The neutral kaon system

The neutral kaon is the most instructive two-state system in particle physics, because in it the two questions "what did you make?" and "what propagates and decays?" have different answers. Produced by the strong force, $K^0(d\bar{s})$ and $\bar{K}^0(s\bar{d})$ are the lightest strangeness-carrying mesons — flavour eigenstates with $S=\pm1$ , made copiously in reactions like $\pi^- p \to \Lambda K^0$ . But strangeness is not conserved by the weak interaction, and the same force that lets these particles decay also connects $K^0$ to $\bar{K}^0$ through second-order box diagrams. A kaon born as pure $K^0$ does not stay pure: it oscillates into $\bar{K}^0$ and back, and the states carrying a definite mass and lifetime are not the flavour states but two superpositions of them. Three ideas organize everything that follows. First, the combinations $K_1$ and $K_2$ are eigenstates of $CP$ with eigenvalues $+1$ and $-1$ ; since two pions are $CP$ -even and three $CP$ -odd, exact $CP$ would force $K_1\to\pi\pi$ and $K_2\to\pi\pi\pi$ exclusively. Second, the vast phase-space difference between these channels ( $Q\approx218$ vs $83$ MeV) makes the $CP$ -even state decay some 570 times faster, splitting the kaon into a short-lived $K_S\approx K_1$ and long-lived $K_L\approx K_2$ . Third — the payoff — a beam left to propagate becomes pure $K_L$ , and if $CP$ were exact it could never make two pions. In 1964 it did: 45 events of $K_L\to\pi^+\pi^-$ in 22 700 decays announced that $CP$ is not a symmetry of nature.

Key takeaways

$K^0(d\bar{s})$ and $\bar{K}^0(s\bar{d})$ are strong-interaction flavour eigenstates ( $S=\pm1$ ), mixed by the weak interaction through $K^0\leftrightarrow\bar{K}^0$ box diagrams.
The $CP$ eigenstates are $K_1=\tfrac{1}{\sqrt2}(K^0+\bar{K}^0)$ ( $CP=+1$ ) and $K_2=\tfrac{1}{\sqrt2}(K^0-\bar{K}^0)$ ( $CP=-1$ ); two pions are $CP$ -even, three pions $CP$ -odd.
The physical states are $K_S$ ( $\tau_S\approx0.90\times10^{-10}$ s, $\to\pi\pi$ ) and $K_L$ ( $\tau_L\approx5.1\times10^{-8}$ s, $\to\pi\pi\pi$ ); the ratio $\tau_L/\tau_S\approx571$ is set almost entirely by phase space.
The mass eigenstates are nearly degenerate, $\Delta m\approx3.48\times10^{-6}$ eV, and the "coincidence" $\Delta m\,\tau_S\approx0.47$ makes strangeness oscillations observable within a $K_S$ lifetime.
In 1964 Christenson, Cronin, Fitch and Turlay saw $K_L\to\pi^+\pi^-$ at the $2\times10^{-3}$ level — the first evidence of $CP$ violation, so $K_S\ne K_1$ and $K_L\ne K_2$ exactly.
Key idea: what you produce (flavour states $K^0,\bar{K}^0$ ) and what propagates and decays (mass/lifetime eigenstates $K_S,K_L$ ) are different bases related by weak mixing; comparing them reveals that $CP$ is not an exact symmetry of the weak interaction.

The core insight

What you produce (the flavour states $K^0$ , $\bar{K}^0$ ) and what propagates and decays (the mass/lifetime eigenstates $K_S$ , $K_L$ ) are different bases related by weak mixing; comparing them — most sharply through the appearance of $K_L\to\pi\pi$ — reveals that $CP$ is not an exact symmetry of the weak interaction.

Where this lives today

The neutral kaon remains physics' most exacting laboratory for discrete symmetries. Fixed-target experiments at CERN (NA48) and Fermilab (KTeV) measured direct $CP$ violation, $\mathrm{Re}(\varepsilon'/\varepsilon)=(1.66\pm0.23)\times10^{-3}$ , showing $CP$ is broken in the decay amplitude and not only the mixing — which "superweak" models could not accommodate. At Frascati, KLOE ran on the $\phi$ -factory DA $\Phi$ NE producing entangled $K_S K_L$ pairs, an EPR experiment where one kaon's decay fixes the other's. The kaon also gives the most stringent $CPT$ test in physics, bounding $|m_{K^0}-m_{\bar{K}^0}|/m_K<10^{-18}$ . And the mixing parameter $\varepsilon_K$ tightly constrains the CKM unitarity triangle, while the ultra-rare "golden mode" $K_L\to\pi^0\nu\bar{\nu}$ — its rate calculable directly from CKM parameters — is hunted by KOTO at J-PARC and $K^+\to\pi^+\nu\bar{\nu}$ by NA62, where any excess signals new physics.

