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

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$

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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: