### 物理代写|量子计算代写Quantum computer代考|PHYS14

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 物理代写|量子计算代写Quantum computer代考|Basic Operations with Qubits

Consider basic operations with qubits.
A quantum gate effect on qubit $|\psi\rangle$ occurs by applying the quantum-mechanical operator, e.g. $U|\psi\rangle[1-3]$. Operators can be represented as unitary matrices. In particular, the evolution of a single qubit is described by a unitary matrix of size $2 \times 2$

The consistent application of a number of operators $U_{1}, U_{2}, \ldots, U_{n}$ to one qubit is equivalent to the effect of some operator $W$ in the form
$$W|\psi\rangle=U_{n}\left(U_{n-1}\left(\ldots\left(U_{2}\left(U_{1}|\psi\rangle\right)\right) \ldots\right)\right)=\left(U_{1} U_{2} \ldots U_{n}\right)|\psi\rangle$$
Its matrix $M_{W}$ is a product of matrices of $U_{i}, i=1,2, \ldots, n$, in the reverse order [4]:
$$M_{W}=M_{U_{n}} M_{U_{n-1}} \ldots M_{U_{1}} .$$
Such an operator $W$ is called a product of operators $U_{1}, U_{2}, \ldots, U_{n}$. Due to the noncommutativity of the matrix multiplication operation, the order in which quantum gates are applied is generally important.

Example 3.1 Let us show that the application of the operator $\sigma_{3}$ (see definition (2.1)) to a qubit in the state $|\psi\rangle=u|0\rangle+v|1\rangle$ brings it to the state $\left|\psi^{\prime}\right\rangle=u|0\rangle-v|1\rangle$.
Proof.
Write a qubit $|\psi\rangle$ in matrix notation:
$$|\psi\rangle=\left[\begin{array}{l} u \ v \end{array}\right] \text {. }$$
Let us define the action of $\sigma_{3}$ on this quantum state:
$$\left|\psi^{\prime}\right\rangle=\sigma_{3}\left[\begin{array}{l} u \ v \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \ 0 & -1 \end{array}\right]\left[\begin{array}{l} u \ v \end{array}\right]=\left[\begin{array}{c} u \ -v \end{array}\right]=u\left[\begin{array}{l} 1 \ 0 \end{array}\right]-v\left[\begin{array}{l} 0 \ 1 \end{array}\right]=u|0\rangle-v|1\rangle \text {. }$$

## 物理代写|量子计算代写Quantum computer代考|The graphic representation of quantum operations

The graphic representation of quantum operations in the form of circuits or diagrams (quantum circuits) is widely used.

A quantum circuit, or network, is an ordered sequence of elements and communication lines connecting them, i.e. wires. Usually, only acyclic circuits are considered in which data flow in one direction-from left to right, and the wires do not return bits to the previous position in the circuit. Input states are attributed to the wires coming from the left. In one time step, each wire may enter data in no more than one gate. The output states are read from the communication lines coming out of the circuit on the right.

As you can see, the circuit solves a problem of a fixed size. Unlike the quantum circuit, quantum algorithms are defined for input data of any size [5].

A quantum-mechanical operator $U$, which converts a one-qubit gate, is presented as follows:
$$\left|\psi_{\text {in }}\right\rangle-U$$
The sequence of the quantum algorithm steps corresponds to the direction on the circuit from left to right.

Table $3.1$ lists the frequently used gates that convert one qubit and the matrix notations of these gates.

Let us show the computation method of the quantum operation matrix based on its effect on basis vectors.
The Hadamard ${ }^{1}$ gate converts the system state in accordance with the rule:
\begin{aligned} &|0\rangle \rightarrow \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle), \ &|1\rangle \rightarrow \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle) . \end{aligned}
Consequently, the arbitrary state $|\psi\rangle$ will change in this case as follows:
\begin{aligned} |\psi\rangle &=\left[\begin{array}{c} u \ v \end{array}\right]=u\left[\begin{array}{l} 1 \ 0 \end{array}\right]+v\left[\begin{array}{l} 0 \ 1 \end{array}\right] \ & \rightarrow u \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)+v \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle)=\frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & 1 \ 1 & -1 \end{array}\right]\left[\begin{array}{l} u \ v \end{array}\right] \end{aligned}
Thus, the Hadamard gate, or the Hadamard element, corresponds to the matrix $\frac{1}{\sqrt{2}}\left[\begin{array}{cc}1 & 1 \ 1 & -1\end{array}\right]$.

Of course, to execute complex algorithms, the qubits must interact with each other and exchange information. In this regard, logical operations involving two or more qubits are of special importance. In Table $3.2$ are listed the most important gates that transform the state of two qubits.

## 物理代写|量子计算代写Quantum computer代考|The following quantum state is set

$$|\psi\rangle=\frac{1}{\sqrt{2}}|0\rangle+\frac{1}{\sqrt{2}} e^{\mathrm{i} \varphi}|1\rangle,$$
where $\varphi$ is a real number. Let us define the result of measuring this state in bases $B_{1}=$ ${|0\rangle,|1\rangle}$ and $B_{2}={|+\rangle,|-\rangle}$, where $|+\rangle=\frac{1}{2}(|0\rangle+|1\rangle)$ and $|-\rangle=\frac{1}{2}(|0\rangle-|1\rangle)$.
Solution.
First of all, we consider the basis $B_{1}$. According to the measurement postulate (see Appendix A on page 105), the probabilities of getting 0 or 1 as a result of the measurement of the state $u|0\rangle+v|1\rangle$ is equal to $|u|^{2}$ and $|v|^{2}$, respectively. In our example, $u=\frac{1}{\sqrt{2}}$ and $v=\frac{1}{\sqrt{2}} e^{i \varphi}$, therefore the result of the measurement of $M$ is
$$M=\left{\begin{array}{l} 0 \text { with a probability of } p_{0}=\frac{1}{2}, \ 1 \text { with a probability of } p_{1}=\frac{1}{2} \end{array}\right.$$
Note that information about the value of $\varphi$ using a measurement in the basis $B_{1}$ cannot be obtained.

