# Probability densities at the LHC

Let $\sigma$ be a cross-section for the collision of two protons as given by

$$\sigma = \intop_0^1 \mathrm{d}x_1 \intop_0^1 \mathrm{d}x_2 \, \sum_{a,b}f_a(x_1,Q^2) f_b(x_2,Q^2) \frac{1}{2\hat{s}} \intop \mathrm{d}\Phi ~ \overline{\left|M_{ab}\right|^2},$$

where $f_i$ are the parton density functions, $\Phi$ is the available phase-space and $M_{ab}$ ist the (let's say tree-level) Matrix element of the partonic process.

What I fail to understand is which of the integrands can be understood as probability densities in the sense that the integral over the whole domain equals $1$.

I have for example always thought of $M_{ab}$ as a probability amplitude, so $\overline{\left|M_{ab}\right|^2}$ would be a probability density. But in which variables? It can't be the momenta of the particles or else the integral over the full phase-space would always be $1$.

Also $f_i(x_1,Q^2)$ is said to give the probabilty of finding parton $i$ with momentum fraction $x_1$ inside the proton at scale $Q$, but does that mean that

$$\intop_0^1 \mathrm{d}x_1 f_i(x_1,Q^2) = 1?$$

Finally people usually say that $\sigma$ represents the probabilty of a process taking place, but since it has a dimension ($\text{GeV}^{-2}$), this can't be true in the mathematical sense.

Can anyone help me connect the dots here?

None of it is a probability density in that sense. As you've noticed, the end result is dimensionful, so it is not a probability either.

The cross-section is proportional to probability for any given process. To get something more like a probability, you have to multiply by a quantity called the "luminosity." The luminosity effectively lumps together all the "collider" details into one quantity, as opposed to the "physics" details that are taken care of in the cross section.

Even this doesn't quite give you a probability- it gives you an expected event rate or number of events in a given interval. Of course, if the time interval is sufficiently short that the expected number of events is much less than one, you can treat this effectively as a probability.

As for $\intop_0^1 \mathrm{d}x_1 f_i(x_1,Q^2) \stackrel{?}{=} 1$, almost. The parton distribution functions are weighted by the expected number of each parton in the nucleon. So, for instance, the function for an up quark in a proton would integrate to more than the function for the down quark.

• Thanks that's great! Do you possibly know anything about the squared Matrix element? – Lxndr Jan 19 '18 at 11:19
• @Lxndr It's a thing you integrate over phase space to get the cross-section :p More seriously, the matrix element squared is simply not a probability density. You can see this most easily in that fact that the matrix element itself is dimensionless, but it is integrated over phase space, so clearly the result cannot be a probability. You can think of it as kind of like a probability in the sense that bigger $\mathcal{M}$ means bigger probability, though. – Chris Jan 19 '18 at 11:40
• But $M$ is the transition amplitude right? And in Quantum mechanics the transition amplitudes can actually be understood as probabilities for the transition, can't they? So what happens in QFT that that is no longer true? – Lxndr Jan 19 '18 at 11:49
• @Lxndr It has to do with the fact that the spectrum of eigenvalues is continuous. When there are discrete eigenvalues, the transition amplitude (squared) can be interpreted as a probability, but it's different with a continuous spectrum. Keep in mind that if you're trying to talk about the probability to end up in a specific eigenstate of momenta that eigenstates of momentum in free space are not physically realizable, normalizable states, so it's not as easy to interpret what the transition amplitude corresponds to physically. – Chris Jan 19 '18 at 12:07
• Okay thanks! I will try and figure out the details by myself. – Lxndr Jan 19 '18 at 12:16