# Proof of QFT for a Periodic Function

For Mosca Keynes, ex 7.1.5:

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

where

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

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?

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