1
$\begingroup$

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

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

with

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

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.?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you for the comment. But how specifically do I relate complex conjugation and the Pauli matrices to symmetry operators? $\endgroup$ – Tom Nov 9 '18 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 '18 at 4:13
  • $\begingroup$ Thank you, but I still do not understand how to make symmetry operators out of them $\endgroup$ – Tom Apr 15 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 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 at 8:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.