CP violation in the Standard Model

CP violation is the statement that the laws of physics are not quite the same for matter and antimatter: run a weak process forward and run its mirror-reflected antiparticle version, and the two rates differ by a hair. For decades this looked like a fragile curiosity, seen only in the decays of neutral kaons at the level of a few parts per thousand. The triumph of the Standard Model is to trace every quark-sector instance of it — kaons, $B$ mesons, $D$ mesons — back to a single number: the one irreducible complex phase that survives in the $3\times3$ unitary CKM matrix once every quark field has been rephased as far as it will go. Two generations of quarks give a real mixing matrix and exact CP symmetry; three generations leave exactly one phase, and that phase is the whole source. This article is the synthesis of the weak-flavour chapter. It takes the CKM matrix and its Wolfenstein expansion from aq-pp-ckm-matrix, the kaon parameter $\varepsilon$ from aq-pp-cp-violation-kaons, and the golden mode $B^0\to J/\psi\,K_S$ from aq-pp-b-meson-physics, and shows how unitarity turns the single phase into a piece of geometry — a closed triangle in the $\bar\rho$ – $\bar\eta$ plane. The power of the picture is over-determination: $|\varepsilon_K|$ , $\sin2\beta$ , $\Delta m_d$ and $|V_{ub}|$ each carve an independent curve through that plane, and they all cross at one point. The triangle closes; one phase explains it all. And yet — the closing note of the whole subject — that same phase is roughly ten orders of magnitude too feeble to have made the matter-dominated Universe we live in.

Key takeaways

All CP violation in the quark sector of the Standard Model comes from one irreducible complex phase in the unitary CKM matrix; two generations give a real matrix and exact CP.
In the Wolfenstein parametrisation the phase lives in $\bar\eta$ , entering only $V_{ub}\approx A\lambda^3(\rho-i\eta)$ and $V_{td}\approx A\lambda^3(1-\rho-i\eta)$ .
Unitarity $V^\dagger V=I$ gives closure relations such as $V_{ud}V_{ub}^{*}+V_{cd}V_{cb}^{*}+V_{td}V_{tb}^{*}=0$ , which draw as a closed unitarity triangle in the $\bar\rho$ – $\bar\eta$ plane.
The rephasing-proof measure is the Jarlskog invariant $J\approx A^2\lambda^6\eta\approx3\times10^{-5}$ ; it equals twice the area of the triangle and vanishes if any mixing angle or the phase is zero.
Independent constraints from $|\varepsilon_K|$ , $\sin2\beta\approx0.69$ , $\Delta m_d$ and $|V_{ub}|$ over-determine $(\bar\rho,\bar\eta)\approx(0.13,0.345)$ and all agree — the triangle closes.
SM CP violation is real but far too small to explain the cosmic matter-antimatter asymmetry, one of the sharpest open problems in physics.
Key idea: all CP violation in quark physics comes from one irreducible complex phase in the CKM matrix; unitarity turns this into a closed triangle in the (ρ,η) plane, which independent experiments over-determine and find consistent.

The core insight

All CP violation in quark physics comes from one irreducible complex phase in the CKM matrix; unitarity turns this into a closed triangle in the $(\bar\rho,\bar\eta)$ plane, which independent experiments over-determine and find consistent — while remaining far too small to explain the matter-antimatter asymmetry of the Universe.

Where this lives today

The unitarity triangle is a living experiment, not a settled diagram. LHCb at CERN and the Belle II experiment at KEK in Japan are pushing every side and angle to sub-degree precision, with LHCb already measuring $\gamma$ to a few degrees and pinning down $B_s^0$ mixing and its CP phase $\phi_s$ . The strategy is over-determination taken to the limit: measure the same apex through tree-level "sides" that are theoretically clean and loop-level "angles" sensitive to new virtual particles, and look for any crack where they fail to meet. A discrepancy between the tree- and loop-determined apex would be a discovery of physics beyond the Standard Model. So far the triangle stubbornly closes, tightening the SM. The deepest open question sits just past the edge of the diagram: because the CKM phase gives a baryon asymmetry $\sim10^{10}$ too small, the matter-antimatter imbalance of the Universe demands new sources of CP violation — perhaps the PMNS phase $\delta_{CP}$ in the lepton sector, perhaps the neutron electric dipole moment, perhaps unseen heavy particles. The CKM triangle is the template; the search is for the second one.