The physical picture

A neutral kaon lives a double life. The strong interaction that makes it, and the semileptonic decays that later "tag" it, speak flavour: $S=+1$ for $K^0$ , $S=-1$ for $\bar{K}^0$ . But nothing forces the particle travelling down the beam pipe to have definite strangeness — the weak Hamiltonian governing propagation and decay does not commute with strangeness, so its eigenstates are mixtures. Diagonalizing it gives two states with sharp masses and lifetimes, $K_S$ and $K_L$ . The first figure lays the two bases side by side, with the $CP$ eigenstates as the bridge.

Figure

Two bases for one particle: the strong interaction produces flavour eigenstates K⁰ and K̄⁰ (strangeness ±1), but the weak interaction propagates and decays the mass/lifetime eigenstates K_S and K_L; the CP eigenstates K₁ (CP +1) and K₂ (CP −1) bridge them, and the two masses differ by only Δm ≈ 3.48×10⁻⁶ eV

The right-hand column is what a beam delivers. Because $K_S$ decays about 570 times faster, a beam that has flown a few decay lengths past its source has lost its $K_S$ component to $\pi\pi$ decay and is, to one part in ten thousand, a pure $K_L$ beam — and the whole logic of the experiments is to exploit that separation. Meanwhile the flavour composition oscillates: the second figure plots a beam born as pure $K^0$ , with the probability of finding a $\bar{K}^0$ rising and beating against the decay, its frequency set by $\Delta m$ .

Figure

Strangeness oscillation of a beam born as pure K⁰: the probability of finding K⁰ (blue) and K̄⁰ (red) versus proper time in units of the K_S lifetime, from the interference of the K_S and K_L amplitudes; the beat frequency is set by the mass splitting through Δm·τ_S = 0.47, and both curves damp toward ¼ once the K_S component has decayed away

The core equations

The formalism rests on distinguishing two orthonormal bases and writing the weak evolution in whichever is convenient. The first equation defines the $CP$ eigenstates, in the convention (phase $\zeta=\pi$ ) where $\hat{C}\hat{P}\,|K^0\rangle=+|\bar{K}^0\rangle$ and $\hat{C}\hat{P}\,|\bar{K}^0\rangle=+|K^0\rangle$ .

|K_1\rangle=\tfrac{1}{\sqrt2}\big(|K^0\rangle+|\bar{K}^0\rangle\big),\qquad |K_2\rangle=\tfrac{1}{\sqrt2}\big(|K^0\rangle-|\bar{K}^0\rangle\big),\qquad \hat{C}\hat{P}\,|K_{1,2}\rangle=\pm|K_{1,2}\rangle \qquad (1)

where: $|K^0\rangle,|\bar{K}^0\rangle$ — flavour eigenstates with $J^P=0^-$ ; $|K_1\rangle,|K_2\rangle$ — $CP$ eigenstates with eigenvalues $+1,-1$ . If $CP$ were exact these would be the states that decay with definite lifetimes. They are not, quite — hence everything below. A decaying two-state system is not Hermitian — probability leaks out through decay. The Weisskopf–Wigner approximation packages the coupled flavour amplitudes $a(t),b(t)$ (the $K^0$ and $\bar{K}^0$ content) into a Schrödinger-like equation with a non-Hermitian effective Hamiltonian.

i\frac{\mathrm{d}}{\mathrm{d}t}\begin{pmatrix}a\\ b\end{pmatrix}=\mathbf{H}\begin{pmatrix}a\\ b\end{pmatrix},\qquad \mathbf{H}=\mathbf{M}-\frac{i}{2}\boldsymbol{\Gamma} \qquad (2)

where: $\mathbf{M}$ — Hermitian mass matrix (dispersive, set mainly by the $\Delta S=2$ box diagram); $\boldsymbol{\Gamma}$ — Hermitian decay matrix (absorptive, from channels common to $K^0$ and $\bar{K}^0$ , dominantly $\pi\pi$ ). $CPT$ forces the diagonals equal, $M_{11}=M_{22}$ , $\Gamma_{11}=\Gamma_{22}$ . Diagonalizing $\mathbf{H}$ gives the physical states — the combinations of $K^0$ and $\bar{K}^0$ that evolve as pure exponentials, with mixing ratio $q/p$ fixed entirely by the off-diagonal elements.

