7.1 Mass and Quantum Numbers of the PCC Hadronic Molecule
All right, two minutes. All of the problems are one-liners. You solved them online, but you have to know what to do.
Let’s discuss quickly question number one. First, the mass of the states. This appears as the particle, right? In this condition, neglecting inelastic channels, this particle is formed by the cascade C and D. This particle can travel, can leave. It is a regular particle, but its internal structure is the hadronic molecule.
What is the mass of this particle? It is the sum. Indeed, the mass of the PCC particle—that’s how we call them—is roughly the mass of the constituents:
M_{\text{PCC}} = M_C + M_D.
That’s the way we define the binding energy:
E_{\text{binding}} = (M_C + M_D) - M_{\text{PCC}}.
Binding Energy Clarification: The binding energy represents the difference between the sum of the constituent masses (M_C + M_D) and the mass of the bound state (M_{\text{PCC}}). A small binding energy implies the state sits near the threshold.
Shouldn’t it be PCC? No, it’s \overline{PCC}. Very good. It used to be called PC, and last year we updated the notations. Now it’s PCC; before it was PC because, with the C, you indicate that it contains the charm. This is better—it’s really clear.
As indices, you list heavy quarks because they kind of conserve quantities with the decay. Light quarks are… we don’t miss them. You can always recover them from the capital letter. If I say \Sigma, you know, that’s a capital. The two light quarks, depending on the charge, are either UU, DD, or UD. Then SAP means pentaquark—5 quarks. SACC is CC plus 3 others. For PCC, this is UD.
Likely, the PCC delta logs we see in this particular channel are of this type—mass of the… This was easy. Neglecting binding energy, you really get the mass of two particles; that’s called the threshold. The language we use is saying that the state is… Since the binding energy is small, the state sits at the threshold:
M_{\text{PCC}} \approx M_C + M_D.
The hadronic threshold is there. How much is it? Approximately like 10 MeV, 100 MeV. What’s the scale? 10. Less than 10. More than 1. The mass of the \Sigma is 2.5 or so; the mass of the D is around 2 GeV. The sum is 4.5 GeV. That’s roughly where one would search for such formations.
Now, quantum numbers, spin algebra—my favorite exercise in this course. We have \frac{1}{2} and \frac{3}{2} instructions. Parity should be - - because we have s-wave. Exactly. Very good. So you expect two states:
J^P = \left(\frac{1}{2}^-\right) \quad \text{or} \quad \left(\frac{3}{2}^-\right).
They have a cool lower bound on lifetime. Where does the lower bound really come from? It’s the binding of the two constituents. Any idea why this state would ever decay and how it would decay actually? When we talk about lifetime, the particle decays, right? That’s what happens with all particles.
How does this particle decay? I would love to say no, but… The energy of the system is conserved. Once you split it into two, the minimal energy in the system is their mass—M_C + M_D. This is already the energy. If they fly apart, the energy is even higher. This is not allowed—energy must be conserved.
We sit in the rest frame of this particle, and all the energy you have is the mass of this particle. That’s it.
Energy Conservation Constraint: The decay into free constituents is forbidden if the binding energy is positive (M_{\text{PCC}} < M_C + M_D), as the system lacks sufficient energy to separate.
7.2 Charm Quark Annihilation, Decay Channels, and Bound State Lifetimes in Heavy Mesons
Because the c and the \bar{c} undergo electromagnetic processes. Annihilation of these two quarks is possible, but this is a rare process. Since it’s an electromagnetic interaction, the probability of this process is much smaller than the process I would like you to consider.
Why have c \bar{c}? Can they annihilate via the strong interaction? No—even if they have different color, they can produce other quarks. This is actually suppressed for heavy particles. The heavier the particles, the smaller the chance for them to annihilate. So this is also strongly suppressed.
The three clones appear only for the color-neutral objects. The three clones we need to construct the color-neutral object here are living in the color octet. It’s not singular, so they don’t. The strict vertex is a different suppression.
The big decay—that’s what happens with this object. If we neglect inelastic channels, the decay of this particle is one of the c quarks transitioning to s, emitting the W: c \rightarrow s + W^+ This is a big decay.
If the structure of the object is like that, we can even understand the probability of this decay happening. Big decay is one thing, but there is another transition. This object decays weakly, so it will decay quickly. We can compute the lifetime due to that.
But there is another process. This state is not the ground state. In the spectrum, we combine quarks 1, 2, and \frac{1}{2}. We have 1 and 0. The lower one is called D, and the upper one is called D^*. The transition D^* \rightarrow D is possible by emitting \pi^0 or \gamma: D^* \rightarrow D + \pi^0 \quad \text{or} \quad D^* \rightarrow D + \gamma
The most likely decay channel is the radiative decay of the D. The width is smaller than the width of the D, so the lifetime is bigger than the lifetime of the D^*. That’s a bound.
You combine two objects, and the object lives as long as the constituents—it would live infinitely if these particles are stable. Like the deuteron is made of a proton and neutron. The proton is stable, and the neutron lives long. The deuteron is stable because the neutron is so bound inside that it cannot decay.
The binding mechanism suppresses the phase space of the decay and extends the lifetime of the particle. The D meson here would live longer than the isolated D^* meson. The width of the state is smaller than the width of the D, and since the lifetime is the inverse of the width: \tau = \frac{1}{\Gamma} The lifetime of this object is bigger than the lifetime of the D.
Will the \Sigma_c^* also survive or be an atomic molecule? It just goes down an energy step instead of falling apart.
Let’s try to understand your question. You asked about the cascade \Sigma_c^* and the \Sigma_c + D^0 molecule. Does this exist? It does. So the molecule can make steps down one energy step—but it can also dissociate into \Sigma_D + \pi or \Sigma_D + \gamma, because the energy for this is a lot: \Sigma_c + D^0 \rightarrow \Sigma_D + \pi \quad \text{or} \quad \Sigma_D + \gamma
The binding energy and the situation of the decays are actively discussed in the field. At every conference, you see roughly five talks on this—it’s a really hot subject.
7.3 Invariant Variables and Kinematics in Four-Legged Processes
Let’s move to item number two. Understand? The plane is the plane of invariance.
For any reaction with a block—an open direction centered on four legs going out (X, A, B, C)—you can define invariant variables that characterize the kinematics.
The first variable is the mass squared of A and B: S = (p_A + p_B)^2 The mass of A and B is the square root of S, so \sqrt{S} is the mass of the system of particles A and B.
Then: T = (p_B + p_C)^2 is the mass squared of the system with particles B and C.
And: U = (p_A + p_C)^2 is the mass squared of the system with A and C.
We figured out earlier that two variables are enough to characterize the kinematics of the four-legged process. This comes from counting degrees of freedom.
The two variables could be:
- S and T
- S and U
- or any pair.
On the plane, they are connected because the sum of the three invariants satisfies the Mandelstam relation: S + T + U = m_A^2 + m_B^2 + m_C^2 + m_X^2
The scattering amplitude (or transition amplitude) depends on only two variables—S and T, S and U, or any pair.
This amplitude is defined as a function on the plane of invariants, and different domains on this plane describe different processes.
The Mandelstam relation shows the interdependence of S, T, and U, constraining the scattering amplitude to depend on only two independent variables.
Before proceeding, let’s focus on the kinematics and see where A, B, C, and D are located.
The way I want to solve this is by checking the signs:
- For each process, is S greater or less than zero?
- What about T?
- And U?
By identifying these, we can place the kinematics.
Let’s consider process A.
- Is S positive? Yes. Why? Because S is the mass squared of two real particles, and this mass is at least the sum of their masses: S \geq (m_A + m_B)^2 In the rest frame, the energy of the system is greater than the sum of the masses if they have momentum.
For A:
- S > (m_A + m_B)^2
- T \geq (m_B + m_C)^2
- U \geq (m_A + m_C)^2
The domain for the first kinematics is where S, T, and U are all greater than their respective thresholds.
For process B:
- T is the physical mass squared of two particles, so T > 0 and T \geq (m_B + m_C)^2.
To discuss the kinematics of cross-channel reactions (like A, M, D, E particles), we can swap momenta.
For particles moving to the other side, we modify the definition of the invariants by flipping the sign of their momenta.
This is what Ares mentioned—you could redefine the invariants accordingly.
7.4 Kinematic Domains, Dalitz Plots, and the Kibble Function in Scattering Amplitudes
Two different directions, and then the sum minus S would appear in the first place there. We could start with this direction. The place for this is when T is more than zero, because T is the physical mass of two particles. And then S is under A and B. This is kind of an impossible combination. They are on different sides, and the variable S is going to be less than zero, and then U is less than zero. So it’s another impossible combination, and the domain, roughly speaking, is where U is below zero. And then S is above, T is above zero.
Here’s T. So T is this direction. And then S is above zero. This and U is above zero. This color has, so here this is the minimum. So this one is B, and same on the C. And then another one, D, where is D? A + C, X + B. A and C give us the U-channel. So this is the positive. U is positive.
The Kibble/Klein function describing physical domain contours: \Phi(s,t,u) = s^2 + t^2 + u^2 - 2st - 2tu - 2us This single function can generate all boundary lines when solved for \Phi(S, T, U) = 0.
And now to see real physical contours for the kinematics. One has to calculate what are the physical ranges of scattering variables. Like if you work in the cosine of the scattering angle, by placing the restriction that cosine must be from -1 to 1, you identify that the border is such that not all points here are allowed, but only a certain region. And then the true border for the scattering is often given by this sixth-order polynomial.
In the center is this canvas. This is our Dalitz plot. This is here. And then another direction could be here. So 1, 2, 3, 4. Four regions for physical reactions that are happening. It’s important to realize that connection because there’s just a single amplitude matrix element that describes all four. So if you get this function and constrain it precisely, it describes all four. Just define it on this domain, and for every point you can compute, it gives you the complex number 1 + 3i. And then this is the value of the quantum transition amplitude. The matrix element you can compute at a different place. Now we have a complex transition amplitude for the decay. I think that’s super cool.
It works very well in QED. When you consider Compton scattering, photon-electron going to photon-electron, this is a crossed process. And it’s exactly the same matrix element that describes the electron-positron annihilation to two photons. And it’s exactly the same process that describes two-photon production of an electron-positron pair in quantum electrodynamics.
In hadron physics, it’s a little bit more complicated. Since we have perturbation theory, we always model. So what we are going to do when we describe the Dalitz plot, we model it in terms of the chunk of the resonances. And then you find that once you want to compute this function right here, the scattering amplitude, it blows up. It has infinities, so it has unphysical behavior. The reason is because we employ a finite range of finite number of resonances, and the physics is more complicated analytically.
The fact that this amplitude is related to that domain tells you already that you have to put an infinite number of resonances. Like we saw the lines on the dispersion, these are all resonances. And in order to relate this in this domain, you have to operate with the infinite sums, and then in infinite terms they would compensate each other and you have a reasonable amplitude.
There has been an effort for now, 30 years, 50 years, of finding such a set of functions that works everywhere nicely and is analytic and reasonable. What’s most difficult is to come up with something that describes the data, because there is Regge theory. You might have heard of Regge theory, and this is one of the approaches. So you come up with a complex function that has all the nice properties—it works reasonably on scattering domains and it works in the decay domain. However, of course it lacks exact understanding of the resonance properties. Like it would put resonances there that have zero width in Regge theory. And one of the interesting developments is to implement resonances with widths.
One more thing. If you like playing with functions, there is a very simple expression that describes contours of the physical domain. This is called the Kibble function. I think it’s worth giving. There is just one function. If you solve \Phi(S, T, U) = 0, you find this line, that line, that line, and that line. Just one function—put it into a solver, and for any given value of S… So let’s fix S. Where is our S? Let’s fix S to be -50. We’re going to get two solutions. One of them… Let’s put S = +20. Then you have two solutions using this.
This function is really easy to type into code. This is the Klein function or Kibble function. It’s almost a complete polynomial, but it misses more terms of this table. So it’s:
\Phi(s, t, u) = s^2 + t^2 + u^2 - 2st - 2tu - 2us
This is called the Kibble or Klein function.
The Kibble function is the combination of three Kibble functions, where every \Lambda_1, \Lambda_2, and \Lambda_3 corresponds to a different channel. So \Lambda_2 is then for B—it’s \Lambda(T, m_0^2, m_X^2). And here is the… what’s missing? Mass and T. And then \Lambda_3 is \Lambda(U, m_X^2, m_B^2).
So then you go to Wolfram Alpha and say, “Give me a contour plot.” And then you form them and you find all this function and discussion on that in the Pikkuharju-Kalenti book. Have we discussed this book? The best book on particle kinematics is written by two authors with unspellable family names. Unspellable. And this is—if I got it right—see Bikulov-Janti. No, cycling? No. Yes, correct me. But the book is excellent. So if you get a chance to get yourself this, you know everything about particle kinematics, particularly very peculiar properties of the Kibble function and the Challenger function.
7.5 Unitarity as a Constraint on Scattering Amplitudes
We are ready to move to the topic of today: unitarity. Unitarity is a constraint on the scattering amplitude. In heavy physics, we don’t compute the scattering amplitude from first principles but rather model it. Our guiding principles are what this amplitude can be and what it should be.
You cannot just write down an arbitrary expression for the amplitude that fits the data. There are some principles, and one of them is probability conservation. Probability conservation translates into the mathematical statement on the amplitude known as unitarity.
You might have seen earlier in the particle scattering course the optical theorem, which relates the imaginary part of the amplitude to the total cross section. It essentially states that the imaginary part is equal to the coefficient of the total cross section:
\text{Im}\, \mathcal{M}(s, t=0) = 2k\sqrt{s}\, \sigma_{\text{tot}}(s),
where:
- \mathcal{M} is the scattering amplitude,
- s is the Mandelstam variable,
- k is the momentum,
- \sigma_{\text{tot}} is the total cross section.
This is a consequence of unitarity—a consequence of probability conservation. It is so important, this very powerful statement, that just this principle alone allows you to derive a decent low-energy scattering amplitude that describes all of the resonance phenomena.
You see the bump in the spectrum due to probability conservation. You know what expression you should take to describe this phenomenon. Moreover, since unitarity is about the analytic properties of the amplitude, it tells you what the analytic structure of the amplitude is.
In the last lecture, we briefly discussed analytic functions and what singularities they might have, like cuts or branch points. Unitarity also tells you the location of the singularities in the complex plane. This comes from the fact that we are dealing with real unitary functions—functions for which the imaginary part determines the locations of the cuts.
Unitarity condition for the S-matrix: S^\dagger S = I where S is the scattering matrix and I is the identity matrix.
Partial wave expansion and unitarity (for elastic scattering): \text{Im}\, f_l(s) = k |f_l(s)|^2 where f_l(s) is the partial wave amplitude for angular momentum l.
Branch cut structure due to unitarity: \text{Disc}\, \mathcal{M}(s) = 2i\, \text{Im}\, \mathcal{M}(s) where \text{Disc}\, \mathcal{M}(s) is the discontinuity across the branch cut.
7.6 Unitarity Constraints on Scattering Amplitudes and Partial Wave Expansion
Let’s get there step by step, starting with the scattering amplitude and partial waves. For the sake of time, let me first state where we arrive, and then we can derive it. We will derive the following three equations, all dealing with scattering amplitudes.
The first equation describes how unitarity acts on the full amplitude. The scattering amplitude A describes a process, and unitarity tells us:
A - A^* = i \int A^* A \, d\Phi,
where d\Phi is the phase space of the intermediate state. Diagrammatically, this represents the relationship between the amplitude and its complex conjugate.
For partial waves, the amplitude A becomes a function of a single variable. We can simplify the phase space and find:
A - A^* = i \rho(s) |A|^2,
where \rho(s) is the phase space factor. This tells us that the imaginary part of the amplitude is related to the squared amplitude.
Now, let’s derive these expressions. We start with the scattering amplitude for a 2-to-2 elastic scattering process. On the board, I draw a diagram where a blob represents the interaction between two particles. Below it is the kinematic representation in the center-of-mass frame, where the total momentum is zero.
In the center-of-mass frame, the momenta of the particles are equal, even if their masses differ. The momentum is given by:
p = \frac{\sqrt{\lambda(s, m_1^2, m_2^2)}}{2\sqrt{s}},
where \lambda is the Källén function. After the interaction, the particles have the same masses and momenta, but the scattering angle changes.
The observable for these interactions is the angular distribution. For 2-to-2 scattering, once the energy is fixed, the only variable is the scattering angle \theta. The amplitude depends on \theta and the energy s.
The scattering amplitude is defined as the matrix element:
\langle p_1', p_2' | p_1, p_2 \rangle,
with energy-momentum conservation enforced by delta functions. The phase space for two-body scattering is:
d\Phi_2 = \frac{1}{(2\pi)^2} \frac{p}{4\sqrt{s}} \, d\cos\theta \, d\phi.
The identity operator integrates over all possible angles, ensuring normalization. The two-body phase space covers all directions on the sphere, representing the 4π solid angle.
Unitarity is a statement about the scattering operator. We separate the trivial part (identity) from the interaction part T, defining the scattering amplitude. The unitary condition for T is:
T - T^\dagger = i T^\dagger T.
This leads to the general unitarity relation:
A - A^* = i \int A^* A \, d\Phi.
For partial waves, we expand the amplitude in Legendre polynomials:
A(s, \theta) = \sum_{j=0}^\infty (2j + 1) A_j(s) P_j(\cos\theta).
The partial wave amplitudes A_j(s) are constrained by unitarity:
\text{Im}\, A_j(s) = \rho(s) |A_j(s)|^2.
This expansion converges well for strong interactions, as high-j contributions are suppressed by the size of hadrons. In experiments, only a few partial waves are needed to describe the data—often fewer than ten. This makes the partial wave expansion a powerful and practical tool.
7.7 Properties and Independence of Partial Wave Amplitudes
Another thing that appears here is the partial wave amplitude. This amplitude is a function of a single variable (k, the momentum). For every partial wave, there is one function of one variable. Each partial wave has fixed quantum numbers (l, the angular momentum quantum number).
The partial wave amplitude for angular momentum l is given by: f_l(k) = \frac{e^{i\delta_l(k)} \sin \delta_l(k)}{k}, where \delta_l(k) is the phase shift for partial wave l.
A large advantage of partial waves is that they do not interact with each other. They don’t influence each other. Every partial wave is independent.
Since quantum numbers in scattering are conserved, partial waves in the initial state are only related to partial waves in the final state with the same quantum numbers. In the unitarity constraint, you will see that it’s actually a single partial wave that relates to its output.
The unitarity constraint shows their independence: |S_l(k)|^2 = 1 \quad \text{or} \quad S_l(k) = e^{2i\delta_l(k)}, where S_l(k) is the S-matrix element for partial wave l.
There is no overlap between waves. They don’t mix.
The partial wave amplitude for angular momentum l is given by: f_l(k) = \frac{e^{i\delta_l(k)} \sin \delta_l(k)}{k}, where f_l(k) is the scattering amplitude, k is the momentum, and \delta_l(k) is the phase shift.
The unitarity constraint shows their independence: |S_l(k)|^2 = 1 \quad \text{or} \quad S_l(k) = e^{2i\delta_l(k)}.
The cross section in terms of partial waves is: \sigma = \frac{4\pi}{k^2} \sum_{l=0}^\infty (2l + 1) \sin^2 \delta_l(k).
These formulas highlight how partial waves are functions of a single variable (k), have fixed quantum numbers (l), and remain independent under unitarity constraints.
7.8 Resonance Phenomenon and Partial Wave Expansion in Scattering Amplitudes
And the way we can proceed is to insert this here and then simplify the phase space. For this, you will need one magic formula, and I don’t have time to derive it. It’s actually present in many references. I will send you the details.
But there is a really cool relation that I would like you to see: when we insert it there, the initial state is like this. It has the angle zero. That means its state is like this. Then, the final state is like that.
For the final state, the Legendre polynomial P_J is a function of \theta or the cosine. \theta is the same as the big Wigner D-function for D^J_{00}(\cos\theta). Write it first differently.
We know the Wigner D-function of J of P vector is still important in Q vector, and it’s equal to D(P) \times D(Q). So, D^J_{0\lambda}—such a cool and powerful expression.
Here, the capital G where \lambda is 00 is equal to e^{-i\theta}. It’s just the definition of the capital G function: the first argument is the phase, the second (or last) argument is the phase, but here it’s zero.
The Wigner D-function relation shows how angular momentum states transform: D^J_{0\lambda}(\theta) = P_J(\cos\theta) where P_J is the Legendre polynomial.
And then the indices—for this, what happens is that you have to integrate over all possible intermediate steps, and both of the amplitudes are expanded in the partial waves. The way to do this is to expand the cosine between the first and second and the last into the composition of intermediate steps. And that’s exactly the functions we need to relate this amplitude to the partial wave expansion.
But the expression we arrived at in partial wave theory—well, you are not surprised by the numerical coefficients because it’s simply the base case. So, what’s done now is to divide the right part by… multiply both parts with \frac{1}{a_J}, so this vanishes. Then this one is \frac{1}{a_J^*}. And then the minus N comes from that.
And then the imaginary part of the partial amplitude is just i^{-L}. It’s amazing because it tells us exactly how the inverse amplitude—how the imaginary part of the inverse amplitude—looks like.
So, let’s maybe plot it. Here’s s. Here is the imaginary part of a_J and the phase space function.
The phase space factor we have here is the function that starts from threshold and approaches \frac{1}{16\pi} at infinity. At threshold, it has a square root behavior: \sqrt{s - (M_1 + M_2)^2}. So it starts as a square root but then approaches a constant, \frac{1}{16\pi}, or \frac{1}{8\pi} \frac{\lambda^{1/2}}{s}.
Since P is a channel function, then this is… and this is something that goes to one at high energy.
And now, I think I managed to tell you the last thing: the modeling. We know the imaginary part. We do not know the real part. The real part of the amplitude is something that is genuine to the interaction.
What we have done so far is the same for:
- Electromagnetic interactions
- Strong interactions
- Gravity interactions—whatever interactions.
Any interaction that has unitarity must have an imaginary part related to the phase space. But the real part of the amplitude is something you have to compute or model.
People working on theory extract the real part of the amplitude by analyzing the lattice data. People working in experiments extract the real part by analyzing the harmonic data.
And the simplest model—or actually, what we have to write—is the inverse amplitude is equal to… some real function that we do a… Is it legal? Is this correct? Yes, any real function.
And then, one of the options for this real function is the pole—one of the simplest approximations. So this is a real function. We take it.
The amplitude computed assuming the real part is a single pole is called the Breit-Wigner amplitude, and this is a relativistic Breit-Wigner amplitude.
Now, I’m looking… so let’s do K-matrix poles—the amplitude that… Actually, modeling the amplitude in terms of pole terms is often called the K-matrix approach.
Let me plot this unperturbed amplitude. Here, I’m going to have the absolute value of A on the Y-axis and L. And M is a parameter that I put in. So these are both real numbers.
It gives me the meaning of the least real numbers that I put in my real function. The numerator, the g, determines how broad the peak is going to appear. And then M is often called the bare mass—the mass at which the peak appears.
And that’s the simplest resonance amplitude I can come up with. So this describes a resonant phenomenon, describes the phenomenon where what is inside of the block is an intermediate resonance.
Particles come toward each other, smash into each other, form an intermediate resonance for a short moment, and then decay. Also, it’s not clear why we call this a resonance phenomenon.
But imagine you have two particles, and you can bring them together, let them interact, and measure the probability for them to interact. And you have a handle to tune the energy of the particles.
