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

Question

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

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.

This post imported from StackExchange Physics at 2014-03-09 08:42 (UCT), posted by SE-user K-boy