Symmetries; Parity · Elementary Particle Physics

Symmetries are sorted here by what they pay out: a conserved charge, a family of states with the same mass, a multiplicative label, or a rule that one interaction keeps and another breaks.

🎯 Why this matters

Whether a symmetry holds is a question about an interaction rather than about the universe. Parity survives the strong and electromagnetic forces and fails in the weak one, so every conservation statement in this book carries an implicit “in which interaction?”.

Chapters 1 and 2 were about things — first instruments, then particles. This chapter is the first one about rules, and it is the hinge of the book: Chapters 4, 7, 8 and 9 are largely applications of what is defined here. The organising idea arrives once, in §3.1, and is then used relentlessly — a symmetry of the Lagrangian is a conservation law, and the pattern of which interaction respects which symmetry is the entire taxonomy of particle physics. This page sets up that idea and then spends the rest of its length on the first and simplest of the discrete symmetries: parity.

📐 Physics you need first — Lagrangians, Noether, and why any of this is a theorem

The book assumes you have met the Noether theorem in classical mechanics. If you have not, here is the whole of it in the language you already have.

A physical system is summarised by one scalar function, the Lagrangian $\mathcal{L}$ , built from the fields and their derivatives. The equations of motion are whatever makes $\int \mathcal{L}\, \mathrm{d}^4x$ stationary — the system’s trajectory is the extremum of a cost functional, exactly as a fitted model is the extremum of a loss. You never need to solve that variational problem on this site; you only need to know that $\mathcal{L}$ contains all the dynamics, so a statement about $\mathcal{L}$ is a statement about everything the system can do.

Noether’s result is then: if a continuous transformation with $n$ parameters leaves $\mathcal{L}$ unchanged, there exist $n$ quantities that do not change with time. Invariance under time translation gives energy; under space translation, momentum; under rotation, angular momentum. Those three are the content of “the laws are invariant under the Poincaré group”, which is assumed everywhere in this book and never re-derived.

The part that is genuinely new to an engineer is the converse habit of mind. In the physics you have met, conservation laws are postulates you are handed. Here they are consequences, and every one of them can be traced back to a specific invariance. That is why a chapter about symmetry is also a chapter about which reactions are allowed to happen.

⚙️ Engineer’s bridge — symmetry is an invariant, a conservation law is an assertion

You already write code in exactly this shape. A loop invariant is a predicate the loop body is not allowed to disturb; you prove it once at the top and then get to assume it everywhere below, which is what makes reasoning about the loop tractable at all. A conservation law is the same contract: electric charge in equals electric charge out, at every vertex, in every process, forever.

The analogy is tighter than it sounds. A selection rule — “this transition cannot happen because it would change $S$ by two units” — is a type error: the transition is not merely unlikely, it is not expressible. Quantum numbers are typed, additive tags, and the composition rules of the type system are the conservation laws.

Where it breaks. In software you choose the invariants and the compiler enforces them. Here nature chose, and the compiler is an experiment. Section 3.1 is careful about this: some quantum numbers are conserved because a symmetry of the Lagrangian forces it, and others are conserved only because nobody has ever seen them fail. The two look identical in a table and are completely different in status — which is the point of the next four subsections.

3.1 Four kinds of symmetry

Bettini sorts symmetries into four categories before doing anything with them. Read the table as a taxonomy you will keep returning to, not as something to memorise now.

The four categories of §3.1, and what each buys you

Category	Continuous?	The book’s example	Where it lands
Space-time (Poincaré)	yes	Lorentz invariance	assumed everywhere; §1.1–1.5
Gauge	yes	U(1) → electric charge; SU(3) → colour; SU(2)⊗U(1) → electroweak	ch05, ch06, ch09
Dynamical	yes	isospin SU(2) — charge independence of the nuclear force	§3.8–3.10, then SU(3) in ch04
Discrete multiplicative	no	parity P, conjugation C, time reversal T	§3.2–3.5, ch07, ch08

A fifth entry, <strong>symmetry breaking</strong>, is not another kind of symmetry but a statement about the ones above: any of them may hold only approximately, or hold for the Lagrangian and fail for the ground state.

Gauge symmetries and absolutely conserved charges

A quantum number is additive when the value for a system is the sum of the values of its parts. Every charge of a fundamental interaction is of this kind — the electric charge, the three colour charges, the weak charges — and each is absolutely conserved, because the Lagrangian of that interaction is invariant under a local unitary group. The group is called the gauge group :

U(1) for the electromagnetic interaction → electric charge;
SU(3) for the strong interaction → colour;
SU(2) ⊗ U(1) for the electroweak interaction → the weak charges.