The physical picture

Picture the flavour-changing weak vertices as a web of complex amplitudes. Each $W$ -boson coupling between an up-type and a down-type quark carries a factor $V_{ij}$ , a complex number whose magnitude sets the transition strength and whose phase is invisible on its own — you can always rotate the quark fields to move a phase from one place to another. What cannot be rotated away is a relative phase between two interfering paths. CP conjugation flips the sign of every such phase (it sends $V_{ij}\to V_{ij}^{*}$ ), so an observable built from the interference of two amplitudes with a genuine relative phase comes out different for particle and antiparticle. The bookkeeping of which phases are real and which are physical is exactly the bookkeeping of the unitarity triangle: draw the three complex terms of a closure relation head to tail, and a non-degenerate triangle — one with area — is the geometric signature that a physical phase has survived.

Figure

The unitarity triangle in the ρ̄–η̄ plane: dividing the d–b closure relation by V_cd V_cb* pins the base from 0 to 1, the apex at (0.13, 0.345), sides R_u = 0.369 and R_t = 0.936, and interior angles β = 21.6°, γ = 69.4°, α = 89.0° that sum to 180°

The triangle is not a metaphor — it is a testable prediction, because each of its sides and angles is a different physical observable, and unitarity demands they fit together. The next figure is the one experimentalists actually draw: instead of assuming the apex, they measure it four different ways. The length of the left side follows from the semileptonic ratio $|V_{ub}/V_{cb}|$ ; the right side follows from the $B_d^0$ oscillation frequency $\Delta m_d$ , which fixes $|V_{td}|$ ; the angle $\beta$ follows directly from the time-dependent CP asymmetry in $B^0\to J/\psi\,K_S$ ; and the kaon parameter $|\varepsilon_K|$ traces a hyperbola. If the Standard Model's single-phase story is right, all four bands must cross at one point. They do.

Figure

Four independent measurements pin the same apex: the V_ub side sets R_u = 0.369 as a circle about the origin, B_d mixing Δm_d fixes R_t = 0.936 as a circle about (1,0), and sin 2β = 0.69 sets the β ray from (1,0) — all crossing at (ρ̄, η̄) = (0.13, 0.345), the over-determination that confirms one phase

The core equations

The starting object is the CKM matrix, written in the Wolfenstein expansion in powers of $\lambda=\sin\theta_C\approx0.225$ . This form makes the hierarchy of quark mixing manifest — diagonal near unity, off-diagonal falling in powers of $\lambda$ — and isolates the CP-violating phase into two corner elements.

V_{\text{CKM}}=\begin{pmatrix} V_{ud}&V_{us}&V_{ub}\\ V_{cd}&V_{cs}&V_{cb}\\ V_{td}&V_{ts}&V_{tb}\end{pmatrix}=\begin{pmatrix} 1-\tfrac{\lambda^2}{2}&\lambda&A\lambda^3(\rho-i\eta)\\[2pt] -\lambda&1-\tfrac{\lambda^2}{2}&A\lambda^2\\[2pt] A\lambda^3(1-\rho-i\eta)&-A\lambda^2&1\end{pmatrix}+\mathcal{O}(\lambda^4) \qquad (1)

where: $\lambda\approx0.2253$ is the sine of the Cabibbo angle; $A\approx0.811$ sets the third-generation couplings; $\rho,\eta$ are the CP-relevant coordinates; and the phase enters only through the combination $\rho-i\eta$ . Reading it: the imaginary part $\eta$ is the sole home of CP violation, and it appears at order $\lambda^3$ — which is why CP violation is a small effect. Isolating the two elements that carry the phase makes the physics transparent: CP violation is confined to transitions involving the top and up quarks in the extreme corners of the matrix.

