Dicke state ($|D_m^n⟩$)

The $n$-qubit Dicke state with $m$ excitations is the normalized equal superpositionwhere $|x|$ is the Hamming weight in the computational basis. The binomial coefficient counts the mutually orthogonal summands, proving normalization. Every permutation of the qubits leaves the vector unchanged. Splitting the sum according to the first bit and using $\left(\genfrac{}{}{0ex}{}{n-1}{m}\right)/\left(\genfrac{}{}{0ex}{}{n}{m}\right)=(n-m)/n$ givesTerms with zero coefficient are omitted at the endpoints. This recursion is useful for partial traces and for distributing a logical qubit among many symmetric physical qubits.

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.

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 $\mathrm{diag}((n-m)/n,m/n)$.