# “Slightly off-shell”?

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
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

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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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