Arrange the amplitudes of $|\phi_1\rangle$ as the matrix
$$
C_1=\begin{pmatrix}a&b\\c&d\end{pmatrix}.
$$
The second state has $C_2=C_1X$, where $X$ swaps the $A_2$ basis states. If Bob receives $A_1$, his two <reduced density matrices> are
$$
\rho_{A_1}^{(1)}=C_1C_1^\dagger,
\qquad
\rho_{A_1}^{(2)}=C_2C_2^\dagger
=C_1XX^\dagger C_1^\dagger,
$$
and are identical. No measurement on $A_1$ contains any information about the shared state.

If Bob instead receives $A_2$,
$$
\rho_{A_2}^{(1)}
=\begin{pmatrix}
a^2+c^2&ab+cd\\
ab+cd&b^2+d^2
\end{pmatrix},
\qquad
\rho_{A_2}^{(2)}=X\rho_{A_2}^{(1)}X.
$$
They differ because $a^2+c^2<b^2+d^2$. Bob can therefore distinguish them with better-than-random success. For equal priors,
$$
\boxed{D(\rho_{A_2}^{(1)},\rho_{A_2}^{(2)})
=b^2+d^2-a^2-c^2},
$$
so the optimal success probability is $\frac12(1+D)$. It is generally below one, so a single copy does not permit certain identification.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 323 / 5 / b / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-323/5/b/ii/solution