ices of PA = IA/2, PB = IB/3, and PAB = IAB/6, respectively, with the property PAB = PA X PB. We therefore S(PA) = S(pB) = t[(I/3)log(I/3)]-3g1.58 (21.54) S(PAB) = which shows that S(PAB) = S(PA) + S(PB) and illustrates that entropy is additive for subsystems with uncorrelated information (namely satisfying PAB = PA X PB or 1/A AB) = (VA) 0/VB)), or, equivalently, whose quantum states are not entangled. Example 21.2: Subsystems with correlated information Consider, for instance, a composite system AB with the joint state (21.55) 4 For each subsystem, it is possible to choose the eigenstate basis {|ii} in which each component (v/i) is in phase with |i>, hence (vi) > 0, meaning that the example where all components are positive is actually not restrictive. Quantum information theory which corresponds to the density operator PAB - + (21.56) It is straightforward to check that tr(PAB) = 1/2 < 1, which illustrates that (VAB) is not a pure state. It is also a basic exercise to show that (VAB), like other states having similar decomposition with any other nonzero coefficients, cannot be decomposed into a tensor product of two subsystem states (VA), /VB). Because of this, we conclude that the information in the two subsystems is correlated, or that the subsystem states are entangled. So far, we only know that the quantum information in system AB is: S(PAB) = -2x(1/6)log(1/6)-0xlog(0) - - 1 X (4/6) log(4/6) (21.57) = 1.251. Given the above assumptions, can we get any knowledge of the quantum information in subsystems A, B? The answer is yes, if we use the tool of partial tracing. Indeed, we may rewrite Eq. (21.55) as a sum of tensor products: PAB 1 x x |1) (1|B). (21.58) Applying partial tracing over A from the definition in Eq. (21.24), while using the property tr(|i) (i|) = 1, we obtain: (P)410 |1) (1/A x |1) (1|B) = + 4tr A (1) (1) A x (0|B) + tr A (|1) (1|A X (1|B)] 4tr A(|1)(1|A) X |0) A(|1) (1|A) X |1) (1|B] (10)(0B+40)(0B+1)(1B) (21.59) The operator PB = trA(PAB) is referred to as the reduced density operator of su