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The Hamiltonian \begin{equation} H=J\sum_{i,j}\vec{n}_i\cdot\vec{n}_j \end{equation} is invariant under a global rotation $\vec{n}_i\rightarrow R\vec{n}_i$, where $\vec{n}$ is a $N$ component rotor with fixed length $|\vec{n}|=1$ and $R$ is the $N\times N$ rotational matrix. Since $R$ satisfies $R^{T}R=1$, the determinant $|R|$ can be $+1$ or $-1$. Therefore, $R$ can take values from $SO(N)$ group which describes the rotation of a rigid rotator or from the $O(N)$ group which includes also improper rotations. It seems the above Hamiltonian can be called either $O(N)$ non-linear sigma model(mostly used) or $SO(N)$ non-linear model(used in some literatures). Then there are some problems.

  1. Is there any difference between $O(N)$/$SO(N)$ non-linear sigma model? Is the classical Heisenberg model about an $O(3)$ rotor or a $SO(3)$ rotor?
  2. If I want given dynamic to the Hamiltonian above by the local rotation $R_{i}$, what is the difference if $R_i \subset SO(3)$ and $R_i \subset O(3)$?
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  • $\begingroup$ @ hongchaniyi For the classical system, it should be $O(3)$; for the quantum one, it should be $SO(3)$, since the transformations should preserve the commutation relations in quantum mechanics. $\endgroup$ – Kai Li May 27 '14 at 23:00
  • $\begingroup$ So you have reduced your dilemma to effectively mapping (a,b,c,d,...) to (-a,b,c,d,...). What can you conclude? $\endgroup$ – Cosmas Zachos Feb 28 '17 at 20:41

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