Finding chiral-symmetric degenerate states numerically

Question

I am dealing with a Chiral-symmetric Hamiltonian such that

$$ 𝑆𝐻𝑆^{−1}=−𝐻. $$

Two of its eigenstates have zero eigenvalue and fulfill $𝑆∣𝜓_{\alpha}⟩=𝑒^{𝑖𝜙_{\alpha}}∣𝜓_{\alpha}⟩$, while the rest have finite eigenvalues and are pairwise symmetric. When I diagonalize my Hamiltonian numerically however, the zero energy states get mixed and the resulting eigenstates are therefore not symmetric.

When I have a symmetry that commutes with my Hamiltonian, $[𝐻,𝑈]=0$, and two degenerate eigenstates, to obtain the symmetric ones I can just add a small perturbation to the Hamiltonian proportional to $𝑈$ which breaks the degeneracy and does not modify the eigenstates. Is there a similar trick one can do to obtain symmetric degenerate eigenstates for symmetries that anticommute with the Hamiltonian?

Answer 1

Can you not apply $S$ to your numerical eigenstates? If so $(1+S)/2 $ and $(1-S)/2$ project out the $S\to \pm 1$ eigenstates.