5
$\begingroup$

When you produce a long-lived particle (like charged pions, muons), it enters your detector. When you produce a short-lived particle (like the higgs, tau), the production of the particle is only seen as an increase of a certain process when the collision's center of mass energy is near the particle's mass. But there are also medium-lifetime particles (lifetime ~1ns), where they never get anywhere near the detector, but the detector's resolution allows us to identify that the decay products of the particle didn't come from the collision point, but rather a small distance away from the collision point.

I'll keep it simple and try to deal with a theory with just two scalar fields - a lighter one and a heavier one. A tree level diagram involved in calculating the scattering of two lighter particles is this:

enter image description here

Obviously, the scattering will increase when the center of mass collision energy approaches the mass of the heavy particle, because the propagator's denominator will go to zero. But another phenomenon that we expect to happen near resonance is displaced vertices. The typical description of this process goes something like this: "as the center of mass energy of the 2x light particle collision approaches the heavy particle's mass, you start to produce the on-shell heavy particle. It flies some distance, then decays, giving you a displaced vertex."

This all seems a bit handwavy to me, and I'd like to get a more rigorous understanding of the process. Is there some genuine way to calculate the cross section for displaced vertices of a certain distance? Or do Feynman diagrams for some reason only describe processes that happen in one point in space? How do we properly understand the transition from

  1. one Feynman diagram (far off resonance) involved in calculating the cross section for the light particles scattering off one another, mediated by an exchange of a virtual heavy particle
  2. to one Feynman diagram used to calculate the cross section for producing a short lived particle that decays (as described by another, separate Feynman diagram)?
$\endgroup$
14
  • 2
    $\begingroup$ Your are misunderstanding what Feynman diagrams are about. I suggest you pick up an introduction to quantum field theory. $\endgroup$ Commented Mar 30, 2023 at 11:33
  • 2
    $\begingroup$ I don't think I am. What exactly in this question betrays a misunderstanding of Feynman diagrams, and what description of Feynman diagrams resolves this question? Just saying "read an entire 1000 page book" (I already have read what I believe are the relevant parts) isn't really helpful, and isn't really in the spirit of this website. $\endgroup$
    – AXensen
    Commented Mar 30, 2023 at 11:36
  • 2
    $\begingroup$ Feynman diagrams are a visual representation of different terms in a perturbation expansion, i.e. they are a mathematical concept. Speaking of displaced vertices and distance does not make sense. $\endgroup$ Commented Mar 30, 2023 at 11:39
  • 3
    $\begingroup$ What you're saying seems to just reiterate that my confusion is valid. The heart of my question is - since Feynman diagrams don't seem to be capable of describing a displaced vertex, but displaced vertices are a real and observable phenomenon in accelerators that should in principle be described by quantum field theory, how do we calculate the probability of an observed displaced vertex. And how do we understand the transition from "two light particles scattering by the exchange of a virtual heavy particle" to "the heavy particle is produced, and decays some distance away from the beam pipe" $\endgroup$
    – AXensen
    Commented Mar 30, 2023 at 11:43
  • 2
    $\begingroup$ Or do Feynman diagrams for some reason only describe processes that happen in one point in space? -- They do not account for position in any way., neither if ithe interaction happens in a point or in (almost) all of space-time. Feynman-Diagrams are used to perturbatively calculate the overlap between asymptotic states. $\endgroup$ Commented Mar 30, 2023 at 12:35

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.