done in twodifferent ways as their combined state could be writtenas | ψ i ⊗ | φ i = X i =1 2 X j =1 ( | ψ i i ⊗ | φ i j ⊗ | φ i j ) (1)= X i =1 2 X j =1 ( | ψ i i ⊗ | φ i j ⊗ | φ i j ) . In the first case, | ψ i i ’s forms four orthogonal two-qubitmeasurement outcomes of the dealer and | φ i j ’s formstwo two-qubit orthogonal measurement outcomes of theintermediate party. In the second case, | ψ i i ’s formsthe four orthogonal three qubit measurement outcomesof the dealer and | φ i j ’s forms the two, single qubit mea-surement outcomes of the intermediate party. The dealercan split the outcome of his/her measurement using a(2,2)-threshold classical secret sharing (CSS). If the par-ticipants co-operate, the receiver can covert | φ i j to | ψ i by applying an appropriate unitary operation.Note that all distributions of qubits among the partiesdon’t yield successful QIS protocols. Further, one needsto choose an entangled channel with appropriate num-ber of qubits given the number of qubits in the secretand the number of parties among whom this secret in-formation has to be split. In fact, it has been observedthat a four-qubit linear cluster state could not be usedfor the QIS of an unknown two-qubit state while a fivequbit linear cluster state could be used for this purpose[6]. In this Letter, we study the entanglement propertiesof the physical system which mediates this QIS proto-col and quantify the number of ways in which quantuminformation could be split among a given number of par-ties. This quantification provides a recipe for the designa QIS protocol with minimal entangled resources. Basedon this study we define splitting efficiency of a quantumchannel, which provides an alternative method for char-acterizing quantum states based on their usefulness forQIS. The introduced model seems to be naturally appli-cable for many quantum networking protocols [13] thatrelies on entanglement and teleportation [14]. We estab-lish the quantification first and then the efficiency of QISin the later part of the letter. Protocol count-
Following theorems yield the proto-col count for the QIS of an unknown n -qubit state | ψ n i = X i ,...i n =0 α i ,...i n | i , ...i n i , where α i ,...i n ∈ C and P | α i ,...i n | = 1. Theorem 1
If Alice, Bob(s) and Charlie share an N qubit entangled state and Alice has a arbitrary n qubitstate | ψ n i that she wants the Bobs and Charlie to share,then Alice needs to possess a minimum of n qubits forthis purpose.Proof: We conflate all Bobs and Charlie into a singleagent Dolly, equipped with Hilbert space H D , which isthe tensor product of the Hilbert spaces of the Bobs andCharlie. Information splitting can be considered as aquantum teleportation from Alice to Dolly. Teleporta-tion with unit fidelity will not be deterministically pos-sible unless maximal entanglement exists between Aliceand Dolly [16]. By virtue of Schmidt decomposition fora bipartite pure state, Dolly’s density operator will bemaximally mixed in a 2 n -dimensional subspace of H D . Ifwe let Alice possess m qubits in the entangled quantumnetwork then it can be proved that m ≥ n from a quan-tum encryption perspective as follows: After Alice’s jointmeasurement in H x ⊗ H A , but before her classical com-munication to Dolly, the no-signalling theorem demandsthat Dolly’s density operator should not have changed,i.e., it must remain maximally mixed within the rele-vant subspace. Here, H x and H A refer to the respectiveHilbert spaces of the unknown secret information and Al-ice’s part of the entangled network respectively. On theother hand, we know by the way deterministic telepor-tation works (for the considered, maximally entangledstates) that Dolly’s state has become: T : | ψ i −→ P X j =1 U j | ψ ih ψ | U † j (2)Alice’s classical communication will be the number j which will allow Dolly to apply operation U j that restores her object’s state to | ψ i . We require the minimal num-ber P in Eq. (2) such that for an arbitrary input state | ψ i , we obtain T ( | ψ i ) = I /D , where I is the unit matrixand D = 2 n is dim( H B ). According to Ref. [17], whichprovides a protocol for classically encrypting a quantumstate, P = D . This in turn means that Alice’s classicalcommunication must be at least log( D ) = 2 n bits long.In turn this means that the outcome of Alice’s measure-ment, which could be encoded into ( m + n ) bits, mustsatisfy m + n ≥ n , or, m ≥ n , as required. (cid:4) It is worth mentioning that Theorem 1 also follows as aconsequence of the Choi-Jamiolkowski Isomorphism [15].