|K_S\rangle=p\,|K^0\rangle+q\,|\bar{K}^0\rangle,\qquad |K_L\rangle=p\,|K^0\rangle-q\,|\bar{K}^0\rangle,\qquad \frac{q}{p}=\sqrt{\frac{M_{12}^{*}-\tfrac{i}{2}\Gamma_{12}^{*}}{M_{12}-\tfrac{i}{2}\Gamma_{12}}} \qquad (3)

where: $p,q$ — complex mixing coefficients, $|p|^2+|q|^2=1$ . If $CP$ holds, $q/p=1$ and $K_S,K_L\to K_1,K_2$ exactly. The physical states are non-orthogonal, $\langle K_S|K_L\rangle=|p|^2-|q|^2\ne0$ . The eigenvalues of $\mathbf{H}$ are complex — real parts the masses, imaginary parts the half-widths. This one line contains both the mass splitting $\Delta m$ that drives oscillation and the width splitting $\Delta\Gamma$ that separates the two lifetimes.

\mu_{S,L}\equiv m_{S,L}-\frac{i}{2}\Gamma_{S,L}=\Big(M_{11}-\tfrac{i}{2}\Gamma_{11}\Big)\pm\sqrt{\Big(M_{12}-\tfrac{i}{2}\Gamma_{12}\Big)\Big(M_{12}^{*}-\tfrac{i}{2}\Gamma_{12}^{*}\Big)} \qquad (4)

where: $\Delta m=m_L-m_S\approx3.48\times10^{-6}$ eV; $\Delta\Gamma=\Gamma_S-\Gamma_L\approx\Gamma_S$ . The diagonal term sets the common mass and mean width; the square root of the off-diagonal product sets the splitting between the two physical kaons. Because $K_S$ and $K_L$ differ in both mass and lifetime, a beam of definite flavour beats between $K^0$ and $\bar{K}^0$ . Projecting the evolved state onto the flavour basis and squaring gives the strangeness-oscillation law of the second figure.

P(K^0\!\to\bar{K}^0;\,t)=\frac{1}{4}\Big[e^{-\Gamma_S t}+e^{-\Gamma_L t}-2\,e^{-(\Gamma_S+\Gamma_L)t/2}\cos(\Delta m\,t)\Big] \qquad (5)

where: $t$ — proper time; the $\cos(\Delta m\,t)$ term is pure quantum interference between the $K_S$ and $K_L$ amplitudes. The dimensionless $x\equiv\Delta m\,\tau_S\approx0.47$ sets how many oscillations fit inside a $K_S$ lifetime — about half a cycle, so the effect is measurable but gentle. Finally, $CP$ violation is quantified by the ratios of $K_L$ to $K_S$ amplitudes into the same $CP$ -even two-pion final states.

\eta_{+-}=\frac{\langle\pi^+\pi^-|K_L\rangle}{\langle\pi^+\pi^-|K_S\rangle}=\varepsilon+\varepsilon',\qquad \eta_{00}=\frac{\langle\pi^0\pi^0|K_L\rangle}{\langle\pi^0\pi^0|K_S\rangle}=\varepsilon-2\varepsilon',\qquad |\varepsilon|\approx2.23\times10^{-3} \qquad (6)

where: $\varepsilon$ — indirect $CP$ violation, the $K_1$ admixture in $K_L$ from mixing; $\varepsilon'$ — direct $CP$ violation in the decay amplitude, $\mathrm{Re}(\varepsilon'/\varepsilon)\approx1.7\times10^{-3}$ . The near-equality $\eta_{+-}\approx\eta_{00}\approx\varepsilon$ is why the effect was first read as mixing alone; resolving $\varepsilon'$ took three more decades.

Derivation ADiagonalizing the effective Hamiltonian to find K_S and K_L

The goal is to start from the two-component weak evolution of equation (2) and show that its eigenstates are the mixed states of equation (3), its eigenvalues the complex masses of equation (4), and — most importantly — that in the exact- $CP$ limit the whole apparatus collapses back onto the $CP$ eigenstates $K_1$ and $K_2$ . This turns "the kaon mixes" into the concrete machinery of $q/p$ , $\Delta m$ and $\Delta\Gamma$ , and pins down which small quantity measures the $CP$ violation.

Step 1. Write the effective Hamiltonian in the flavour basis using $CPT$ ( $H_{11}=H_{22}\equiv D$ ) and Hermiticity of $\mathbf{M},\boldsymbol{\Gamma}$ (so $M_{21}=M_{12}^{*}$ , $\Gamma_{21}=\Gamma_{12}^{*}$ ): $\mathbf{H}=\begin{pmatrix}D & H_{12}\\ H_{21} & D\end{pmatrix}$ with $H_{12}=M_{12}-\tfrac{i}{2}\Gamma_{12}$ and $H_{21}=M_{12}^{*}-\tfrac{i}{2}\Gamma_{12}^{*}$ . Why: $CPT$ is the one symmetry we insist on keeping, and it makes the diagonal entries equal — so any difference between the two physical kaons must come from the off-diagonal mixing, not an intrinsic $K^0$ – $\bar{K}^0$ mass difference.

