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Is there any physical reason/symmetry prohibiting me from studying the $e^-\rightarrow e^-\gamma$ interaction? Because I haven't seen any QFT textbook dealing with such a simple process. It is the simplest QED interaction, after all...

I have tried writing down a Feynman amplitude for such process $$i\mathcal{M}=ie\bar{u}(\vec{p}')\gamma^{\mu}u(\vec{p})\epsilon_{\lambda\mu}(\vec{k})$$ where $p',p$ are the final and initial momenta of the electron, whereas $k$ is the momentum of the emitted photon. $\lambda$ denotes the polarization of the latter. I can obtain the square of this amplitude, sum with respect to final spin (and polarization) states and average with respect to initial ones to obtain $$\frac{1}{2}\sum_{s,s',\lambda}|\mathcal{M}|^2=4e^2 (2m^2-p\cdot p')$$ I can not help, but to notice that the final result for the squared Feynman amplitude does not depend on the emitted photon. Is this correct?

Now, let's see what happens when we try to calculate something measurable, like the decay rate for the interaction at hand. The latter (in the reference frame of the initial electron), is defined as $$\Gamma=\frac{1}{2m}\int\frac{d^3\vec{p}'}{(2\pi)^3}\frac{1}{2p'^0} \int\frac{d^3\vec{k}}{(2\pi)^3}\frac{1}{2k^0} (2\pi)^4\delta^{4}(p-p'-k)\frac{1}{2}\sum_{s,s',\lambda}|\mathcal{M}|^2$$ which results in something rather peculiar, namely $$\Gamma=2e^2\int\frac{d\Omega_{\vec{k}}}{(2\pi)^3}\int_0^{\infty}d|\vec{k}| |\vec{k}|^2\frac{1}{2k^0}\frac{1}{2\sqrt{|\vec{k}|^2+m^2}} \bigg(1-\sqrt{|\vec{k}|^2+m^2}\bigg) (2\pi)\delta\bigg(m-k^0\sqrt{|\vec{k}|^2+m^2}\bigg)$$ Is this result meaningful? Or should the peculiarity of my answer indicate that something wrong is going on. If this is not the way to study the interaction at hand using QED, then how do I do that?

Any help will be appreciated.

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This decay is not allowed by conservation of energy. Imagine this interaction in the rest frame of the electron. Before the interaction, the electron is not moving and therefore has energy $m$. After the interaction, the electron and the photon must move in order to have 0 momentum. However, this implies that the electron gained energy kinetic energy simply by emitting a photon (losing energy) which violates energy conservation. It is easy to see that in this case, there is no way the final energy and initial energy is equal.

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  • $\begingroup$ Oh, yes, you are right, thanks $\endgroup$
    – schris38
    Commented Apr 10, 2023 at 16:49

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