Higher-order topological superconductors from Weyl semimetals

We thank the referee for the further clarification, and indeed our previous response was not completely satisfactory.

The key question is that whether one should necessarily require $[T, R_{4z}]=0$. To this end, we would like first invoke a familiar conclusion — a two-fold symmetry necessarily commutes or anti-commutes with time-reversal symmetry. Indeed, in several well-known works on classifying topological crystalline insulators (TCI), e.g., the work by Shiozaki and Sato, the classification of a TCI with mirror symmetry M depends on whether $MT= TM$ or $ MT=-TM$.

In fact, here $MT=\pm TM$ are the only two options. This is due to the fact that $M, T$ and$ MT$ are two-fold symmetries. Since $MT$ is an anti-unitary symmetry, we can write $MT = U K$, where $U$ is a unitary operator and $K$ is complex conjugation. Then we have $(MT)^2 = UU^* = U(U^{-1})^T = \phi$. Since $MT$ is a two-fold symmetry, $\phi$ must be a diagonal operator. Therefore, $U = \phi U^T \phi$. Explicitly expressing the components, this is only possible if $\phi=\pm 1$. We thus have $MTMT=\pm 1$. Further assuming $M^2 =1$, and given that $T^2 = \pm 1$, we have $MT = \pm TM$.

However, to the best of our knowledge, no such constraints exist for a generic symmetry, including our four-fold symmetry $R_{4z}$. A counterexample is a generic spin rotation symmetry $S = \exp(i\sigma_z\theta/2)$, and a spinless time-reversal symmetry $T = K$. For a generic angle $\theta$, $ S$ and $T$ neither commute nor anti-commute. But this scenario is absolutely physical.

Going back to our case, time-reversal symmetry is broken in general but is restored at high-symmetry points. First, as we mentioned, the two $R_{4z}$ eigenvalues at every $K$ point in Fig. 5 are not TR partners. Instead each of them is invariant under TR. Second, because $R_{4z}$ does not commute or anti commute with T, its eigenvalues does not need to be real. In fact, our particular form of $R_{4z}$ in the BdG Hamiltonian comes from the fact that the $p+ip$ pairing order parameter carries an orbital angular momentum 1. In order to maintain rotation symmetry, a rotation in the pseudospin degree of freedom $\vec\tau=(\tau_x,\tau_y)$ needs to be incorporated into the rotation operator, i.e., $R_{4z} \propto \exp(i\tau_z \pi/4)$. This makes our case quite similar to the example with S above.

It is instructive to look at $R_{4z}^2=C_2$, which is a twofold symmetry. Its eigenstate $C_2 |Ψ\rangle = λ |Ψ\rangle$ satisfies the following relation: $ C_2 T |Ψ\rangle = \pm T C_2 |Ψ\rangle = \pm λ^* T |Ψ\rangle,$ where the $\pm$ depends on whether $C_2$ and $T$ commute or anti-commute. We see that if $T |Ψ\rangle = |Ψ\rangle$, then $λ$ needs to be either real (if $C_2$ and $T$ commute) or imaginary (if they anti-commute). In our case, $C_2 = i\tau_x$ and $T=K$, and we have $C_2 T = -T C_2$. Therefore $C_2$ eigenvalues are required to be imaginary. Indeed, by squaring the eigenvalues in Fig. 5, all $C_2$ eigenvalues are imaginary, consistent with the requirement of time-reversal symmetry.

We hope this addresses the referee’s insightful question.