Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?)

share|cite|improve this question
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 2:15
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:… – James Jul 24 '12 at 21:59
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 3:33

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, , 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

share|cite|improve this answer
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
@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
@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
@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
@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

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.

share|cite|improve this answer
If you boost the higgs, you time dilate its decay rate. – Jerry Schirmer Jul 24 '12 at 14:46
@JerrySchirmer, Thanks! I modified the answer to include that effect. – FrankH Jul 24 '12 at 16:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.