eft as a relatively easy exercise to show that: (21.32) where IA, I B are the identity matrices of the subsystems A, B. Using this property in Eq. (21.31), and also the additive and commutative properties of the trace (see Eq. (17.32)) we obtain: S(PAB) (21.33) We must now make sense of the traces involved in the right-hand side of this equation. Recall first that PAB is the density operator of the composite system in state 1/s ). From the discussion in the previous section, and in particular the result in Eq. (21.6), we have learnt that the expectation value of any observable U, is given by (U) AB = tr(PABU). There is no reason to assume that the same outcome would be obtained by the same measurement applied just to subsystem A separately, namely (U) A = tr(PAU) # tr(PABU) = (U) AB. This is because in the joint system AB, the subsystem B may also contribute to positive outcomes for the same observable. On the other hand, intuition dictates that there must exist an operator U' such that the expected value of U' in the joint system AB and that of U in the subsystem A do coincide, i.e., (21.34) It is a fine academic issue to show that U' =UQIB, which expands U to the full system space, is, in fact, the only candidate to solve such a measurement reconciliation problem. Here, we may just accept it as an intuitive postulate. Hence, (21.35) = = -S(PB), to be followed by substitution in Eq. (21.33) to yield the inequality S(PAB) < S(PA) + S(pB), as claimed in Eq. (21.28), which is now established. As also claimed earlier, the equality in Eq. (21.28) stands when (and only when) PAB = PA X PB, meaning that the entropies of the subsystems are additive, according 21.2 Relative, joint, and conditional entropy, and mutual information S(PAB)=S(PA)+S(pB). (21.36) Proving this equality is straightforward. Indeed, since PAB = PA D PB, we can now write S(PAB) = -tr[pABlog(p)@pB)], and as we have seen from the results in Eqs. (21.32) to (21.35), the right-hand side is identical to S(PA) + S(PB). Quantum mutual information As we hav