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## 物理代写|量子计算代写Quantum computer代考|Quantum Fourier Transform

Let us recall the main definitions associated with the classical Fourier ${ }^{1}$ transform known from the course of mathematical analysis.

The Fourier transform is one of the central methods of modern applied mathematics [1,2].

Consider a function $s: \mathbb{R} \rightarrow \mathbb{C}$. Let $s(t)$ be absolutely integrable on its domain of definition, that is $\int_{-\infty}^{\infty}|s(t)| d t<\infty$.

The Continuous Fourier Transform of the function $s(t)$ is determined by the relation
$$S(f)=\int_{-\infty}^{\infty} s(t) e^{2 \pi i f t} d t$$
Function $S(f)$ is called an image of the original function $s(t)$.
Note. Traditionally, $s(t)$ in physical applications is considered as a signal, and $S(f)$ is considered as a signal spectrum. Furthermore, the variable $t$ has the meaning of time and $f$ has the meaning of linear frequency. Note that the designation $S(f)=$ $\mathcal{F}[s(t)]$ is widely used.

In computational problems, the considered function is usually defined not analytically, but in the form of discrete values on some grid, usually uniform.
In practical problems, the sample length
$$\mathbf{x}=\left(\ldots, x_{-2}, x_{-1}, x_{0}, x_{1}, x_{2}, \ldots\right)$$
is usually bounded, so let us consider the transition from a continuous integral transform $\mathcal{F}[s(t)]$ to an approximate integral sum:$$(\mathcal{F}[s(t)]){n}=\int{-\infty}^{\infty} s(t) e^{2 \pi i f_{n} t} d t \rightarrow \sum_{k=-\infty}^{\infty} x_{k} e^{2 \pi i f_{n} t_{k}} \delta=\delta \sum_{k=0}^{N-1} x_{k} e^{2 \pi i k n / N}$$
Thus, we come to the definition of the Discrete Fourier Transform (DFT) of $N$ values $x_{k}$, where $k=0,1, \ldots, N-1$ :
$$y_{n}=\sum_{k=0}^{N-1} x_{k} e^{2 \pi i k n / N}, 0 \leqslant n \leqslant N-1$$
or, using the designation for the root of unity $\omega=e^{2 \pi i / N}$,
$$y_{n}=\sum_{k=0}^{N-1} \omega^{k n} x_{k}, 0 \leqslant n \leqslant N-1 .$$

## 物理代写|量子计算代写Quantum computer代考|The Inverse Discrete Fourier Transform

The Inverse Discrete Fourier Transform is determined by the formula
$$x_{n}=\frac{1}{N} \sum_{k=0}^{N-1} \omega^{-k n} y_{k}, 0 \leqslant n \leqslant N-1 .$$
In a vector notation, the considered relations have the form $\mathbf{y}=\mathcal{F}[\mathbf{x}]$ and $\mathbf{x}=$ $\mathcal{F}^{-1}[\mathbf{y}]$

The sequential application of direct and inverse DFT to an arbitrary vector $\mathbf{x}$ does not change its components: $\forall \mathbf{x} \in \mathbb{R}^{n} \mathcal{F}^{-1}[\mathcal{F}[\mathbf{x}]]=\mathbf{x}$ and $\mathcal{F}\left[\mathcal{F}^{-1}[\mathbf{x}]\right]=\mathbf{x}$. Prove, for example, the first of these equations.
$$\left(\mathcal{F}^{-1}[\mathcal{F}[\mathbf{x}]]\right){n}=\frac{1}{N} \sum{k=0}^{N-1} \omega^{-k n}(\mathcal{F}[\mathbf{x}]){k}=\frac{1}{N} \sum{k=0}^{N-1} \omega^{-k n} \sum_{i=0}^{N-1} \omega^{i k} x_{i}$$
Let us change the summation order:
$$\left(\mathcal{F}^{-1}[\mathcal{F}[\mathbf{x}]]\right){n}=\sum{i=0}^{N-1} x_{i}\left(\frac{1}{N} \sum_{k=0}^{N-1} \omega^{k(i-n)}\right)$$
Further, use the properties of the values $\omega=e^{2 \pi i / N}$, considered in Exercise 75 (see also [3]):
$$\frac{1}{N} \sum_{k=0}^{N-1} \omega^{k(i-n)}= \begin{cases}1, & \text { or } i=n \ 0, & \text { or } i \neq n\end{cases}$$
Finally, we obtain $\left(\mathcal{F}^{-1}[\mathcal{F}[\mathbf{x}]]\right){n}=x{n} \forall n=0,1, \ldots, N-1$. The equality $\mathcal{F}\left[\mathcal{F}^{-1}[\mathbf{x}]\right]=\mathbf{x}$ is proved in the same way.

