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I 'Real particle and virtual particle interactions' Part 2...

  1. Aug 25, 2016 #1
    Hi folks,

    I number of years ago, I asked a question about real particles interacting with virtual particles - since then, I've learned a little bit (maybe not too much :-), but I'm still curious about one thing. Original thread here:

    https://www.physicsforums.com/threads/interactions-between-real-and-virtual-particles.367842/

    Since virtual particles appear in a particular perturbative expansion approach, as in my original question, could you ever have a 'real' electron, for example interacting (annihilating) with the positron of a virtual electron-positron pair? That is, in the perturbative expansion, would this ever actually be one of the terms involved? Since the 'real' electron is effectively on shell, and the virtual positron os off shell, shouldn't terms of this type not be added?

    I ask in part because of this posting:

    https://profmattstrassler.com/artic...ysics-basics/virtual-particles-what-are-they/

    Wouldn't the real electron interaction with the virtual electron be 'incompatible'? Or would you still find these types of terms showing up in the perturbative expansion approach?

    (and I realize there's perhaps no exact line between what's virtual and what's real)

    Thanks all! Sorry to dwell on the same sort of question...
     
  2. jcsd
  3. Aug 25, 2016 #2
    Yes virtual particles come from a perturbative expansion in Quantum Field Theory (QFT). You can expand the path integral. Then you will get many Feynman diagrams, and for the case of Quantum Electrodynamics (QED), the possible interaction vertex is electron-positron-photon vertex, if you take a particle travelling backgrounds in time to be the antiparticle.

    A (connected) connected Feynman diagram will have a certain number of external lines (also called legs). Each external line corresponds to a real particle. Each internal line corresponds to what we call a virtual paticle. It is just a name. In a given interaction, particles from the far future and from the far past are considered as distinct legs. This is why a scattering diagram will have four external legs, corresponding to two particles in the far past and two particles in the far future.

    You can have a real electron travelling fowards in time interacting with a virtual positron travelling fowards in time. Just take a Feynman diagram in which a positron and photon scatter, the result is a virtual positron which interacts with an incoming real electron, and produces a photon. Unfortunatelly I can't draw this here.
     
  4. Aug 25, 2016 #3
    Thanks Lucas SV - helpful!

    In the last situation you describe, would you always need a real incoming positron initially?

    I guess I'm wondering if there's a Feynman diagram where a single electron goes in (one leg) annihilates with a virtual positron (not an incoming leg) producing a photon, and then, somehow, that photon interacts with something else to produce an electron... actually, maybe this is just silly. The main question is, can you have a Feynman diagram where one real particle goes in (let's say a fermion) and the same type of particle comes out, and yet, internally, the incoming particle went through some type of annihilation?
     
  5. Aug 25, 2016 #4
    Good question. It depends on the theory. Since in QED, there is essentially only one kind of vertex, If a real electron moving foward in time is incoming in a vertex, this necessarily implies that there will be an outgoing arrow and a wigly line also hitting this vertex. Now the answer depends on what we mean by annihilation. You could think of an internal annihilation as meaning a vertex in which the electron and positron are moving foward in time, so the fermionic arrow goes from the past towards the vertex and then goes back to the past, while the photon goes to the future. If we insist that the incoming positron is virtual, this means it originates in a vertex, as opposed to the far past. This is where I got the idea I said before, and it corresponds to this new vertex being a photon scattering of a positron. The new vertex can also be pair production.

    This 'internal annihilation' term is not standard as far as I know, however. Usually annihilation means that there are two external legs from the past, one is a positron and one is an electron, and two photons somewhere along the diagram. More precisely, If you cut these two photon lines, the diagram splits into two, the past diagram which starts with a positron and an electron (past legs) and ends with two photons (future legs), and a future diagram, in which starts with two photons (past legs) and can end in an arbitrary way.
     
    Last edited: Aug 25, 2016
  6. Aug 26, 2016 #5

    A. Neumaier

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    The process of renormalization, which is needed to make sense of perturbation theory in relativistic quantum field theories, is needed precisely to eliminate this kind of behavior. In technical terms, renormalized perturbation theory features only 1-particle irreducible graphs.
     
  7. Aug 26, 2016 #6

    vanhees71

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    More specifically what you described above in #3 is a "self-energy" contribution to the electron/positron field, which in leading order is given by a diagram, where you have an incoming and an outgoin electron-positron line connected by a loop consisting of an internal electron-positron and an internal photon line. The latter are "off shell" or "virtual", but the diagram tells you the precise meaning which integral to write down. You find that such types of diagrams lead, however, to useless results because the integrals are divergent, a phenomenon well known from classical electron theory used by Lorentz. The solution of this is, as A. Neumaier stressed in the previous posting, what's known as "renormalization theory".
     
  8. Aug 27, 2016 #7
    Thanks A. Neumaier! So (just so I'm clear), if your using relativistic QFT, you need to eliminate the "self-energy" contributions to eliminate the divergent integrals? (so 1-particle irreducible graphs only).

    (and thanks to vanhees71 too!)
     
  9. Aug 28, 2016 #8

    A. Neumaier

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    In principle yes. However, the self-energy contributions reappear in resummed form in the renormalized propagators. They are essential for getting correct predicitons. Vertex functions also need to be renormalized.
     
  10. Aug 28, 2016 #9
    However,, renormalized perturbation theory still always features only 1-particle irreducible graphs?
     
