# Troubles with Haldane Shastry Spin Chain

I'm reading the article "Exact solution of an S=1/2 Heisenberg antiferromagnetic chain with long-ranged interactions", which shows how to solve the problem of a long range-inverse squared interacting spin chain. I'm having trouble with equations 5 and 6 of the article. I tried everything but could not get the same result. Does anyone know how it works?

• It would be best to type the equations into the question (using Mathjax not an image -images of equations and pages are strongly discouraged on Physics SE and Mathjax is the standard ). May 14, 2019 at 13:17
• Could you show your work? It isn't clear what having "tried everything" means... May 16, 2019 at 13:20

\begin{align} &\prod_{q\in Q} a^{\dagger}_{q}=\frac{1}{\sqrt{N^M}}\prod_{q\in Q} \sum_{n=1}^{N}e^{iqn}a^{\dagger}_n\\ &=\frac{1}{\sqrt{N^M}}\sum_{n_1,...n_M=1}^N(\prod_{i=1}^Me^{iq_in_i})a^{\dagger}_{n_1}...a^{\dagger}_{n_M}\\ &=\frac{1}{M!\sqrt{N^M}}\sum_{n_1,...n_M=1}^N\sum_{\sigma\in S_M}(\prod_{i=1}^Me^{iq_in_{\sigma(i)}})a^{\dagger}_{n_{\sigma(1)}}...a^{\dagger}_{n_{\sigma(M)}}\\ &=\frac{1}{M!\sqrt{N^M}}\sum_{n_1,...n_M=1}^N\sum_{\sigma\in S_M}(\prod_{i=1}^Me^{iq_in_{\sigma(i)}})sgn(\sigma)a^{\dagger}_{n_1}...a^{\dagger}_{n_M}\\ &=\frac{1}{M!\sqrt{N^M}}\sum_{n_1,...n_M=1}^N\det(e^{iq_in_j})a^{\dagger}_{n_1}...a^{\dagger}_{n_M}\\ &\bar{P}_G\prod_{p\in(-K)^c}a^{\dagger}_{p\uparrow}\prod_{q\in Q}a^{\dagger}_{q\downarrow}|0\rangle\\ &=\frac{\bar{P}_G}{(M!)^2N^M}\sum_{n_1,...n_M=1}^N\sum_{m_1...m_M=1}^N \det(e^{ip_in_j})\det(e^{iq_im_j})a^{\dagger}_{n_1\uparrow}...a^{\dagger}_{n_M\uparrow}a^{\dagger}_{m_1\downarrow}...a^{\dagger}_{m_M\downarrow}|0\rangle\\ &=\frac{1}{(M!)^2N^M}\sum_{n_1,...n_M=1}^N\sum_{\sigma\in S_M}\det(e^{ip_in_j})\det(e^{iq_in_{\sigma(j)}})a^{\dagger}_{n_1\uparrow}...a^{\dagger}_{n_M\uparrow}a^{\dagger}_{n_{\sigma(1)}\downarrow}...a^{\dagger}_{n_{\sigma(M)}\downarrow}|0\rangle \\ &=\frac{1}{(M!)^2N^M}\sum_{n_1,...n_M=1}^N\sum_{\sigma\in S_M}\det(e^{ip_in_j})\det(e^{iq_in_{\sigma(j)}})sgn(\sigma)a^{\dagger}_{n_1\uparrow}a^{\dagger}_{n_1\downarrow}...a^{\dagger}_{n_M\uparrow}a^{\dagger}_{n_{M}\downarrow}|0\rangle\\ &=\frac{1}{M!N^M}\sum_{n_1,...n_M=1}^N\det(e^{ip_in_j})\det(e^{iq_in_{j}})a^{\dagger}_{n_1\uparrow}a^{\dagger}_{n_1\downarrow}...a^{\dagger}_{n_M\uparrow}a^{\dagger}_{n_{M}\downarrow}|0\rangle \end{align}