Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

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

+ 1 like - 0 dislike
3557 views

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
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, $IGG=Z_2$. 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 $SU(2), U(1), Z_2$ three kinds of IGGs? From the math side, for example, can $IGG=Z_4$?

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

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysics$\varnothing$verflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...