 Journal of Quantum Information Science, 2013, 3, 6-9 http://dx.doi.org/10.4236/jqis.2013.31002 Published Online March 2013 (http://www.scirp.org/journal/jqis) Re-Formulation of Mean King’s Problem Using Shannon’s Entropy Masakazu Yoshida, Hideki Imai Graduate School of Science and Engineering, Chuo University, Tokyo, Japan Email: masakazu-yoshida@imailab.jp Received November 23, 2012; revised December 30, 2012; accepted January 8, 2013 ABSTRACT Mean King’s problem is formulated as a retrodiction problem among noncommutative observables. In this paper, we reformulate Mean King’s pr oblem using Shannon ’s entropy as the first step of introducing qu antum uncertainty relatio n with delayed classical information. As a result, we give informational and statistical meanings to the estimation on Mean King problem. As its application, we give an alternative proof of nonexistence of solutions of Mean King’s prob- lem for qubit system without using entanglement. Keywords: Mean King’s Problem; Q uantum Retrodiction Problem; Quantum Estimation Problem; Shannon’s Entropy 1. Introduction In 1987, Vaidman, Aharonov, and Albert [1] introduced Mean King’s problem as a challenge to an uncertainty principle among noncommutative observables. The pro- blem can be interpreted as a kind of quantum estimation (or states discrimination) problem with a delayed clas- sical information. In the proposed setting, two players King and Alice play their roles: King asks Alice to prepare qubit system in an arbitrary state. King measures the system with a projective measurement relevant to one of observables , y , and . Alice is permitted to measure the post measurement state once in an arbitrary measurement. After Alice’s measurement, King reveals the kind of observable employed by him to Alice. Then, Alice should retrodict King’s outcome by using her outcome and the kind of observable. It is a problem to construct a pair of an initial quantum state and a meas- urement e mployed by Alice such that she estimates King’s outcome with Probability 1, in which case we say that there exists a solution to the problem. In the original work [1], it is shown that there is such a pair provided that Alice uses an entanglement: That is, Alice prepares not only one qubit system but also an ancillary qubit system secretly in the Bell state. Then, Alice gives one of the systems to King and the other system is kept by herself. She measures the bipartite sys- tem in the post measurement state and then can retrodict the King’s ou tput after King ’s reveal of his measurement kinds. Mean King’s problem has been generalized concern- ing the prepared quantum system and King’s measure- ments [2-7]. In particular, it has been proved [2-5] that Alice can estimate King’s outcome by using a maximally entangled state in a setting that King measures one of the systems with one of projective measurements constructed from Mutually Unbiased Basis [8,9]. On the other hand, Alice cannot retrodict the outcome with certainty without using entangled states in the setting [6,7]. In the refer- ence, an upper bound of the success probability is also introduced. In this paper, we reformulate Mean King’s problem from a viewpoint of Shannon’s entropy. We can naturally characterize the solution by means of the zero condi- tional entropy of King’s outcome given Alice’s out- come an d ki nd o f Kin g’ s measurement. As its application, we give an alternative proof of nonexistence of solu tions of Mean King’s problem for qubit setting without using any entangled state. This paper is organized as follows. In the next section, we introduce a general setting of Mean King’s problem. In Section 3, we reformulate the problem using Shan- non’s conditional entropy. In Section 4, we give an alter- native proof of nonexistence of solutions. Finally, in Sec- tion 5, we summarize this paper. 