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## 物理代写|量子计算代写Quantum computer代考|Phase Evaluation Algorithm

In a number of quantum-mechanical problems, you need to define eigenvalues of some unitary operator $U$ :
$$U|\psi\rangle=\lambda|\psi\rangle,$$
where $|\psi\rangle$ is a known eigenvector of operator $U, \lambda$ is an eigenvalue, which needs to be defined. Due to the unitarity of $U$, the equality $|\lambda|=1$ is fulfilled, and therefore the eigenvalue can be represented as $\lambda=e^{2 \pi i} \varphi_{r}$, where $\varphi$ is a real number (as it is said, the transformation phase) satisfies the condition $0 \leqslant \varphi<1$.
Let the bit representation of the value $\varphi$ is finite:
$$\varphi=0 . \varphi_{1} \varphi_{2} \ldots \varphi_{n} .$$
In the phase estimation algorithm, there will be used 2 quantum registers, the first of which contains $n$ qubit, originally pre-set in $|0\rangle$. The second register is originally in the state $|\psi\rangle$, the number of qubits in this register depends on the specific type of the operator $U$. The algorithm consists of the following steps.

1. For $N=2^{n}$, we compute the Direct Fourier Transform, applied to $|\underbrace{00 \ldots 0}{n q u b i t s}||\psi\rangle$. The result is equal to $\sum{q=0}^{N-1}|q\rangle|\psi\rangle$.
2. Apply $U$ to the second register $q$ times:
$$|q\rangle|\psi\rangle \rightarrow|q\rangle U^{q}|\psi\rangle=e^{2 \pi i \varphi q}|q\rangle|\psi\rangle$$
The first $n$ qubits form the state
$$\frac{1}{\sqrt{N}} \sum_{q=0}^{N-1} e^{2 \pi i \varphi \varphi q}|q\rangle=\mathcal{F}\left[\left|2^{n} \varphi\right\rangle\right] .$$
3. Calculate the inverse Fourier Transform $\mathcal{F}^{-1}$, applied to the first $n$ qubits and perform the measurement procedure. We obtain $\left|2^{n} \varphi\right\rangle=\left|\varphi_{1} \varphi_{2} \ldots \varphi_{n}\right\rangle$ with a probability, equal to one.

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

The search for information in some data structure is an important task in the field of computer sciences. Traditionally, it is considered a simple data structure-an array with integer elements. Generalization of search algorithms into more complex structures, as a rule, does not cause any difficulties [1, 2].

Let us briefly remind you how the search problem is solved by classical methods.
Let there be a problem of determining the index of the array element equal in value to this particular value, which is called target. Sequential search algorithm is applied to work with an unordered array and operates in the following way: sequentially looks through the array elements starting from the first one and compares them with the target one. If the sought element is found, the index of this element is returned, otherwise the result must be a number that does not correspond to any element, for example, $-1$.

Analysis of the sequential search algorithm shows that when the target element is located at the end of an array, the number of required element operations is $N$, where $N$ is the array size. For the complexity of this algorithm in the average case, we get the expression
$$A(N)=\frac{N+1}{2}=\Theta(N)$$
which uses the asymptotic notation $\Theta(g(N))$ is the class of functions growing at the same rate as $g(N)$ (see note on page 47 ). Thus, the search in an unordered array has an asymptotic complexity linear in the size of input data.

Consider the quantum analogue of the search algorithm. We formulate a quantum search problem in the following form.
There is specified an unordered set of non-negative integers
$${0,1,2, \ldots, N-1}$$
exactly one of which is the sought number, let us denote such sought number by $q_{0}$. To check if any value is appropriate, there is used the following function:
$$F:{0,1,2, \ldots, N-1} \rightarrow \mathbb{B}$$

which operates in accordance with the rule:
$$F(q)= \begin{cases}1, & \text { if } q=q_{0} \ 0, & \text { if } q \neq q_{0}\end{cases}$$