Step 2. Solve the $2\times2$ eigenproblem: eigenvalues $\mu_{S,L}=D\pm\sqrt{H_{12}H_{21}}$ (equation 4) and eigenvectors $(p,\pm q)^{T}$ with $q/p=\sqrt{H_{21}/H_{12}}$ (equation 3). Why: for equal diagonals the eigenvectors are the symmetric and antisymmetric flavour combinations weighted by $q/p$ , so all the mixing is compressed into one complex ratio $q/p$ .

Step 3. Take the $CP$ -symmetric limit. $CP$ invariance requires $H_{12}=H_{21}$ ( $M_{12},\Gamma_{12}$ real in this convention); then $q/p=1$ , $|K_S\rangle=\tfrac{1}{\sqrt2}(|K^0\rangle+|\bar{K}^0\rangle)=|K_1\rangle$ and $|K_L\rangle=|K_2\rangle$ . Why: the anchor — the mass eigenstates coincide with the $CP$ eigenstates precisely when $CP$ is good, so any measured departure of $q/p$ from unity is a direct, model-independent signal of $CP$ violation.

Step 4. Parametrize the departure. With $\varepsilon=(p-q)/(p+q)$ the physical states become $|K_S\rangle\propto|K_1\rangle+\varepsilon|K_2\rangle$ and $|K_L\rangle\propto|K_2\rangle+\varepsilon|K_1\rangle$ : each carries a small "wrong- $CP$ " impurity $\varepsilon$ . Why: this is the mechanism behind $K_L\to\pi\pi$ — the long-lived state is mostly $CP$ -odd $K_2$ but contains a fraction $\varepsilon\approx2.2\times10^{-3}$ of $CP$ -even $K_1$ , and it is that fraction which decays to two pions. *As a check: the eigenvalues $\mu_{S,L}=D\pm\sqrt{H_{12}H_{21}}$ and eigenvectors $(p,\pm q)$ reduce exactly to $K_1,K_2$ when $q/p=1$ , and $\varepsilon\to0$ in the $CP$ -conserving limit.*

Go deeper · Derivation B · 1 minStrangeness oscillations from the two-lifetime evolution

Derivation BStrangeness oscillations from the two-lifetime evolution

The aim is to derive the oscillation formula (5) from scratch: take a state prepared as pure $K^0$ , expand it in the physical basis whose time dependence we know, evolve it, and project back onto $\bar{K}^0$ . The result shows two nearly-degenerate decaying states interfering, giving a strangeness beat whose frequency is the mass splitting and whose envelope is the decay — a two-level quantum beat with complex frequencies. We work in the $CP$ -conserving approximation ( $K_S=K_1$ , $K_L=K_2$ ), exact to $10^{-3}$ here.

Step 1. Express the initial flavour state in the physical basis: $|K^0\rangle=\tfrac{1}{\sqrt2}(|K_1\rangle+|K_2\rangle)=\tfrac{1}{\sqrt2}(|K_S\rangle+|K_L\rangle)$ . Why: only $K_S,K_L$ have simple exponential time dependence, so a beam of definite flavour must be resolved into them before it can be evolved.

Step 2. Evolve each with its own complex frequency, $\mu=m-\tfrac{i}{2}\Gamma$ , giving $|K^0(t)\rangle=\tfrac{1}{\sqrt2}\big(e^{-i\mu_S t}|K_S\rangle+e^{-i\mu_L t}|K_L\rangle\big)$ . Why: the imaginary part of $\mu$ shrinks each amplitude as $e^{-\Gamma t/2}$ while the real part advances its phase as $e^{-imt}$ — decay and oscillation on the same exponential.

Step 3. Project onto $\bar{K}^0$ using $\langle\bar{K}^0|K_S\rangle=+\tfrac{1}{\sqrt2}$ and $\langle\bar{K}^0|K_L\rangle=-\tfrac{1}{\sqrt2}$ : the amplitude is $\langle\bar{K}^0|K^0(t)\rangle=\tfrac12\big(e^{-i\mu_S t}-e^{-i\mu_L t}\big)$ . Why: the opposite signs are the crux — the $\bar{K}^0$ amplitude is a difference of the two exponentials, vanishing at $t=0$ (pure $K^0$ ) and growing only as the eigenstates dephase.

