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 features!

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

123 submissions , 104 unreviewed
3,547 questions , 1,198 unanswered
4,552 answers , 19,366 comments
1,470 users with positive rep
411 active unimported users
More ...

By saying a physical state has some 'symmetry', what do we really mean?

+ 1 like - 0 dislike
41 views

Here our arguments are restricted to the realm of the Projective Symmetry Group(PSG) proposed by Prof. Wen,

Quantum Orders and Symmetric Spin Liquids. Xiao-Gang Wen. Phys. Rev. B 65 no. 16, 165113 (2002). arXiv:cond-mat/0107071.

and the following notations are the same as those in my previous question, Two puzzles on the Projective Symmetry Group(PSG)?.

When we say the projected physical spin state $P\Psi$ has some 'symmetry', e.g., translation symmetry, there will be two understandings:

(1) After a translation of the mean-field Hamiltonian $H(\psi_i)$, say $DH(\psi_i)D^{-1}$, the physical spin state is unchanged, say $P\Psi'\propto P\Psi$, where $\Psi'$ is the ground state of the translated Hamiltonian $DH(\psi_i)D^{-1}$.

(2) $D(P\Psi)\propto P\Psi$.

I would like to know: are the above understandings equivalent to each other? Thanks in advance.

This post imported from StackExchange Physics at 2014-03-09 08:42 (UCT), posted by SE-user K-boy
asked Sep 20, 2013 in Theoretical Physics by Kai Li (975 points) [ no revision ]

1 Answer

+ 1 like - 0 dislike

I just found that I asked a naive question and I can answer it by myself now.

(1) and (2) are equivalent to each other. Because if $\Psi$ is a ground state of $H(\psi_i)$, then $\Psi'=D\Psi$ is the ground state of $DH(\psi_i)D^{-1}$, and $[P,D]=0$, therefore $D(P\Psi)=P\Psi'$.

Remark: More generally, when we talk about any kind of symmetry of the physical state, the identity $[P,A]=0$ is the reason for the equivalence between (1) and (2) statements. Where the unitary(or antiunitary, e.g. time-reversal) operator $A$ represents the corresponding symmetry.

This post imported from StackExchange Physics at 2014-03-09 08:42 (UCT), posted by SE-user K-boy
answered Sep 20, 2013 by Kai Li (975 points) [ no revision ]

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$\hbar\varnothing$sicsOverflow
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
...