Sunday 31 May 2020

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


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 $SU(2)$.


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.



Answer



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.


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...