# What symmetry class does 1D spinless $p$-wave superconductor belongs to?

$Z_{2}$ 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 $$H=\sum_k \phi_k^\dagger \begin{pmatrix} \xi(k) & 2i\Delta \sin(k)\\ -2i\Delta \sin(k ) & -\xi(k)\end{pmatrix}\phi(k)$$

• @AndrewMcAddams I don't understand what do you mean by spontaneous symmetry breaking here? – 12sa Jan 2 '15 at 20:27

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.
• 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? – 12sa Jan 6 '15 at 15:53
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.