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Context. $\newcommand{\qr}[1]{\left|#1\right\rangle}$ A passage from a lecture by Scott Aaronson: "As an example, instead of writing out a vector like $$(0,0,3/5,0,0,0,4/5,0,0),$$ you can simply write $$\frac{3}{5}\qr{3} + \frac{4}{5}\qr{7},$$ omitting all of the 0 entries."

Source: https://www.scottaaronson.com/democritus/lec9.html

Question. Where does the $\qr{3}$ and $\qr{7}$ come from? I do understand the amplitudes $3/5$ and $4/5$, but I'm looking for help on understanding the $3$ and $7$.

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A vector is a list of elements that represent the weights for the sum of some basis. This basis can in this case be the set $\{|1\rangle,|2\rangle,|3\rangle,|4\rangle,|5\rangle,|6\rangle,|7\rangle,|8\rangle,|9\rangle\}$. The weights are zero for $\{|1\rangle,|2\rangle,|4\rangle,|5\rangle,|6\rangle,|8\rangle,|9\rangle\}$ and therefore the corresponding terms omitted in the summation.

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    $\begingroup$ $\newcommand{\qr}[1]{\left|#1\right\rangle}$ So, this means that the basis is (in "base 2") $\qr{0001}, \qr{0010}, \qr{0011} = \qr{3}, ..., \qr{0111} = \qr{7}, ..., \qr{1001}$? (Thank you. I get what you mean about the weights.) $\endgroup$ Commented Oct 25, 2023 at 23:31
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    $\begingroup$ Yes, assuming you adopt MSB first format. $\endgroup$ Commented Oct 25, 2023 at 23:33
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    $\begingroup$ @user1145880 take note many people (including me) start counting from $0$ instead of from $1$. Thus I would call Aaronson's vector $\frac{3}{5}|2\rangle + \frac{4}{5}|6\rangle$. $\endgroup$ Commented Oct 26, 2023 at 0:19
  • $\begingroup$ Thank you, @MarkSpinelli. Your comment made a big difference here. Thanks a lot. (Also thanks to gIS for writing a better question.) $\endgroup$ Commented Oct 26, 2023 at 21:13

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