Past exam of the laser science and quantum information processing course of the University of Oxford / 2015 / 8 / a

Normalization requires $2p+2q=1$, so $q=1/2-p$ and $0\le p\le 1/2$. With logarithms to base two, the Shannon entropy isThe convention $0\log 0=0$ gives $H=1$ at both endpoints. In the interior,Hence the unique maximum is $H=2$ bits at $p=q=1/4$. The curve is concave and symmetric about $p=1/4$.

Choose the message assignment $x_1\leftrightarrow {\Psi }^{+}$, $x_2\leftrightarrow {\Psi }^{-}$, $x_3\leftrightarrow {\Phi }^{+}$ and $x_4\leftrightarrow {\Phi }^{-}$. Starting from $|{\Psi }^{+}⟩$, apply respectively $I,Z,X,XZ$ to the first qubit. This produces those four orthonormal Bell states, up to a physically irrelevant overall sign in the last case. The ensemble density operator isin computational order $00,01,10,11$, where ${\Pi }_B=|B⟩⟨B|$. The cancellation of off-diagonal terms follows from the equal probabilities within each sign pair; the individual encoded states remain entangled.

A Bell basis measurement reads the label perfectly because the four code states are orthogonal. Thus this is equivalent to a classical error-free encoding when unrestricted joint measurement is allowed. Each conditional encoded state is pure, soThe second formula is the classical-quantum mutual information, here equal to attainable classical information. The first says there is no uncertainty about the joint pure code state once the message is known; it does not say that measuring its individual qubits has deterministic outcomes.