Symmetric-arm reduction of an exchange Hamiltonian ($HF=F{H}_T,|s_k⟩=q^{-1/2}{\sum }_a|a,k⟩$)

Consider $q$ identical length-$n$ arms joined to one hub, with XY exchange interactions of strength $J_0/\sqrt{q}$ from the hub to each first site and strength $J_k$ between sites $k,k+1$ on each arm. Let $|s_0⟩$ be the hub excitation and $|s_k⟩=q^{-1/2}{\sum }_{a=1}^q|a,k⟩$. This orthonormal symmetric single-excitation subspace is invariant, and its restricted Hamiltonian operator iswith endpoint terms omitted. At the first link the $q$ hub contributions add to $J_0$; all other links act identically on the arms. Thus the linear isometry $F|k⟩=|s_k⟩$ intertwines $H$ with the path Hamiltonian operator $H_T={\sum }_{k=0}^{n-1}{J}_k(|k⟩⟨k+1|+|k+1⟩⟨k|)$: $HF=F{H}_T$. Expanding the matrix exponential gives $e^{-iHt}F=F{e}^{-i{H}_{T}t}$. Perfect quantum state transfer from zero to $n$ on the path therefore produces an equal excitation superposition at the $q$ arm tips. Since all other qubits are zero, the tip state is a pure W state, up to the transfer's common phase.