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W and Z bosons are observed/discovered. But as force carrying bosons they should be virtual particles, unobservable? And also they require to have mass, but if they are virtual they may be off-shell, so are they virtual or not.

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If you were in a system above the electroweak temperature you would be surrounded by a sea of very real W and Z bosons. –  user346 Feb 1 '11 at 6:11
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Really? They would become stable? –  Vladimir Kalitvianski Feb 1 '11 at 11:20
    
They have been observed in particle accelerators, therefore they can certainly be real. –  Noldorin Feb 1 '11 at 20:13
    
@Noldorin: You might want to be careful in connecting "observability" to "real" in this context. We reconstruct on-shell weak bosons by their decay products, which is also how we reconstruct off-shell weak bosons. –  dmckee Feb 1 '11 at 22:50
    
@dmckee: but there would be an energy level where the decay process would essentially be reversible, right? At that point, seeing the Z would be as probable as seeing the electron/position pair that it would decay to. –  Jerry Schirmer Feb 1 '11 at 23:59

5 Answers 5

up vote 6 down vote accepted

Seems to me there is a confusion between various concepts, let me try to clear it up:

  1. Virtual particle is one that doesn't live forever, at some stage it gets converted to something else. As Jeff points out, none of us lives long enough to tell the difference, so the distinction between virtual and non-virtual is a matter of degree. Particles that live for a long time are declared "real", and particles that decay quickly are called "virtual". These are just names, there is no implication that "virtual" particles don't really exist, like white unicorns and other mythical creatures, those are all real measurable effects you can see with your own eyes...

  2. Any particle can be either real or virtual, whether or not it is massive, whether or not it is bosonic force carrier, or fermionic matter. There is a sense in which massive particles tend to live shorter life (because they have more opportunities to decay), but this is just a rule of thumb.

  3. Off-shell can be taken here to be synonymous with "virtual".

Hope that helps.

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Thank you Moshe. –  pho Feb 2 '11 at 2:14
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You confuse virtual particles and unstable particles. I just wrote my own answer to the question that explains details. –  Arnold Neumaier Mar 7 '12 at 19:08
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This answer is wrong! "Virtual" particles has nothing to do with decaying quickly. There are virtual electrons and virtual photons! –  whistles Mar 8 '12 at 11:09
    
He didn't say that it had to decay - it could also annihilate or something. –  gn0m0n Jun 28 at 8:53

Photons are force carrying bosons and come in both virtual and real varieties. There is nothing wrong with that.

Virtual means off-shell, and real means on-shell.

Even on-shell weak bosons decay very quickly, however, because there are plenty of modes with the right quantum numbers and much lower total mass (and thus lots of phase space).


I want to emphasize that on- versus off-shell is a concept that applies to unstable particles, event those that are too highly unstable to be experimentally isolated.

The lifetime of the $\mathrm{Z}^0$ boson is measured by observing the with of it's mass distribution in high energy experiment such as those conducted at the Tevatron and the LHC. That mass is found to be approximately 91 GeV with a width of about 2.5 GeV corresponding to a lifetime on order of $10^{-24}$ seconds.

These Z's come from a variety of processes such as tau decay and are on-shell (but obviously very unstable).

Now we travel to the nuclear energy regime and measure the asymmetry of polarized electron scattering off of a proton target at experiments like JLAB's $G^0$. The asymmetry can be attributed to $e + p \overset{Z}{\to} e' + p'$ (almost all of the cross-section comes from $e + p \overset{\gamma}{\to} e' + p'$, but that is all symmetric). These experiments were carried out at squared momentum transfers in the neighborhood of 1 GeV. That is that the asymmetry come from Z scattering events where $|p_Z|^2 \approx 1\text{ GeV}^2$, so these Z's are very much off-shell.

