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Define orthonormal symmetric vectorsThe symmetric-arm reduction of an exchange Hamiltonian here has two arms. To verify it directly, the link between radial distances $j$ and $j+1$ has strength $J_j$ on either side, while the two central links have strength $J_0/\sqrt{2}$. The calculation of part (a) therefore givesIn particular this subspace is invariant and contains the initial vector. Define the linear isometry $F|j⟩=|s_j⟩$. The displayed matrix entries prove $H_{e}F=F{H}_T$, or $F^{†}{H}_{e}F=H_T$. Induction gives $H_e^{m}F=F{H}_T^m$ for every nonnegative integer $m$; hence the convergent exponential series proves $e^{-i{H}_{e}t}F=F{e}^{-i{H}_{T}t}$. This establishes the mapping for the entire evolution, rather than just its first two powers. The supplied endpoint transfer also forces every $J_j$ to be nonzero, since a broken link would disconnect the two endpoints. Successive powers applied to $|s_0⟩$ then generate all $|s_j⟩$, so this is exactly the cyclic subspace governing the initial state.

Using the stated perfect quantum state transfer identity at time $\pi /2$,The two terms have the excitation at sites $0$ and $2N$, respectively, and all intermediate qubits are zero in both terms. The full state thus factors into a Bell state at the endpoints and the vacuum on the interior. Its partial trace isThe coherent cross terms survive the trace because the interior vectors coincide. We have used the transfer phase convention supplied in the question; a common transfer phase would leave this density operator unchanged.

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.