Author

Mikhail Mikhasenko

16.1 Electron-Photon Scattering, Form Factors, and Baryon Magnetic Moments

Today’s lecture will start by reviewing electron-photon scattering, then continue with the magnetic moment. Later, we will discuss different experiments and how they produce various hadrons.

The electron-photon spectrum was discussed last time. In this context, remember three cross sections:

  • The Rosenfeld cross section
  • The Mott cross section
  • The Rosenbluth cross section

What’s important here are the charges. We neglected the spins of the particles initially. The logical extension is the Mott cross section, where \beta is the velocity and the electron spin is considered. It is essentially the Rutherford cross section multiplied by a correction term:

\left(\frac{d\sigma}{d\Omega}\right)_{\text{Mott}} = \left(\frac{d\sigma}{d\Omega}\right)_{\text{Rutherford}} \cdot \left(1 - \beta^2 \sin^2\left(\frac{\theta}{2}\right)\right)

This term suppresses scattering at backward angles. At 180 degrees, the cross section is lower due to conservation laws for relativistic particles.

Experiments show the cross section falls more steeply than predicted by the Mott formula. This discrepancy arises because the target is not point-like. The finite size is accounted for by a form factor.

The form factor is the Fourier transform of the charge density:

F(Q^2) = \int \rho(r) e^{i \vec{q} \cdot \vec{r}} d^3r

For a point-like particle (\rho(r) = \delta(r)), the form factor is a constant. For an exponential charge density (e.g., a proton), it takes a dipole form:

G_E(Q^2) = \left(1 + \frac{Q^2}{\Lambda^2}\right)^{-2}, \quad \Lambda \approx 16.71\,\text{GeV}^2

For a nucleus with a diffuse edge, the form factor exhibits oscillatory behavior.

The Rosenbluth cross section incorporates electric and magnetic form factors:

\frac{d\sigma}{d\Omega} = \left(\frac{d\sigma}{d\Omega}\right)_{\text{Mott}} \left[ \frac{G_E^2(Q^2) + \tau G_M^2(Q^2)}{1 + \tau} + 2\tau G_M^2(Q^2) \tan^2\left(\frac{\theta}{2}\right) \right]

where \tau = Q^2/4m_p^2 and Q^2 is the momentum transfer.

To extract the proton radius, plot the measured cross section divided by the Mott cross section versus Q^2. The slope at Q^2 \to 0 gives the radius:

\langle r_p^2 \rangle = -6 \hbar^2 \left.\frac{dG_E(Q^2)}{dQ^2}\right|_{Q^2=0}

The electric form factor of the proton follows a dipole distribution, while the neutron’s magnetic form factor starts at -1.91 at Q^2 = 0. The values at Q^2 = 0 correspond to:

  • The proton’s charge (1)
  • The magnetic moments of the proton and neutron

Measuring at low Q^2 is challenging because it requires small scattering angles. Experiments like those at the Mainz accelerator use spectrometer arms to detect electrons and protons. New experiments, such as MUSE at MESA, aim to improve these measurements.

Historically, discrepancies arose between proton radii measured via:

  • Electron scattering (\sim 16.86 fm)
  • Atomic spectroscopy (Lamb shift), which suggested a smaller radius

New experiments, like AMBER at CERN (using muons), seek to resolve this.

These experiments revealed the proton’s substructure. Now, let’s discuss spin, flavor symmetry, and calculating baryon magnetic moments.

The total baryon wave function must be antisymmetric. Color wave function antisymmetry implies the spin-flavor part must be symmetric. For a proton (spin-1/2), the symmetric spin-flavor wave function is:

|p \uparrow \rangle = \frac{1}{\sqrt{2}} \left( \chi_S \psi_S + \chi_A \psi_A \right)

where \chi_S and \psi_S are symmetric spin and flavor states, and \chi_A, \psi_A are antisymmetric.

The magnetic moment of a quark is:

\mu_q = \frac{e_q \hbar}{2m_q}

For up (e_u = +2/3) and down (e_d = -1/3) quarks:

\mu_u = +\frac{2}{3} \frac{e\hbar}{2m_u}, \quad \mu_d = -\frac{1}{3} \frac{e\hbar}{2m_d}

The proton’s magnetic moment (spin-up) sums contributions from its quarks:

\mu_p = \langle p \uparrow | \sum_{i=1}^3 \mu_i \sigma_z^{(i)} | p \uparrow \rangle = \frac{4}{3}\mu_u - \frac{1}{3}\mu_d

For the neutron, swapping up and down quarks gives:

\frac{\mu_n}{\mu_p} \approx -\frac{2}{3} \quad \text{(if } m_u \approx m_d \text{)}

Experimentally, this ratio is close to -1.91/2.79 \approx -16.68, consistent with the quark model.