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Every photon you see is virtual. Just draw the Feynman diagram starting with emission of the photon from some atomic system and ending with the absorption by your retina. Of course some photons are closer to being on-shell than others. –  pho Feb 1 '11 at 14:15
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I disagree. Photons I see are free, real ones. They propagate. The "virtual" particles do not propagate. Even the short-living resonances are more real than the "virtual" particles. –  Vladimir Kalitvianski Feb 1 '11 at 16:06
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@Noldorin: Virtual photons are the same as off-shell photons. They have four momentum $q$ with $q^2 \ne 0$. Real photons have $q^2=0$. These are the usual physics definitions. Wikipedia says "virtual particle is a particle that exists for a limited time and space." If a photon was emitted by an atomic system and then absorbed by your eye it existed for a limited time and space. You can also check that $q^2$ is not exactly zero. It is therefore virtual. Photons you see are of course "real" in the vernacular sense, but I was using a precise definition of real and virtual as described above. –  pho Feb 1 '11 at 21:50
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@Noldorin: Exactly how do you know my statement is wrong? Perhaps some things you think you know are not actually correct. The fact that you have never heard the term "off-shell" might be a clue. I am making a subtle point, and that is that "real" particles as they are usually defined are an idealization. –  pho Feb 1 '11 at 22:12
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@dmckee: I agree, but I think my pedantry does have some point in that it is clear that some people think of virtual particles as mysterious things which are somehow completely different from the "real" particles they observe while in fact it is all a matter of degree and using various experiments we observe particles which are off-shell to various degrees. Anyway, the horse is dead and I will stop beating it. –  pho Feb 1 '11 at 23:13

They can be real, no problem with that. However, all Z and W observed are virtual. And yes, they are off-shell, whatever that means to you. We actually measure their width, for instance, http://arxiv.org/pdf/0909.4814 (it's more or less 2 GeV around a mass of more or less 80-90 GeV).

What is observed is a pole in the S matrix for some final states in particle collisions. So, really, what is measured is a certain differential cross section, to which is fit a distribution with this pole structure. From the values, both the mass and the width are read out.

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You confuse virtual particles and unstable particles. I'll write in a moment my own answer to the question that explains details. –  Arnold Neumaier Mar 7 '12 at 18:20

All observed particles are real particles in the sense that, unlike virtual particles, their properties are verifiable by experiment. In particular, W and Z bosons are real but unstable particles at energies above the energy equivalent of their rest mass. They also arise as unobservable virtual particles in scattering processing exchanging a W or Z boson, though the existence of a corresponding exchange diagram is visible experimentally as a resonance.

Virtual particles and unstable (i.e., short living) particles are conceptually very different entities. Since there seems to be a widespread confusion about the meaning of the terms (and since Wikipedia is quite unreliable in this respect) let me give precise definitions of some terms:

A stable, observable (and hence real in the sense specified above) particle has a real mass $m$ and a real 4-momentum $p$ satisfying $p^2=m^2$; one also says that it is on-shell. For such particles one can compute S-matrix elements, and according to quantum field theory, only for such particles. In perturbative calculations, stable particles correspond precisely to the external lines of the Feynman diagrams on which perturbation theory is based. Only a few elementary particles are stable, and hence can be associated with such external lines. (However, in subtheories of the standard model that ignore some interactions, particles unstable in Nature can be stable; thus the notion is a bit context dependent.)

A virtual particle has real momentum with $p^2\ne m^2$ (one also says that they are off-shell), and cannot exist as it would violate energy conservation. In perturbative calculations, virtual particles correspond precisely to the internal lines of the Feynman diagrams on which perturbation theory is based, and are only visual mnemonic for integrations over 4-momenta not restricted to the mass shell. In nonperturbative methods for calculating properties of particles, there is no notion of virtual particles; they are an artifact of perturbation theory.

Virtual particles are never observable. They have no properties to which one could assign in any formally meaningful way a dynamics, and hence some sort of existence in time. In particular, it is meaningless to think of them as short-living objects. (Saying they pop in and out of existence for a time allowed by the uncertainty pronciple has no basis in any dynamical sense - it is pure speculation based on illustrations for the uneducated public, and from a widespread misunderstanding that internal lines in Feynman diagrams describe particle trajectories in space-time).

All elementary particles may appear as internal lines in perturbative calculations, and hence possess a virtual version. For a more thorough discussion of virtual particles, see Chapter A8: Virtual particles and vacuum fluctuations of my theoretical physics FAQ.

An unstable observable (and hence real in the sense specified above) particle has a complex mass $m$ and a complex 4-momentum $p$ satisfying $p^2=m^2$. (One shouldn't use the term on-shell or off-shell in this case as it becomes ambiguous). The imaginary part of the mass is relatied to the halflife of the particle. At energies below the energy $E= Re\ mc^2$, unstable elementary particles are observable as resonances in cross sections of scattering processes involving their exchange as virtual particle, while at higher energies, they are observable as particle tracks (if charged) or as gaps in particle tracks; in the latter case identifiable by the tracks of their charged products.

