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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 电子工程代写|光子简介代写Introduction to Photonics代考|Energy Exchange Field/Matter

The average electric vacuum field energy density of a stationary field is constant, since
$$\left\langle\frac{\partial}{\partial t} \frac{\varepsilon_{0} \mathbf{E} \cdot \mathbf{E}}{2}\right\rangle=\left\langle\varepsilon_{0} \mathbf{E} \cdot \frac{\partial \mathbf{E}}{\partial t}\right\rangle=\frac{1}{2} \operatorname{Re}\left[-\mathrm{j} \omega \varepsilon_{0} \tilde{\mathbf{E}}(\omega) \cdot \tilde{\mathbf{E}}^{*}(\omega)\right]=0 ;$$
the same applies to the magnetic vacuum field energy, so that, according to Eq. (1.54) (and not surprisingly), the average energy flux through a closed surface vanishes in vacuum. In the presence of a polarizable medium, this is generally not the case because of the last term in Eq. (1.54), which is the product of the polarization current density $\partial \mathbf{P} / \partial t$ and the electric field. In complex notation, $\mathbf{P}(t)=\operatorname{Re}[\tilde{\mathbf{P}}(t)]$ with
$$\tilde{\mathbf{P}}(t)=\tilde{\mathbf{P}}(\omega) \mathrm{e}^{\mathrm{j} \omega t} ;$$
according to Eq. (1.7),
$$\tilde{\mathbf{P}}(\omega)=\varepsilon_{0} \tilde{\chi} \tilde{\mathbf{E}}(\omega),$$
where we assume the susceptibility to be scalar but complex, $\tilde{\chi}=\chi^{\prime}+\mathrm{j} \chi^{\prime \prime}$ (implying a phase shift between the electric field and the polarization); the polarization current density is therefore
$$\frac{\partial \tilde{\mathbf{P}}(t)}{\partial t}=\mathrm{j} \omega \varepsilon_{0} \tilde{\chi} \tilde{\mathbf{E}}(t)$$
and
$$\left\langle\mathbf{E} \cdot \frac{\partial \mathbf{P}}{\partial t}\right\rangle=\frac{1}{2} \operatorname{Re}\left[\tilde{\mathbf{E}}(\omega) \cdot\left[\mathrm{j} \omega \varepsilon_{0} \tilde{\chi} \tilde{\mathbf{E}}(\omega)\right] \right]=-\chi^{\prime \prime} \frac{\omega \varepsilon_{0} \tilde{\mathbf{E}}(\omega) \cdot \tilde{\mathbf{E}}^{}(\omega)}{2}$$

## 电子工程代写|光子简介代写Introduction to Photonics代考|Polarization States

An important property of an optical wave is its polarization state, i.e., the orientation of the electric field vector in space; it influences, among other things, the reflection and transmission behavior at interfaces between different media. According to Eq. (1.67), the electric field vector lies in a plane normal to $\mathbf{k}$ and has two degrees of freedom. ${ }^{5}$ In general, the electric field vector of a harmonically oscillating wave describes an ellipse in this plane, rotating with a period of $2 \pi / \omega$; depending on the sense of rotation, this state is called left or right elliptically polarized. If the ellipse degenerates to a line, the state is linearly polarized; another special case is circularly polarized light.

It is convenient to describe these states in a cartesian coordinate system whose $z$-axis is chosen to be parallel to $\mathbf{k}$. The electric field can then be represented by a two-dimensional vector; the general case Eq. (1.32) is given by
\begin{aligned} &E_{x}(z, t)=E_{0 x} \cos (\omega t-k z) \ &E_{y}(z, t)=E_{0, y} \cos (\omega t-k z+\Delta \phi) \end{aligned}
for convenience, the origin of the time coordinate is chosen such that $\phi_{(x)}=0$ and $\phi_{(y)}=\Delta \phi$.

If the two field components are in phase ( $\Delta \phi=0$ ), $\mathbf{E}$ oscillates along a line oriented under the angle $\varphi=\arctan \left(E_{0, y} / E_{0, x}\right)$ in respect to the $x$-axis; such a field is called linearly polarized.

If the phase difference is $\Delta \phi=\pm \pi / 2$ and $E_{0, x}=E_{0, y}=E_{0}$, then the field vector in a given plane $z=0$ describes a circle
\begin{aligned} &E_{x}(t)=E_{0} \cos \omega t \ &E_{y}(t)=\mp E_{0} \sin \omega t \end{aligned}
and the wave is called circularly polarized. For an observer facing the light wave, the temporal rotation is clockwise (cw) for $\Delta \phi=\pi / 2$ and counterclockwise (ccw) for $\Delta \phi=-\pi / 2$, respectively; the two states are called right (cw) or left (ccw) circularly polarized and denoted by the symbols $\sigma^{+}, \sigma^{-}$. A snapshot $(t=0)$ of the spatial trace of the field vector of right (left) polarized light shows a right (left)handed helix
\begin{aligned} &E_{x}(z)=E_{0} \cos k z \ &E_{y}(z)=\pm E_{0} \sin k z \end{aligned}
with a pitch length of $\lambda=2 \pi / k$ (Fig. 1.6).

# 光子简介代考

## 电子工程代写|光子简介代写Introduction to Photonics代考|Energy Exchange Field/Matter

$$\left\langle\frac{\partial}{\partial t} \frac{\varepsilon_{0} \mathbf{E} \cdot \mathbf{E}}{2}\right\rangle=\left\langle\varepsilon_{0} \mathbf{E} \cdot \frac{\partial \mathbf{E}}{\partial t}\right\rangle=\frac{1}{2} \operatorname{Re}\left[-\mathrm{j} \omega \varepsilon_{0} \tilde{\mathbf{E}}(\omega) \cdot \tilde{\mathbf{E}}^{*}(\omega)\right]=0$$

$$\tilde{\mathbf{P}}(t)=\tilde{\mathbf{P}}(\omega) \mathrm{e}^{\mathrm{j} \omega t} ;$$

$$\tilde{\mathbf{P}}(\omega)=\varepsilon_{0} \tilde{\chi} \tilde{\mathbf{E}}(\omega)$$

$$\frac{\partial \tilde{\mathbf{P}}(t)}{\partial t}=\mathrm{j} \omega \varepsilon_{0} \tilde{\chi} \tilde{\mathbf{E}}(t)$$
$$\left\langle\mathbf{E} \cdot \frac{\partial \mathbf{P}}{\partial t}\right\rangle=\frac{1}{2} \operatorname{Re}\left[\tilde{\mathbf{E}}(\omega) \cdot\left[j \omega \varepsilon_{0} \tilde{\chi} \tilde{\mathbf{E}}(\omega)\right]\right]=-\chi^{\prime \prime} \frac{\omega \varepsilon_{0} \tilde{\mathbf{E}}(\omega) \cdot \tilde{\mathbf{E}}(\omega)}{2}$$

## 电子工程代写|光子简介代写Introduction to Photonics代考|Polarization States

$$E_{x}(z, t)=E_{0 x} \cos (\omega t-k z) \quad E_{y}(z, t)=E_{0, y} \cos (\omega t-k z+\Delta \phi)$$

$$E_{x}(t)=E_{0} \cos \omega t \quad E_{y}(t)=\mp E_{0} \sin \omega t$$

$$E_{x}(z)=E_{0} \cos k z \quad E_{y}(z)=\pm E_{0} \sin k z$$

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