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Are virtual particles really there?

  1. May 11, 2005 #1
    As we know, when we do calculations in QFT, we write down the ampitude, and find that it can be explaind as Feynman diagrams, which is easier to work out. Then we use Feynman diagrams as a tool.

    Now the question is, is the propagator really exists in the small distance, i.e. the Feynman diagrams really represent the physical process in the small distance, or it is only a mathematical tool, while no virtual actually formed, or they really formed, but different from the Feynman diagrams we see?

    Perhaps as we mainly test the QFT by scattering, it's difficult for us to know what is really happening inside the reaction, but i am curious :)

  2. jcsd
  3. May 12, 2005 #2
    https://www.physicsforums.com/journal.php?s=&action=view&journalid=13790&perpage=10&page=3 [Broken]

    Scroll down to the "what are virtual particles" entry

    You should be able to find out there are TWO distinct ways in which virtual particles are created. After reading the text, can you distinguish between those two ?

    Last edited by a moderator: May 2, 2017
  4. May 12, 2005 #3


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    Yes,it's true.Feynman diagrams involve 2 types of virtual particles.Indeed,we're only considering scattering states (in and out),so "what happens inbetween" is of no interest to experimentalists.

  5. May 12, 2005 #4


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    I think it is a matter of taste, but I'd be inclined to call it a mathematical tool which has some physical suggestive value :shy:

    The reason is that Feynman diagrams are a tool to do a calculation which is a series devellopment. Now, if somehow we'd know how NOT to go through that series devellopment, but calculate the correlation functions of the interacting theory directly, suddenly there wouldn't be any Feynman diagrams anymore. So what's the physics behind it then, if their existance depends on your way of solving a mathematical problem ?
    That said, Feynman diagrams (especially in the electroweak sector) are highly suggestive, and do give you some sort of physical feeling of what's going on. But that's probably true because of the smallness of the coupling constant, which means that tree diagrams already contain the bulk of the correct answer. The mess you get in QCD is probably more indicative of the limitedness of feynman diagrams as "true physics going on", and not as a memory-aid for the terms in a series devellopment leading you to a solution.

  6. Nov 28, 2010 #5
    Very old post, but I like it, it is formulated very clearly.

    My question here is then, when pertubation theory, with its Feynman diagrams and its propagators/ virtual particles are just a tools, just a calculation schemes, that sould not be equated with physical reality, what then?

    When two electrons scatter, how do they interact? Not by virtual photons, they do not exist, they are just tools from pertubation theory. But by what then?
  7. Nov 28, 2010 #6

    Vanadium 50

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    The electromagnetic field.
  8. Nov 28, 2010 #7
    Is not a quantized electromagnetic field made of photons??? Even if the number is completely uncertain for a Coloumb field, we have a quantised field here, don't we? So how does it transmit momentum between the two charges?
    Last edited: Nov 28, 2010
  9. Nov 28, 2010 #8
    Found a beautiful post of selfadjoint on PF about virtual particles.

  10. Nov 29, 2010 #9
    So what now? Are virtual particles really there?

    If not, what else to explain momentum transfer between two electrons? The classical electromagnetic field with forces at a distance as Vanadium 50 suggests? I don't think so.

    By the way the question how forces are transmitted between two particles is not some random question, but one of the most important question in quantum field theory.
  11. Nov 29, 2010 #10
    Well, that's not quite on the money. I'm sure experimentalists would be very interested, but by definition, virtual particles can only exist for extremely short periods of time, as determined by the uncertainty principle, and are thus untestable.

    However, technically, the particles given off as Hawking radiation from Black Holes would have started off as virtual particles. Not sure if they count.
  12. Nov 29, 2010 #11
    Right, they are untestable. But also crucial for QFT to work! All the spectular QFT calculation depend on them. That's why I find it kurious that on this forum (as I read in the archive on some earlier threads) they often have been denied any reality or are called even silly.

