At the proposed boundary $p_0=1/(d+1)$, the identity from part (iv) yields
$$
\rho_{p_0}
=\frac{d}{d+1}\sigma
+\frac1{d+1}\frac Dd.
$$
Both $\sigma$ and $D/d$ are <convex combinations> of <product states>, so $\rho_{p_0}$ is a <separable quantum state>. For $0\leq p\leq p_0$, the state $\rho_p$ is a convex combination of $\rho_{p_0}$ and the maximally mixed product state $I/d^2$, and is therefore separable.

For the converse, the <partial transpose> of the maximally entangled projector is $F/d$, where $F$ is the <swap operator>. Therefore
$$
\rho_p^{T_B}=\frac pdF+\frac{1-p}{d^2}I.
$$
On the antisymmetric subspace, $F$ has eigenvalue $-1$, so the corresponding eigenvalue of $\rho_p^{T_B}$ is
$$
\frac{1-p}{d^2}-\frac pd,
$$
which is negative exactly when $p>1/(d+1)$. The <positive partial transpose criterion> then proves that $\rho_p$ is entangled. Thus
$$
\boxed{\rho_p\text{ is separable exactly when }p\leq\frac1{d+1}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 323 / 2 / v / Solution

{title2=$F$}
{wiki}

The swap operator exchanges two tensor factors: $F(|\psi\rangle\otimes|\phi\rangle)=|\phi\rangle\otimes|\psi\rangle$. It has <eigenvalue> $+1$ on the symmetric subspace and $-1$ on the antisymmetric subspace, and the <partial transpose> of $|\Phi^+\rangle\langle\Phi^+|$ is $F/d$.

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 Incoming links: Swap operator