Other quantum numbers are additive too — the quark flavours, the baryon number , the lepton flavours and the total lepton number — but no gauge symmetry stands behind them, so nothing guarantees them. As it happens, the quark and lepton flavours are not conserved: the weak interaction changes them, which is what Chapters 7, 8 and 10 are about.

⚙️ Engineer’s bridge — gauge invariance is a local redundancy in the representation

You already accept one gauge freedom without noticing: the zero of electric potential. Grounding a circuit at a different node changes every node voltage and no measurable current. The potential carries one more degree of freedom than the physics does, and the extra one is pure bookkeeping.

A gauge symmetry says the same thing, with the redundancy allowed to differ at every point in space-time independently — that is what local means, and it is a much stronger demand. Insisting that the physics survive an arbitrary per-point relabelling turns out to force the existence of a field to carry the relabelling between neighbouring points. That field is the photon, the gluon, the W and the Z. This site will not derive that (§5.1 and §6.3 do); the thing to carry forward now is the shape of the argument: demand a redundancy, get a force.

Where it breaks. A software abstraction barrier is a convention you can violate at a cost. A gauge redundancy cannot be violated at all — states related by a gauge transformation are not two states, they are one state with two names.

Dynamical symmetries

A dynamical symmetry is continuous and unitary, like a gauge symmetry, but not local and not attached to a force carrier. What it does is sort particles into multiplets whose members behave alike, and thereby set the structure of the mass spectrum. The example the chapter spends seven pages on is the charge independence of the nuclear force, whose symmetry group is SU(2) and whose quantum number is isospin — §3.8–3.10.

⚙️ Engineer’s bridge — a multiplet is a degenerate eigenspace

Take any system with a symmetry — a ring of identical coupled oscillators, a symmetric matrix with a repeated eigenvalue — and its modes come in degenerate sets. The symmetry does not tell you the eigenvalue; it tells you the dimension of the block. Break the symmetry slightly, by detuning one oscillator, and the degenerate set splits into closely spaced levels whose spacing measures the size of the breaking.

That is exactly the structure of an isospin multiplet. The proton and neutron would have identical masses if isospin were exact; they differ by 1.3 MeV out of 939, about one part in 700, and that ratio is the size of the electromagnetic and quark-mass breaking. A multiplet is a repeated eigenvalue, and a mass splitting is a perturbation lifting it.

Where it breaks: degenerate perturbation theory assumes the splitting is small compared with the gap to the next level, so the multiplet stays recognisable. Isospin obliges — a few MeV on a GeV — but flavour SU(3) does not: the strange quark is 93 MeV heavier, the decuplet rungs are 145 MeV apart, and the “perturbation” is comparable to the structure it is perturbing. That is why isospin predicts ratios to about 1 % and SU(3)_f only to tens of per cent, and why treating both as the same kind of statement is the standard way to be wrong in this chapter.

Discrete multiplicative symmetries

Parity P, particle–antiparticle conjugation C, and time reversal T. None can be assembled out of infinitesimal steps, so Noether has nothing to say about them, and the conserved quantity is a multiplicative sign rather than an additive count. P and C supply quantum numbers that label particles; T does not, for a reason deferred to §3.4. The rest of this page is about P.

Symmetry breaking, in two flavours that must not be confused

The two ways a symmetry fails, as two potentials. On the left the symmetry is absent from the theory. On the right it is present in the theory and absent from the answer — and the dashed second minimum, the one the system did not choose, is the reason this case has consequences the left one does not.

Explicit breaking means some interaction simply does not respect the symmetry, or respects it only approximately. Then only the interactions that do respect it conserve the corresponding quantum number — and which those are is always an experimental question, never a theoretical one. That sentence is the licence for the whole of the next widget.

Spontaneous breaking means the interaction does respect the symmetry — the Lagrangian is invariant — but the state of minimum energy, which physicists call the vacuum , is not. Bettini gives four analogies, and they are worth keeping because each highlights a different feature:

A shoal of fish, weightless, far from surface and sea bed. Every direction is equivalent. One fish turns, the shoal follows, and the rotational symmetry is gone. Feature: the breaking is collective — it needs many identical components.
A rectangular metal plate stood on its short edge with a perfectly vertical load applied to the top. Left and right faces are equivalent. Raise the load and at one calculable value the plate buckles — left or right. Feature: there is a critical value of a control parameter, below which nothing happens.
A drop of water floating in a space station, cooled below 0 °C. Perfectly symmetric under any rotation, until it freezes into a crystal with axes pointing somewhere. Feature: you cannot predict the axes by inspecting the liquid.
A ferromagnet cooled through its Curie point. Above it the atomic moments point randomly and the two directions along a microcrystal’s easy axis are equivalent; below it, Weiss domains form and each has chosen one. Feature: the choice is local — different domains choose differently.

Spontaneous breaking happens at the fundamental level too, discovered by Nambu and collaborators in 1960, because in relativistic quantum mechanics the vacuum is not empty but a busy, dynamical state. Goldstone showed in 1961 that spontaneously breaking an exact continuous symmetry produces a number of massless bosons — Goldstone bosons . If the symmetry is also broken explicitly, those bosons acquire mass, the more the larger the explicit breaking, and are called pseudo-Goldstone bosons. That is the case of chiral symmetry in §6.8–6.9, and the same mechanism, applied to a gauge symmetry, becomes the Higgs mechanism in §9.12.

⚙️ Engineer’s bridge — spontaneous breaking is a pitchfork bifurcation

The buckling plate is the textbook supercritical pitchfork: a control parameter (the load) crosses a critical value, one stable fixed point becomes unstable, and two stable ones appear symmetrically on either side. Draw the bifurcation diagram of $\dot{x} = \mu x - x^3$ and you have drawn the right-hand panel above, with $\mu$ the load and $x$ the direction of bow.

Everything the physics says follows from that picture. The equations retain their symmetry — $x \to -x$ leaves $\dot{x} = \mu x - x^3$ unchanged — while every solution past the critical point breaks it. Nothing chose; the symmetric solution merely stopped being stable. And the flat direction connecting the two minima, the one that costs no energy to move along, is the Goldstone mode.

Where it breaks. In a one-dimensional bifurcation the two branches are discrete. When the broken symmetry is continuous, the minima form a whole circle (or sphere) of equivalent states, and it is the existence of that zero-cost direction around the circle that produces a massless particle rather than merely a second option.

The payoff: which interaction conserves what

Everything above exists to make one table meaningful. The book poses it as Problem 3.1 and never prints the answer, so this site is the answer. It is the single most-reused artefact in the chapter — nearly every “is this reaction allowed?” question in Chapters 3, 4, 7 and 8 is settled by reading one cell.

What each interaction conserves — the answer to Problem 3.1

quantum number	strong	electromagnetic	weak
Iisospin
I_zthird component of isospin
Sstrangeness
flavourthe other quark flavours (C, B̃, T)
Bbaryon number
Llepton number (total)
L_e, L_μ, L_τlepton flavours
Pparity
Cparticle–antiparticle conjugation
Ttime reversal
Jangular momentum
J_zthird component of angular momentum
Qelectric charge

parity · weak · violated

Violated maximally, not slightly. This is Chapter 7, and it is the reason a neutrino parity cannot even be defined: neutrinos have only weak interactions.

Y* marks a law that holds in every collision and decay yet fails somewhere else — lepton flavour fails for neutrinos in flight, and T fails only in the neutral-kaon system. Click any cell for the reason and the measurement behind it.

Two rows deserve a second look now, because they are the ones the taxonomy above was built to distinguish. Q is conserved because U(1) gauge invariance forces it. B and L look identical in the table — additive, conserved by everything — but no gauge symmetry protects them, so their conservation is a measurement, and one that people keep trying to break (§3.6–3.7).

3.2 Parity

The parity operation $\mathcal{P}$ inverts the three spatial coordinate axes. Note the dimension-dependence: in two dimensions inverting both axes is the same as rotating by 180°, so it is nothing new. In three dimensions it is not a rotation at all — inverting three axes equals inverting one axis followed by a 180° rotation, and no amount of rotating gets you a mirror image.

Why parity is interesting in three dimensions and trivial in two. The determinant of the transformation is $(-1)^{d}$ — in even dimensions it is $+1$ and the map sits inside the rotation group, in odd dimensions it is $-1$ and it does not.

The five-line scheme, and where the classification comes from

The book gives the bookkeeping in five lines: $\mathcal{P}$ (1) inverts the space coordinates, $\mathbf{r} \Rightarrow -\mathbf{r}$ ; (2) leaves the time alone, $t \Rightarrow t$ ; consequently (3) inverts momenta, $\mathbf{p} \Rightarrow -\mathbf{p}$ ; (4) leaves angular momenta alone, $\mathbf{r} \times \mathbf{p} \Rightarrow \mathbf{r} \times \mathbf{p}$ ; (5) including spin, $\mathbf{s} \Rightarrow \mathbf{s}$ .

More generally: scalars are unchanged, pseudoscalars change sign, vectors change sign and axial vectors do not. That four-way classification is the working vocabulary of every matrix element in the rest of the book — so rather than asserting it, derive it. Every row below is the product of the signs of the pieces the quantity is built from.

How P, C and T act — and where the four-way classification comes from

quantity	P	C	T	it is a…
ttime	+	+	−	scalar
rposition	−	+	+	vector
plinear momentum	−	+	−	vector
Lorbital angular momentum	+	+	−	axial vector
sspin	+	+	−	axial vector
Eenergy	+	+	+	scalar
Qelectric charge	+	−	+	scalar
jcurrent density	−	−	−	vector
𝐄electric field	−	−	+	vector
𝐁magnetic field	+	−	−	axial vector
σ·p̂helicity	−	+	+	pseudoscalar
L·sspin–orbit term	+	+	+	scalar
𝐄·𝐁the E·B invariant	−	+	−	pseudoscalar
s·𝐄electric dipole moment term	−	−	−	pseudoscalar

σ·p̂ — helicity · built from axial vector · polar vector

The most important row on the list. An axial vector dotted into a polar one is a PSEUDOscalar: rotationally invariant, but it changes sign in a mirror. A left-handed particle looks right-handed to its own reflection, which is why "the weak interaction is left-handed" and "the weak interaction violates parity" are the same sentence.

Compose a term — the signs multiply

s·pP −C +T +pseudoscalar

P-odd. A matrix element containing this term changes sign in a mirror, so its interference with any P-even term produces an observable that must vanish if parity is conserved. Measuring that observable to be zero is how the limits in this chapter were set; measuring it to be large is how parity violation was discovered.

The C and T columns belong to §3.3 and §3.4; they are shown here so the table is met once and only once. Nothing in this widget is tabulated data — every sign is the product of the signs of the pieces the quantity is built from, which is what "P, C and T are multiplicative" means.

💡 What this really says — two facts — r flips and t does not — and the rest is arithmetic

There are only two independent facts here — $\mathbf{r}$ flips and $t$ does not — plus one more for C, that charge flips. Everything else is arithmetic. Momentum is a displacement per unit time, so it flips. Angular momentum is a cross product of two things that both flip, so it does not. A magnetic field is a cross product too, so it does not either, while an electric field is a force per charge, so it does.

The classification is then just a two-by-two of (does it have a direction?) against (does it flip?). A quantity with no direction that nevertheless flips is the strange one, and it has a name — pseudoscalar — because it is the alarm bell: helicity $\boldsymbol\sigma\cdot\hat{\mathbf{p}}$ is a pseudoscalar, so an interaction that cares about helicity cannot conserve parity. That single observation is Chapter 7 in one line.

Which states have a parity at all

You may speak of the parity of a state only if it is an eigenstate of $\mathcal{P}$ . The vacuum is one, and its parity is defined to be positive.

A single particle can be a parity eigenstate only if it is at rest — a moving particle has a momentum, and momentum flips. The eigenvalue in that rest frame is called the intrinsic parity , or simply the parity.

Now the subtlety that separates bosons from fermions. Inverting the axes twice returns the original coordinates, so $\mathcal{P}^2$ is the identity — or a rotation by $2\pi$ about some axis, which for scalars, vectors and tensors is the same as no rotation at all. Hence for a boson $P^2 = 1$ , so $P = \pm 1$ .

A fermion is described by a spinor $\psi$ , and a $2\pi$ rotation sends $\psi \to -\psi$ (§2.5). To get back to $\psi$ you must apply $\mathcal{P}$ four times, and that changes the answer:

\htmlClass{t-p4}{P^4} = 1 \quad\Longrightarrow\quad \htmlClass{t-P}{P} = \htmlClass{t-root}{\sqrt[4]{1}} = \pm 1,\ \pm i

Bosons obey P² = 1 and so P = ±1. Fermions obey P⁴ = 1, which admits two more values — and §2.9's Majorana fermion is obliged to use them.

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.

🔢 Worked example — watching the spinor sign appear

The claim “a $2\pi$ rotation costs a spinor a minus sign” is the only input, and it is one matrix multiply. A rotation by $\theta$ about $z$ acts on a spin-½ spinor as $\exp(-i\theta\sigma_z/2)$ ; the half-angle is the whole story.

Reproduce it

import numpy as np

def rot_z(theta):                       # rotation of a spin-1/2 spinor about z
    return np.array([[np.exp(-1j*theta/2), 0], [0, np.exp(+1j*theta/2)]])

psi = np.array([1, 0])                  # spin up
for turns in (1, 2):
    factor = (rot_z(2*np.pi*turns) @ psi)[0].real
    print(f"{turns} full turn(s) of 2pi: psi -> {factor:+.0f} psi")

prints

1 full turn(s) of 2pi: psi -> -1 psi
2 full turn(s) of 2pi: psi -> +1 psi

One turn gives $-1$ , two turns give $+1$ . So $\mathcal{P}^2$ may equal a $2\pi$ rotation $=-1$ on a spinor, which forces $P^4 = 1$ rather than $P^2 = 1$ , and opens the door to $P = \pm i$ . A boson would show $+1$ after a single turn, and the door stays shut.

