Because the two flags $E_C$ and $F_C$ are <orthogonal projections>, the flagged state is block diagonal. If $h(\lambda)$ denotes the <binary entropy>, then
$$
S(\rho_{ABC})
=h(\lambda)+\lambda S(\rho'_{AB})
+(1-\lambda)S(\rho''_{AB}),
$$
and
$$
S(\rho_{BC})
=h(\lambda)+\lambda S(\rho'_B)
+(1-\lambda)S(\rho''_B).
$$
Its unflagged marginals are $\rho_{AB}=\lambda\rho'_{AB}+(1-\lambda)\rho''_{AB}$ and $\rho_B=\lambda\rho'_B+(1-\lambda)\rho''_B$. Substitution in cancels the two binary-entropy terms and gives
$$
\boxed{
H(A|B)_{\lambda\rho'+(1-\lambda)\rho''}
\geq
\lambda H(A|B)_{\rho'}
+(1-\lambda)H(A|B)_{\rho''}}.
$$
This is exactly the <concavity of quantum conditional entropy>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 323 / 3 / ii / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-323/3/ii/b/solution