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What happens in the middle of a Feynmann Diagram?

  1. Dec 4, 2006 #1
    I'm a high school student and I've been recently learning about feynmann diagrams and I've always wondered what happens in the intermediate stage in a feynmann diagram. i.e the step in between when a neutron turns into a proton and the emergance of a boson. I've tried asking my physics teacher but he says he doesn't know so it will be interesting to learn some theories about what could be happening.
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  3. Dec 4, 2006 #2
    Are you referring to a beta-decay?

    http://upload.wikimedia.org/wikipedia/en/b/b5/Beta-Decay.png [Broken]
    Last edited by a moderator: May 2, 2017
  4. Dec 4, 2006 #3


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    Er... what process exactly are you describing here? A neutron going into a proton and a "boson"? What boson?

    This isn't a beta decay, where a neutron becomes a proton, electron, and anti-neutrino. So you need to explain what particular process you are refering to.

  5. Dec 4, 2006 #4


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    Well, with some indulgence, one can consider that it is a weak process where we have the neutron turning in a proton and a (virtual) W- boson, which then immediately (or almost so) turns into an electron+anti-neutrino, in lowest order.

    That said, on this level I don't think that there is any distinguishable means to see any effect of the W- ; in other words a point interaction would be just as good.
  6. Dec 4, 2006 #5


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    Excellent question. The real answer is we don't know. In fact, a basic assumption of particle physics(and Feynman diagrams) is that so-called three point interactions are the basic driving force behind particle physics. That is, we assume that particle -> particle transformations can occur, and then see what the consequences are. This assumption of transformations is, of course, empirically based.

    Reilly Atkinson
  7. Dec 4, 2006 #6


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    Go down to the local university bookstore and buy Feynman's very short, inexpensive, and non mathematical introduction to quantum electrodynamics:

    It is very easy to read, and you will find that it gives a very good introduction to what is going on in Feynman's variety of quantum mechanics.
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  8. Dec 4, 2006 #7
    We don't assume that at all. Four point interactions are also present (as well as mass terms). The Feynman diagram is a perfectly well defined mathematical expression within perturbation theory.

    It is simply a transition from an initial state to a final state, as according to the principle of least action.
  9. Dec 4, 2006 #8
    Are you really sure you want to say 'mathematically well defined'. Dyson has shown that the perturbation series is most probably divergent. And I have not met a mathematician yet who says that renormalisation is mathematically well defined.

    Well I know that this is definately not correct.
  10. Dec 5, 2006 #9


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    Satisferborum, I think that you actually want to understand the processes of particle creation and destruction as processes continuous in time. This cannot be understood from the usual Feynman diagrams because they describe the S-matrix, i.e., the transition amplitude from t=infinity to t=-infinity. However, you can describe it continuously in the Schrodinger picture by writing the quantum state as
    [tex]|\Psi(t)\rangle = \sum_a c_a(t) |\Psi_a\rangle[/tex]
    where [tex]|\Psi_a\rangle[/tex] are normalized states with a definite number of particles of different species. The Hamiltonian of the system governs the time evolution of the coefficients [tex]c_a(t)[/tex]. However, the Hamiltonian does not describe it completely, beacuse there is also a "collapse" of the state due to a quantum measurement. What exactly this "collapse" is depends on the interpretation of quantum mechanics you use. For example, the Bohmian interpretation offers a completely continuous description of an effective "collapse":
    Last edited: Dec 5, 2006
  11. Dec 5, 2006 #10


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    First, I neglected other than 3-pt interactions to keep things simple. Second, we know that particle transformations do occur provided the appropriate conservation laws are upheld. I'm thinking of E&M radiation or pion exchange in pion-nucleon scattering, as examples.

    You tell me why an accelerating charged particle radiates? Whence the photon, where does it come from, how does it get outside the particle? (Part of the issue is that we know that QFT says that a real electron is pretty complicated, with clouds of photons, all manner of pairs, and so on. we don't know much about the details of the electron's structure, so we are lacking quite fundamental info. on the behavior of charged particles.

    Re dynamics and finite times: just use standard time dependent perturbation theory, or resolvants -- a particularly good account of finite time dynamics can be found in Atom-Photon Interactions by Cohen-Tannoudji et al. Also see Dirac's Quantum Mechanics.
    Reilly Atkinson
  12. Dec 5, 2006 #11
    Perturbation series undoubtedly are divergent, but this is not the issue. The Feynman diagrams themselves are well defined - I can sit down and work them out and get one and only one answer. At the loop level I have to worry about a renormalization procedure, but using counterterms, that again is perfectly well defined. The only part which is on relatively unsure footing is the derivation of the Feynman rules themselves via perturbation theory, but since they work so phenomenally well, I don't think we should be so fussy.

    Would you care to explain why, or are you just going to make baseless assertions?

    An accelerating charged particle radiates in order to preserve the U(1) local symmetry. Admitedly I don't have an answer as to why the universe wants to preserve the U(1) symmetry, but that would be a rather tall order. Your second question is not well defined - what do you mean by 'inside the particle'?

    On the last issue, it seems to me that we know quite a lot about the electron structure. See for example, http://www.slac.stanford.edu/spires/find/hep/www?j=POSCI,HEP2005,061 [Broken]
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  13. Dec 5, 2006 #12


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    Severian -- Exactly, U1 is after the fact. 'Outside' the particle -- analogy with photoelectric effect. Crudely speaking, one can imagine a photon in the electron's cloud getting kicked out -- just to be clear, I mean this to be a metaphore.

    Thanks for the reference. For my thesis I computed radiative corrections to electron --nucleon scattering experiments for some of the experiments that Robert Hofstadter's group used to map out nucleon form factors. In those days, just before SLAC, high energy typically meant a few hundred MEV.

    The experimental work in your reference is pretty cool. My how things have changed since the early 1960's; physics marches on.
    Reilly Atkinson

  14. Dec 19, 2006 #13
    This may not be the right thread, but "electron acceleration" has been mentioned, so maybe it is. I thought Maxwell started this one, but he certainly had not heard about QFT!

    What is bugging me is
    a)what is the energy distribution of photons being emitted from such acceleration

    And b) In an antenna excited by alternating current, are the photns radiated with the energy/wavelength/frequency of the excitation?

    And c)If so, why the heck??

    PS I'm an elderly engineer whowas once taught "Electriicity and magnetism", and I now need to unlearn it

    Thanks folk

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