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In the single-excitation subspace, take a connected real nearest-neighbor hopping matrix $A$ with every coupling nonzero and endpoint control $B=\mathrm{diag}(-1,1,\dots ,1)$. Endpoint commutators isolate the first transition, then propagate along the chain to generate both off-diagonal skew-Hermitian directions and adjacent diagonal differences. These generate $\mathfrak{su}(N)$. For $N\ge 3$, $\mathrm{Tr}B=N-2\ne 0$ supplies the central direction, giving $\mathfrak{u}(N)$. For $N=2$ the algebra is $\mathfrak{su}(2)$, which still gives full state controllability but omits an arbitrary global phase. A broken coupling separates an unreachable chain component.

For a periodic spin-one-half Heisenberg model $H=-(J/{\hslash }^2){\sum }_{n=1}^N{\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$, $J>0$, the fully aligned state has $E_0=-NJ/4$. In its single-excitation subspace, let $|n⟩$ denote a down spin at site $n$ in an otherwise up background. Direct action of each bond givesThe normalized translation eigenstates are $|q⟩=N^{-1/2}{\sum }_n{e}^{iqna}|n⟩$, with $q=2\pi \ell /(Na)$ modulo $2\pi /a$. Their energy eigenvalues are $E_0+J[1-\cos (qa)]$. This is an exact one-magnon calculation, not a many-magnon harmonic approximation. Near $q=0$ the energy above $E_0$ is $J{a}^2{q}^2/2$.

The wavenumber restriction comes from requiring a periodic plane-wave coefficient pattern, equivalently requiring this vector to be an eigenvector of a one-site translation with eigenvalue $e^{-iqa}$. The identity $T^N=I$ alone imposes no restriction on a general state: it already holds throughout the ring Hilbert space. An arbitrary finite sum with coefficients $e^{iqna}$ still returns after $N$ translations, but need not be a one-site translation eigenstate unless $e^{iqNa}=1$.

For physical spin ladder operators, $S_j^{\pm }=S_j^x\pm i{S}_j^y$. Operators at distinct sites commute. Expanding the two products cancels their cross terms and givesAdding the remaining component provesFor spin one-half, $S^{+}|\downarrow ⟩=\hslash |\uparrow ⟩$, $S^{-}|\uparrow ⟩=\hslash |\downarrow ⟩$, and raising an up spin or lowering a down spin gives zero. These formulas restore the physical $\hslash$ factors implicit in the dimensionless ladder hint.

Let $|n⟩$ abbreviate the state with the only down spin at site $n$, keeping $|{\psi }_0⟩$ for the fully up state. If $n$ lies on neither end of bond $(j,j+1)$, that bond contains two up spins. The ladder terms vanish, while $S_j^z{S}_{j+1}^z$ gives ${\hslash }^2/4$. ThusFor a down spin at $j$, the $z$ product is instead $-{\hslash }^2/4$, and $S_j^{+}{S}_{j+1}^{-}$ moves it to $j+1$ with amplitude ${\hslash }^2$. For a down spin at $j+1$, the other ladder product moves it back. ThereforeEvery bond in $|{\psi }_0⟩$ is aligned, so summing its $N$ bond energies givesThis is indeed a ground-state energy: a pair of spin-one-half particles has ${\mathbf{S}}_j\cdot {\mathbf{S}}_{j+1}$ eigenvalues ${\hslash }^2/4$ and $-3{\hslash }^2/4$, so each ferromagnetic bond is bounded below by $-J/4$, simultaneously attained by the fully aligned state. The isotropic Heisenberg model has a whole maximum-total-spin ground multiplet, not a unique all-up ground state.

In the single-excitation subspace, only the two bonds adjacent to the down spin differ from the aligned background. Their diagonal energy increase is $J$, and each neighboring hop has amplitude $-J/2$. Hence, with periodic site indices,This is an exact finite-dimensional hopping problem. Choose the active one-site translation convention $T|n⟩=|n+1⟩$. A periodic plane-wave coefficient pattern requires $e^{iq(n+N)a}=e^{iqna}$, soThere are $N$ independent such wavevectors, for example one set of $N$ consecutive integers $\ell$ in a Brillouin zone. The normalized one-magnon Bloch states on a ring are $|q⟩=N^{-1/2}{\sum }_{n=1}^N{e}^{iqna}|n⟩$. Reindexing with the periodic coefficient condition gives $T|q⟩=e^{-iqa}|q⟩$ and hence $T^N|q⟩=|q⟩$.

