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 $T_{fi}$ 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 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.

### $\mathrm{d}LIPS$

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 $\mathrm{d}p_1$ from $\mathrm{d}^3\mathbf{p}_1$? 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 $\mathbf{p}_1$ 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 $\left\vert p_1\right\vert$ 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 post. 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 $\sim 2\,\text{GeV}$, 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

$\mathcal{BR}\left(W\to\text{hadrons}\right)=\dfrac{\Gamma\left(W\to\text{hadrons}\right)}{\Gamma(W)}\simeq 67$%.

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

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

## 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 $b$ ($p_i^*$ in the centre-of-mass frame) in addition to that of the particles 1 and 2 ($p_f^*$), and that we’re now dealing with a total centre-of-mass energy $\sqrt{s}=E_a^*+E_b^*$ (rather than just the mass squared of the decaying particle, in the previous case). Surely enough, we find:

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: