Solution

Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 57 / 1 / b / Solution

{c}
{title2=$E|\psi\rangle=\alpha|D_k^{2k+1}\rangle+\beta|D_{k+1}^{2k+1}\rangle$}

For odd $n=2k+1$, a symmetric <quantum teleportation> resource can encode an unknown <qubit> into $E(\alpha|0\rangle+\beta|1\rangle)=\alpha|D_k^n\rangle+\beta|D_{k+1}^n\rangle$. Begin with $|D_{k+1}^{n+1}\rangle$, apply a <controlled-NOT gate> from the input to the sender's resource qubit, apply a <Pauli X gate> to that resource qubit and a <Hadamard gate> to the input, and measure those two qubits. The four unnormalized remaining vectors are $E|\psi\rangle/2$, $EX|\psi\rangle/2$, $EZ|\psi\rangle/2$ and $EXZ|\psi\rangle/2$. Each outcome has probability $1/4$. Since $X^{\otimes n}$ interchanges $|D_k^n\rangle$ and $|D_{k+1}^n\rangle$, and $Z^{\otimes n}$ acts as logical $Z$ up to the common phase $(-1)^k$, local Pauli corrections make the output independent of the result. The <one-qubit reduction of Dicke-state superpositions> gives
$$
\rho=\frac1n\begin{pmatrix}k+|\alpha|^2&(k+1)\alpha\beta^*\\(k+1)\alpha^*\beta&k+|\beta|^2\end{pmatrix},\qquad \langle\psi|\rho|\psi\rangle=\frac{k+1+2k|\alpha|^2|\beta|^2}{n}.
$$
For $n>1$ this distributes imperfect copies, as required by the <no-cloning theorem>. For $n=1$ it reduces to exact <quantum teleportation>.

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 Incoming links: Dicke-resource telecloning