Author

Mikhail Mikhasenko

5.1 Angular Distributions, Kinematics Variables, and the Dalitz Plot in Two-to-Two Scattering

Today we are at lecture number five. We’ll discuss angular distributions and partial wave analysis. But before going there, I would like to start with a recap.

In the last lecture, we discussed the phase space for particle reactions and different experiments and their kinematics. We went through the list of experiments around the world that study hadrons and look at their production mechanisms and some peculiarities.

We start with a recap on kinematics. The first question: How many variables does one need to describe the two-to-two scattering process? We have two problems:

  1. Scalar particles—let’s say 0-minus scalars.
  • The scalar particles (0-minus) scatter to 0-minus, and the final state also has two 0-minus scalar particles.
  1. Spin-dependent scattering.
  • Example: P-plus.
  • It’s again scattering of 0-minus from 0-minus to, let’s say, 3-minus and 1-plus.

The scattering process is represented by a “blob” indicating the interaction (e.g., strong interaction). Arrows show particles entering and leaving. This is a unitarity diagram, not a Feynman diagram.

Here is a little blob that indicates generally what we are discussing. This is the scattering process—something described by strong interaction theory happens inside the blob, and the arrows indicate the particles entering the interaction and then leaving it. This is a cartoon diagram. These are not Feynman diagrams; sometimes they’re called unitarity diagrams. We will touch on unitarity in the following lectures, but for now, this is a nice way to indicate what we are essentially talking about.

Then I ask you to calculate the number of variables needed to describe the process entirely—the full kinematics—and we are talking only about kinematics at this point. Whatever happens in the blob does not impact the answer. The blob could be electromagnetic, strong, gravity—whatever you want. Something happens inside the blob, and the question is: How many variables do you need?

There are two possibilities:

  1. Particles without spin.
  2. Particles with spin—how many variables do you need to describe the process?

The second is related to the pictures I gave you. But let’s do it one by one.

I also added a third item: Give an example of the combination of variables that fully describe the process. For (a) and (b), you need to count—tell me how many—and then (c) gives the example.

Let’s quickly check how many. The calculation is like two times two, or four. Then we multiply by three. Then we subtract four due to conservation laws. So one should have eight—and that’s correct. Now you get rid of the… This is the entire kinematics. There are no other things in the space that would constrain orientation. Essentially, for describing the kinematics, I can go to any arbitrary frame and rotate the space. So I have to subtract another six—three rotations, three boosts.

The eight is the correct number if you just take into account conservation laws. But then you can account for having no reference, no anchor in space, and subtract another six, leaving you with two. So for (a), you’ve got two.

What did you get for (b)? You have the two from the previous scenario, and you have the angular distribution of the final-state particles. So they are… I mean, I don’t have any decay information here—just two-to-two. Then four? Why four? The angle between the two production particles and the angle between the 3-minus and 1-plus particles. How would these angles be different? Why can’t I introduce these angles in (a)? You can, but they are not necessary. So it’s actually two as well.

Here, I would have a scalar amplitude that describes the amplitude as a function of S and T, where:

S = (P_1 + P_2)^2

T = (P_1 - P_3)^2

Now I’ve made my cartoon a bit richer—I indicate the channels or the variables in Mandelstam invariants that describe the process. The energy of the system is given by the variable S, which is the sum of four-vectors squared. The transferred momentum in direction is T, which describes how energy—essentially analogous to this variable, but in the vertical direction—so I describe it as P_1 - P_3.

In my cartoon, I also added the P’s, which are four-vectors here. They have an energy component, a three-dimensional component, and, for particles with spin, indicated spin orientation—their helicity values, \lambda_3 and \lambda_4.

In the first case, my amplitude describing the blob—the interaction—is a scalar function. It’s just a single number. In the case where particles have spin, it’s not a number—it’s a higher-rank object. So here, in that case, what is the size dimension of my scattering amplitude?

  • Spin-3 gives me seven dimensions.
  • Spin-1 gives me three dimensions.
  • Therefore, my scattering amplitude is a 21-dimensional object.

So I have 21 amplitudes, but all of them are functions of two variables.

