Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 2 / 35A (Principles of Quantum Mechanics)

Orthonormality gives $⟨{\mathcal{B}}_1|{\mathcal{B}}_i⟩={\delta }_{1i}$.

Tracing out $B$ gives ${\rho }_A=I/2$. Its Von Neumann entropy is $-\mathrm{tr}{\rho }_A{\log }_2{\rho }_A=1$, and $\mathrm{tr}{\rho }_A^2=1/2$, so it is mixed.

No. A pure product state has a pure reduced density matrix, whereas part (b) gives $I/2$.

The values are $+1,+1,-1,-1$ for ${\mathcal{B}}_1,\dots ,{\mathcal{B}}_4$. Thus $z$-spin outcomes are perfectly correlated for the first pair and perfectly anticorrelated for the second.

Using $T=2\mathrm{SWAP}-I$, the symmetric triplet Bell state have eigenvalue $1$, while the antisymmetric singlet has eigenvalue $-3$. With the ordering shown, ${\mathcal{B}}_1,{\mathcal{B}}_2,{\mathcal{B}}_3$ are triplet combinations and ${\mathcal{B}}_4$ is the singlet.

Applying $I\otimes I,X,Z,$ or $XZ$ to one qubit permutes the Bell basis up to phases. Composing two such Pauli operations maps any chosen Bell state to any other, proving local equivalence.