The Parity of the Pions · Elementary Particle Physics

Both measurements on this page work by elimination rather than by observation: the apparatus removes every possibility except one, and whatever survives is the result.

🎯 Why this matters

An answer reached by elimination is only as strong as the list it eliminated from. That is why the state count here is done exhaustively rather than argued — an overlooked possibility would not weaken the result, it would reverse it.

§3.2 and §3.3 built the machinery. This page is the chapter’s experimental heart: two measurements that turn P from a rule into a number, one for the charged pion and one for the neutral. Neither measures parity directly — parity is not a thing you can point an instrument at. Both instead arrange a situation in which only one answer is consistent, which is the argument style that dominates the rest of the book. The second one matters twice over: the book says explicitly that §9.15 determines the spin and parity of the Higgs boson the same way.

The charged pion: stop it in deuterium

The reaction is capture at rest,

\htmlClass{t-pi}{\pi^-} + \htmlClass{t-d}{d} \;\to\; \htmlClass{t-n}{n} + \htmlClass{t-n}{n}

(3.20)

Bettini p. 113. A low-energy π⁻ beam is stopped in liquid deuterium; the pions lose their energy by ionisation and come to rest in the liquid.

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.

💡 What this really says — the reaction is chosen so that the only unknown is the pion’s parity

Everything about this reaction is chosen so that the only unknown is the pion’s parity. The initial state is fixed: a pion at rest in an $l = 0$ orbit, so $J = 1$ from the deuteron’s spin alone, and an initial parity of $P(\pi)\cdot P(d)\cdot(-1)^0 = P(\pi)$ . The final state is fixed too, because two identical fermions with $J = 1$ have exactly one available configuration.

So the equation reads: if this reaction happens at all, the pion’s parity equals the parity of that one final state. Observing the reaction is the measurement. There is nothing to fit and no rate to model — the answer is carried by the existence of the process, not by its size.

📐 Physics you need first — why a pion always ends up in an S orbit

The argument above assumed $l = 0$ for the captured pion, and that assumption carries the whole result. It is not obvious, and it took a piece of atomic physics (Day, Snow and Sucher, 1960) to establish.

A mesic atom is tiny. Replace an electron by a π⁻ and the Bohr radius shrinks by the mass ratio, $a \propto 1/m$ : the pion is 273 times heavier than the electron, so the orbit is 273 times smaller. That one number drives everything else.

Capture happens from S states. The pion cascades down to $n \approx 7$ within about a picosecond. At those levels only the $S$ orbitals have appreciable wave function at the nucleus — the same $l = 0$ property that makes electron capture possible in nuclear physics — so a pion that reaches an $S$ state is absorbed almost at once.

And the non-S pions get there anyway. This is the subtle step. Because the mesic atom is so small, it is electrically neutral on the outside and drifts into other molecules, where it meets the intense, non-uniform field near another nucleus. That field Stark-mixes the levels of fixed $n$ — the states of different $l$ are degenerate in a pure Coulomb potential, so an external field mixes them freely — and the mixture repopulates the $S$ waves, which are then absorbed immediately.

The conclusion: practically all captures occur from $l = 0$ . It was checked, not assumed, by measuring the X-rays from the atomic cascade.

🪜 From a stopped pion to P(π) = −1

Step 1 of 6 — Fix the initial angular momentum

J_i = s_d \oplus s_\pi \oplus l = 1 \oplus 0 \oplus 0 = 1

Why you may do this: The deuteron has spin 1, the pion spin 0, and capture is from l = 0 (the prerequisite box above). Nothing is left to choose: the initial state has J = 1.

Bettini pp. 113–114. Six steps, and the only experimental input is the last one: that the reaction happens at all.

🔢 Worked example — the state count, done exhaustively

The step that carries the argument is “there is only one choice”. That deserves a machine check rather than trust, because the whole result rests on no other $J=1$ state having been overlooked.