2. Setting of General Mean King’s Problem We introduce a general setting of Mean King’s problem. The problem is constructed from the following steps: 1) By the King’s order, Alice prepares a quantum system S, described by a d-dimensional Hilbert space , in an arbitrary initial state. 2) King performs one of measurements C opyright © 2013 SciRes. JQIS  M. YOSHIDA, H. IMAI 7 0,0,1,, m kk jj Mk m constructed from mea- surement operators on the system and obtains an out- come . [Remind that 0 m k j M satisfies † 0 mkk jj j M I , pro- bability of obtaining an outcome from a measure- ment of a state is given by †kk jj trM M , then the post measurement state is given by ††kk kk jj jj MM trMM [10,11]]. 3) Alice performs a POVM (Positive Operator Valued Measure) measurement on the system in the 0 n ii RR post measurement state and obtains an outcome . i [Remind that is called a POVM if 0 n ii R0 n i i RI and hold for any i [10,11]]. 0 i R 4) After Alice’s measurement, King reveals the kind of measurement k to Alice. 5) Alice tries to estimate King’s outcome perfectly with her measurement outcome i and King’s meas- urement . k With given and measurements d 0 m kk j MM , we say that a solution to the Mean King’s problem exists if and only if a pair of an initial state and a measurement employed by Alice exist such that she estimates King’s outcome with Probability 1. Notice that Alice can utilize an entanglement: In step 1), she secretly prepares an ancilla system A and chooses an appropriate entangled state on the bipartite system S A. In Step 3), she performs a general measurement (POVM measurement) on the bipartite system. SS In the next section, we reformulate the problem using Shannon’s cond itional entro py. 3. Re-Formulation of the Problem Let be random variables expressing the kind of the measurements employed by King, the outcomes obtained by King, and the outcomes obtained by Alice’s measurement R, respectively. Then, we can reformulate the Mean King’s problem using the conditional entropy as follows: ,,KJI Find an initial state a nd a measurem ent such that R ,HJIK0, (1) where H denotes Shannon’s cond itional entropy. Note that HJI is generally strictly positive, other- wise Alice can guess King’s outcome without a delayed information K. By the chain rule of the conditional entropy, Equation (1) is equivalent to the following relation: ,HKJI HKI . (2) Let ,, ,, KJI Pkji be a joint probability of , ,,KJI and let ,,, , KI KJI j P kji ,,Pki be the marginal joint probability of and . We find that Equation (2) is equivalent to I ,,,, , ,,0or,,, , KJIKJI KI Pkji PkjiPki (3) for each . Indeed, by the definition of conditional entropy, we can rewrite Equation (2) as follow: ,,KJI ,, ,, ,, ,, , ,, log, ,log , KJI KJI kji KI KI ki PkjiPkj Pki Pki i (2’) where P denotes a conditional probability corre- sponding to the random variables. If ,, ,,, 0 IKJI kj PiPkji holds, using the monoto- nically increasing property ,, , log, ,log, KJI KI Pkji Pki, Equation (2’) is equi- valent to ,, ,, ,, , ,, log,, ,, log, , KJI KJI KJI KI Pk jiPk ji PkjiPki (2”) for any . Noting that ,,kji 0 I Pi holds if and only if ,, 0 KJI ji,,Pk for any , Equation (2’) is equivalent to (2”) also in this case. Therefore, we have obtained the equivalence between Equations (2) and (3). In our setting, a solution to the Mean King’s problem is to find an initial state ,ki and a measurement such that Conditions (1), (2), or (3) holds. R 4. Nonexistence of Solutions in Qubit Setting In this section, we give an alternate proof of nonexis- tence of solutions to Mean King’s problem withou t using entanglement in qubit system. In the setting, Alice pre- pares not bipartite system but one qubit in a state . Recall that qubit is described by 2-dimensional complex vector space . King employs one of three projective measurements, 2 0,1 :0, kkkk jjj j MM k ,1,2 , where 0,1 k jj are three kinds of orthonormal basis on , e.g., three pairs of eigenvectors cor responding to Paulli matrices 2 , y , and . The post measurement state is k j if King chooses and obtained an Kk outcome from