$Z_{2}$ topological invariant exist for Kitaev model.

What symmetries does it conserve? And to what symmetry class it belongs to? The hamiltonian for kitaev model can be written as $$ H=\sum_k \phi_k^\dagger \begin{pmatrix} \xi(k) & 2i\Delta \sin(k)\\ -2i\Delta \sin(k ) & -\xi(k)\end{pmatrix}\phi(k) $$

  • $\begingroup$ @AndrewMcAddams I don't understand what do you mean by spontaneous symmetry breaking here? $\endgroup$
    – 12sa
    Jan 2, 2015 at 20:27

2 Answers 2


It belongs to the symmetry class of no symmetry. i.e. the only symmetry is the fermion-number-parity conservation $Z_2^f$, which is always the symmetry of fermionic systems. See my paper http://arxiv.org/abs/1111.6341 for a discussion on the full-symmetry group $G_f$ for fermion systems.


The Kitaev model belongs to class D of the Altland-Zirnbauer classification. Here's the periodic table of non-interacting (gapped) fermionic topological systems.

enter image description here

The one circled in red corresponds to the 1D $p$-wave superconductor (or Kitaev chain). As you can see from the symmetry columns, it only possesses particle-hole symmetry ($\Xi$), while the time-reversal symmetry ($\Theta$) and the so-called chiral symmetry ($\Pi = \Theta \Xi$) are explicitly broken.

  • $\begingroup$ the time reversal symmetry for spinless hamiltonian can be checked using equation $h(-k)=h(k)^*$ and by using this equation it seems that above hamiltonian is time reversal symmetric so what I am missing here? $\endgroup$
    – 12sa
    Jan 6, 2015 at 15:53
  • $\begingroup$ Just the fact that you are studying a spinless system implies time-reversal symmetry is broken. Electrons are spinful particles. However, electrons can be effectively made spinless by spin-polarizing them with a magnetic field (internal or external). In general, some form of magnetic order needs to be present in order to create a Kitaev chain. Magnetic order, by definition, breaks time-reversal symmetry. $\endgroup$
    – NanoPhys
    Jan 7, 2015 at 2:16
  • $\begingroup$ here we are not considering spin degree of freedom so we can't talk about magnetic order or spin degree of freedom ? $\endgroup$
    – 12sa
    Jan 10, 2015 at 22:32
  • $\begingroup$ Yes, we have to freeze out the spin degree of freedom. There are many ways to do that: external magnetic field, intrinsic magnetism, etc. For example, in the Kouwenhoven experiment (dx.doi.org/10.1126/science.1222360), they used an external magnetic field, whereas in the Yazdani experiment (dx.doi.org/10.1126/science.1259327), they used a ferromagnetic chain of iron atoms. $\endgroup$
    – NanoPhys
    Jan 11, 2015 at 3:00
  • $\begingroup$ @NanoPhys ,Isn't The hamiltonian as written in the question belong to BDI class ? How do i compare it with $\endgroup$ Mar 18, 2015 at 12:57

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