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qitd352 - qitd352 Dense Coding Teleportation No Cloning...

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qitd352 Dense Coding, Teleportation, No Cloning Robert B. Gri ffi ths Version of 8 February 2012 References: NLQI = R. B. Gri ffi ths, “Nature and location of quantum information” Phys. Rev. A 66 (2002) 012311; http://arxiv.org/archive/quant-ph/0203058 QCQI = Quantum Computation and Quantum Information by Nielsen and Chuang (Cambridge, 2000). Look up “superdense coding”, “teleportation”, “no-cloning” in the index. Add p. 187 to the teleportation references. Contents 1 Fully Entangled States 1 2 Dense Coding 2 3 Teleportation 4 4 No Cloning 7 1 Fully Entangled States As previously noted, entangled states on a tensor product are peculiarly quantum in the sense that there is no good classical analog for them. Dense coding and teleportation are two processes which make use of entangled states, and for this reason appear somewhat strange from an everyday “classical” perspective. We shall later introduce a measure of entanglement for pure states, but for the moment all we need are fully entangled (or maximally entangled ) states. Let H a and H b be two Hilbert spaces of the same dimension d a = d b = d . Any state on | Ψ on H = H a H b can be written in the Schmidt form: | ψ = j λ j | a j | b j , (1) where {| a j } and {| b k } are suitable orthonormal bases (which depend upon | ψ ). A fully entangled state is one for which all the λ are equal (or equal in magnitude if one does not impose the condition λ j > 0), and thus equal to 1 / d if | ψ is normalized. In the case of two qubits, d = 2, the Bell states | B 0 = ( | 00 + | 11 ) / 2 , | B 1 = ( | 01 + | 10 ) / 2 , | B 2 = ( | 00 - | 11 ) / 2 , | B 3 = ( | 01 - | 10 ) / 2 , (2) are examples of fully-entangled states which form an orthonormal basis. Fully entangled states can also be characterized in the following way. Let the reduced density operators for a normalized | ψ on H a and H b be defined in the usual way: ρ a = Tr b ([ ψ ]) , ρ b = Tr a ([ ψ ]) . (3) For a fully entangled state, ρ a = I/d = ρ b . (4) 1
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Note that we are assuming that the reduced density matrices come from a pure state | ψ on H a H b , and not from a mixed state represented by a density operator. (Entanglement for mixed states is a complex problem which is far from well understood at the present time.) Exercise. Show that only one of the equalities in (4) is actually needed, as the second is a consequence of the first (and vice versa). Fully entangled states do not have a unique Schmidt decomposition. Given any orthonormal basis {| a j } of H a , there is an orthonormal basis {| b k } of H b , one that depends both on {| a j } and on | ψ , such that (1) holds (with λ j = 1 / d ). Exercise. Prove this assertion by expanding | ψ in the form j | a j | β j , and using (4). Given two normalized fully entangled states | ψ and | φ on H a H b one can always find unitary operators U and V on H a and H b such that | φ = ( U V ) | ψ . (5) That there is some unitary operator W on H a H b mapping | ψ to | φ is a consequence of the fact that they have the same norm. What is special about (5) is that W is of the form U V . It will be convenient to refer to call such an operator a local unitary . The idea of “local” is that one thinks of the subsystems
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