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eavesdropping Optimal in quantum cryptography with six states Dagmar Bru ISI, Villa Gualino, Viale Settimio Severo 65, 10133 Torino, Italy (Received May 7, 1998) A generalization of the quantum cryptographic protocol by Bennett and Brassard is discussed, using three conjugate bases, i.e. six states. By calculating the optimal mutual information between sender and eavesdropper it is shown that this scheme is safer against eavesdropping on single qubits than the one based on two conjugate bases. We also address the question for a connection between the maximal classical correlation in a generalized Bell inequality and the intersection of mutual informations between sender/receiver and sender/eavesdropper. 03.65.Bz, 03.67.-a, 03.67.Dd quant-ph/9805019 7 May 1998 In 1984 Bennett and Brassard [1] suggested a quantum cryptographic protocol, in the following called BB84, which enables two parties to establish a secret key, using principles of quantum mechanics. In this scheme the sender of the quantum information, usually called Alice, transmits quantum bits in the basis | 0 , | 1 or the conjugate basis | , | , 0 1 de ned by | = 0 | = 1 1 (| 0 2 1 (| 0 2 +|1 ) |1 ) (1) to the receiver Bob, who performs measurements in these bases. After classical communication via a public channel a secret key can be established by using only those cases in which the bases of Alice and Bob coincide. In this paper we want to discuss a generalized scheme which is based on the use of three rather than two bases. The third one used in addition to the previous ones is denoted by | , | and de ned by 0 1 | = 0 | = 1 1 (| 0 2 1 (| 0 2 + i| 1 ) i| 1 ) . (2) In the Bloch vector picture a density matrix is written as = 1 (1 + s ), with s being the Bloch vector and l 2 the Pauli matrices. The six states can be viewed as Bloch vectors pointing along the positive and negative x-, y- and z-directions. Alice sends one out of these, denoted as | in , with equal probability. Such a scenario is a straightforward extension of the traditional protocol and its possibility has been mentioned at various occasions [2]. Our main purpose is to point out that this generalized scheme is principally more secure than the one in [1]. This is due to the fact that the optimal strategy an eavesdropper, traditionally called Eve, can design to gather information by performing some unitary transformation on the quantum bit in transit, gives her in our scenario less information for a xed disturbance of Bob s qubit. As Alice increases the set of inputs it is more di cult for Eve to learn something in transit. It was conjectured in [3] and shown in [4] that in the BB84 scenario the disturbance corresponding to Bob and Eve possessing the same information with respect to Alice exhibits a connection to the CHSH inequality. After deriving Eve s optimal strategy we will ask whether in the generalized protocol the crossing point between the two relevant mutual informations has a connection to a generalized Bell inequality where Alice and Bob use the observables ai a and bi b , {i = 1, 2, ..., n}, respectively, with the Bloch vectors ai , bi spanning not only a plane, but the Bloch sphere. After this introduction and outline of the paper let us derive the eavesdropping strategy that is optimal with respect to the mutual information between Alice and Eve, I AE . We do not consider collective or coherent attacks, but only interaction with single qubits. The most general unitary transformation Eve can design is of the form U | 0 | X = F| 0 | A + 1 F| 1 | B (3) U | 1 | X = F| 1 | C + 1 F| 0 | D . (4) The rst qubit is the one sent to Bob and acted on by Eve. Eve s initial state is | X , and | A , | B , | C , | D refer to her normalized states after the interaction. It was shown in [5] that it is su cient for Eve to use two qubits in order to extract the maximal information. The delity of Bob s bit is F and is taken to be in the interval 1/2 F 1. We assume Eve to be clever enough to treat all six possible states in the same way (i.e. with same disturbance for Bob) - otherwise Alice and Bob could nd out about her existence by comparing error rates in di erent bases. This assumption results in three constraints which the scalar products of Eve s states have to ful ll: 1 B |D = 0 , Re C |A = 2 A |B + D |C = 0 . Unitarity of the matrix U means A |D + B |C = 0 . The mutual information between Alice and Bob is given by I AB = 1 + D log D + (1 D) log(1 D) , where D is the disturbance of Bob s qubit, de ned by D = 1 F = 1 in | B 1 , F (5) (6) (7) | in , (8) and B is the right hand side of equations (3) and (4), traced over Eve s bits. All logarithms are taken to base 2. By construction Bob s disturbance