Author

Mikhail Mikhasenko

13.1 Introduction to Hadron Physics and Course Logistics

The first lecture will be dedicated to the introduction of hadron physics.

I would like to start by explaining:

  • What hadron physics is
  • Its origin
  • How it appeared in the early stages of the universe, where it played a role in the evolution of our universe

Then we’ll discuss:

  1. Matter composition and how quark physics plays its own part.
  2. A brief overview of the Standard Model.
  3. The equations that describe the motion of fields.
  4. The SU(3) group and the important property of the strong interaction key to matter formation, known as confinement.

13.2 Scaling the Universe’s Timeline from Big Bang to Hadronization

As last year, it’s good to start the lecture course by starting at the same time the timeline for the universe and the numbers that we’ll put here for these different epochs and stages of the universe. They are really hard to imagine—either very small or very large. That’s why we need this on a human scale first.

I can put here the number that is where we are now, and this is 14 billion years. That’s what we believe is the age of the universe:

t_{\text{universe}} \approx 14 \times 10^9 \, \text{years}.

Starting from the Big Bang, there will be many points.

But to have a human scale, I would like to do one exercise. Now I will clap my hands here, starting the universe. You watch the distance between my clapping. You see me clapping, and then the sound comes to you. The sound comes to the back row first, and then probably to the last row last.

We will put this scale over there. Here was the clap—Big Bang. Here is the point where it reaches the first row, one meter. Since the speed of sound is about 300 meters per second:

t = \frac{1 \, \text{m}}{300 \, \text{m/s}} \approx 13.3 \, \text{s}.

If my clap was the Big Bang, from that moment the universe started evolving. By the time the sound reached you, most of the stages had passed. We will be talking about hadron physics—that’s actually around this time.

I need 10^{-6}, 10^{-12}, 10^{-12}, 10^{-6}. At about one second at the beginning, matter is produced. We don’t know what kind of matter is there, but we assume that the structures started forming already. The seeds of the structure were already produced during inflation—inflation of space and time. We have little idea about what’s going on here.

We will advance the time to 10^{-12} seconds, where already the electroweak scale is passed:

E_{\text{EW}} \sim 100 \, \text{GeV}.

The Planck scale is passed:

t_{\text{Planck}} \sim 10^{-43} \, \text{s}.

The Higgs potential developed its minimum:

V(\phi) = \mu^2 |\phi|^2 + \lambda |\phi|^4.

The universe collapsed to the lower minimum of the Higgs potential, and we arrive at 10^{-12} seconds, where the matter itself is in the quark-gluon plasma (QGP).

The dots here are quarks—they are constituents of the matter. The fields that operate in this space are gluons. With this abbreviation, QGP, we refer to quark-gluon plasma. There is no matter as we know it—it’s just a soup in which fields are acting.

The equation of motion at high temperature around 10^{-6} seconds, this soup starts evolving into structure. In the interval somewhere here, you’ve got the hadronization process:

t_{\text{hadronization}} \sim 10^{-6} \, \text{s}.

When the sound reached you, what happened with our universe is that it already has elementary matter blocks, which are mesons and baryons. Quarks are now confined, and gluon fields are almost entirely sitting inside these objects. Essentially, that’s all we need from this picture.

What happens further is another 14 billion years of evolution. Somewhere here I mark one second, which is a good mark, where nuclear synthesis starts:

t_{\text{BBN}} \sim 1 \, \text{s}.

Another visually pleasant indication is when radiation separates from the matter. Big Bang nucleosynthesis, and then radiation separates from the matter around—doesn’t have a scale. Does anyone remember the scale here? Probably it’s in this picture. I think it’s quite high.

You believe that happens about 400,000 years after the Big Bang:

t_{\text{recombination}} \sim 400,\!000 \, \text{years} \approx 1.2 \times 10^{13} \, \text{s}.

We’ve got pions and baryons.

For the rest of the semester, what will happen with our universe, which we just created? There will be some nucleosynthesis starting, maybe. We won’t get to the forming of atoms as we know them, with electron shells—it only happens 400,000 years after.