Relative parities, and the conventions

The absolute parity of a boson can be defined without ambiguity, and §3.5 shows how it is measured for the pion. Fermions cannot be treated that way: they carry half-integer spin, so angular-momentum conservation forces them to be produced in pairs, and only relative parities are ever accessible. Quantum field theory supplies the constraint:

\htmlClass{t-pf}{P_f}\,\htmlClass{t-pfb}{P_{\bar f}} = -1

(3.1)

A field-theory result, not a convention: the product is fixed even though neither factor separately is.

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 — only the product of two parities is physical

Only the product is physical. You are free to name one of the two parities; the equation then names the other, and any observable depends on the pair.

Two consequences, which are the same statement read in two directions:

if $P$ is real ( $\pm 1$ ), then fermion and antifermion have opposite parity — $P_f = +1 \Rightarrow P_{\bar f} = -1$ ;
if $P$ is imaginary ( $\pm i$ ), then $P_f P_{\bar f} = (\pm i)^2 = -1$ with the same value on both sides, so fermion and antifermion have equal parity.

For a Dirac fermion the intrinsic parity is completely arbitrary and is chosen real. For a Majorana fermion there is no choice: the particle is its own antiparticle, so $P_f = P_{\bar f}$ , and (3.1) then demands $P^2 = -1$ — the parity must be imaginary. That is the same $\pm i$ the $P^4 = 1$ argument allowed, now compulsory rather than merely permitted, and it is §2.9 continued.

The parity conventions this book uses, and which of them are choices

Particle	Parity	Status
proton	+1
antiproton	−1	forced by (3.1) once the proton is fixed
other non-strange baryons	relative to the proton	measured
charged leptons	+1
neutrinos		not a convention — impossible in principle
Λ hyperon	+1
all quarks	+1	definition
all antiquarks	−1	follows from (3.1)
photon		derived

The <em>Status</em> column is the one to read. A convention can never be measured wrong; a derived value can. Confusing the two is the most common way to misread a parity argument.

The parity of a two-particle system

This is the workhorse result of the chapter — every parity argument in §3.5, Chapter 4 and §9.18 runs through it. It takes three lines, and each line is a statement about spherical harmonics rather than about particles.

🪜 From spherical harmonics to P = P₁P₂(−1)ˡ

Step 1 of 5 — Set up the two bases(3.2)

|p, l, m\rangle = \sum_{\theta,\phi} Y_l^{*m}(\theta,\phi)\, |\mathbf{p}, -\mathbf{p}\rangle

Why you may do this: In the CM frame the two particles have opposite momenta, so one direction (θ, φ) fixes the configuration completely. The angular-momentum basis |p, l, m⟩ and the direction basis |θ, φ⟩ describe the same states, and the spherical harmonics are the change-of-basis matrix between them — the amplitude for a state of definite l to point in a given direction.

Why the CM frame: a system with net momentum is not a parity eigenstate at all, since P flips that momentum. Only here is the question well posed.

The derivation is short because the physics is entirely in step 3. Everything before it sets up the basis; everything after it multiplies.

\htmlClass{t-P}{P} = \htmlClass{t-p1}{P_1}\htmlClass{t-p2}{P_2}\,(-1)^{\htmlClass{t-l}{l}}

(3.4)

Two constants and one integer. Nothing about the interaction that binds the pair enters.

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 — two sources of handedness: what the particles are, and how they orbit

A two-particle state has two independent sources of “handedness”: what the particles are (their intrinsic parities, fixed once and for all) and how they orbit (the shape of the spatial wave function). The formula says the two multiply and nothing else contributes.

The orbital factor is the interesting one, and it is the same fact that makes a $p$ orbital look different from an $s$ orbital in a chemistry textbook. An $s$ wave ( $l = 0$ ) is spherical — it looks identical in a mirror, so $(-1)^0 = +1$ . A $p$ wave ( $l = 1$ ) has two lobes of opposite sign, so reflecting through the origin swaps them and the wave function changes sign: $(-1)^1 = -1$ . Higher $l$ just alternates.

Engineer’s reading: $(-1)^l$ is the parity of the $l$ -th basis function of an expansion on the sphere, exactly as $(-1)^n$ is the parity of the $n$ -th Chebyshev or Legendre polynomial on a line. Even index → even function.

🔢 Worked example — the $(-1)^l$ factor, checked numerically

Step 3 above is the whole derivation, so it is worth confirming rather than trusting. Inversion sends $\theta \to \pi - \theta$ and $\phi \to \pi + \phi$ ; for $m = 0$ the azimuth drops out and the ratio should be exactly $(-1)^l$ at every angle.