V_{ub}=A\lambda^3(\rho-i\eta),\qquad V_{td}=A\lambda^3(1-\rho-i\eta) \qquad (2)

where: $V_{ub}$ governs $b\to u$ semileptonic decays and $V_{td}$ governs $B_d^0$ mixing. Reading it: a non-zero $\eta$ in either element forces a non-zero phase; if $\eta=0$ the whole matrix is real and CP is exact. Because $V$ is unitary, its columns are orthonormal. The orthogonality of the down and bottom columns gives the closure relation whose three terms are all of the same order, $A\lambda^3$ , so none dominates and the triangle they form is non-degenerate.

V_{ud}V_{ub}^{*}+V_{cd}V_{cb}^{*}+V_{td}V_{tb}^{*}=0 \qquad (3)

where: each term is a complex number of magnitude $\sim A\lambda^3\approx9\times10^{-3}$ . Reading it: three complex numbers summing to zero, drawn head to tail, close a triangle in the complex plane — the unitarity triangle. Dividing equation (3) by the (nearly real) middle term $V_{cd}V_{cb}^{*}$ rescales the base to unit length and places the apex at the point conventionally called $(\bar\rho,\bar\eta)$ , with the two non-trivial sides $R_u$ and $R_t$ .

\bar\rho+i\bar\eta=-\frac{V_{ud}V_{ub}^{*}}{V_{cd}V_{cb}^{*}},\qquad R_u=\sqrt{\bar\rho^2+\bar\eta^2}=\left|\frac{V_{ud}V_{ub}^{*}}{V_{cd}V_{cb}^{*}}\right|,\qquad R_t=\sqrt{(1-\bar\rho)^2+\bar\eta^2}=\left|\frac{V_{td}V_{tb}^{*}}{V_{cd}V_{cb}^{*}}\right| \qquad (4)

where: $\bar\rho=\rho(1-\tfrac{\lambda^2}{2})$ , $\bar\eta=\eta(1-\tfrac{\lambda^2}{2})$ are the properly-normalised apex coordinates. Reading it: $R_u$ comes from $|V_{ub}|$ , $R_t$ comes from $|V_{td}|$ (hence $\Delta m_d$ ), and the apex height $\bar\eta$ is a direct measure of CP violation — a flat triangle ( $\bar\eta=0$ ) means no CP violation. The interior angles are rephasing-invariant arguments of ratios of CKM products, so each is a genuine observable measurable in a specific interference experiment.

\beta=\arg\!\left(-\frac{V_{cd}V_{cb}^{*}}{V_{td}V_{tb}^{*}}\right),\qquad \gamma=\arg\!\left(-\frac{V_{ud}V_{ub}^{*}}{V_{cd}V_{cb}^{*}}\right),\qquad \alpha=\pi-\beta-\gamma \qquad (5)

where: $\beta$ is measured from $B^0\to J/\psi\,K_S$ , $\gamma$ from $B\to DK$ tree decays, $\alpha$ from $B\to\pi\pi,\rho\rho$ . Reading it: $\tan\beta=\bar\eta/(1-\bar\rho)$ and $\tan\gamma=\bar\eta/\bar\rho$ ; the constraint $\alpha+\beta+\gamma=180°$ is the closure condition, a direct test of three-generation unitarity. Finally, the single convention-independent number that says "CP is violated, full stop" is the Jarlskog invariant — the common area of every unitarity triangle, independent of how you rephase the quark fields.

J=\operatorname{Im}\!\left(V_{us}V_{cb}V_{ub}^{*}V_{cs}^{*}\right)\approx A^2\lambda^6\,\eta\approx 3\times10^{-5} \qquad (6)

where: $J$ is the Jarlskog invariant; twice the area of any unitarity triangle equals $J$ . Reading it: $J\neq0$ if and only if CP is violated, and its tininess ( $\sim10^{-5}$ ) is the seed of the baryogenesis shortfall discussed below.

Derivation ACounting the one irreducible CP phase

The goal is to show by pure parameter counting why three generations of quarks are the minimum needed for CP violation and why exactly one CP-violating phase survives. We begin from a completely general $N\times N$ unitary mixing matrix, subtract the parameters a real rotation would already carry, then subtract the phases that can be rotated away by redefining the quark fields, and read off the irreducible remainder — zero for two generations, one for three. This is the calculation Kobayashi and Maskawa did in 1973, and it is the logical foundation of everything else in this article.

