C003 derivation

On the two-level subspace $\{|{\downarrow}\rangle,|e\rangle\}$ of each ion, $\sigma_{x,de}=|{\downarrow}\rangle\langle e|+|e\rangle\langle{\downarrow}|$. The collective operator is $$S_{x,e}=\sigma_{x,de}\otimes\mathbb1+\mathbb1\otimes\sigma_{x,de}.$$

Restrict to the closed three-dimensional subspace spanned by $$|0\rangle\equiv|{\downarrow}{\downarrow}\rangle,\quad |1\rangle\equiv\tfrac{1}{\sqrt2}(|{\downarrow}e\rangle+|e{\downarrow}\rangle),\quad |2\rangle\equiv|ee\rangle,$$ which is the symmetric ($J=1$) ladder of two spin-$1/2$'s built from $\sigma_{x,de}$. In this basis $S_{x,e}$ is the spin-1 $J_x$: $$S_{x,e}\to\sqrt2\begin{pmatrix}0&1&0\\1&0&1\\0&1&0\end{pmatrix}\cdot\tfrac{1}{\sqrt2}? \;\;\Longrightarrow\;\; S_{x,e}|0\rangle=\sqrt2\,|1\rangle,\;\; S_{x,e}|1\rangle=\sqrt2(|0\rangle+|2\rangle),\;\; S_{x,e}|2\rangle=\sqrt2\,|1\rangle.$$ Hence $$S_{x,e}^2|0\rangle = \sqrt2\,S_{x,e}|1\rangle = 2(|0\rangle+|2\rangle),\qquad S_{x,e}^2|2\rangle = 2(|0\rangle+|2\rangle).$$ Within the two-dimensional space $\{|0\rangle,|2\rangle\}$, $S_{x,e}^2$ acts as $2\begin{pmatrix}1&1\\1&1\end{pmatrix}=2(\mathbb1+X)$ where $X$ swaps $|0\rangle\leftrightarrow|2\rangle$ (the $|1\rangle$ component vanishes because $S_{x,e}^2|0\rangle$ has none). Thus on this 2D space $$U_A(\Phi)=e^{-i\Phi S_{x,e}^2}=e^{-2i\Phi}\,e^{-2i\Phi X}=e^{-2i\Phi}\big(\cos(2\Phi)\,\mathbb1 - i\sin(2\Phi)\,X\big).$$ Therefore $$\langle ee|U_A(\Phi)|{\downarrow}{\downarrow}\rangle = \langle2|U_A|0\rangle = -i\,e^{-2i\Phi}\sin(2\Phi),\qquad |\langle ee|U_A|{\downarrow}{\downarrow}\rangle|^2=\sin^2(2\Phi).$$ At $\Phi=\pi/4$, $\sin^2(\pi/2)=1$: complete population transfer $|{\downarrow}{\downarrow}\rangle\to|ee\rangle$. Verified.