### 物理代写|量子力学代写quantum mechanics代考|PHYC30018

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (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|量子力学代写quantum mechanics代考|Model Problem

Consider two particles moving in one dimension along the $x$-axis. Particle one is free to move in a large circle as in Fig. 2.1, so it satisfies periodic boundary conditions. It has the eigenfunctions and eigenvalues of Eqs. (2.31) and (2.32)
\begin{aligned} \psi_{n}(x) &=\frac{1}{\sqrt{L}} e^{2 \pi i n x / L} & & ; n=0, \pm 1, \pm 2, \cdots \ E_{n} &=\frac{(2 \pi \hbar)^{2}}{2 m L^{2}} n^{2} & & ; \text { particle one } \end{aligned}
The eigenfunctions satisfy the orthonormality condition in Eq. (2.33).
The second particle is confined to a box of a much shorter length as in Fig. 3.1, and it has the eigenfunctions and eigenvalues of Eqs. (3.12) and $(3.13)^{1}$
\begin{aligned} \psi_{n}(x)=\sqrt{\frac{2}{L}} \sin \left(\frac{n \pi x}{L}\right) & ; n=1,2,3, \cdots \ E_{n}=\frac{(\pi \hbar)^{2}}{2 m L^{2}} n^{2} & ; \text { particle two } \end{aligned}
These eigenfunctions are illustrated in Figs. $3.2$ and $3.3$. They are also

orthonormal. It is assumed here that the box is completely transparent to the first particle, which passes right through it. ${ }^{2}$

The starting hamiltonian and general solution for this two-particle system are then
\begin{aligned} H_{0} &=\frac{p_{1}^{2}}{2 m_{1}}+\frac{p_{2}^{2}}{2 m_{2}}+V_{\mathrm{box}}\left(x_{2}\right) \ \Psi_{0}\left(x_{1}, x_{2}, t\right) &=\sum_{n_{1}, n_{2}} c_{n_{1}, n_{2}}^{0}(t) \psi_{n_{1}}\left(x_{1}\right) e^{-i E_{n_{1}} t / \hbar} \psi_{n_{2}}\left(x_{2}\right) e^{-i E_{n_{2}} t / \hbar} \end{aligned}

## 物理代写|量子力学代写quantum mechanics代考|Golden Rule

Now, as previously, we are in a position to iterate these equations and obtain a power series in $H^{\prime}$. Since the r.h.s. of Eqs. (5.12) is already linear in $H^{\prime}$, we can just make use of our previous coefficients $c_{n_{1}^{\prime}, n_{2}^{\prime}}^{0}(t)=c_{n_{1}^{\prime}, n_{2}^{\prime}}^{0}$

on the r.h.s.! This gives
$$i \hbar \frac{d c_{n_{1}, n_{2}}(t)}{d t}=\sum_{n_{1}^{\prime}, n_{2}^{\prime}}\left\langle n_{1}, n_{2}\left|H^{\prime}\right| n_{1}^{\prime}, n_{2}^{\prime}\right\rangle e_{n_{1}^{\prime}, n_{2}^{\prime}}^{0} e^{i\left(E_{n_{1}}+E_{n_{2}}-E_{n_{1}^{\prime}}-E_{n_{2}^{\prime}}\right) / \hbar}+\cdots$$
Suppose it is the state $\psi_{n_{1}^{0}}\left(x_{1}\right) \psi_{n_{2}^{0}}\left(x_{2}\right)$ that is occupied at the initial time $t=0$, so that
$$c_{n_{1}^{\prime}, n_{2}^{\prime}}^{0}=\delta_{n_{1}^{\prime}, n_{1}^{0}} \delta_{n_{2}^{\prime}, n_{2}^{0}} \quad \text {; given initial state }$$
Then, at a later time, the amplitude for finding the system in a different two-particle state satisfies
$$\begin{array}{r} i \hbar \frac{d c_{n_{1}, n_{2}}(t)}{d t}=\left\langle n_{1}, n_{2}\left|H^{\prime}\right| n_{1}^{0}, n_{2}^{0}\right\rangle e^{i\left(E_{n_{1}}+E_{n_{2}}-E_{n_{1}^{0}}-E_{n_{2}^{0}}\right) t / \hbar} \ ;\left(n_{1}, n_{2}\right) \neq\left(n_{1}^{0}, n_{2}^{0}\right) \end{array}$$
Integration of this relation between the initial time $t=0$, and the total elapsed time $t=T$, gives
$$c_{n_{1}, n_{2}}(T)=-\frac{1}{\hbar}\left\langle n_{1}, n_{2}\left|H^{\prime}\right| n_{1}^{0}, n_{2}^{0}\right\rangle \frac{1}{\omega}\left(e^{i \omega T}-1\right)$$
where the initial and final energies of the pair, and energy differences, are defined by
\begin{aligned} E_{0} & \equiv E_{n_{1}^{0}}+E_{n_{2}^{0}} \ E & \equiv E_{n_{1}}+E_{n_{2}} \ \hbar \omega & \equiv E-E_{0} \end{aligned}