## 物理代写|量子计算代写Quantum computer代考|Answers, Hints, and Solutions

1. Answer. $\left[\sigma_{1}, \sigma_{2}\right]=2 i \sigma_{3},\left[\sigma_{2}, \sigma_{3}\right]=2 i \sigma_{1},\left[\sigma_{3}, \sigma_{1}\right]=2 i \sigma_{2}$.
2. Solution. $\sigma_{1} \sigma_{2} \sigma_{3}=i I$, where $I$ is the identity matrix of size $2 \times 2$.
3. Answer. $\sigma_{1}^{2}+\sigma_{2}^{2}+\sigma_{3}^{2}=3 I$, where $I$ is the identity matrix of the second order.
4. Solution.
We will use the Taylor ${ }^{9}$ formula to decompose the exponents into a power series:
$$e^{t}=\sum_{j=0}^{\infty} \frac{t^{j}}{j !}$$
In this formula, we substitute $t=\mathrm{i} \sigma_{k} \varphi$ for $k \in{1,2,3}$ and $\varphi \in \mathbb{R}$. We obtain
$$\exp \left(\mathrm{i} \sigma_{k} \varphi\right)=\sum_{j=0}^{\infty} \frac{\left(\mathrm{i} \sigma_{k} \varphi\right)^{j}}{j !}=\sum_{j=0}^{\infty} \frac{\mathrm{i}^{j} \sigma_{k}^{j} \varphi^{j}}{j !}$$
Note that $\sigma_{k}^{2}=I$ (see formula (2.3)), which leads to the validity of $\sigma_{k}^{2 j}=I$ and $\sigma_{k}^{2 j+1}=\sigma_{k}$ for all non-negative integer $j$. Next, we present the decomposition in the Taylor series with two sums, by odd and even degrees:
\begin{aligned} \exp \left(\mathrm{i} \sigma_{k} \varphi\right) &=\sum_{j=0}^{\infty}\left(\frac{\mathrm{i}^{2 j} \sigma_{k}^{2 j} \varphi^{2 j}}{(2 j) !}+\frac{\mathrm{i}^{2 j+1} \sigma_{k}^{2 j+1} \varphi^{2 j+1}}{(2 j+1) !}\right) \ &=\sum_{j=0}^{\infty} \frac{(-1)^{j} I \varphi^{2 j}}{(2 j) !}+\sum_{j=0}^{\infty} \frac{(-1)^{j} \mathrm{i} \sigma_{k} \varphi^{2 j+1}}{(2 j+1) !} \ &=I \sum_{j=0}^{\infty} \frac{(-1)^{j} \varphi^{2 j}}{(2 j) !}+\mathrm{i} \sigma_{k} \sum_{j=0}^{\infty} \frac{(-1)^{j} \varphi^{2 j+1}}{(2 j+1) !} \ &=I \cos \varphi+\mathrm{i} \sigma_{k} \sin \varphi \end{aligned}
Obtained sums $\sum_{j=0}^{\infty} \frac{(-1)^{j} \varphi^{2 j}}{(2 j) !}$ and $\sum_{j=0}^{\infty} \frac{(-1)^{j} \varphi^{2 j+1}}{(2 j+1) !} \operatorname{converge}$ to $\cos \varphi$ and $\sin \varphi$, respectively. As a result, we get:
$$\exp \left(\mathrm{i} \sigma_{k} \varphi\right)=I \cos \varphi+\mathrm{i} \sigma_{k} \sin \varphi \text { for } k \in{1,2,3}, \varphi \in \mathbb{R} \text {. }$$

## 物理代写|量子计算代写Quantum computer代考|Phase Evaluation Algorithm

$$U|\psi\rangle=\lambda|\psi\rangle$$

$$\varphi=0 . \varphi_{1} \varphi_{2} \ldots \varphi_{n}$$

1. 为了 $N=2^{n}$ ，我们计算直接傅里叶变换，应用于 $|\underbrace{00 \ldots 0} n q u b i t s||\psi\rangle$. 结果等于 $\sum q=0^{N-1}|q\rangle|\psi\rangle$.
2. 申请 $U$ 到第二个㝯存器 $q$ 次:
$$|q\rangle|\psi\rangle \rightarrow|q\rangle U^{q}|\psi\rangle=e^{2 \pi i \varphi q}|q\rangle|\psi\rangle$$
首先 $n$ 量子比特形成状态
$$\frac{1}{\sqrt{N}} \sum_{q=0}^{N-1} e^{2 \pi i \varphi p \rho q}|q\rangle=\mathcal{F}\left[\left|2^{n} \varphi\right\rangle\right]$$
3. 计算傅里叶逆变换 $\mathcal{F}^{-1}$, 应用于第一个 $n$ 量子比特并执行测量程序。我们获得 $\left|2^{n} \varphi\right\rangle=\left|\varphi_{1} \varphi_{2} \ldots \varphi_{n}\right\rangle$ 概率，等于一。