    Even if non-pertubative quantum field theory would work for, say, electromagnetic interactions, and there where no need for Feynman diagrams, the question remains how is momentum transfered betwenn two static charges if not by processes that violate the dispersion relation but at the same time can't be detected due to the time-energy uncertainty relation.
  13. Nov 29, 2010 #12
    No, photons are DISTURBANCES in the field.
  14. Nov 29, 2010 #13
    The classical Coulomb field deviates from the classical 1/r2 radial dependence at distances less than about 1 electron Compton wavelength because of virtual charged particles in the lowest-order Feynman diagram (vacuum polarization) in Coulomb scattering off the nucleus. The vacuum polarization effect (virtual electron and positron) shifts the atomic energy levels of negative muons in muonic atoms as much as ≈ 1%. Muonic atom transition energies have been measured and compared to theory with very high precision.

    Bob S
  15. Nov 29, 2010 #14
    If they are 'really there' and Hawking radiation is somehow proven would this not violate some fundamental rule that you cannot create something out of nothing?

    I never liked the idea of Hawking radiation - its like creating a perpetual motion machine. Black holes would be constantly adding mass to the universe out of virtual particles. Unless the particles come from some pool/aether or some unknown available source.

    So if virtual particles are detected someday the energy has to come from somewhere. In they are only useful functional figments of QM - thats cool too.

    I may be over my head in these comments!
  16. Nov 30, 2010 #15
    I have a similar viewpoint as vanesch. I don't think that it makes sense to talk about how 'real' a virtual photon is.

    I personally think that although QFT techniques such as Feynman diagrams allow us to play with the perturbation theory more systematically, the old QM time-independent perturbation theory sometimes can give us more insight. After all, QFT is just a special kind of QM.

    For example, in QED, we have three terms in the Lagrangian (or Hamiltonian), which are (Free bare electron term) + (Free bare photon term) + (Coupling term). Here, I want to describe what a physical electron is, in terms of bare electrons and bare photons, at the level of QM time-independent perturbation theory.

    At the zeroth order in the coupling, a physical electron is simply a bare electron. The coupling term allows it to emit a bare photon and change its momentum. If this process were energy conserving (in terms of bare quantities!), it would have an actual transition amplitude, but it isn't. Now, those energy non-conserving photon emission processes(or off-shell processes, if we move to frequency domain, as is done when considering Feynman diagrams) give the first order correction to our bare electron state. That is, up to the first order in the coupling, the following statement holds.

    (quantum state corresponding to one physical electron) =
    (quantum state corresponding to one bare electron) + \sum (small coefficient)*(quantum state corresponding to one bare electron + one bare photon),

    where all the states in the sum should have the same total momentum as the 0th order bare electron, and the small coefficients are given by the time-independent perturbation theory (those involving energy denominators).

    What we call a 'virtual photon' is basically a bare photon state that appears in correction terms on the bare electron state.

    Coulomb interaction can also be explained in this language. At the lowest order, it originates from the coupling between the zeroth order part of a physical electron (which is just a bare electron) with the first order part of another physical electron. (which consists of a bare electron plus a bare photon, and this photon part is responsible for the coupling).

    In short, I think we shouldn't take some expressions like 'emitting a virtual photon' too seriously. It just means bare photon states that appears in the perturbation series describing a physical electron state.
    Last edited: Nov 30, 2010
  17. Nov 30, 2010 #16
    As long as the ejected particle carries away mass and energy, which in turn causes a like reduction in black hole's mass... where's the "perpetual"? HR shouldn't be confused with vacuum fluctuations, or other examples... it's really very odd and abstract in the math.
  18. Dec 1, 2010 #17
    Everbody who doubts the necessity of virtual paticles to explain nature, I recommend reading Anthony Zee's Quantum field in a nutshell, especially part 1 of the book. On page 27 he says for example, "that the exchange of a particle can produce a force was one of the most profund conceptual advances in physics."
    Last edited: Dec 1, 2010
  19. Dec 1, 2010 #18

    Vanadium 50

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    But Tony Zee is well aware of the S-matrix formalization.
  20. Dec 1, 2010 #19
    He proabably also knows, that in the real world we're not actually looking at asymptotic states that scatter all the way to infinity, and thus any particle you approximate as on-shell when computing S-matrices can be thought of as very slightly off-shell.
  21. Dec 1, 2010 #20
    Right... we're NOT actually looking at that (probably), and thus when you remove the math that uses them, you lose the virtual particles; they have no existence off-paper, forget "slightly off-shell".
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