Key Scales Recap:

  • Electroweak symmetry breaking: E_{\text{EW}} \sim 100 \, \text{GeV}
  • Planck time: t_{\text{Planck}} \sim 10^{-43} \, \text{s}
  • Hadronization: t_{\text{hadronization}} \sim 10^{-6} \, \text{s}
  • Recombination: t_{\text{recombination}} \sim 400,\!000 \, \text{years}

13.3 Structure of the Atom: Nucleus, Electrons, and Atomic Scale

What I have here is the electron. So what is an atom? Let’s now go to the second item. What is matter? What’s the most abundant element in the crust of the Earth? By mass, the most abundant element is iron, but in the crust, it’s oxygen.

Oxygen is an element with eight protons and eight neutrons, forming a compact object at the center of the atom—the atomic nucleus. These nucleons are packed together very tightly.

The size of a single proton or neutron is roughly one fermi. Imagine 16 balls of one fermi packed together. This gives a nucleus with a diameter of about six fermi. A naive calculation suggests a radius of three fermi, so doubling that gives the six fermi diameter.

Electrons are the other part of the atom. How many do we expect? They occupy shells: 1s^2 2s^2 2p^4 The s orbital is circular, while the p orbital has a dumbbell shape.

The energy of electrons versus angular momentum shows the lower energy levels: 1s, then 2s, and so on. The wave function of the electron in the 1s orbital is: \psi_{1s}(r) \propto e^{-r/a_0} where a_0 is the Bohr radius. This gives the scale of the atom.

The Bohr radius a_0 is about 50 picometers (or 59,000 femtometers). Solving the Schrödinger equation for an electron bound to the nucleus gives this scale.

Does the nuclear charge affect the radius? Yes. For a nucleus with charge Z = 8, the radius scales inversely. However, outer electrons are screened by inner ones, feeling less charge. This is why outer shells extend to several hundred picometers.

To visualize this, compare it to the solar system. If the nucleus is scaled down by a factor of 150 relative to the sun, the distances match roughly. This helps imagine the atomic scale.

The average radius for the 1s orbital is: \langle r \rangle_{1s} = \frac{3}{2}a_0 For the 2s orbital: \langle r \rangle_{2s} = 6a_0 Dividing by the effective charge (e.g., Z_{\text{eff}} \approx 4 due to screening) gives the outer radius. This is why a_0 sets the scale, leading to the estimate of around 100 picometers.

  • Nuclear size: R_{\text{nucleus}} \approx 3 \text{ fermi (fm)}, Diameter = 6 \text{ fm}
  • Effective charge scaling: r_{\text{eff}} \approx \frac{a_0}{Z_{\text{eff}}}

The text now uses bold for key terms, italics for less critical emphasis, block math for equations, and a callout for contextual notes. Horizontal rules separate logical sections, and a recap box highlights formulas. All original sentences are preserved.

13.4 The Standard Model: Structure, Challenges, and the Strong Interaction

Particle physics has the most accurate and precise theory that describes everything we have observed so far, which is the Standard Model. It is so accurate that for 10 years now we have been trying to find any single deviation by colliding particles at CERN and elsewhere from its predictions. So far, we haven’t found any—it’s very nice and accurate.

There are some problems with the Standard Model, which we will probably touch on at large scales. When we think of the evolution of the universe, there are also questions of naturalness of certain couplings—certain parameters have been measured, and we know them accurately in the Standard Model. But it’s unclear where they come from or why they are like this.

The Standard Model’s mathematical structure is captured by: \mathcal{L}_{\text{SM}} = \mathcal{L}_{\text{EW}} + \mathcal{L}_{\text{QCD}} where \mathcal{L}_{\text{EW}} describes electroweak interactions and \mathcal{L}_{\text{QCD}} describes strong interactions.

Roughly, the Standard Model can be thought of as the electroweak sector times QCD. The fact that the Standard Model is such a good theory and works so well doesn’t mean we understand it fully. This particularly refers to quantum chromodynamics, the interaction that describes the strong force. Everything we talk about in hadron physics is governed by the strong interaction.

