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  Is the spin-rotation symmetry of Kitaev model $D_2$ or $Q_8$?

+ 3 like - 0 dislike
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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 $\pi$ rotation about the three spin axes, and in some recent papers, the authors give the "group"(see the Comments in the end) $G=\left \{1,e^{i\pi S_x}, e^{i\pi S_y},e^{i\pi S_z} \right \}$, where $(e^{i\pi S_x}, e^{i\pi S_y},e^{i\pi S_z})=(i\sigma_x,i\sigma_y,i\sigma_z )$, with $\mathbf{S}=\frac{1}{2}\mathbf{\sigma}$ and $\mathbf{\sigma}$ being the Pauli matrices.

But how about the quaternion group $Q_8=\left \{1,-1,e^{i\pi S_x}, e^{-i\pi S_x},e^{i\pi S_y},e^{-i\pi S_y},e^{i\pi S_z}, e^{-i\pi S_z}\right \}$, with $-1$ representing the $2\pi$ spin-rotation operator. On the other hand, consider the dihedral group $D_2=\left \{ \begin{pmatrix}1 & 0 &0 \\ 0& 1 & 0\\ 0&0 &1 \end{pmatrix},\begin{pmatrix}1 & 0 &0 \\ 0& -1 & 0\\ 0&0 &-1 \end{pmatrix},\begin{pmatrix}-1 & 0 &0 \\ 0& 1 & 0\\ 0&0 &-1 \end{pmatrix},\begin{pmatrix}-1 & 0 &0 \\ 0& -1 & 0\\ 0&0 &1 \end{pmatrix} \right \}$, and these $SO(3)$ matrices can also implement the $\pi$ spin rotation.

So, which one you choose, $G,Q_8$, or $D_2$ ? Notice that $Q_8$ is a subgroup of $SU(2)$, while $D_2$ is a subgroup of $SO(3)$. Furthermore, $D_2\cong Q_8/Z_2$, just like $SO(3)\cong SU(2)/Z_2$, where $Z_2=\left \{ \begin{pmatrix}1 & 0 \\ 0 &1\end{pmatrix} ,\begin{pmatrix}-1 & 0 \\ 0 &-1 \end{pmatrix} \right \}$.

Comments: The $G$ defined above is even not a group, since, e.g., $(e^{i\pi S_z})^2=-1\notin G$.

Remarks: Notice here that $D_2$ can not be viewed as a subgroup of $Q_8$, 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
asked Dec 30, 2013 in Theoretical Physics by Kai Li (980 points) [ no revision ]
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@ Matthew TItsworth You are welcome. $D_2$ appears in the 3rd paragraph on page 1 of this paper.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user K-boy
@K-boy. Look at the order of the elements in your $D_4$. They are $\{1,2,4,4,4,4\}$. The The dihedral group of order $8$ has two elements of order $4$ and five elements of order $2$. See also here,here,and here. Trimok is right.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user Matthew Titsworth
@ Matthew TItsworth Thank you very much. I will rethink about it.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user K-boy
Maybe I made a mistake and my $D_4$ is not isomorphic to the dihedral group of rank 8.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user K-boy
@ Trimok Thanks for your corrections.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user K-boy
Most recent comments show all comments
Of course it is. It helps if I pull the $i$ out. It also helps if I don't comment before I've had coffee. Apologies.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user Matthew Titsworth
Take the closure of $D_2$ under multiplication. This obviously gives you the set $D_4$. So the sets $D_2$ and $D_4$ generate the same group. This was the reason for the statement about notation and asking which paper you are referring to when you say "in some recent papers, the authors give..." I searched through Kitaev's "Anyons..." paper, but there is no mention of it there. There's also no mention of $D_2$ in the Yao and Lee paper. I don't have a copy of the Baskaran paper immediately available. It would be helpful if you could clarify the context from which your question is drawn.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user Matthew Titsworth

1 Answer

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The set $G$ gives the representation of the identity and generators of the abstract group of quaternions as elements in $SL(2,\mathbb C)$ which are also in $SU(2)$. Taking the completion of this yields the representation $Q_8$ of the quaternions presented in the question.

From the description of the symmetry group as coming from here, consider the composition of two $\pi$ rotations along the $\hat x$, $\hat y$, or $\hat z$ axis. This operation is not the identity operation on spins (that requires a $4\pi$ rotation). However, all elements of $D_2$ given above are of order 2.

This indicates that the symmetry group of the system should be isomorphic to the quaternions and $Q_8$ is the appropriate representation acting on spin states. The notation arising there for $D_2$ is probably from the dicyclic group of order $4\times 2=8$ which is isomorphic to the quaternions.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user Matthew Titsworth
answered Dec 30, 2013 by Matthew Titsworth (200 points) [ no revision ]
It probably worth mentioning that the quaternion group $Q$ is one of the two Schur covers of the Klein four-group $K$. The other one is $D_4$, the dihedral group of degree 4.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user Isidore Seville
@ Matthew TItsworth Thanks for your clear summary.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user K-boy

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