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For $|\Omega ⟩=\alpha |D_m^n⟩+\beta |D_{m+1}^n⟩$, $|\alpha {|}^2+|\beta {|}^2=1$ and $0\le m<n$, the reduced density matrix of any one qubit isUse the Dicke state recursion. The rest-of-system vectors of different Hamming weights are orthogonal. For the diagonal terms this leaves the probabilities of the distinguished bit being zero and one. In the cross term, only the rest vector $|D_m^{n-1}⟩$ occurs in both states, with coefficients $\sqrt{(n-m)/n}$ multiplying $|0⟩$ and $\sqrt{(m+1)/n}$ multiplying $|1⟩$. Thus ${\mathrm{Tr}}_{\mathrm{rest}}(|D_m^n⟩⟨D_{m+1}^n|)=\sqrt{(n-m)(m+1)}|0⟩⟨1|/n$, proving the matrix. A single Dicke state alone gives the diagonal matrix $\mathrm{diag}((n-m)/n,m/n)$.

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.

The $n$-qubit W state is the Dicke state with one excitation:For three qubits it is $(|001⟩+|010⟩+|100⟩)/\sqrt{3}$. Its reduced density matrix on any one qubit is $\mathrm{diag}((n-1)/n,1/n)$, from the Dicke state recursion. For $n>1$ both diagonal entries are nonzero, so that qubit is entangled with the rest: a pure product state would have a pure marginal. A symmetric-arm reduction of an exchange Hamiltonian supplies a dynamical preparation from one initial excitation.