It is known that the Kitaev Hamiltonian and its spin-liquid ground state both break the $SO(3)$ 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 $SU(2)$.

Supplementary: As an example, consider a two spin-1/2 system. We want to gain some insights that what kinds of wavefunctions preserves the $Q_8$ spin-rotation symmetry from this simplest model. For convenience, let $R_\alpha =e^{\pm i\pi S_\alpha}=-4S_1^\alpha S_2^\alpha$ represent the $\pi$ spin-rotation operators around spin axes $\alpha=x,y,z$, where $S_\alpha=S_1^\alpha+ S_2^\alpha$. Therefore, by saying a wavefunction $\psi$ has $Q_8$ spin-rotation symmetry, we mean $R_\alpha\psi=\lambda_ \alpha \psi$, with $\left |\lambda_ \alpha \right |^2=1$.

After a simple calculation, we find that a $Q_8$ spin-rotation symmetric wavefunction $\psi$ could only take one of the following 4 possible forms:

$(1) \left | \uparrow \downarrow \right \rangle-\left | \downarrow \uparrow \right \rangle$, with $(\lambda_x,\lambda_y,\lambda_z)=(1,1,1)$ (Singlet state with full $SU(2)$ spin-rotation symmetry), which is annihilated by $S_x,S_y,$ and $S_z$,

$(2) \left | \uparrow \downarrow \right \rangle+\left | \downarrow \uparrow \right \rangle$, with $(\lambda_x,\lambda_y,\lambda_z)=(-1,-1,1)$, which is annihilated by $S_z$,

$(3) \left | \uparrow \uparrow \right \rangle-\left | \downarrow \downarrow \right \rangle$, with $(\lambda_x,\lambda_y,\lambda_z)=(1,-1,-1)$, which is annihilated by $S_x$,

$(4) \left | \uparrow \uparrow \right \rangle+\left | \downarrow \downarrow \right \rangle$, with $(\lambda_x,\lambda_y,\lambda_z)=(-1,1,-1)$, which is annihilated by $S_y$.

Note that any kind of superposition of the above states would no longer be an eigenfunction of $R_\alpha$ and hence would break the $Q_8$ spin-rotation symmetry.

  • 1
    $\begingroup$ When talking the rotational symmetry, people tends to refer to the SO(3) group and its subgroups. So the symmetry group here is $\tilde{D}_2$. $D_4$ is its projective representation (or projective symmetry group). $\endgroup$ Dec 30, 2013 at 15:49
  • 3
    $\begingroup$ To which papers are you referring? There's a notational irritation that always occurs when dealing with the Dihedral groups where some people write $D_n$ and some write $D_{m}$ - where $n$ is the number of edges and vertices of the n-gon and $m=2n$ is the number of group elements - for the same group. However, I don't think either of your $D_2$ or $D_4$ are groups. In the case of your $D_4$, $e^{i\pi S_x}e^{i\pi S_y}$ is not an element of your set. However, if both were taken to generate all elements of the group, it should be apparent that they would generate the same group here. $\endgroup$ Dec 30, 2013 at 15:56
  • $\begingroup$ @ Matthew TItsworth I have added the expressions for $S_x,S_y,S_z$ to my question, and it's direct to show that $e^{i\pi S_x}e^{i\pi S_y}=e^{-i\pi S_z}\in D_4$ and hence $D_4$ is of course a group. While $D_2$ is not a group as I have mentioned in the Comments at the end of my question. $\endgroup$
    – Kai Li
    Dec 30, 2013 at 18:28
  • 3
    $\begingroup$ @K-boy : As you noticed, your "$D_2$" is not a group, so your notation is wrong, and your $\tilde D_2$ is the true $D_2$, the dihedral group of rank $4$. And your "$D_4$" is not the dihedral group of rank $8$, but the quaternion group $Q= Q_4$ of rank $8$ (the first in the family of the dicyclic groups $Q_{2n}$, of rank $4n$). And the true dihedral $D_4$ group is not isomorphic to $D_2 \times Z_2$ (while $D_2$ is isomorphic to $Z_2 \times Z_2$). Ref: Ramond, Group Theory, Cambridge, pages $13-17$. Finally, (abstract) groups, and representations of groups, are two different things. $\endgroup$
    – Trimok
    Dec 30, 2013 at 20:18
  • 1
    $\begingroup$ @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. $\endgroup$ Dec 30, 2013 at 20:43

2 Answers 2


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.

  • 1
    $\begingroup$ 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. $\endgroup$ Dec 30, 2013 at 22:26
  • $\begingroup$ @ Matthew TItsworth Thanks for your clear summary. $\endgroup$
    – Kai Li
    Dec 30, 2013 at 22:30

The right answer was given by Everett You, see the first comment below my question: Using the notations in the question, the true spin symmetry of Kitaev model is $D_2=Q_8/Z_2$ rather than $Q_8$, since $Q_8$ contains a $Z_2$ gauge transformation. Just like the true spin symmetry of Heisenberg model is $SO(3)=SU(2)/Z_2$ rather than $SU(2)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.