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I've read everywhere that in QFT, this case QED, the interaction between particles is seen as a exchange of gauge bosons wich transmit the interaction, in this case photons.

But i want to know what is the "property" of the photon which cause the repulsion of two electrons or the atraction of an electron and a positron. Or it's just a misconception to think on "classical" electric repulsion or atraction at this scales? Is happening something else in the interaction? The interpretation is incomplete?

Julian Ar.
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  • The repulsion is easy. When you shoot someone with a bullet, you get a recoil and the victim is pushed by the bullet. The momentum you and he get is equal per the conservation law. So you equally repell. The attraction is a bit trickier and requires the exchanged virtual photon to have some exotic parameters that are only possible for virtual particles. I will leave this part to the experts. Note that hypothetical bosons with an even spin (0 for the Higgs and 2 for graviton) can mediate only attraction. The spin of 2 means, rotate by a half turn (360/2) and nothing changes, so no repulsion. – safesphere Sep 13 '17 at 05:09
  • I guess you want an intuitive explanation. The best I know: the second question on http://math.ucr.edu/home/baez/physics/Quantum/virtual_particles.html. –  Sep 13 '17 at 08:39
  • The specific property of the photon that you are looking for is its spin (s=1 for the photon). In general, QFT with interactions mediated by gauge bosons has mechanism for determining whether the interations are attractive or repulsive that are depemndent on the boson spin. You just have to do the math. – Lewis Miller Sep 13 '17 at 15:12

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First we need to state, yet again, that virtual particles don't exist. The way we calculate particle interactions is using perturbation theory, and for each term in the perturbative expansion we need to integrate an equation called the propagator. The Feynman diagrams are graphical representations of the equation for the propagator and do not show actual particles. The virtual particles shown in the diagrams are an artefact of the perturbative expansion.

With that out of the way, when we do the calculation the spin of the virtual particle that we are using in the calculation determines whether the interaction energy is negative or positive. With a zero or spin two particle the interaction energy for like charges is negative and the interaction energy for opposite charges is positive, so like charges attract and unlike charges repel. For a spin one virtual particle the opposite is true so like charges repel and unlike charges attract.

Photons have spin one, so like electrical charges repel and opposite electrical charges attract.

Gravitons have spin two so like charges attract, hence masses gravitationally attract each other. If negative masses existed (and we are fairly sure they don't) then a positive and negative mass would repel each other.

I think what you are asking is whether the virtual photons are different in the Feynman diagrams for like and unlike electrical charges interacting, but this is a meaningless question as virtual photons don't exist. When we calculate the propagator represented by the diagram of the virtual photon it has a different sign in the two cases.

If you're feeling brave the stuff I've outlined above is described in more detail in the answers to Why do same/opposite electric charges repel/attract each other, respectively? and How does one show using QED that same/opposite electric charges repel/attract each other, respectively?.

John Rennie
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    Ok, i get it, thank you. But can i ask you something more? So what's the motivation for talk about virtual particles in first place? I mean, why this concept exist, if are just an artifact for calculations? – Julian Ar. Sep 13 '17 at 21:32