CP Violation · Elementary Particle Physics

Until this experiment nobody could tell a distant civilisation which of their two kinds of particle we call matter. A rate difference of two parts in a thousand settles it.

🎯 Why this matters

An absolute statement about matter needs a law that is not symmetric, and this is the only one available. Two parts in a thousand is small enough to have been missed for decades and absolute enough to serve as a definition.

§8.3’s charge-asymmetry plot had a loose end: at late times, when only $K_L$ survive, the asymmetry should relax to zero and does not. This section is what that means.

Before the experiment, the taxonomy — because the chapter’s remaining six sections are organised by it and the three kinds are constantly conflated.

Three ways to violate CP, and they are independent

Bettini p. 324. Any one can occur without the others. Keeping them apart is the single most useful thing to carry out of this chapter.

kind↕	what is wrong↕	needs↕	where↕
in the mixing<br/>(indirect, "in the wave function")	the propagating states are not CP eigenstates — $K_S$ contains a little $K_2^0$ and $K_L$ a little $K_1^0$	mixing	this section — the parameter $\varepsilon$
in the decay<br/>(direct)	$\|A(M\to f)\| \neq \|A(\bar M\to\bar f)\|$ — the amplitudes themselves differ	two interfering weak amplitudes and a strong phase difference	§8.8 ( $\varepsilon'$ ), §8.9, §8.10 — the only kind that works for charged mesons
in the interference	decaying directly and decaying after mixing carry different phases	mixing, and a final state both flavours can reach	§8.6 — the B system's cleanest measurement

So: violation in the mixing , violation in the decay , and violation in the interference .

The third is the subtle one: it occurs even if CP is conserved both in the mixing and in the decay, provided the two paths differ in phase. And the second is the only kind available to a charged meson, which is why §8.10 exists and why mixing and CP violation are genuinely separate phenomena.

The 1964 experiment

Christenson, Cronin, Fitch and Turlay set out to test whether

K_L \to \pi^+\pi^-

happens at all. CP says it cannot. Everything about the apparatus follows from the fact that if it does happen, it is rare — and swamped by three-body decays that also give two charged tracks.

🛠️ Fig. 8.4 — the Brookhaven two-arm spectrometer

Click a numbered marker for what that piece does.

Bettini Fig. 8.4, redrawn. The geometry is set by the decay it is looking for: two charged tracks, roughly coplanar with the beam, sharing the beam's total momentum.

The two cuts, and why the control regions are the argument

A $K_L \to \pi^+\pi^-$ decay has two properties that a three-body decay cannot fake, because in the three-body case an unseen neutral carries away momentum:

the vector sum of the two track momenta points along the beam, so $\cos\theta = 1$ where $\theta$ is the angle between that sum and the beam direction;
the invariant mass $m(\pi^+\pi^-)$ equals the kaon mass.

Both hold only for a genuine two-body decay. So the analysis plots $\cos\theta$ in three slices of $m(\pi^+\pi^-)$ — one on the kaon mass, and two just below and just above it:

484 < m(ππ) < 494 MeV — below
494 < m(ππ) < 504 MeV — on the K mass
504 < m(ππ) < 514 MeV — above

Fig. 8.5, redrawn schematically from the published distributions (Fitch, Nobel Lecture 1980). The forward peak appears in the central mass slice and in neither control region. The excess was 45 ± 9 events on a background of about 11.

⚙️ Engineer’s bridge — the sidebands are the measurement; the peak is only where you look

Panels (a) and (c) of Fig. 8.5 are not decoration and they are not padding. They are the experiment.

A peak at $\cos\theta = 1$ in the central mass bin, on its own, proves very little. Detectors have acceptance edges; forward-going tracks are reconstructed better than wide-angle ones; a spectrometer built to accept a particular kinematics will naturally pile events up at the edge of that acceptance. Any of those would put a spike exactly where the signal is expected.

What kills all of them at once is that the same spike does not appear in the mass bins on either side. Acceptance effects do not know where the kaon mass is. Neither do neutron interactions, nor the reconstruction of three-body decays. Only something that genuinely has $m(\pi\pi) = m_K$ and $\cos\theta = 1$ populates the central bin and no other — and the only such thing is a two-body decay of a kaon.

This is the sideband, and it is the backbone of every rare-signal search since. Define a signal region; define control regions where the signal cannot be but every background can; show the excess lives in one and not the others. The control regions carry no signal by construction, which is exactly what makes them informative.

An engineer knows the same move by other names. A negative control in an assay. A holdout arm in an A/B test, where the point of the arm that receives nothing is that it receives nothing. A canary that is deliberately excluded from the change. Measuring a baseline on an idle system before attributing a regression to your patch.

The general principle: an anomaly is only as convincing as the places you checked where it should have been absent. The 1964 result is 45 events. It stood because of the two panels that contained nothing.

