For Mosca Keynes, ex 7.1.5:

You are asked to prove:

$\text{QFT}^{-1}_{mr}|\phi_{r,b}\rangle = \frac{1}{\sqrt{r}}\sum_{k=0}^{r-1}e^{-2\pi i \frac{b}{r}k}|mk\rangle$


$|\phi_{r,b}\rangle = \frac{1}{\sqrt{m}}\sum_{z=0}^{m-1}|zr + b\rangle$

with period $r$, shift $b$ and $m$ repetitions.

I have an answer, I don't want to write my full workings so not to ruin the exercise for others, but I am looking to clarify a step in my workings to make sure I didn't just 'force' the proof.

I get to a point where I can factor to QFT result into two parts where get:

$\frac{1}{m\sqrt{r}}\sum_{z=0}^{m-1}\sum_{k=0}^{r-1}e^{-2\pi izk}e^{-2\pi i \frac{b}{r}k}|mk\rangle$

To get the final result I assume that:

$\sum_{z=0}^{m-1}\sum_{k=0}^{r-1}e^{-2\pi izk} = m$, given that $e^{-2\pi izk} = 1$ where $z,k\in\mathbb{Z}$, is this final stage of my proof correct or have I gone in completely the wrong direction?


1 Answer 1


If you assume that $e^{−2\pi z k} = 1$ then the sum is $ \sum_{z=0}^{m-1} \sum_{k=0}^{r-1} 1 = \sum_{z=0}^{m-1} r = mr, $


However, at the same time

$$ \begin{split} &\frac{1}{m\sqrt{r}}\sum_{z=0}^{m-1}\sum_{k=0}^{r-1}e^{-2\pi izk}e^{-2\pi i \frac{b}{r}k}|mk\rangle \\ =& \frac{1}{m\sqrt{r}}\sum_{z=0}^{m-1}\sum_{k=0}^{r-1}e^{-2\pi i \frac{b}{r}k}|mk\rangle \\ =& \frac{1}{m\sqrt{r}}m\sum_{k=0}^{r-1}e^{-2\pi i \frac{b}{r}k}|mk\rangle \end{split} $$ the $m$s cancel and it looks like you get your result.

  • $\begingroup$ Sorry yeah that was my typo including the sum over r as well, I was just trying to get the sum over m to vanish, which seems to work! Thanks for clarifying! $\endgroup$
    – Sam Palmer
    Jun 9, 2020 at 19:51

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