 Solution

Use one-bit remote preparation of real qubit states. Alice measures her qubit in the orthonormal basisBecause this change of basis is real and orthogonal, the shared Bell state has the expansionHer projective measurement gives each outcome with probability one half. She sends $m=0$ for the first outcome and $m=1$ for the second. Bob's conditional pure state is respectively $|{\alpha }_{\theta }⟩$ or $|{\beta }_{\theta }⟩$.

For $m=0$ Bob does nothing. For $m=1$ he applies the fixed unitary operatorwhere $Y$ is the Pauli Y gate. Bob need not know $\theta$, because this correction is the same for all real $\theta$. Both outcomes produce the exact density operator $|{\alpha }_{\theta }⟩⟨{\alpha }_{\theta }|$, using one classical bit and only local operations.

This remote state preparation consumes the shared entangled state. Before receiving the bit, Bob has the mixture $(|{\alpha }_{\theta }⟩⟨{\alpha }_{\theta }|+|{\beta }_{\theta }⟩⟨{\beta }_{\theta }|)/2=I/2$, independent of $\theta$, as required by quantum no-signalling. The real-state restriction is what permits a fixed unitary correction of the orthogonal outcome.