The 3/2⁺ Baryons · Elementary Particle Physics

Nine particles fill three rows and leave one corner of the triangle empty. The hole is the result of this section, and it is worth more than the nine states that are there.

🎯 Why this matters

A classification that only organises can absorb whatever turns up next. One with a hole cannot: the missing state’s charge, strangeness, spin and mass are all fixed before anyone looks, so the scheme is either paid or broken.

§4.1 built two tools. This section uses both, on the same family of particles, and the choice of tool is forced by something entirely practical: what you can make a beam of. The Δ(1232) is found by formation because a pion beam on hydrogen has exactly its quantum numbers. The Σ(1385) and Ξ(1530) are found by production because no beam of Λ or Ξ exists and never will.

By the end there are nine particles arranged in three rows, with a conspicuous gap where a tenth should be. §4.8 fills it.

What a formation experiment can reach

To form a baryon you need a baryon target and a meson beam, and both lists are short. The target must be an elementary particle held in bulk: liquid hydrogen gives free protons, and since free neutrons cannot be held at all, the simplest nucleus containing one — deuterium — has to stand in. The beam must live long enough to be transported, which among mesons means pions and kaons and nothing else.

The entire menu of formation experiments, and what each reaches

beam	target	$S$	what it forms	result
π⁺, π⁻	p (liquid H₂), n (via D₂)	0		Several dozen N(xxxx) and Δ(xxxx) resonances
K⁻	p, n	−1	I = 0 and I = 1 states, since 2 ⊗ 2 = 1 ⊕ 3	The Λ(xxxx) and Σ(xxxx) resonances
K⁺	p, n	+1	S = +1 baryons — if any existed
Λ, Σ, Ξ, or any meson but π and K	—	—	—

Everything in this section follows from this table. <strong>The Δ is formed, the Σ and Ξ are produced, and the K⁺N null result is a discovery in its own right.</strong>

💡 What this really says — the one entrance channel that produced no resonances at all

“No resonance exists in the K⁺–nucleon system” is a stranger statement than it looks. Every other entrance channel available in the 1960s produced dozens of resonances; this one produced none, at any energy anyone could reach.

A K⁺N state would be a baryon ( $\mathcal{B} = 1$ ) with strangeness $+1$ . Nothing in Chapter 3 forbids such an object — baryon number and strangeness would both be perfectly conserved in its formation and decay. It simply is not there.

Hold onto that, because it is one of the quark model’s cleanest predictions in reverse. Three quarks can carry $S = 0, -1, -2$ or $-3$ , depending on how many are strange. Positive strangeness needs an antiquark, and an antiquark makes the baryon number wrong. The absence in the table is not an experimental gap; it is a structural impossibility, and §4.8 will say so in one line.

The Δ(1232), by formation

π⁺p total
π⁺p elastic

Fig. 4.4, schematic — a sum of Breit–Wigners on a smooth background, drawn to reproduce the book's figure rather than digitised from it. Three things are real and worth reading: at low momentum the two curves COINCIDE, because elastic is the only open channel; the Δ(1232) is an enormous peak reaching ~200 mb; and at high momentum the total settles near 25 mb and creeps slowly upward. The two marked higher resonances are barely visible here — which is the point of the next figure.

💡 What this really says — the two curves separating is a channel count

At the lowest momenta the total and elastic cross-sections are the same curve. That is not an approximation: below the threshold for making an extra pion, the only thing a π⁺ and a proton can do is scatter elastically, so $\sigma_{\text{tot}} = \sigma_{\text{el}}$ exactly.

As the energy rises, channels open one by one — $\pi^+p \to \pi^+\pi^0p$ , then $\pi^+\pi^+\pi^-p$ , then $\pi^+K^+K^-p$ — and the elastic fraction falls. That falling fraction is the elasticity $\eta$ of §4.1, seen from the outside: $\eta^2$ is roughly the ratio of the two curves, and the widget there showed what a small $\eta$ does to a resonance’s visibility.

So the gap between these two curves is the reason the Δ(1232) is the only resonance in this plot you can see by eye.

🔬 Experiment card — partial wave analysis of πN scattering, the 1960s

Apparatus

Positive and negative pion beams on liquid hydrogen (and deuterium for the neutron), at accelerator laboratories worldwide. No single machine — this is a decade of experiments whose value lies in being combined.

What is measured

The differential cross-section

d\sigma/d\Omega

, channel by channel, in small steps of beam energy, with the systematic uncertainties controlled well enough that data from different laboratories can be merged. Isospins come from comparing charge states, exactly as in Example 3.7; spins and parities come from decomposing the angular distributions into partial waves.