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

$$A(N)=\frac{N+1}{2}=\Theta(N)$$

$$0,1,2, \ldots, N-1$$

$$F: 0,1,2, \ldots, N-1 \rightarrow \mathbb{B}$$

$$F(q)=\left{1, \quad \text { if } q=q_{0} 0, \quad \text { if } q \neq q_{0}\right.$$

## 物理代写|量子计算代写Quantum computer代考|Answers, Hints, and Solutions

1. 回答。 $\left[\sigma_{1}, \sigma_{2}\right]=2 i \sigma_{3},\left[\sigma_{2}, \sigma_{3}\right]=2 i \sigma_{1},\left[\sigma_{3}, \sigma_{1}\right]=2 i \sigma_{2}$.
2. 解决方安。 $\sigma_{1} \sigma_{2} \sigma_{3}=i I$ ， 在哪里 $I$ 是大小的单位矩阵 $2 \times 2$.
3. 回答。 $\sigma_{1}^{2}+\sigma_{2}^{2}+\sigma_{3}^{2}=3 I$ ，在哪里 $I$ 是二阶单位矩阵。
4. 解决方案。
5. 我们将使用泰勒 ${ }^{9}$ 将指数分解为草级数的公式:
6. $$7. e^{t}=\sum_{j=0}^{\infty} \frac{t^{j}}{j !} 8.$$
9. 在这个公式中，我们代入 $t=\mathrm{i} \sigma_{k} \varphi$ 为了 $k \in 1,2,3$ 和 $\varphi \in \mathbb{R}$. 我们获得
10. $$11. \exp \left(\mathrm{i} \sigma_{k} \varphi\right)=\sum_{j=0}^{\infty} \frac{\left(\mathrm{i} \sigma_{k} \varphi\right)^{j}}{j !}=\sum_{j=0}^{\infty} \frac{\mathrm{i}^{j} \sigma_{k}^{j} \varphi^{j}}{j !} 12.$$
13. 注意 $\sigma_{k}^{2}=I$ (见公式 (2.3) )，这导致了有效性 $\sigma_{k}^{2 j}=I$ 和 $\sigma_{k}^{2 j+1}=\sigma_{k}$ 对于所有非负整数 $j$. 接下来，我们以奇数度和偶数度呈现泰 勒级数中的分解，其中包含两个和:
14. $$15. \exp \left(\mathrm{i} \sigma_{k} \varphi\right)=\sum_{j=0}^{\infty}\left(\frac{\mathrm{i}^{2 j} \sigma_{k}^{2 j} \varphi^{2 j}}{(2 j) !}+\frac{\mathrm{i}^{2 j+1} \sigma_{k}^{2 j+1} \varphi^{2 j+1}}{(2 j+1) !}\right) \quad=\sum_{j=0}^{\infty} \frac{(-1)^{j} I \varphi^{2 j}}{(2 j) !}+\sum_{j=0}^{\infty} \frac{(-1)^{j} \sigma_{k} \varphi^{2 j+1}}{(2 j+1) !}=I \sum_{j=0}^{\infty} \frac{(-1)^{j} \varphi^{2 j}}{(2 j) !}+\mathrm{i} \sigma_{k} \sum_{j=0}^{\infty} \frac{(-1)^{j} \varphi^{2 j+1}}{(2 j+1) !} 16.$$
17. 所得款项 $\sum_{j=0}^{\infty} \frac{(-1)^{j} \varphi^{2 j}}{(2 j) !}$ 和 $\sum_{j=0}^{\infty} \frac{(-1)^{j} \varphi^{2} \varphi^{2 j 11}}{(2 j+1) !} \operatorname{converge}$ 至 $\cos \varphi$ 和 $\sin \varphi ，$ 分别。结果，我们得到：
18. $$19. \exp \left(\mathrm{i} \sigma_{k} \varphi\right)=I \cos \varphi+\mathrm{i} \sigma_{k} \sin \varphi \text { for } k \in 1,2,3, \varphi \in \mathbb{R} . 20.$$

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## MATLAB代写

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