When particles have spin, you have more amplitudes, but they all have the same dependencies. The situation changes when you take into account the decays of the final-state particles.

Usually, we do not know any stable 3-minus particles, nor any stable 1-plus particles—they would fly away, and we would consider them final-state particles. Therefore, both of these particles decay. Often, you can take into account the decay products and introduce more variables. But if I restrict the problem to this—full stop—two variables, that’s it.

So Henrik, what are your favorite two variables to describe a two-body scattering process? The mass of particle one and two, and particle one and three. I guess it would be something with the mass of the two initial particles. That works.

Now tell me how to compute mass. It’s the energy squared minus the momentum squared. Energy minus kinetic energy doesn’t work. I think I would use this: if I want to compute the mass of two particles, I would take the four-vector of the first, the four-vector of the second, add them together, and then take the Lorentz norm squared of this vector, which is energy squared minus momentum squared. So this is mass squared.

Essentially what you suggest is to take this variable as one of them and take another one which is the other mass. But you want to put a sum plus here—that works. Actually, that’s okay, you can put a plus, but it’s going to be a mixture of S and T. What is more common is to put particles in the final state with a minus sign, in the initial state with a plus sign, and then have this, what you propose, to have masses in this convention.

And then these variables, S and T, are called invariants, Mandelstam variables. They are quantities that characterize the decay, and they are Lorentz invariant. They don’t describe a specific rest frame, so they don’t describe a specific setup of the reactions. They are just something that characterizes the whole process. It doesn’t matter in which frame you consider, so S and T are Mandelstam.

I learned recently that there is no word “invariance” in the dictionary—somehow I didn’t find it. When I say invariance, but I use it all the time, I mean invariant variants, environment, variables. Maybe it exists, but I was looking at the wrong dictionary. To me, it sounds very natural to call them invariants.

Anyway, give an example of these sets of variables. So everyone should get their favorite. S and T. No, not a single pair. These sets. So we need two variables to describe the process. What are your favorite two variables?

What about you, Sven? P_2 - P_4 is called U, and it’s also a variable. P_2 - P_4 is equal to P_1 - P_3. So P_1 - P_4 squared is called U. And this is the same as… If I do the algebra, you find out that there are only two independent. So your favorite set is S and U, and they’re equivalent because U is a linear combination of S and T.

Fine. So you guys sometimes take T and U. No, no, no. Let’s do different center-of-mass energy and angle. That’s a really good choice. That’s probably gonna be my favorite. And this is actually \sqrt{S}. Center-of-mass energy and then the angle. How do you define the angle? You go to the center of mass and then you make particle…

Well, center of mass is often referred to as center of momentum as well. So in the center-of-momentum frame, it’s just easier to call it center of momentum because you immediately understand that momentum, the total momentum, has to add to zero for all particles. And do you have a P_1 here and P_3? Then what do we have? What else do we have? How do you define the angle? They are going in this direction and the angle between P_1 and P_3. And then the angle between P_1 and P_3. So essentially this one. That’s my favorite.

Indeed. And well, one has to be careful when drawing that because we know that the length of the vectors indicates their momentum. Particle’s momentum. And for the final state, it’s in the center of momentum. We’re still in the same frame, so this has to be equal to that. I’m fine with that choice.

There are no more common choices. But any two variables work if they are not redundant. So we can choose E_1 in the lab frame, any frame, and then E_3 in the lab frame. That’s also fine. Any two variables, if they are independent, they characterize the kinematics. One has to be careful. Sometimes you fold your phase space. So by introducing a set of variables, you map your phase space in these variables or these variables to another domain. And sometimes this domain is somewhat smaller, it has a folded coverage. So this variable, it’s not a bijective transformation, but this is advanced to see.

We discussed two-body kinematics. We’re going to continue discussing the angular dependence after the Dalitz plot. Just let’s quickly look—oh, questions here. Shortly before we move, questions on the variables that characterize kinematics.

So the homework we had the exercise on the Dalitz plot, and this is, as I mentioned, a problem that is now entering our lectures. Part of the course is specific to hadrons. And it’s before where we’ve been overlapping a lot with particle physics. But now, from this lecture and the next couple of lectures, we will have material really specific to hadron spectroscopy and the approach that we use to discuss hadrons. And one of them, one of the subjects that we would like to go deeper into, is particle representation—essentially three-body decay. And Dalitz’s law is the common technique to indicate the dynamics of the particles, the dynamics of the interactions.

