Are the symmetry operators well defined in the context of Projective Symmetry Group(PSG)? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-08-25T20:25:49Z https://physics.stackexchange.com/feeds/question/77642 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/77642 3 Are the symmetry operators well defined in the context of Projective Symmetry Group(PSG)? Kai Li https://physics.stackexchange.com/users/21487 2013-09-17T13:12:52Z 2013-10-07T07:38:42Z <p>Consider the Schwinger-fermion approach $\mathbf{S}_i=\frac{1}{2}f_i^\dagger\mathbf{\sigma}f_i$ to spin-$\frac{1}{2}$ system on 2D lattices. Just as Prof.Wen said in his <a href="http://prb.aps.org/abstract/PRB/v65/i16/e165113" rel="nofollow">seminal paper on PSG</a>, the enlarged Hilbert space and gauge redundancy complicate our symmetry analyses. </p> <p>Now let's take the translation-symmetry as an example. The unitary translation-symmetry operator $D$ is defined as $D\psi_iD^{-1}=\psi_{i+a}$, where $\psi_i=(f_{i\uparrow},f_{i\downarrow}^\dagger)^T$ and $a$ is the lattice vector. As we know, the transformation $\psi_i\rightarrow \widetilde{\psi_i}=G_i\psi_i(G_i\in SU(2))$ doesn't change the spin operators and the projective opearator $P=\prod_{i}(2\hat{n}_i-\hat{n}_i^2)$(Note here $P\neq \prod _i(1-\hat{n}_{i\uparrow}\hat{n}_{i\downarrow})$). Similarly, in the new basis $\widetilde{\psi_i}$, we can define another translation-symmetry operator $\widetilde{D}$ as: $\widetilde{D}\widetilde\psi_i\widetilde{D}^{-1}=\widetilde\psi_{i+a}$. But $D\widetilde\psi_iD^{-1}=G_i\psi_{i+a}\neq \widetilde\psi_{i+a}$, which means that <strong>$\widetilde{D}\neq D$, the translation operators depend on the 'fermion basis' we choose</strong>. Does this imply the translation operators unphysical?</p> <p>But the translation operators should be physical, so are they equivalent in the physical subspace, say for any physical spin-state $\phi=P\phi$, does $\widetilde{D}\phi=D\phi$? If this is true, then how to prove it?</p> <p>Thanks in advance.</p> https://physics.stackexchange.com/questions/77642/-/78218#78218 1 Answer by Kai Li for Are the symmetry operators well defined in the context of Projective Symmetry Group(PSG)? Kai Li https://physics.stackexchange.com/users/21487 2013-09-22T17:52:01Z 2013-09-22T18:59:31Z <p>Luckly, I just found that I can answer this question by myself now, and the answer is 'Yes', the base-dependent symmetry operators become the <em>same</em> in the physical subspace, here is the proof (The notations used here are the same as those in <a href="https://physics.stackexchange.com/questions/77758/two-puzzles-on-the-projective-symmetry-grouppsg">Two puzzles on the Projective Symmetry Group(PSG)?</a>):</p> <p>Let $A$ be the symmetry operator(e.g., lattice translation, rotation, and parity symmetries, and time-reversal symmetry). First of all, $A$ should make sense in the physical subspace, in the sense that if $\phi$ is a physical state, then $A\phi$ should also be a physical state, this is true due to the fact $[P,A]=0$. Secondly, after a gauge rotation $\psi_i\rightarrow\widetilde{\psi_i}=R\psi_iR^{-1}=G_i\psi_i$, the symmetry operator $A$ defined in $\psi_i$ basis would changes to $\widetilde{A}=RAR^{-1}$ defined in $\widetilde{\psi_i}$ basis, now use the identity $PR=RP=P$ in <a href="https://physics.stackexchange.com/questions/77758/two-puzzles-on-the-projective-symmetry-grouppsg">Two puzzles on the Projective Symmetry Group(PSG)?</a>, it's easy to show that $\widetilde{A}P=AP$, and hence for any physical state $\phi$, we have $\widetilde{A}\phi=A\phi$, which means that the symmetry operator $A$ is well defined in the physical subspace.</p> <p>Note that $R$ is the local $SU(2)$ <strong>gauge rotation</strong> instead of <strong>spin rotation</strong>, and in the above proof we have used $[P,A]=[P,\widetilde{A}]=0$.</p> <p><strong>Remark:</strong> The <em>spin-rotation</em> symmetry operator is a little special in the sense that it is <em>basis independent</em> (This is obvious due to the SU(2) gauge structure of Schwinger-fermion representation).</p>