Suppose that we use the Schwinger-fermion ($\mathbf{S_i}=\frac{1}{2}f_i^\dagger\mathbf{\sigma}f_i$) mean-field theory to study the Heisenberg model on 2D lattices, and now we arrive at the mean-field Hamiltonian of the form $H_{MF}=\sum_{<ij>}(\psi_i^\dagger u_{ij}\psi_j+H.c.)$ with $u_{ij}=t\sigma_z$($t>0$), where $\psi_i=(f_{i\uparrow},f_{i\downarrow}^\dagger)^T$, and $\sigma_z$ is the third Pauli matrix.

Now let's find the IGG of $H_{MF}$, by definition, the pure gauge transformations in IGG should satisfy $G_iu_{ij}G_j^\dagger=u_{ij}\Rightarrow G_j=\sigma_zG_i\sigma_z $ on link $<ij>$—(1) . Specifically, consider the IGGs on the following different 2D lattices:

(a)Square and honeycomb lattices(unfrustrated): These two lattices can be both viewed as constituted by 2 sublattices denoted as $A$ and $B$. Due to Eq.(1), it's easy to show that for **both** of these two lattices the gauge transformations $G_i$ in **the same sublattice** are *site-independent* while those in different sublattices differ by $G_A=\sigma_zG_B\sigma_z$ and $IGG=SU(2)$.

(b)Triangular and Kagome lattices(frustrated):Due to Eq.(1), it's easy to show that for **both** of these two lattices the gauge transformations $G_i$ are *global* (site-independent) and $G_i=\bigl(\begin{smallmatrix}
e^{i\theta }& 0\\
0& e^{-i\theta }
\end{smallmatrix}\bigr)$ which means that $IGG=U(1)$.

So my question is: The same *form* mean-field Hamiltonian $H_{MF}$ may has *different* $IGGs$ on *different* lattices?

This post imported from StackExchange Physics at 2014-03-09 08:43 (UCT), posted by SE-user K-boy