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For example, why is the semileptonic $B$ decay $B \to X\ell\nu$ inclusive?

I can't find any definition of these frequently used terms, strange.

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+1 great question! –  David Z Nov 22 '10 at 22:10

2 Answers 2

An experimental take

Exclusive implies that you have measured the energy and momenta of all the products (well, with an exception I'll discuss below). Inclusive means that you may have left some of the products unmeasured.

This applies to scattering processes as well as decays.

Some things to note:

  • Exclusive measurements allow you to nail down one, well defined physics process, while inclusive measurements may tell you about a collection of processes
  • It is generally difficult to measure neutral particles
  • If there are more than a couple of products it begins to require a lot of instrumentation to reliably collect them all and (crucially) to know how well you have done so

In the process you are asking about the neutrino is necessarily unobserved rendering the measurement inclusive, further an $X$ in the final state is often used to indicate unmeasured and unspecified stuff (i.e. it means the measurement is inclusive by design). Here unspecified includes case in high acceptance instruments where you consider all events with the specified products: those for which we know $X$ is empty those for which $X$ is non-empty and well characterized, and those for which $X$ is ill-characterized.

Theoretical view

I'm less sure of how theorist use these terms, but I believe there is a parallel. Something like: exclusive means one and only one process, while inclusive means all processes that include the specified products.

Convergence of theory and experiment

Of course, we haven't really learned anything until we get theory and experiment together, which is sometimes traumatic for both communities. Still exclusive measurements and calculations are clearing talking about the same thing, and inclusivity can be made to agree with some care in building the experiment and assembling the theoretical results.

Experimenters cheating on exclusivity

Sometimes in nuclear physics we talk about scattering measurements as exclusive when there is an unmeasured, heavy, recoiling nucleus involved. The assumption being that that it carries a small fraction of the total energy and momentum involved and can be neglected, though there is some risk from this if the nucleus is left in a highly excited state.

In particular my dissertation project was on $A(e,e'p)$ reaction (elastic scattering of protons out of a stationary nuclear target where the beam was characterized and both the proton and outgoing electron were observed), and we assumed that the remnant nucleus was left largely undisturbed and recoiling with a momentum opposite the Fermi motion of the target proton.

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That's always struck me as an odd choice of terminology - intuitively "exclusive" seems like it should have been the case where some particles are excluded from the measurement. –  David Z Nov 22 '10 at 22:16
    
@David: I think the take is "we've exclude the possibility that it is anything else...". But don't quote me on it. –  dmckee Nov 22 '10 at 22:20
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"Exclusive" means we consider exclusively this given set of particles. In "inclusive" processes we include several specific channels into a single analysis. It all becomes a bit more tricky in semi-inclusive processes, for example, in semi-inclusive DIS, SIDIS: $e+p\to e'+h+X$. Here you focus on distribution of a specific final hadron $h$ and sum over all the hadrons that can accompany it. –  Igor Ivanov Nov 23 '10 at 0:46
    
Thanks everyone! That certainly clarifies it. For the same reason then $B\rightarrow X\gamma$ must be inclusive, since X is undetected stuff. I'm glad I know the difference now. I'm new to this board and was skeptical asking, but I'm glad to see so many knowledgeable and helpful users :) –  Qrius Nov 23 '10 at 16:37
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In particle physics $X$ is not necessarily undetected. It can be detected as well, but we just don't care about what's inside $X$, we don't use the information the detector gives us on what's inside $X$. –  Igor Ivanov Nov 23 '10 at 20:16

Some remarks on theoretical issues involved in inclusive/exclusive processes.

First of all, in the inclusive vs. exclusive vs. semi-inclusive classification, we care only about final hadrons.

Typically, exclusive processes are much more difficult to calculate than inclusive ones. When you calculate the cross section of an inclusive process, it is usually sufficient to calculate it at the quark/gluon level. You know that these quarks and gluons will eventually hadronize into some hadrons, but in this calculation you don't care how exactly this will happen. So, we effectively integrate over the full final phase space of hadrons by setting it equal to the phase space of quarks/gluons, which you know how to calculate.

As you move from fully inclusive to less inclusive processes (jet production, semi-inclusive processes, exclusive processes), you are trying to gain some insight into the structure of the hadronic final state. And the more details you want to see, the more difficult is the calculation. It is not sufficient anymore to calculate production of quarks/gluons. You now need to specify their probability to hadronize into a given hadron or set of hadrons (e.g. via fragmentation functions). You need to consider differential distributions w.r.t. more kinematic variables than before. You often need generalization of partonic distributions, but factorization theorems become much harder to prove for less exclusive processes. All this makes exclusive processes difficult to calculate, at least at high energies, when quark and gluon degrees of freedom are fully important.

At the same time, exclusive processes often give you a deeper insight into the structure of hadrons than the fully inclusive ones. For example, inclusive deep-inelastic scattering (DIS) is sensitive only to very few partonic densities, even when polarization degree of freedom is taken into account. Semi-inclusive and exclusive processes in DIS (such as SIDIS, various diffractive processes, DVCS = deeply virtual compton scattering) are sensitive to many new partonic distributions, which reflect more subtler aspects of the proton's structure. That's why they are studied despite being so difficult.

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