Reproduce it

import numpy as np
Y = {0: lambda t: 0.5*np.sqrt(1/np.pi) * np.ones_like(t),          # Y_0^0
     1: lambda t: 0.5*np.sqrt(3/np.pi) * np.cos(t),                # Y_1^0
     2: lambda t: 0.25*np.sqrt(5/np.pi) * (3*np.cos(t)**2 - 1)}    # Y_2^0
th = np.linspace(0.1, 3.0, 7)
for l, f in Y.items():
    ratio = np.unique(np.round(f(np.pi - th) / f(th), 9))
    print(f"l={l}:  Y(pi-theta)/Y(theta) = {ratio[0]:+.0f}   (-1)^l = {(-1)**l:+d}")

prints

l=0:  Y(pi-theta)/Y(theta) = +1   (-1)^l = +1
l=1:  Y(pi-theta)/Y(theta) = -1   (-1)^l = -1
l=2:  Y(pi-theta)/Y(theta) = +1   (-1)^l = +1

Exactly $(-1)^l$ , at all seven angles, with no residual — the ratio is a constant, which is what “eigenvalue” means. The lobe diagrams on the reference page colour the harmonics by sign, so the same statement can be read off by eye.

Two cases that are used constantly

\begin{aligned} P(\htmlClass{t-mm}{m_1, m_2}) &= (-1)^{\htmlClass{t-l}{l}} \\[2pt] P(\htmlClass{t-ff}{f\bar f}) &= (-1)^{\htmlClass{t-l}{l}+1} \end{aligned}

(3.5, 3.6)

Both are Eq. (3.4) with the product P₁P₂ evaluated: +1 for identical-parity mesons, −1 for any fermion–antifermion pair.

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 — for two pions, parity alternates strictly with spin

For two pions, which are spinless, the orbital angular momentum is the total: $J = l$ . So the accessible states are $J^P = 0^+,\ 1^-,\ 2^+,\ \ldots$ — parity strictly alternating with spin.

That list holds provided the two pions are different. If they are identical, Bose statistics demands a symmetric state, which forces $l$ — and therefore $J$ — to be even, leaving only $J^P = 0^+,\ 2^+,\ \ldots$ The odd- $J$ entries are not suppressed; they do not exist. This is a first taste of the argument style that dominates §3.5 and Chapter 4: count the states statistics allows, and the answer is often unique.

For a fermion–antifermion pair, the extra sign means an $S$ wave has negative parity, which is why the ground state of positronium — and of every $q\bar q$ meson — is $J^P = 0^-$ or $1^-$ rather than $0^+$ . Every pseudoscalar meson in the particle tables ( $\pi^\pm$ , $K^\pm$ , $\eta$ ) is this one minus sign.

🔢 Worked example — Example 3.1, positronium in the $S$ and $P$ waves

Problem. Find the possible $J^P$ for a spin-½ particle and its antiparticle in an $S$ wave and in a $P$ wave. (Positronium, the $e^+e^-$ atom, is exactly this system.)

Setup. The two spins combine to $s = 0$ (singlet) or $s = 1$ (triplet). $S$ wave means $l = 0$ , $P$ wave means $l = 1$ . Then $J$ runs over $|l-s|, \ldots, l+s$ , and the parity comes from (3.6). The site writes these in spectroscopic notation $^{2s+1}L_J$ , which is the label used from here on.

Reproduce it

L = ['S', 'P', 'D']
for l in (0, 1):
    for s in (0, 1):
        P = (-1)**(l + 1)                        # Eq. (3.6): P(f fbar) = (-1)^(l+1)
        for J in range(abs(l - s), l + s + 1):
            print(f"{2*s+1}{L[l]}{J}   l={l} s={s} J={J}   J^P = {J}{'+' if P > 0 else '-'}")

prints

1S0   l=0 s=0 J=0   J^P = 0-
3S1   l=0 s=1 J=1   J^P = 1-
1P1   l=1 s=0 J=1   J^P = 1+
3P0   l=1 s=1 J=0   J^P = 0+
3P1   l=1 s=1 J=1   J^P = 1+
3P2   l=1 s=1 J=2   J^P = 2+

Answer. $S$ wave: $^1S_0$ has $J^P = 0^-$ and $^3S_1$ has $1^-$ , both negative parity. $P$ wave: $^1P_1$ has $1^+$ , and the triplet gives $^3P_0 = 0^+$ , $^3P_1 = 1^+$ , $^3P_2 = 2^+$ — all positive. Six states, matching the book exactly.

Sanity check. Parity depends on $l$ alone, so a whole wave has one sign; the spin only redistributes $J$ . That is why the two $S$ states are both negative and all four $P$ states are positive. Adding the C eigenvalue to this table is §3.3, and it is what turns it into Table 3.1 and into the idea of an exotic $J^{PC}$ .

