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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|量子光学代写Quantum Optics代考|Time-Harmonic Fields

In particular for linear materials one often deals with light excitations oscillating with a single frequency $\omega$, and because of the linearity of the material response the system oscillates after an initial transient with the same frequency $\omega$. We here follow the notation of the introduction chapter and assume electromagnetic fields of the form
$$\boldsymbol{E}(\boldsymbol{r}, t)=e^{-i \omega t} \boldsymbol{E}(\boldsymbol{r}), \quad \boldsymbol{B}(\boldsymbol{r}, t)=e^{-i \omega t} \boldsymbol{B}(\boldsymbol{r})$$

A few words of caution are at place. First, the true electromagnetic fields are of course real quantities. Whenever we are interested in these real fields we have to take the real parts of the corresponding expressions. Second, we use the same symbols for $\boldsymbol{E}(\boldsymbol{r}, t)$ and $\boldsymbol{E}(\boldsymbol{r})$. In general, this should not cause too much confusions because we will always clearly state whether we work in the time domain or use time-harmonic fields. For time-harmonic fields and a linear material response, Maxwell’s equations can be written in the form
Maxwell’s Equations for Time-Harmonic Fields
$$\begin{array}{rc} \nabla \cdot \varepsilon \boldsymbol{E}=\rho, & \nabla \times \boldsymbol{E}=i \omega \boldsymbol{B} \ \nabla \cdot \boldsymbol{B}=0, & \nabla \times \frac{1}{\mu} \boldsymbol{B}=\boldsymbol{J}-i \omega \varepsilon \boldsymbol{E} . \end{array}$$
We have not explicitly indicated that $\rho, \boldsymbol{J}$ correspond to external charge and current distributions, and will suppress this dependence from here on whenever it is clear from the context.

物理代写|量子光学代写Quantum Optics代考|Sinusoidal Waves

Any electric field can be decomposed into plane waves by means of a Fourier transformation
$$\boldsymbol{E}(\boldsymbol{r})=\int e^{i k \cdot \boldsymbol{r}} \boldsymbol{E}_{\boldsymbol{k}} \frac{d^3 k}{(2 \pi)^3}$$

Here $k$ denotes a wavevector and $\boldsymbol{E}{\boldsymbol{k}}$ is a Fourier component. A corresponding decomposition can be obtained for the magnetic field. Consider next a single sinusoidal wave $$E(\boldsymbol{r})=E_k e^{i k \cdot r}, \quad B(r)=B_k e^{i k \cdot r} .$$ If we are interested in the propagation of a more complicated field, we can always decompose the field into plane waves, using Eq. (2.36), propagate the sinusoidal components separately, and put together the final result at the end. The action of the nabla operator on a plane wave becomes $\nabla \rightarrow i \boldsymbol{k}$, see Exercise 2.10. Thus, in absence of sources $\rho=J=0$ Maxwell’s equations take the following form: Maxwell’s Equations for a Plane Wave \begin{aligned} \varepsilon \boldsymbol{k} \cdot \boldsymbol{E}_k &=0, & \boldsymbol{k} \times \boldsymbol{E}_k &=\omega \boldsymbol{B}{\boldsymbol{k}} \ \boldsymbol{k} \cdot \boldsymbol{B}k &=0, & \boldsymbol{k} \times \boldsymbol{B}_k &=-\omega \mu \varepsilon \boldsymbol{E}_k . \end{aligned} A few important things can be directly inferred from these equations. First, electromagnetic waves are transverse waves where $\boldsymbol{E}{\boldsymbol{k}}, \boldsymbol{B}{\boldsymbol{k}}$ are perpendicular to the wavenumber $\boldsymbol{k}$, as can be seen from $\boldsymbol{k} \cdot \boldsymbol{E}{\boldsymbol{k}}=\boldsymbol{k} \cdot \boldsymbol{B}{\boldsymbol{k}}=0$. Second, $\boldsymbol{E}{\boldsymbol{k}}, \boldsymbol{B}{\boldsymbol{k}}$, and $\boldsymbol{k}$ form a right-handed orthogonal basis. Finally from $$\boldsymbol{k} \times \boldsymbol{k} \times \boldsymbol{E}{\boldsymbol{k}}=-k^2 \boldsymbol{E}{\boldsymbol{k}}=\omega \boldsymbol{k} \times \boldsymbol{B}{\boldsymbol{k}}=-\omega^2 \mu \varepsilon \boldsymbol{E}_{\boldsymbol{k}}$$

量子光学代考

物理代写|量子光学代写Quantum Optics代考|时谐场

.

$$\boldsymbol{E}(\boldsymbol{r}, t)=e^{-i \omega t} \boldsymbol{E}(\boldsymbol{r}), \quad \boldsymbol{B}(\boldsymbol{r}, t)=e^{-i \omega t} \boldsymbol{B}(\boldsymbol{r})$$

$$\begin{array}{rc} \nabla \cdot \varepsilon \boldsymbol{E}=\rho, & \nabla \times \boldsymbol{E}=i \omega \boldsymbol{B} \ \nabla \cdot \boldsymbol{B}=0, & \nabla \times \frac{1}{\mu} \boldsymbol{B}=\boldsymbol{J}-i \omega \varepsilon \boldsymbol{E} . \end{array}$$

有限元方法代写

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MATLAB代写

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