Past exam of the introduction to quantum information course of the University of Oxford / 2021 / 3

The partial trace removes the other orthonormal factor in the Schmidt decomposition:For instance, tracing $|a_k{b}_k⟩⟨a_{\ell }{b}_{\ell }|$ over $B$ gives $⟨b_{\ell }|b_k⟩|a_k⟩⟨a_{\ell }|$, which vanishes unless $k=\ell$. Consequently the nonzero eigenvalues of the two reduced density operators agree, with multiplicity. If the subsystem dimensions differ, the orthonormal Schmidt families can be completed to full bases and the extra eigenvalues are zero.

No: the advertised pure joint state is incompatible with these marginals. Their eigenvalue multisets are $\{0.9,0.1\}$ and $\{0.7,0.3\}$, which differ. Part a shows that a bipartite pure state must give the same nonzero Schmidt spectrum on both sides.

This excludes Eve's purity claim, not every possible joint state or every possibility of entanglement. A mixed state can have unequal marginal spectra; the product ${\rho }_A\otimes {\rho }_B$, for example, has exactly the stated marginals.

By linearity, the isotropic qubit map acts asIt contracts or expands every direction by the same factor and reverses the vector when $r<0$. More generally ${\mathcal{E}}_r(A)=rA+(1-r)\mathrm{tr}(A)I/2$, so the map preserves trace and Hermiticity for every real $r$.

The output of a state of Bloch radius $u$ has eigenvalues $(1\pm ru)/2$. These are nonnegative for every $0\le u\le 1$ exactly whenIf $|r|>1$, a pure input already gives a negative eigenvalue. Conversely, $|r|\le 1$ preserves the Bloch ball, hence positivity for states and, by positive scaling, for all positive operators. This is positivity of the map alone; physical implementation on part of an entangled system needs complete positivity.

Use the unnormalized Choi matrix convention $J_r={\sum }_{j,k=0}^1|j⟩⟨k|\otimes {\mathcal{E}}_r(|j⟩⟨k|)$, which has trace two. In the ordered basis $|00⟩,|01⟩,|10⟩,|11⟩$,Equivalently $J_r=2r|\Omega ⟩⟨\Omega |+(1-r)I_4/2$. The convention obtained by applying the channel to one half of the normalized Bell state gives $J_r/2$; its positivity criterion is the same.

The Bell state $|\Omega ⟩$ is an eigenvector of $J_r$ with eigenvalue $(1+3r)/2$. Its three-dimensional orthogonal complement has eigenvalue $(1-r)/2$. The Choi criterion for complete positivity therefore givesFor sufficiency one can also write a Kraus representation explicitly:In this range the four coefficients are nonnegative and sum to one, giving a physical mixture of Pauli conjugations. This is the complete positivity range of the depolarizing channel in the Bloch-scaling parameterization.

Write the unknown common eigenbasis as an unknown unit Bloch vector $\mathbf{n}$. Then ${\rho }_A$ has vector $0.8\mathbf{n}$, while ${\rho }_B$ has vector $-0.4\mathbf{n}$. A single deterministic channel working for every possible basis must therefore map $0.8\mathbf{n}$ to $-0.4\mathbf{n}$ for every unit $\mathbf{n}$.

By the basis-independent conversion of qubit Bloch radii, linearity forces that channel to be ${\mathcal{E}}_{-1/2}$. Explicitly, add the conditions for $\mathbf{n}$ and $-\mathbf{n}$ to get $\mathcal{E}(I)=I$, then use the three coordinate directions to fix its action on every Pauli matrix. But $-1/2<-1/3$, so part iv excludes complete positivity. Thus the forward conversion cannot be implemented reliably without knowing the basis.

The reverse conversion would require ${\mathcal{E}}_{-2}$, which is not even positive by part ii. Therefore the reverse conversion is also impossible. Unitary operations interspersed with physically allowed depolarizing maps cannot evade these restrictions: their composition remains a physical quantum channel. The question concerns one basis-independent operation on the supplied qubit, not an additional resource of many copies for learning its basis.

With the basis known, choose a unitary operator that interchanges its two basis states. On the Bloch sphere this is a rotation by $\pi$ about an axis perpendicular to $\mathbf{n}$, sending $\mathbf{n}$ to $-\mathbf{n}$. Follow it by ${\mathcal{E}}_{1/2}$, which is completely positive. ThenThus ${\rho }_A$ can be converted to ${\rho }_B$ using the allowed operations and a suitable physically available depolarizing parameter.

The reverse direction would have to increase Bloch radius from $0.4$ to $0.8$. Unitaries preserve radius, and every completely positive ${\mathcal{E}}_r$ has $|r|\le 1$, so any composition of the allowed operations cannot increase it. Hence ${\rho }_B$ still cannot be converted to ${\rho }_A$ using these resources. This conclusion uses the stated unitary/depolarizing resources; an unrestricted state-preparation channel would be a different allowance.