Where it breaks: a negative control excludes the backgrounds you thought to control for, and nothing else. A process that peaked in the same mass window and the same angular bin would have passed both panels untouched — which is exactly what the three-body decays K_L → πππ and K_L → πℓν had to be shown not to do, by kinematics rather than by the controls themselves.

And the controls carry the statistics of the signal: with 45 events in the peak, the sidebands constrain a fake rate only to their own Poisson accuracy. A holdout arm with five users in it is not a holdout arm. The control tells you the shape of what you already imagined; it is silent about anything shaped like your signal.

🔬 Experiment card — Christenson, Cronin, Fitch and Turlay, Brookhaven AGS, 1964

Apparatus

A neutral beam from the AGS: protons on a target, a sweeping dipole to remove charged particles, a collimator to select what went straight. Several metres of flight so that only

K_L

remain, through a decay volume filled with helium rather than evacuated — a compromise that suppresses both beam interactions and, critically,

K_S

regeneration. Downstream, a two-arm magnetic spectrometer, spark chambers before and after a bending magnet in each arm, triggered by Čerenkov counters.

What is measured

For every two-track event: the momentum and charge of each track, hence the invariant mass

m(\pi^+\pi^-)

and the angle

\theta

between the summed momentum and the beam. The observable is the distribution of $\cos\theta$ in slices of $m(\pi^+\pi^-)$ — one slice on the kaon mass and two flanking it.

The result

A forward peak of

45 \pm 9

events at

\cos\theta = 1

, present only in the central mass slice. In the two control slices the distribution is flat. The branching ratio is

$\mathrm{BR}(K_L\to\pi^+\pi^-) \approx 2\times10^{-3}$

What it proved

That CP is violated. The

K_L

decays to a CP-even final state, so it cannot be the CP eigenstate

K_2^0

; the states of definite mass and lifetime are not the states of definite CP.

The effect is tiny — about one $K_L$ in 200 000 takes the forbidden route — but it is not zero, and “not zero” is a qualitative statement about the laws of physics. Together with C violation (§7.6) it means no symmetry relates matter to antimatter exactly, which is a precondition for the universe containing more of one than the other (Chapter 12).

What it does to the states

If $K_S$ and $K_L$ are not $K_1^0$ and $K_2^0$ , write in the impurity — one complex number, ε :

|K_S\rangle = \frac{1}{\sqrt{1+|\varepsilon|^2}}\left(|K_1^0\rangle + \htmlClass{t-e}{\varepsilon}|K_2^0\rangle\right), \qquad |K_L\rangle = \frac{1}{\sqrt{1+|\varepsilon|^2}}\left(\htmlClass{t-e}{\varepsilon}|K_1^0\rangle + |K_2^0\rangle\right)

(8.27)

Bettini p. 326. One complex number describes the whole of CP violation in the mixing — and the same ε appears in both states, which is what CPT requires.

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 — ε is a rotation angle, not a new ingredient

Nothing was added to the system. There are still exactly two states, still built from the same $K^0$ and $\bar K^0$ . What Eq. (8.27) says is that the basis which propagates is turned very slightly away from the basis which has definite CP — by an angle of about $|\varepsilon| = 2.2\times10^{-3}$ radians, or 0.13 degrees.

An engineer will recognise the shape immediately. Two orthogonal modes; a coupling matrix that is almost diagonal in the basis you like; and the true eigenvectors therefore almost — but not exactly — the ones you named. The leakage is first order in the off-diagonal term, and a mode that was forbidden by symmetry now appears with amplitude $\varepsilon$ and rate $|\varepsilon|^2$ . Cross-talk between two nearly-decoupled channels, a filter whose stopband is deep but finite, a rotation matrix you approximated as the identity.

Two things follow, and both matter.

The rate is the square. $|\varepsilon| = 2.2\times10^{-3}$ is an amplitude. The observable rate ratio is $|\varepsilon|^2 = 5\times10^{-6}$ — one $K_L$ in 200 000. Reading the amplitude as though it were a probability is the single easiest mistake here.

The same $\varepsilon$ appears in both states, with the same sign, and that is not a simplification. An arrangement with $+\varepsilon$ in one and $-\varepsilon$ in the other is exactly what CPT forbids. So the symmetric form is a prediction being carried along, and §8.5’s closing aside shows it paying off: the phase of $\varepsilon$ turns out to be fixed by quantities that have nothing to do with CP.