## 物理代写|量子力学代写quantum mechanics代考|Density of Final States

Suppose we are doing a scattering experiment in our simple model. We can prepare the target in a given state with energy $E_{n_{2}^{0}}$, and we can prepare an incident beam with a well-defined energy $E_{n_{1}^{0}}=\hbar^{2} k_{0}^{2} / 2 m_{1}$, where $k_{0}=$ $2 \pi n_{1}^{0} / L_{1}$. We certainly can achieve the energy resolution to determine that the target ends up in another state with discrete energy $E_{n_{2}}$; however, with the scattered particle, the situation is more complicated. Let us, for simplicity, call the size of the big region in which the first particle moves $L_{1} \equiv L$. The final particle energy is $E_{n_{1}}=\hbar^{2} k^{2} / 2 m_{1}$ with $k=2 \pi n_{1} / L$, and as $L$ becomes very large, these energies are very closely spaced. Thus no matter how small our resolution $d k$ is on the final particle, many final states will lie within this resolution! For large $L$, the number of these states $d n_{f}$ is
$$d n_{f}=\frac{L}{2 \pi} d k \quad ; L \rightarrow \infty$$
Thus all of these states will get into our final detector, and the transition rate that we actually measure is of necessity
$$R_{f i} d n_{f}=R_{f i}\left(\frac{L}{2 \pi} d k\right) \quad ; \text { measured rate }$$
Equation (5.28) then reads
$$R_{f i} d n_{f}=\frac{2 \pi}{\hbar}\left|\left\langle n_{1}, n_{2}\left|H^{\prime}\right| n_{1}^{0}, n_{2}^{0}\right\rangle\right|^{2} \delta\left(E-E_{0}\right)\left(\frac{L}{2 \pi} d k\right)$$
Multiply and divide this expression by $d E$. It is then possible to immediately do the integral over $E$ using Eq. (5.27), where we have summed over all of the energy-conserving events that get into our detector. Hence ${ }^{4}$
$$R_{f i} d n_{f}=\frac{2 \pi}{\hbar}\left|\left\langle n_{1}, n_{2}\left|H^{\prime}\right| n_{1}^{0}, n_{2}^{0}\right\rangle\right|^{2} \rho_{E}$$
where $\rho_{E}$ is known as the density of final states
$$\rho_{E}=\frac{L}{2 \pi}\left(\frac{d k}{d E}\right) \quad ; \text { density of final states }$$

## 物理代写|量子力学代写quantum mechanics代考|Model Problem

ψn(X)=1大号和2圆周率一世nX/大号;n=0,±1,±2,⋯ 和n=(2圆周率⁇)22米大号2n2; 粒子一

ψn(X)=2大号罪⁡(n圆周率X大号);n=1,2,3,⋯ 和n=(圆周率⁇)22米大号2n2; 粒子二

H0=p122米1+p222米2+在b○X(X2) Ψ0(X1,X2,吨)=∑n1,n2Cn1,n20(吨)ψn1(X1)和−一世和n1吨/⁇ψn2(X2)和−一世和n2吨/⁇

## 物理代写|量子力学代写quantum mechanics代考|Golden Rule

Cn1′,n2′0=dn1′,n10dn2′,n20; 给定初始状态

Cn1,n2(吨)=−1⁇⟨n1,n2|H′|n10,n20⟩1ω(和一世ω吨−1)

## 物理代写|量子力学代写quantum mechanics代考|Density of Final States

dnF=大号2圆周率dķ;大号→∞

RF一世dnF=RF一世(大号2圆周率dķ); 实测率

RF一世dnF=2圆周率⁇|⟨n1,n2|H′|n10,n20⟩|2d(和−和0)(大号2圆周率dķ)

RF一世dnF=2圆周率⁇|⟨n1,n2|H′|n10,n20⟩|2ρ和

ρ和=大号2圆周率(dķd和); 终态密度

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