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# 物理代考| Normal Mode Expansion of the Electromagnetic Field 量子力学代写

## 物理代写

6.6 Normal Mode Expansion of the Electromagnetic Field

The next challenge is to express the free electromagnetic field in normal modes, that is, as a set of uncoupled simple harmonic oscillators. We work in a big cubical box of volume $\Omega=L^{3}$, and apply periodic boundary conditions.

Quantum Electrodynamics

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The total energy in the free electromagnetic field in the box is obtained through the sum of the squares of the electric and magnetic fields. In SI units it is given by

$$E=\frac{1}{2} \int_{\Omega} d^{3} x\left(\varepsilon_{0} \vec{E}^{2}+\frac{1}{\mu_{0}} \vec{B}^{2}\right) \quad ; \text { field energy }$$

One has the freedom of choosing a gauge for the electromagnetic potentials, ${ }^{7}$ and here we work in the Coulomb gauge. This gauge has the great advantage that, when quantized, there is a one-to-one correspondence of the resulting quanta with physical photons. For free fields, the Coulomb gauge is defined by

$\vec{\nabla} \cdot \vec{A}=0 \quad ; \Phi=0 \quad ;$ Coulomb gauge $(6.35)$

The electric and magnetic fields are then given by

$$\vec{E}=-\frac{\partial \vec{A}}{\partial t} \quad ; \vec{B}=\vec{\nabla} \times \vec{A}$$

With periodic boundary conditions, the normal modes are given by plane waves

\begin{aligned} q_{\vec{k}}(\vec{x}, t) &=\frac{1}{\sqrt{\Omega}} e^{i\left(\vec{k} \cdot \vec{x}-\omega_{k} t\right)} & ; \vec{k} &=\frac{2 \pi}{L}\left(n_{x}, n_{y}, n_{z}\right) \ \omega_{k} &=|\vec{k}| c & & ; n_{i}=0, \pm 1, \pm 2, \ldots \end{aligned}

Once again, we have an infinite, discrete set of wavenumbers, and the nor- mal modes are orthonormal $$\int_{\Omega} d^{3} x q_{\vec{k}}^{\star}(\vec{x}, t) q_{\vec{k}^{\prime}}(\vec{x}, t)=\delta_{\vec{k}, \vec{k}^{\prime}}$$

Now introduce a set of orthogonal, transverse unit vectors $\vec{e}$ as shown in Fig. 6.1. They satisfy $\vec{e}{\vec{k} s} \cdot \vec{k}=0$ ${ }{\vec{k} s} \cdot \vec{e}{\vec{k} s^{\prime}}=\delta{s, s^{\prime}}$$; s=1,2 52 The vector potential can now be expanded in normal modes as follows \vec{A}(\vec{x}, t)=\sum_{\vec{k}} \sum_{s=1}^{2}\left(\frac{\hbar}{2 \omega_{k} \varepsilon_{0} \Omega}\right)^{1 / 2}\left[a_{\vec{k} s} \vec{e}{\vec{k} s} e^{i\left(\vec{k} \cdot \vec{x}-\omega{k} t\right)}+a_{\vec{k} s}^{\star} \vec{e}{\vec{k} s} e^{-i\left(\vec{k} \cdot \vec{x}-\omega{k} t\right)}\right] where we have chosen particular amplitudes for the normal modes which will make the energy come out nicely. This expansion has the following features to recommend it: • This expression is real, giving rise to real (\vec{E}, \vec{B}) ; • Since only the transverse polarization vectors are used in the expansion, one has ensured that s. ## 物理代考 6.6 电磁场的常模扩展 下一个挑战是将正常模式下的自由电磁场表示为一组未耦合的简单谐振子。我们在一个体积 \Omega=L^{3} 的大立方体盒子中工作，并应用周期性边界条件。 量子电动力学 51 盒子内自由电磁场的总能量是通过电场和磁场的平方和得到的。在 SI 单位中，它由下式给出$$ E=\frac{1}{2} \int_{\Omega} d^{3} x\left(\varepsilon_{0} \vec{E}^{2}+\frac{1}{\mu_{0 }} \vec{B}^{2}\right) \quad ; \text { 场能量 } $$可以自由选择电磁势的规范 { }^{7}，这里我们使用库仑规范。该规范具有很大的优势，即在量化时，产生的量子与物理光子一一对应。对于自由场，库仑规范定义为 \vec{\nabla} \cdot \vec{A}=0 \quad ; \Phi=0 \quad ; 库仑规 (6.35) 电场和磁场由下式给出$$ \vec{E}=-\frac{\partial \vec{A}}{\partial t} \quad ; \vec{B}=\vec{\nabla} \times \vec{A} $$在周期性边界条件下，正常模式由平面波给出$$ \开始{对齐} q_{\vec{k}}(\vec{x}, t) &=\frac{1}{\sqrt{\Omega}} e^{i\left(\vec{k} \cdot \vec{x }-\omega_{k} t\right)} & ; \vec{k} &=\frac{2 \pi}{L}\left(n_{x}, n_{y}, n_{z}\right) \ \omega_{k} &=|\vec{k}| C ＆ ＆ ; n_{i}=0, \pm 1, \pm 2, \ldots \end{对齐} $$再一次，我们有一个无限的离散波数集，并且正常模式是正交$$ \int_{\Omega} d^{3} x q_{\vec{k}}^{\star}(\ vec{x}, t) q_{\vec{k}^{\prime}}(\vec{x}, t)=\delta_{\vec{k}, \vec{k}^{\prime}} $$现在引入一组正交的横向单位向量 \vec{e}，如图 6.1 所示。它们满足 \vec{e}{\vec{k} s} \cdot \vec{k}=0 { }{\vec{k} s} \cdot \vec{e}{\vec {k} s^{\prime}}=\delta{s, s^{\prime}}$$; s=1,2$
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• 这个表达式是实数，产生实数 $(\vec{E}, \vec{B}) ;$
• 由于在扩展中仅使用横向偏振矢量，因此已确保

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