Parity conservation is not a law of nature

Strong and electromagnetic interactions conserve parity; weak interactions do not — that is Chapter 7, and it was the biggest surprise in the field’s history. But “strong interactions conserve parity” is an experimental claim, and this section is about how it was bounded.

The trick is that a parity-violating amplitude is not directly visible. Write a matrix element as a scalar plus a pseudoscalar piece:

\htmlClass{t-M}{M} = \htmlClass{t-MS}{M_S} + \htmlClass{t-MPS}{M_{PS}}

(3.7)

Eq. (3.7): the decomposition every parity test in this book is built on.

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 — you cannot see a term, only its interference

Square it and there are three pieces:

$|M|^2 = |M_S|^2 + |M_{PS}|^2 + 2\,\mathrm{Re}\!\left(M_S^{*} M_{PS}\right).$

The first two are each invariant under parity — squaring a pseudoscalar gives a scalar, so either term alone is undetectable as a parity violation. Only the cross term changes sign, and only it can betray the pseudoscalar piece.

This is a completely familiar situation. It is a homodyne measurement: a weak signal in quadrature with a strong local oscillator is invisible in the power spectrum but shows up linearly in the beat term. Same algebra, same payoff — the sensitivity goes as $|M_{PS}|/|M_S|$ , the amplitude ratio, rather than as its square, so a strong “reference” amplitude buys you sensitivity you could not get any other way.

The observable that carries the cross term must itself be P-odd, and that means the experiment must measure a pseudoscalar — a helicity, a $\langle\boldsymbol\sigma\cdot\mathbf{p}\rangle$ , or a decay that a P-even matrix element simply cannot produce. Bettini’s example is the third kind, and it is the cleanest: find a decay that is forbidden unless parity is violated, and then fail to find it.

🔬 Experiment card — Tonner 1957, the parity-forbidden decay of ²⁰Ne*

Apparatus

A proton beam of tunable energy on a fluorine target. Capture forms the compound nucleus ²⁰Ne in an excited state; a detector downstream counts

\alpha

particles and ¹⁶O recoils. Sweeping the beam energy sweeps the excitation energy of the compound nucleus, so a state that decays this way appears as a resonance in the yield at one energy.

🛠️ Searching for a forbidden resonance

Click a numbered marker for what that piece does.

The logic is entirely in callout 2: the initial and final states have opposite parity, so the transition can only proceed through a parity-violating amplitude.

What is measured

The yield of ¹⁶O + α as a function of proton energy — specifically, whether a resonance appears at the energy corresponding to the

J^P = 1^+

state at

Q = 13.2

MeV. This is a null experiment: the quantity of interest is the size of a peak that should not be there.

The result

The resonance was not found (Tonner 1957). Translated into amplitudes, that null result sets the limit for strong interactions at

|M_{PS}/M_S|^2 \le 10^{-8}

— Eq. (3.8) — so the parity-violating amplitude is below

10^{-4}

of the parity-conserving one.

What it proved

That “the strong interaction conserves parity” is not an assumption but a measured statement, good to one part in

10^4

at the amplitude level. Every parity argument later in this book — the

\pi^-

and

\pi^0

parities of §3.5, the

J^{PC}

assignments of Chapter 4, the spin-parity of the Higgs in §9.18 — leans on this being true of the production mechanism.

Erratum — Eq. (3.8) is printed inverted

The book prints $|M_S/M_{PS}|^2 \le 10^{-8}$ . That is the wrong way up. $M_S$ is the parity-conserving amplitude and $M_{PS}$ the parity-violating one, as (3.7) defines them two paragraphs earlier — so as printed, the equation states that the conserving amplitude is at most $10^{-4}$ of the violating one, i.e. that the strong interaction violates parity almost maximally. That contradicts the sentence immediately above it. The limit is $|M_{PS}/M_S|^2 \le 10^{-8}$ , and that is what this page uses.

🔢 Worked example — what a $10^{-8}$ limit means in the laboratory

The limit is on a squared amplitude ratio, but nobody measures amplitudes. What an experiment measures is the P-odd asymmetry — the size of the cross term relative to the total. With $r \equiv |M_{PS}/M_S|$ and the two amplitudes in phase, the fractional cross term is $A = 2r/(1+r^2)$ .

