In the latest reports, Referee Dr.~Frank Schindler reiterated that from a physical point of view, the roto-inversion operator $\mathsf{R}_{4z}$ and time-reversal $\mathsf{T}$ must commute. According to the referee, while in certain examples, the commutation between the spatial symmetry operator and time-reversal operator may be violated, it can be restored if the operators are chosen to their ``physical" forms. Referee Prof.~Titus Neupert very nicely summarized that the disagreement stemmed from two different conventions of defining symmetry operators, and asked us to elucidate the symmetry operations from the perspective of electronic orbitals.

We fully agree with both referees' point of view, and we thank them for helping us better understand their concerns. It turns out our initial result is fully consistent with the symmetry representation of orbital degrees of freedom, which can be clarified by a modification of notation.

Interpretation of roto-inversion symmetry in the normal state

Let us first consider the normal state on which time-reversal is represented as $\mathsf T = K$, and and rotoinversion (which is a combination of $C_ {4z}$ and $M_ z$) operator is given by $\mathsf R_ {4z} = \hat f_ 1(0) \sigma_ x + \hat f_ 3(0) \sigma_ z$, where $\sigma$ represents our psudespin degree of freedom, and $\hat f^2_ 1(0) + \hat f^2_ 3(0) = 1$, as given in Eq. (8) in our paper. Importantly, here $\mathsf R_ {4z}$ and $ \mathsf{T}$ do commute.

Such a choice of symmetry operators can indeed be understood from the perspective of orbital degrees of freedom. Without loss of generality, let us consider the special case of $\hat f_ 1(0) = 0$, and $\hat f_ 3(0) = 1$, such that $\mathsf R_ {4z} = \sigma_ z$. In this case, we have two orbitals per unit cell both located at rotoinversion invariant points. The $|\sigma_ z = 1\rangle$ orbital can be chosen to be , e.g., an $s$-orbital, while the $|\sigma_ z = -1\rangle$ can be a $d_ {xy}$-orbital. This choice is not unique; for example $|\sigma_ z = 1\rangle=|d_ z^2\rangle$ and $|\sigma_ z = - 1\rangle = |p_ z\rangle$ would also work.

For general values of $\hat f_ 1(0)$ and $\hat f_ 3(0)$, the physical interpretation of $|\sigma_ z = \pm 1\rangle$ orbitals can be obtained as a linear superposition of $|s\rangle$ and $|d_ {xy}\rangle$ via a unitary transformation. Generally, the $\mathsf R_{4z} $ degree of freedom can be realized by, e.g., a (spin-polarized) $s$-orbital or a $d_ {z^2}$-orbital, while the $|\mathsf R_ {4z} =-1\rangle$ degree of freedom can be realized by, e.g. a $d_ {xy}$-orbital or a $p_ z$-orbital.

Non-commutation between roto-inversion symmetry and time-reversal symmetry in the BdG Hamiltonian

Now we move on to the superconducting state with the $p+ip$ pairing term $H_ {\Delta} = \int_ {k} \Delta^{\alpha \beta} (k) c^\dagger_ {\alpha}(k) c^\dagger_ {\beta}(- k)$, which transforms under rotoinversion in the non-trivial one-dimensional representation as $\mathsf R_ {4z} : H_ {\Delta} \mapsto - i H_ {\Delta} $, since it carries angular momentum $1$. Consequently, the mean field Hamiltonian $H = H_ n + H_ {\Delta} + H^\dagger_ {\Delta}$, is not invariant under the action of operator $\mathsf R_ {4z} = \hat f_ 1(0) \sigma_ x + \hat f_ 3(0) \sigma_ z$.

We note that the additional factors of $i$ in front of the pairing terms that spoil the symmetry can be removed by an additional $U(1)$ transformation: $U(\frac{\pi}{4}) : c^\dagger_a (k) \rightarrow e^{i\frac{\pi}{4}} c^\dagger_a (k)$, and $U(\frac{\pi}{4}) : c_a (k) \rightarrow e^{-i\frac{\pi}{4}} c_a (k)$. This $U(1)$ operation leaves the normal part of the Hamiltonian unchanged. Therefore, we define a new operator $\tilde R_ {4z} = U(\frac{\pi}{4}) \mathsf R_ {4z} $. (In the paper we had an abuse of notation in which we did not give this new combined symmetry a new symbol. We admit this is confusing, and have now fixed this.) We now have $[\tilde R_ {4z}, \mathsf T] \neq 0$ because of the added $U(1)$ factor. Crucially, even if $\tilde R_ {4z}$ is not the ``physical" roto-inversion operator, its action on the spatial part of the Hamiltonian is identical to roto-inversion. Indeed, we relied on the eigenvalues of $\tilde R_ {4z}$ to find the Wannier representation of the BdG Hamiltonian by treating it as that of an insulator.

In the Nambu space, this $U(1)$ operator is represented by $e^{-i\frac{\pi}{4} \tau_ z}$, and thus we have $\tilde R_ {4z} = e^{i\frac{\pi}{4} \tau_ z} \mathsf R_ {4z} = e^{-i\frac{\pi}{4} \tau_ z} (\hat f_ 1(0) \sigma_ x + \hat f_ 3(0) \sigma_ z) $, as given in the paper. We note that constructing the symmetry operator for the BdG Hamiltonian starting form the symmetry operator in the normal state and the one-dimensional representation of the pairing terms in the way we just described follows closely the constuction given in Eqs. (7), (8), and (9) of [Trifunovic 2019] (https://arxiv.org/pdf/1910.11271.pdf).

We also note that this additional unitary operator in the Nambu space plays a similar role as $M_ y$ does in the example given by Referee Titus Neupert in obtaining $[M_ x,\tilde T]\neq 0$, although there the modification was on the time-reversal operator $\tilde T \equiv M_ y T$.

Finally, we also note that our choice of time-reversal symmetry $\mathsf{T}=K$ is, strictly speaking, also not ``physical", since the physical time-reversal symmetry operator squares to $-1$. As is well understood and similar to what was mentioned by Referee Neupert, here $\mathsf{T}=K$ can be understood as a composite of the physical time-reversal symmetry $\mathsf{T}' = -is_ y K$ and a spin rotation $is_y$. In this context, we assume the band electrons we consider is fully spin polarized.