Put $\rho=p\rho_1+(1-p)\rho_2$. The <Löwner order>[operator inequality] $\rho\geq p\rho_1$ and the <operator monotonicity of logarithm> give, on the support of $\rho_1$,
$$
\log\rho\geq\log(p\rho_1)
=(\log p)I+\log\rho_1.
$$
Consequently,
$$
-p\operatorname{Tr}(\rho_1\log\rho)
\leq pS(\rho_1)-p\log p.
$$
The corresponding inequality from $\rho\geq(1-p)\rho_2$ yields
$$
-(1-p)\operatorname{Tr}(\rho_2\log\rho)
\leq(1-p)S(\rho_2)-(1-p)\log(1-p).
$$
Adding these inequalities and using $S(\rho)=-\operatorname{Tr}(\rho\log\rho)$ proves the <entropy bound for a binary mixture>:
$$
\boxed{S(\rho)\leq pS(\rho_1)+(1-p)S(\rho_2)+H(p)}.
$$
Singular states follow by adding a positive multiple of the identity and taking a <limit>; the endpoint cases use $0\log0=0$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 323 / 4 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-323/4/b/solution