the projection postulate. After that, Alice measures qubit in the post measurement state with a POVM measurement . Then, we obtain 0,1 ii RR Copyright © 2013 SciRes. JQIS  M. YOSHIDA, H. IMAI 8 the following joint probability, ,, ,, kkk KIJ k jjji PPkji Rkj , (4) where K Pk denotes the pro bability that King chooses the projective measurement k . For a fixed , we ob- serve that there are three types A, B, C of the joint prob- abilities satisfying Equation (3) characterized as follows: k Type A: There uniquely exists a pair of outcomes ,ji such that ,, ,, 0 KJI Pkji holds. ,, ,,0 KJI Pkji holds for any ,,ji ji . Type B: There uniquely exists an out come such that ,, ,, 0 KJI Pkji holds for any . i ,, ,, 0 KJI Pkji holds for and any i. jj Type C: ,, ,, 0 KJI Pkji, ,, ,, 0 KJI Pkji , , and ,,KJI Pk ,, 0ji ,,KJ ,, 0Pkji I hold for ii and . jj In Figure 1, we show a complete classification of probabilities for each type, wh ere the number of kinds of the probabilities satisfying Type A, Type B, and Type C is 8. Now, we try to find and such that each three joint probabilities for satisfies any of the above 8 kinds of the probabilities. R , 20,1k By using Equation (4), we obtain the equivalent rela- tions for each types and the joint probability ,, ,, KJI Pkji as follows: The joint probability satisfies Type A if and only if 00 0 ,, kk MR M 1 1 k R M or 00 11 0 ,, kk MRMR M k or 1001 1 ,, kk MRMR M k or 10 11 0 ,, kk MRMR M k hold. The joint probability satisfies Type B if and only if 0 1 , kk MM or k hold. 000 11 ,,, kkk MR MMR 100 110 1 ,,,, kkkk MRMMR MM The joint probability satisfies Type C if and only if 010 01 ,, , kk k MMR M 1 k R M or 010 11 0 ,, , kkk k MMRMR M hold. Let us focus on two probabilities ,, ,, KJI Pkji and ,, ,, KJI Pkj k i with . If both are Type A, kk 0 or 1 k holds for and 0 kk or 1 k holds for. Therefore, we cannot construct the prob- ability satisfying a pair of (Type A, Type A). This fact is also derived from construction of 0 and 1. In a simi- lar way, we cannot construct the probability satisfy- ing any of pairs of types (Type B, Type B), (Type C, Type C), (Type A, Type B), (Type B, Type A), (Type C, Type A), and (Type A, Type C). On the other hand, there are k R R Figure 1. Three types of the probabilities. and such that the probabilities satisfy any of pairs of (Type B, Type C) and (Type C, Type B). For instance, R 0001 1 ,, kk MRM RM k s atisfies (Type B, Type C). According to the above fact, we obtain ,0HJIK for two kinds of the projective measurements k and k . However, it turns out that Alice cannot find a solu- tion for three kinds of projective measurement as fol- lows: First, from the above discussion, candidates of possibly pairs are (Type B, Type C, Type B) and (Type C, Type B, Type C) corresponding to . However, 0,1,2k the first one, 0 0 or 0 1 holds for Type B of 1st term and 2 0 or 2 1 holds for Type B of 3rd term. Therefore, the first one is ruled out of the candidate. In a similar way, the second one is also ruled out of the can- didate from a viewpoint of the measurement . Thus, we can conclude that R ,HJIK0 dose not hold for three kinds of the measurements. 5. Conclusion We reformulated Mean King’s problem and gave new insight to the problem from a view point of Shannon’s entropy. As its application, we gave an alternative proof of nonexistence of solutions for qubit setting without using entanglement. We expect that new insights from viewpoints of quantum probabilistic theory, quantum Copyright © 2013 SciRes. JQIS  M. YOSHIDA, H. IMAI Copyright © 2013 SciRes. JQIS 9 communication, and so on will be given to Mean King’s problem by using the reformulation given in this paper. REFERENCES [1] L. Vaidman, Y. Aharonov and D. Z. Albert, “How to Ascertain the Values of σ x, σ y, and σ z of a Spin-1/2 Particle,” Physical Review Letters, Vol. 58, No. 14, 1987, pp. 1385-1387. doi:10.1103/PhysRevLett.58.1385 [2] B.-G. Englert and Y. 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