The constituent quark masses include:

  • Valence quarks
  • Surrounding gluons/quark-antiquark pairs

This explains their larger values (\sim 300 MeV) compared to bare quark masses.

Magnetic moments are measured via:

  • Spin splitting in magnetic fields (e.g., Stern-Gerlach)
  • Radiative decays in photon-proton collisions for short-lived baryons (e.g., \Delta^{+})

For the \Omega^- baryon (three strange quarks), the magnetic moment is:

\mu_{\Omega^-} = 3\mu_s = 3\left(-\frac{1}{3}\frac{e\hbar}{2m_s}\right) = -\frac{e\hbar}{2m_s}

This concludes the recap of electron-photon scattering and magnetic moments. Next, we’ll explore further applications of flavor symmetry.

16.2 Light Baryon Spectroscopy and Hadron Production Methods

We already discussed the proton. Now let’s think about the light baryons. We have seen you can use SU(3) for flavor with the up, down, and strange quark, and combine it with SU(2) spin. This gives us SU(6) for spin-flavor symmetry:

\text{SU(6)} \supset \text{SU(3)}_{\text{flavor}} \otimes \text{SU(2)}_{\text{spin}}

As discussed earlier, for SU(3) flavor, we have the baryon decuplet and octet. Combining flavor and spin into SU(6), we get the 20, 70, and 56-plets. The 56-plet is antisymmetric, while the others are mixed symmetric. With these considerations, we can arrange the known light baryons into these multiplets. I won’t cover all of them, but let’s look at some mass differences as examples.

For instance, we have:

  • The radial excitation number
  • Orbital angular momentum L
  • The spin, which can be 1/2 or 3/2

For spin 1/2, we get the nucleons, and for 3/2, we enter the decuplet, which includes the \Delta. I’ll focus on the N^* and \Delta^* baryons, excluding those with strange quarks. These can have excited states from radial or orbital angular momentum excitations. For L=1, the spin can still be 1/2 or 3/2.

Let me give an example of radial excitation. If we arrange these states in patterns:

  • S denotes L=0
  • The N=1 state corresponds to N=0, which is just the proton or nucleon
  • The next band contains five states, with masses ranging from 1500 to 1700 MeV

One state might be slightly lower, and others could include D-states, and so on.

The key point here is the mass difference of about 600 MeV, arising from spin-orbit and spin-spin interactions:

\Delta M \approx 600 \text{ MeV}

These states are well-documented up to certain energies in the PDG, with star ratings indicating their reliability. A goal of light baryon spectroscopy is to identify how many states exist and their patterns, based on quark model predictions. Are these correct, or is there different dynamics? We need to study the states’ properties, which leads me to discuss experimental methods for producing hadrons.

Question about 1S and 2S: I’m surprised you labeled this as 3S. What’s the difference between 1S and 2S? The radial wave function changes, but no orbital angular momentum is added—just the hadron’s size increases. How does this introduce more states? I was checking the quark model’s predictions. Why three states? I’d expect one, like the \rho resonance.

The answer lies in multiplets from different LS combinations. For n=1, l=0, there’s a spin combination. Some papers claim more states should exist, including \Delta resonances, but these are distinct. Some studies don’t separate them clearly.

For radial excitations (increasing N), the number of states in the multiplet shouldn’t change because no extra degrees of freedom are added. The 2P multiplet has the same number of states as 1P. Higher excitations introduce L=2, altering the count.

Agreed—for N=1, 2, 3, the state count remains the same. Experimentally, only the first band is well-known; the rest are uncertain. Some states match quark model predictions, others don’t, leaving gaps in light baryon spectroscopy.

Next, let’s discuss hadron production. Electromagnetic probes like electron or photon beams can be used, with colliding or fixed-target setups. Hadron beams (e.g., pions or protons) are another option.

Starting with electromagnetic probes, consider the BESIII facility in China. It’s a symmetric e^+e^- collider with beam energies of 1–2.47 GeV. The annihilation produces a virtual photon, which can form a J/\psi, decaying into other hadrons.

The center-of-mass energy \sqrt{s} = 2E_e determines the producible particles. For BESIII, this covers masses up to ~2 GeV. The cross-section shows peaks like J/\psi and \psi(2S), with decays like J/\psi \to \gamma \eta \eta' or 3\pi.

At higher energies (e.g., Belle, \sqrt{s} \approx 10 GeV), bottomonium states like \Upsilon are produced.