Step 1. A general $N\times N$ unitary matrix is fixed by $N^2$ real parameters, because unitarity $V^\dagger V=I$ imposes exactly $N^2$ real constraints on the $2N^2$ real numbers of a general complex $N\times N$ matrix. Why: this total is the pool from which the physical mixing parameters must be drawn.

Step 2. Of those $N^2$ parameters, a real orthogonal rotation would use $\tfrac{1}{2}N(N-1)$ Euler angles; the remainder, $N^2-\tfrac12 N(N-1)=\tfrac12 N(N+1)$ , are complex phases. Why: the angles are CP-conserving mixing that survives even in a real matrix, so any CP violation must hide among the phases.

Step 3. The quark fields can be rephased freely: there are $N$ up-type and $N$ down-type fields, giving $2N$ phase rotations, but one overall common rephasing leaves $V$ unchanged, so $2N-1$ of the phases are unphysical and can be removed. Why: only relative phases between physical amplitudes are observable; the rest are convention.

Step 4. The number of physical, irremovable phases is therefore $\tfrac12 N(N+1)-(2N-1)=\tfrac12(N-1)(N-2)$ . For $N=2$ this is $0$ ; for $N=3$ it is exactly $1$ . Why: two generations give a real Cabibbo rotation and exact CP symmetry, while three generations leave one irreducible phase $\delta$ — the single origin of CP violation in the Standard Model quark sector. The same count gives three mixing angles for $N=3$ , matching the observed $(\theta_{12},\theta_{13},\theta_{23},\delta)$ .

Go deeper · Derivation B · 1 minFrom unitarity to the closed triangle

Derivation BFrom unitarity to the closed triangle

The goal is to convert the abstract statement " $V$ is unitary" into the concrete geometric object that experiments test — a closed triangle whose sides and angles are separate observables. We take one orthogonality relation among the columns of $V$ , evaluate each term in the Wolfenstein expansion to show they share a common order $A\lambda^3$ , normalise by the middle term to fix a unit base, and identify the apex, sides, and angles with physical quantities so that closure becomes a falsifiable prediction of three-generation unitarity.

Step 1. Orthogonality of the $d$ and $b$ columns of $V$ reads $\sum_i V_{id}V_{ib}^{*}=0$ , i.e. $V_{ud}V_{ub}^{*}+V_{cd}V_{cb}^{*}+V_{td}V_{tb}^{*}=0$ . Why: unitarity is not optional in the SM — it follows from there being exactly three generations with a single unitary mixing matrix, so any failure of this relation signals new physics.

Step 2. Insert the Wolfenstein forms: $V_{ud}V_{ub}^{*}\approx A\lambda^3(\rho+i\eta)$ , $V_{cd}V_{cb}^{*}\approx -A\lambda^3$ , and $V_{td}V_{tb}^{*}\approx A\lambda^3(1-\rho-i\eta)$ . All three are $\mathcal{O}(A\lambda^3)$ , so no term is negligible. Why: three comparable complex numbers summing to zero form a genuine triangle with area, unlike the other unitarity relations (e.g. among the first two rows) where one term is far smaller and the triangle is squashed almost flat.

Step 3. Divide through by $V_{cd}V_{cb}^{*}$ to normalise. The base runs from $0$ to $1$ along the real axis; the apex sits at $\bar\rho+i\bar\eta=-V_{ud}V_{ub}^{*}/(V_{cd}V_{cb}^{*})$ , and the third side closes back from $(1,0)$ to the apex with length $R_t=\sqrt{(1-\bar\rho)^2+\bar\eta^2}$ . Why: normalising removes the convention-dependent overall phase and scale, leaving a triangle whose shape is pure physics.

Step 4. Identify observables: the left side $R_u=\sqrt{\bar\rho^2+\bar\eta^2}$ is set by $|V_{ub}/V_{cb}|$ , the right side by $\Delta m_d\propto|V_{td}|^2$ , the angle $\beta$ by the $B\to J/\psi K_S$ asymmetry, and the apex height $\bar\eta$ by $|\varepsilon_K|$ . Closure requires all of them to meet at one apex. Why: this is the crux — one phase predicts that four independent measurements land on a single point, and the measured over-determination (Section "The physical picture") confirms it to within errors.

