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Transactional Nature of Quantum Information Subhash Kak Computer Science, Oklahoma State Univ © Subhash Kak, June 2009 1 Quantum information and processing • Fourier transform in (log n)2 rather than n log n • Database search in n1/2 rather than n [Grover] • Factorization in log n rather than n1/2 [Shor] 2 Qubits versus bits |Ψ> = a |0> + b |1> where |a|2 + |b|2 = 1 |Ψ> 3 Qubit dynamics A quantum computer (U) rotates a state |Ψ> 4 Quantum register Each cell has a qubit. Number of states is 2n 1 2 3 4 5 . . . . . n 5 Information and Entropy Classical information I(x) = - log p(x) Information is additive Classical Entropy H(X) = - ∑i p(xi) log p(xi) Von Neumann entropy Sn ( ) tr( log ) = - ∑i λi log λi 6 Von Neumann entropy Entropy of the mixed state p 0 is equal to 0 1 p p log p (1 p) log( 1 p) 7 Mixed states 3 1 | | 00 | | 11 | 4 4 3 / 4 0 0 1 / 4 • Its von Neumann entropy equals 0.81 bits. This mixed state can be viewed to be generated from a variety of ensembles of states. 8 An Entangled State 1 | (| 00 | 11 ) 2 .5 0 0 .5 0 0 0 0 0 .5 0 0 0 0 0 .5 9 The von Neumann entropy is zero. But the two objects individually have entropy of 1 bit. .5 0 0 .5 Von Neumann entropy is not additive. An entangled state is pure and its entropy is zero, but the state of its components is mixed and their entropy is non=zero! 10 Multiple copies of a state Consider multiple copies of the qubit | a | 0 b | 1 1 (tr( ) I tr( X ) X tr(Y )Y tr( Z ) Z ) 2 where tr(Xρ) etc are average values 11 A more constrained example Preparer of states A trying to communicate the ratio a/b = m to B, where a and b are real and the state is either pure or a mixture a m 1 m2 b 1 1 m2 It is the pure state: | a | 0 b | 1 or it is the mixture of the states: | 1 a | 0 b | 1 | 2 a | 0 b | 1 12 Informational entropy A proposed measure for informational entropy: Sinf ( ) ii log ii i It depends only on the diagonal terms and therefore it reflects the mutual relationship between the sender and the recipient 13 0 ≤ Sinf(ρ) ≤ n Sinf ( A, B) Sinf ( A) Sinf ( B) Sinf ( pi i ) pi Sinf ( i ) i i Sinf (ρ) ≥ S n ( ) 14 Example 0.71 0.15 0.15 0.29 Sinf(ρ) = -0.71 log2 .71 – 0.29 log2 .29 = 0.868 bits The eigenvalues of ρ are 0.242 and 0.758 and, therefore, the von Neumann entropy is: =-0.242 log2 .242 – 0.758 log2 .758 =.2422.047 + .758.400=0.798 bits 15 Von Neumann entropy is less than Inf. Entropy by 0.07 bits Example (contd.)-- Partial Information case 1 We know that the ensemble consists of a pure and mixed component as follows: .5 .5 .8 0 0.3 0.7 . 5 . 5 0 . 2 Entropy is 0.3 + 0.70.722= 0.805 bits 16 Example (contd.)-- Partial Information case 2 We know that the ensemble consists of a pure and mixed component as follows: 0.316 2 3 2 3 2 0 3 0.684 0.731 0 0 . 269 1 3 Entropy is 0.3160.918+0.6840.84 =0.290+0.575=0.865 bits 17 Quantum cryptography protocol 18 Concluding Remarks • If information cannot be defined independent of the experimenter, quantum computing may be harder to implement • Amongst other things, it implies greater attention to errors • Is transactional nature of quantum information of relevance to other fields of physics? 19 References • S. Kak, Prospects for quantum computing. arXiv:0902.4884v1 • S. Kak, Quantum information and entropy, International Journal of Theoretical Physics 46, pp. 860-876, 2007. • S. Kak, Information complexity of quantum gates, International Journal of Theoretical Physics, vol. 45, pp. 933-941, 2006. • S. Kak, Three-stage quantum cryptography protocol, Foundations of Physics Letters, vol. 19, pp. 293-296, 2006. • S. Kak, General qubit errors cannot be corrected, Information Sciences, vol. 152, pp. 195-202, 2003 20