Now, photo-production experiments: fixed-target setups like:

  • A2 (Mainz, 16.8–1.6 GeV)
  • CBELSA/TAPS (Bonn)
  • GlueX (US, 6–12 GeV)

Photon beams are generated via Bremsstrahlung from electron beams:

\frac{dN}{dE_\gamma} \propto \frac{1}{E_\gamma}

For \gamma p collisions, the center-of-mass energy is:

s = m_p^2 + 2E_\gamma m_p

With E_\gamma \leq 3.2 GeV, resonances up to 2.7 GeV (e.g., \Delta) are accessible.

Can hyperons be produced here? No—flavor conservation forbids strangeness change in s-channel. However, t-channel processes (e.g., \gamma p \to K^+ \Sigma) are possible, as planned for the INSIGHT experiment.

Quantum number selection rules apply. For E1 photons (L=1, parity -):

J^P = \frac{1}{2}^- \text{ or } \frac{3}{2}^-

Resonances are labeled like S_{11}, where:

  • S: L=0 (meson orbital momentum)
  • First subscript: 2 \times isospin (1 for nucleons, 3 for \Delta)
  • Second subscript: 2 \times total angular momentum

Final note: Higher energies allow meson spectroscopy, but we’ll cover that next time.

16.3 SU(2) Algebra and Delta Baryon Wave Functions

We’ve been talking to theorists yesterday. There will be advanced theoretical hadron physics next semester. Everyone is invited to take it, and it will be completely complementary to this course—it will go in-depth into group theory.

One thing we discussed is that in this course, we cover SU(2) well. In the theory course, this SU(2) is considered trivial, so it goes to SU(3) in depth, looking at the irreducible representations. But for this course, we focus on SU(2), the groups we discussed for the flavor, for the isospin, and for the spin, as Farah demonstrated today.

It’s indeed simple once you understand it, and understanding requires practice. I mentioned last time, Farah mentioned today, and we will mention yet another time how the SU(2) algebra works. But I wanted to invite you to engage with a little exercise.

Key SU(2) Algebra Formulas:

  • Commutation relations: [J_i, J_j] = i \hbar \epsilon_{ijk} J_k or [I_i, I_j] = i \epsilon_{ijk} I_k (with \hbar = 1).
  • Raising/lowering operators: J_{\pm} = J_x \pm i J_y (similarly for I_{\pm}). J_- |j, m\rangle = \sqrt{j(j+1) - m(m-1)} \, |j, m-1\rangle. J_+ |j, m\rangle = \sqrt{j(j+1) - m(m+1)} \, |j, m+1\rangle.

I’ll quickly write a wave function—it goes nowhere—and I invite you to write down the answer, leave it on the table right here. You don’t even need to write your name on it. We just want to gauge how well you understand what’s going on.

So, I want to consider a delta that has three quarks in isospin states. There are four states: \Delta^{++}, \Delta^+, \Delta^0, and \Delta^-. I would like to build this spin-flavor wave function.

For example, the \Delta^+ state would be |uud\rangle with spin projection J_z = \frac{3}{2}. On this little piece of paper, write:

  1. What is J_- acting on |\Delta^+ (J=\frac{3}{2}, J_z=\frac{3}{2})\rangle?
  2. What is J_+ acting on |\Delta^+ (J=\frac{3}{2}, J_z=\frac{3}{2})\rangle?
  3. What is I_+ acting on |\Delta^+ (I=\frac{3}{2}, I_z=\frac{1}{2})\rangle?
  4. What is I_- acting on |\Delta^+ (I=\frac{3}{2}, I_z=\frac{1}{2})\rangle?

You have two minutes. I have a meeting now, so I won’t provide solutions—it’s trivial once you understand the logic. Just give it a try.

A reminder: the total isospin is the sum of the isospins of the three quarks, \mathbf{I} = \mathbf{I}_1 + \mathbf{I}_2 + \mathbf{I}_3. The same applies for \mathbf{J}. You don’t need much derivation if you know the answer. We just want to check if you remember how the lowering operators act.

Exercise Solutions (for reference):

  • J_- |\Delta^+ (J=\frac{3}{2}, J_z=\frac{3}{2})\rangle = \sqrt{3} \, |\Delta^+ (J=\frac{3}{2}, J_z=\frac{1}{2})\rangle
  • J_+ |\Delta^+ (J=\frac{3}{2}, J_z=\frac{3}{2})\rangle = 0 (max J_z state)
  • I_+ |\Delta^+ (I=\frac{3}{2}, I_z=\frac{1}{2})\rangle = \sqrt{3} \, |\Delta^{++} (I=\frac{3}{2}, I_z=\frac{3}{2})\rangle
  • I_- |\Delta^+ (I=\frac{3}{2}, I_z=\frac{1}{2})\rangle = \sqrt{2} \, |\Delta^0 (I=\frac{3}{2}, I_z=-\frac{1}{2})\rangle