is the same no matter which state was sent by Alice. The procedure to calculate the mutual information between Alice and Eve is more involved. We expand | A = A | 00 + A | 10 + A | 01 + A | 11 , where the complex coe cients have to satisfy | A |2 + | A |2 + | A |2 + | A |2 = 1 (10) (9) and similarly for | B , | C , | D . We are free to choose | B as one of the four basis vectors, e.g. | B = | 00 and can ful ll the rst constraint in equation (5) by setting | D = | 11 , without loss of generality. We then nd for the mutual information the form I AE = 1 + 1 ( [F | A |2 + (1 F ), F | C |2 ] + 2 [F | A |2 , F | C |2 ] + [F | A |2 , F | C |2 ] + [F | A |2 + (1 F ), F | C |2 ]) , where we de ne [x, y] = x log x + y log y (x + y) log(x + y) . (12) (11) Note that [x, 1 x] is the entropy function. The task is to maximize I AE with the constraints of eq. (5) and (6). The method of Lagrange multipliers leads to a set of equations which can not be simultaneously ful lled unless A = C = 0 and A = C = 0. This means that the best solution for Eve is to use states such that A |B = 0 = C |D , which one would have expected. Now we have only two parameters left, A and C , and can write I AE = 1 + 1 F ( [| A |2 , | C |2 ] + [(1 | A |2 ), (1 | C |2 )]) , 2 which is a convex function. Here we have used [F x, F y] = F [x, y] . (14) (13) It is straightforward to write down the system of equations which has to be ful lled in order to maximize I AE . Due to their high symmetry one can nd one solution easily, namely | A |2 = 1 | C |2 , and thus I AE = 1 + F [| A |2 , 1 | A |2 ] . 2 (16) (15) By checking the higher derivatives one con rms that this is a maximum, which is, due to convexity, the absolute maximum. Inserting into the second line of eq. (5) allows us to nd the best relative phase between | A and | C and thus leads to the solution for the highest mutual information that Eve can extract from measuring her two qubits, I AE = 1 + (1 D) {f(D) log f(D) + (1 f(D)) log(1 f(D))} , 1 f(D) = 1 1 + D(2 3D) . 2 1 D (17) This function is shown in gure 1, where we also give the corresponding mutual information for BB84, taken from [4], for the purpose of comparison. (I AB is identical in both cases.) Note that our curve lies everywhere below the one for the BB84 case. The six state protocol is therefore more secure against on eavesdropping single qubits. In our case both bits of Eve carry mutual information, unlike the one described by [4]. If she would either measure only one of her two bits, or if she would use a 1-bit probe from the beginning, her maximal information would be I AE,1bit = 1 + f1 (D) log f1 (D) + (1 f1 (D)) log(1 f1 (D)) , f1 (D) = 1 2 1+D+ D(2 3D) , (18) which is the lowest curve in gure 1. The calculation for the 1-bit probe follows the same line as explained above for the 2-bit probe, but is less involved. Note that in order to maximize her mutual information in the six-state scheme Eve necessarily needs two qubits as resource, whereas for BB84 a 1-bit probe is su cient to reach optimality [6]. The scheme described in [1] can also be realized by Alice and Bob sharing a singlet, i.e. a maximally entangled state. This was discussed in [7,8]. In this case, which we will consider for the rest of this article, Alice and Bob can test for eavesdropping by calculating the correlation coe cient in the CHSH inequality. Without any disturbance S, of Bob s bit they will nd | S | = 2 2. This value is decreased when Eve interacts unitarily with Bob s bit. As was shown in the intersection of the two curves for I AB and I AE corresponds to | S | = 2, i.e. at disturbances [4], 1 D 2 (1 1/ 2) the CHSH inequality (between Alice and Bob) is not violated. The natural question arises whether the corresponding intersection for the generalized scheme is related to a generalized Bell inequality. In the six-state protocol the reduced density matrix of Alice and Bob after Eve s interaction reads D 0 0 0 1 0 1 D 2D 1 0 AB = (19) 0 2D 1 1 D 0 2 0 0 0 D where the matrix elements are written in the order 00,10,01,11. For any number of measurement directions that Alice and Bob can use to test a Bell inequality we nd | S(D) | = | Sq | (1 2D) (20) where Sq denotes the correlation for D = 0, i.e. the undisturbed singlet. Thus in our case the measurement directions that are optimal for the singlet are also optimal for D = 0, i.e. a mixed state. This does not hold in general [9]. We will refer to the disturbance where | S(D) | = | Sc |, i.e. where S reaches the classical limit, as Dc . Let us rst look at the case where Alice and Bob are using two measurement directions each that do not necessarily lie in a plane. Here the inequality for a model with local hidden variables reads | S | 2. We can make use of Cirel