https://haifei.pro/tags/jaynes-cummings/Recent content in Jaynes-Cummings on Haifei's HomeHugozh-CNhfwang132@gmail.com (hfwang132)hfwang132@gmail.com (hfwang132)This work is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License.Sun, 13 Apr 2025 01:52:37 +0800https://haifei.pro/post_20250413_%E4%BB%8E%E9%87%8F%E5%AD%90%E5%9C%BA%E8%AE%BA%E5%88%B0%E8%85%94%E9%87%8F%E5%AD%90%E7%94%B5%E5%8A%A8%E5%8A%9B%E5%AD%A6-%E6%9B%B4%E9%AB%98%E6%9B%B4%E5%A6%99%E7%9A%84%E7%94%B5%E5%8A%A8%E5%8A%9B%E5%AD%A6-5/Sun, 13 Apr 2025 01:52:37 +0800hfwang132@gmail.com (hfwang132)https://haifei.pro/post_20250413_%E4%BB%8E%E9%87%8F%E5%AD%90%E5%9C%BA%E8%AE%BA%E5%88%B0%E8%85%94%E9%87%8F%E5%AD%90%E7%94%B5%E5%8A%A8%E5%8A%9B%E5%AD%A6-%E6%9B%B4%E9%AB%98%E6%9B%B4%E5%A6%99%E7%9A%84%E7%94%B5%E5%8A%A8%E5%8A%9B%E5%AD%A6-5/<p>在本篇文章中，我们将从 QED 的拉氏量出发：</p> \[\begin{aligned} \mathcal{L}= \bar{\psi}(\mathrm{i}\gamma^\mu \partial_\mu -m)\psi - eA_\mu \bar{\psi} \gamma^\mu \psi -\frac{1}{4} F_{\mu\nu} F^{\mu\nu} \end{aligned} \]<p>这个拉氏量非常复杂：它不仅考虑了电子的反粒子——正电子，并且耦合项 \(- eA_\mu \bar{\psi} \gamma^\mu \psi\) 还是一个三次项，可以描述电子-正电子对产生/湮灭等多种过程。由于三次项的存在，这个拉氏量没有解析解，只能用量子场论的微扰法求解。</p>