For unstable particles one can compute S-matrix elements only in approximate theories where the particle is treated as stable, or by analytic continuation of the standard formulas for stable particles to complex energies and momenta.

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Your distinction between a "virtual particle" ($p^2 \ne m^2$ for real $m$ and $p$) and an "unstable observable" ($p^2 = m^2$, but $m$ complex) seems to be without a difference. And where does a free neutron fit into this picture? It is unstable, but it is clearly real and has a very precisely measurable mass (and the proton even more so if it is in fact unstable). –  dmckee Mar 7 '12 at 20:05
    
There is no difference between a real and a complex mass?? - A neutron is an unstable, nonelementary particle. As every unstable particle it is a real particle, consistent with what I wrote. Its mass is almost real, as it is quite long-living, but has a slight imaginary part. en.wikipedia.org/wiki/Particle_decay –  Arnold Neumaier Mar 7 '12 at 20:18
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You claim there are two distinct categories here, but their experimental signature is the same (they decay in time given by Heisenberg and conserve $E$ and $p$). How do I know which category a particle belongs to? Do it become virtual when it's lifetime is below $10^{-5}$ s, or is the muon real? How about $10^{-10}$ s? That would make the $K^0$ real in the long form but virtual in the short; $10^{-12}$ s makes the tau virtual. But it gets worse...the top quark's lifetime is comparable to that of the weak bosons. Is it the only unreal quark? –  dmckee Mar 7 '12 at 21:38
    
In other words the "complex" mass bit looks like a bookkeeping trick. You can define what ever you want, but you have to show me a different experimental behavior. –  dmckee Mar 7 '12 at 21:41
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Please give me a definition of what it means that a virtual particle has a life-time of $10^{-12}s$. It cannot be defined consistent with the usual definition of a virtual particle as an internal line of a Feynman diagram. Times can be associated meaningfully only to objects that have a state, so that one can form probabilities, and virtual particles lack such a state. –  Arnold Neumaier Mar 8 '12 at 8:03

Let me give a second, more technical answer.


Observable particles. In QFT, observable (hence real) particles of mass $m$ are conventionally defined as being associated with poles of the S-matrix at energy $E=mc^2$ in the rest frame of the system (Peskin/Schroeder, An introduction to QFT, p.236). If the pole is at a real energy, the mass is real and the particle is stable; if the pole is at a complex energy (in the analytic continuation of the S-matrix to the second sheet), the mass is complex and the particle is unstable. At energies larger than the real part of the mass, the imaginary part determines its decay rate and hence its lifetime (Peskin/Schroeder, p.237); at smaller energies, the unstable particle cannot form for lack of energy, but the existence of the pole is revealed by a Breit-Wigner resonance in certain cross sections. From its position and width, one can estimate the mass and the lifetime of such a particle before it has ever been observed. Indeed, many particles listed in the tables http://pdg.lbl.gov/2011/reviews/contents_sports.html by the Particle Data Group (PDG) are only resonances.


Stable and unstable particles. A stable particle can be created and annihilated, as there are associated creation and annihilation operators that add or remove particles to the state. According to the QFT formalism, these particles must be on-shell. This means that their momentum $p$ is related to the real rest mass $m$ by the relation $p^2=m^2$.
More precisely, it means that the 4-dimensional Fourier transform of the time-dependent single-particle wave function associated with it has a support that satisfies the on-shell relation $p^2=m^2$. There is no need for this wave function to be a plane wave, though these are taken as the basis functions between the scattering matrix elements are taken.

An unstable particle is represented quantitatively by a so-called Gamov state (see, e.g., http://arxiv.org/pdf/quant-ph/0201091.pdf), also called a Siegert state (see, e.g., http://www.cchem.berkeley.edu/millergrp/pdf/235.pdf) in a complex deformation of the Hilbert space of a QFT, obtained by analytic continuation of the formulas for stable particles. In this case, as $m$ is complex, the mass shell consists of all complex momentum vectors $p$ with $p^2=m^2$ and $v=p/m$ real, and states are composed exclusively of such momentaum vectors. This is the representation in which one can take the limit of zero decay, in which the particle becomes stable (such as the neutron in the limit of negligible electromagnetic interaction), and hence the representation appropriate in the regime where the unstable particle can be observed (i.e., resolved in time).