Theorem 2
It is necessary for the recipient’s system tobe in a maximally mixed state, but not for that of anyintermediate party P .Proof: By application of the Agrawal-Pati theorem [16],the n qubits with the recipient should be maximallymixed. The reduced density operator for any interme-diate party P , however, need not be maximally mixed inits support. To see this, consider the scenario when threeparties Alice, Bob and Charlie share an entangled stateof the form | ζ i ≡ cos θ | + i B | ψ − i AC + sin θ |−i B | ψ + i AC ,where, |±i = √ ( | i ± | i ) and θ ∈ [0 , π/ θ = π/
4. SinceAlice’s and Bob’s measurements commute, and Bob doesnot use Alice’s classical communication, Bob might aswell measure first. If he does to obtain the outcome |±i clearly Alice is able to deterministically teleport to Char-lie with fidelity 1. This is alternatively confirmed by see-ing that Alice’s and Charlie’s reduced density operatorfor | ζ i is maximally mixed (in its support). These obser-vations give us Theorem 2. (cid:4) Lemma 1
The maximum number of protocols one canconstruct for this purpose is ( N − n ) .Proof: We let the third person (say Charlie) have thelast n qubits, on which he will apply a suitable localunitary transformation and reconstruct the unknown n qubit information. Therefore Charlie will possess, ( N − n + 1) th qubit to the N th qubit. Now, the first ( N − n )qubits need to be distributed among Alice and Bob. Thiswould correspond to ( N − n −
1) protocols. However, fromTheorem 1, all the protocols, in which Alice possesses lessthan n qubits fail. Hence, the total number of protocolsthat one can construct is ( N − n ). For at least oneprotocol to work out, we have N ≥ (2 n + 1), yieldinga recipe for the design of a QIS protocol with minimalentangled resources. (cid:4) Corollary.
By substituting N = 4 and n = 2 in thisformula, we can deduce that four qubit states cannot beused for the QIS of an unknown two qubit state | ψ i .This shall be illustrated below as follows. We let Alicepossess the unknown two qubit state | ψ i and qubit 1,Bob possess qubit 2 and Charlie 3,4 in the four qubit clus-ter state [6], | C i = ( | i + | i + | i−| i ) respectively. Alice can perform a three partite measure-ment on | ψ i as in the above protocol. For instance,if Alice performs a three-qubit measurement and ob-tains the outcome ( | i + | i + | i − | i ), then,the Bob-Charlie system collapses to α ( | i + | i ) + α ( | i + | i )+ α ( | i−| i )+ α ( | i−| i ).However, from the above state one cannot obtain | ψ i , byperforming another measurement on some of its qubits toget | ψ i . This could be seen more clearly as follows, IfBob performs a single particle measurement and obtainsthe outcomes √ ( | i ± | i ), then Charlie’s state collapsesto α ( | i + | i ) + α ( | i + | i ) + α ( | i − | i ) + α ( | i − | i ). If Charlie can apply an operation U andobtain | ψ i , it can be clearly seen using simple matrixalgebra that there exists no unitary operator U , which isindependent of the coefficients that carries out the task.Since, the coefficients of | ψ i are unknown, we concludethat this protocol fails illustrating the usefulness of thetheorem.We define P k ( N ) as the number theoretic partitions, Q k ( N ) as the ordered partition of N restricted to k terms, C to be the binomial coefficient and in general Q k ( N ) ≥ P k ( N ). On account of Theorem 1, the dealermust have at least n qubits to initiate the QIS protocoland the receiver must have precisely n qubits (ignoringextraneous qubits) to be able to reconstruct the secret.Depending on the pre-shared entanglement chosen, theprotocol may be symmetric between the Bobs, in whichcase interchanging two Bobs is equivalent or completelyasymmetric between all of them. Theorem 3 If k (3 ≤ k ≤ N − n + 2) parties sharean N qubit entangled state and the first party has anarbitrary n qubit state that he/she wants the remain-ing members to share, then the maximum number ofprotocols that can be constructed for this purpose is P N − nj = k − P k − ( j ) in the symmetric case, and boundedabove by P N − nj = k − Q k − ( j ) = N − n C k − in the generalcase. Proof.
