Timeline for Diagrammatic language for fault-tolerance: I don't understand the difference between preparation correctness and propagation properties
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 6, 2022 at 18:29 | comment | added | Marco Fellous-Asiani | Overall, PCP+PPP means "the preparation does the job it is supposed to do AND errors "somehow" do not propagate". Only "somehow" because of my example (i) in which propagation is present. Those are two independent properties. But the most "important one" for a computer would be PCP I guess (however for proof purposes we might need PPP as well). | |
| Mar 6, 2022 at 18:28 | comment | added | Marco Fellous-Asiani | Thanks. Would you agree with the following? (i) PPP doesn't always ensure that errors do not propagate. For instance for $s=t=1$, I could imagine that my preparation created a logical error (in this case one fault induced many errors but the resulting space being in the code space, PPP will be verified). (ii) PCP says that for $s \leq t$, the preparation "does the job it is supposed to do". Then, PPP does not imply PCP from my example (i). Also, PCP does not imply PPP because in PCP we could have one fault inducing two errors conceptually possible. | |
| Mar 6, 2022 at 15:40 | comment | added | Jahan Claes | @StarBucK if a SINGLE error occurs early on in a state preparation, and the state preparation procedure happens to spread errors, then a single error early on could result in, say, a high-weight Pauli error on the overall state. PPP and PCP are precisely the conditions that say this doesn’t happen. | |
| Mar 6, 2022 at 15:38 | comment | added | Jahan Claes | @Starbuck you’re assuming that the preparation step doesn’t spread errors badly. But this is precisely what PPP and PCP are trying to enforce. | |
| Mar 6, 2022 at 12:20 | comment | added | Marco Fellous-Asiani | Hello. Thank you for your answer. Actually, I totally agree with the first part of your message, and this is why I am even more confused. Both in PPP and PCP, there is the condition $s \leq t$, where $t$ is the number of errors that the code can correct. For me, this condition exactly ensures that if you tried to prepare the logical $|0\rangle$, and have $s$ errors during its preparation, you are within $s$ errors of $|0\rangle$ but "further away" from $|1\rangle$. Hence I don't see how we could be closer to another codeword than the one we prepared. Hence, for me PPP $\Rightarrow$ PCP | |
| Mar 6, 2022 at 3:24 | history | answered | Jahan Claes | CC BY-SA 4.0 |