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I've been trying to self-learn how to do basic QFT calculations and I'm a little bit confused as to what's considered "an interaction". If I want to model an electron releasing a photon I can use a bunch of feynman diagrams, some of which will have off-shell particles in them, but in all those diagrams the outgoing photon is always on shell - the off shell terms are integrated over. Similarly, when two electrons scatter there will be some photons on the internal lines of the feynman diagrams that are integrated over, all of which have off-shell internal photons.

Now when an electron from across the room releases a photon that electrons in my eye absorb, is that photon on-shell or off-shell? Why are observed particles on-shell, when particles are always interacting with other particles? Which particles are real outgoing particles of some interaction, and which particles are virtual internal particles to some interaction? I can think of a few answers, but I'm not sure which one is correct. (if any are correct).

(Answer 1): Virtual particles are just mathematical abstractions used to perturbatively calculate the dynamics of real particles, which are a-priori on shell. No particles are virtual, electron-electron scattering is a fundamentally different process than electrons releasing photons and other electrons absorbing those photons because they describe outcomes of different types of experiments. (This answer is what I've seen a few times, including when I've asked my friends who are particle physics grad students)

(Answer 2): This is just the measurement problem - if I try to measure whether an electron releases a photon, then that photon will be real, but if instead I measure the scattering between electrons they'll exchange a bunch of virtual photons. This is weird, but not more weird than other quantum phenomena. (I've not seen this answer anywhere, but it seems like a logical explanation to me?)

(Answer 3): Observing a particle requires that particle to have some spatio-temporal extent which causes the off-shell terms to decohere into something that looks on-shell. So only purely free photons are on-shell, every photon we actually ever observe are necessarily off-shell, since no photons are free. Virtual particles have the same relationship to real particles that 'real waves' have to plane waves. (This is the answer I've seen on most pop-science sites, usually phrased in terms of the uncertainty principle)

(Answer 4): Both real and virtual particles are mathematical abstractions that come from doing perturbative calculations - at the end of the day there is some very complicated algebra of observables and the spacetime-evolution of those observables requires a generalized conservation of energy (like the ward-takahashi identity) via noether's theorem. Virtual particles exist for the same reason that bound particles do - particles are just greene's functions for certain operators and noether's theorem does not necessarily require all particles to be on-shell.

Are any of these explanations outright accepted or dismissed, or is this a contended matter of interpretation, like we have with a lot QM questions? I've seen a lot of discussion of interpretational issues of QM but very little about interpretation in QFT - is there a "standard" interpretation for these matters in the same way we have the copenhagen interpretation? Are there any measurable, physical experiments that give us hints to what's happening here - in the same way quantum decoherence experiments help us refine our intuitions about similar problems there?

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  • $\begingroup$ Part of me wonders if I'm confusing myself by conflating "off shell" and virtual here? Im not sure if these two notions are entirely equivalent since all non-free particles are necessarily off shell? $\endgroup$
    – Opisthokont
    Commented Mar 20 at 11:45
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    $\begingroup$ may be my answer here physics.stackexchange.com/q/185110 would help you understand the difference between observations and predictions by a physics mathematical theory $\endgroup$
    – anna v
    Commented Mar 20 at 13:47
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    $\begingroup$ Virtual by definition means that there is no track. Only in the mathematics as seen in the Feynman diagram given in the reference I gave, the quantum numbers are carried and can be inferred by conservation laws. There are no tracks of virtual particles. $\endgroup$
    – anna v
    Commented Mar 20 at 19:51
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    $\begingroup$ Yes, electron electron scattering happens to first order with a virtual exchange of an off mass shell photon, as in the diagram. An electron can generate a real photon in a brehms strahlung , en.wikipedia.org/wiki/Bremsstrahlung . Then that photon can scatter off another electron , with a different Feynman diagram. $\endgroup$
    – anna v
    Commented Mar 20 at 20:12
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    $\begingroup$ They are not exchanging quantum mechanically, there are two different quantum mechanical scattering interactions. $\endgroup$
    – anna v
    Commented Mar 21 at 4:48

2 Answers 2

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The reason external lines are real particles is because they are defined to be real particles, since they are the observables in the interaction. Each one gets 4-momentum label, and maybe a spin.

Internal lines are off-shell because 4 momentum is conserved at all vertices, so they can't be on-shell...except maybe if it's $g \rightarrow gg$, since massless particles can be on shell with

$$ p^{\mu}_1 = p^{\mu}_2+p^{\mu}_3$$

..but..who thinks about on-shell gluons?

