Entanglement asymmetry in periodically driven quantum systems

We thank the referee for a careful reading of our manuscript and pointing out some important issues. We have now rewritten parts of the draft to make our calculation method clear and added an appendix. We think that these additions will address referee's concerns. Below, We respond to their comments in details. \

1. The trace of $n^{\rm th}$ power of the symmetry projected reduced density matrix has the following expression: \begin{align} {\rm Tr}(\rho_{A,Q}^{n}) = \int^\pi_{-\pi} ..... \int^\pi_{-\pi} \frac{d\alpha_1 .... d\alpha_n}{(2\pi)^n} {\rm Tr} \Big[\prod_{j=1}^n \rho_A \exp[i\alpha'_j Q_A] \Big] \end{align} where $\alpha'_j$ =$ \alpha_j-\alpha_{j+1} $ and $\alpha_{n+1}=\alpha_{1}$ so that $\sum_{j=1}^{n}\alpha'_{j} = 0$. In our previous response, we have mistakenly stated that this would reduce to the Goldstein and Sela's answer. We apologize for the misleading statement. We do agree to the referee that our set-up is different to that used by Goldstein and Sela for computing symmetry resolved entanglement entropy. Below, we provide the reason for this. We have now corrected several statements in the manuscript in Secs. 1, 4.1 and 5 and added an appendix to show the details of our calculation.

2. In our work, we have implicitly assumed that the defect line is not topological. The main justification of this comes from the fact $[\rho,Q]\neq 0$ in our case ( and hence $[\rho_A, Q_A] \ne 0$) since we are working on a strip which hosts boundary state: \begin{align} (T(z)-\bar{T}(\bar{z}))|B\rangle=0 \; \text{at} \; z=\bar{z}. \end{align} A solution of the above equation is a conformal boundary state $|B\rangle$, which can be written in terms of linear combination of Ishibashi states. By definition, an Ishibashi state is constructed out of a primary and taking sum of all level Virasoro descendants acting on that primary with a specific holomorphic structure(for reference, see equation (2.8) of arXiv:1412.6226). This kind of state is expected to break the symmetry of the charge sector since $H|B\rangle \neq 0$. We have computed the Renyi asymmetry in this state. Note that computing $n$-point bulk correlation function in a boundary state is equivalent to $2n$-point holomorphic correlation function in the full complex plane without defect. This `method of image trick' is a consequence of the Ward identity in BCFT, where one need to impose $T(z)=\bar{T}(\bar{z})$ at the boundary(a review of this is given in appendix (D) of arXiv:1907.08763). This further implies in the presence of boundary defect, each Riemann sheet (dressed with $\exp[i \alpha'_j Q_A]$) will contribute to the stress tensor as shown in Eq. (36) of our work and we got (2) as the conformal dimension of composite twist operator.

3. Regarding how to ensure that the asymmetry vanishes for $[\rho_{A},Q_{A}]=0$, we note the following. As the Referee correctly pointed out, for a topological defect operator, it is possible to fuse the defect lines into the same Riemann sheet (For reference, see JHEP05 (2024) 059). In such a case, the composite twist dimension should reduce to the same as referee has suggested (this can also be seen from expression of $X_n$) and yield \begin{align} d_{n}=\frac{c}{24}(n-\frac{1}{n}) + \Delta(\sum_{j=1}^{n}\alpha'_{j})/n^2 ---- (2) \end{align} Since $\sum_{j=1}^n \alpha'_j=0$, $\Delta=0$ in this case; hence the asymmetry will vanish. This will happen if the initial (ground) state respects the symmetry: $[\rho,Q]=0$. In this case, by definition, $[\rho_{A},Q_{A}]=0$ and we can fuse the defect line into one sheet. Using (3), we can easily see the entanglement asymmetry will vanish in this case. We thank the referee for pointing this out.

4. We note that from Eqs. (36) and (37) of our work, we end up with \begin{align} d_{n}=\frac{c}{24}(n-\frac{1}{n}) + \sum_{j=1}^{n-1}\Delta(\alpha'_{j})/n^2 \end{align} where $\Delta$ is a function of $\alpha'_j$ i.e. $\Delta(\alpha'_{j})$= $K (\alpha'_{j})^{2}$= $K (\alpha_{j}-\alpha_{j+1})^{2}$ for a class of CFTS. This form is dictated by the fact that for us $[\rho_A, Q_A]\ne 0$. This is central to obtaining the final answer as we claimed in equation (39) and (40) of our work.

We hope that these changes will address referee's concern regarding the computations in Sec. 4.1.

We thank the referee for their remark and for supporting publication.