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I'm not new to QFT, yet there are some matters which are quite puzzling to me. I often come across the statement that real particles (the ones we actually measure in experiments, not virtual ones) are "slightly off-shell". What does this actually mean? To my knowledge, something being off-shell means that it violates the relativistic energy-momentum relation. But how can this be possible for particles we actually consider to be "physical"? Please fill me in on the mathematical/experimental backing for such a statement to any degree of detail necessary.

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Please add a reference to the "slightly off-shell" statement. –  BjornW Nov 17 '11 at 0:10
    
To be honest, I haven't seen it anywhere in the literature, which is one reason why I am a little confused, because it goes against what I have learnt so far in my studies of QFT. I have only heard it from people who I thought should have known it better in the context of the discussion of the reality of virtual particles. So is there any truth to it or should I disregard it as nonsense? –  Frederic Brünner Nov 17 '11 at 0:15
    
you mean e.g. electron in hydrogen atom is slightly off-shell because of the binding energy $m\alpha^2$? –  pcr Nov 17 '11 at 0:48
    
Maybe it is an aspect of the problem, but I was more thinking about relativistic particles in colliders. Is there any reason to consider for example an incoming proton before a collision "off-shell"? –  Frederic Brünner Nov 17 '11 at 0:54
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I see. According to the discussion in this thread, "slightly off-shell" is a misleading reference to the "problem" that in reality, one can never measure asymptotic in- and out-states because detectors are not placed infinitely far away from the area of scattering. This answer seems alright, however I'm not satisfied by Susskind quotes. Even if he is right and every particle is on the way from one interaction to another, for all practical purposes one still has to make a distinction between real incoming and outgoing particles and virtual ones(which within the formalism are just a construct). –  Frederic Brünner Nov 17 '11 at 1:25

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See for instance the comments on my answer to Are W & Z bosons virtual or not?.

Basically the claim is that the observed particle represents a path internal to some Feynman diagram and accordingly there is a integral over it's momentum.

I'm not a theorist, but as far as I can tell the claim is supportable in a pedantic way, but not very useful.

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This aspect of quantum field theories also finds expression in the "infraparticle" approach. The basic idea is to discuss the soft photon field and a "bare" particle together, and call it a "dressed" particle. The propagator of a dressed electron can be computed in the infrared to be of the form $\frac{k\cdot\gamma+m}{(k^2-m^2+\mathrm{i}\epsilon)^{1-\alpha/\pi}}$, instead of the bare particle propagator $\frac{k\cdot\gamma+m}{k^2-m^2+\mathrm{i}\epsilon}$. In the ultraviolet, however, at short distances, where perhaps one might want this point of view to hold more than at large distances, this approximation falls apart.

The "slightly off-shell" that you speak of is equivalent to the fact that the effective propagator is not (the Feynman propagator version of) a delta function in momentum space.

A lot of this was worked out in the 50s and 60s, but my understanding is that it has been displaced by the success of the mathematics of the renormalization group. You could look at section II of Thomas Appelquist and J. Carazzone, Phys. Rev. D 11, 2856–2861 (1975), "Infrared singularities and massive fields", which is a brief, quite readable review at that time (which is where I found the propagator above). Still, there are recent references on the Wikipedia page.

I'm a little out of my depth here, but this is as well as I can express it here. Of course, according to Feynman we're all a little out of our depth with QFT.

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Isn't this propagator behavior gauge-dependent? –  Vladimir Kalitvianski 18 hours ago

The reason people say this is because all particles you see are absorbed after a finite time, and the notion of on-shell is asymptotic. The finite time means that they are really internal lines in a diagram, and so ever-so-slightly off shell. The exactly on-shell S-matrix is an asymptotic quantity, relevant only in the holographic limit.

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Isn't that a little weird? We have never observed a "slightly-off shell" particle. Why would the formalism have those particles as the one's that hit the detector, if we know that off-shell particles do not hit the detector? –  kηives Jan 9 '13 at 3:22
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@kηives: You misunderstood--- it's the opposite, we have never observed an exactly on-shell particle, since when they are absorbed by the detector, they are internal, and therefore off-shell by a tiny amount. The reason on-shell is more fundamental is holography and holography alone, all other physics, that is space-time physics, makes off-shell particles fundamental and on-shell stuff just an unnatural limit. This is why S-matrix theory is a weird and important idea, it's holography in proto-form. –  Ron Maimon Jan 9 '13 at 17:16
    
I suppose what I mean by "off-shell" is a violation of $p^2=m^2$. When you say (paraphrase) "all we see are particles that violate this slightly," I don't think that statement is true. Even statistically, if we detected enough particles and the deviation is slight, we should eventually detect a small deviation from $p^2=m^2$, correct? Yet, as far as I know, that has yet to happen. –  kηives Jan 9 '13 at 20:11
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@kηives: It always happens, because of the finite time between emission and absorption. The off-shellness is the degree to which the particle has an uncertain energy in the traditional pre-Feynman formulation, and this is limited to the inverse between the time of production and absorption by the time-energy uncertainty relation. It's completely academic, of course we wouldn't be able to see the off-shellness for macroscopic propagation of a photon from the Andromeda galaxy. But it's in-principle important, because it is why people were wary of on-shell formalisms philosophicaly. –  Ron Maimon Jan 9 '13 at 21:20
    
There remains a problem here, though. If we accept that slightly off-shell particles are all that we detect, we are confronted by the fact that, in perturbation theory at a given order, we may alter the value of an amplitude that corresponds to an off-shell process by effecting a field redefinition. For on-shell processes this isn't possible since the field redefinitions, at on-shell values, by the renormalization conditions. And it doesn't matter if we're only off-shell by an infinitesimal amount. Field redefinitions can be effected with arbitrarily large effect on the off-shell amplitude. –  MarkWayne Mar 29 '13 at 6:39

For a real particle to be off-shell, it is sufficient to be in an external field of some sort. It is not necessary to be "absorbed". "Absorbed" are virtual particles, for example, "virtual photons" who describe non propagating fields like a Coulomb one.

For a Compton scattering, the real electron is "off-shell" during interaction with the real photon, i.e., while being in its field.

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