Troubles with Haldane Shastry Spin Chain

Question

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?

Answer 1

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}