Reproduce it

L, ok, dead = 'SPDF', [], []
for l in range(4):
    for s in (0, 1):
        for J in range(abs(l - s), l + s + 1):
            term = f"{2*s+1}{L[l]}{J}"
            if (l + s) % 2:                        # exchange = (-1)^l * (-1)^(s+1) != -1
                dead.append(term); continue
            ok.append((term, l, s, J, (-1)**l))    # P = (+1)(+1)(-1)^l for two nucleons

print("two identical neutrons: l+s must be EVEN (antisymmetric overall)")
for term, l, s, J, P in ok:
    tag = "   <-- the only J=1 state" if J == 1 else ""
    print(f"   {term}   l={l} s={s}  J={J}   P={'+' if P > 0 else '-'}{tag}")
one = [t for t, l, s, J, P in ok if J == 1]
print(f"states with J=1: {one}  -> parity {[P for t, l, s, J, P in ok if J == 1][0]}")
print("deleted by Fermi statistics (l+s odd): " + ", ".join(dead))

prints

two identical neutrons: l+s must be EVEN (antisymmetric overall)
 1S0   l=0 s=0  J=0   P=+
 3P0   l=1 s=1  J=0   P=-
 3P1   l=1 s=1  J=1   P=-   <-- the only J=1 state
 3P2   l=1 s=1  J=2   P=-
 1D2   l=2 s=0  J=2   P=+
 3F2   l=3 s=1  J=2   P=-
 3F3   l=3 s=1  J=3   P=-
 3F4   l=3 s=1  J=4   P=-
states with J=1: ['3P1']  -> parity -1
deleted by Fermi statistics (l+s odd): 3S1, 1P1, 3D1, 3D2, 3D3, 1F3

Eight states up to $l = 3$ , exactly one of them with $J = 1$ , and six more deleted before they could compete. Two of the deleted ones — $^3S_1$ and $^1P_1$ — would have had $J = 1$ : without Fermi statistics the argument would have had three candidates and no conclusion. Switch the widget in §3.3 to its Majorana mode to see the identical deletion happen there.

⚙️ Engineer’s bridge — a constraint problem with exactly one solution

Read the whole argument as a constraint satisfaction over a finite space:

variables: $l$ , $s$ of the final state;
constraints: antisymmetry under exchange ( $l + s$ even), and $J = 1$ from the initial state;
objective: none — you are not optimising, you are checking for uniqueness.

The solution set has exactly one element, and that is the measurement. No rate was computed, no matrix element was estimated, nothing was fitted. The physics input is a pair of integer constraints, and the output is a sign.

This is the same move as deriving a value from an invariant rather than measuring it: if a checksum, a parity bit or a type constraint admits only one consistent completion, you do not need to observe the missing field — you reconstruct it. The engineering instinct that “over-constrained systems are informative” is precisely why the deuteron was the right target.

Where it breaks: the reconstruction is only as good as the constraint list. Miss one allowed state and the solution stops being unique — which is why the exhaustive enumeration above is worth running rather than trusting.

Where it breaks: exhaustive enumeration is only decisive while the constraints are exact. Every rule used here — parity conservation, angular momentum, Bose symmetry — is a strong-interaction statement, and the weak interaction respects none of the first. So the argument establishes the pion’s parity given that the capture proceeds strongly, and that premise is doing real work: the same enumeration applied to a weak process would return an empty solution set and be wrong rather than informative.

🔬 Experiment card — Panofsky et al. 1951, π⁻ capture in deuterium

Apparatus

The target is the apparatus: a low-energy π⁻ beam brought into liquid deuterium, slow enough that a large fraction of the pions lose all their energy to ionisation and stop inside the liquid. Everything after that — the atomic capture, the cascade, the Stark mixing — happens on its own, in picoseconds.

What is measured

Not a spectrum and not an angle: whether the reaction happens, and whether its rate is suppressed. A parity-forbidden capture would be absent or drastically rarer than the competing channels; an allowed one is not.

