Particle Decay | Bobak Hashemi's Blog

In collider events, we often see what’s called Initial State Radiation or (ISR). Particle physicists use this term to identify events where one of the particles in some collision produced a boson before there was any interaction. I’d like to give a brief discussion as to what this means for the physics, that is, what’s different about an event with initial state radiation and the same processes without it?

The tree level diagrams of the standard model, courtesy of Wikipedia.

First off, ISR is typically a photon, a Z boson, or a gluon. You can see why by looking at this wonderful image to the right that shows a list of the “tree level” standard model decays. Notice that there are mostly diagrams which have three particles at each vertex, to be honest, I don’t know whether the physicists consensus is that we have ruled out vertices between a higher number of standard model particles, but I seem to recall some argument about renormalizability that might exclude such processes… In any case, such decays are certainly not a part of the standard model Lagrangian (a topic for another post), so if we are working within the context of the standard model, they don’t exist. Another point to further convince you that other decays are less relevant is that even if those decays were possible, with each particle you attach to a vertex, you lose some probability of having all the ingredients you need to make it happen, so even if higher vertex diagrams exist, they would be have a lower probability of occurring (we say they would be suppressed) by that fact as well.

These are the most likely decays because they all happen to first order in perturbation theory, that is to say, they don’t have any loops. So these are the most important processes to keep in mind when working in standard model physics; they are going to happen most frequently.

When we speak about ISR, we are typically talking within the context of another physical processes. For instance, below is a diagram of an electron and a positron annihilating to make a photon (the star means it’s an “off-shell” photon with mass), then our final state is a charm /anti-charm pair.

A positron and electron combine into a massive photon which then decays into a charm/anti-charm pair. Taken from here.

In the context of this decay, we have the ISR of another photon. So if we want to talk about initial state radiation, we are talking about a decay where we can tack on another decay and still have the same physics process. If you look at the chart of the standard model interactions, you’ll notice that the only way a fermion can come into and out of a decay as the same particle is when it is coupled to one of the boson I listed above.

Now that we’ve shown that ISR (in the sense we’ve defined it here) has to be a gluon, a photon, or a Z boson, we can get to real point of this post… what happens to an event when you add in some ISR?

The first thing is the most obvious, the rate of the decay is reduced. Each vertex in your Feynman diagram adds some factor, less than 1, to the amplitude associated with that diagram. So no matter what decay you choose, since you’ve added a vertex to your diagram, you’ve reduced the probability of the process occurring.

Unfortunately, I am not positive how to compute the precise variation of the probability of measuring an event under the constraint of additional ISR particles. I’ll explain why. With two ISR bosons, you have to consider at least one extra diagram at tree level. For instance, if we wanted to add another $\gamma$ to the electron-positron annihilation, we could either attach that particle to the electron or the positron, so we end up with 3 indistinguishable initial states that lead to our charm-anticharm pair with two particles of ISR. Both bosons are attached to the positron, both are attached to the electron, or they are split. Accounting in quantum field theory is still very confusing to me, so unfortunately the best I can do is say another ISR piece probably reduces the probability of the event occurring by about the same percentage as the first. This follows the same line of logic as before, since there is only one diagram without any ISR, and two diagrams with ISR, one for each choice of lepton to which we attach the $\gamma$ .

The next, and most physically relevant effect is the so-called boost of the off-shell photon. Consider a point in our diagram where the $\gamma *$ exists in the case with and without ISR. When there is no ISR, the center of mass of the system is equal to the rest frame of the $\gamma *$ , but when there is ISR, the center of mass of the system is now somewhere in between the two. Notice that center of mass is a misleading term in particle physics, what we really mean the frame in which the particles have 0 net momentum, the ISR photon in this system has momentum equal to its energy, but no mass. Why is this important?

Let’s think about what effect this has on the momentum of the charm quarks. In the case of the decay without any ISR, there are a lot of nice things that are true about the momenta of the charms. First, if this is a collider, then there should be essentially no transverse momentum in initial electron-positron system. That is to say, there’s a beam line in which the electrons are moving, so the leptons should have no component of momentum in the plane at a right angle to that beam line. That means that if you add up the momentum vectors of the charm pair, it should also have no transverse momentum. With an ISR boson however, it can be seen easily from momentum conservation that the momentum of the charm pair is different from the momentum of the electron-positron pair by the exactly the ISR photon’s momentum.

$p_e + p_{\bar{e}} + p_{\gamma^{ISR}} = p_{\gamma *} = p_c + p_{\bar{c}}$

which implies that

$p_c + p_{\bar{c}} - p_e - p_{\bar{e}} =p_{\gamma^{ISR}}$

This means that the transverse momentum of the charm pair need not be 0, since it’s equal and opposite to the transverse momentum of the emitted ISR photon. In another post, I’d like to discuss the probability distribution for the direction of the emitted photon, or more generally, how to compute that probability in arbitrary decays.

The other point to address is that in particle collisions, we typically have a lot of symmetry in the initial momentum distribution. If we were at an electron+positron collider like in the example above, in most cases, the net momentum along the beam line would be approximately 0 for the e+p system by construction (normally we build colliders to accelerate two beams to equal momentum). This means that the $\gamma *$ would have the highest probability to be produced at rest in the lab frame. In the case of ISR, however, this is clearly not the case, the $\gamma *$ instead has the highest probability to be boosted with respect to the lab frame by 3-velocity equal the ISR particle’s momentum over the mass of the $\gamma *$ . That is

$v_{\gamma *} \rightarrow v_{\gamma *} \frac{p_{\gamma}}{m_{\gamma *}}$ .

This means two things for the momentum vectors of the charm quarks. First, their components will typically be larger. Why? Because their kinetic energy will be larger as they were produced by a particle that was in motion with respect to the lab. Secondly, they will tend to be closer together, since under a Lorentz boost the momentum vectors of the two particles will be smooshed together. I’ve tried to illustrate this below.

The smooshing together of vectors under a Lorentz boost. In the rest frame of the off-shell photon, we have a ruler at rest with respect to the lab which has its graduation lines smooshed together due to Lorentz contraction of the lab. Then we choose any two vectors for the momentum of the daughter particles on one half of the plane (to balance an ISR vector, not shown, pointing to the left). Consider points along the trajectory of the momenta A and A’, after unsmooshing the ruler, we inspect A and A’ in the new coordinate system. The vertical distance to the ruler has not changed, but we see the angle between the two particles is now smaller.

So let’s recap.

ISR is typically spoken about in the context of a known decay that can be found without the emission of a Z, $\gamma$ , or gluon.
By looking for a decay with ISR, you reduce the rate at which that decay occurs.
ISR boosts the entire decay system with respect to the lab causing the transverse momentum of decay products to be non-zero, and adding kinetic energy to the system.
Boosted decay products will be smooshed together in the lab frame.

Thanks for reading!

-Bobak