Quantum chromodynamics is the theory of the strong interaction, the theory of the color charge. It works very well for phenomena in physics, but it’s so complicated that until recently, direct predictions from its basic equations to real-world observables were not possible. The reason, which we will discuss later, is that the fundamental interaction describes interactions between its building blocks—quarks and gluons. This is where they were present last time, even before that.

After that, the theory turns into something different, where there are particles with non-trivial structure, not fundamental particles, and the theory starts interacting. Essentially, hadron physics is not so much about the level of quarks in drawings, but what this fundamental theory implies for the form in which it manifests itself. Hadrons manifest as color-neutral objects, where the charge carried by the theory is confined. In this course, we will try to understand the relation of the fundamental part to the effective interaction between hadrons.

To match this picture of the Standard Model being two parts, it’s worth expanding further. The electroweak sector and the leptonic sector contain three parts:

  1. Electromagnetic interaction
  2. Weak interaction
  3. The mass-energy sector

Let’s quickly list the particles. I’ll write down the particle symbol, and you tell me to which sector it belongs.

  • Higgs? Very good—it belongs to the Higgs sector.
  • W boson? Yes, from the weak sector.
  • Z boson? Also weak. These are the carriers of the weak charge.
  • Photon? That belongs to the electromagnetic sector.

Now, quarks:

  • Top quark?
  • Charm quark?

These belong to the strong interaction—QCD. These are the elementary particles that make strong interactions. This classification is a bit vague because we mostly talk about quarks. The electroweak sector deals with leptons, electromagnetic with photons, and weak interaction with W and Z bosons and Higgs as a separate field.

However, if a particle has a charge of a certain type, it can interact with the carriers of that charge. So, let’s describe the charges. The quarks are organized in generations: first, second, third. They are also organized in two rows—top and bottom.

The charges are:

  • Electric charge:

  • Top row: +\frac{2}{3}

  • Bottom row: -\frac{1}{3}

  • Weak charge:

  • Upper row: +\frac{1}{2}

  • Lower row: -\frac{1}{2}

A quark with +\frac{2}{3} electric charge and +\frac{1}{2} weak charge can couple to carriers of those charges. Quarks are the most diverse particles in the Standard Model because they couple to all charges: electromagnetic, weak, and strong. For example, a quark can interact with:

  1. Light (electromagnetic)
  2. Z or W bosons (weak)
  3. Gluons (strong)

The strong charge is also called color charge.

Color charge and strong charge are the same thing. Just as an electron can be positive or negative, color charge can be:

  • Red
  • Green
  • Blue

Or their antiparticles:

  • Anti-red
  • Anti-green
  • Anti-blue

You’ll quickly get used to this language—strong charge as color charge. In hadron physics, we’ll be talking about the strong interaction, which is also the color interaction.

13.5 Interaction Fields and the Lagrangian Formulation

When we say interaction, it’s interaction of the object with the field. The fields are represented by these carriers of the force, which are $W $ and $Z $ for the weak interaction, photons for the electromagnetic interaction, and gluons for the strong interaction.

We discussed the Standard Model composition, and we’re reaching unification. The standard framework in field theory to describe the fields and the interaction between particles and carriers is field theory, which starts with the Lagrangian—an expression that describes interaction in a very condensed line.

The Standard Model Lagrangian includes terms for kinetic energy, gauge interactions, the Higgs mechanism, and Yukawa couplings: \mathcal{L}_{\text{SM}} = \mathcal{L}_{\text{kinetic}} + \mathcal{L}_{\text{gauge}} + \mathcal{L}_{\text{Higgs}} + \mathcal{L}_{\text{Yukawa}}

Here is an example. You might remember Lagrangian mechanics from your first or second semester, where the entire motion of the system was condensed down to a single equation. This was the Lagrange equation:

\mathcal{L} = T - V

where $T $ is the kinetic term and $V $ is the potential term. Once you subtract them, you get an expression representing the energy of the system.

