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

Interpret the imposed measurement restriction as direct measurement of the encoded two-qubit Bell state in the computational basis, without a preceding entangling Bell decoder. For the uniform messages the conditional probabilities areEach outcome has probability $1/4$, so $H(Y)=2$ bits, while $H(Y|X)=1$ bit. ThereforeThe outcome reveals even versus odd parity, but none of the plus/minus label. Each outcome is compatible with two equally probable messages.

Now insert an arbitrary single-qubit unitary operator $U$ on one wire immediately before measurement. Measuring that wire is equivalent to measuring along a Bloch vector $\mathbf{n}$, with the other wire still measured along $z$. All four Bell states have zero local means, and $⟨X\otimes Z⟩=⟨Y\otimes Z⟩=0$, whereas $⟨Z\otimes Z⟩=+1$ for ${\Phi }^{\pm }$ and $-1$ for ${\Psi }^{\pm }$. HenceThe sign within each pair is still inaccessible, andIt stays one for a computational-axis measurement, but a Hadamard gate makes $n_z=0$ and reduces it to zero. Thus a single local gate can change the information, by reducing it, and cannot increase it above one bit.

There is an important scope qualification: if Bob is allowed the full two-qubit decoder from part b before his final computational measurement, those measurements already identify all four messages and give two bits. The one-bit result refers to restricting the Bell readout itself to product computational measurements. Merely specifying the final detector basis without restricting the decoder would not impose that one-bit limit.