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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 seminal paper on PSG, the enlarged Hilbert space and gauge redundancy complicate our symmetry analyses.

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 $\widetilde{D}\neq D$, the translation operators depend on the 'fermion basis' we choose. Does this imply the translation operators unphysical?

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?

Thanks in advance.

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Please link to the arxiv free paper –  Trimok Sep 17 '13 at 15:30

1 Answer 1

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 same in the physical subspace, here is the proof (The notations used here are the same as those in Two puzzles on the Projective Symmetry Group(PSG)?):

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 Two puzzles on the Projective Symmetry Group(PSG)?, 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.

Note that $R$ is the local $SU(2)$ gauge rotation instead of spin rotation, and in the above proof we have used $[P,A]=[P,\widetilde{A}]=0$.

Remark: The spin-rotation symmetry operator is a little special in the sense that it is basis independent (This is obvious due to the SU(2) gauge structure of Schwinger-fermion representation).

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