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  Different invariant gauge groups (IGG) on different lattices with the same form mean-filed Hamiltonian?

+ 1 like - 0 dislike
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Suppose that we use the Schwinger-fermion (Si=12fiσfi) mean-field theory to study the Heisenberg model on 2D lattices, and now we arrive at the mean-field Hamiltonian of the form HMF=<ij>(ψiuijψj+H.c.) with uij=tσz(t>0), where ψi=(fi,fi)T, and σz is the third Pauli matrix.

Now let's find the IGG of HMF, by definition, the pure gauge transformations in IGG should satisfy GiuijGj=uijGj=σzGiσz 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 Gi in the same sublattice are site-independent while those in different sublattices differ by GA=σzGBσz 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 Gi are global (site-independent) and Gi=(eiθ00eiθ) which means that IGG=U(1).

So my question is: The same form mean-field Hamiltonian HMF 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
asked Sep 8, 2013 in Theoretical Physics by Kai Li (980 points) [ no revision ]
retagged Mar 9, 2014

1 Answer

+ 1 like - 0 dislike

Yes, it is possible. In my understanding, the most easy way to identify IGG is to study the loop of ansatz with the same basepoint. Starting from SU(2) formulation, if loops are colinear, then IGG=SU(2). If loops are coplanar, then IGG=U(1). Otherwise, IGG=Z2. The reason to consider loop is because only gauge flux (more accuracy, Wilson loop) is observable. You can easily see the difference between loops in honeycomb/square lattice with loops in Kagome/triangular lattice. And this is the true reason for your question. However, there is one thing to be notice: there may exists hidden symmetry beyond the formulation. For example, even if we use U(1) slave boson formulation, we can still have SU(2) spin liquid. However, it is very difficult to extract the hidden symmetry, I think.

This post imported from StackExchange Physics at 2014-03-09 08:43 (UCT), posted by SE-user Shenghan Jiang
answered Sep 10, 2013 by Shenghan Jiang (25 points) [ no revision ]
@ Shenghan Jiang, thanks for your good answer. But I'm studying PSG at the very beginning and got a lot of puzzles, like by 'loop of ansatz ', what does it mean? And are there only SU(2),U(1),Z2 three kinds of IGGs? From the math side, for example, can IGG=Z4?

This post imported from StackExchange Physics at 2014-03-09 08:43 (UCT), posted by SE-user K-boy
@ Shenghan Jiang, from the math viewpoint, we can construct many subgroups of SU(2) (math.stackexchange.com/questions/488309/…), so physically, I think all these subgroups may become the possible IGGs in proper physical models. What's your opinion? Thanks.

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

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