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 ...

A simple question on $SU(2)$ gauge transformations in Wen's papers on projective symmetry group (PSG)?

+ 1 like - 0 dislike
313 views

Recently I am studying the projective symmetry group (PSG) and the associated concept of quantum order first proposed by prof.Wen.

In Wen's paper, see the last line of Eq.(8), the local SU(2) gauge transformation for spinor operators is defined as $\psi_i\rightarrow G_i\psi_i$, where $\psi_i=(\psi_{1i},\psi_{2i})^T$ are fermionic operators and $G_i\in SU(2)$. Why we define it like this?

Since as we know, the Shcwinger fermion representation for spin-1/2 can be written as $\mathbf{S}_i=\frac{1}{4}tr(\Psi_i^\dagger\mathbf{\sigma}\Psi_i)$, where $\Psi_i=\begin{pmatrix} \psi_{1i} & -\psi_{2i}^\dagger \\ \psi_{2i} & \psi_{1i}^\dagger \end{pmatrix}$, and $G_i\Psi_i$ which is the same as the above transformation $\psi_i\rightarrow G_i\psi_i$ is in fact a spin rotation of $\mathbf{S}_i$, while $\Psi_iG_i$ does not change spin $\mathbf{S}_i$ at all.

So in Eq.(8), why we define the SU(2) gauge transformation as $G_i\Psi_i$ rather than $\Psi_iG_i$?

This post imported from StackExchange Physics at 2014-03-09 08:43 (UCT), posted by SE-user K-boy
asked Sep 2, 2013 in Theoretical Physics by Kai Li (975 points) [ no revision ]
retagged Mar 9, 2014

2 Answers

+ 3 like - 0 dislike

a) For a rotation of axis $\vec n$ and angle $\theta$, the transformation for a spin $1/2$ representation is :

$$\psi \to e^{i \frac{\theta}{2} \vec\sigma. \vec n}\psi = G(\vec n,\theta)\psi\tag{1}$$ where $\psi$ is a 2-component row complex spinor. Here $G(\vec n,\theta)$ is a member of $SU(2)$. The action of a matrix on a 2-component row object is always at the left of the object.

b) Beginning with : $\vec S=\frac{1}{4}tr(\Psi^\dagger \vec \sigma\Psi)$, suppose we have a transformation $\Psi \to \Psi G$. As you noticed, it would not change the value of $\vec S$, because :

$$tr(G^\dagger\Psi^\dagger \vec \sigma\Psi G)=tr(\Psi^\dagger \vec \sigma\Psi GG^\dagger) = tr(\Psi^\dagger \vec \sigma\Psi) \tag{2}$$

But $\vec S$ is a vector (it is in the fundamental or vectorial representation of $SO(3)$, or, if you prefer, in the adjoint representation of $SU(2)$), so it has to change under a rotation. So the transformation $\Psi \to \Psi G$ is non-valid and irrelevant.

This post imported from StackExchange Physics at 2014-03-09 08:43 (UCT), posted by SE-user Trimok
answered Sep 2, 2013 by Trimok (950 points) [ no revision ]
@ Trimok, thanks for your answer. But I am still unclear that why we define the SU(2) gauge transformation as $G_i \Psi_i$ rather than $\Psi_i G_i$? Maybe I'm not familiar with gauge theory.

This post imported from StackExchange Physics at 2014-03-09 08:43 (UCT), posted by SE-user K-boy
The transformation law, for the spinor representation, correspond to equation (1), so the transformation does not act on a matrix, but on a 2-row spinor. So, it is mandatory, in equation (1), that $G$ is at the left. Now, using $\Psi$, is a "trick" to present the spinor as a pseudo "matrix" to be used in a trace expression, it is not a fundamental presentation for the spinor representation. Now, you may consider the matrix $\Psi$ as $2 *2-$row spinors, so you may imagine that you apply $G$ on the first-column spinor, then on the second-column spinor, so $G$ must stay at the left, too.

This post imported from StackExchange Physics at 2014-03-09 08:43 (UCT), posted by SE-user Trimok
@ Trimok, ok. And in the gauge transformation $\psi_i \rightarrow G_i \psi_i$, why we restrict $G_i$ belong to $SU(2)$ rather than $U(2)$? Thanks.

This post imported from StackExchange Physics at 2014-03-09 08:43 (UCT), posted by SE-user K-boy
@K-boy : Spinor representations are some special representations of orthogonal groups $SO(p,q)$ which cannot be constructed with vectors or tensors (see Clifford algebra). If one is interesting with spinor representations of $SO(3)$, we have to look at the double cover of $SO(3)$, which is $SU(2)$ (this means that $S0(3) = SU(2)/Z_2$). So, we have to look at the representations of $SU(2)$. Each representation of $SU(2)$ is labelled by an integer or a half-integer. The representation corresponding to $1/2$ corresponds to a spinor representation. You may go further and considering the Lorentz

This post imported from StackExchange Physics at 2014-03-09 08:43 (UCT), posted by SE-user Trimok
@K-boy : .....group $SO(3,1)$, wich has a double cover $SL(2,C)$. The representations of $SL(2,C)$ are labelled by $2$ numbers $(p,q)$ which can be integers or semi-integers. A left-handed spinor corresponds to the representation $(1/2,0)$, a right-handed spinor correspond to the representation $(0,1/2)$. They transform in the same way under rotations, but not with boosts.

This post imported from StackExchange Physics at 2014-03-09 08:43 (UCT), posted by SE-user Trimok
+ 0 like - 0 dislike

God, I just found that I made a very foolish misunderstanding of prof.Wen's paper. In fact, the operator $\psi_i=(\psi_{1i},\psi_{2i})^T=(f_{1i},f_{2i}^\dagger)^T$ and $\mathbf{S_i}=\frac{1}{2}f_i^\dagger\mathbf{\sigma}f_i$, but I misunderstood $\mathbf{S_i}=\frac{1}{2}\psi_i^\dagger\mathbf{\sigma}\psi_i$ before(Please see Eq.(8) in this paper http://prb.aps.org/abstract/PRB/v65/i16/e165113 at the very beginning).

Now everything goes well. In fact, the $SU(2)$ gauge transformations $\psi_i \rightarrow G_i\psi_i$ are totally equivalent to $\bigl(\begin{smallmatrix} f_{1i} & -f_{2i}^\dagger\\ f_{2i} & f_{1i}^\dagger \end{smallmatrix}\bigr)\rightarrow \bigl(\begin{smallmatrix} f_{1i} & -f_{2i}^\dagger\\ f_{2i} & f_{1i}^\dagger \end{smallmatrix}\bigr)G_i$. And the $SU(2)$ matrices being on the left or right depends on the notation of spinon operartors that you define, which is not the key point here.

Note: The operators $\psi_i$ that appear in all of Wen's papers on PSG in 2002 are not the direct annihilation operators for Schwinger-fermions $f_i$, so please be very careful!

This post imported from StackExchange Physics at 2014-03-09 08:43 (UCT), posted by SE-user K-boy
answered Sep 7, 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$\hbar$ysi$\varnothing$sOverflow
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
...