Reproduce it

r = 1e-4                                 # |M_PS / M_S| allowed by Tonner's null result
A = 2*r / (1 + r**2)                     # the P-odd interference asymmetry it would produce
print(f"|M_PS/M_S|^2 <= {r**2:.0e}   ->   asymmetry <= {A:.1e}")

prints

|M_PS/M_S|^2 <= 1e-08   ->   asymmetry <= 2.0e-04

So the bound corresponds to an observable effect of at most 2 parts in 10 000. Note how much the interference bought: a rate measurement sensitive to $|M_{PS}|^2$ alone would need to reach $10^{-8}$ , four orders of magnitude harder. The curve below is that trade, and the vertical line is where Tonner left the strong interaction.

interference (what is measured)
rate only, ∝ r² (no interference)

Why parity tests are interference experiments. At the Tonner limit the interference observable is 2×10⁻⁴ while the rate excess is 10⁻⁸ — the gap between the two curves is the entire sensitivity of the method.

🔬 Experiment card — Wu and Shaknov 1950, testing Eq. (3.1) itself

Apparatus

Positronium — an

e^+e^-

atom — formed and allowed to annihilate. It does so from the

^1S_0

state into two photons, whose linear polarizations are then compared by Compton scattering off two separate analysers.

What is measured

The correlation between the two photon polarizations: whether they prefer to be parallel or perpendicular. That correlation is a direct read-out of the

J^P

of the annihilating state, because the two options correspond to different combinations of the photon polarization vectors.

The result

The correlation was characteristic of an initial state with

J = 0

and odd parity — that is,

J^P = 0^-

, exactly what (3.6) predicts for

^1S_0

(-1)^{0+1} = -1

What it proved

Eq. (3.1) — the statement

P_f P_{\bar f} = -1

, which was a field-theory result rather than a measurement. The electron and positron really do have opposite intrinsic parity. Keep this experiment in mind: the same observable, the angle between two planes fixed by photon polarizations, measures the

\pi^0

parity in §3.5 and the spin-parity of the Higgs boson in §9.18. One idea, three times, sixty years apart.

⚠️ Natural units

Nothing on this page carries a unit that needs restoring — parities, $l$ and $J$ are pure numbers, and $\hbar = c = 1$ only shows up in writing angular momenta as integers rather than as multiples of $\hbar$ . The one dimensionful quantity quoted, the ²⁰Ne excitation $Q = 13.2$ MeV, is a laboratory energy in the ordinary sense. See the notation page if the convention is still new.

🔑 If you remember only three things

Not every state has a parity at all. The label is defined only for eigenstates of the operator, so “what is its parity?” is sometimes a question with no answer.
A convention is not a result. Some parities are assigned rather than measured, and the physics lives entirely in the combinations that no choice of convention can move.
Symmetry turns some conservation laws into theorems and leaves others as data. The ones that follow from the Lagrangian never need testing; the interesting experiments are aimed at the ones that do not.

Where this goes next

§3.3 repeats this entire structure one octave up for particle–antiparticle conjugation $C$ , and the combination with what you have just read produces Table 3.1 — the $J^{PC}$ list — and with it the idea of quantum numbers that no $q\bar q$ state can reach. §3.5 then puts the two-particle parity formula to work on real measurements: the $\pi^-$ parity from capture in deuterium, and the $\pi^0$ parity from the angle between two conversion planes. And Chapter 7 demolishes the assumption this page has been careful to state as a measurement rather than a law.

✅ Check yourself — symmetries and parity

0/6 answered · 0 correct

1.Electric charge $Q$ and baryon number $B$ are both additive and both conserved by every interaction in the table. What is the difference between them?
Hint: Look at the Status idea from the conventions table: which conservation law is forced and which is merely observed?
2.Why does a magnetic field $\mathbf{B}$ come out as an axial vector while an electric field $\mathbf{E}$ comes out as a polar vector? (Try building both in the ParityLab above.)
Hint: Ask what each one is made of, and count how many factors flip.
3.Two identical pions are in a state of orbital angular momentum $l$ . Which spin-parity values $J^P$ can the pair have?
Hint: Pions are spinless, so $J = l$ . Then remember what Bose statistics demands of two identical bosons.
4.Positronium annihilating from the $^3S_1$ state: what is its $J^P$ ?
Hint: Spectroscopic notation is $^{2s+1}L_J$ , so read off $s$ , $l$ and $J$ — then use Eq. (3.6).
5.Tonner looked for the decay $^{20}\mathrm{Ne}^* (1^+) \to {}^{16}\mathrm{O}\,(0^+) + \alpha\,(0^+)$ and did not find it. Why does that particular non-observation bound parity violation?
Hint: Work out the parity the final state is forced to have, given that $J$ must be conserved.
6.The Tonner limit is $|M_{PS}/M_S|^2 \le 10^{-8}$ , yet the experiment did not need $10^{-8}$ sensitivity. Predict what the plot in the worked example shows about why.
Hint: The observable is the interference term, not the squared parity-violating amplitude.