I'm half way through the excellent "Student Friendly Quantum Field Theory" and I read that single vertex Feynman diagrams in QED are "not physical" because their corresponding amplitudes are zero. For example the diagram $$ e_{\mathbf{p_1}}^- + e_{\mathbf{p_2}}^+ \to \gamma_{\mathbf{k_1}} $$ has a probability amplitude that (when you calculate it) includes a factor of $\delta^{(4)}(k_1 - p_1 - p_2)$, where the boldface p's and k are 3-momenta and the normal typeface are 4-momenta. The argument goes that since the photon is massless we must have $k_{1\mu}k_1^{\mu} = 0$, but if you work out $(p_1+p_2)_{\mu}(p_1 + p_2)^{\mu}$ it turns out non-zero, therefore we can't find a real photon momentum that makes the dirac delta, and consequently the amplitude $\langle\gamma_{\mathbf{k_1}}\lvert e_{\mathbf{p_1}}^- e_{\mathbf{p_2}}^+\rangle$ non zero. A similar reasoning shows that every other single vertex Feynman diagram (e.g. $e_{\mathbf{p_1}}^-\to \gamma_{\mathbf{k_1}} + e_{\mathbf{p_2}}^-$) are also "non physical".

So my questions are:

  1. If these diagrams are non-physical what's the simplest diagram that generates a photon that is physical. Or, "where do all the photons come from?"
  2. Are there any interpretations (for the fact the amplitude for single vertex diagrams are zero) other than that they are "non-physical"? For example, perhaps photons with $k^2 \ne 0$ are possible, but live too short a time to be observed.

Please be gentle, I'm not actually a student, just an enthusiast and this is my lockdown reading!

  • 1
    $\begingroup$ Can you give the page where this issue is discussed? $\endgroup$ – proton Apr 28 '20 at 10:11
  • 1
    $\begingroup$ It's page 218 in my edition. Also see Box 8-1: "Off the Mass Shell" - Why a Single Vertex Interaction is Not Physical. Cheers. $\endgroup$ – Alex Zeffertt Apr 28 '20 at 10:48

The reason you state for why the amplitude $e^+e^-\to\gamma$ vanishes is correct. But I would like to simplify it a bit. Mainly because it is not a consequence of QFT but of Special Relativity.

Suppose you are in the center-of-mass frame of the electron-positron pair. Momentum conservation tells you that in this frame the resulting particle will be produced at rest and will have a mass of $M^2 = (p_1+p_2)^2 = (E_1+E_2)^2>4m_e^2$. The photon is never at rest and its mass squared is zero. So there is no way to conserve momentum.

Note that there simply cannot exist a photon with $k^2\neq 0$. Such a photon would not travel at the speed of light, so it's inherently a contradiction. Also note that, by the same argument as above done for the opposite process, we can show that photons cannot decay. And the same conclusions hold for all particles with zero mass.

The simplest process that gives photons is $e^+e^-\to2\gamma$. The presence of two photons makes it kinematically viable because we can have a nonzero invariant mass in the center-of-mass frame of the two.

  • $\begingroup$ Thanks for the response. I can see how $e^+e^-\to2\gamma$ would solve the problem, but it must be relatively rare. I don't suppose my lightbulb uses positrons to make photons. Is there a more common case? Also, when you say there simply cannot exist a photon with $k^2 \ne 0$ are we essentially just making a definition: if $k^2 = 0$ it's a photon, otherwise it's a "virtual" photon. It seems odd to me that some intermediate particles in Feynman diagrams are "virtual" and can't exist as initial or final state whilst other intermediate particles can! $\endgroup$ – Alex Zeffertt Apr 28 '20 at 11:00
  • $\begingroup$ Electron-positron annihilation is not something going on in your lightbulb because antimatter is quite rare. But it a fairly common process in hep. I don't know the cross section though. $\endgroup$ – MannyC Apr 28 '20 at 11:08
  • 1
    $\begingroup$ Don't be fooled by the concept of "virtual particles." They are not real particles but just a convenient notion to talk about propagators in Feynman diagrams. When people say that virtual particles are created/annihilated, that's very imprecise and I don't particularly like that terminology. $\endgroup$ – MannyC Apr 28 '20 at 11:08
  • $\begingroup$ By the way, nearly all processes that produce light in everyday life do not involve the decay of a particle. So there is no need to disturb QFT to explain that. At the fundamental level it's just electrons in atoms that get excited by some levels and then come back and emit light. In that case the photon is semiclassical. $\endgroup$ – MannyC Apr 28 '20 at 11:14
  • $\begingroup$ I get the mass related argument why the virtual photons are not physical, or at least not the same kind of thing as a normal photon. Although this doesn't seem to apply to all virtual particles (e.g. the electron/positron pair in a loop). Maybe I'm just reading too much into the phrase "not physical". If all it means is "you won't get this experimental result" then that's okay. In that case it just raises exactly the same philosophical questions as the double slit experiment. $\endgroup$ – Alex Zeffertt Apr 28 '20 at 11:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.