## 物理代写|光学代写Optics代考|PHS2062

2023年2月6日

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Rather than writing $\varepsilon(t)$ with an amplitude and phase, as in Eq. (3.14), we may write it as a complex number with a real and imaginary component:
$$\varepsilon(t)=i \varepsilon^1 \frac{1}{\sqrt{2}}(Q+i P)$$
where $Q$ and $P$ are dimensionless real numbers. The $i$ and $\frac{1}{\sqrt{2}}$ in Eq. (3.15) are present by convention. $\varepsilon^1$ is a constant with units of electric field. $Q$ and $P$ are real numbers and are dynamical variables describing the time-dependence of the field. Equation (3.15) is summarized in Fig. 3.3.
Substituting the complex field amplitude, Eq. (3.15), into Eq. (3.5) gives
\begin{aligned} \boldsymbol{E}(\boldsymbol{r}) & =i \varepsilon \varepsilon^1\left(\frac{1}{\sqrt{2}}(Q+i P) e^{i \boldsymbol{k} \cdot \boldsymbol{r}}-\frac{1}{\sqrt{2}}(Q-i P) e^{-i \boldsymbol{k} \cdot \boldsymbol{r}}\right) \ & =\boldsymbol{\varepsilon} \sqrt{2} \varepsilon^1\left(-Q \frac{\left(e^{i \boldsymbol{k} \cdot \boldsymbol{r}}-e^{-i \boldsymbol{k} \cdot \boldsymbol{r}}\right)}{2 i}-P \frac{\left(e^{i \boldsymbol{k} \cdot \boldsymbol{r}}+e^{-i \boldsymbol{k} \cdot \boldsymbol{r}}\right)}{2}\right) \ & =-\boldsymbol{\varepsilon} \sqrt{2} \varepsilon^1[Q \sin (\boldsymbol{k} \cdot \boldsymbol{r})+P \cos (\boldsymbol{k} \cdot \boldsymbol{r})] \end{aligned}
where we have used the Euler relation. Alternatively, we may write
$$\boldsymbol{E}(\boldsymbol{r})=-\boldsymbol{\varepsilon} \sqrt{2} \varepsilon^1 \sqrt{Q^2+P^2} \sin (\boldsymbol{k} \cdot \boldsymbol{r}+\varphi)$$
where $\varphi=\tan ^{-1}(P / Q)$. Hence, according to Eq. (3.16), $Q$ and $P$ are two components of the field that are out of phase by $\pi / 2$. $Q$ and $P$ are known as the quadrature components of the field. The field associated with a single mode of radiation may be described by two independent components-its magnitude and phase in Eq. (3.5) or, alternatively, by its two quadrature components $(Q$ and $P$ ) in Eq. (3.16).

## 物理代写|光学代写Optics代考|Classical Hamiltonian

The classical expression for the energy of an electromagnetic field is
$$H=\frac{1}{2} \int_V\left(\epsilon_o|\boldsymbol{E}|^2+\frac{1}{\mu_o}|\boldsymbol{B}|^2\right) d V$$
where $\epsilon_o$ is the permittivity of free space, $\mu_o$ is the permeability of free space, and $V$ is the volume of integration. $\epsilon_o$ and $\mu_o$ satisfy the relation $c=\left(\epsilon_o \mu_o\right)^{-1 / 2}$ where $c$ is the speed of light in vacuum. Here, $\frac{\epsilon_o}{2}|\boldsymbol{E}|^2$ and $\frac{1}{2 \mu_o}|\boldsymbol{B}|^2$ is the energy density (energy per unit volume) of the electric and magnetic fields, respectively. Using Eqs. (3.5) and (3.6), the electric field and magnetic field contributions to the energy are equal, giving
$$H=\epsilon_o \int_V|\boldsymbol{E}|^2 d V$$
which is a well-known result for classical electromagnetic waves. Equation (3.17) gives
$$H=2 \epsilon_o\left(\varepsilon^1\right)^2\left(Q^2+P^2\right) \int_V \sin ^2(\boldsymbol{k} \cdot \boldsymbol{r}+\varphi) d V$$
We need to choose a volume for the integration of Eq. (3.20). The volume $V$ could be a real volume (light confined in a cavity by two mirrors), the finite volume associated with a wavepacket or pulse of light, or waves confined in a fictitious box with periodic boundary conditions. The latter is a commonly used trick in quantum optics (indeed, throughout physics) and is discussed below.

To evaluate the integral in Eq. (3.20), we imagine the light confined in a fictitious box with side length $L$. The traveling wave satisfies periodic boundary conditions for the field, $\boldsymbol{E}(x=0)=\boldsymbol{E}(x=L)$, as shown for a few modes in Fig. $3.4$ (similarly for the $y$ and $z$ directions). We imagine the field approaching the right side of the box and “wrapping around” to the left side of the box. We can approach a continuous range of wavelengths by choosing a very large box (large $L$ ). With these boundary conditions, the integration of $\sin ^2(\boldsymbol{k} \cdot \boldsymbol{r}+\varphi)$ over the volume of the box becomes $\frac{1}{2}$, giving
$$H=\epsilon_o V\left(\varepsilon^1\right)^2\left(Q^2+P^2\right)$$
where $V=L^3$ is the volume of the box.

# 光学代考

$$\varepsilon(t)=i \varepsilon^1 \frac{1}{\sqrt{2}}(Q+i P)$$

$$\boldsymbol{E}(\boldsymbol{r})=i \varepsilon \varepsilon^1\left(\frac{1}{\sqrt{2}}(Q+i P) e^{i \boldsymbol{k} \cdot \boldsymbol{r}}-\frac{1}{\sqrt{2}}(Q-i P) e^{-i \boldsymbol{k} \cdot \boldsymbol{r}}\right) \quad=\varepsilon \sqrt{2} \varepsilon^1$$

$$\boldsymbol{E}(\boldsymbol{r})=-\varepsilon \sqrt{2} \varepsilon^1 \sqrt{Q^2+P^2} \sin (\boldsymbol{k} \cdot \boldsymbol{r}+\varphi)$$

## 物理代写|光学代写Optics代考|Classical Hamiltonian

$$H=\frac{1}{2} \int_V\left(\epsilon_o|\boldsymbol{E}|^2+\frac{1}{\mu_o}|\boldsymbol{B}|^2\right) d V$$

$$H=\epsilon_o \int_V|\boldsymbol{E}|^2 d V$$

$$H=2 \epsilon_o\left(\varepsilon^1\right)^2\left(Q^2+P^2\right) \int_V \sin ^2(\boldsymbol{k} \cdot \boldsymbol{r}+\varphi) d V$$

$$H=\epsilon_o V\left(\varepsilon^1\right)^2\left(Q^2+P^2\right)$$

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## MATLAB代写

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