It is known that the Kitaev Hamiltonian and its spin-liquid ground state both break the SU(2) spin-rotation symmetry. So what's the spin-rotation-symmetry group for the Kitaev model?
It's obvious that the Kitaev Hamiltonian is invariant under π rotation about the three spin axes, and in some recent papers, the authors give the "group"(see the Comments in the end) G={1,eiπSx,eiπSy,eiπSz}, where (eiπSx,eiπSy,eiπSz)=(iσx,iσy,iσz), with S=12σ and σ being the Pauli matrices.
But how about the quaternion group Q8={1,−1,eiπSx,e−iπSx,eiπSy,e−iπSy,eiπSz,e−iπSz}, with −1 representing the 2π spin-rotation operator. On the other hand, consider the dihedral group D2={(100010001),(1000−1000−1),(−10001000−1),(−1000−10001)}, and these SO(3) matrices can also implement the π spin rotation.
So, which one you choose, G,Q8, or D2 ? Notice that Q8 is a subgroup of SU(2), while D2 is a subgroup of SO(3). Furthermore, D2≅Q8/Z2, just like SO(3)≅SU(2)/Z2, where Z2={(1001),(−100−1)}.
Comments: The G defined above is even not a group, since, e.g., (eiπSz)2=−1∉G.
Remarks: Notice here that D2 can not be viewed as a subgroup of Q8, just like SO(3) can not be viewed as a subgroup of SU(2).
This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user K-boy