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The Pauli matrices obey $X|0⟩=|1⟩$, $X|1⟩=|0⟩$, $Y|0⟩=i|1⟩$, and $Y|1⟩=-i|0⟩$. Thus $X\otimes X$ and $Y\otimes Y$ contribute opposite amplitudes on $|00⟩,|11⟩$ and equal amplitudes on $|01⟩,|10⟩$. For the XY exchange interaction $E=(X\otimes X+Y\otimes Y)/2$,Equivalently $E=|01⟩⟨10|+|10⟩⟨01|$. It transfers an excitation across a link without creating or destroying excitations, so a sum of such links preserves the single-excitation subspace.

Replace the two arms by three identical arms of length $N$ attached to one central qubit. Label the hub $c$ and the sites on arm $a$ by $(a,j)$, $a=1,2,3$, $j=1,\dots ,N$. Use XY exchange interactions with coupling $J_0/\sqrt{3}$ on each hub-to-first-site link, and coupling $J_j$ on each arm link $(a,j)$ to $(a,j+1)$. Initially excite only the hub: this is the product state $|1{⟩}_c$ with every arm qubit in $|0⟩$.

In the symmetric single-excitation subspace, define $|s_0⟩=|c⟩$ and $|s_j⟩=3^{-1/2}{\sum }_{a=1}^3|a,j⟩$. The hub coupling isand the reverse coupling has the same coefficient. All other radial links have coefficients $J_j$, just as in part (c). Thus the symmetric-arm reduction of an exchange Hamiltonian is again exactly $H_T$. At time $\pi /2$ its evolution isEvery nontip qubit is zero, so tracing them out leaves the three tips in the pure W stateThe essential change is the additional identical arm and the normalization of each hub coupling by $\sqrt{3}$ instead of $\sqrt{2}$.

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.