# Why can we drop bits in front of the decimal in QFT?

In Preskill's notes on quantum information, he includes a section on the quantum Fourier transform (QFT) for period finding. Starting from the classical Fast FT over bitstrings, we can express any ket $$|x\rangle \in |\{0,1\}^n\rangle$$ as it's binary expansion $$x = \sum_i 2^i x_i\equiv (x_{n-1}x_{n-2}\cdots x_0.);\\ i \in \{0, ..., n-1\}$$

Then, the FFT phase factor, $$xy/N$$ where $$N\equiv 2^n$$. We define $$[N]\equiv\{0,...,N-1\}$$ and evaluate the phase factor as: $$\frac{xy}{N} =\frac{1}{N}\big(\sum_{i\in[n]}2^ix_i\big) y \\ =\big(\sum_{i\in[n]}\frac{1}{2^n}2^i x_i\big) y \\ =\big(\sum_{i\in[n]}2^{(i-n)} x_i \big) y \\ =\frac{1}{2^n}x_0y + \frac{1}{2^{n-1}}x_1 y +\cdots+\frac{1}{2}x_{n-1}y \\ =x_0 (.y_{n-1}\cdots y_{0}) + x_1 (y_{n-1}.y_{n-2}\cdots y_0) +\cdots+x_{n-1}(y_{n-1}y_{n-2}\cdots y_1.y_0)$$

Then, Preskill (and others) make the claim that the terms in the product $$\frac{xy}{N}$$ when exponentiated are equivalent to: $$=x_0 (.y_{n-1}\cdots y_{0}) + x_1 (.y_{n-2}\cdots y_0) +\cdots+x_{n-1}(.y_0)$$ Basically saying the terms in front of the decimal make no contribution to $$\exp(2\pi i x y/N)/N$$.

This is not at all obvious to me and I don't see anywhere that motivates this. It must be the result of the periodicity of the exponential about powers of two but I don't see it worked out anywhere and the index hauling gets unwieldy quickly. What calculation do I need to perform in order to get to this result?

Recall that $$\exp(2\pi i x y /N)$$ is $$N$$-periodic:
$$\exp(2\pi i x y / N) = \exp(2 \pi i (x y + N)/N).$$
Preskill's notes uses $$N = 2^n$$ and so in the product of $$xy$$ you can freely discard the terms that are multiples of $$2^n$$. Try to carry out the calculation yourself given the expressions for $$x$$ and $$y$$ in (6.51).
If the case with general $$n$$ is getting messy just try $$n = 3$$ and observe the pattern.