## 物理代写|量子计算代写Quantum computer代考|The Fast Fourier Transform

According to the definition, the computation of the vector $\mathbf{y}$ for the given $\mathbf{x}$ requires $O\left(N^{2}\right)$ complex multiplications. ${ }^{2}$ There is a way to significantly reduce the asymptotic complexity of the DFT. The Fast Fourier Transform algorithm, or FFT, requires only $O\left(N \log {2} N\right)$ multiplication operations; there are also ways to parallelize the FFT [6]. In further consideration, we will assume that $N=2^{m}$ for some natural number $m$. This limitation is not fundamental, because one of the following approaches can be used if you need to implement the Fast Fourier Transform algorithm for $N \neq 2^{m}$ : (1) fill in several cells of the array representing a signal with zeros, so that $N$ becomes equal to the nearest power of two; (2) use more complex generalizations of the FFT (see, for example, [7]). To distribute FFT operations among computational nodes, the following approach [8] is used: the vector $\mathbf{x}=\left(x{0}, \ldots, x_{N-1}\right)$, representing input signal counts, is considered as a two-dimensional array of size $N_{1} \times N_{2}$, where $N_{1}=2^{m_{1}}, N_{2}=2^{m_{2}}$ for some $m_{1}, m_{2} \in \mathbb{N}$, such that $2^{m_{1}+m_{2}}=N$. Then the index $i$ of an arbitrary element of vector $\mathbf{x}$ can be written in the form
$$i=L N_{1}+l \text {, where } 0 \leqslant l \leqslant N_{1}-1,0 \leqslant L \leqslant N_{2}-1 .$$
Components of Fourier transform $\mathbf{y}=\mathcal{F}[\mathbf{x}]$ are calculated according to the formula
$$y_{n}=\sum_{k=0}^{N_{1} N_{2}-1} e^{\frac{2 \pi i k n}{N_{1} N_{2}}} x_{k}$$
for all $0 \leqslant n \leqslant N_{1} N_{2}-1$, or
$$y_{\widetilde{I} N_{2}+\tilde{L}}=\sum_{L=0}^{N_{2}-1} \sum_{l=0}^{N_{1}-1} e^{\frac{2 \pi i}{N_{1} N_{2}}}\left(\widetilde{\left.I N_{2}+\tilde{L}\right)\left(L N_{1}+l\right)} x_{L N_{1}+l} .\right.$$
After algebraic transformations, let us write the components of the vector $\mathbf{y}$ in the following form:
$$y_{I N_{2}+\tilde{L}}=\sum_{l=0}^{N_{1}-1} e^{\frac{2 \pi i \bar{I} I}{N_{1}}}\left(e^{\frac{2 \pi i l \tilde{L}}{N_{1} N_{2}}}\left(\sum_{L=0}^{N_{2}-1} e^{\frac{2 \pi i L \bar{L}}{N_{2}}} x_{L N_{1}+l}\right)\right) .$$

## 物理代写|量子计算代写Quantum computer代考|Quantum Fourier Transform

$$S(f)=\int_{-\infty}^{\infty} s(t) e^{2 \pi i f t} d t$$

$$\mathbf{x}=\left(\ldots, x_{-2}, x_{-1}, x_{0}, x_{1}, x_{2}, \ldots\right)$$

$$(\mathcal{F}[s(t)]) n=\int-\infty s^{\infty}(t) e^{2 \pi i f_{n} t} d t \rightarrow \sum_{k=-\infty}^{\infty} x_{k} e^{2 \pi i f_{n} t_{k}} \delta=\delta \sum_{k=0}^{N-1} x_{k} e^{2 \pi i k n / N}$$

$$y_{n}=\sum_{k=0}^{N-1} x_{k} e^{2 \pi i k n / N}, 0 \leqslant n \leqslant N-1$$

$$y_{n}=\sum_{k=0}^{N-1} \omega^{k n} x_{k}, 0 \leqslant n \leqslant N-1$$

## 物理代写|量子计算代写Quantum computer代考|The Inverse Discrete Fourier Transform

$$x_{n}=\frac{1}{N} \sum_{k=0}^{N-1} \omega^{-k n} y_{k}, 0 \leqslant n \leqslant N-1 .$$

$$\left(\mathcal{F}^{-1}[\mathcal{F}[\mathbf{x}]]\right) n=\frac{1}{N} \sum k=0^{N-1} \omega^{-k n}(\mathcal{F}[\mathbf{x}]) k=\frac{1}{N} \sum k=0^{N-1} \omega^{-k n} \sum_{i=0}^{N-1} \omega^{i k} x_{i}$$

$$\left(\mathcal{F}^{-1}[\mathcal{F}[\mathbf{x}]]\right) n=\sum i=0^{N-1} x_{i}\left(\frac{1}{N} \sum_{k=0}^{N-1} \omega^{k(i-n)}\right)$$

$$\frac{1}{N} \sum_{k=0}^{N-1} \omega^{k(i-n)}={1, \quad \text { or } i=n 0, \quad \text { or } i \neq n$$

## 物理代写|量子计算代写Quantum computer代考|The Fast Fourier Transform

$$i=L N_{1}+l, \text { where } 0 \leqslant l \leqslant N_{1}-1,0 \leqslant L \leqslant N_{2}-1 .$$

$$y_{n}=\sum_{k=0}^{N_{1} N_{2}-1} e^{\frac{2 \pi k_{n}}{N 1 N_{2}}} x_{k}$$

$$y_{\tilde{I} N_{2}+\mathscr{L}}=\sum_{L=0}^{N_{2}-1} \sum_{l=0}^{N_{1}-1} e^{\frac{2 \pi^{2}}{\bar{H}{1} N{2}}}\left(I N_{2}+\widetilde{\tilde{L})\left(L N_{1}+l\right) x_{L N_{1}+l}}\right.$$

$$y_{I N_{2}+L}=\sum_{l=0}^{N_{1}-1} e^{\frac{22 i I I}{N_{1}}}\left(e^{\frac{2 \pi i l}{N 1 N 2}}\left(\sum_{L=0}^{N_{2}-1} e^{\frac{2 \pi i L L}{N_{2}}} x_{L N_{1}+l}\right)\right)$$

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