Go deeper · Derivation C · 1 minThe Jarlskog invariant as the rephasing-proof measure

Derivation CThe Jarlskog invariant as the rephasing-proof measure

The goal is to build the one number that certifies CP violation without reference to any parametrisation, and to expose exactly what must be non-zero for CP to be violated. A drawback of the phase $\delta$ or the coordinate $\eta$ is that their numerical values shift when you rephase quark fields; we want an invariant. We define the Jarlskog determinant, show it equals twice the area of every unitarity triangle, reduce it to $A^2\lambda^6\eta$ in the Wolfenstein form, and read off the necessary conditions — all three generations, all mixing angles, and the phase must participate.

Step 1. Define $J$ through $\operatorname{Im}\!\left(V_{ij}V_{kl}V_{il}^{*}V_{kj}^{*}\right)=J\sum_{m,n}\varepsilon_{ikm}\varepsilon_{jln}$ , a quantity manifestly unchanged by $V_{ij}\to e^{i\phi_i}V_{ij}e^{-i\phi_j}$ . Why: every quark rephasing cancels between the four factors, so $J$ is the convention-independent content of the phase — a real physical number.

Step 2. Geometrically, $|J|$ is twice the area of any of the (unnormalised) unitarity triangles; a triangle with zero area means $J=0$ means no CP violation. In the standard parametrisation $J=c_{12}c_{23}c_{13}^2 s_{12}s_{23}s_{13}\sin\delta$ , and in Wolfenstein $J\approx A^2\lambda^6\eta$ . Why: this ties the abstract invariant to the picture — CP violation is literally the area of the triangle.

Step 3. Read off the vanishing conditions: $J=0$ if any mixing angle is $0$ or $90°$ , if $\delta=0$ or $180°$ , or if any two same-charge quark masses are degenerate (which would let a rotation remove the phase). Why: CP violation is not generic — it needs three non-degenerate generations, all three mixing angles non-trivial, and a non-trivial phase, all at once. This is why it took a third generation to accommodate it.

Step 4. Numerically, with $A=0.811$ , $\lambda=0.2253$ , $\eta=0.345$ : $J=A^2\lambda^6\eta=(0.658)(1.31\times10^{-4})(0.345)\approx3.0\times10^{-5}$ . Why: the smallness of $J$ — a product of a sixth power of $\lambda$ and the already-small $\eta$ — is not a detail but the headline: the Universe's CP violation, as the SM knows it, is a $10^{-5}$ effect, and that is the seed of the baryogenesis problem.

Limiting cases

Case	Result	Meaning
Two generations	$J=0$ , matrix real	Cabibbo mixing alone conserves CP exactly; no phase survives rephasing
Any mixing angle $\to0$ or $90°$	$J=0$	one generation decouples; CP restored even with three families
Degenerate up- or down-quark masses	$J=0$	a rotation absorbs the phase; mass non-degeneracy is required
Phase $\delta\to0$ or $\pi$ (so $\eta\to0$ )	flat triangle, $J=0$	matrix effectively real; the triangle collapses to a line
Full three-generation SM	$J\approx3\times10^{-5}$ , $\bar\eta\approx0.34$	non-degenerate triangle; CP violated in $K$ , $B$ , $D$ systems
Cosmological requirement	need effective CP of $\mathcal{O}(1)$	SM $J$ is $\sim10^{10}$ too small; baryogenesis needs new sources

Common trapEvery complex phase in the CKM matrix is physical and CP-violating

It is not. A general $3\times3$ complex matrix has nine phases, but eight of them are pure convention: you can rotate the phases of the six quark fields (five independent rephasings, since one overall phase does nothing) to sweep them away. Exactly one phase — the Jarlskog-invariant combination — cannot be removed by any field redefinition, and only that one is physical. This is why counting matters: the naive "there are lots of complex entries, so lots of CP violation" is wrong. It is also why CP violation is impossible with two generations even though a $2\times2$ unitary matrix has complex entries — all of its phases are removable, leaving the real Cabibbo rotation. The physical content is captured by rephasing-invariant quantities: the angles $\alpha,\beta,\gamma$ , the apex $(\bar\rho,\bar\eta)$ , and above all $J$ . If a calculation's answer depends on a phase you could have rotated away, it is not an observable.