Step 4. Square the modulus, $|z_1-z_2|^2=|z_1|^2+|z_2|^2-2\mathrm{Re}(z_1 z_2^{*})$ , to obtain $P=\tfrac14[e^{-\Gamma_S t}+e^{-\Gamma_L t}-2e^{-(\Gamma_S+\Gamma_L)t/2}\cos(\Delta m\,t)]$ , equation (5). Why: the cross term carries $\cos(\Delta m\,t)$ , so the beat frequency is a direct read-out of the mass difference — historically how $\Delta m$ was first extracted. *As a check: $P(K^0\!\to\bar{K}^0;0)=0$ (the beam starts pure $K^0$ ) and $P\to\tfrac14 e^{-\Gamma_L t}$ for $t\gg\tau_S$ (only $K_L$ survives); the survival probability $P(K^0\!\to K^0)$ carries $+\cos(\Delta m\,t)$ , and the two probabilities sum to $\tfrac12(e^{-\Gamma_S t}+e^{-\Gamma_L t})$ , as required by unitarity.*

Go deeper · Derivation C · 1 minWhy K₁ makes two pions and K₂ makes three

Derivation CWhy K₁ makes two pions and K₂ makes three

The purpose here is to establish what gives the two lifetimes their enormous ratio: the $CP$ quantum numbers of the pionic final states and the phase space of each. We compute the parity and charge conjugation of $\pi\pi$ and $\pi\pi\pi$ , match them to the $CP$ eigenvalues of $K_1$ and $K_2$ , and show that the far larger $\pi\pi$ phase space makes the $CP$ -even kaon short-lived — where the labels " $K_S$ " and " $K_L$ " acquire their physical identity.

Step 1. Find $CP$ of the two-pion state. The kaon has spin zero, so the pions emerge with $L=0$ . Parity gives $P=P_\pi^2(-1)^L=+1$ ; charge conjugation gives $C=(-1)^L=+1$ (both $\pi^+\pi^-$ and $\pi^0\pi^0$ ). Hence $CP(\pi\pi)=+1$ . Why: each pion's negative parity squares away, and with $L=0$ nothing makes the pair $CP$ -odd — two pions are unavoidably $CP$ -even.

Step 2. Find $CP$ of the three-pion state. For the dominant low- $L$ configuration, $P=P_\pi^3(-1)^{L}=-1$ and $C=+1$ , so $CP(\pi\pi\pi)=-1$ ; the exactly $CP$ -odd $3\pi^0$ case is guaranteed by Bose symmetry. Why: the extra pion flips the parity, so three pions are $CP$ -odd where two are $CP$ -even — the channels select opposite $CP$ eigenstates.

Step 3. Match to the kaons. $K_1$ ( $CP=+1$ ) can decay to $\pi\pi$ ; $K_2$ ( $CP=-1$ ) cannot, and is driven to $\pi\pi\pi$ . Why: $CP$ conservation ties each eigenstate to a final state of matching $CP$ , forbidding the fast $\pi\pi$ mode for $K_2$ .

Step 4. Compare phase space. The energy release is $Q_{\pi\pi}=m_K-2m_\pi\approx218$ MeV against $Q_{3\pi}=m_K-3m_\pi\approx83$ MeV, and the three-body mode is further suppressed by the extra final-state particle. The upshot is $\Gamma(K_1)\gg\Gamma(K_2)$ . Why: a decay rate rises steeply with available energy and falls with each extra particle, so the $CP$ -even kaon empties into $\pi\pi$ hundreds of times faster — giving $\tau_L/\tau_S\approx571$ and $K_S\approx K_1$ , $K_L\approx K_2$ . *As a check: $CP(\pi\pi)=+1$ and $CP(\pi\pi\pi)=-1$ ; the energy releases $Q_{\pi\pi}\approx218\gg Q_{3\pi}\approx83$ MeV confirm the phase-space hierarchy, and the measured $\tau_L/\tau_S=571$ follows directly from it.*

Limiting cases

Case	Result	Meaning
Exact $CP$ ( $\varepsilon\to0$ )	$K_S=K_1,\ K_L=K_2$ ; $K_L\to\pi\pi$ forbidden	mass and $CP$ bases coincide; the 1964 signal would vanish
Degenerate masses ( $\Delta m\to0$ )	$\cos(\Delta m\,t)\to1$ , no strangeness beat	oscillation frequency is the mass splitting; kill it and mixing shows only in decay
No shared channel ( $\Gamma_{12}\to0$ )	mixing purely dispersive, $\Delta\Gamma\to0$	the two lifetimes coincide; the $K_S$ / $K_L$ split needs a common final state
Long beam ( $t\gg\tau_S$ )	$K_S$ gone, essentially pure $K_L$	the experimental workhorse — a clean $CP=-1$ beam to test against
Equal $u,c$ masses (GIM)	box amplitude cancels, $\Delta m\to0$	$\Delta m_K$ is a probe of the charm mass — how $m_c\sim1.5$ GeV was predicted
$CPT$ imposed	$m_{K^0}=m_{\bar{K}^0}$ , $\Gamma_{11}=\Gamma_{22}$	forbids an intrinsic flavour mass difference; all splitting is off-diagonal

