Why is the excitation energy of antiferro 1/2 spin XXZ chain associated with the $\psi_{n+2}$ and $\psi_{n-2}$

Question

I do not understand the equation on page 5 of nagaoso's book "quantum field theory in strongly correlated electronic systems". The Hamiltonian is shown in this form $H=\frac{J_\bot}{2}\sum_{i}(S_i^+S_{i+1}^-+S_i^-S_{i+1}^+)+\frac{J_z}{2}\sum_iS_i^zS_{i+1}^z$ and the ground state is $|singlet\rangle=\frac{1}{\sqrt{2}}(|\frac{1}{2},-\frac{1}{2}\rangle-|-\frac{1}{2},\frac{1}{2}\rangle)$. Assume the Ising limit $J_z\gg J_{\bot}>0$ and consider the state containing one domain wall. Calling $\Psi_n$ the wave function of the state where a domain wall is present bewteen site n and n+1, then obtain $(H-E_{Neel})\Psi_n=\frac{J_z}{2}\Psi_n+\frac{J_{\bot}}{2}(\Psi_{n+2}+\Psi_{n-2}).$

Question: I do not understand why $\Psi_{n+2}$ and $\Psi_{n-2}$ come in the equation. The figure below is the screenshot of the content on page 5.

Answer 1

When the off-diagonal terms in the Hamiltonian (those proportional to $J_\perp$) hit $\Psi_n$ you have only two terms surviving. When $S^-\otimes S^+$ hits the sites $(n-1,n)$ you get $\Psi_{n-2}$. When $S^+\otimes S^-$ hits the sites $(n+1,n+2)$ you get $\Psi_{n+2}$.