I am reading how Hadamard is performed in surface code at https://arxiv.org/pdf/1208.0928.pdf
At some point, they apply H on all data qubits, and they claim that it converts X to Z stabilizers and vice versa, and I still do not understand why this identity is true:
Appendix J, page 49, bulletpoint 4:
We perform the key to this process, executing physical Hadamards on all the data qubits in the patch. As this changes the eigenbases from $\hat Z$ to $\hat X$ and vice versa, we change the identity of the measure qubits from $\hat X$ to $\hat Z$ stabilization and vice versa, as shown in Fig. 27e. Also, the $\hat X_L$ and $\hat Z_L$ logical operators swap their identities. This step, and the two that follow, are done in between two cycles of the surface code (so the measure-X and measure-Z qubits do not perform any stabilization on the isolated patch during these three steps).
Perhaps these is 2 different operations? 1) Apply H on data qubits, 2) change stabilizers
After this operation, we still have a stabilized logical qubit, so it can not be in +L state, but just in one of 0L or 1L. So it feels like the H did nothing. What am I missing?