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This question already has an answer here:

I have recently been reading about Quantum Electrodynamics which I found very interesting, but even more confusing.

I understand photons mediate the electromagnetic force and interactions between charges can be described by virtual photon exchange. And that in Quantum Mechanics particles paths can be evaluated by sum over all paths...

An electron travelling from one point to another will take all paths, emitting and reabsorbing virtual photons. Does the electrons take all paths including those that involve faster than light velocities?

Is this question null, Does it mean anything to ask about properties of virtual particles?

Do I have the wrong idea about what virtual particles are?

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marked as duplicate by ACuriousMind, Jim, Kyle Kanos, DavePhD, John Rennie Nov 20 '14 at 15:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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The term virtual is used in other places in physics. For example in virtual images in a a mirror : we see an object in great verisimilitude, even ourselves. Why is the image called virtual and not real? Because it has the optical properties of the imaged object but not a large number of other attributes, mass being the simplest. In addition, its existence is not independent of the object, a prime attribute of reality macroscopically.

The term virtual coupled to the term particle comes from the shorthand of Feynman diagrams. Before Feynman invented his diagrams the study of elementary particle interactions was very complicated, involving convoluted integrals in series expansions .

electron scattering

electron electron scattering, time upward in y.

Feynman diagrams allowed to order the perturbation terms according to complexity, which led to seeing easily the more dominant parts of the expansion in series, ordered in "orders". This is a first order diagram and the calculation has within the integral a pole related to an exchange of a "particle" that has all the attributes of a photon, except its mass. It is off mass shell, i.e. the four vector that transfers energy and momentum from one electron to the other looks like a photon, the way the optical image of you looks like you, but is not an on mass shell photon, so it is called virtual.

Feynman diagrams can be very complicated, like these that allow the calculation of top production:

top production

top production one of diagrams.

In this diagram for the calculation of the crossection for top production real , i.e. on mass shell, are the incoming on the left and the outgoing on the right particles. All the rest are virtual. Note the exchange of W which has a mass close to 100GeV on shell, it can have any mass consistent to the maths of the diagram, but it carries the quantum numbers of a W boson.

An electron travelling from one point to another will take all paths, emitting and reabsorbing virtual photons. Does the electrons take all paths including those that involve faster than light velocities?

Here you are mixing the Feynman diagram approach of calculating interaction crossections with the path integral approach where the term "virtual particle" has little meaning. In this calculational approach this article might help which explains simply about fields and the real particles being excitations of fields.

In the Feynman diagram approach one needs an interaction, the electron running along is not interacting with anything that can be framed as a Feynman diagram, therefore it has no meaning to say it is emitting and absorbing virtual photons let alone their velocity.

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    $\begingroup$ To Anna: I am also interested in this question. I read your explanation which is very illuminating, but I don't see the answer to the part: virtual photons run faster-than-light? So, have you an answer to this part? $\endgroup$ – Sofia Nov 20 '14 at 14:45
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    $\begingroup$ @Sofia To have a virtual particle you need a Feynman diagram. Feynman diagrams are within the dimensions of the Heisenberg uncertainty principle so no definition of velocity can have meaning in the sense of motion from point a to point b at a time interval t. Now theoretically one could say that since the electromagnetic potential falls as 1/r , i.e. not zero, an electron on the earth could scatter with an electron on the moon with a very-very small measurable crossection and use the Feynman diagram representation. Can this happen with a velocity larger than c? the electron on the moon move $\endgroup$ – anna v Nov 20 '14 at 15:24
  • $\begingroup$ faster than a laser beam started at the same time? ( let us suppose the electron is on an earth satellite in vacuum, and the other one on a moon peak in vacuum) My intuition says no, otherwise we would have action at a distance, but my mathematical abilities do not stretch to prove my statement. I think the link I gave withe the fields explanation deals with this. $\endgroup$ – anna v Nov 20 '14 at 15:27
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Yes, you are right. The four momentum of a virtual photon needn't to lie on the mass shell. Thus the zeroth component of the four momentum of a virtual photon is independent of its spatial components. The reason for this is that the zeroth component of the four momentum of a virtual photon arises from the Fourier transform of the step function. See S. Weinberg, The quantum theory of fields, vol 1, section 6.2 for detail.

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