11: Photons: quantization of a single electromagnetic field mode
A one-dimensional cavity along z-axis
服从Maxwell Equation
$$ \begin{aligned}\nabla \times \mathbf{E}& \begin{aligned}=\frac{\partial\mathbf{B}}{\partial t}\end{aligned} \\ \nabla\times\mathbf{B}& =-\mu_0\varepsilon_0\frac{\partial\mathbf{E}}{\partial t} \\\nabla\cdot\mathbf{B}& =0 \\\nabla·\mathbf{E}& =0. \end{aligned} $$
the field is assumed to be polarlized along x-direction. 两无限大平行平面组成波导电场方向垂直于平面法向向量
两无穷大导体平面之间
平面电磁波解：
边界条件为界面切向电场为零，法向磁场为零。延z轴传播的平面电磁波只能延y轴偏振，不能延x轴偏振
驻波解，板面积$F$：
边界条件不变。不妨设驻波延x轴偏振，在z方向上形成驻波
A running transverse wave
$$ \mathbf{E}=\hat{\mathbf{x}} A\left(\alpha(t) e^{i k z}+\alpha^{*}(t) e^{-i k z}\right) $$
线偏振推导：
[Helmholz equation] 由$z=0$处的边界条件知$\sin kz$，由$z=L$处知$k$
the linearly polarized 线偏振的 eletric field oscilates 振动 in the x direction with the classical expression.
$q(t)$ is a time dependent function, to be determined. Assume that $q(t)$ is dimensionless.
$$ \boldsymbol {E}(\boldsymbol x,t )=\hat{\mathbf{x}} p(t) A \sin (k z).\qquad[E]=[p][A][1]\\ k=n \pi / L\qquad[1/m] $$
Choosing one special value of $k$→a special standing wave $(n)$→a mode of the field
**[Faraday’s law ]**then it follows from the Maxwell equation (Faraday’s law)
$$ \boldsymbol {B}(\boldsymbol x,t)=-\hat{\mathbf{y}} A s(t) k \cos k z, \quad \text { where } \quad \dot{s}(t)=p(t) $$
[the other Maxwell equation ]Assume that the magnetic field has no static component 假设磁场其他分量为零 ,
$$ \dot{\boldsymbol {E}}=c^{2} \nabla \times \boldsymbol {B}=-c^{2} s(t) A k\left|\begin{array}{ccc} \hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \\ \partial_{x} & \partial_{y} & \partial_{z} \\ 0 & \cos k z & 0 \end{array}\right| $$
we get:【由Maxwell方程组解出时间项】
$$ \begin{aligned}\hat{\mathbf{x}}\dot{p}\left(t\right)A\sin{kz}& =-\hat{\mathbf{x}}c^2s(t)Ak^2\sin kz \\\dot{p}\left(t\right)& =-c^2k^2s(t)=-\omega^2s(t) \\\ddot{p}\left(t\right)+\omega^2p(t)&=0\\p(t)&=p_{0} \cos \left(\omega t+\varphi_{0}\right) \\ s(t)&=\frac {p_0} {\omega}\sin (\omega t+\varphi_0) \end{aligned} $$
$q$ is defined as the canoical momentum by
$$ q:=M \dot{p}=-M \omega^{2} s(t) $$
Consider this solution as one correpsonding to a fictitous harmonic oscillator of mass $M$ and circular frequncy $\omega$, that can be obtained frome a classical Hamiltonian:
$$ \mathcal{H}=\frac{q^2}{2M}+\frac{M\omega^2p^2}2 $$
Calculate the energy $W$ carried by the electromagnetic field in the cavity
$$ \begin{aligned} \boldsymbol{W} & =F\int_0^L\left(\frac12\varepsilon_0\mathbf{E}^2+\frac1{2\mu_0}\mathbf{B}^2\right)dz= \\ &=F\frac12A^2(\varepsilon_0p^2\underbrace{\int_0^L\sin^2kzdz}{=L/2}+s^2(t)\frac1{\mu_0}\frac{\omega^2}{c^2}\underbrace{\int_0^L\cos^2kzdz}{=L/2}) \\ &=\frac{FL}2A^2\frac{\varepsilon_0}2\left(p^2+\omega^2s^2\right) \\&=\frac{VA^2}2\frac{\varepsilon_0}2\left(p^2+\frac{q^2}{M^2\omega^2}\right)=\frac{VA^2}2\varepsilon_0\frac1{M\omega^2}\left(\frac{M\omega^2p^2}2+\frac{q^2}{2M}\right) \end{aligned} $$
IF we made the following identification, we see that the dynamics of a given mode is the same as that of a harmonic oscillator.
The potential energy of the oscilllaor is that of the electric field while the kinetic energy of the oscillator corresponds to the energy if the magnetic field.
This analogy was realized in the end of the 19th century by Lord Rayleigh and J. Jeans, and this line of thought was also kept by M. lanck when he derived his law of the blackbody radiation. When doing so, however, he prescribed that the many modes in the cavity exchange their energy in quanta, whose energy is $hν=ℏω$
energy in quanta ⇒ quantum mechanics of the oscillator→ the mode
$$ M=\frac{\varepsilon_{0} V A^{2}}{2 \omega^{2}}\\ A=\sqrt{\frac{2 \omega^{2}}{MV\varepsilon_{0} }} $$
$H=\frac{1}{2}(\frac{Q^2}{M}+M\omega^2P^2),$
为方便计算，给出驻波解为以下形式
$$ E_x(z,t)=\left(\frac{2\omega^2}{V\varepsilon_0}\right)^{1/2}q(t)\sin(kz) $$
$$ B_{y}\left(z,t\right)=\left(\frac{\mu_{0}\varepsilon_{0}}{k}\right)\left(\frac{2\omega^{2}}{V\varepsilon_{0}}\right)^{1/2}\dot{q}\left(t\right)\cos\left(kz\right). $$
$$ H=\frac{1}{2}(Q^2+\omega^2P^2), $$