to the (maximum possible) VN entropy of originator's source, max[S(p)] = Xideal, as discussed earlier and also formally established in the exercise. Substituting p1 = 0.5 into Eq. (23.61), we obtain the following analytical form of the channel capacity: (23.62) 23.3 Capacity of some quantum channels Substituting p = 0 in the above, we get the upper bound for the channel capacity, which corresponds to the originator's maximum source entropy max[S(p)]: Xideal = max S(p) (23.63) This example, thus, provides an illustration of the HSW maximization problem, leading, in this case, to a tractable analytical solution. It also illustrates the property x(p) < Xideal = max S(p), which can be used to guide the maximization problem solving. Example 23.2: Bit-flip channel The quantum operation of this channel is defined in Eq. (23.13). Assuming the orthogonal symbols P1=10)(0| = and P2 = |1) (1), we obtain (23.64) and the corresponding VN entropy 1=ste(Emm) (p+p1-2pp1)log(p+p1-2pp1) - (23.65) =f(p+p1-2pp1). Quantum channel noise and channel capacity Next, we calculate the quantum operation E onto P1 and P2 along with the corresponding entropies: e(p1)= pXp1X+(1-p) (23.66) S[E(p1)] =-[(1-p)log(1-p)+plogp] = f(1-p) = f(p) E(P2) = pXp2X + (1 p)p2 (23.67) S[E(P2)] = f(p). Hence, we obtain B = (PiS[8(p)) =pif(p)+(1-pi)f(p) (23.68) =f(p). Combining the results in Eqs. (23.65) and (23.68), we finally obtain the channel capacity x(p,p1)=max(x) max(A-B) (23.69) max[f(p+p1-2pp1)-f(p)]. We note that the argument satisfies the property x'(p, p1) = x'(p, 1 - p1). The max- imization problem can be solved analytically by studying the partial derivatives ax'(p,p1)/dp and but we may observe, again, that the capacity cannot exceed the amount Xideal = max S(p). Using the definition of the symbol density operator p = pipi + (1-p1)p2, - we clearly have S(p) = f(p1), and Xideal = max S(p), yielding p1 = 1/2, as expected (maximum VN entropy is obtained for a uniform message-symbol distribution). Thus, we have for all p the channel capacity: x(p)=x'(