Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 57 / 3

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.

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.

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 $\sqrt{2}$.