Worked examples

1 · Building the unitarity triangle from the fit (fully worked)

A global fit gives $\lambda=0.2253$ , $A=0.811$ , $\rho=0.13$ , $\eta=0.345$ . Find the apex, both side lengths, and all three angles, and check that they sum to $180°$ .

Step 1. Apex coordinates. To the order needed, $\bar\rho\approx\rho=0.13$ and $\bar\eta\approx\eta=0.345$ (the correction factor $1-\lambda^2/2=0.975$ shifts them to $\bar\rho=0.127$ , $\bar\eta=0.336$ ; we keep the leading values). So the apex sits at $(0.13,0.345)$ , comfortably above the real axis — already a signal that the triangle has area and CP is violated.

Step 2. Side lengths. The left side $R_u=\sqrt{\rho^2+\eta^2}=\sqrt{0.0169+0.1190}=\sqrt{0.1359}=\mathbf{0.369}$ . The right side $R_t=\sqrt{(1-\rho)^2+\eta^2}=\sqrt{0.7569+0.1190}=\sqrt{0.8759}=\mathbf{0.936}$ . These are the lengths the $|V_{ub}|$ and $\Delta m_d$ measurements must reproduce.

Step 3. Angles. From the origin, $\tan\gamma=\eta/\rho=0.345/0.13=2.654$ , so $\gamma=\arctan(2.654)=\mathbf{69.4°}$ . At $(1,0)$ , $\tan\beta=\eta/(1-\rho)=0.345/0.87=0.3966$ , so $\beta=\arctan(0.3966)=\mathbf{21.6°}$ . Closure fixes $\alpha=180°-69.4°-21.6°=\mathbf{89.0°}$ .

Step 4. Cross-check with the measured golden mode: $\sin2\beta=\sin(43.3°)=\mathbf{0.685}$ , in excellent agreement with the $B$ -factory world average $\sin2\beta\approx0.69$ . The triangle closes: $\alpha+\beta+\gamma=89.0°+21.6°+69.4°=180.0°$ , and its independently-measured angle $\beta$ matches $\sin2\beta$ from data.

2 · The Jarlskog invariant and the baryogenesis shortfall

Compute $J$ from the same fit, relate it to the triangle area, and compare the Standard Model's CP violation with what cosmology needs.

Step 1. Jarlskog invariant. $J\approx A^2\lambda^6\eta$ . Evaluate the pieces: $A^2=0.811^2=0.658$ ; $\lambda^6=(0.2253^2)^3=(0.05076)^3=1.31\times10^{-4}$ ; times $\eta=0.345$ . So $J=0.658\times1.31\times10^{-4}\times0.345=\mathbf{2.97\times10^{-5}}$ , consistent with the PDG value $J\approx3.1\times10^{-5}$ .

Step 2. Area interpretation. The normalised triangle has base $1$ and height $\bar\eta=0.345$ , so its area is $\tfrac12\times1\times0.345=\mathbf{0.173}$ . The invariant area of the unnormalised triangle is $J/2\approx1.5\times10^{-5}$ ; the normalisation by $|V_{cd}V_{cb}^{*}|^2\approx(A\lambda^3)^2$ restores the $0.173$ . Either way, non-zero area means CP violation.

Step 3. The cosmological comparison. The observed baryon-to-photon ratio is $\eta_B=n_B/n_\gamma\approx6.1\times10^{-10}$ — a small but non-zero matter excess. Sakharov's conditions say generating it requires baryon-number violation, C and CP violation, and a departure from equilibrium. The SM has all three in principle, but the CP violation available is set by $J\approx3\times10^{-5}$ , further suppressed by ratios of quark masses over the electroweak temperature.