The result

Several dozen resonances in the πN and K⁻N systems, each with a mass, a width, an isospin and a

J^P

. The Δ(1232) is the

P_{33}

wave:

L = 1

I = 3/2

J = 3/2

, and its Argand trajectory is a textbook elastic Breit–Wigner riding the unitarity circle.

What it proved

J^P = 3/2^+

for the Δ. The parity follows from

L = 1

in one line — and the method, not the particle, is the point: the vast majority of these resonances are invisible in the cross-section and exist only as loops in the Argand plane.

P(\Delta) = (-1)^{\htmlClass{t-L}{L}}\,\htmlClass{t-Ppi}{P(\pi)}\,\htmlClass{t-PN}{P(N)} = (-1)^1 \times (-1) \times (+1) = \htmlClass{t-res}{+1}

Bettini p. 137. Once the partial wave analysis says L = 1, the parity is arithmetic — and it uses the two intrinsic parities measured in §3.2 and §3.5.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

the partial-wave amplitude in the complex plane

energy, in half-widths from the peak: (E − m)/(Γ/2) = 0.00elasticity η = 1.00

phase shift δ	90.0°
\|f\|, in units of 1/q	1.000
\|f\|², relative to the peak	1.000

|f|², the cross-section — what a scan shows you
δ in degrees ÷ 180 — what the Argand loop shows you

The same resonance, twice. The peak (solid) is what a cross-section scan measures and it shrinks as η² — drop η to 0.25 and it nearly vanishes into a background. The phase (dashed) sweeps 45° → 90° → 135° across one width whatever η is, which is why the Argand plane finds resonances the cross-section hides.

Fig. 4.2(a). The only open channel is the one you came in through, so the trajectory rides the unitarity circle itself — the scattering probability reaches its maximum allowed value at the peak.

⚙️ Engineer’s bridge — three poles, one plot, and only one visible peak

Fig. 4.5 puts the Δ(1232), Δ(1600) and Δ(1920) on a single Argand diagram, and the contrast with Fig. 4.4 above is the lesson of this section.

In the cross-section, the Δ(1232) is a 200 mb mountain and the other two are bumps you would not bet on. In the complex plane all three are loops. The reason is the one the widget makes tangible: the modulus goes as $\eta^2$ and the phase does not. A resonance that sends most of its decay into other channels is quadratically suppressed in the elastic cross-section while its phase sweep is untouched.

The engineering habit is identical to reading a Bode plot of a system with several poles. A weak, well-separated pole is a barely visible ripple in $|H(\omega)|$ and an unmistakable feature in $\arg H(\omega)$ — and if the resonance sits on a large smooth background, the magnitude plot is worse than useless because the background dominates it. Fitting the phase is how you find poles you cannot see.

Where it breaks: a real partial-wave analysis has to fit all the waves at once with a background, and the loops overlap and interfere. The idealised single circle of the widget is the textbook case; the Δ(1232) happens to be close to it, which is why it is the one everybody draws.

Where it breaks: the idealised circle assumes one resonance dominating one partial wave with a constant background. The Δ(1232) is close to that and almost nothing else is: overlapping resonances of the same $J^P$ interfere, so their circles distort into shapes from which the individual parameters cannot be read without a coupled-channel fit. The plot is a diagnostic for the clean case and a warning for the rest — which is why partial-wave analysis is a research field rather than a procedure.

The Σ(1385), by production

No $\Lambda$ beam exists, so the Σ* has to be made and then reconstructed. Alvarez and collaborators (1963) used a 1.5 GeV/c K⁻ beam from the Bevatron into the 72″ hydrogen bubble chamber of §1.13b:

K^- + p \to \Lambda + \pi^+ + \pi^-, \qquad \begin{aligned} K^-p &\to \htmlClass{t-a}{\Sigma^{*+}}\pi^- \to \Lambda\pi^+\pi^- \\ K^-p &\to \htmlClass{t-b}{\Sigma^{*-}}\pi^+ \to \Lambda\pi^+\pi^- \end{aligned}

(4.11, 4.12)

Bettini p. 138. One final state, two possible intermediate states — which is why the analysis needs two axes rather than one histogram.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

Fig. 4.6 — K⁻p → Λπ⁺π⁻ at p_K = 1.5 GeV/c

resonance strength = 1.00show the projections onto each axis

Two perpendicular bands, each at a fixed value of one squared mass, and each projecting to a peak on its own axis. Turn the strength to zero to see what the same kinematics looks like with no resonance at all.

