How do we distinguish between virtual particle exchange and particle decay?

Question

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?)

Answer 1

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.

Answer 2

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