The observable is the ratio of the two transition amplitudes into the same final state:

\htmlClass{t-n}{\eta_{+-}} \equiv |\eta_{+-}|\htmlClass{t-p}{e^{i\phi_{+-}}} = \frac{\htmlClass{t-a}{A(K_L\to\pi^+\pi^-)}}{\htmlClass{t-b}{A(K_S\to\pi^+\pi^-)}}, \qquad |\eta_{+-}|^2 = \frac{\Gamma(K_L\to\pi^+\pi^-)}{\Gamma(K_S\to\pi^+\pi^-)}

(8.28)

Bettini p. 327, Eqs. (8.28) and (8.29). A ratio of amplitudes into the SAME final state, which is why every hadronic complication cancels and a pure number is left.

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.

If the violation is entirely in the mixing then $\eta_{+-} = \varepsilon$ . That is checkable directly from branching ratios and lifetimes:

🔢 Worked example — ε from four measured numbers

The numerator is the 1964 result; the denominator is the $K_S$ ‘s main decay. Both branching ratios must be divided by the corresponding lifetime to get rates:

\Gamma(K_L\to\pi^+\pi^-) = \frac{1.967\times10^{-3}}{51160\;\text{ps}} = 3.845\times10^{-8}\;\text{ps}^{-1}

\Gamma(K_S\to\pi^+\pi^-) = \frac{0.692}{89.54\;\text{ps}} = 7.728\times10^{-3}\;\text{ps}^{-1}

|\eta_{+-}|^2 = \frac{3.845\times10^{-8}}{7.728\times10^{-3}} = 4.97\times10^{-6} \quad\Longrightarrow\quad |\eta_{+-}| = \mathbf{2.230\times10^{-3}}

against the quoted $|\varepsilon| = (2.232 \pm 0.011)\times10^{-3}$ — agreement to 0.1 %. The two lifetimes are doing most of the work: the branching ratios differ by a factor 350, but the lifetimes differ by 571 in the other direction, and the rate ratio is their product.

Note what the smallness means. $|\varepsilon| \approx 2\times10^{-3}$ is an amplitude; the rate ratio is $|\varepsilon|^2 = 5\times10^{-6}$ . About one $K_L$ in 200 000 decays the forbidden way.

the four ε numbers, and whether they close

tS, tL = 89.54, 51.16e3                       # ps
BR_L, BR_S = 1.967e-3, 0.6920
eps, ReEps, phi, dL = 2.232e-3, 1.596e-3, 43.52, 3.320e-3
import numpy as np

GL, GS = BR_L/tL, BR_S/tS
print("|eta_+-| from branching ratios and lifetimes:")
print(f"  Gamma(K_L -> pipi) = {BR_L:.3e} / {tL:.0f} ps = {GL:.3e} /ps")
print(f"  Gamma(K_S -> pipi) = {BR_S:.4f}    / {tS} ps = {GS:.3e} /ps")
print(f"  |eta|^2 = {GL/GS:.3e}  ->  |eta_+-| = {np.sqrt(GL/GS):.3e}")
print(f"  book (8.30): {eps:.3e}      agreement {abs(np.sqrt(GL/GS)/eps-1)*100:.1f}%")

print("\nthe book quotes FOUR numbers related by THREE approximate relations.")
print("do they close?")
pred = eps*np.cos(np.radians(phi))
print(f"  |eps| cos(phi) = {eps:.3e} x cos({phi}) = {pred:.3e}")
print(f"  quoted Re(eps)                          = {ReEps:.3e}      off by {abs(pred/ReEps-1)*100:.1f}%")
print(f"  2 Re(eps)      = {2*ReEps:.3e}")
print(f"  measured delta_L                        = {dL:.3e}      off by {abs(2*ReEps/dL-1)*100:.1f}%")
print("\nboth relations are first order in eps, and both residuals sit at the")
print("few-per-cent level -- which is the size of the terms dropped (eps^2,")
print("epsilon-prime, and Delta S = Delta Q violation).  they close as well")
print("as they are entitled to, and no better.")

print("\nthe size of the effect:")
print(f"  amplitude impurity  |eps|   = {eps:.2e}")
print(f"  rate ratio          |eps|^2 = {eps**2:.2e}")
print(f"  -> about 1 K_L in {1/eps**2:.0f} decays the CP-forbidden way")

prints

|eta_+-| from branching ratios and lifetimes:
Gamma(K_L -> pipi) = 1.967e-03 / 51160 ps = 3.845e-08 /ps
Gamma(K_S -> pipi) = 0.6920    / 89.54 ps = 7.728e-03 /ps
|eta|^2 = 4.975e-06  ->  |eta_+-| = 2.230e-03
book (8.30): 2.232e-03      agreement 0.1%

the book quotes FOUR numbers related by THREE approximate relations.
do they close?
|eps| cos(phi) = 2.232e-03 x cos(43.52) = 1.618e-03
quoted Re(eps)                          = 1.596e-03      off by 1.4%
2 Re(eps)      = 3.192e-03
measured delta_L                        = 3.320e-03      off by 3.9%

both relations are first order in eps, and both residuals sit at the
few-per-cent level -- which is the size of the terms dropped (eps^2,
epsilon-prime, and Delta S = Delta Q violation).  they close as well
as they are entitled to, and no better.

