Timeline for Hadamard in surface code not understood
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 1, 2023 at 12:39 | vote | accept | Ron Cohen | ||
| Feb 1, 2023 at 12:39 | |||||
| Feb 1, 2023 at 12:06 | comment | added | DaftWullie | If it was in the state $|0\rangle_L$ to start with, that means it was in the $+1$ eigenstate of $Z_L$. However, when we apply hadamards everywhere, $Z_L\rightarrow X_L$. Hence, the stabilized state is stabilized in the $|+\rangle_L$ state. | |
| Feb 1, 2023 at 11:12 | comment | added | Ron Cohen | But 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? | |
| Jan 31, 2023 at 10:43 | history | answered | DaftWullie | CC BY-SA 4.0 |