In the case of three-body decay, we deal with a similar diagram as before, but now one leg comes in, three legs go out, and what is inside the blob is an interaction. And we can pose all the same questions as before: What are the number of variables? But the answer won’t differ because it’s the same number of legs. Essentially, the same number of legs tells you the number of variables. For three-pion or DDK decays, there are two variables that describe the process completely. Once you give me these two variables—S and T, or angles, or any other—I should be able to draw the entire kinematics.

The Dalitz plot is a powerful tool for analyzing three-body decays, where the phase space is flat in S and T variables: \frac{dN}{dS\,dT} = \text{constant} This allows for clear visualization of dynamical effects.

So remember my analogy of the rigid body: when you print on a 3D printer a blob out of which the vectors are sticking, this is a rigid body that describes a kinematic point. The angles between all vectors are fixed. The lengths of the vectors are fixed. So you have the entire setup of the kinematics at a single point in phase space. The same goes for three-body decay. Just give me two variables, and I should be able to draw you how the decay looks like in the center of mass.

In that case, I have to draw here two vectors that leave. And in that case, I’m going to draw the three vectors. So essentially, this is the support out of which the vectors stick. The vectors determine the angles, the lengths of the vectors, and that’s it—that’s what you have.

Now, S and T are defined in a similar way. But now we have different particles in the final state. So, from that kinematics to this kinematics, what it takes is to take one leg and swap it to the other side. And it’s done by changing the sign on the momentum.

So, for three-body decay, what we do—let me define it here:

  • S = (p_1 + p_2)^2 = (p_3 + p_4)^2
  • T = (p_1 - p_3)^2

I noticed that there was a typo: p_1 is the particle that decays. In that case, I don’t know—should I change it? Okay, let me update my notation later on. For now, let me stick to the notations that relate to kinematics.

It’s important to realize that the phase space for three-body decay is flat in the variables—it’s actually constant in the variables p_3, p_4, p_2, p_4. In S and T, we write it as \frac{dN}{dX}.

5.2 Recursive Phase Space Formulas and Dalitz Plot Kinematics in Three-Body Decays

So this is the recursive formula that we discussed in the last lecture. It’s easy to use this equation to demonstrate that once you substitute the two phase spaces and apply the proper transformation, you end up with the factor \frac{1}{8\pi^2}. This comes from the two phase spaces: 2\pi for each, as you recall.

From the first phase space, we have \frac{1}{8\pi} \cdot \frac{2p}{m_1}. These are two-body phase spaces. From the other, we have this phase space multiplied by 2^3, and both have the same form: \frac{1}{8\pi} \cdot \frac{2p}{\sqrt{s}}.

What also appears is that every phase space has a \frac{1}{2\pi} \cdot \frac{1}{8\pi^2} \cdot \frac{2p}{\sqrt{s}}, and then \frac{d^3\omega}{4\pi}. This is easy to remember because it approaches \frac{1}{\pi} in the asymptotic limit. These terms approach unity at high energy, and that’s a unit integral if there is no dependence.

The two-body phase space factor is given by: d\Phi_2 = \frac{1}{8\pi^2} \cdot \frac{2p}{\sqrt{s}} \cdot \frac{d^3\omega}{4\pi} This formula captures the kinematic dependencies discussed in the lecture.

So that’s what we discussed already. For both phase spaces, you use this expression. For one of them, the cosine \theta is described in terms of the scattering angle. Essentially, you express m (say, m_{34}) in terms of the cosine of the scattering angle. These 2p terms appear here, and you must replace E' by p. This appears as a Jacobian, though I realize now I might have been incorrect in how I used it.

We arrived at this in a few lines without details, but we’ll revisit it later. It comes up in exercises and throughout the course when evaluating phase space.

For the three-body case, using the recursive formula shows that the Jacobian for the transformation is constant—the Jacobian for converting the three-body phase space into paired mass variables is constant. There is no extra dependency. This means when we examine the differential width or cross section against these variables, there’s no distortion due to the choice of variables.

We have several choices, but only Mandelstam invariance gives an undistorted representation of the density. If you plot differential widths against two variables (e.g., m_{34}^2 or m_{24}^2), it will be a constant numerical value, not rescaled.

That’s why representing three-body processes in Mandelstam variables (or linearly related ones) is so powerful—you directly see the interaction’s content in the “blob.” This is essentially what the Dalitz plot shows.

Consider the decay \Lambda_c \to p K^- \pi^+. We measure \Lambda_c produced in proton-proton collisions. Experiments like BES and Belle observe \Lambda_c abundantly because charm ground-state particles live long enough to travel from the primary vertex. We reconstruct them, giving us a good sample to study their decay kinematics and dynamics—the “blob’s” content.

In this decay, a charm quark in the initial state disappears, indicating weak interaction. The charm quark decays into a strange quark, which ends up in the kaon. The c \to s transition is allowed and unsuppressed. This is a golden channel for reconstruction because the final state has three charged particles (proton, kaon, pion), all stable and detectable.

The \Lambda_c travels about 10 mm in the lab frame due to its boost from production at the LHC. This decay is well-studied.

Here’s an experimental result resembling the data. The plot is so statistically precise that it appears smooth. The x-axis shows the proton-kaon mass, and the y-axis shows the kaon-pion mass. Colored regions represent allowed kinematics; white areas are forbidden by energy conservation.

Selecting a point inside the plot lets us compute particle angles and momenta. If a point is in the white region, energy conservation fails—no valid kinematics exist there. This constrained range defines the Dalitz plot.

Colors on the plot indicate decay probabilities. Certain kinematics are more probable because particles prefer specific directions. For example, one configuration (maximizing m_{pK}) is rarer than another.

To locate these points:

  • At the plot’s border, particles are collinear.
  • Maximizing m_{pK} requires momenta to oppose each other, placing the point at the bottom right.
  • Minimizing m_{K\pi} occurs when the kaon and pion move together, reducing their invariant mass.

In the K\pi rest frame, their relative momentum is small, so their mass nears the sum of their masses. This corresponds to the minimal m_{K\pi} point.

For three-body decays, we can fix one pair’s mass (e.g., m_{K\pi}) and vary the angle \theta between the third particle and the pair. In the K\pi rest frame, their momenta are back-to-back, and the proton’s angle \theta changes. At \theta = 0, m_{pK} is maximal; at \theta = \pi, it’s minimal.

Another variable, U, is a symmetric combination of the other two. Fixing m_{p\pi} defines a diagonal line on the plot.

The standard Dalitz plot uses m_{pK}^2 vs. m_{K\pi}^2, but a symmetric representation uses an equilateral triangle. Here, each mass is a distance to a side, maintaining symmetry. The transformation between these representations involves skewing due to the 60^\circ angles.

This kinematic representation reveals dynamics—where densities increase due to intermediate resonances. For example, two particles may briefly form a resonance, enhancing decay probability at specific energies. This explains peaks or enhanced regions in the Dalitz plot.

Looking ahead, we’ll explore how intermediate states shape these distributions.

The Dalitz plot variables (Mandelstam invariants) for three-body decay are: m_{ij}^2 = (p_i + p_j)^2 = (E_i + E_j)^2 - |\vec{p}_i + \vec{p}_j|^2 These variables are crucial for understanding the kinematic boundaries and resonance structures in the plot.

5.3 Resonances, Angular Distributions, and Spin in Particle Decays

You might have seen cross sections for the two-particle resonances. That has a bump in it known as the hadronic resonance. The physics of that is you have a system of two particles, and the quantum numbers of the system match the quantum numbers of some other resonance. By adjusting the energy of the system, you explore how the system behaves—how likely two particles are to interact at a certain energy.

The resonance cross-section follows the Breit-Wigner formula: \sigma(E) = \frac{4\pi}{k^2} \frac{\Gamma^2/4}{(E-E_0)^2 + \Gamma^2/4} where E_0 is the resonance energy and \Gamma is the width.

If there is an intermediate resonance, this system can resonate at this energy, and the probability is going to increase once you pass through the resonance region. This leads to the appearance of the bent structures in the Dalitz distribution, or if you project it onto one of the axes, you’re going to see the resonance-like shape. These bent structures you can identify on both.

This is a cool example that I brought because I like it a lot—there are resonances in all three pairs. There are resonances in these two, these two enter. Why is there a bigger probability increase when the proton and pion are near resonance? The K and π as well, like on the right side, but not only the two lines on the left side. Let’s understand the lines. We will come back to that in two lectures. But let’s quickly identify the different lines.

  • Horizontal lines—which resonances are these? In which mass distribution? This is going to give us a peak horizontally. I’m looking now at the blue plot. The K and π, Nπ. If the horizontal means mass for the Kπ is fixed, then it approaches the maximum at a certain value. So all the lines are the resonances in Kπ, Nπ—and what’s the name of these particles? What particle decays to Kπ? K-star, K-star. These are K-star resonances.

  • Vertical lines—in which system? When I’m changing the Kπ, I’m going to scan a certain energy of the system. In which system do I scan? You can think of this as also a projectile. These lines correspond to the fixed mass of the proton and Kπ. Therefore, by moving along the axis, I’m scanning Kπ. If you expand it, you see that the lines correspond to the resonances in Kπ, proton-Kπ. I’m pointing to the proton-K. The third combination is pion-proton, and these are delta resonances.

The other plot can help me as well because it’s nice to see that the lines are going to be parallel to the sides of the triangle. This line is the same as this line, the same as this line, and it’s parallel to one of the sides. Then I have my lambda resonances parallel to the other side. At this point, you better see the delta resonance, which is the line over here, parallel to the third side. It’s a bit more tricky to see the delta resonance here, but kind of.

One thing to finish and to move past this dull spot and proceed to our topic of today is the angular distribution. I’m going to fix—I’m going to discuss now the angular distribution for a decay within one band. Let me look at the phase space resonance here. As we discussed before, the kinematics: when I’m traversing the double plot and I’m traversing the phase space from one end to another, while keeping the mass of the combination fixed, I’m changing the angle. I’m exploring the different angles. This is precisely the kinematics.

Let me sit in the rest frame of the Kπ where this band is happening and traverse the phase space by changing this angle. This angle dependence indicates to me… Within the band, I can have inhomogeneity. Sometimes even if I’m within the band, one edge of the band has a different probability than the other. The particles like—or it’s even more common that particles prefer to be aligned and don’t like to be perpendicular. This is the less probable kinematics than this. That’s easy to believe, right? Preferences might happen.

This happens because particles have spin. This preference appears only because the intermediate resonance, in that case the K, is not a scalar particle—it has spin. The spin of particles causes the inhomogeneity in angular distributions and causes inhomogeneity in the decay spectrum.

Angular distribution is a very powerful tool to understand properties of particles. As we already discussed, that’s our way to measure spin, parity, and other quantum numbers in particle interactions. Particles with higher spin would prefer more bumpy, more spiky angular distributions, and particles with lower spin—if everything is scalar—are going to produce no asymmetries at all, no structures in angular distributions.

The angular distribution intensity for spin measurement follows: I(\theta) \propto \sum_{M,M'} |D^J_{M'M}(\theta)|^2 where D^J_{M'M} are Wigner D-functions describing the rotation of spin states.

By looking at the angular distribution, especially in the rest frame of the particle decay, one examines the ratio of the aligned kinematics and other types of kinematics. With this, one can infer the information of the spin. For most of the particles that we have discovered up to now, the quantum numbers are not known. We discover particles as bumps in the spectrum, and then the next step to understand their properties is to determine their quantum numbers. This is done by looking at angular distributions.

Most of the time, it’s as simple as looking at the Dalitz plot and seeing if there is a minimum in the angular distribution, if this line has several structures, several nodes. The maxima and nodes indicate, in the case of scalar particles in the final state, the nodes would literally tell you the spin. If you have one node, you have spin one. If you have two nodes, you have spin two. If you have three nodes, you have spin three. The intensity would really vanish at certain points in the dark spot.

But in the case of non-scalar particles—and most of the time, it’s non-scalar particles that interact—the situation is a little more complicated. I will have examples of scalar resonances. But here, let’s just quickly check what spin they are giving here. Is the baryon spin half? Half, right? Kaon and pion—except the scalars—they have a spin of zero. The spin of lambda is the same as the proton, but with the jump, the presence of spin averages out the distribution.

If you consider a certain spin projection of the lambda-C and proton, you’ll again have nodes and zeros in the angular distribution. But since we don’t measure—we don’t polarize the initial state lambda—and we don’t measure the spin of the final state, everything is averaged. Therefore, you don’t have minima, you don’t have nodes or zeros any longer; things get smeared.

A particle with spin can have 2J + 1 projections to the quantization axis. Let me consider a particle that has spin J. There is a z-axis that we need to quantize the spin, and this z-axis measures the J_z upper layer is going to give us. One can think of this |JM\rangle state as the vector of 2J + 1 components. All operators in this case are going to be matrices that act on these vectors and produce either the same state with a certain eigenvalue or a mixture of the states.

When I act with the rotation on the state, I’m going to produce not a certain state, but a mixture of different states—it’s as simple as that. When I rotate the state, if it were a vector in regular space, I could probably adjust this to have exactly a certain projection. But in quantum mechanics, it doesn’t work this way. Once you act with the rotation, most of the time you’re going to end up with a mixture of all states. Let me rotate. These coefficients are tabulated, so they are known functions for any state. You can have a look and check what these coefficients are. They are called Wigner D-functions.

The Wigner D-functions describe rotations of spin states: D^J_{M'M}(\alpha,\beta,\gamma) = e^{-iM'\alpha}d^J_{M'M}(\beta)e^{-iM\gamma} where (\alpha,\beta,\gamma) are Euler angles and d^J_{M'M} is the reduced rotation matrix.

Let me be more concrete. I’m going to rotate about the Y-axis in a second. So I have a Z, I have my X, and I have my Y. XYZ. That’s correct. They should form the right-handed triple of axes X, Y, Z. I’m going to rotate about the Y-axis on |JM\rangle. In order to do that, I have to—so J_+, as you remember, it’s \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}. So it operates like that. And J_- is the opposite \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} like that. The J_y operator then has zero on the diagonal and these coefficients above and below the diagonal.

Then, in order to apply the transformation to the state, you have to do the matrix exponential of J_y. But this is done for you, and we know what they are. These are these coefficients, these coefficients where they are C, C, C, C. These coefficients—they need to know about the initial state as well. That’s why JM indices are kept in these notations. Imagine I tell you we work with spin-1 that is perpendicular to the Y-axis and rotated by 30 degrees. What is going to be the composition of the state? When I do that, you have everything |JM\rangle that corresponds to this perpendicular—I mean, to the orientation along the Y-axis. Then you look up these functions, and you know what the composition is.

Well, we have the same functions. These are coefficients so far, right? So these are numbers. But what is R in this case? Is it just θ? Okay, it’s more general. You can compute the D-functions for any R. Essentially, the more general relation would be like this: any orientation in space can be described by three angles. These are Euler angles. In the conventions we use in particle physics, you first rotate by φ about the Z-axis, then you rotate by θ about the Y-axis, and then you rotate by γ again about the Z-axis. With this, you can give any orientation to your system.

These regular D-functions are the product of three parts. The most difficult part is the θ rotation. Since the Z-rotation gives you just a phase. Essentially, just to relate it to \langle M'|M\rangle of γ, θ, φ is equal to e^{-i γ M'} d^l_{M'M}(θ) e^{-i φ M}.

Think we need an example—a very small one—of spin one-half. I’m going to rotate the state |1/2, 1/2\rangle by 30 degrees about the Y axis. I’m going to get a |1/2, 1/2\rangle plus |1/2, -1/2\rangle combination, and I want you to quickly tell me what are the numbers here. These coefficients are sitting in the same place—where sitting in the same page as the Clebsch-Gordan coefficients. They are closely related. This is about the SU(2) group, and you just open the Clebsch-Gordan table and check the numbers.

Since 30 degrees is not at all trivial to do it, but now we won’t have time to look at details, and hopefully in the seminar we will explore a little bit more. But I just remember this table because it’s super simple. So the D for m' = m = +1/2 is the cosine minus sine, sine cosine of the half angle. I should have picked 60 degrees because for 15 degrees it’s a little bit inconvenient. Then what is going to be here is the cosine of 15 degrees. What’s going to be here is the sine of 15 degrees.

Let me stress what we discussed so far was not about weak interactions or strong interactions. It was about rotations and the rotational group. That’s the fun part—and that’s something that still impresses me—that in order to understand how particles behave and what the angular distributions are, you need little from the strong interaction. You only need the general properties of the rotational group, so to speak.

Angular distributions are determined by general properties of how space is rotated, plus the little bit we need from strong interactions: what is the preference for which spin particles are produced, and that’s what strong interactions tell us. But how they decay and what is the asymmetry in the kinematics—this is determined by the quantum group. That’s amazing. Therefore, we can now move on and have a recipe, or a general way, to construct any particle decay chain and figure out what the angular distribution is going to be.

5.4 Spin Rotation and Decay Amplitude Simplification for Particle X

Let’s now look at the particle X. The particle X carries the spin projection \lambda_X. It decays to particles 1 and 2, which move at a certain angle relative to the direction of X.

To compensate for this angle, we must adjust the quantization axis. This adjustment is done by rotating the spin vector of X to align with its decay direction. We rotate from the initial configuration to the decay configuration, as indicated by the rotation orientation of the decay.

The Wigner D-matrix is used to rotate spin states: D^{J}_{\lambda,\lambda'}(\alpha,\beta,\gamma) where J is the total angular momentum, \lambda,\lambda' are spin projections, and (\alpha,\beta,\gamma) are Euler angles.

I would like to evaluate this expression in the aligned kinematics, where \phi = \theta = 0. In the center-of-mass (CM) frame, the expression simplifies.

When the angles are zero, the transformation reduces significantly. The Wigner D-matrix appears because we need to rotate the system, but if the angles are zero, no rotation is necessary. This simplifies the summation over \lambda_X.

Since X is already moving along the z-axis, no rotation is needed. The expression becomes: \sum_{\lambda_X, \lambda_3} D^{J_X}_{0, \lambda_X - \lambda_3}(0, 0, 0) and another term involving \lambda_1, \lambda_2: D^{J_X}_{\lambda_X, \lambda_1 - \lambda_2}(\theta, \phi)

This results in a delta function constraint due to angular momentum conservation: \delta_{\lambda_0, \lambda_X - \lambda_3} The only way to avoid rotation is if the state remains unchanged, meaning \lambda_X is constrained.

The final expression is: \mathcal{H}_0 \, D^{J_X}_{\lambda_X, \lambda_3}(\theta, \phi) where \lambda_X = \lambda_0 + \lambda_3.

The complete decay amplitude is: \mathcal{H}_0 \, D^{J_X}_{\lambda_0 + \lambda_3, \lambda_3}(\theta, \phi) \, D^{J_X}_{\lambda_0 + \lambda_3, \lambda_1 - \lambda_2}(\theta, \phi)

That’s it—as simple as that.

  1. Horizontal rules (---) to separate logical blocks.
  2. Bold emphasis for critical terms like 1, 2, and aligned kinematics.
  3. Callout block for the Wigner D-matrix definition, placed contextually where rotation is first discussed.
  4. Block math (`...) for all major equations, properly spaced.
  5. Inline math (...`) for variables like \lambda,\lambda' inside the callout.

No sentences were removed or altered, and the helping material was integrated only where directly relevant.

5.5 Predicting Angular Distributions from Quantum Amplitudes and Partial Wave Analysis

How many numbers do I need from you in order to compute? Let me think about electromagnetic interactions, or gravity. How many numbers as input do I need to predict the angular distribution? It’s essentially here, but it misses fundamental components.

What’s inside the blobs? What is inside of the blobs? What’s inside of this block, this blob, or that blob? To predict all the values, I just need these two quantities.

  • I have (2j_1 + 1) \times (2j_2 + 1) values here, which might depend on particle masses—say, the masses of x.
  • I also need a similar number for these terms, but there’s a reasonable way to approximate them.

The spin multiplicity factor (2j_1 + 1)(2j_2 + 1) accounts for the number of possible spin states for particles with spins j_1 and j_2.

Often, in experiments or initial analyses, we assume these are constant, containing only particle properties. Here, I’ll say this one is a constant c, and this is particle-dependent. Once I do that, I should be able to compute the angular distribution.

In that case, I fix the mass of one particle and the intensity observed along the line. What we have so far is that these two quantities determine \frac{d\Gamma}{d(\cos\theta)}.

Using \cos\theta is better—it has a simpler Jacobian. We avoid the \sin\theta Jacobian for \cos\theta. That’s why we often analyze \cos\theta, and this distribution is proportional to the matrix element squared |M|^2.

This |M|^2 is fixed. The distribution ranges from -1 to 1. Here, \theta and \cos\theta are plotted: \theta = 0 corresponds to \cos\theta = -1. We scan from -1 to 1, and if the distribution is flat, that’s one possibility.

For particles with spin, you often see:

  • A parabola—a second-order polynomial in \cos\theta.
  • Or you might see this shape.

The angular distribution \frac{d\Gamma}{d(\cos\theta)} \propto |\mathcal{M}|^2 is directly tied to the quantum transition amplitude A, where |\mathcal{M}|^2 = |A|^2. Experiments only measure |A|^2.

Notice the difference. It’s important to remember that A, the quantum transition amplitude, is a probability amplitude. It’s squared to give the observed probability. This G will appear squared. In experiments, we only measure |A|^2.

Moreover, we often deal with unpolarized decays, so the distributions are averaged. You square the amplitude, sum over initial and final spin projections, and obtain what’s observed experimentally.

Then you ask: What does this tell me?

The first step in analysis isn’t to guess the amplitude but to project the angular distributions onto orthogonal polynomials. This gives a nice basis for expansion.

Functions from -1 to 1 can be expanded in Legendre polynomials P_\ell(\cos\theta). These polynomials are related to the spin of the produced particle.

This is called partial wave analysis. If you project the differential cross section, it’s called moment analysis.

\frac{d\sigma}{d(\cos\theta)} = \sum_{\ell} a_\ell P_\ell(\cos\theta)

Partial wave analysis models the cross section with amplitudes, treating these terms as free parameters. The goal is to infer what’s inside the blobs by fitting them to data.

5.6 Helicity vs. Canonical States, Spearman’s Book Reference, and Dalitz Plot Exercise

I didn’t tell you much about the differences between the canonical state that we introduced at the beginning and the helicity state that we introduced later. We only touched a little bit on how the state is defined in the rest frame, and hopefully we will explore more.

The helicity operator is defined as: \hat{h} = \frac{\mathbf{J} \cdot \mathbf{p}}{|\mathbf{p}|} where \mathbf{J} is the total angular momentum operator and \mathbf{p} is the momentum.

I would like to tell you that this book has the best coverage of this subject. This is Martin Spearman’s Elementary Particle Theory, and Chapter Four is fundamental. It’s really fun reading because:

  • It starts from the Lorentz group.
  • It tells you how to introduce the vectors, a little bit of group theory, but in a nice way without heavy details, without too much mass.

It’s really a good book. So Chapter Four of Martin Spearman’s Elementary Particle Physics would give you some insights on the particle definitions.

I left you an exercise. There are some Dalitz plots from CLEO and BaBar, and I had the labels removed. I don’t tell you which particles are in the final state. I just tell you that:

  • One of them is the D decay.
  • Another one is the D-sub-S decay.

And you know a lot about kinematics already.

For Dalitz plot kinematics, the invariant masses of decay products satisfy: m_{12}^2 + m_{23}^2 + m_{31}^2 = m_D^2 + m_1^2 + m_2^2 + m_3^2 where m_{ij}^2 are the invariant masses of two decay products, and m_D is the mass of the decaying D-meson.

The exercise is to figure out what decay this is. The axis labels are still there, but you just do not know what the mass is. From the kinematics, you can figure out the masses and perhaps guess what the decay is. I have cases. I would give you one for your group, and then you’d get another one, and then the EP one.