= One-qubit reduction of Dicke-state superpositions
{title2=$\rho_{01}=\alpha\beta^*\sqrt{(n-m)(m+1)}/n$}
For $|\Omega\rangle=\alpha|D_m^n\rangle+\beta|D_{m+1}^n\rangle$, $|\alpha|^2+|\beta|^2=1$ and $0\leq m<n$, the <reduced density matrix> of any one qubit is
$$
\rho=\frac1n\begin{pmatrix}(n-m)|\alpha|^2+(n-m-1)|\beta|^2&\sqrt{(n-m)(m+1)}\,\alpha\beta^*\\\sqrt{(n-m)(m+1)}\,\alpha^*\beta&m|\alpha|^2+(m+1)|\beta|^2\end{pmatrix}.
$$
Use 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 $|D_m^{n-1}\rangle$ occurs in both states, with coefficients $\sqrt{(n-m)/n}$ multiplying $|0\rangle$ and $\sqrt{(m+1)/n}$ multiplying $|1\rangle$. Thus $\operatorname{Tr}_{\mathrm{rest}}(|D_m^n\rangle\langle D_{m+1}^n|)=\sqrt{(n-m)(m+1)}|0\rangle\langle1|/n$, proving the matrix. A single <Dicke state> alone gives the diagonal matrix $\operatorname{diag}((n-m)/n,m/n)$.
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