the size of the effect:
amplitude impurity  |eps|   = 2.23e-03
rate ratio          |eps|^2 = 4.98e-06
-> about 1 K_L in 200730 decays the CP-forbidden way

Telling an extraterrestrial what “matter” means

The charge asymmetry of §8.3 now has a value:

\delta_L = \frac{N(K_L\to\pi^-\ell^+\nu) - N(K_L\to\pi^+\ell^-\bar\nu)}{N(K_L\to\pi^-\ell^+\nu) + N(K_L\to\pi^+\ell^-\bar\nu)} = (3.32\pm0.06)\times10^{-3} = 2\,\mathrm{Re}\,\varepsilon

and the book draws the consequence better than any paraphrase. Suppose you must explain to a distant correspondent which of the two charges we call positive, with no shared object to point at. Every other definition fails: charge conjugation, parity and their product were all thought to be exact, so any recipe built from them is ambiguous.

This one is not. Prepare a neutral kaon beam. Travel far enough from the production point that only the long-lived component remains. Count the decays with a lepton of each charge. The charge of the more common one — by about three parts in a thousand — is what we call positive.

The recipe works anywhere in the universe, and it works because CP violation is real. The book’s closing line is worth keeping: if your correspondent reports the opposite answer and then comes to visit, be careful — apologise, but do not shake their hand.

Aside — the phase of ε is not a free parameter, and the book does not say so

Eq. (8.35) gives $\phi = 43.52° \pm 0.05°$ and the book observes only that it is “about $\pi/4$ ”. It is much better than that: the phase is predicted, and the prediction is one of the sharpest tests of CPT in physics.

If CPT holds and the $2\pi$ channel dominates the width difference — both true here — then $\varepsilon$ is forced to the so-called superweak phase

\phi_{SW} = \arctan\!\left(\frac{2\htmlClass{t-dm}{\Delta m}}{\htmlClass{t-dg}{\Delta\Gamma}}\right)

Supplied — the book quotes the measured 43.52° and observes only that it is 'about π/4'. It is far better than that: the phase is PREDICTED from two independently measured lifetimes, and the agreement is one of the sharpest CPT tests in physics.

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.

Put the measured numbers in. $\Delta m = 5.293\times10^{-3}$ ps⁻¹ and $\Delta\Gamma = \Gamma_S - \Gamma_L = 1/89.54 - 1/51160 = 1.1149\times10^{-2}$ ps⁻¹, so

\phi_{SW} = \arctan(0.9496) = \mathbf{43.52°}

against a measured $43.52°$ . Four significant figures, from two lifetimes and an oscillation frequency, none of which is a CP-violating quantity.

That is worth pausing on. The magnitude $|\varepsilon|$ is a free parameter of nature and had to be measured. The phase is not: it is fixed by quantities already known from §8.2 and §8.3, and its agreement with experiment tests CPT rather than measuring anything new. A deviation would be a discovery.

🔑 If you remember only three things

There are three independent ways to violate CP and this page finds one. Mixing, decay and their interference can each fail separately, and none implies another.
The number is tiny and the conclusion is absolute. Two parts in a thousand suffices, because the question is only whether a difference exists at all.
It was found in the one system where it could be seen. With lifetimes differing by 571, a forbidden decay stands out against almost nothing.

Where this goes next

Everything so far is CP violation in the mixing — one complex number $\varepsilon$ , describing states that are not quite CP eigenstates. Whether the decay amplitudes also violate CP is a separate question, and a much harder one: the answer is $\varepsilon'$ , smaller than $\varepsilon$ by another factor of a thousand, and extracting it needs the double ratio of §8.8.

First, though, §8.6 changes system. The $B^0$ has no 571-fold lifetime ratio to exploit and no $\varepsilon$ -sized effect — instead it has CP violation in the interference, which is order one rather than $10^{-3}$ , and a clean prediction: the asymmetry is a pure sinusoid of amplitude $\sin2\beta$ .

✅ Check yourself — the discovery of CP violation

0/6 answered · 0 correct

1.Of the three kinds of CP violation, select every one that requires the meson to mix with its antiparticle. (More than one.)
2.Why are the two flanking mass slices in Fig. 8.5 the most important part of the figure?
3.|ε| ≈ 2.2 × 10⁻³. How rare is the CP-violating decay?
4.The book quotes |ε|, Re ε, φ and δ_L. Checking them against each other, the relations close only to a few per cent. What does that mean?
5.φ = 43.52° is 'about π/4'. Is that a coincidence?
6.Why is the decay volume of the 1964 experiment filled with helium rather than evacuated?