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The difference between virtual particles and unstable particles is discussed at length in this question (namely, virtual particles correspond to internal lines in Feynman diagrams and are not associated with any measurable physical state). So what is going in the case of, say, Higgs production at the LHC, where the Higgs does not live long enough to reach the detector? What are the calculational and experimental differences between producing a Higgs via quark fusion, which then decays into a pair of leptons (for example) and quark-quark to lepton-lepton scattering, proceeding via Higgs exchange?

(To put it another way, we'll see an increase in lepton-lepton production from virtual Higgs exchange, but how is this distinct from seeing particles produced from the decay of an unstable Higgs?)

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James
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  • The Higgs is always virtual because it is unstable. This was a debate in the 1960s, about whether unstable particles count as particles because they are always virtual. It's ultimately a convention, and people settled on a sensible definition that every independent propagating field in the Lagrangian is a particle. – Ron Maimon Jul 24 '12 at 02:15
  • So would you like to comment on the Arnold Neumaier's answer from that question I linked, in which he clearly distinguishes between the two? – James Jul 24 '12 at 12:52
  • Arnold Neumaier's answer is describing the signature of a new particle on the scattering, it's an additional two fermion-two fermion scattering with a nonlocal vertex. The Higs is still virtual in the answer. – Ron Maimon Jul 24 '12 at 13:12
  • I think that may be referring to his answer here. I meant his answer here: http://physics.stackexchange.com/questions/4349/are-w-z-bosons-virtual-or-not – James Jul 24 '12 at 21:59
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    That answer you link to is mathematically probably ok (I didn't read it carefully), but physically incorrect. It is known since Schwinger and Feynman that the diagrams have an interpretation as a particle process. It is difficult, but possible, to give this heuristic picture a better mathematical grounding, but regardless, it is correct that you can do Feynman graphs as sum over space-time paths, without any modification of the formalism. The answer is unfortunate. – Ron Maimon Jul 25 '12 at 03:33

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The standard model predicts that the Higgs boson has a lifetime on the order of $10^{-22}$ seconds. That means that if the Higgs were moving close to the speed of light, it could move something like $34\gamma$ times the diameter of a proton (on average) before it decays. $\gamma$ is the time dilation factor from special relativity which is $$\gamma = \sqrt{\frac{1}{1-\frac{v^2}{c^2}}}$$ So, theoretically, if we could precisely aim two quarks at each other with sufficient precision and measure the vertex of the two outgoing leptons with sufficient precision, we could actually measure how far the Higgs traveled (on average) before it decayed. Note that this, like all radioactive decays, follows an exponential fall off with respect to time, so it could travel significantly longer than $34\gamma$ proton diameters, but the probability of this rapidly approaches zero.

Now, this distance of $34\gamma$ proton diameters is far to small to actually be measured at the LHC or any other proposed accelerator. But this lifetime is measurable by measuring the width of the "bump" that the Higgs creates in the cross sections that can be measured at accelerators.

This bump in the cross section and this "significant" distance between the vertices will only occur when the total energy center of mass energy of the two incoming quarks is close to the Higgs mass (of 125 GeV - this is called an "on shell" Higgs production). You are correct - when the incoming quarks have a mass that is significantly different that 125 GeV ("off shell"), the Higgs will still contribute to the cross section for two quarks to create two leptons via a virtual Higgs exchange, but in this cases the incoming quark and outgoing lepton vertices will be VERY close to each other - nothing like $34\gamma$ proton diameters apart you get for on-shell Higgs production. I am only guessing, but I bet the vertices would be much less than 1 proton diameter apart for these far "off shell" lepton production processes.

Of course, as you change the energy of the collision from far off shell to exactly on shell, the distance between the two vertices will continuously change from near zero to the average of $34\gamma$ proton diameters, but there is no particular point where you can say there is uniquely a "real" Higgs in this case versus a "virtual" Higgs in that case. However, there is still a dramatic difference between the exactly on-shell versus far off-shell results.

FrankH
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Decaying particles are described by complex energies, the imaginary part of which encodes life-time information. They are observable; in case of very short-lived particles such as the Higgs in the form of resonances, http://en.wikipedia.org/wiki/Resonance_(particle_physics) , i.e., a peak in the production rate of products of Higgs decays. The decay itself would be visible only at much better time resolution, i.e., far higher energies.

In contrast, virtual particles have real energies with 4-momentum violating the equation $p^2=(mc)^2$. They are unobservable.

