Timeline for Is the Pauli group for $n$-qubits a basis for $\mathbb{C}^{2^n\times 2^n}$?
Current License: CC BY-SA 4.0
6 events
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| Sep 4, 2021 at 14:45 | history | edited | biryani | CC BY-SA 4.0 |
fixed grammar
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| Sep 26, 2020 at 10:08 | comment | added | Enrico | @biryaniTo prove that the new set you obtained is linearly independent, wouldn't you need to prove that $Tr(AB)=0$ for every $A$ and $B$ of the new set? In that case, I haven't understood how the trace propriety with respect to the tensor product comes into play. When you compute $Tr(A_1 \otimes ... \otimes A_n) = Tr(A_1)...Tr(A_n)$ you are computing the trace of a new element of the set but not the trace of the product of the elements of that set $Tr(AB)$. In the latter case, it's not allowed to write $Tr(AB)=Tr(A)Tr(B)$ and then expand the two traces as you proposed, so I missed a step. | |
| Jul 11, 2018 at 9:42 | vote | accept | Josu Etxezarreta Martinez | ||
| Jul 11, 2018 at 9:12 | comment | added | biryani | Yes. In the case of error correction, general errors are decomposed into linear combination of Pauli errors and corrected. A more detailed explanation of how this is done can be found in theory.caltech.edu/people/preskill/ph229/notes/chap7.pdf. | |
| Jul 11, 2018 at 8:48 | comment | added | Josu Etxezarreta Martinez | Thanks for the answer. Does this then imply that by discretization of errors the consideration of the Pauli group as the set of all possible errors, then all the errors are considered too when designing an error correction code? | |
| Jul 11, 2018 at 8:26 | history | answered | biryani | CC BY-SA 4.0 |