Let us turn to measurement using the second basis $B_{2}={|+\rangle,|-\rangle}$. We express the vectors of the computational basis through $|+\rangle$ and $|-\rangle}$ :
\begin{aligned} |0\rangle &=\frac{1}{\sqrt{2}}(|+\rangle+|-\rangle), \ |1\rangle &=\frac{1}{\sqrt{2}}(|+\rangle+|-\rangle) . \end{aligned}
In this regard, we obtain
\begin{aligned} |\psi\rangle &=\frac{1}{\sqrt{2}}|0\rangle+\frac{1}{\sqrt{2}} e^{\mathrm{i} \varphi}|1\rangle \ &=\frac{1}{2}(|+\rangle+|-\rangle)+\frac{1}{2} e^{\mathrm{i} \varphi}(|+\rangle+|-\rangle) \ &=\frac{1}{2}\left(1+e^{\mathrm{i} \varphi}\right)|+\rangle+\frac{1}{2}\left(1-e^{\mathrm{i} \varphi}\right)|-\rangle . \end{aligned}
The probability of finding the system in the state $|+\rangle$ is
\begin{aligned} p_{+} &=\frac{1}{2}\left(1+e^{\mathrm{i} \varphi}\right) \times\left(\frac{1}{2}\left(1+e^{\mathrm{i} \varphi}\right)\right)^{*} \ &=\frac{1}{4}\left(1+e^{\mathrm{i} \varphi}\right)\left(1+e^{-\mathrm{i} \varphi}\right)=\frac{1}{4}\left(2+e^{\mathrm{i} \varphi}+e^{-\mathrm{i} \varphi}\right) \end{aligned}

## 物理代写|量子计算代写Quantum computer代考|Basic Operations with Qubits

$$W|\psi\rangle=U_{n}\left(U_{n-1}\left(\ldots\left(U_{2}\left(U_{1}|\psi\rangle\right)\right) \ldots\right)\right)=\left(U_{1} U_{2} \ldots U_{n}\right)|\psi\rangle$$

$$M_{W}=M_{U_{n}} M_{U_{n-1}} \ldots M_{U_{1}} .$$

$$|\psi\rangle=[u v] .$$

$$\left|\psi^{\prime}\right\rangle=\sigma_{3}\left[\begin{array}{ll} u & v \end{array}\right]=\left[\begin{array}{llll} 1 & 0 & 0 & -1 \end{array}\right]\left[\begin{array}{ll} u & v \end{array}\right]=\left[\begin{array}{ll} u-v \end{array}\right]=u\left[\begin{array}{ll} 1 & 0 \end{array}\right]-v\left[\begin{array}{ll} 0 & 1 \end{array}\right]=u|0\rangle-v|1\rangle .$$

## 物理代写|量子计算代写Quantum computer代考|The graphic representation of quantum operations

$$\left|\psi_{\text {in }}\right\rangle-U$$

$$|0\rangle \rightarrow \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle), \quad|1\rangle \rightarrow \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle) .$$

$$|\psi\rangle=[u v]=u\left[\begin{array}{ll} 1 & 0 \end{array}\right]+v\left[\begin{array}{ll} 0 & 1 \end{array}\right] \quad \rightarrow u \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)+v \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle)=\frac{1}{\sqrt{2}}\left[\begin{array}{llll} 1 & 1 & -1 \end{array}\right][u v]$$

## 物理代写|量子计算代写Quantum computer代考|The following quantum state is set

$$|\psi\rangle=\frac{1}{\sqrt{2}}|0\rangle+\frac{1}{\sqrt{2}} e^{\mathrm{i} \varphi}|1\rangle$$

$$|0\rangle=\frac{1}{\sqrt{2}}(|+\rangle+|-\rangle),|1\rangle \quad=\frac{1}{\sqrt{2}}(|+\rangle+|-\rangle) .$$

$$|\psi\rangle=\frac{1}{\sqrt{2}}|0\rangle+\frac{1}{\sqrt{2}} e^{\mathrm{i} \varphi}|1\rangle \quad=\frac{1}{2}(|+\rangle+|-\rangle)+\frac{1}{2} e^{\mathrm{i} \varphi}(|+\rangle+|-\rangle)=\frac{1}{2}\left(1+e^{\mathrm{i} \varphi}\right)|+\rangle+\frac{1}{2}\left(1-e^{\mathrm{i} \varphi}\right)$$

$$p_{+}=\frac{1}{2}\left(1+e^{\mathrm{i} \varphi}\right) \times\left(\frac{1}{2}\left(1+e^{\mathrm{i} \varphi}\right)\right)^{*}=\frac{1}{4}\left(1+e^{\mathrm{i} \varphi}\right)\left(1+e^{-\mathrm{i} \varphi}\right)=\frac{1}{4}\left(2+e^{\mathrm{i} \varphi}+e^{-\mathrm{i} \varphi}\right)$$

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