The boundary and the band positions are exact — √s = 2.0217 GeV from a 1.5 GeV/c K⁻ on a proton at rest, and the axes run from (m_Λ+m_π)² = 1.576 to (√s−m_π)² = 3.543 GeV², which is the ellipse in the book's figure. The POINTS ARE SIMULATED from that kinematics and the quoted Σ(1385) parameters; they are not the published Alvarez events. Slide the strength to zero to see the same reaction with no resonance in it.

💡 What this really says — why the axes are squared masses

The book’s figure plots $m^2$ and not $m$ , and the reason is the whole point of the next section: the phase-space volume element is uniform in the squared masses. A three-body final state with no dynamics in it fills this plot evenly. So a region of higher density is not a kinematic effect that has to be divided out — it is the matrix element, directly visible.

That is why the strength slider is the useful control here. At zero the plot is a featureless ellipse of dots: energy and momentum conservation and nothing else. Turn it up and two perpendicular bands appear, one at fixed $m^2(\Lambda\pi^+)$ and one at fixed $m^2(\Lambda\pi^-)$ , each projecting to a peak on its own axis.

Both bands sit at 1385 MeV with the same width, ≈ 35 MeV. Same mass, two charge states, and Λπ is a pure $I = 1$ state — so this is a Σ, and §4.3 is where the uniformity claim gets proved.

🔢 Worked example — the boundary of Fig. 4.6, from first principles

Everything about the frame of that plot is fixed by two masses and a beam momentum, with no reference to any resonance.

Reproduce it

import numpy as np
mK, mp, mL, mpi = 0.493677, 0.938272, 1.115683, 0.13957
hbar, c = 6.582119569e-22, 2.99792458e8          # MeV s, m/s
pK = 1.5
EK = np.hypot(pK, mK)
rs = np.sqrt(mK**2 + mp**2 + 2*EK*mp)
print(f"K- p at p_K = {pK} GeV/c")
print(f"   E_K = {EK:.4f} GeV -> sqrt(s) = {rs:.4f} GeV")
print(f"   m^2(L pi) runs from (m_L + m_pi)^2 = {(mL+mpi)**2:.4f} to "
      f"(sqrt(s) - m_pi)^2 = {(rs-mpi)**2:.4f} GeV^2")
print("   the book's plot is framed 1.50 to 4.00, with the contour spanning 1.58 to 3.54")
print(f"Sigma(1385): band at m^2 = {1.3852**2:.4f} GeV^2, and both bands sit there")
for nm, G in (("Sigma(1385)", 35.0), ("Xi(1530)", 7.2)):
    tau = hbar / G
    if nm.startswith("Sigma"):
        print(f"   Gamma = {G:.0f} MeV -> tau = {tau:.2e} s, c*tau = {tau*c*1e15:.1f} fm")
    else:
        print(f"{nm}: Gamma = {G:.1f} MeV -> tau = {tau:.2e} s, c*tau = {tau*c*1e15:.1f} fm")
print("   even the narrowest resonance here decays within 30 fm of where it was made")

prints

K- p at p_K = 1.5 GeV/c
 E_K = 1.5792 GeV -> sqrt(s) = 2.0217 GeV
 m^2(L pi) runs from (m_L + m_pi)^2 = 1.5757 to (sqrt(s) - m_pi)^2 = 3.5425 GeV^2
 the book's plot is framed 1.50 to 4.00, with the contour spanning 1.58 to 3.54
Sigma(1385): band at m^2 = 1.9188 GeV^2, and both bands sit there
 Gamma = 35 MeV -> tau = 1.88e-23 s, c*tau = 5.6 fm
Xi(1530): Gamma = 7.2 MeV -> tau = 9.14e-23 s, c*tau = 27.4 fm
 even the narrowest resonance here decays within 30 fm of where it was made

The last line is worth pausing on. The Ξ(1530) is a narrow resonance by the standards of this chapter — 7 MeV against the Δ’s 117 — and it still travels only 27 fm. There is no version of this measurement in which the two vertices are separated; production and reconstruction is the only route.