There is a logical qualification to the condition stated in the paper. On an $N$-site ring $T^N=I$ already holds for every state. In particular, the finite sum written with an arbitrary real $q$ also returns to itself after $N$ translations; that condition alone does not quantize $q$. The displayed allowed values are the ones for which the ansatz is a single one-site translation eigenstate, equivalently a consistently periodic plane wave. This is the additional condition needed for the subsequent energy-eigenstate argument. Other $q$ values generally describe superpositions of the allowed momentum modes.

Apply the hopping Hamiltonian operator and shift the summation index using periodic boundary condition. The coefficients from $|n-1⟩$ and $|n+1⟩$ are respectively $e^{iqa}$ and $e^{-iqa}$ times the original coefficient. Thusand the nearest-neighbour ferromagnetic magnon dispersion isThe unnormalized state in the question has the same energy. Distinct allowed momenta are orthogonal, since $N^{-1}{\sum }_n{e}^{i(q-q^{\prime })na}={\delta }_{q{q}^{\prime }}$ over one set of representatives.

For large $N$ the sampled wavevectors become dense, and the requested sketch is the periodic curve $[E(q)-E_0]/J=1-\cos (qa)$ over $-2\pi <qa<2\pi$. It has a zero at $q=0$ and tends to the equivalent zero at either open endpoint; maxima $2J$ occur at $q=\pm \pi /a$. Near any equivalent zero,using $q$ measured from that zero. The low-momentum ferromagnetic magnon is quadratic, with group velocity ${\hslash }^{-1}dE/dq=(Ja/\hslash )\sin (qa)$. The $q=0$ state is proportional to the total spin lowering operator acting on $|{\psi }_0⟩$, so it belongs to the same ground multiplet and has exactly zero excitation energy. Extending the plot beyond one reciprocal period repeats states rather than adding new independent modes. The lattice spacing $a$ is the nearest-neighbor chord length specified in the paper; $Na$ is the translation length in this lattice description, not necessarily the continuous circumference of a circular arrangement.

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.

In the single-excitation subspace, part (a) turns each coupling into a hopping coefficient between adjacent sites. With $|\tilde{n}⟩$ denoting the excitation at site $N+n$, the two central links giveFor $N\ge 2$ the links from these neighbors to the next sites have strength $J_1$. Applying the Hamiltonian operator a second time yieldsThe central coefficient is $2(J_0/\sqrt{2}{)}^2=J_0^2$. On the comparison path, direct multiplication givesThus the two calculations match under $|0⟩\mapsto |\tilde{0}⟩$ and $|j⟩\mapsto (|\widetilde{-j}⟩+|\tilde{j}⟩)/\sqrt{2}$ for $j=1,2$. If $N=1$, the terms involving $J_1$ and distance two are absent on both sides.

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 $\sqrt{2}$.

Both the drift and local $Z_k$ control commute with $S$, so $[H_S+f(t)Z_k,S]=0$ at every time. The propagator consequently obeys $U(t{)}^{†}SU(t)=S$, as follows by differentiating this expression. Thus every reachable unitary operator preserves the magnetization sectors of a spin chain. If $P_q$ is a spectral projector of $S$, then $U$ commutes with $P_q$ and $\mathrm{Tr}(P_q\rho (t))=\mathrm{Tr}(P_q\rho (0))$.

The eigenspaces are nontrivial proper subspaces: $|0\cdots 0⟩$ has eigenvalue $N$, while $|10\cdots 0⟩$ has eigenvalue $N-2$. One cannot transfer the first pure state to the second, although their density operators have the same spectrum. Nor can one implement a unitary that mixes these two sectors. Hence $\boxed{\text{the full system has neither operator, density, nor pure-state controllability}}$. Its Dynamical Lie algebra lies in the block-diagonal commutant of $S$. Controllability within a fixed sector, such as the single-excitation subspace, is still possible and must be tested separately.