So essentially, in the experiment, you do the following: you smash particles and count in how many cases they interact—they do scatter. Then you’re going to observe something like that, where the X-axis would be the probability for them to interact, and the Y-axis is the energy with which you collide them.
If you collide at this energy, the probability to interact is small. But once you tune the energy to the mass of the resonance frequency—the mass of the resonance—you have a huge cross-section, and then it dies out again.
So this resonance phenomenon indicates that there is something interesting inside the plot of my interactions—namely, a particle, an intermediate particle that can be formed.
7.9 Poles, Zeros, and Unitarity in Scattering Amplitudes
Now, what happens from the two-part 2 \times K matrix is the two poles. The cross section looks now something like that. It has two peaks:
- The first peak corresponds to the first pole.
- The second peak corresponds to the second pole.
There is a zero in between, which comes from the fact that K vanishes.
The lecturer corrects a typo:
- Originally written: K
- Corrected to: K^{-1}
The average expression should be 1 - \rho \cdot (K^{-1}). Then, K multiplies this expression.
If K vanishes, the entire expression vanishes, therefore two poles. If you call K itself, it has one singularity and then it gives another set. In order to reach the second singular asymptotics, it needs to cross the zero.
So K vanishes in between the poles, and the zero propagates with the amplitude squared. What is this amplitude? Zero, not plus zero. Very good. This is zero on this axis.
Well, another thing to note is that the values of the energy where amplified peaks occur are not exactly these masses. You see, they enter an expression in a rather complicated form. Therefore:
- The peak happens very close to this value.
- But not exactly at that value.
That’s the reason why this mass is sometimes called bare masses. This process of putting into the expression is addressing of bare particle masses with the propagator within the dressing process.
Let’s have another look at this expression:
A = \frac{K}{1 - \rho \cdot K}
When we expand this in the Taylor series, it’s going to be:
A = K + K \cdot \rho \cdot K + K \cdot \rho \cdot K \cdot \rho \cdot K + \cdots
What we see is an infinite series of terms that allows us to give a meaning to the terms K and \rho:
- \rho comes from the two-particle phase space and indicates two particles propagating.
- The K-term is the elementary interaction—K is the point-like interaction.
Diagrammatically:
- \rho is the 2-particle parameter.
- K is the elementary interaction.
The vertex amplitude is a complex function. What we’ve discussed so far is how the length of this complex vector looks like. But what we haven’t discussed is the angle—the scattering phase, the argument of the scattering amplitude.
The scattering phase also has interesting behavior. It actually makes a circle in the complex plane.
What you see here is that from threshold, the amplitude increases. If you draw it in the complex plane:
- It starts small.
- Then increases to its largest value.
- Then goes down and arrives again at zero.
- Then makes a second shape like that—where the variable A here is the imaginary part of A.
What I’m showing here is the function of s. The maximal value is approached around m_1^2. Then there is the second circle that the amplitude makes, completing this shape. This is called an Argand diagram.
There is more in the homework sheet I’ve set. The Argand diagram is simply plotting the amplitude or amplitude times \rho.
F = A \cdot \rho
That’s just a more convenient quantity to plot in the complex plane.
To summarize:
- Unitarity is an important constraint that comes from probability conservation.
- It practically gives us a tool to model the amplitude.
- It tells us that the imaginary part of the amplitude is fixed, and we only have to worry about the real part.
Moreover, this real part has the meaning of a point-like interaction that has to be resumed to all orders. The way to model this—there are many techniques to model the real part, and one of them is at the end discussed now.
Another one is the modeling with the polynomial, and this is called the scattering length approximation. The first term would be the scattering length. The second—so you can write:
K^{-1} = a^{-1} + \frac{r_s r_p}{2}k^2
and this is called the effective range approximation.
Key formulas discussed:
- Amplitude: A = \frac{K}{1 - \rho \cdot K}
- Unitarity condition: \text{Im}\, A = \rho |A|^2
- Effective range approximation: K^{-1} = a^{-1} + \frac{r_s r_p}{2}k^2