The result

π⁻d → nn proceeds, with a cross-section that is not suppressed.

What it proved

P(\pi^-) = -1

. The pion is a pseudoscalar , and with

J = 0

it is the archetype of the

J^P = 0^-

mesons that fill the tables in Chapter 4. The measurement is entirely a yes/no observation converted into a sign by the state counting above.

The neutral pion: spin and parity from two photons

A meson of nearly the same mass as the charged $\pi^\pm$ was seen decaying as

\htmlClass{t-pi}{\pi^0} \to 2\htmlClass{t-g}{\gamma}

(3.21)

Bettini p. 114. To call this particle the neutral pion, it has to be shown to be a pseudoscalar like the charged ones — and the two-photon final state is what makes that possible.

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.

Step one: the spin cannot be 1

The argument uses no dynamics at all. In the centre-of-mass frame there are only three vectors available — the photon momentum $\mathbf q$ and the two transverse polarizations $\mathbf e_1$ , $\mathbf e_2$ — and the amplitude must be built from them, linear in each polarization because each photon appears once.

For $J^P = 1^-$ the amplitude must read $M \propto \mathbf V\cdot\mathbf S$ with $\mathbf V$ a vector; for $J^P = 1^+$ , $M \propto \mathbf A\cdot\mathbf S$ with $\mathbf A$ an axial vector, where $\mathbf S$ is the π⁰ polarization. Two conditions then apply to whatever is built:

\underbrace{\htmlClass{t-ex}{\mathbf e_1 \leftrightarrow \mathbf e_2,\quad \mathbf q \leftrightarrow -\mathbf q}}_{\text{Bose symmetry}} \qquad\qquad \underbrace{\htmlClass{t-tr}{\mathbf e_1\cdot\mathbf q = 0,\quad \mathbf e_2\cdot\mathbf q = 0}}_{\text{transversality}}

(3.22, 3.23)

Bettini p. 114. Two constraints, and between them they exhaust the possibilities.

Every symbol, one at a time

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Everything you can build, and what happens to it

candidate	what it is	under (3.22)	under (3.23)	verdict
$(\mathbf e_1 \cdot \mathbf e_2)\,\mathbf q$	vector — the J^P = 1⁻ candidate		survives	FAILS Bose symmetry
$\mathbf e_1 \times \mathbf e_2$	axial vector — the J^P = 1⁺ candidate		survives	FAILS Bose symmetry
$(\mathbf e_1\cdot\mathbf q)\,\mathbf e_2 \pm (\mathbf e_2\cdot\mathbf q)\,\mathbf e_1$	vector, either sign	either		FAILS transversality
$\mathbf e_1 \cdot \mathbf e_2$	scalar — the J^P = 0⁺ candidate		survives	ALLOWED
$(\mathbf e_1 \times \mathbf e_2)\cdot\mathbf q$	pseudoscalar — the J^P = 0⁻ candidate		survives	ALLOWED

Once (3.23) has removed everything containing e·q, only two independent structures remain — e₁·e₂ and e₁ × e₂ — and both are antisymmetric when dressed with the extra q a spin-1 amplitude needs. That is the whole of the book's “no vector or axial vector can be built”: not a failure of imagination, an exhausted list.

💡 What this really says — the spin-1 exclusion is a counting argument, not a dynamical one

The spin-1 exclusion is a counting argument about available structures, and it is worth separating from the parity question that follows.

A spin-1 parent needs an amplitude carrying one free vector index, to contract with its own polarization $\mathbf S$ . Building that index out of $\mathbf e_1, \mathbf e_2, \mathbf q$ requires an odd number of leftover $\mathbf q$ ‘s — and $\mathbf q$ is exactly the object that flips sign when the two identical photons are swapped. So a spin-1 amplitude is forced to be antisymmetric, and Bose statistics forbids it.

A spin-0 parent needs no index at all, so no leftover $\mathbf q$ is required, and the two surviving structures both come out symmetric. Spin 0 is allowed and spin 1 is not, for reasons that never mention the pion. This is the same statement as the Landau–Yang theorem, and it is why the discovery of a two-photon resonance at 125 GeV immediately told the LHC experiments that the Higgs boson is not spin 1 (§9.15).

Step two: scalar or pseudoscalar

Both surviving structures are legal, and they differ by exactly one thing — how they want the two polarizations oriented.

\begin{aligned} M &= \htmlClass{t-aS}{a_S}\;\htmlClass{t-dot}{\mathbf e_1 \cdot \mathbf e_2} && \text{scalar } (J^P = 0^+) \\[3pt] M &= \htmlClass{t-aP}{a_P}\;\htmlClass{t-cross}{\mathbf e_1 \times \mathbf e_2 \cdot \mathbf q} && \text{pseudoscalar } (J^P = 0^-) \end{aligned}

(3.24)

Bettini p. 115. Two amplitudes, one observable difference: a dot product is maximal for parallel polarizations, a cross product for perpendicular ones.

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.

💡 What this really says — parity reduced to a geometry question about two polarization vectors

Parity has been reduced to a geometry question about two polarization vectors: does the decay prefer them parallel or perpendicular?

The reason a cross product means “pseudoscalar” is the classification from §3.2: $\mathbf e_1 \times \mathbf e_2$ is an axial vector , and dotting an axial vector with the polar vector $\mathbf q$ gives a quantity that changes sign under inversion. It is the same construction as $\boldsymbol\sigma\cdot\mathbf p$ in the ParityLab, and the same one that will identify parity violation in Chapter 7. Build the combination in that widget and read the sign yourself.

The catch is that photon polarization is not measurable at 67 MeV. There is no polarimeter for a gamma ray of that energy, so the elegant statement above is, by itself, unusable — and the rest of the section is about how the experiment got at it anyway.

Step three: read the polarization off a pair of planes

Rarely — with a branching ratio of $3.4 \times 10^{-5}$ — both photons are virtual and convert internally:

\htmlClass{t-pi}{\pi^0} \to \gamma^* + \gamma^* \to \bigl(\htmlClass{t-p1}{e^+e^-}\bigr)_1 + \bigl(\htmlClass{t-p2}{e^+e^-}\bigr)_2

(3.25)

Bettini p. 115 — double internal conversion. A virtual photon may carry a non-zero mass and turns straight into an e⁺e⁻ pair whose invariant mass is that mass.

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.

💡 What this really says — polarization → plane normal → measurable angle

The chain is: polarization → plane normal → measurable angle. You cannot measure a photon’s polarization, but you can measure the directions of four charged tracks, and the plane each pair defines remembers the polarization of its parent. So the abstract preference of (3.24) — parallel or perpendicular polarizations — becomes a histogram of one angle φ between two planes, peaking at $\varphi = 0$ for a scalar and $\varphi = \pi/2$ for a pseudoscalar.

It costs a factor $3.4\times10^{-5}$ in rate to convert an unmeasurable quantity into a measurable one. That trade — accept a rare channel because it carries information the common one destroys — is one of the field’s recurring moves, and it appears again in every “golden channel” of Chapter 9.

The π⁰ parity · the angle between two e⁺e⁻ planes

angle between the planes φ = 90°

0⁺ scalar · wants ∥ polarizations	0.53
0⁻ pseudoscalar · wants ⟂ polarizations	1.47
ratio 0⁻ / 0⁺ at this φ	2.77

Relative rates from dN/dφ ∝ 1 + a cos 2φ with a = ∓0.47. The two hypotheses are furthest apart at φ = 0 and φ = 90°, and indistinguishable at φ = 45° — where cos 2φ vanishes and every parity gives the same answer. An experiment that only measured near 45° would learn nothing.

