*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 (I don’t have a working link yet - comment?).*

## 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 $T_{fi}$ and the density $\rho(E_i)$ of energy states, can be rewritten as an integral over *accessible states* $n$ as

Right off the bat, we point out the *leitmotiv* of this tutorial : $\Gamma_{fi}$ 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 $T_{fi}$; and the integration measure encodes the kinematics of the process.

### Two-body decay

Let us then look at a generic $a\to 1+2$ 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 $T_{fi}$ with the more suitable **Lorentz-invariant matrix element** $\mathcal{M}_{fi}$ (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 $\mathcal{M}_{fi}$ from wavefunctions with Lorentz-invariant normalisation and the fact that the other two are just 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**.

### $\mathrm{d}LIPS$

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 $\mathrm{d}^3\mathbf{p}$ 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 $z$-axis only, with some $\gamma$ and $\beta=v_z/c$ factors :

where I’ve introduced an extra factor of $\mathrm{d}p_z/\mathrm{d}p_z$ simply because I can, but also to retrieve $\mathrm{d}^3\mathbf{p}$. Thus we see that only if there is no transformation will $\mathrm{d}^3\mathbf{p}$ be Lorentz-invariant, or in other words, it isn’t. It fails to be by the derivative $\mathrm{d}p_z’/\mathrm{d}p_z$, which we will now compute. From the transformation formulae

we get

The tricky part here is to remember that $E$ is a function of $p_z$ 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 $E_a=m_a$ and $\mathbf{p}_a=\mathbf{0}$ - i.e. the decaying particle is at rest. The two daughter particles will be produced *back to back*, with $\mathbf{p}_1=\mathbf{p}^*=-\mathbf{p}_2$ (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 $\mathrm{d}^3\mathbf{p}_2$ therefore fixes $\mathbf{p}_2=-\mathbf{p}_1$. We keep the notation $E_2$, but be careful that we have now $E_2^2=p_1^2+m_2^2$ :

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 $p_1$… 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 $p_1^2$ 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 $r$, we write :

where $\Omega$ is the usual solid angle. We now have :

Three terms contain $p_1$ : a delta-function, some other function, and a measure. This should ring a bell, *and if it doesn’t, check out this article*. We write :

with :

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 $\mathcal{M}_{fi}$ 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 $\Gamma_{fi}$ 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 $m_a$ (the mass of the decaying particle) and $p^*$, 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 $\Omega$… This has consequences on how we think about experimental particle physics experiments, and is tied with the useful quantity $\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega}$, 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* $\Gamma$ of the W boson was $\sim 2$ GeV, and it could decay to hadrons (well, quarks) or a pair of lepton and corresponding neutrino. Instead of quoting the individual $\Gamma$ of these processes, it is more convenient to talk about the **branching ratio**

And since branching ratios must sum up to unity, we have $\mathcal{BR}\left(W\to\ell^+\nu\right)\simeq 33\%$.

## One for the road

Can you show that, in the centre-of-mass frame, the following expression for $p^*$ 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 $a+b\to 1+2$. The only differences should be that we’re also considering the momenta of the particles $a$ and $b$ ( 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 :

### Addendum

*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 $\Gamma$ for this new $a+b\to 1+2$ process, but this isn’t too hard; repeating the first few steps we did before, we find :

We combine the $(v_a+v_b)$ denominator of $\sigma$ with the $4E_aE_b$ of $\Gamma$ to form a quantity called the **Lorentz-invariant flux factor** $F$. In the centre-of-mass frame, it is :

Still in the centre-of-mass frame, we have $\mathbf{p_a}+\mathbf{p_b}=\mathbf{0}$, and $E_a+E_b=\sqrt{s}$ ; 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 :