These are notes from a PHY421 tutorial I gave today, essentially mirroring the discussion in Thomson’s Modern Particle Physics (first few sections of chapter 3) and in Prof. Costanzo’s lecture notes.

Particle decays and phase space

Fermi’s golden rule

We start by recalling that Fermi’s golden rule, i.e. the transition rate

expressed in terms of the transition matrix element and the density of energy states, can be rewritten as an integral over accessible states as

Right off the bat, we point out the leitmotiv of this tutorial: is a physical quantity of interest, and should be made “as Lorentz invariant as possible”; all the physics (read, particle-dependent physics) are contained in the transition matrix element ; and the integration measure encodes the kinematics of the process.

Two-body decay

Let us then look at a generic process, for which we write the previous equation as

As explained in this post, the delta-function enforces conservation of energy. In our ongoing effort to make everything nice and Lorentz invariant (because we want to be able to make frame-independent calculations), let’s replace the transition matrix element with the more suitable Lorentz-invariant matrix element (this is one of the main objects of interest of SM QFT, as you will see in QED and QCD):

Plugging everything in, and moving out of the integral only the terms that are not being integrated over:

By construction of from wavefunctions with Lorentz-invariant normalisation and the fact that the other two are just delta functions, we know that the first three terms under the integral sign are Lorentz-invariant; quid of the last two? Feel free to skip the next section, which is a simple proof of the transformation properties of the so-called Lorentz-invariant phase space measure.

The aim of this section is to prove that the following measure

is indeed Lorentz-invariant. As is often the case in QFT, there are two ways of solving a given problem. The Feynman method (or Feynman algorithm, coined by Nobel prize winning theoretical physicist Murray Gell-Mann) is as follows: write down the problem; think real hard; write down the solution. You may find the second method more applicable, and that is trial and error. The first step towards getting the Lorentz-invariant formulation of is to determine by how much it fails to be Lorentz-invariant in the first place.

With that in mind, consider a transformation along the -axis only, with some and factors:

where I’ve introduced an extra factor of simply because I can, but also to retrieve . Thus we see that only if there is no transformation will be Lorentz-invariant, or in other words, it isn’t. It fails to be by the derivative , which we will now compute. From the transformation formulae

we get

The tricky part here is to remember that is a function of and so should also be differentiated. Rearranging the left- and right-most terms of the equation, we conclude

which is the definition of a Lorentz-invariant quantity. Back to the main calculation.

Two-body decay (cont’d)

Now that we’ve made sure everything inside the integral is nice and Lorentz-invariant, let’s get down to business. In the centre-of-mass frame, we have and - i.e. the decaying particle is at rest. The two daughter particles will be produced back to back, with (the star notation refers to the specific momentum that satisfies conditions we’re going to describe in a bit). Let’s then make the appropriate simplifications where we can:

It seems like we’re performing two integrals, one over all the possible momenta of particle 1, and similarly for particle 2, but remember that the delta-function is imposing momentum conservation! Integrating over therefore fixes . We keep the notation , but be careful that we have now :

At this stage, we’d like to get rid of the second delta-function: after all, it’s only a function of the single variable … The problem is, how do we extract from ? To answer this question we have to think about a “momentum sphere”. This is a sphere of radius in momentum space, where a vector can be built from its three components and specifies a unique point on the sphere. Going back to first year maths, we know that we can also give it spherical polar coordinates! Bearing in mind that will play the role of , we write:

where is the usual solid angle. We now have:

Three terms contain : a delta-function, some other function, and a measure. This should ring a bell, and if it doesn’t, check out this post. We write:


Finally, putting everything together, we obtain the desired result:

A few points of interest

As stated at the beginning, we’ve extracted out of the integral everything we could that wasn’t directly related to the nitty-gritty of a specific decay (i.e. which particle decaying to which daughters?) - we’re left with only inside the integral. This is the statement that the matrix element encodes all the information about the physical process. Hence why it’s so important to come up with a consistent QFT to be able to calculate such matrix elements.

Furthermore, the formula we’ve just derived for is valid for any decay process. The phase-space factor is common to any such process and can be calculated only once. Note also that the two free parameters associated to it are (the mass of the decaying particle) and , the unique momentum assignment that ensures conservation of energy - this should make physical sense from a classical point of view.

It is also interesting to observe that the integral of matrix elements is no longer over momentum, but the solid angle … This has consequences on how we think about experimental particle physics experiments, and is tied with the useful quantity , the differential cross-section, which we will see in a second.

Final point: a particle can decay to various pairs of daughters, dubbed decay channels. For instance, last time I checked, the total width of the W boson was , and it could decay to hadrons (well, quarks) or a pair of lepton and corresponding neutrino. Instead of quoting the individual of these processes, it is more convenient to talk about the branching ratio


And since branching ratios must sum up to unity, we have %.

One for the road

Can you show that, in the centre-of-mass frame, the following expression for holds?

Hint: start with conservation of energy, and square.

Interaction cross-section

The derivation of this formula is somewhat redundant, but a similar argument as before can be made for the scattering process . The only differences should be that we’re also considering the momenta of the particles and ( in the centre-of-mass frame) in addition to that of the particles 1 and 2 (), and that we’re now dealing with a total centre-of-mass energy (rather than just the mass squared of the decaying particle, in the previous case). Surely enough, we find:


For completeness, the derivation of the above equation. The interaction cross-section is the number of interactions per unit time per target particle normalised by the incident flux, i.e.

We have to find the for this new process, but this isn’t too hard; repeating the first few steps we did before, we find:

We combine the denominator of with the of to form a quantity called the Lorentz-invariant flux factor . In the centre-of-mass frame, it is:

Still in the centre-of-mass frame, we have , and ; everything reduces nicely to an integral we’ve already calculated. We are left with:

(where I’ve grouped the terms by where they come from). Simplifying the constants yields the desired result: