# 物理代写|量子光学代写Quantum Optics代考|Fresnel Coefficients

#### Doug I. Jones

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## 物理代写|量子光学代写Quantum Optics代考|Fresnel Coefficients

We start with Faraday’s law for time-harmonic fields $\nabla \times \boldsymbol{E}=i \mu \omega \boldsymbol{H}$. Taking the curl on both sides of the equation and using Ampere’s law with Maxwell’s displacement current, we are led to
$$\nabla \times \frac{1}{\mu} \nabla \times \boldsymbol{E}=i \omega(\nabla \times \boldsymbol{H})=i \omega(-i \omega \varepsilon \boldsymbol{E})$$
Here and in the following we consider media with no charge and current distributions, $\rho=\boldsymbol{J}=0$. From Eq. (8.19) and by taking in a similar fashion the curl of Ampere’s law we are led to the wave equations for the electromagnetic fields
\begin{aligned} & \mu \nabla \times \frac{1}{\mu} \nabla \times \boldsymbol{E}-\omega^2 \mu \varepsilon \boldsymbol{E}=0 \ & \varepsilon \nabla \times \frac{1}{\varepsilon} \nabla \times \boldsymbol{H}-\omega^2 \mu \varepsilon \boldsymbol{H}=0 . \end{aligned}
These two wave equations exhibit a kind of duality in the sense that one equation is obtained from the other one by replacing $\boldsymbol{E} \leftrightarrow \boldsymbol{H}, \varepsilon \leftrightarrow \mu$ [20]. This duality is not deep in a physical sense and does not reflect an underlying symmetry, in fact other replacements can be found that transform one equation to the other one. However, we will find such replacement useful below.

Consider the setup depicted in Fig. 8.9 where a plane wave impinges on an interface between two materials with $\varepsilon_1, \mu_1$ and $\varepsilon_2, \mu_2$. We can distinguish two situations for the polarization.

Transverse Electric (TE). Here the electric field $\boldsymbol{E}$ vector is parallel to the interface, or, in other words, transverse to the symmetry axis (here the $z$-axis) of the layer.
Transverse Magnetic (TM). Here the magnetic field vector $\boldsymbol{H}$ is parallel to the interface. It has been shown in Eq. (2.43) that a general wave can always be decomposed into its TE and TM components.

We first ponder on the TE polarization case, the results for TM polarization will then directly follow from the duality principle. Without loss of generality, we assume that the propagation direction of the wave is in the $x z$-plane and the electric field vector points along $y$, such that $\boldsymbol{E}=E_y \hat{\boldsymbol{y}}$. From Gauss’ law we then find
$$\nabla \cdot \varepsilon(z) \boldsymbol{E}=\frac{\partial}{\partial y} \varepsilon(z) E_y=0 \Longrightarrow \varepsilon(z) \frac{\partial E_y}{\partial y}=0,$$
where we have used that for a stratified medium $\varepsilon(z)$ only depends on the $z$ coordinate. We see that $E_y$ does not depend on $y$. Using this and working out the spatial derivatives in Eq. (8.20), we are led to
$$\left(\frac{\partial^2}{\partial x^2}+\mu(z) \frac{\partial}{\partial z} \mu^{-1}(z) \frac{\partial}{\partial z}+\omega^2 \mu \varepsilon\right) E_y=0 .$$

## 物理代写|量子光学代写Quantum Optics代考|Transfer Matrices

We now turn to the stratified medium depicted in Fig. 8.10. Let us introduce a two-component vector where the components $e_m^{+}$and $e_m^{-}$describe the fields that propagate in medium $m$ into the positive and negative $z$-direction, respectively. For simplicity, in the following we drop the subscript $y$ for the electric field component and the superscript $T E$ for the Fresnel coefficients. Let us consider the first interface. The field in medium 1 is $e_1^{+}+e_1^{-}$and the field in medium 2 is $e_2^{+}+e_2^{-}$. Submitting these fields to the boundary condition of Eq. (8.23) gives
$$e_1^{+}+e_1^{-}=e_2^{+}+e_2^{-}, \quad \frac{k_{1 z}}{\mu_1}\left(e_1^{+}-e_1^{-}\right)=\frac{k_{2 z}}{\mu_2}\left(e_2^{+}-e_2^{-}\right)$$

