= Symmetric-arm reduction of an exchange Hamiltonian
{title2=$HF=FH_T,\quad |s_k\rangle=q^{-1/2}\sum_a|a,k\rangle$}
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 $|s_0\rangle$ be the hub excitation and $|s_k\rangle=q^{-1/2}\sum_{a=1}^q|a,k\rangle$. This <orthonormal> symmetric <single-excitation subspace> is invariant, and its restricted <Hamiltonian operator> is
$$
H|s_0\rangle=J_0|s_1\rangle,\qquad H|s_k\rangle=J_{k-1}|s_{k-1}\rangle+J_k|s_{k+1}\rangle,
$$
with 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> $F|k\rangle=|s_k\rangle$ intertwines $H$ with the path <Hamiltonian operator> $H_T=\sum_{k=0}^{n-1}J_k(|k\rangle\langle k+1|+|k+1\rangle\langle k|)$: $HF=FH_T$. Expanding the <matrix exponential> gives $e^{-iHt}F=Fe^{-iH_Tt}$. <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.
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