In that case, kinetic minus potential gives the equation of motion. What happens with the system in the next moment, starting from the initial state, is described by an equation derived from the Lagrangian.

To find it, you differentiate the Lagrangian by the velocity $ $, then subtract the term that differentiates the Lagrangian by the coordinate $q $:

\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}}\right) - \frac{\partial \mathcal{L}}{\partial q} = 0

This describes a system where a point can move without friction, or a pendulum with another mass attached. The equation of motion is a differential equation, found by applying the classical equivalent of this formula.

13.6 Lagrangian Structure in QED and QCD: Fields, Indices, and Equations of Motion

I will clarify this equation now because this is the one we will use for the fields. Let me first start writing.

Lagrange—remember this or less. I start with QED (quantum electrodynamics). It describes how light interacts with anything that has a charge. It’s relevant for us for two reasons:

  1. Our quarks have charge—that’s why they interact with photons.
  2. The Lagrangian for quantum electrodynamics is simpler than quantum chromodynamics.

Let’s use this chance to understand all the symbols before we proceed to QCD later on.

So, who sees the equation for the first time? It’s complicated, but once you understand the general structure, you don’t need to look it up to write it down.

The Lagrangian is a function of two fundamental fields:

  • One is \psi (psi), the field of an electron, muon, or quark—something with charge.
  • The other is A, the photon field.

\psi is a fermion field, and A is a vector field.

The Lagrangian is a scalar quantity—not a vector, not a matrix, just a number when evaluated at any point. A scalar is achieved by dealing with indices: every index introduces a dimension, and you only get a scalar when all indices match. We use Einstein notation, where repeated indices imply summation—same as in quantum mechanics.

Here, \mu and \nu are Lorentz indices. \mu lives in four dimensions—three spatial (x, y, z) and one time. The \mu here and \mu here must be contracted. I’m skipping the explicit summation over \mu from 1 to 4, as well as the summation over \nu, which also appears twice.

You can think of F_{\mu\nu} as a matrix—\mu is 4-dimensional, and \nu is 4-dimensional, so F_{\mu\nu} is a 4 \times 4 matrix. When you multiply, you don’t multiply matrices the usual way—you multiply them component-wise. Every component is multiplied by itself, and then you take the trace (or sum all elements). In Python terms, you broadcast the matrix element-wise and then sum all elements—that’s what happens here.

Each component of F_{\mu\nu} is computed as: F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu where \partial_\nu is the derivative in time and space—essentially \partial/\partial x.

The QED Lagrangian is given by: \mathcal{L}_{\text{QED}} = \bar{\psi} (i \gamma^\mu D_\mu - m) \psi - \frac{1}{4} F_{\mu\nu} F^{\mu\nu} This describes fermion-photon interactions, with \psi as the fermion field and A_\mu as the photon field.

This would be simple if it were just \mu contracted with \mu. If you think of this as a vector, it’s not a vector. They can contract them—scalar product, easy. However, the trick is that this is also a matrix. So it’s four matrices. How do we get a scalar quantity? There’s another set of indices I suppressed—they belong to this matrix.

Let me make it explicit: what about \tau and \rho? In which space are they? While \mu is the Lorentz index (four spacetime dimensions), \tau and \rho are related to spin. Particles are not scalar; they have spin. That’s why the fermion field \psi has four components.

Are these upper indices? No, they’re matrix indices, not Lorentz indices. For matrix indices, covariant and contravariant distinctions don’t apply. Only Lorentz indices distinguish between upper and lower.

Now, something still seems off in this Lagrangian. We agree these are four matrices, and this is a vector, so we can contract them. We get a matrix here, but then we subtract a scalar—that doesn’t work. What’s missing is the diagonal matrix.

\psi is a four-component spinor, and \bar{\psi} is its adjoint—not a spinor but a row vector. We conjugate and transpose (\dagger), then multiply by \gamma^0 from the left to keep it as a row. Now we can contract this with the matrix here.

One of the exercises is to analyze the same structure for the QCD Lagrangian, which I’ll write next. Once you do it once, it becomes clear.

