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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?

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    $\begingroup$ 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 ). $\endgroup$ May 14, 2019 at 13:17
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    $\begingroup$ Could you show your work? It isn't clear what having "tried everything" means... $\endgroup$
    – Anyon
    May 16, 2019 at 13:20

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I'm not sure if this is still relevant to you, but I've been reading this paper recently. I'll just paste my derivation from my note. Please remind me if there's any error.

\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}

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