A second representation in terms of normalizable states of real mass is given by a superposition of scattering states of their decay products, involving all energies in the range of the Breit-Wigner resonance. In this standard Hilbert space representation, the unstable particle is never formed; so this is the representation appropriate in the regime where the unstable particle reveals itself only as a resonance.

The 2010 PDG description of the Z boson, http://pdg.lbl.gov/2011/reviews/rpp2011-rev-z-boson.pdf discusses both descriptions in quantitative detail (p.2: Breit-Wigner approach; p.4: S-matrix approach).

(added March 18, 2012): All observable particles are on-shell, though the mass shell is real only for stable particles.


Virtual (or off-shell) particles. On the other hand, virtual particles are defined as internal lines in a Feynman diagram (Peskin/Schroeder, p.5, or Zeidler, QFT I Basics in mathematics and physiics, p.844). and this is their only mode of being. In diagram-free approaches to QFT such as lattice gauge theory, it is even impossible to make sense of the notion of a virtual particle. Even in orthodox QFT one can dispense completely with the notion of a virtual particle, as Vol. 1 of the QFT book of Weinberg demonstrates. He represents the full empirical content of QFT, carefully avoiding mentioning the notion of virtual particles.

As virtual particles have real mass but off-shell momenta, and multiparticle states are always composed of on-shell particles only, it is impossible to represent a virtual particle by means of states. States involving virtual particles cannot be created for lack of corresponding creation operators in the theory.

A description of decay requires an associated S-matrix, but the in- and out- states of the S-matrix formalism are composed of on-shell states only, not involving any virtual particle. (Indeed, this is the reason for the name ''virtual''.)

For lack of a state, virtual particles cannot have any of the usual physical characteristics such as dynamics, detection probabilities, or decay channels. How then can one talk about their probability of decay, their life-time, their creation, or their decay? One cannot, except figuratively!


Virtual states. (added on March 19, 2012): In nonrelativistic scattering theory, one also meets the concept of virtual states, denoting states of real particles on the second sheet of the analytic continuation, having a well-defined but purely inmaginary energy, defined as a pole of the S-matrix. See, e.g., Thirring, A course in Mathematical Physics, Vol 3, (3.6.11).

The term virtual state is used with a different meaning in virtual state spectroscopy (see, e.g., http://people.bu.edu/teich/pdfs/PRL-80-3483-1998.pdf), and denotes there an unstable energy level above the dissociation threshold. This is equivalent with the concept of a resonance.

Virtual states have nothing to do with virtual particles, which have real energies but no associated states, though sometimes the name ''virtual state'' is associated to them. See, e.g., https://researchspace.auckland.ac.nz/bitstream/handle/2292/433/02whole.pdf; the author of this thesis explains on p.20 why this is a misleading terminology, but still occasionally uses this terminology in his work.


Why are virtual particles often confused with unstable particles? As we have seen, unstable particles and resonances are observable and can be characterized quantitatively in terms of states. On the other hand, virtual particles lack a state and hence have no meaningful physical properties.

This raises the question why virtual particles are often confused with unstable particles, or even identified.

The reason, I believe, is that in many cases, the dominant contribution to a scattering cross section exhibiting a resonance comes from the exchange of a corresponding virtual particle in a Feynman diagram suggestive of a collection of world lines describing particle creation and annihilation. (Examples can be seen on the Wikipedia page for W and Z bosons, http://en.wikipedia.org/wiki/Z-boson.)

This space-time interpretation of Feynman diagrams is very tempting graphically, and contributes to the popularity of Feynman diagrams both among researchers and especially laypeople, though some authors - notably Weinberg in his QFT book - deliberately resist this temptation.

However, this interpretation has no physical basis. Indeed, a single Feynman diagram usually gives an infinite (and hence physically meaningless) contribution to the scattering cross section. The finite, renormalized values of the cross section are obtained only by summing infinitely many such diagrams. This shows that a Feynman diagram represents just some term in a perturbation calculation, and not a process happening in space-time. Therefore one cannot assign physical meaning to a single diagram but at best to a collection of infinitely many diagrams.


