# Why can $(0,0,3/5,0,0,0,4/5,0,0)$ be written as $\frac35|3\rangle+\frac45|7\rangle$?

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."

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 $$7$$.

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.
• $\newcommand{\qr}[1]{\left|#1\right\rangle}$ 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.) Commented Oct 25, 2023 at 23:31
• @user1145880 take note many people (including me) start counting from $0$ instead of from $1$. Thus I would call Aaronson's vector $\frac{3}{5}|2\rangle + \frac{4}{5}|6\rangle$. Commented Oct 26, 2023 at 0:19