Common trapThe mass eigenstates are the CP eigenstates

It is tempting — and almost every first pass slips here — to say $K_S$ is $K_1$ and $K_L$ is $K_2$ . They are only approximately equal. The long-lived kaon is predominantly the $CP$ -odd $K_2$ but carries a small $CP$ -even admixture, $|K_L\rangle\propto|K_2\rangle+\varepsilon|K_1\rangle$ with $|\varepsilon|\approx2.2\times10^{-3}$ , and it is exactly that impurity which lets $K_L$ decay to two pions. To equate the bases is to declare $CP$ exact and miss the discovery entirely. Two further traps ride alongside. First, the assignment $K_S\approx K_1$ comes only from decay phase space — the $CP$ -even state happens to have the faster channel — not from any deep asymmetry between the $CP$ eigenstates. Second, "indirect" $CP$ violation ( $\varepsilon$ , the wrong- $CP$ admixture in the state) is distinct from "direct" $CP$ violation ( $\varepsilon'$ , an asymmetry in the decay amplitude itself); the kaon shows both.

Worked examples

1 · Making a pure K_L beam (confidence builder)

A neutral kaon beam is produced with momentum $p=10$ GeV/ $c$ . Using $\tau_S=0.895\times10^{-10}$ s and $\tau_L=5.12\times10^{-8}$ s, find the two decay lengths in the laboratory and estimate how far the beam must travel before it is $99.99\%$ pure $K_L$ .

Step 1. Boost. With $m_K=497.6$ MeV and $p=10$ GeV/ $c$ , $E=\sqrt{p^2+m_K^2}=10.012$ GeV, so $\gamma=E/m_K=\mathbf{20.1}$ and $\beta\approx1$ dilates every proper time by this factor.

Step 2. Proper decay lengths. $c\tau_S=(2.998\times10^8)(0.895\times10^{-10})=\mathbf{2.68\ cm}$ and $c\tau_L=(2.998\times10^8)(5.12\times10^{-8})=\mathbf{15.3\ m}$ — a ratio of 571.

Step 3. Lab decay lengths. Multiply by $\gamma\beta\approx20.1$ : $L_S=\mathbf{0.54\ m}$ (where the $K_S$ intensity falls by $1/e$ ) and $L_L=\mathbf{308\ m}$ .

Step 4. Purity. The surviving $K_S$ fraction after $\ell$ is $e^{-\ell/L_S}$ ; suppressing it to $10^{-4}$ needs $\ell/L_S=\ln 10^4=9.2$ , i.e. $\ell\approx\mathbf{5\ m}$ , over which $K_L$ barely changes ( $e^{-5/308}=0.984$ ). After about 5 m the beam is a clean 99.99% $K_L$ beam — the setup Cronin and Fitch used to hunt the forbidden $K_L\to\pi\pi$ .

2 · Reading the mass difference off the oscillation

The mass splitting is quoted as $\Delta m=3.48\times10^{-6}$ eV. Show that this makes $\Delta m\,\tau_S\approx0.47$ , and interpret what that number says about the strangeness oscillation.

Step 1. Convert to a frequency. The oscillation angular frequency is $\Delta m/\hbar$ with $\hbar=6.582\times10^{-16}$ eV·s, so $\Delta m/\hbar=(3.48\times10^{-6})/(6.582\times10^{-16})=\mathbf{5.29\times10^{9}\ s^{-1}}$ .

Step 2. Form the product. $x=\Delta m\,\tau_S/\hbar=(5.29\times10^{9})(0.895\times10^{-10})=\mathbf{0.474}$ .

Step 3. Interpret. The phase gained in one $K_S$ lifetime is $0.47$ rad, only $\sim8\%$ of a cycle, before the interfering $K_S$ amplitude decays away. This is the "coincidence" $\Delta m\approx\tfrac12\Gamma_S$ : mass splitting and decay rate are comparable — exactly the regime where the interference term of equation (5) is neither frozen nor washed out, so $\Delta m$ is measurable precisely because $x\sim1$ .

3 · The size of CP violation from the branching ratios

The observed branching ratios are $\mathrm{BR}(K_L\to\pi^+\pi^-)=1.97\times10^{-3}$ and $\mathrm{BR}(K_S\to\pi^+\pi^-)=0.692$ . Extract $|\eta_{+-}|$ and compare with $|\varepsilon|$ .