The true meaning of virtual particles. For anyone still tempted to associate a physical meaning to virtual particles as a specific quantum phenomenon, let me note that Feynman-type diagrams arise in any perturbative treatment of statistical multiparticle properties, even classically, as any textbook of statistical mechanics witnesses.

More specifically, the paper http://homepages.physik.uni-muenchen.de/~helling/classical_fields.pdf shows that the perturbation theory for any classical field theory leads to an expansion into Feynman diagrams very similar to those for quantum field theories, except that only tree diagrams occur. If the picture of virtual particles derived from Feynman diagrams had any intrinsic validity, one should conclude that associated to every classical field there are classical virtual particles behaving just like their quantum analogues, except that (due to the lack of loop diagrams) there are no virtual creation/annihilation patterns. But in the literature, one can find not the slightest trace of a suggestion that classical field theory is sensibly interpreted in terms of virtual particles.

The reaon for this similarity in the classical and the quantum case is that Feynman diagrams are nothing else than a graphical notation for writing down products of tensors with many indices summed via the Einstein summation convention. The indices of the results are the external lines aka ''real particles'', while the indices summed over are the internal lines aka ''virtual particles''. As such sums of products occur in any multiparticle expansion of expectations, they arise irrespective of the classical or quantum nature of the system.


(added September 29, 2012)

Interpreting Feynman diagrams.

Informally, especially in the popular literature, virtual paricles are viewed as transmitting the fundamental forces in quantum field theory. The weak force is transmitted by virtual Zs and Ws. The strong force is transmitted by virtual gluons. The electromagnetic force is transmitted by virtual photons. This ''proves'' the existence of virtual particles in the eyes of their afficionados.

The physics underlying this figurative speech are Feynman diagrams, primarily the simplest tree diagrams that encode the low order perturbative contributions of interactions to the classical limit of scattering experiments. (Thus they are really a manifestation of classical perturbative field theory, not of quantum fields. Quantum corrections involve at least one loop.)

Feynman diagrams describe how the terms in a series expansion of the S-matrix elements arise in a perturbative treatment of the interactions as linear combinations of multiple integrals. Each such multiple integral is a product of vertex contributions and propagators, and each propagator depends on a 4-momentum vector that is integrated over. In additon, there is a dependence on the momenta of the ingoing (prepared) and outgoing (in principle detectable) particles.

The structure of each such integral can be represented by a Feynman diagram. This is done by associating with each vertex a node of the diagram and with each momentum a line; for ingoing momenta an external line ending in a node, for outgoing momenta an external line starting in a node, and for propagator momenta an internal line between two nodes.

The resulting diagrams can be given a very vivid but superficial interpretation as the worldlines of particles that undergo a metamorphosis (creation, deflection, or decay) at the vertices. In this interpretation, the in- and outgoing lines are the worldlines of the prepared and detected particles, respectively, and the others are dubbed virtual particles, not being real but required by this interpretation. This interpretation is related to - and indeed historically originated with - Feynman's 1945 intuition that all particles take all possible paths with a probability amplitute given by the path integral density. Unfortunately, such a view is naturally related only to the formal, unrenormalized path integral. But there all contributions of diagrams containing loops are infinite, defying a probability interpretation.

According to the definition in terms of Feynman diagrams, a virtual particle has specific values of 4-momentum, spin, and charges, characterizing the form and variables in its defining propagator. As the 4-momentum is integrated over all of $R^4$, there is no mass shell constraint, hence virtual particles are off-shell.

Beyond this, formal quantum field theory is unable to assign any property or probability to a virtual particle. This would require to assign to them states, for which there is no place in the QFT formalism. However, the interpretation requires them to exist in space and time, hence they are attributed by inmagination with all sorts of miraculous properties that complete the picture to something plausible. (See, for example, the Wikipedia article on virtual particles.) Being dressed with a fuzzy notion of quantum fluctuations, where the Heisenberg uncertainty relation allegedly allows one to borrow for a very short time energy from the quantum bank, these properties have a superficial appearance of being scientific. But they are completely unphysical as there is neither a way to test them experimentally nor one to derive them from formal properties of virtual particles.

