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# 物理代考| Classical Optics 量子力学代写

## 物理代写

1.1 Classical Optics
Consider a non-dispersive wave which is the real part of
$$\Psi(x, t)=e^{i(k x-\omega t)}=e^{i k(x-c t)} \quad ; \omega=k c$$
Here $c$ is the velocity of the wave, and the frequency and wavelength are related by
$$\omega=2 \pi \nu=k c=2 \pi \frac{c}{\lambda}$$
As we have seen, this could be an electromagnetic wave in vacuum, a transverse wave on a string under tension, or the sound wave in a medium. This wave satisfies the wave equation
$$\frac{\partial^{2} \Psi(x, t)}{\partial x^{2}}=\frac{1}{c^{2}} \frac{\partial^{2} \Psi(x, t)}{\partial t^{2}} \quad ; \text { wave equation }$$
We have also seen that a linear combination of two such waves with slightly different wavenumbers $k$, produces an amplitude modulated signal. A more general linear combination can produce a localized wave packet, or pulse.

Huygen’s principle states that each point on a wavefront acts as a source of an outgoing spherical wave. From this, and its generalizations, one derives single-slit diffraction, two-slit and multi-slit interference, and most of classical wave optics. ${ }^{1}$
${ }^{1}$ See Probs. 1.1-1.2. For a more detailed discussion here, see [Wiki (2021)].
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Introduction to Quantum Mechanics
1.2 Planck Distribution
Early in the twentieth century, Planck was studying the distribution of energy as a function of frequency for the electromagnetic radiation in a cavity. Normal modes are uncoupled simple harmonic oscillators. The classical equipartition theorem says that the energy of a simple harmonic oscillator at an absolute temperature $T$ is
where $k_{B}$ is Boltzmann’s constant $$k_{B}=1.381 \times 10^{-23} \mathrm{~J} /{ }^{\circ} \mathrm{K} \quad ; \text { Boltzmann’s constant }$$ the frequency, this classical result says there should be an ever-increasing energy as a function of frequency for the radiation in a cavity, the so-called ultraviolet catastrophe. ${ }^{2}$

To fit his data, Planck employed an empirical expression of the form
$$\langle\varepsilon(\nu)\rangle=\frac{h \nu}{e^{h \nu / k_{B} T}-1} \quad ; \text { Planck distribution }$$
where $h$ is a constant obtained from the fit, now known as Planck’s constant
$$\frac{h}{2 \pi} \equiv \hbar=1.055 \times 10^{-34} \mathrm{Js} \quad ; \text { Planck’s constant }$$
Note that at low frequency, the Planck distribution reproduces the equipartition result
$$\frac{h \nu}{e^{h \nu / k_{B} T}-1} \rightarrow k_{B} T \quad ; h \nu \ll k_{B} T$$
while at high frequency, it now disappears exponentially
$$\frac{h \nu}{e^{h \nu / k_{B} T}-1} \rightarrow h \nu e^{-h \nu / k_{B} T} \quad ; h \nu \gg k_{B} T$$
$$\frac{h \nu}{e^{h \nu / k_{B} T}-1} \rightarrow h \nu e^{-h \nu / k_{B} T} \quad ; h \nu \gg k_{B} T$$
One can ask where this empirical Planck distribution might come from. Suppose that in each mode with frequency $\nu$ in the cavity it is possible to have any number $n$ of photons, each with energy
‘ See Prob. 1.3.
3 Then the mean energy in the mode at the temperature an elementary statistical calculation with the Boltzmann $e^{-n h \nu / k_{B} T}$ as $$\begin{array}{l}\text { Motivation } \ \qquad \varepsilon(\nu)\rangle=\frac{\sum_{n=0}^{\infty}(n h \nu) e^{-n h \nu / k_{B} T}}{\sum_{n=0}^{\infty} e^{-n h \nu / k_{B} T}} \ \qquad=-\frac{d}{d\left(1 / k_{B} T\right)} \ln \left(\sum_{n=0}^{\infty} e^{-n h \nu / k_{B} T}\right) \ \text { The photons of light each have an energy } \ \text { The sum is just a geometric series }^{3} \ \qquad \sum_{n=0}^{\infty} x^{n}=\frac{1}{1.3}\end{array}$$

We know the momentum flux in an electromagnetic wave is $1 / c$ times the energy flux, and hence each photon in light also has a momentum
$$p=\frac{h \nu}{c} \quad ; \text { photon }$$
Photons are now observed every day in the laboratory as single events in low-intensity radiation detectors.
${ }^{3}$ Note $e^{-n h \nu / k_{B} T}=\left(e^{-h \nu / k_{B} T}\right)^{n} .$

## 物理代考

1.1 经典光学

$$\Psi(x, t)=e^{i(k x-\omega t)}=e^{i k(x-c t)} \quad ; \omega=k c$$

$$\omega=2 \pi \nu=k c=2 \pi \frac{c}{\lambda}$$

$$\frac{\partial^{2} \Psi(x, t)}{\partial x^{2}}=\frac{1}{c^{2}} \frac{\partial^{2} \Psi (x, t)}{\partial t^{2}} \quad ; \text { 波动方程 }$$

${ }^{1}$ 见问题。 1.1-1.2。有关此处更详细的讨论，请参阅 [Wiki (2021)]。
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1.2 普朗克分布
20 世纪初，普朗克正在研究腔内电磁辐射的能量分布与频率的函数关系。正常模式是非耦合的简单谐振子。经典均分定理说，简单谐振子在绝对温度 $T$ 下的能量为

$$\langle\varepsilon(\nu)\rangle=\frac{h \nu}{e^{h \nu / k_{B} T}-1} \quad ; \text { 普朗克分布 }$$

$$\frac{h}{2 \pi} \equiv \hbar=1.055 \times 10^{-34} \mathrm{Js} \quad ; \text { 普朗克常数 }$$

$$\frac{h \nu}{e^{h \nu / k_{B} T}-1} \rightarrow k_{B} T \quad ; h \nu \ll k_{B} T$$

$$\frac{h \nu}{e^{h \nu / k_{B} T}-1} \rightarrow h \nu e^{-h \nu / k_{B} T} \quad ; h \nu \gg k_{B} T$$
$$\frac{h \nu}{e^{h \nu / k_{B} T}-1} \rightarrow h \nu e^{-h \nu / k_{B} T} \quad ; h \nu \gg k_{B} T$$

‘见概率。 1.3.
3 然后在温度下模态的平均能量用玻尔兹曼 $e^{-nh \nu / k_{B} T}$ 作为 $$\begin{array}{l}\text { Motivation } \ \qquad \varepsilon(\nu)\rangle=\frac{\sum_{n=0}^{\infty}(nh \nu) e^{-nh \nu / k_{B} T}}{\ sum_{n=0}^{\infty} e^{-nh \nu / k_{B} T}} \ \qquad=-\frac{d}{d\left(1 / k_{B} T\ right)} \ln \left(\sum_{n=0}^{\infty} e^{-nh \nu / k_{B} T}\right) \ \text { 每个光子都有一个能量} \ \text { 总和只是一个几何级数 }^{3} \ \qquad \sum_{n=0}^{\infty} x^{n}=\frac{1}{1.3}\end {数组}$$

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