J^P = 0⁺ scalar, a = +0.47
J^P = 0⁻ pseudoscalar, a = −0.47

The whole parity measurement, as one picture: two curves in antiphase, separated by 2a at φ = 0 and 90°, crossing at 45° and 135°. The π⁰ is the lower curve at φ = 0.

what statistics buys

events collected N = 30,511

σ(⟨a⟩) = 0.015 · the two hypotheses are 64.6σ apart — settled beyond argument.

Scaled from the error Samios actually published, σ = 0.24 on 112 events, as 1/√N. At 112 this gives 3.9σ against the 3.3σ the book quotes, because the published figure came from a full likelihood-ratio analysis over the measured pair masses rather than from ⟨a⟩ alone — so read the curve as the shape of the gain, not as a reproduction of either result.

Erratum — Eq. (3.26) is missing its constant term

The book prints the φ distribution as

\frac{dN}{d\varphi\,dm_1\,dm_2} \propto a(m_1, m_2)\cos 2\varphi ,

which cannot be right: with $a < 0$ — the pseudoscalar case the section is arguing for — the right-hand side is negative over half the range, and a number of events cannot be negative. The intended expression carries a constant term,

\frac{dN}{d\varphi\,dm_1\,dm_2} \propto 1 + a(m_1, m_2)\cos 2\varphi ,

and the book’s own Fig. 3.1 confirms it: both plotted curves oscillate about roughly 600 events per bin and never come near zero. With this form, $a > 0$ peaks at $\varphi = 0$ (scalar) and $a < 0$ at $\varphi = \pi/2$ (pseudoscalar), which is exactly what the text says two paragraphs earlier. Everything downstream — including $\langle a\rangle = -0.41 \pm 0.24$ against expectations of $\pm 0.47$ — is unaffected.

KTeV fit, J^P = 0⁻ (a ≈ −0.36 as plotted)
J^P = 0⁺ expectation, same amplitude
ideal 0⁻ shape, a = −0.47 (no acceptance)

Fig. 3.1 redrawn — the two fitted curves, not the 50 data points, which the printed figure does not support digitising. The KTeV points sit on the continuous 0⁻ curve; the dashed 0⁺ curve is in antiphase with it, and the two are furthest apart exactly where the polarizations are parallel or perpendicular.

📏 Why the plotted amplitude is 0.36 and not 0.47

Read the amplitude straight off Fig. 3.1: the curves run between about 385 and 815 events per bin, so they oscillate about 600 with an amplitude near 215 — a fractional modulation of $215/600 = 0.36$ , not the 0.47 the theory quotes for $\langle a \rangle$ .

Nothing is wrong. $\pm 0.47$ is the ideal form-factor average; what a detector plots is that shape after acceptance, resolution and the $m_1, m_2$ dependence of the form factor have all diluted it. A real angular distribution is almost always flatter than the theory curve, because every effect that mixes bins pushes the modulation toward zero and none pushes it up.

The practical consequence is that a measured amplitude cannot be compared with a theoretical one until the dilution is modelled — which is why the published result is a likelihood ratio over the full $(\varphi, m_1, m_2)$ distribution rather than a number read off a histogram.

🔬 Experiment card — Samios et al. 1962, 112 events in a bubble chamber

Apparatus

A hydrogen bubble chamber (§1.13b) — the four charged tracks of reaction (3.25) are photographed directly, and their momenta come from curvature in the chamber’s field.

What is measured

Per event: the angle φ between the two pair planes, and both pair masses

m_1

m_2

. The masses matter because the form factor depends on them, so each event carries a different amount of information.

The result

A likelihood ratio over all 112 events favouring the pseudoscalar hypothesis by 3.3 standard deviations, expressible as a weighted average form factor

\langle a\rangle = -0.41 \pm 0.24

against expectations of

+0.47

(scalar) and

-0.47

(pseudoscalar).

What it proved

The π⁰ is a pseudoscalar — at 3.3σ, which in 1962 was persuasive and by modern standards is evidence, not observation. Its quantum numbers

J^P = 0^-

match the charged pions, which is what licenses calling it the third member of the isospin triplet in §3.8.

