In this paper the author found that the ground state of the following Hamiltonian

$$ H = \sum_{i=1}^{L-1} [S_i \cdot S_{i+1} - \beta (S_i \cdot S_{i+1 })^2] , $$

where $\beta $ is a real parameter, is generally fourfold degenerate (as long as $L$ is large enough). It is a numerically finding. But why? Why is it degenerate? If so, why is it four? Some people say at each end of the chain, effective there is a spin-$1/2$. But how can you get spin-$1/2$ out of spin-1?

  • $\begingroup$ That Hamiltonian is in the so-called 'Haldane phase' which is a well-known example of a Symmetry Protected Topological (SPT) phase. These are characterized by how certain symmetries (e.g. the $SO(3)$ of a spin-$1$) are projectively represented on the edge of the system (e.g. as the $SU(2)$ representation, i.e. a 'spin-$\frac{1}{2}$'). The mechanism by which this happens is 'symmetry fractionalization'. This concept is for example explained in a different context (the Kitaev chain) here: physics.stackexchange.com/questions/199647/… $\endgroup$ – Ruben Verresen May 28 '17 at 10:02
  • $\begingroup$ You can get spin-1/2 out of spin-1 by symmetry fractionalization, which is a key feature of SPT phase. $\endgroup$ – Everett You Jun 14 '17 at 21:25

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