Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.