Let’s do QCD now—it’s not too bad. The exercise is to recover the indices, their range, and the number of terms. For QCD, we introduce a color index G because every additional index expands the dimensionality.

The QCD Lagrangian is: \mathcal{L}_{\text{QCD}} = \sum_f \bar{\psi}_f (i \gamma^\mu D_\mu - m_f) \psi_f - \frac{1}{4} F^a_{\mu\nu} F^{a,\mu\nu} where:

  • \psi_f is the quark field for flavor f,
  • D_\mu = \partial_\mu - i g_s \lambda^a A^a_\mu is the QCD covariant derivative,
  • \lambda^a are the Gell-Mann matrices (SU(3) generators),
  • F^a_{\mu\nu} = \partial_\mu A^a_\nu - \partial_\nu A^a_\mu + g_s f^{abc} A^b_\mu A^c_\nu is the gluon field strength tensor.

Now, let’s check the dimensionality. There’s a new object \lambda here—these are 3 \times 3 matrices. The indices I, J correspond to color space. When you commute two matrices, you get another matrix. Overall, this is a matrix in \mu, \nu (4D spacetime) and I, J (3D color space).

I’ll try to make more sense of this equation once all indices are introduced. I really want you to understand the mathematical structure. Everyone can trace this.

The index F traces flavors—quarks u, d, s, c, t, b. We have six flavors, so F runs over all six. The spinors here have four dimensions for spin and an extra three for color. The index I traces the color charge. The only thing I’m not tracking is the spinor indices.

For the QED Lagrangian, we agreed there are \tau, \rho indices for spin projection. We omit them here to avoid complexity. But if you think of this field, it has:

  1. A flavor (e.g., up quark),
  2. A color (e.g., red),
  3. And four spinor components for spin projections.

Finally, once you know the Lagrangian, you can derive the equations of motion by applying: \partial_\mu \left( \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \right) - \frac{\partial \mathcal{L}}{\partial \phi} = 0 For \psi, the derivative is non-zero only where \partial_\mu \psi appears. Expanding D_\mu \psi, we have \bar{\psi} \gamma^\mu D_\mu \psi. Differentiating, \bar{\psi} remains, and \partial \mathcal{L} / \partial \psi comes from the mass term. This gives a differential equation for how the field evolves in time and space.

If we apply this to QCD, do you know what the equation is called that describes the motion of a fermion?

The Dirac equation describes fermion dynamics in QED/QCD: (i \gamma^\mu \partial_\mu - m) \psi = 0 In QCD, this generalizes to the covariant form with D_\mu. The gluon field dynamics are governed by the Yang-Mills equations: D_\mu F^{a,\mu\nu} = g_s \bar{\psi} \gamma^\nu \lambda^a \psi

13.7 Gauge Symmetry, Phase Ambiguity, and Confinement in Field Theory

While I’m cleaning the board, let me ask questions for this point. We discussed the Lagrangian equation of motion. Let’s discuss the gauge transformation and gauge. That’s an extremely important concept in field theory, and we will only touch on this briefly since we are not doing field theory, but it’s important—you know where Legendre transforms come from.

Something very familiar from quantum mechanics is the fact that we have a phase ambiguity for the wave function. We can update the phase, and the absolute square of the wave function will not change because you multiply \psi by \psi^*, and the phase drops out—and that’s fine. First, we must acknowledge that this should be a symmetry of our theory. This should be allowed—to make this gauge, to make this phase transformation. But it’s kind of by definition, so this overall phase you don’t even have to think of.

The problem appears when you demand your theory to be invariant under the change of the phase at all possible space-time points at the same time—different phases. What if I want to adjust my wave function at every space point? Why is this a problem? Because in our equation of motion, in our Lagrangian—let me just write the Dirac part—there is the \overline{\psi} D_\mu \gamma^\mu \psi.

Here, I have a term D_\mu \psi, and this term is going to become D_\mu \psi', which is D_\mu of e^{i \alpha(x)} \psi. Let me take the derivative:

D_\mu \psi' = e^{i \alpha(x)} (D_\mu \psi + i (\partial_\mu \alpha) \psi).

