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Unformatted text preview: qitd352 Dense Coding, Teleportation, No Cloning Robert B. Griffiths Version of 8 February 2012 References: NLQI = R. B. Griffiths, Nature and location of quantum information Phys. Rev. A 66 (2002) 012311; http://arxiv.org/archive/quantph/0203058 QCQI = Quantum Computation and Quantum Information by Nielsen and Chuang (Cambridge, 2000). Look up superdense coding, teleportation, nocloning 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 Full y Entangl ed 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  on H = H a H b can be written in the Schmidt form:  = j j  a j  b j , (1) where { a j } and { b k } are suitable orthonormal bases (which depend upon  ). A fully entangled state is one for which all the are equal (or equal in magnitude if one does not impose the condition j > 0), and thus equal to 1 / d if  is normalized. In the case of two qubits, d = 2, the Bell states  B = (  00 +  11 ) / 2 ,  B 1 = (  01 +  10 ) / 2 ,  B 2 = (  00  11 ) / 2 ,  B 3 = (  01  10 ) / 2 , (2) are examples of fullyentangled 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  on H a and H b be defined in the usual way: a = Tr b ([ ]) , b = Tr a ([ ]) . (3) For a fully entangled state, a = I/d = b . (4) 1 Note that we are assuming that the reduced density matrices come from a pure state  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  , such that (1) holds (with j = 1 / d )....
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This note was uploaded on 02/15/2012 for the course PHYS 3101 taught by Professor Staff during the Spring '08 term at Pittsburgh.
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