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

+ 1 like - 0 dislike

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 (980 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 (980 points) [ no revision ]

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