Assume $c_1{c}_2\ne 0$, as required for a connected chain and the supplied divisions. The single-excitation subspace has dimension three, so its Dynamical Lie algebra ${\mathfrak{g}}_1={\mathrm{Lie}}_{\mathbb{R}}\{iA,iB\}$ is contained in $\mathfrak{u}(3)$. The intended conclusion from nine independent skew-Hermitian directions is $\boxed{{\mathfrak{g}}_1=\mathfrak{u}(3)}$, giving unitary operator controllability, density operator controllability and pure-state controllability on this sector. Here is a direct generation argument that does not assume an independent identity control or rely on the normalization of the printed intermediate matrices.

Here $c_n$ denotes the hopping entry of the supplied reduced $A$. With the stated Pauli matrices, restricting $c_n(X_n{X}_{n+1}+Y_n{Y}_{n+1})$ instead gives hopping $2c_n$; absorbing that common factor into the reduced coupling does not change the Dynamical Lie algebra.

Write $R_{mn}=E_{mn}-E_{nm}$ and $Y_{mn}=i(E_{mn}+E_{nm})$ using matrix units. The physical generators giveNext isolate the second transition and generate its other direction:Finally $R_{13}=[R_{12},R_{23}]$ and $Y_{13}=[R_{12},Y_{23}]$. The three $R_{mn}$, three $Y_{mn}$ and two diagonal differences are eight independent traceless skew-Hermitian matrices, hence span $\mathfrak{su}(3)$. Because $\mathrm{Tr}(iB)=i\ne 0$, subtracting the traceless part of $iB$ also supplies $i{I}_3$, completing $\mathfrak{u}(3)$. This is the trace-direction step in identity augmentation in quantum controllability and the three-site instance of Endpoint control of an XX spin chain. If a coupling vanishes, the chain disconnects and the full-sector conclusion fails.

There is a redundant direction in the printed final pair of commutators: with the printed intermediate definitions, both are proportional to $E_{13}-E_{31}$. Thus that pair cannot itself establish the stated independence. The mixed commutator $[R_{12},Y_{23}]=i(E_{13}+E_{31})$ above provides the missing direction and establishes the required conclusion independently.

Let ${\mathfrak{g}}_{\mathrm{full}}$ be the algebra on the full eight-dimensional space. Restriction to the invariant single-excitation sector is a surjective Lie algebra homomorphism $R:{\mathfrak{g}}_{\mathrm{full}}\to \mathfrak{u}(3)$. If $\dim {\mathfrak{g}}_{\mathrm{full}}=9$, the rank-nullity theorem gives $\ker R=0$, soThis is an abstract algebra isomorphism; on the full Hilbert space it is a reducible eight-dimensional representation preserving sectors of dimensions $1,3,3,1$. The sector actions are linked by the same nine generators, rather than independently controllable blocks. In particular, this is not $\mathfrak{u}(8)$ or $\mathfrak{su}(8)$, so it does not imply full-system controllability. The full spin Hamiltonians are traceless, so their algebra is embedded in $\mathfrak{su}(8)$ even though its abstract type is $\mathfrak{u}(3)$; its central direction is not an unrestricted overall phase on the full space.

For a network of $N$ spin-one-half sites with $Z|0⟩=|0⟩$ and $Z|1⟩=-|1⟩$, the single-excitation subspace has exactly one spin in state $|1⟩$ and dimension $N$. Excitation-conserving exchange interactions preserve it. Its restricted Hamiltonian is a hopping matrix whose entries connect the corresponding sites. Controllability within this subspace need not imply controllability on the full $2^N$-dimensional space.

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.

The isotropic XY exchange interaction between two qubits is $E=(X\otimes X+Y\otimes Y)/2$. Its action isIndeed, $X|0⟩=|1⟩$, $X|1⟩=|0⟩$, $Y|0⟩=i|1⟩$, and $Y|1⟩=-i|0⟩$. The two terms cancel on equal bits and add on unequal bits. Thus a network Hamiltonian operator $H={\sum }_{a,b}{K}_{ab}{E}_{ab}$ preserves excitation number; in the single-excitation subspace, it is the weighted adjacency matrix, moving an excitation from $a$ to $b$ with hopping coefficient $K_{ab}$.