I can hardly remember, so please correct if I am mistaken. In path integral formulation, the probability of detecting a quantum particle at a given point in space is calculated through the integral of all possible paths leading to this point, while each path has its own amplitude, there are interference effects, and so on. In a sense, one can say that between emission and absorption, a particle moves along all possible paths.

How does this generalize to the case of particle interactions in the context of Feynman diagrams? For example, in the scattering of two particles, integration is carried out through the addition of all possible diagrams leading to a given interaction result? In other words, in a sense, we can say that between the initial and final states particles simultaneously do not interact/interact in all possible ways (in this case, the amplitude of each diagram depends on the value of the coupling constant and the number of interaction vertices)?


1 Answer 1



In something like:

$$ e^-+\mu^-\rightarrow e^-+\mu^- $$

you start with two free particle states, and end with 2 free particle states. In between, the electron and electromagnetic fields (and perhaps the weak field) go through all possible configurations that conserve energy an momentum. That is something that cannot be solved exactly, so it is approximated with Feynman diagrams:

enter image description here

Note that this only includes connected diagrams. In theory you need to consider a field configuration with 2 separate diagrams, say diagram (a) plus a photon loop off to the side, in practice they are included by exponentiating the sum of connected diagrams. (How that works, is technical).

Also, if you look at the higher order diagrams, say (d), there can be an arbitrary four momentum running around that electron-position loop in the photon line, so it in fact represents an infinite number of field configurations, but since you just add diagrams, that's what the integral does.

Each electron-photon vertex does include a factor of $-i(-e)$, but that's not the only contribution to the diagram. You simply have to evaluate them to find the total contribution. (A positron-photon vertex has $-i(+e)$, which may not make sense if you watch youtube videos where attraction is described as exchanging a boomerang between 2 row boats, while repulsion is a basketball...that is simply putting too much classical reality into virtual particle exchange).

Several things to note:

In the tree level diagram ($t$-channel), the exchanged photon is highly virtual. For instance, in the center of mass frame, with incoming energy $E \gg m_{\mu}$, and scattering at 180 degrees, it's 4-momentum is:

$$ q^{\mu} = p^{\mu}-p'^{\mu} = (E,0,0,E)-(E,0,0,-E) = (0,0,0,2E)$$

That is, it has zero energy and lots of momentum. As a particle, it is space-like (hence:virtual). It is not emitted by the muon and the absorbed by the electron, nor is it emitted by muon and then absorbed by the electron. It's just "exchanged".

Meanwhile, in $s$-channel $e^-e^+$-annihilation, you find:

$$ q^{\mu} = p_1^{\mu}+p_2^{\mu} = (E,0,0,E)+(E,0,0,-E) = (2E,0,0,0)$$

and the photon has lots of mass and zero momentum. Again: not a property of real photons.

Nevertheless, virtual photons have longitudinal and transverse polarizations (depending only on the kinematics of the external lines), which can be used to separate the electric and magnetic responses of nuclear targets at fixed momentum transfer. It makes them seem quite like real particles, but they are indeed virtual.

For identical particles in the final state, you need to include $u$-channel processes:

enter image description here

that swap the final state four-momenta.

Finally, all these diagrams are amplitudes. They interfere. At tree-level, the exchanged particle could be a photon, and it could be a $Z^0$, hence both processes contribute, and their interference term is parity violating. (This fact is used to probe the spin of the strange quark sea via elastic scattering of polarized electrons off protons).

The point is: all diagrams contribute, and you have to add them all up to get the amplitude. Only then do you square it to get a probability, which is converted into a differential cross-section to compare to experiment.

Moreover, virtual particles are perturbative approximations to intermediate field configurations, that's it. They are not actually particles, but it can useful to treat them as such in oder to interpret scattering experiments.

  • $\begingroup$ Are diagrams b) and c) 1PI? $\endgroup$ Jan 7, 2021 at 3:08
  • $\begingroup$ what's the name for "connected"? $\endgroup$
    – JEB
    Jan 7, 2021 at 3:52
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    $\begingroup$ In theory you need to consider a field configuration with 2 separate diagrams, say diagram (a) plus a photon loop off to the side I don't understand, to be honest. That is, the photon loop and the particles from diagram (a) do not interact, they are not connected in any way? Why should they be considered separately? And the second question: is the whole process the sum of all diagrams, or is the process described in only one of the diagrams being implemented? $\endgroup$ Jan 7, 2021 at 10:11
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    $\begingroup$ @ArmanArmenpress I would recommend learning the basics of Feynman diagrams. All orders of Feynman diagrams should be included in the final amplitude, as an asymptotic series (practically, the series is terminated after some order and we obtain an approximation). Secondly (and this is evident in the path integral formalism), the contribution of disconnected diagrams gets cancelled out while obtaining physical correlation functions/the non-trivial parts of the S-matrix, so there is really no need to consider them in this case. $\endgroup$ Jan 8, 2021 at 6:09
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    $\begingroup$ @АрманГаспарян I know what the path integral is. I'm saying that your message has almost nothing to do with my comment towards ArmanArmenpress. JEB is talking about additional "vacuum bubble" diagrams in their answer, which I'm saying getting cancelled out as a direct consequence of the linked-cluster theorem (which is not trivial). So your "Obviously, this is the same as in the case of the path integral" comment is a non-sequitur. $\endgroup$ Jan 8, 2021 at 10:17

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