Why EM wave generated by wiggling charge is real, not a virtual photon?

Question

In electrostatics, charged particles attract or repel via exchange of virtual photons. But when charges wiggle (or more generally accelerate), they generate electromagnetic waves which are often called "real" photons. For such a photon to be a real particle, its dispersion relation must sit on-shell. This requires the precise measurement of its momentum and energy. However, uncertainty principle says we can't get precise measurement unless the field is indefinite in time and space.

In my head, a massless spin-1 photon field exists indefinitely without any beginning or end. On the other hand, the EM field generated by a wiggling charge has a beginning so that it can't be infinite in time and space. So, why is this EM field called a "real" photon? Or is it an asymptotically real photon but indeed virtual? Or am I misunderstanding something?

I read the Wikipedia article about virtual particle. According to the article, the Casimir effect is due to virtual photons interacting with metal plates. I don't understand why the photon here is virtual. Can anybody explain it, too?

For more clarification, the notion of real particle conflicts in my head. I think it is on-shell particle and it has either a beginning or end or none of these in a Feynman diagram. However these two concepts are conflicting because for the particle newly annihilated or created, the energy and momentum relation must remain vague due to uncertainty principle. I can say a particle or field is on-shell only when it exists indefinitely like a planewave. Can anybody resolve it?

To me it honestly seems like you are mixing QM, Classical Electrodynamics and QFT. Could you edit your question and make it clear in which framework this discussion should be? — Sanya, Oct 29 '16 at 15:31
Possible duplicate of Virtual photons, what makes them virtual? — Robin Ekman, Oct 29 '16 at 16:00
@Sanya Thank you for the comment. I think this question is in QFT regime, but I'm afraid I don't know how to bring classical EM wave to this regime and I think the explanation of a physical phenomenon can be present regardless of framework. — Liberty, Oct 29 '16 at 16:10
@ACuriousMind Thank you for comment. Yes, but I remember the outgoing particles/ fields are onshell when we treat them as momentum eigenstate at asymptotic infinity.... However when I know the history and onset of EM wave generated from wiggling charge, is it the same story? If there is uncertainty both in energy and momentum, How can I say its dispersion relation...? With the aid of the last answer, I concluded that the error is negligible to use in dispersion relation. — Liberty, Oct 31 '16 at 01:15

score 0 · Accepted Answer · edited Apr 13 '17 at 12:40

To start with there is no Heisenberg uncertainty between energy and momentum; the HUP is between momentum and position, there do exist additional uncertainty relations , but not between energy and momentum. The mass is a fixed real number , the "length" of the four vector in special relativity. So particles which are on mass shell are called real, because they do have an energy and momentum constraining the four vector to the mass of a given particle.

Secondly virtual particles are in the realm of quantum mechanics , quantum field theory and the attendant Feynman diagrams. Please see my answer here on what a virtual particle is.

In electrostatics, charged particles attract or repell via exhange of virtual photon.

That is not correct. Electrostatics emerges from the underlying quantum mechanical level. The interactions between atoms and molecules constituting the macroscopic surfaces studied can be described quantum mechanically with virtual photon exchanges, but the build up to the classical static field is not as simple addition.

You may get an idea of the complexity by reading this blog, of how classical electromagnetic waves emerge from the quantum substrate. ( My answer here might help in visualizing how photons build up a classical wave). The "wiggling charge" is releasing real photons, shich can be measured one photon at a time.

So, how this EM field is called "real" photon? Or is it asymptotically real photon but indeed virtual?

The EM field as a classical light wave is not called a "photon". The field theoretical photon field, which fills all space and is acted upon by creation and annihilation operators to model the motion of a photon is the one you are confusing with the EM field.

From the linked blog entry one can understand the complexity of going from the atoms and molecules ( 10^23 per mole) to a charged surface, it is not a simple addition of virtual photons that will give the coulomb field macroscopically.

I appreciate your answer. However, what I intended for uncertainty in energy and momentum is not these two's but energy-time and momentum-position separately. I think we need sufficient time to determine energy exactly and enough space for the precise momentum. Since the newly generated field or particle lives in finite time and localized in space, I kind of felt vague saying they are onshell. I read the blogs and pages you linked which took so long time but really helpful. However I really didn't get the idea that coulomb interaction is not the exchange of virtual particle. — Liberty, Oct 30 '16 at 16:18
Indeed, I'm not talking about the photon field which is object of qft, but classical EM wave, a coherent state of many photons according to the blog you linked. Some articles say the EM wave is real photon. But I think you don't agree with it. Can I get the correct notion? I'm sorry for asking too much. Again I appreciate your clear answer. — Liberty, Oct 30 '16 at 16:19
Yes , the electromagnetic wave is made up of real photons as the double slit experiment with single photons at a time demonstrates clearly ( last link). The classical interference is built up by real photons. The coulomb interaction is built up by a lot of virtual partilce exchanges. Two different mechanisms — anna v, Oct 30 '16 at 16:23
The single photon at a time double slit fulfills the momentum-(x,y) HUP because the size of the points is microns and if you put in numbers the HUP is satisfied. The same is true for the energy time, because our delta(t) is large enough together with the delta(E) to be ok with the HUP. The mass is fixed by the four vector"length" — anna v, Oct 30 '16 at 16:25
Thank you. Last comment helped my understanding and I think I can now legally say these are onshell and real. — Liberty, Oct 30 '16 at 16:30

score -1 · Answer 2 · answered Oct 29 '16 at 15:42

-1

The total EMF contains two addenda (see the solution in a textbook): a "near field" (like a Coulomb one) and a "radiated field". Their behavior, as a function of the distance, differs. The "near field" does not propagate even though it is wiggling too. In terms of its total flux, it decays with distance.

The radiated field total flux does not decay. Far away from the source the radiated field can be represented as a superposition of plane waves, which are associated with real photons. The near field is then called "virtual photons".

answered Oct 29 '16 at 15:42

Vladimir Kalitvianski

13,885

No. The constituent plane waves are some idealization. In fact, a photon has some $10^4$ or more wavelengths. It is already sufficient to speak of a certain frequency. – Vladimir Kalitvianski Oct 29 '16 at 16:52
What do you mean "a photon has some 10$^4$ or more wavelengths"? Does that number represent anything real? There is clearly some uncertainty in the wavelength of a photon, but how does that mean it has more than one? – Peter Shor Oct 30 '16 at 17:26
You have to find out what the real photon is. It is a low intensity and high number of wavelength thing. In other words, it is of finite distance in space. – Vladimir Kalitvianski Oct 30 '16 at 18:52
A photon is an elementary particle. Photons of the same position and momentum are indistinguishable; photons simply don't have enough parameters to carry 10$^4$ wavelengths. – Peter Shor Oct 30 '16 at 18:59
@PeterShor: Certain frequency implies an infinite number of the wavelengths. In practice there happen uncertainty in emitting and absorbing a photon due to recoil, etc.; thus finite (but large) number of the wavelengths. – Vladimir Kalitvianski Nov 01 '16 at 07:27

Why EM wave generated by wiggling charge is real, not a virtual photon?

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