son s inequality [10] in which the norm of the operator C = a1 a b1 b + a2 a b1 b + a2 a b2 b a1 a b2 b (21) is shown to obey ||C|| 2 2. (Here ai refer to Alices directions of measurement and bi to those of Bob.) This means that the maximal value the quantum correlation can take is | Sq | = 2 2, no matter whether the measurement directions span a plane or a sphere. This value is reached in the CHSH scenario. One can intuitively understand this in the following way: in order to maximize the sum of scalar products of the measurement directions their relative angles have to be as small as possible, i.e. they have to lie on a great circle of the sphere. Thus we cannot nd a ratio for | Sq /Sc | that is higher than 2 and therefore we cannot establish a Bell inequality in the sphere that corresponds to the intersection of I AE with I AB for the generalized protocol, because here Dc is larger than in BB84. We can generally exclude such a correspondence for n measurements by each party, i.e. chained Bell inequalities [11]: the inequality reads now | S | 2n 2. The relevant operator C for this case can be written as a sum of operators of the form used in Cirel son s inequality which we call C1 , ..., Cn 1. Due to the inequality 3 ||C|| = ||C1 + C2 + ... + Cn 1|| ||C1|| + ... + ||Cn 1|| (n 1) 2 2 (22) we know an upper limit of the quantum correlation. Thus we nd | Sq /Sc | 2 as in the paragraph above and can generally exclude the mentioned connection. Note that inequalities like the original Bell inequality and a recent suggestion by Ardehali [12] where two directions of measurement coincide cannot be used for our purpose: the eavesdropping interaction causes the expectation value a a a b to be smaller than 1 if D > 0. To summarize, we have discussed a quantum cryptographic protocol based on six quantum states and shown that it is safer against eavesdropping on single qubits than the BB84 scheme, because Eve s maximal mutual information is smaller than in the BB84 scenario. Furthermore, in order to reach the maximal mutual information the eavesdropper needs to use a two-bit probe and thus has to perform a more complicated transformation than in BB84. If her resource consists of only one qubit, she gains even less information. We have to mention some practical disadvantage: in order to establish a key one will here loose 2/3 of the signals rather than 1/2 in the BB84 scenario, when using equal probabilities for all states. We have also shown that the best way to test a CHSH inequality is to use measurement directions that lie in a plane. In the six-state protocol there is no natural relation between the classical limit of a Belltype correlation coe cient and the intersection of the information curves. We hope that this cryptographic scheme may reach practical relevance in the light of recent suggestions to produce maximally entangled pairs of distant atoms [13]. Valuable critics and comments by Asher Peres and inspiring discussions with Alois W rger are gratefully acknowlu edged. [1] C. H. Bennett and G. Brassard, in Proceedings of the IEEE International Conference on Computers, Systems, and Signal Processing, Bangalore, India (IEEE, New York, 1984), pp. 175-179. [2] C. A. Fuchs, private communication; N. Gisin, Talk given at Torino Workshop, July 1997; A. Peres, private communication. [3] N. Gisin and B. Huttner, Phys. Lett. A 232, 463 (1997). [4] C. A. Fuchs, N. Gisin, R. B. Gri ths, C.-S. Niu, and A. Peres, Phys. Rev. A 56, 1163 (1997). [5] C. A. Fuchs and A. Peres, Phys. Rev. A 53, 2038 (1996). [6] C.-S. Niu and R. B. Gri ths, unpublished; E. Biham and T. Mor; Phys. Rev. Lett. 79, 4034 (1997). [7] A. Ekert, Phys. Rev. Lett. 67, 661 (1991). [8] C. H. Bennett. G. Brassard and N. D. Mermin, Phys. Rev. Lett. 68, 557 (1992). [9] R. Horodecki, P. Horodecki, and M. Horodecki, Phys. Lett. A 200, 340 (1995). [10] B. S. Cirel son, Lett. Math. Phys. 4, 93 (1980); see also A. Peres, Quantum Theory: Concepts and Methods, Kluwer, Dordrecht, 1995, p. 175. [11] L. Braunstein and C. M. Caves, Ann. Phys. (NY) 202, 22 (1990). [12] M. Ardehali, Phys. Rev. A 57, 114 (1998). [13] H.-J. Briegel, W. D r, S.J. van Enk, J.I. Cirac, and P. Zoller, quant-ph/9712027. u 4 1 I^AB I^AE I^AE,1bit I^AE,4states 0.8 0.6 mut. Inf. 0.4 0.2 0 0.05 0.1 0.15 0.2 0.25 D 0.3 0.35 0.4 0.45 0.5 FIG. 1. Maximal mutual information I AE between Alice and Eve as function of Bob s disturbance D. The upper curve holds for BB84 [4] and is shown for the purpose of comparison. The lower curves refer to the six-state protocol. Their analytic forms are shown in equations (17) and (18). The mutual information between Alice and Bob is in both scenarios given by the curve I AB . 5

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