Multiplying the first equation with $k_{1 z} / \mu_1$ and adding or subtracting the second one leads us to
$$T_{12} e_1^{+}=e_2^{+}+R_{12} e_2^{-}, \quad T_{12} e_1^{-}=R_{12} e_2^{+}+e_2^{-},$$
where we have explicitly indicated the dependence of the Fresnel coefficients on the media indices. We can thus relate the field components on both sides of the interface through a matrix equation
$$\left(\begin{array}{c} e_1^{+} \ e_1^{-} \end{array}\right)=\frac{1}{T_{12}}\left(\begin{array}{cc} 1 & R_{12} \ R_{12} & 1 \end{array}\right)\left(\begin{array}{c} e_2^{+} \ e_2^{-} \end{array}\right) .$$
We next introduce a matrix
$$\overline{\bar{M}}{m-1, m}=\frac{1}{T{m-1, m}}\left(\begin{array}{cc} 1 & R_{m-1, m} \ R_{m-1, m} & 1 \end{array}\right)$$
that describes how the fields are connected over an interface through the boundary conditions of Maxwell’s equations. In a multilayer system we additionally have to consider the phases $e^{ \pm i k_{m z} d_m}$ acquired by the fields when propagating through the $m$ ‘th layer with thickness $d_m$. This can be done by introducing the propagation matrices $\overline{\bar{P}}m$ $$\overline{\bar{P}}_m=\left(\begin{array}{cc} e^{-i k{m z} d_m} & 0 \ 0 & e^{i k_{m z} d_{m t}} \end{array}\right)$$

# 量子光学代考

## 物理代写|量子光学代写Quantum Optics代考|Fresnel Coefficients

$$\nabla \times \frac{1}{\mu} \nabla \times \boldsymbol{E}=i \omega(\nabla \times \boldsymbol{H})=i \omega(-i \omega \varepsilon \boldsymbol{E})$$

$$\mu \nabla \times \frac{1}{\mu} \nabla \times \boldsymbol{E}-\omega^2 \mu \varepsilon \boldsymbol{E}=0 \quad \varepsilon \nabla \times \frac{1}{\varepsilon} \nabla \times \boldsymbol{H}$$

$\boldsymbol{E} \leftrightarrow \boldsymbol{H}, \varepsilon \leftrightarrow \mu[20]$ 。这种对偶性在物理意义上并不 深刻，也没有反映出潜在的对称性，事实上，可以找到 其他替代方法，将一个方程式转换为另一个方程式。但 是，我们会在下面发现这种替换很有用。

$$\nabla \cdot \varepsilon(z) \boldsymbol{E}=\frac{\partial}{\partial y} \varepsilon(z) E_y=0 \Longrightarrow \varepsilon(z) \frac{\partial E_y}{\partial y}=0$$
$\varepsilon(z) z E_y y$. 使用它并计算出方程式中的空间导数。 (8.20)，我们被引导到
$$\left(\frac{\partial^2}{\partial x^2}+\mu(z) \frac{\partial}{\partial z} \mu^{-1}(z) \frac{\partial}{\partial z}+\omega^2 \mu \varepsilon\right) E_y=0$$

## 物理代写|量子光学代写Quantum Optics代考|Transfer Matrices

$$\nabla \cdot \varepsilon(z) \boldsymbol{E}=\frac{\partial}{\partial y} \varepsilon(z) E_y=0 \Longrightarrow \varepsilon(z) \frac{\partial E_y}{\partial y}=0$$
$\varepsilon(z) z E_y y$. 使用它并计算出方程式中的空间导数。 (8.20)，我们被引导到
$$\left(\frac{\partial^2}{\partial x^2}+\mu(z) \frac{\partial}{\partial z} \mu^{-1}(z) \frac{\partial}{\partial z}+\omega^2 \mu \varepsilon\right) E_y=0$$$\overline{\bar{P}}m=\left(\begin{array}{llll}e^{-i k m z d_m} & 0 & 0 & e^{i k{m z} d_{m t}}\end{array}\right)$

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