The minimum number of qubits with theBobs is ( k −
2) and the maximum number, ( N − n ).Fix the number of qubits with all Bobs together to be j . The number of protocols in the symmetric case isthe number of ways j can be partitioned into k − P k − ( j ). Summing over all possible j ’s gives thetotal number of protocols in the symmetric case. Ifthe state is such that each Bob is inequivalent toany other, then clearly the number of protocols is P N − nj = k − Q k − ( j ) = P N − nj = k − j − C k − = N − n C k − , forit can be shown that Q l ( m ) = m − C l − . In general, thisis an upper bound on the number of protocols, as there will be partial symmetry among the Bobs. In order foratleast one protocol to work out we have N − n ≥ k − (cid:4) While Theorem 1 specifies the threshold size of Al-ice’s system to be useful for QIS of completely unknownquantum states, one may ask about the threshold sizerequired for the QIS of specific types of quantum statesof the form | φ n i = α | i ⊗ n + β | i ⊗ n . This leads us toTheorem 4. Theorem 4
If Alice, Bob and Charlie share an N -qubitentangled state and Alice has an (entangled) n -qubit en-tangled state of the form | φ n i = α | i ⊗ n + β | i ⊗ n thatshe wants Bob and Charlie to share, then Alice needs topossess only one qubit for this purpose. Proof:
Consider a scenario where Alice, Bob and Char-lie share a symmetric N -qubit GHZ state of the form, | φ GHZ i = √ ( | i N + | i N ), where N = 2 + n , with thefirst qubit with Alice, second with Bob, and the remain-ing with Charlie. Alice can perform a ( n + 1) particlemeasurement on her qubits and conveys its outcome toCharlie (or Bob) using ( n +1) classical bits. Her outcomeand the corresponding Bob-Charlie states are:Alice’s Outcome Bob-Charlie State | ψ i α | i ⊗ ( n +1) + β | i ⊗ ( n +1) | ψ i α | i ⊗ ( n +1) − β | i ⊗ ( n +1) | ψ i α | i ⊗ ( n +1) + β | i ⊗ ( n +1) | ψ i α | i ⊗ ( n +1) − β | i ⊗ ( n +1) where √ | ψ , i = | i ⊗ ( n +1) ± | i ⊗ ( n +1) and √ | ψ , i = | i ⊗ n ⊗ | i ± | i ⊗ n ⊗ | i form mutual orthogonal out-comes of measurement. Using Alice’s 2-bit communica-tion, Charlie (or Bob) can apply the single-qubit Paulioperation I, Z, X or Y to bring their entangled state tothe form α | i ⊗ ( n +1) + β | i ⊗ ( n +1) . By measuring each ofhis/her qubit in a suitable basis to obtain the outcome |±i and communicating his 1-bit outcome to Charlie, thelatter can reconstruct | φ n i . If Alice’s 2-bit outcome isknown, Bob or Charlie possess partial information about α or β , and hence about | φ n i . As before, this partialinformation can be eliminated by having Alice split her2-bit information between Bob and Charlie using a (2,2)-threshold CSS scheme. (cid:4) Lemma 2
The number of protocols one can construct forthis purpose is N − n C k − + n − . Proof.
It was shown in Lemma 1 that one can construct( N − n ) protocols for k = 3 in the case where Alice hadto possess at least n qubits. But, for the present casethe protocols in which Alice possesses less than n qubitsalso work for n ≥
2. Given that the recipient has n qubits, the remaining ( N − n ) qubits is to be split amongthe remaining parties, this can be done in ( N − n − k ≥
3, one can construct N − n C k − + n − (cid:4) Efficiency of Information Splitting-
The above theoremsnaturally lead to questions about the efficiency of a quan-tum channel for QIS. We define the ”splitting efficiency”( η ) of a quantum channel as, η = n n max P n max n =1 nζ n P n max n =1 nζ ′ n , (3)where ζ n refers to the number of protocols that can beconstructed for splitting up of | ψ n i among k parties for agiven entangled channel, ζ ′ n refers to the maximum num-ber of protocols that can be constructed for splitting upof | ψ n i among k parties as given by Theorem 3. Here, n max = ⌊ N − k +22 ⌋ does not depend on the particular chan-nel, but on N and k only and n is the largest size of asecret (in qubit units) that can be split with N qubitentangled channel among k parties. To illustrate this,we consider a maximally entangled N qubit GHZ stategiven by, | φ GHZ i From Theorems 1 and 2, we can notethat | φ GHZ i can be used for the QIS of only unknownsingle qubit state | ψ i . From Eq. (3), we can calculateits the splitting efficiency as η GHZ = (1 /n max ). For a N qubit linear cluster state [18], which can be used for theQIS of both arbitrary single and two qubit systems, thesplitting capacity is given by, η lc = (1 /n max ) if N = 3 , /n max ) if N ≥
5. As it can be seen, η lc > η GHZ for
N > N clusterstates ( Box states) [19] exhibit even more interestingproperties than the linear cluster states as all the possi-ble protocols work out for QIS among three parties. Thesplitting capacity of box states is given by, η Box = 1. Adetailed comparison of the splitting efficiencies for theseentangled channels is shown in Fig. 2.
Conclusions-
We have established a formal under-standing of the properties of a physical system whichmediates an information splitting protocol and quantifiedthe number of ways of achieving the same. The study isapplicable to a wide range of systems which need not bemaximally entangled (as given by Theorem 2) and is nat-urally applicable for different kinds of quantum network-ing protocols that relies on teleportation. Subsequently,the study allows the characterization of multi particleentangled states in terms of its ‘information splitting ef-ficiency’. This result helps us to design an optimal quan-tum network with minimal resources and paves a way to-wards the understanding of the usefulness of higher-qubitentangled states for quantum networking protocols. ∗ Electronic address: [email protected] S p li tt i ng e ff i c i en cy Splitting efficiency of different types of entangled states GHZ statesLinear cluster statesBox states0 5 10 15 20 25 30 35 40 45 5000.511.5 Number of parties (k) M a x i m u m S p li tt i ng e ff i c i en cy Maximum splitting efficiency vs Number of parties GHZ statesLinear cluster statesBox states
FIG. 2:
The variation of splitting capacity with a) the numberof qubits for three parties, b) the number of parties [1] Bruce Schneier,
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