\langle \htmlClass{t-a}{a} \rangle = \htmlClass{t-val}{-0.41 \pm 0.24} \qquad\text{against}\qquad \htmlClass{t-th}{+0.47}\ (0^+), \quad \htmlClass{t-th}{-0.47}\ (0^-)

(3.27)

Bettini p. 115. One number, one sign, and an error bar large enough that the sign is the only thing it establishes.

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.

🔬 Experiment card — KTeV 2008 at Fermilab, 30 511 events

Apparatus

A dedicated kaon-beam experiment with a magnetic spectrometer and calorimetry — the same machine built to measure CP violation in neutral kaons (§8.5), which produces π⁰ mesons by the billion as a by-product.

What is measured

The same angle φ, on 272 times as many events. At a branching ratio of

3.4 \times 10^{-5}

, 30 511 candidates imply of order

10^9

reconstructed π⁰ decays.

The result

The φ distribution of Fig. 3.1, following the

0^-

curve and in antiphase with the

0^+

one. Scaling Samios’ published error as

1/\sqrt N

gives

\sigma(\langle a\rangle) \approx 0.015

, so the two hypotheses sit some 60 standard deviations apart.

What it proved

Nothing new — and that is the point. The same method, the same angle and the same two curves, with the statistics turned up by a factor of 272, convert a 1962 indication into a settled fact. Move the N slider in the widget above to watch the separation grow as

\sqrt N

⚙️ Engineer’s bridge — this is matched filtering, and $\sqrt N$ is the whole story

Strip the physics away and the measurement is a familiar signal-processing task: decide which of two known shapes generated a noisy dataset.

The two hypotheses are the templates $1 \pm 0.47\cos 2\varphi$ . They are known in advance, with no free parameters — the ideal case for a likelihood-ratio test, which is the optimal detector for exactly this problem.
Every event is one sample. The uncertainty on the fitted amplitude falls as $1/\sqrt N$ , so separation between the hypotheses grows as $\sqrt N$ : 112 events give a few σ, and 272 times more gives $\sqrt{272} \approx 16.5$ times the separation.
Measure where the templates differ most. The two curves cross at $\varphi = 45°$ and $135°$ , where an event carries no information at all, and separate maximally at $0°$ and $90°$ . Slide the φ control in the widget to those points and watch the ratio of the two rates go to 1 and to 2.8.

That last item is experiment design in one sentence, and it is the same optimality principle as choosing where to place samples when fitting a known model — put your measurements where the candidate curves disagree.

Where it breaks: a matched filter assumes the templates are exact. Here they are diluted by acceptance (the 📏 note above), so the real analysis fits the dilution too rather than assuming the ideal shapes.

Where it breaks: a matched filter is optimal against known noise with a known template, and here neither is quite given. The acceptance dilutes the shapes, so the real fit must model the dilution as well — meaning the template depends on a detector simulation, and a wrong simulation biases the answer rather than merely widening it. Optimality is a property of the model you filter with, not of the technique.

🔢 Worked example — what 112 events buy, and what 30 511 do

The book quotes 3.3σ for Samios and gives no number for KTeV. Both can be estimated from the one published error, $\sigma(\langle a\rangle) = 0.24$ on 112 events, scaled as $1/\sqrt N$ .