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  • $\begingroup$ Is it correct to think about "Interaction collapse" in this sense? I.e. when particles get close together a bunch of things can happen, but when we measure one thing actually happening the field somehow (atemporally?) "collapses" into a state described by one specific path integral calculatable via a bunch of feynman diagrams? $\endgroup$
    – Opisthokont
    Commented Mar 21 at 2:08
  • $\begingroup$ The definition aspect makes sense, but I guess the thing that slightly bothers me instead of like, directly describing the dynamics of two electrons hitting each other, we instead have to ask "what are all the possible asymptotically free particles two electrons can produce" and calculate the probability of each individual outcome. Why can't we describe the dynamics of what the field does directly like we can with QM? $\endgroup$
    – Opisthokont
    Commented Mar 21 at 2:12
  • $\begingroup$ I understand asking "what actually happened" is a bit of a silly question in the quantum world, but in QM I can at least answer "what did the wavefunction do at all times and places?", is that question also answerable when discussing particle creation and scattering? $\endgroup$
    – Opisthokont
    Commented Mar 21 at 2:14
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    $\begingroup$ QM and QFT are different. Diagrams are for the S-matrix formalism: we have initial and final free particle states, and we calculate the probability to go from initial -> final, which involves a path integral, which is perturbative expanded into terms that look like virtual particles. There really is no quantum woo to interpret. $\endgroup$
    – JEB
    Commented Mar 21 at 3:48
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    $\begingroup$ @Opisthokont idk, my QFT and QCD teachers were Preskill and Feynman, respectively, and what I learned is that I'm no expert, but...the intermediate state is a (coherent) sum of all field configuration that conserve the usual suspects (which includes loops: integrals over $d^4p$), so QED with $\alpha\approx 1/137$ we can expand that in power of alpha and the diagrams look like virtual particles: tree, 1-loop, 2-loop (oh and there's a trick to include disconnected diagrams we forget about)..while in QCD ($\alpha_s \approx 1$) that approach is hopeless. $\endgroup$
    – JEB
    Commented Mar 21 at 14:15
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Well, unless you have "virtual eyes" which can scan false vacuum for fluctuations (joking 😆), your eyes intercepts real photons.

Interacting with virtual particles requires a very special conditions, like in Casimir effect it requires ultra-short distances between plates to feel vacuum pressure OR it requires HEP (High-Energy-Physics) to promote virtual particles into existence in some effect like Hawking_radiation either Unruh_effect.

Virtual particles is NOT something like a "Tales from the Crypt" :-), because as I've pointed out in a couple of effects,- they manifests themselves in quite real ways. Basic point of them is that vacuum is never an empty space, because nature doesn't like "nothing", it always fills nothing with "something".

EDIT

More rigorous explanation why eyes or other EM radiation detectors, can't register virtual photons. Virtual_photon is not a free particle, it's a electromagnetic force career. It exists just as "a glue" or "repellent" between interacting charged particles.

While ordinary light particle,- photon is emitted from some atom and is free to travel until something disturbs it's way. It's not bounded to any system and does not mediate any interaction between particles.

You can also see that difference between real photon Feynman diagram (annihilation for example): enter image description here

And virtual photon which couples pair of scattering electrons :

enter image description here

HTH!

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    $\begingroup$ I'm confused by your answer. Can't all interactions whose time evolution has a pertubative expansion be described as being mediated by virtual particles? $\endgroup$
    – Opisthokont
    Commented Mar 20 at 11:32
  • $\begingroup$ It is interesting though I guess? I usually think of vacuum weirdness here as relating to the the fact that there exist nonequivalent cyclic representations of C* algebras. But does this also mean that "on-shell" ness in general isn't well-defined in relativistic settings the same way stone-von neumann doesn't apply anymore? $\endgroup$
    – Opisthokont
    Commented Mar 20 at 12:05
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    $\begingroup$ All detected photons are real. You can't detect virtual photons by direct means, only indirectly as a manifestation of fact that some charges were pushed or pulled by other charges. So to say virtual photon is electromagnetic force manifestation. $\endgroup$
    – Agnius Vasiliauskas
    Commented Mar 20 at 12:35
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    $\begingroup$ Why isnt that true of real photons as well? I thought we could only detect photons by their electromagnetic interaction as well. I guess one way of saying it is that virtual particles dont leave trails in cloud chambers? But there are always some "real" particles that dont leave trails as well, right? $\endgroup$
    – Opisthokont
    Commented Mar 20 at 18:26
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    $\begingroup$ I've only read a couple books (zee and griffiths) to get a rough idea of what these calculations are. Ive been wanting to read haag - I have math background in operator algebras, but not much physics background so I've just been trying to wrap my head around how physicists practically think about these things which these answers have been very helpful with. My plan now is to read haag and see if the formalism makes things more clear to me before reformulating this question. $\endgroup$
    – Opisthokont
    Commented Mar 21 at 9:04

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