Consider an arbitrary 1D chain (of length $N$) of fermions with an arbitrary quadratic Hamiltonian of the form

$$\mathcal{H}=\hat{\Psi}^\dagger H \hat{\Psi}$$


$$\hat{\Psi}=\left(a_1, a_2, ...,a_N,a_1^\dagger, a_2^\dagger, ...,a_N^\dagger \right)^T$$

a vector of fermionic operators where $a_n^\dagger$ creates a fermion at site $n$.

Are there some straight forward recipes for determining whether the Hamiltonian has any symmetries, specifically chiral, time-reversal, and particle-hole symmetry etc.?


1 Answer 1


Just use the eyeball technique: the form of $\hat{\Psi}$ suggests that you express the single particle hamiltonian $H$ as a $2 \times 2$-block operator and look for relations between the blocks. Hence, start with complex conjugation, the three Pauli matrices and products of Pauli matrices and complex conjugation.

  • $\begingroup$ Thank you for the comment. But how specifically do I relate complex conjugation and the Pauli matrices to symmetry operators? $\endgroup$
    – Tom
    Nov 9, 2018 at 10:55
  • $\begingroup$ These are natural candidates for your symmetries, because they naturally square to $\pm 1$ and give relations between $a_j$ and $a_j^{\dagger}$. $\endgroup$
    – Max Lein
    Nov 12, 2018 at 4:13
  • $\begingroup$ Thank you, but I still do not understand how to make symmetry operators out of them $\endgroup$
    – Tom
    Apr 15, 2019 at 7:33
  • $\begingroup$ You have 3 Pauli matrices and complex conjugation, which gives you 7 candidates for discrete symmetries. Given a hamiltonian, you need to see which, if any, (anti)commutes. In practice it is easier to compute $U H U^{-1}$ and compare that to $\pm H$. Conversely, if you are looking for a hamiltonian with particular symmetries (or that breaks particular symmetries), you select the candidates (e. g. an odd antiunitary). Then comparing $U H U^{-1}$ to $\pm H$ gives relations between the four block operators. This allows you to pick a model that preserves or breaks the relevant symmetries. $\endgroup$
    – Max Lein
    Apr 17, 2019 at 0:59
  • $\begingroup$ Thanks again, but I still do not understand how this relates to the symmetries of the system. Ok so I can form some $U$ that is some combination of the Pauli matrices, but what actually is it. What does it represent $\endgroup$
    – Tom
    Apr 18, 2019 at 8:17

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