The Ξ(1530), and an isospin measured from a ratio

The $S = -2$ resonances can only be produced, and the lowest was found in a 1.8 GeV/c K⁻ beam in the same 72″ chamber, in two reactions:

\begin{aligned} K^-p &\to \htmlClass{t-x}{\Xi^{*0}}K^0 \to \htmlClass{t-c1}{\Xi^-\pi^+}K^0 \\ K^-p &\to \htmlClass{t-x}{\Xi^{*0}}K^0 \to \htmlClass{t-c2}{\Xi^0\pi^0}K^0 \end{aligned}

(4.13, 4.14)

Bettini p. 138. The same resonance seen in two of its decay channels — and the ratio of the two peak heights is the measurement.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

Breit–Wigner fit: m = 1529 ± 5 MeV, Γ = 7.2 ± 2 MeV
non-resonant background (the book’s dashed line)

Fig. 4.7 redrawn as the fitted curve and its background — the 72″ bubble chamber's Ξ⁻π⁺ mass distribution. The reaction is almost entirely resonant: the background under the peak is a couple of events per bin against a peak of twenty-five. The Ξ⁰π⁰ distribution shows the same peak at about half the height, and that factor of two is the next argument.

🪜 Reading the isospin off a branching ratio

Step 1 of 5 — Fix what is already known

\Xi^{*0}: \ Q = 0,\ \mathcal{B} = 1,\ S = -2 \ \Rightarrow\ Y = -1,\ I_z = Q - Y/2 = +\tfrac12

Why you may do this: Gell-Mann–Nishijima (3.39) gives the third component with no new measurement. And since it decays to Ξπ — an I = 1/2 with an I = 1 — the total isospin can only be 1/2 or 3/2.

So the question is binary, and a single ratio can settle it.

Bettini pp. 139–140, Eqs. (4.16)–(4.17). Two hypotheses, two Clebsch–Gordan decompositions, one measured ratio.

🔢 Worked example — both hypotheses, from the same Clebsch–Gordan routine

The two decompositions (4.16) and (4.17) are the site’s lib/cg.ts evaluated at $j_1 = 1/2$ , $j_2 = 1$ — no new machinery and no table lookup.

Reproduce it

from math import sqrt, factorial as f
def cg(j1, m1, j2, m2, J, M):                    # Condon-Shortley, = src/lib/cg.ts
    if m1 + m2 != M or abs(M) > J or J > j1 + j2 or J < abs(j1 - j2): return 0.0
    pre = sqrt((2*J+1)*f(int(j1+j2-J))*f(int(j1-j2+J))*f(int(-j1+j2+J))/f(int(j1+j2+J+1)))
    pre *= sqrt(f(int(j1+m1))*f(int(j1-m1))*f(int(j2+m2))*f(int(j2-m2))*f(int(J+M))*f(int(J-M)))
    s = 0.0
    for k in range(40):
        d = [j1+j2-J-k, j1-m1-k, j2+m2-k, J-j2+m1+k, J-j1-m2+k]
        if any(x < 0 or x != int(x) for x in d): continue
        s += (-1)**k/(f(k)*f(int(d[0]))*f(int(d[1]))*f(int(d[2]))*f(int(d[3]))*f(int(d[4])))
    return pre * s

print("Xi*0 has Iz = +1/2 and decays to Xi(I=1/2) + pi(I=1)")
for I, tag in ((1.5, "3/2"), (0.5, "1/2")):
    a = cg(0.5, -0.5, 1, 1, I, 0.5)              # Xi- pi+
    b = cg(0.5,  0.5, 1, 0, I, 0.5)              # Xi0 pi0
    verdict = "the neutral peak would be TWICE as high" if b*b > a*a else "the neutral peak is HALF as high"
    print(f"  I = {tag} :  <Xi- pi+| = {a:+.4f} -> {a*a:.4f}    <Xi0 pi0| = {b:+.4f} -> {b*b:.4f}")
    print(f"             ratio Xi0pi0 / Xi-pi+ = {b*b/(a*a):.3f}   -> {verdict}")
print("measured: about 1/2  ->  I(Xi*) = 1/2, and I = 3/2 is wrong by a factor of 4")
print(f"sqrt(1/3) = {sqrt(1/3):.4f}, sqrt(2/3) = {sqrt(2/3):.4f} — the coefficients the book prints")

prints

Xi*0 has Iz = +1/2 and decays to Xi(I=1/2) + pi(I=1)
I = 3/2 :  <Xi- pi+| = +0.5774 -> 0.3333    <Xi0 pi0| = +0.8165 -> 0.6667
           ratio Xi0pi0 / Xi-pi+ = 2.000   -> the neutral peak would be TWICE as high
I = 1/2 :  <Xi- pi+| = -0.8165 -> 0.6667    <Xi0 pi0| = +0.5774 -> 0.3333
           ratio Xi0pi0 / Xi-pi+ = 0.500   -> the neutral peak is HALF as high
measured: about 1/2  ->  I(Xi*) = 1/2, and I = 3/2 is wrong by a factor of 4
sqrt(1/3) = 0.5774, sqrt(2/3) = 0.8165 — the coefficients the book prints

Note what the measurement did not need: no absolute rate, no efficiency, no luminosity. Two peaks in the same experiment, and their ratio. This is the §3.9 method — cancel everything you cannot compute — applied to a decay instead of a cross-section.

