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Oct 5, 2022 at 0:39 answer added jjgoings timeline score: 4
Oct 3, 2022 at 20:30 comment added jjgoings Yup, seems it needs the extra factors of $\sqrt{2}$ I had left off before. See, e.g., Eq(14) in arxiv.org/abs/2109.01076
Oct 3, 2022 at 20:20 comment added jjgoings @CraigGidney Ah, good point! Maybe then the transformation I should have done is represent $|0\rangle$ as $1/\sqrt{2} \times$ leaf node? Likewise for the $\langle 1|$. Then the probabilities should work out in the simple example?
Oct 3, 2022 at 18:03 comment added Craig Gidney I don't think the scalar value is even supposed to be the probability in the first place? For example, the leaf node on its own has value $\sqrt{2}|0\rangle$ instead of $|0\rangle$ which is not normalized to start with.
S Oct 3, 2022 at 16:38 review First questions
Oct 4, 2022 at 6:11
S Oct 3, 2022 at 16:38 history asked jjgoings CC BY-SA 4.0