Step 1. Relate $\eta_{+-}$ to rates. $|\eta_{+-}|^2=\dfrac{\Gamma(K_L\to\pi^+\pi^-)}{\Gamma(K_S\to\pi^+\pi^-)}=\dfrac{\mathrm{BR}_L}{\mathrm{BR}_S}\cdot\dfrac{\tau_S}{\tau_L}$ , since $\Gamma=\mathrm{BR}/\tau$ .

Step 2. Plug in. $\dfrac{\mathrm{BR}_L}{\mathrm{BR}_S}=\dfrac{1.97\times10^{-3}}{0.692}=2.85\times10^{-3}$ and $\dfrac{\tau_S}{\tau_L}=1.75\times10^{-3}$ , so $|\eta_{+-}|^2=4.98\times10^{-6}$ .

Step 3. Root. $|\eta_{+-}|=\sqrt{4.98\times10^{-6}}=\mathbf{2.23\times10^{-3}}$ . This matches $|\varepsilon|\approx2.23\times10^{-3}$ : the "forbidden" amplitude is two parts in a thousand of the allowed one, and to this accuracy it is entirely indirect (mixing) $CP$ violation. The residual direct piece $\varepsilon'$ is a further $\sim10^{-3}$ smaller — hence NA48 and KTeV to resolve $\mathrm{Re}(\varepsilon'/\varepsilon)=(1.66\pm0.23)\times10^{-3}$ .

Physics of the portrait

The neutral kaon is a rare case where theoretical bravado — predicting a particle mixture no one had asked for — was vindicated, then overturned a symmetry everyone assumed. The portrait belongs not to one person but to the mechanism that opened it: the Gell-Mann–Pais idea that the neutral kaon is a superposition with two distinct lifetimes.

Origins · 1 minGell-Mann-Pais mechanism and how this was discovered

Gell-Mann-Pais mechanism

the 1955 prediction of two neutral-kaon lifetimes · strangeness mixing and CP

Neutral V-particles turned up in cosmic rays in 1947 (Rochester and Butler), but made no sense until Gell-Mann and Nishijima invented strangeness in 1953 to explain why they were produced strongly yet decayed weakly. The leap came in 1955: Gell-Mann and Pais reasoned that if $K^0$ and $\bar{K}^0$ are distinct antiparticles yet share the same $2\pi$ channel, the second-order weak interaction must connect them — so the states of definite lifetime are not $K^0$ and $\bar{K}^0$ but their combinations. From a purely quantum-mechanical argument about particle–antiparticle interference they boldly predicted a second, long-lived neutral kaon when only the short-lived one was known — confirmed in 1956 when Lederman and collaborators found the $K_2$ at Brookhaven. Pais and Piccioni then predicted that a $K_L$ beam crossing matter would regenerate $K_S$ , because $K^0$ and $\bar{K}^0$ scatter differently — soon borne out. The final twist was not in the script: in 1964 Christenson, Cronin, Fitch and Turlay found the long-lived beam making two pions after all, and $CP$ fell. Kobayashi and Maskawa explained it in 1973 by adding a third quark generation — wrung out of a two-part-per-thousand effect in a beam of kaons.

History

1947

— Rochester and Butler photograph neutral V-particle decays in cosmic rays

1955

— Gell-Mann and Pais predict K⁰ mixing and a second, long-lived neutral kaon

1956

— Lederman and collaborators observe the long-lived K₂ at Brookhaven

1964

— Christenson, Cronin, Fitch and Turlay find K_L → π⁺π⁻, discovering CP violation

1973

— Kobayashi and Maskawa trace CP violation to a third quark generation

Going further

Texts. Thomson, Modern Particle Physics, §14.4 and §14.4.1–2 · Perkins, Introduction to High Energy Physics, 4th ed., Ch. 11 · Griffiths, Introduction to Elementary Particles, 2nd ed., §4.4 · Commins & Bucksbaum, Weak Interactions of Leptons and Quarks, Ch. 12 · Branco, Lavoura & Silva, CP Violation, Chs. 6–7 · Bigi & Sanda, CP Violation, Ch. 6. The $\Delta S=2$ box diagram that generates $M_{12}$ , and the GIM/charm suppression that controls its size, are treated in the weak-interaction and CKM articles listed below. The key numerical results used throughout are: $CP(\pi\pi)=+1$ , $CP(\pi\pi\pi)=-1$ ; eigenvalues $\mu_{S,L}=D\pm\sqrt{H_{12}H_{21}}$ reducing to $K_1,K_2$ at $q/p=1$ ; oscillation law (5) with $\Delta m\,\tau_S=0.47$ ; $|\eta_{+-}|=2.23\times10^{-3}$ from the branching ratios; and $L_S=0.54$ m, $L_L=308$ m at $p=10$ GeV/ $c$ . All inputs ( $\tau_S=0.895\times10^{-10}$ s, $\tau_L=5.12\times10^{-8}$ s, $\Delta m=3.48\times10^{-6}$ eV, $m_K=497.6$ MeV) are PDG values.

