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May 17, 2022 at 5:39 comment added quest @CraigGidney I guess I am done with that. Juts confused with np.dot and @ one more time
May 17, 2022 at 1:47 comment added quest @CraigGidney Could you give an example for a 8*8 matrix? (or 4*4 ) So that I can see what I am missing. I am trying do this method for my matrix (which is also 8*8) but probably missed something and I do not want to open a topic for the same question.
Aug 4, 2020 at 13:36 comment added Craig Gidney @vasjain The matrices have size 2^n by 2^n, so multiplying them naively has cost O((2^n)^3) = O(8^n). You do this O(n) times. Everything else is less expensive.
Aug 4, 2020 at 8:01 comment added vasjain I apologize, I didnt follow the complexity analysis. Could you explain it in a more detail as how you reached $O(8^N)$ and $2N$.
Aug 3, 2020 at 19:54 comment added Craig Gidney Note that you do need to check that the Paulis you inferred from the first column actually reproduce the rest of matrix.
Aug 3, 2020 at 19:08 comment added Craig Gidney Look at the first column of the matrix. It should have exactly one non-zero entry. The row of that entry in binary tells you which qubits got Pauli X operations. Then conjugate the matrix with Hadamards and repeat the same trick to get the locations of Pauli Zs. There's a lot of leeway here as it's not nearly as expensive as the matrix multiplication step.
Aug 3, 2020 at 17:59 comment added vasjain How would you check if a matrix is Pauli Product
Aug 3, 2020 at 17:31 history answered Craig Gidney CC BY-SA 4.0