The long list of manifestations of virtual particles mentioned in the Wikipedia article cited are in fact manifestations of computed scattering matrix elements. They manifest the correctness of the formulas for the multiple integrals associated with Feynman diagrams, but not the validity of the claims about virtual particles.

Though QFT computations generally use the momentum representation, there is also a (physically useless) Fourier-transformed complementary picture of Feynman diagrams using space-time positions in place of 4-momentaa. In this version, the integration is over all of space-time, so virtual particles now have space-time positions but no dynamics, hence no world lines. (In physics, dynamics is always tied to states and an equation of motion. No such thing exists for virtual particles.)


Can one distinguish real and virtual photons?

There is a widespread view that external legs of Feynman diagrams are in reality just internal legs of larger diagrams. This would blur the distinction between real and virtual particles, as in reality, every leg is internal.

The basic argument behind this view is the fact that the photons that hit an eye (and this give evidence of something real) were produced by excitation form some distant object. This view is consistent with regarding the creation or destruction of photons as what happens at a vertex containing a photon line. In this view, it follows that the universe is a gigantic Feynman diagram with many loops of which we and our experiments are just a tiny part.

But single Feynman diagrams don't have a technical meaning. Only the sum of all Feynman diagrams has predictive value, and the small ones contribute most - otherwise we couldn't do any perturbative calculations.

Moreover, this view contradicts the way QFT computations are actually used. Scattering matrix elements are always considered between on-shell particles. Without exception, comparisons of QFT results with scattering experiments are based on these on-shell results.

It must necessarily be so, as off-shell matrix elements don't make formal sense: Matrix elements are taken between states, and all physical states are on-shell by the basic structure of QFT. Thus thestructure of QFT itself enforces a fundamental distinction between real particles representable by states and virtual particles representable by propagators only.

The basic problem invalidating the above argument is the assumption that creation and desctruction of particles in space and time can be identified with vertices in Feynman diagrams. They cannot. For Feynman diagrams lack any dynamical properties, and their interpretation in space and time is sterile.

Thus the view that in reality there are no external lines is based on a superficial, tempting but invalid identification of theoretical concepts with very different properties.

The conclusion is that, indeed, real particles (represented by external legs) and virtual particles (represented by internal legs) are completely separate conceptual entities, clearly distinguished by their meaning. In particular, never turns one into the other or affects one the other.

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I think you are right that this is a question of definitions. I have seen this vocabulary fight now many times. Some people have learned that a virtual particle is by definition an internal line in a Feynman diagram. Others have learned that a virtual particle is by definition an off-mass-shell particle. Rather than fighting about the correct definition, it is more helpful to explain the difference between the two definitiions. –  Jim Graber Mar 18 '12 at 19:14
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This answer is not good. Virtual particles are not unphysical, and it is wrong to characterize them this way. The perturbation series can be recast in terms of particle paths, and these paths can be thought of as particles going around and colliding, and this is not wrong, despite Weinberg's distate for it. In condensed matter nonrelativistic field theory, effective particles such as phonons can be virtual, although in this case, the virtual particles obeying the Schrodinger equation are equivalently described by real particles going around on Feynman paths. –  Ron Maimon Jul 25 '12 at 3:39
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@RonMaimon: A single Feynman diagram (thus a diagram that could be interpreted as a real path) gives an infinite contribution to the amplitude once it contains a loop. Physical results are only obtained after renormalization cancelling groups of diagrams. In lattice gauge theories, one cannot even talk about diagrams; so how can they have physical meaning? –  Arnold Neumaier Jul 25 '12 at 10:11
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Oh yes, But the standard model is about the continuum, wheras the lattice is mathematically nearly trivial, compared to the continuum. Moreover, in a lattice model you don't even have a time direction, so you cannot interpret diagrams in terms of paths in time. Imaginary time is physically devoid of meaning. Moreover, a single finite renormalized term is composed already of multiple Feynman diagrams, so a single diagram means nothing. –  Arnold Neumaier Jul 25 '12 at 17:18
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@Arnold Neumaier, but in scattering theory external legs are taken to be on-shell plane waves, which is an approximation. My understanding is that in reality they are slightly off-shell (and of course not plane waves) due to having a finite lifetime (its measured energy can be off-shell, as long as $\Delta E\Delta t<\hbar$) –  user1247 Sep 21 '12 at 16:45

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