 One-qubit reduction of Dicke-state superpositions

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