Reproduce it

import numpy as np
a, s112, N112 = 0.47, 0.24, 112                      # theory value, Samios error, Samios N
sep = 2 * a
print(f"separation between the two hypotheses: |+{a} - (-{a})| = {sep:.2f}")
for N, tag in ((112, 'Samios 1962; book quotes 3.3'), (30511, 'KTeV 2008')):
    s = s112 * np.sqrt(N112 / N)
    print(f"   N = {N:6d}  sigma(<a>) = {s:.4f}  -> {sep/s:4.1f} sigma   ({tag})")
r = 30511 / N112
print(f"   statistics gained: {r:.1f}x events -> {np.sqrt(r):.1f}x separation")
print(f"<a> = -0.41 +- 0.24 is {0.41/0.24:.1f} sigma from zero, "
      f"{(0.47+0.41)/0.24:.1f} sigma from the scalar +{a}")
print(f"KTeV: 30511 candidates at BR 3.4e-05 -> of order {30511/3.4e-5:.1e} pi0 decays behind them")

prints

separation between the two hypotheses: |+0.47 - (-0.47)| = 0.94
 N =    112  sigma(<a>) = 0.2400  ->  3.9 sigma   (Samios 1962; book quotes 3.3)
 N =  30511  sigma(<a>) = 0.0145  -> 64.6 sigma   (KTeV 2008)
 statistics gained: 272.4x events -> 16.5x separation
<a> = -0.41 +- 0.24 is 1.7 sigma from zero, 3.7 sigma from the scalar +0.47
KTeV: 30511 candidates at BR 3.4e-05 -> of order 9.0e+08 pi0 decays behind them

The naive estimate gives 3.9σ where the paper reports 3.3σ, and the gap is instructive: the published figure came from a likelihood ratio over the full $(\varphi, m_1, m_2)$ distribution, where events with unfavourable pair masses carry less weight than the flat average assumes. Simple scaling always flatters — treat it as the shape of the gain, not as a substitute for the analysis.

What the scaling does capture exactly is the shape: 272 times the events buys 16.5 times the separation, never 272 times. Every improvement in a statistics-limited measurement is bought at that square-root exchange rate, which is why the step from evidence to settled took 46 years and a kaon factory.

Erratum — the name of the third author

The running text on p. 113 credits the mesic-atom cascade theory to “T. B. Day, G. A. Snow and J. Sucker”. The name is J. Sucher — Joseph Sucher, third author of Day, Snow and Sucher, Phys. Rev. Lett. 3 (1960) 61. The book’s own reference list has it right; only the running text does not.

🔑 If you remember only three things

The 2008 repeat changed the statistics, not the argument. A hundred and twelve events in 1962 and thirty thousand in 2008, with the same geometry doing the work both times.
Spin has to be settled before parity can be asked about. The exclusion of spin 1 is a separate counting argument, and the parity question is only meaningful once it has been made.
One unproven assumption carries the whole charged-pion result. That the captured pion sits in an S orbit is not obvious, and it needed a theory of its own before the measurement could be trusted.

Where this goes next

§3.6–3.7 turns to the additive counting rules — baryon and lepton number — and to the proton-decay searches that bound them.
§3.8 needs both results from this page: the π⁰ can only join the π⁺ and π⁻ in an isospin triplet because all three share $J^P = 0^-$ .
§4.5 is the multiplet of $0^-$ mesons this measurement admits the pion to, and the search for states whose $J^{PC}$ no $q\bar q$ pair can produce — the lattice in §3.3.
§7.2–7.3 is what happens when parity stops being conserved, so that an argument like this one no longer runs.
§9.15 determines the spin and parity of the Higgs boson from a two-photon final state and an angular distribution. Bettini flags the connection here explicitly: this page is the rehearsal for that one.

✅ Check yourself — the parity of the pions

0/5 answered · 0 correct

1.In $\pi^- d \to nn$ , why does the final state have exactly one option rather than three?
Hint: Which $J = 1$ states would exist if the two neutrons were distinguishable?
2.The pion is captured from an $l = 0$ orbit. What actually guarantees that, given that it starts in a high- $l$ state?
3.Why can a particle decaying to two photons not have spin 1?
4.Set φ = 45° in the PlaneAngleLab above. What happens to the ratio of the two hypotheses' rates, and why does it matter?
Hint: Look at $\cos 2\varphi$ at 45°.
5.KTeV collected 272 times as many events as Samios. Roughly how much better is the separation between the two parity hypotheses?