It is known that the one-dimensional (1D) Ising model can be mapped to a free Majorana model using a Jordan-Wigner transformation and its two degenerated ground states are well interpreted by the two Majorana zero modes at the two ends of the chain in "Majorana language".

It is evident that the ground states of an infinite (without ends) 1D Ising chain are degenerate. How can I understand the degeneracy in its Majorana counterparts which is also infinite (without ends)?

Another question which confuses me is whether we can treat this two-fold degeneracy as two Majorana zero modes in 1D Ising model. If not, what is the relationship between the two-fold degeneracy in the Ising model and the two Majorana zero modes in the free Majorana model?

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    $\begingroup$ We have a paper on this: arxiv.org/abs/1412.5985 "Topological degeneracy (Majorana zero-mode) and 1+1D fermionic topological order in a magnetic chain on superconductor via spontaneous Z2 symmetry breaking" $\endgroup$ Feb 10, 2015 at 21:54
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    $\begingroup$ There is an answer at physicsoverflow.org/27103 $\endgroup$ Mar 4, 2015 at 15:46

1 Answer 1


I think the reason is that when you have an Ising chain with open periodic boundary condition, it can be exactly mapped to an open Kitaev chain, so the degeneracy of the Ising chain becomes the degeneracy of the Kitaev chain, i.e. the Majorana zero modes. But if you have an Ising chain with periodic boundary condition, it cannot be exactly mapped to a periodic Kitaev chain. (The boundary conditions messes up the exact mapping.) So the degeneracy of the periodic Ising chain does not imply the degeneracy of a periodic Kitaev chain.


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