# What's the difference between inclusive and exclusive decays?

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

## 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
"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
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