Nine particles, three rows, one gap

Erratum — the cross-reference on p. 140

Introducing this multiplet the book writes: “The comparison with the $J^P = 1/2^+$ baryons (Fig. 3.1)”. Fig. 3.1 is the KTeV φ-angle distribution of §3.5 — a plot about parity, not a multiplet. The $1/2^+$ baryon multiplets are Fig. 3.2.

The comparison the sentence intends is worth making, and it is the one §4.8 eventually explains: the $1/2^+$ baryons form an octet and these $3/2^+$ ones a decuplet, and why nature builds exactly those two families and not the other ten conceivable ones is the deepest result in the chapter.

Fig. 4.8 — the strongly decaying J^P = 3/2⁺ baryons · J^P = 3/2⁺

Click any member for its quantum numbers — and for the Gell-Mann–Nishijima check I_z = Q − Y/2.

Masses are the book's figure values in MeV. Quark contents are shown here for orientation, but nothing in §4.2 used them — every one of these particles was found and characterised by cross-sections, Dalitz plots and Clebsch–Gordan ratios, years before the quark model existed.

💡 What this really says — the shape is a triangle, and it is missing a corner

Compare this figure with the $J^P = 1/2^+$ octet of §3.8 and the difference is stark:

	1/2⁺ ground states	3/2⁺ resonances
$Y = +1$	a doublet: n, p	a quartet: Δ⁻ Δ⁰ Δ⁺ Δ⁺⁺
$Y = 0$	a singlet (Λ) and a triplet (Σ)	a triplet only — no 3/2⁺ Λ exists
$Y = -1$	a doublet: Ξ⁻ Ξ⁰	a doublet: Ξ⁻ Ξ⁰
total	8	9 so far

This is the decuplet under construction, and two features are begging for an explanation. There is no $I = 0$ , $S = -1$ member — a 3/2⁺ partner of the Λ was looked for and does not exist. And the rows are shrinking by one member each time: 4, 3, 2. A row of 1 would complete a triangle, at $Y = -2$ , $S = -3$ .

The mass spacing points at where it would sit. From 1232 to 1385 is 153 MeV, from 1385 to 1530 is 145 MeV — equal steps, one per unit of strangeness. One more step lands near 1675 MeV.

That particle is the Ω⁻, and §4.8 tells the story of predicting it on a napkin and finding it in a single bubble-chamber photograph. Both features — the missing singlet and the completed triangle — come out of the quark model for free.

🔑 If you remember only three things

The technique was chosen by the beam, not by the physics. Formation needs a beam of the entrance channel, and nobody can build a beam of Σ.
No visible peak is not the same as no state. Three poles share one plot here and only one of them makes a bump in the cross-section.
An isospin can be read off a ratio. The Ξ(1530) got its assignment from how its decays divide, without a multiplet ever being completed.

Where this goes next

§4.3–4.4 proves the claim this section leaned on: that phase space is uniform in the squared masses, so structure on a Dalitz plot is dynamics. It then extracts spin and parity from the shape of that structure.
§4.5 does for mesons what this section did for baryons, with the η and ω found as bumps in a 3π mass distribution.
§4.8 supplies the tenth member of this figure, and explains both the missing 3/2⁺ Λ and why the Δ⁺⁺ needs colour to exist at all.
§6.1 revisits the “cross-section settles near 25 mb” remark, which is the beginning of a different story about what a hadron looks like at high energy.

✅ Check yourself — the 3/2⁺ baryons

0/5 answered · 0 correct

1.No resonance is found in the K⁺–nucleon system at any energy. Why is that a result rather than a gap in the data?
2.In the πp cross-section plot, the total and elastic curves coincide at low momentum and separate as the energy rises. What is that separation?
3.Set the Dalitz plot's resonance strength to zero. What does the resulting picture prove?
Hint: What would you expect to see if the phase-space element were not uniform in these variables?
4.The Ξ*⁰ peak in Ξ⁰π⁰ is about half the height of the one in Ξ⁻π⁺. How does that fix the isospin?
5.The three rows of Fig. 4.8 have 4, 3 and 2 members, at masses 1232, 1385 and 1530 MeV. What does that pattern predict?