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

Include all fixed resource states and all measurement records in an environment, so the entire physical protocol has an linear isometry $U$ from the unknown input to a larger pure output. If every recipient receives a perfect copy, its reduced density matrix is the pure projector $|\psi ⟩⟨\psi |$. A pure marginal forces factorization: write a joint vector in a basis beginning with $|\psi ⟩$; its squared components along every orthogonal basis vector vanish because their marginal probabilities are zero. Applying this successively to all recipients giveswhere $|R⟩$ is the fixed initial resource and $|E_{\psi }⟩$ may depend on the input. This includes Alice's discarded systems and every possible outcome record.

Choose two distinct nonorthogonal pure states, for example $|0⟩$ and $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, and write their overlap as $s$. Preservation of the inner product by the linear isometry requiresNormalized environment vectors have overlap of absolute value at most one. Thus $0<|s|<1$ would imply $|s|\le |s{|}^N$, impossible for $N>1$. Hence perfect transmission of a copy to every recipient is impossible for $N>1$. This proves the no-cloning theorem obstruction even though Alice keeps no copy herself.

Write $N=2k+1$ and abbreviate the Dicke states by $|D_j^n⟩=|{\Psi }_j^n⟩$. Alice's shared resource has decompositionFollow the actual circuit order: a controlled-NOT gate from the input to $q_0$, a Hadamard gate on the input and a Pauli X gate on $q_0$, followed by computational basis measurements. Just before measurement, the joint vector isThe Dicke states in each branch are orthogonal, so each branch has squared norm $1/4$. Conditioning on outcome $00$ and normalizing therefore gives the state of the other $N$ parties:This is the encoding used in Dicke-resource telecloning. The extra Pauli X gate on Alice's resource qubit is essential to the displayed ordering of the amplitudes.

The state in part (b) is symmetric in the receiving qubits. Single out Bob using the Dicke state recursion. For $k\ge 1$, the two vectors areDistinct excitation counts give orthogonal states of the other recipients. Taking their partial trace, the only surviving off-diagonal term comes from the common rest vector $|D_k^{N-1}⟩$. The one-qubit reduction of Dicke-state superpositions is consequentlyPut $p=|\alpha {|}^2$ and $q=|\beta {|}^2$, so $p+q=1$. The overlap with the input isFor $N=1$ there is no system to trace out, and the same matrix is $|\psi ⟩⟨\psi |$, with overlap one. For every $N>1$ the overlap is below one, even at its maximum $p=q=1/2$, so these are imperfect copies.

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.