A much more detailed answer can be found at https://physics.stackexchange.com/a/22064/7924

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Arnold Neumaier
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  • OK, but my question is: virtual Higgs exchange will contribute a peak in the cross-section of dilepton production (as an example) - how is this distinct from the peak from (real, but unstable) Higgs decay? – James Jul 24 '12 at 12:51
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    Virtual Higgs exchange is the computational procedure by which (in a perturbative setting) the peak in the cross-section of dilepton production is predicted. (A lattice gauge theory has no virtual particle concept but could predict the peak, too, at least in principle. This proves that virtual particles are tied to the computational procedure.) - On the other hand, Higgs decay is how Nature actually does it. – Arnold Neumaier Jul 24 '12 at 13:23
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    @ArnoldNeumaier: That's ridiculous. You can't distinguish "decay" from virtual exchange--- what if the Higgs decays at a spacelike separation from its production? Virtual particles are not a calculation procedure. – Ron Maimon Jul 24 '12 at 21:58
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    @RonMaimon: If a Higgs decays at some distance from where it was created, it is a real particle with complex energy, not a virtual particle. Virtual particles have no existence outside of perturbation theory. – Arnold Neumaier Jul 25 '12 at 10:07
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    @ArnoldNeumaier: This is completely incorrect. You can have strong virtual photon exchange between a lepton and a quark in a proton which is nonlocal in space, corresponding to a long-range interaction. The Higgs is no different. Your interpretation is an anti-Feynman anti-Schwinger disease that has been going on since 1945, it's got to stop. The idea that the particle thing only works at perturbations is wrong. – Ron Maimon Jul 25 '12 at 16:48
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    @RonMaimon: Feynman diagrams (hence virtual particles) mean something only in perturbation theory. Not even nonperturbative functional integrals feature virtual particles. On the other hand, perturbation theory of a system of classical anharmonic oscillators features virtual particles, though nobody claims their physical existence. – Arnold Neumaier Jul 25 '12 at 17:20
  • I totally agree with Arnold Neumaier. The concept of (elementary) particle is related to irred. representations of Poincare group. Thus particles have a well defined value of the Casimir invariant $E^2-P^2$, while virtual particles do not. @RonMaimon, could you provide your definition of "particle"? By the way, there are plenty authors who claim that virtual particles are not particles. Could you provide any reference that support your opinion? Thank you. – Diego Mazón Jul 25 '12 at 23:20
  • By "well defined value" I mean every particle has a given value. This is independent of whether the particle has a zero or nonzero decay width (stable or unstable). – Diego Mazón Jul 25 '12 at 23:30
  • For example, one can make experiments with neutrons to determine the relation between their energy and their momentum, and one see that $E^2-P^2$ is a constant. They are hence true particles (although they are not elementary, but this has nothing to do). However, they decay in a finite, observable time, so they are unstable. – Diego Mazón Jul 25 '12 at 23:42
  • @RonMaimon: How many "virtual photons" are exchanged in a Bhabha (electron-positron) scattering? Don't you think that it depends on the approximation (number of loops) you make in the perturbative calculation? Therefore, are "virtual photons" true particles? – Diego Mazón Jul 26 '12 at 00:10
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    @drake: Neutron $E^2-P^2$ can be thought to differ from the square of the neutron mass by the liftime (but it's a ridiculously tiny thing). This is the neutron width. You can't make an experiment on a free neutron to determine the mass better than this, simply because the neutron is virtual. The number of virtual photons in any electromagnetic process is infinite, as is the number of real photons, because of the infrared divergences. This is an issue with all formalisms based on particles, and it requires an infrared regulator. The ultraviolet business is fixed by a lattice. – Ron Maimon Jul 26 '12 at 01:20
  • @RonMaimon: I regret writing the previous example with an unstable particle like the neutron, my mistake. Since tree-level calculations don't involve divergencies, you are interpreting as real only the sum of all Feynman diagrams which contribute to a given process, right? Other point: a virtual photon can have an arbitrary real value of $E^2-P^2$, so it isn't related to a irred. rep. of Poincare and it does not have a defined value of mass in your opinion, right? – Diego Mazón Jul 26 '12 at 01:47
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    @drake: If a virtual photon travels a long time between emission and absorption, the amplitude gets sucked to the pole in the propagator, and the photon looks closer and closer to massless. It's a continuous thing. The particle picture is not without problems, but it can be made consistent (at least on a lattice in a finite box). There are more insidious issues, like can you see instantons using particles? A theta angle will give no contribution at any order of peturbations (naively) so how does it work in particle picture? There is an answer, but it's all hokey and not in the literature. – Ron Maimon Jul 26 '12 at 02:14