Prerequisites. CKM Matrix · Symmetries in QM · Intrinsic Parity of Fermions · Weak Interactions of Quarks. Leads to. Strangeness Oscillations · $CP$ Violation in Kaons · $CP$ Violation in the Standard Model · $B$ -Meson Physics.

Ask the tutor: Ask me how the $\varepsilon$ admixture in $K_L$ turns the 1964 count of 45 two-pion events into a number, or how the GIM box diagram fixes $\Delta m$ and once predicted the charm mass.

Is the long-lived kaon K_L exactly the same as the CP eigenstate K₂?

Almost, but not exactly — and the "not exactly" is the whole story. $K_L$ is predominantly the $CP$ -odd $K_2$ but carries a small $CP$ -even admixture, $|K_L\rangle\propto|K_2\rangle+\varepsilon|K_1\rangle$ with $|\varepsilon|\approx2.2\times10^{-3}$ , and that impurity is why $K_L$ can decay to two pions where a pure $CP$ -odd state cannot. Set $\varepsilon=0$ and you recover exact $CP$ and lose the 1964 discovery. For lifetimes $K_L\approx K_2$ is fine; for anything touching $CP$ , keep the $\varepsilon$ term.

How do experimenters know whether a kaon is a K⁰ or a K̄⁰ at the moment it decays?

Through semileptonic decays and the $\Delta S=\Delta Q$ rule. A $K^0(d\bar{s})$ decays as $K^0\to\pi^-e^+\nu_e$ (positive lepton); a $\bar{K}^0(s\bar{d})$ as $\bar{K}^0\to\pi^+e^-\bar{\nu}_e$ (negative lepton). The lepton charge "tags" the strangeness at the instant of decay, so plotting tagged flavour against proper time maps the oscillation of equation (5) directly — and the charge asymmetry also carries $CP$ -violation information.

What is regeneration, and why does it happen?

When a pure $K_L$ beam (an equal, antisymmetric mix of $K^0$ and $\bar{K}^0$ ) crosses matter, the two flavour components scatter differently: a $\bar{K}^0$ can annihilate on nucleons and produce hyperons where a $K^0$ cannot, so their forward-scattering amplitudes differ. The material attenuates the two pieces unequally, and an unequal mixture is no longer pure $K_L$ — it has acquired a $K_S$ component, which decays to two pions right after the slab. Pais and Piccioni predicted this in 1955; it confirms the mixing picture and was a background the 1964 experiment had to rule out.

Why does the neutral kaon have two lifetimes but the charged kaon only one?

Because only a neutral particle can mix with its own antiparticle. $K^+(u\bar{s})$ and $K^-(s\bar{u})$ carry opposite, exactly-conserved charge, so nothing turns one into the other — there is nothing to diagonalize, and $K^+$ has a single lifetime. The neutral $K^0$ and $\bar{K}^0$ differ only in strangeness, which the weak interaction does not conserve, so it mixes them into two eigenstates of different mass and lifetime. Two lifetimes are the fingerprint of particle–antiparticle mixing, seen again in the $B^0$ , $B_s^0$ and $D^0$ systems.

Is CP violation in kaons a large effect or a small one?

Small — about two parts in a thousand. The forbidden $K_L\to\pi\pi$ amplitude is $|\eta|\approx2.2\times10^{-3}$ of the allowed one, and the direct piece $\varepsilon'$ is a thousand times smaller still. Its importance is out of all proportion to its size: it was the first crack in the mirror-symmetry of physical law, it is one of the Sakharov conditions for a matter-dominated universe, and its explanation forced a third quark generation. A tiny number with enormous consequences.

What is the difference between indirect (ε) and direct (ε′) CP violation?

Indirect $CP$ violation ( $\varepsilon$ ) lives in the mixing: $K_L$ is not a pure $CP$ eigenstate, so it reaches a $CP$ -even final state simply by containing a bit of the wrong- $CP$ kaon. Direct $CP$ violation ( $\varepsilon'$ ) lives in the decay amplitude: even a pure $CP$ eigenstate would decay to $\pi\pi$ at slightly different rates for kaon and antikaon. For years it was unclear whether $\varepsilon'$ existed — a "superweak" theory could fake the data with $\varepsilon$ alone — and its firm measurement, $\mathrm{Re}(\varepsilon'/\varepsilon)\approx1.7\times10^{-3}$ , proved $CP$ violation is genuinely built into the Standard Model weak decay, as Kobayashi and Maskawa required.