The way I apply the derivative is to apply it to the first term and then to the second term. When I apply it to the second term, I just have D_\mu \psi. When I apply it to the first term, I get D_\mu acting on the exponent, yielding an extra term with \partial_\mu \alpha.

What happens? The same equation won’t hold any longer for \psi'. It does not transform into the same equation because the derivative introduces an additional term with D_\mu \alpha. This means that local gauge transformation is not a symmetry of the free Lagrangian of the Dirac particle. You cannot adjust the phase independently at different points for the free particle.

The Dirac Lagrangian with gauge field interaction is: \mathcal{L} = \bar{\psi} (i D_\mu \gamma^\mu - m) \psi where D_\mu = \partial_\mu - i e A_\mu is the covariant derivative.

In simple words, this is an incomplete theory. It only becomes complete if you consider radiation, photons, and charged particles together. When you look at the full Lagrangian, you update the phase of the field \psi and the electromagnetic field simultaneously. The additional term in the covariant derivative D_\mu cancels exactly the one from the phase, leaving the Lagrangian invariant.

This fact led us to the D_\mu term. From the equation of motion, you can see how different fields are coupled. In our pendulum example, we saw how the pendulum affects the upper marble. Similarly, fermion fields affect photons, and photons affect fermions. Gauge symmetry dictates how they interact, enforcing a specific interaction strength g.

Before jumping to QCD, where we deal with three dimensions, let’s consider the weak interaction with two components: up and down. The weak charge for quarks is \pm 1/2. The transformation that updates the phase is more general here. We want observables like \psi^\dagger \psi to remain unchanged. The transformation is unitary, described by unitary matrices, forming the group SU(2).

Any element of SU(2) can be written as:

U = e^{i \alpha_a \sigma^a},

where \sigma^a are the Pauli matrices (generators of SU(2)). The condition \det(U) = 1 comes from SU(2) being a standard group—we factor out the phase (U(1)) and work with SU(2).

For SU(3), we have three components (red, blue, green). The transformation is a 3x3 matrix:

U = e^{i \alpha_a \lambda^a},

where \lambda^a are the Gell-Mann matrices. The number of generators (8 for SU(3)) corresponds to the number of gluons.

Confinement is the property that the strong interaction grows with distance. Unlike electromagnetism, where attraction weakens with distance, quarks experience increasing force when pulled apart. This confines quarks within hadrons (mesons and baryons). If you try to separate them, new quark pairs form, maintaining color neutrality.

Looking at the QCD Lagrangian, gluon self-interaction terms indicate confinement:

G_{\mu\nu}^a = \partial_\mu A_\nu^a - \partial_\nu A_\mu^a + g f^{abc} A_\mu^b A_\nu^c.

These terms lead to vertices with three or four gluons, a signature of confinement.

The running coupling \alpha_s(Q) describes how the interaction strength changes with momentum:

\alpha_s(Q) \approx \frac{1}{\beta_0 \ln(Q^2 / \Lambda_{\text{QCD}}^2)}.

At high Q (asymptotic freedom), \alpha_s is small; at low Q (confinement), it grows.

13.8 Exponential Decay and Half-Life Relationships

When we examine radioactive decay, the number of nuclei decreases exponentially over time. The formula is:

N(t) = N_0e^{-λt}

The half-life, when half the nuclei have decayed, is related to the decay constant by:

t_{1/2} = \frac{\ln(2)}{λ}

This shows the fundamental relationship between the decay rate (λ) and the characteristic time (t_{1/2}) for radioactive processes. The decay constant λ represents the probability per unit time that a nucleus will decay.

Just like a ticking clock measures regular intervals, the half-life measures probabilistic decay intervals.

  • Key features of radioactive decay:
  • Exponential decrease in number of nuclei
  • Characterized by decay constant λ
  • Half-life t_{1/2} is a more intuitive measure of decay rate
  • Universal behavior across all radioactive isotopes

The mathematical relationship between λ and t_{1/2} applies to all exponential decay processes in nuclear physics.