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

Define $E|0⟩=|D_k^N⟩$ and $E|1⟩=|D_{k+1}^N⟩$. From the branch calculation in part (b), outcome $R_0=0,R_1=1$ leavesThus the effective input teleported was $X|\psi ⟩=\beta |0⟩+\alpha |1⟩$. A Pauli X gate flips every bit and hence maps a Dicke state with $j$ excitations to one with $N-j$ excitations. Since $N-k=k+1$, applying $X$ at every receiving party interchanges the two logical basis states and restores $E|\psi ⟩$. ThereforeIn particular Bob's uncorrected reduced density matrix is $X\rho X$, and his correction returns it to the matrix in part (c). The remaining branches can likewise be corrected: $Z^{\otimes N}$ implements logical $Z$ up to the common global phase $(-1{)}^k$, while $X^{\otimes N}$ implements logical $X$. Thus the outcome dependence can be removed by local Pauli gates even though the resulting copies remain imperfect.