### 物理代写|量子力学代写quantum mechanics代考|PHYC90007

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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|量子力学代写quantum mechanics代考|Two-State Mixing

So far in looking at transition rates we have worked to leading order in $H^{\prime}$. We now simplify the problem enough so that we can treat $H^{\prime}$ exactly. We still seek separated solutions to the Schrödinger equation as in Eqs. (2.18)$(2.22)$, so that we have
$\Psi(x, t)=\psi(x) e^{-i E t / \hbar}$
$H \psi(x)=E \psi(x) \quad ; H=H_{0}+H^{\prime}$
The eigenfunction $\psi(x)$ can be expanded in the complete set of solutions to the unperturbed problem
\begin{aligned} \psi(x) &=\sum_{n} a_{n} \psi_{n}(x) \ H_{0} \psi_{n}(x) &=E_{n}^{0} \psi_{n}(x) \end{aligned}
Substitution into the eigenvalue equation, and the use of the orthonormality of the eigenfunctions $\psi_{n}(x)$, gives
$$\sum_{n^{\prime}}\left[\left(E_{n}^{0}-E\right) \delta_{n, n^{\prime}}+\left\langle n\left|H^{\prime}\right| n^{\prime}\right\rangle\right] a_{n^{\prime}}=0$$

This relation is still exact, and there is one equation for each $n$. We are thus faced with an infinite set of coupled algebraic equations for the amplitudes $\left(a_{1}, a_{2}, a_{3}, \cdots\right)$; however, we now make some simplifying assumptions:

• We assume that it is only the mixing of a pair of states $\left(\psi_{1}, \psi_{2}\right)$ that is important for us;
• We assume that the pair is degenerate, with energy $E_{0} ;{ }^{6}$
• We assume that the diagonal elements of $H^{\prime}$ vanish.
• We assume the off-diagonal elements of $H^{\prime}$ are real, with $H_{12}^{\prime}=H_{21}^{\prime}$.

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

The classical hamiltonian for a charged particle in an electromagnetic field with vector and scalar potentials $(\vec{A}, \Phi)$ is given by
$$H=\frac{1}{2 m}[\vec{p}-e \vec{A}(\vec{x}, t)]^{2}+e \Phi(\vec{x}, t)$$
We shall justify this by showing that the classical Hamilton’s equations produce the Lorentz force on the particle
$$\vec{F}=e(\vec{E}+\vec{v} \times \vec{B}) \quad \text {; Lorentz force }$$
\begin{aligned} &\frac{\partial H}{\partial p_{i}}=\frac{d x_{i}}{d t} \ &\frac{\partial H}{\partial x_{i}}=-\frac{d p_{i}}{d t} \end{aligned}
The first of Hamilton’s equations expresses the particle velocity as
$$v_{i}=\frac{d x_{i}}{d t}=\frac{1}{m}[\vec{p}-e \vec{A}(\vec{x}, t)]{i}$$ Differentiation of this relation gives $$\frac{d p{i}}{d t}=m \frac{d^{2} x_{i}}{d t^{2}}+e \frac{d A_{i}(\vec{x}, t)}{d t}$$

## 物理代写|量子力学代写quantum mechanics代考|Interaction With the Radiation Field

To lowest order, the interaction with the radiation field in the hamiltonian in Eq. (6.14) is
\begin{aligned} H^{\prime} &=-\frac{e}{2 m}[\vec{p} \cdot \vec{A}(\vec{x}, t)+\vec{A}(\vec{x}, t) \cdot \vec{p}] \quad ; \vec{p}=\frac{\hbar}{i} \vec{\nabla} \ \vec{A}(\vec{x}, t) &=A(\vec{k}, s) \vec{e}{\vec{k} s} \frac{1}{2}\left[e^{i(\vec{k} \cdot \vec{x}-\omega t)}+e^{-i(\vec{k} \cdot \vec{x}-\omega t)}\right] \end{aligned} Here $A(\vec{k}, s)$ is the amplitude of the vector potential in the classical wave. Since the fields are transverse with $\vec{k} \cdot \vec{e}{\vec{k} s}=0$, we can move $\vec{p}$ through to the right in the first term, and rewrite
$$H^{\prime}=-\hat{\rho} \vec{A}(\vec{x}, t) \cdot \frac{\vec{p}}{m}$$
To leading order in $\vec{A}$, this looks like the classical expression $-e \vec{A}(\vec{x}, t) \cdot \vec{v}$. If we are looking at transitions that put energy into the system, then, as before, we can simply use ${ }^{5}$
$$\vec{A}(\vec{x}, t) \doteq \frac{1}{2} A(\vec{k}, s) \vec{e}{\vec{k} s} e^{i(\vec{k} \cdot \vec{x}-\omega t)} \quad ; E{f}=E_{i}+\hbar \omega$$
We also know that the time-average energy flux in the classical wave is $^{6}$
$$S_{\mathrm{inc}}=\left[\frac{\varepsilon_{0} \omega^{2}}{2} A^{2}\left(\vec{k}_{,} s\right)\right] c \quad ; \text { energy flux }$$

## 物理代写|量子力学代写quantum mechanics代考|Two-State Mixing

Ψ(X,吨)=ψ(X)和−一世和吨/⁇
Hψ(X)=和ψ(X);H=H0+H′

ψ(X)=∑n一个nψn(X) H0ψn(X)=和n0ψn(X)

∑n′[(和n0−和)dn,n′+⟨n|H′|n′⟩]一个n′=0

• 我们假设它只是一对状态的混合(ψ1,ψ2)这对我们很重要；
• 我们假设这对是简并的，有能量和0;6
• 我们假设对角元素H′消失。
• 我们假设非对角元素H′是真实的，与H12′=H21′.

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

H=12米[p→−和一个→(X→,吨)]2+和披(X→,吨)

F→=和(和→+在→×乙→); 洛伦兹力

∂H∂p一世=dX一世d吨 ∂H∂X一世=−dp一世d吨

dp一世d吨=米d2X一世d吨2+和d一个一世(X→,吨)d吨

## 物理代写|量子力学代写quantum mechanics代考|Interaction With the Radiation Field

H′=−和2米[p→⋅一个→(X→,吨)+一个→(X→,吨)⋅p→];p→=⁇一世∇→ 一个→(X→,吨)=一个(ķ→,s)和→ķ→s12[和一世(ķ→⋅X→−ω吨)+和−一世(ķ→⋅X→−ω吨)]这里一个(ķ→,s)是经典波中矢量势的幅度。由于场是横向的ķ→⋅和→ķ→s=0, 我们可以移动p→在第一学期向右，并改写

H′=−ρ^一个→(X→,吨)⋅p→米

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