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

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

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,873 answers , 20,701 comments
1,470 users with positive rep
502 active unimported users
More ...

Symmetry, gauge, and projective symmetry group (PSG)?

+ 3 like - 0 dislike
132 views

My following questions come from the understanding of the relations between the PSGs for two gauge-equivalent mean-field (MF) Hamiltonians (or MF ansatz). Considering the Schwinger-fermion ($f_{\sigma}$) MF approach to spin-liquid phases of spin-1/2 system. Let $H$ and $H'$ be two $SU(2)$ gauge-equivalent MF Hamiltonians, i.e., $H'=RHR^{-1}$, where the unitary operator $R$ represents an $SU(2)$ gauge rotation generated by $J=\frac{1}{2}\psi^\dagger \mathbf{\tau} \psi$ with $\psi=(f_\uparrow,f_\downarrow^\dagger)^T$.

Now, if $G_UU\in PSG(H)$, then simple calculation shows that $RG_UUR^{-1}\in PSG(H')$, where $U$ represents a symmetry operator and $G_U$ is the gauge operator associated with $U$. But we are used to the combined form of a symmetry operator followed by a gauge operator for an element in PSG, thus there are several ways to rewrite the expression $RG_UUR^{-1}$ as: (1) $(RG_UUR^{-1}U^{-1})U$; (2) $(RG_U)U'$ with $U'=UR^{-1}$; or (3) $(RG_UR^{-1})U''$ with $U''=RUR^{-1}$. So how to understand these expressions?

As for (1): The question is whether $RG_UUR^{-1}U^{-1}$ is an $SU(2)$ gauge operator? More specifically, it seems generally impossible to write $UR^{-1}U^{-1}$ as an gauge operator generated by $J=\frac{1}{2}\psi^\dagger \mathbf{\tau} \psi$. However, if we generalize the definition of $SU(2)$ gauge operators $R$ to those satisfying 3 properties A: Unitary; B: $R\psi_iR^{-1}=W_i\psi_i, W_i\in SU(2)$ matrices, which implies that physical spins should be gauge invariant (e.g., $R\mathbf{S}_iR^{-1}=\mathbf{S}_i$); and C: $RP=PR=P$ with projection operator $P$, which implies that physical spin-space should be gauge invariant. Then one can show that $UR^{-1}U^{-1}$ indeed fulfills the above 3 properties A,B,C where $U$ is time reversal, $SU(2)$ spin rotation, or lattice symmetries. (Furthermore, $R$ respects A,B,C $\Rightarrow R^{-1}$ respects A,B,C; $R_1,R_2$ both respect A,B,C $\Rightarrow R_1R_2$ respects A,B,C.) Therefore, the expression $RG_UUR^{-1}U^{-1}$ in (1) is an $SU(2)$ gauge operator in the sense A,B,C.

As for (2) or (3): We may ask: If $U$ represents some symmetry (e.g., time reversal, $SU(2)$ spin rotation, or lattice symmetries), then does $UR$ or $RU$ still represent the same physical symmetry? Where $R$ is an $SU(2)$ gauge operator (in the sense A,B,C mentioned above). One can show that $U'=UR$ or $RU$ represents the same physical symmetry as $U$ in the following sense: $U'\mathbf{S}_iU'^{-1}=U\mathbf{S}_iU^{-1}$ and $U'\phi=U\phi$, where $\phi=P\phi\in$ physical spin space. (Note that $U'\phi=U\phi$ is still a physical spin state due to $[P,U]=[P,U']=0$.) Therefore, the $U'$ and $U''$ in expressions (2) and (3) indeed represent the same physical symmetry as $U$.

Are my understandings correct? Thanks in advance.

A useful formula: Let $G_iU_i\in PSG(H), i=1,2,...,n$, then the $SU(2)$ gauge operator $G_U$ associated with the combined symmetry $U=U_1U_2\cdots U_n$ has the following form $$G_U=G_1U_1G_2U_2\cdots G_{n-1}U_{n-1}G_nU_{n-1}^{-1}\cdots U_2^{-1}U_1^{-1}$$ such that $G_UU\in PSG(H)$.

This post imported from StackExchange Physics at 2015-02-25 16:26 (UTC), posted by SE-user Kai Li
asked Nov 3, 2014 in Theoretical Physics by Kai Li (975 points) [ no revision ]
Another simple formula: If $G_UU\in PSG(H)$, then $G_{U^{-1}}=U^{-1}G_U^{-1}U$ such that $G_{U^{-1}}U^{-1}\in PSG(H)$.

This post imported from StackExchange Physics at 2015-02-25 16:26 (UTC), posted by SE-user Kai Li

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$\hbar$ysicsO$\varnothing$erflow
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).
To avoid this verification in future, please log in or register.




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

Your rights
...