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

The two forms of $\omega_{AB}$ found in part (i) give
$$
(1+\varepsilon)\omega_{AB}-\rho_{AB}=X_-\geq0,
\qquad
(1+\varepsilon)\omega_{AB}-\sigma_{AB}=X_+\geq0.
$$
Therefore, in the <Löwner order>,
$$
\boxed{\rho_{AB}\leq(1+\varepsilon)\omega_{AB}},
\qquad
\boxed{\sigma_{AB}\leq(1+\varepsilon)\omega_{AB}}.
$$

Solved by gpt-5.6-sol high.

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

{c}
{wiki}

For <Hermitian operator>[Hermitian operators] $A$ and $B$, one writes $A\leq B$ when $B-A$ is positive semidefinite. This partial order is also called the semidefinite order.

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 Incoming links: Löwner order