Assume a qubit of general definition (21.10) where |0), |1) are two pure states in the 2D quantum space V2, From the definitions in Eqs. (21.3) and (21.4), we have (21.11) (21.12) S(p) = (21.13) = -plogp-(1-p)log(1-p)=f(p) In the result in Eq. (21.13), we recognize the Shannon entropy of a two-event source X = {0,1}, corresponding to the two possible "states" of a classical bit (see Eqs. (4.13) and (4.14)). The average qubit information is, thus, equivalent to that of two classical bits, the amount depending on the weights Po, p1 in the state mixture. As described in Chapter 4, the function f(p) has a maximum of unity for p = 1/2, corresponding to |a|=|B| = 1/ (phases being arbitrary) or a uniform distribution Po = p1 = 1/2, and for the VN entropy, Smax(p) = log 2 = 1. In this case, the quantum information amounts to exactly one classical bit. The minimum of f(p) is zero, which is reached either when Po = 1, p1 = 0 or when Po = 0, p1 = 1, meaning that the qubit is in a pure state, i.e., 1q) = |0) or 19) = |1), giving Smax(p) = 0. In this case, there is no quantum information in the system, as there is no information in a single, deterministic classical It is clear that the VN entropy of an n-qubit (or quNit) in the quantum space Vn always has the maximum Smax(p) = logn when the system is in the most homogeneous state superposition with a uniform probability distribution pi = 1/n, hence corresponding to maximum quantum information. In the general case, we have 0 < S(p) < logn. It is a generally possible that the state 1/4/ can be a mixture of nonorthogonal states. In this case, the corresponding density matrix p is nondiagonal. How, then, can we determine the VN entropy of the system? The answer is that we know it is possible to express the density operator in the computational basis of eigenstates, where the matrix is diagonal. Given the eigenvalues di (i = 1...n) corresponding to the eigenstates {|2i}}i=1..n, the VN entropy, thus, takes the form (21.14) Quantum information theory As an illustration (see also the