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Quantum Computing

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  • $\begingroup$ Thank you for your clarification! I've also came up with the same decomposition for the CNOT gate. I would like to ask you another question if possible: in the decomposition we use the projective measurement (I+Z)/2, and if I use exactly this operation it seems that everything works well. Instead, if I try to really implement that operation by projecting the state into |0> I do not obtain the correct result for certain expectation values. The difference between the 2 methods is that in the first one I have a non-normalized state. How do I implement such a thing in a real quantum device? $\endgroup$ Commented Sep 30, 2022 at 10:20
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    $\begingroup$ You can fix the second approach if you multiply the projective measurement by the probability of such measurement outcome being obtained, as noted in Eq. (8) of the paper. On an actual quantum device, whenever (1 + $\alpha$ A)/2 appears in the middle of the circuit, we measure the operator A. This yields one of two outcomes, +1 or -1, which define the $\alpha$ associated with this particular circuit execution. For one such execution, no multiplication by the probability is made; this is only done at the end, when we have the statistics from the multiple circuit executions. $\endgroup$ Commented Sep 30, 2022 at 11:47
  • $\begingroup$ Thank you again for your answer, now it is finally clear. I have just one more question to proceed with my work: in this paper( arxiv.org/pdf/1712.09271.pdf ) and in other ones they introduce the C factor to evaluate the sampling overhead (C^(2n)) of the decomposition. It is the sum of the modulus of the coefficients of the decomposition. In the abstract of the paper of the 1st question, they say that the sampling overhead scales as 9^n, where "n" is the number of non-local gates that we decompose. Why 9? In the sum we have 10 terms with coefficient 1/2. So it should be 25(5^2). $\endgroup$ Commented Sep 30, 2022 at 13:48
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    $\begingroup$ In the decomposition derived by Mitarai and Fujii, there are only six terms (cf. Fig. 1 or Fig. 2), each with prefactor with absolute value $1/2$, so $C = 6 \times 1/2 = 3$, in which case the sampling overhead for $n$ such two-qubit operations is $3^{2n} = 9^n$, in agreement with Endo, Benjamin and Li. A more detailed explanation of this scaling is provided by Mitarai and Fujii in the proof of Theorem 5 (cf. page 6 of main text and Appendix E). $\endgroup$ Commented Oct 3, 2022 at 16:31
  • $\begingroup$ Thank you for the answer. So they consider the term inside the sum symbol as a unique term repeated 4 times (for the 4 values of alpha)? I thought I should have considered each term inside the sum separately, and in this way I obtain 10 terms in total. Thank you for the clarification $\endgroup$ Commented Oct 4, 2022 at 9:59