Step 4. The verdict. Carrying the suppression through, the SM predicts a baryon asymmetry of order $10^{-20}$ or smaller — about ten orders of magnitude too little to make $\eta_B\approx6\times10^{-10}$ . The single CKM phase is genuine and beautifully consistent across kaons and $B$ mesons, yet it cannot be the source of the matter-dominated Universe — a sharp, quantitative pointer to CP violation beyond the Standard Model.

Physics of the portrait

CP violation in the Standard Model is a story of a phase that hid in plain sight: predicted from the algebra of unitary matrices, made convention-proof by a single invariant, and finally drawn as a triangle in a plane. The invariant is the work of Cecilia Jarlskog, whose 1985 measure turned "there is a phase" into "here is a number that cannot lie."

Origins · 1 minCecilia Jarlskog and how this was discovered

Cecilia Jarlskog

b. 1941 · the rephasing-invariant measure of CP violation

Cecilia Jarlskog, a Swedish theoretical physicist born in 1941, gave the study of CP violation the tool it had been missing. After Kobayashi and Maskawa showed in 1973 that three generations leave one irremovable phase, a practical problem remained: the numerical value of that phase, and of the Wolfenstein coordinate $\eta$ , depends on how you choose to rephase the quark fields, so neither is directly an observable. In 1985 Jarlskog cut through this by defining a single quantity — now universally called the Jarlskog invariant $J$ — built from a rephasing-invariant combination of four CKM elements, $\operatorname{Im}(V_{ij}V_{kl}V_{il}^{*}V_{kj}^{*})$ . Its power is threefold: it is completely convention-independent, so any two physicists compute the same number; it is non-zero if and only if CP is violated, making it a clean yes-or-no diagnostic; and it has a beautiful geometric meaning as twice the common area of every unitarity triangle. Jarlskog also made explicit the full set of conditions for CP violation — three non-degenerate generations, all three mixing angles non-trivial, and the phase itself non-trivial — packaging them into a single elegant determinant of the quark mass matrices. Her invariant is now the standard way the Particle Data Group quotes the strength of quark-sector CP violation, $J\approx3\times10^{-5}$ , and the language in which every global CKM fit reports whether the triangle truly has area. Jarlskog held professorships at Lund and elsewhere and served on the Nobel physics committee, remaining a forceful advocate for rigour in separating what the field has actually measured from what is mere convention.

History

1964

— Cronin and Fitch observe CP violation in neutral kaon decays

1973

— Kobayashi and Maskawa predict a third generation to accommodate CP violation

1983

— Wolfenstein casts the CKM matrix in its λ-expansion, isolating the phase in η

1985

— Jarlskog defines the rephasing-invariant measure J of CP violation

2001

— BaBar and Belle establish CP violation in B mesons, measuring sin 2β

Going further

Texts. Thomson, Modern Particle Physics, §14.6–14.7 (the unitarity triangle, $\sin2\beta$ , Eq. 14.67 and Fig. 14.25) · Peskin & Schroeder, An Introduction to Quantum Field Theory, §20.3 (CKM matrix and CP violation) · Griffiths, Introduction to Elementary Particles, 2nd ed., §4.6 and Ch. 9 · Halzen & Martin, Quarks and Leptons, Ch. 12 · Schwartz, Quantum Field Theory and the Standard Model, Ch. 29 · Branco, Lavoura & Silva, CP Violation (the standard monograph). The key numerical results to keep in mind: the physical-phase count gives $(N-1)(N-2)/2=1$ for $N=3$ ; the apex is $(\bar\rho,\bar\eta)=(0.13,0.345)$ with sides $R_u=0.369$ and $R_t=0.936$ ; the angles $\beta=21.6°$ , $\gamma=69.4°$ , $\alpha=89.0°$ sum to $180°$ and yield $\sin2\beta=0.685$ ; the Jarlskog invariant evaluates to $J=A^2\lambda^6\eta=3.0\times10^{-5}$ ; and the SM baryon asymmetry falls short of the observed value by $\sim10^{10}$ . The global-fit inputs throughout are $\lambda=0.2253$ , $A=0.811$ , $\rho=0.13$ , $\eta=0.345$ , consistent with PDG-era CKMfitter/UTfit values.

Prerequisites. The CKM Matrix · CP Violation in Kaons · B-Meson Physics · The Neutral Kaon System. Leads to. Matter-Antimatter Asymmetry · Open Questions in Particle Physics · The Three-Flavour PMNS Matrix.

