Past exam of the laser science and quantum information processing course of the University of Oxford / 2016 / 8 / Solution

The transmitted branch is the Kraus operatorFor nonzero loss it is not trace preserving: $\mathrm{tr}(E\rho E^{†})$ is the transmission probability. A complete channel includes an orthogonal vacuum or erasure flag, with a loss operator $L=\sqrt{R}|\mathrm{vac}⟩⟨0|$ and $E^{†}E+L^{†}L=I$. Postselecting only the first outcome gives the single-operator description in the question.

For $|\pm ⟩$, the unnormalized transmitted vector is $(\sqrt{1-R}|0⟩\pm |1⟩)/\sqrt{2}$. Its norm squared and normalized conditional state areThe reduced density matrix of either member of the initial pair has eigenvalues $\alpha ,1-\alpha$. Since the joint state is pure, its Von Neumann entropy is zero. With logarithms to base two and $h_2$ the binary entropy function,The two nonzero Schmidt coefficients show that this pure pair is entangled exactly when $0<\alpha <1$. Its mutual information rises from zero to two bits at $\alpha =1/2$ and returns to zero.

For this known pure family, every positive mutual information signals entanglement. For an otherwise unspecified two-qubit state, positive mutual information can also be classical. The mutual information bound for separable states gives the stronger general witness $I(A:B)>1$ bit. On this curve it holds for approximately $0.110028<\alpha <0.889972$; the rest of the open interval is still entangled, but that generic one-bit witness alone would not certify it.

If Bob's qubit passes through the channel, the unnormalized retained state is $\sqrt{\alpha }|11⟩+\sqrt{(1-\alpha )(1-R)}|00⟩$. ThusFor $0\le R<1$, equal Schmidt coefficients require $\alpha =(1-\alpha )(1-R)$, givingThe retained state is then $(|00⟩+|11⟩)/\sqrt{2}$. At $R=1$, this limiting compensation has zero transmission probability and yields no maximally entangled pairs. Conditioning can increase the entanglement of retained pairs, while the discarded fraction accounts for the cost.