  11. Aug 28, 2016 #10

    vanhees71

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    No, the 1PI graphs are the fundamental building blocks for diagrams that have to be renormalized. For a renormalizable model like QED (or the entire Standard Model) renormalization means to lump a finite set of infinities into unobservable "bare" parameters (wave-function normalization, masses of the particles, and coupling constants) and express everything in terms of observable physical and finite parameters. The S-matrix elements consist of all connected diagrams which are then built with the 1PI diagrams (renormalized proper vertex functions) connected by full propagators. The higher-order renormalized contributions to the self-energies and vertices, containing loops in the corresponding Feynman diagrams, are not simply left out of the calculation but lead to important quantum corrections to the classical approximation (given by the tree-level diagrams) like the Lamb shift in the hydrogen atom or the anomalous magnetic moment of the electron, which are among the best understood quantities ever since the theory prediction agrees with experiments to at least 12 significant digits.
     
  12. Aug 28, 2016 #11
    Ok, but then as A. Neumaier has suggested, you don't get terms diagrams where one real particle goes in (let's say a fermion) and the same type of particle comes out, and yet, internally, the incoming particle went through some type of annihilation...
     
  13. Aug 28, 2016 #12

    vanhees71

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    Of course, you have such diagrams. The 1PI ones are the self-energy diagrams and they contribute to the propagators of the theory. Usually this is depected by drawing a blob at the lines in the Feynman diagrams when using (resummed) perturbation theory including higher-order ("radiative") effects. In the internal lines, where the momentum are off the mass shell you get an effect from this "dressing" of the propagator. On the external lines they provide just wave-function normalization factors which are cancelled by the corresponding free-particle wave amplitudes. So you don't need to dress the external lines and use the usual free-particle wave amplitudes from the free theory.
     
  14. Aug 28, 2016 #13
    Hi vanhees71,

    Thanks for your response. Unfortunately, we've gone well over my head now, alas. If expressing everything in terms of observable physical and finite parameters, how can you have a diagram that involves a fermion (e.g. electron) coming in, interacting with an (unobservable) virtual electron-positron pair, and ultimately leading to an electron coming out... this isn't physically observable (i.e the internal process), and A. Neumaier indicated this situation was something that led to the need for renormalization in the first place. So I'm somewhat confused...
     
  15. Aug 28, 2016 #14

    vanhees71

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    That's exactly the point of renormalization. In relativistic QFT a particle interpretation is possible only in terms of "asymptotic free states". If you have a single electron, there's nothing to interact with and it will stay a single electron forever. The contributions from the self-energy diagrams become visible only indirectly via the dressed propagators in diagrams describing S-matrix elements of scattering processes.
     
  16. Aug 28, 2016 #15

    A. Neumaier

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    Yes, but one doesn't sum only the propagators but also the vertices. The full renormalized diagram story is available in a nice paper by t'Hooft.
     
  17. Aug 28, 2016 #16
    Sorry to beat a dead horse - but (even with the pointer to the t'Hooft paper), I'm still quite confused. So do you, or do you not, after renormalization have terms of the type that I'd originally indicated in #3 above - if we consider a single fermion only, i.e., a Feynman diagram with one leg only (nothing else - no other legs) and the same type of particle comes out (i.e., a single leg out), and yet, internally, the incoming particle went through some type of annihilation? vanhees71, you said you do have such diagrams, but are they altered after renormalization (since the internal processes are unobservable)?

    Thanks - I'll get it eventually :-)
     
  18. Aug 28, 2016 #17

    A. Neumaier

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    One does have such terms in bare perturbation theory, but they come out infinite and are completely meaningless. What is meaningful and finite are the equations (or perturbation series) determining the renormalized objects, as in t'Hooft's paper.

    I don't understand why you want to breath life into a dead thing by speculating about something unrelated to actual practice. If you want to be entertained you just take the diagrams and make up stories about what they mean. But if you want to understand quantum field theory on a more than lay people level you need to forget it all and take the diagrams as what they really are - pictorial shorthand for complicated integrals that make sense only as a sum - after the individual contributions were renormalized. If you try to steer an intermediate path you only stay and/or get confused and get confusing attempts to teach you something you don't want to hear.
     
    Last edited: Aug 29, 2016
  19. Aug 29, 2016 #18

    vanhees71

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    You cannot explain this without the appropriate math.

    It's not true that the self-energy diagrams are just divergent and have no influence on measurable quantities. E.g., for the Lamb shift (together with the vertex corrections) they are contributing to the energy eigenvalues.

    The best way to see this is to look at the effective quantum action. When evaluating it you'll find the (UV) divergences of various kinds. Renormalization theory shows you that you can reorganize the perturbative series of the quantum action such that everything is expressed with observable finite parameters (normalization of the fields, masses, and coupling constants). The divergent diagrams, however also contain finite non-local pieces that cannot be lumped in to these constants, and they make the observed deviations from what's predicted at tree level, and these deviations in some cases (Lamb shift, anomalous magnetic moment of the electron) are known experimentally and theoretical to 12 or more signifificant digits of accuracy, and these values agree. That's why QED is among the best confirmed models we have today!
     
  20. Sep 8, 2016 #19
    Interesting thread!

    This may be too far off topic - but I've read that Loop Quantum Gravity doesn't require renormalization (basically the fundamental length/time limits mean that the associated integrals don't diverge), that is, avoids all UV divergence issues. In this case, do you end up with Feynman diagrams of the type asimov42 first suggested, with one leg in and one one leg out and yet creation / annihilation operations occurring (with virtual particles) in between. Is there a fundamental difference because the perturbative expansion would only contain a finite number of terms?
     
  21. Sep 8, 2016 #20

    A. Neumaier

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    The standard interpretation of Feynman diagram assumes an underlying relativistic quantum field theory, which always requires renormalization. If one alters the rules of the game by dropping space-time, everything needs reinterpretation, including Feynman diagrams and the particle concept. I don't know much about loop quantum gravity, does it have a good scattering theory with asymptotic particles? Without that virtual particles lose their meaning completely.
     
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