Heisenberg Hamiltonian in terms of spinon operators

Question

In Chapter 9 of Xiao-Gang Wen's book Quantum Field Theory of Many-body Systems, the spin operator $\textbf{S}_i$ is represented by $$\begin{equation}\textbf{S}_i=\frac{1}{2}f^{\dagger}_{i\alpha}\boldsymbol\sigma_{\alpha\beta}f_{i\beta}, \end{equation}$$ where $f_{i\alpha}$ denotes the spinon operator at site i with spin $\alpha=\uparrow,\downarrow.$ Then it states that the Heisenberg Hamiltonian \begin{equation} H=\sum_{<ij>}J_{ij}\,\mathbf{S}_i\cdot\mathbf{S}_j, \end{equation} can be written as, in terms of the spinon operators, \begin{equation} H=\sum_{<ij>}-\frac{1}{2}J_{ij}f^{\dagger}_{i\alpha}f_{j\alpha}f^{\dagger}_{j\beta}f_{i\beta}+\sum_{<ij>}J_{ij}\left(\frac{1}{2}n_i-\frac{1}{4}n_in_j\right). \end{equation} So there are 3 terms in total. By using the relation $\boldsymbol\sigma_{\alpha\beta}\cdot\boldsymbol\sigma_{\gamma\epsilon}=2\delta_{\alpha\epsilon}\delta_{\beta\gamma}-\delta_{\alpha\beta}\delta_{\gamma\epsilon}$, I can get the third term, and my confusion lies in the first 2 terms. Why is there a minus sign in the first term as well as an extra second term? From my computation, there should be no minus sign, and no second term at all.

Any help is appreciated.

Answer 1

I think you probably haven't fully taken into account the anticommutation relation $ \{ f_{i\alpha}, f_{j\beta}^\dagger\} = \delta_{ij} \delta_{\alpha\beta}$. After applying the Pauli matrix relation in your post, you should have the third term, as well as $$ \sum_{\langle ij\rangle} J_{ij} \frac{1}{2} f_{i\alpha}^\dagger f_{i\beta} f_{j\beta}^\dagger f_{j\alpha} = \sum_{\langle ij\rangle} J_{ij} \frac{1}{2} f_{i\alpha}^\dagger f_{i\beta} \left( \delta_{\alpha\beta} - f_{j\alpha}f_{j\beta}^\dagger \right) = \sum_{\langle ij\rangle} \left[ -\frac{1}{2} J_{ij} f_{i\alpha}^\dagger f_{j\alpha} f_{j\beta}^\dagger f_{i\beta} + J_{ij} \frac{n_i}{2}\right] $$ which gives you the sign change in the first term, as well as the second term.