Ask the tutor: Ask me to walk through why the counting $\tfrac12(N-1)(N-2)$ gives exactly one phase, or to trace how $\sin2\beta$ from $B^0\to J/\psi K_S$ lands on the same apex as $|V_{ub}|$ and $\Delta m_d$ .

Is CP violation the same thing as matter-antimatter asymmetry?

No — they are linked but distinct. CP violation is a microscopic property of the interactions: a weak process and its CP-mirror (antiparticles, spatially reflected) have slightly different rates. The matter-antimatter asymmetry is a cosmological fact: the Universe today contains matter but essentially no antimatter, quantified by $\eta_B\approx6\times10^{-10}$ . Sakharov showed that generating the second requires the first, along with baryon-number violation and a departure from thermal equilibrium. So CP violation is a necessary ingredient. But the CKM phase supplies far too little of it — about ten orders of magnitude short — so the observed asymmetry needs additional CP violation from beyond the Standard Model. CP violation is the mechanism; the asymmetry is one thing that mechanism, if strong enough, could have produced.

Why does CP violation need all three generations rather than just a complex number somewhere?

Because most complex numbers in the mixing matrix are convention, not physics. You are free to redefine the phase of each quark field, and doing so shuffles phases around the CKM matrix without changing any measurable rate. For two generations there are enough of these free rephasings to make the entire matrix real — so however complex it looks, CP is exactly conserved. For three generations the counting leaves exactly one phase that no rephasing can remove. Equivalently, the Jarlskog invariant, which is built to be rephasing-proof, is automatically zero for two generations and can be non-zero only for three. So what matters is not "having a complex entry" but having a phase that survives every field redefinition — which first happens at three generations.

What exactly is the Jarlskog invariant, in plain terms?

It is the single number that answers "is CP violated, and by how much?" without depending on any arbitrary choices. The phase $\delta$ and the coordinate $\eta$ both change value if you rephase the quark fields, so neither is a clean observable on its own. Jarlskog's insight was to combine four CKM elements in a particular product, $\operatorname{Im}(V_{ij}V_{kl}V_{il}^{*}V_{kj}^{*})$ , chosen so every field rephasing cancels out. The result, $J\approx3\times10^{-5}$ , is the same no matter how anyone sets up their conventions. It has a lovely geometric meaning — it equals twice the area of every unitarity triangle — so "CP is violated" and "the triangle has area" and " $J\neq0$ " are three ways of saying one thing.

How is the angle β actually measured?

Through a time-dependent interference in neutral $B$ mesons. A $B^0$ can decay to $J/\psi K_S$ directly, or first oscillate into a $\bar B^0$ and then decay to the same final state; because $J/\psi K_S$ is accessible to both, the two paths interfere, and the interference carries the phase of the mixing amplitude, which is $2\beta$ . At the $B$ -factories one $B$ meson tags the flavour and the other decays to $J/\psi K_S$ ; the rate difference versus the time between the two decays traces a sinusoid of amplitude $\sin2\beta$ . The world average $\sin2\beta\approx0.69$ gives $\beta\approx22°$ — matching the value the sides of the triangle independently predict.

Does the Standard Model have CP violation anywhere besides the CKM matrix?

In principle, yes — in the strong interaction, through a term proportional to a parameter $\bar\theta$ that would also violate CP. But experiment finds $\bar\theta<10^{-10}$ : for reasons unknown, strong-interaction CP violation is either absent or fantastically small, a puzzle called the strong CP problem. Apart from that, the SM's only established source of CP violation is the single CKM phase. The lepton sector may violate CP through an analogous PMNS phase $\delta_{CP}$ , but that is still being measured. So the "Standard Model of CP violation" is, in practice, the one CKM phase — which is why its inability to explain the cosmic asymmetry is so striking.

If the triangle always closes, what is left to discover?

Precision is the discovery tool. The sides of the triangle come from "tree-level" processes that are theoretically clean and insensitive to new heavy particles; the angles and the kaon constraint come partly from "loop-level" processes where unknown heavy particles could contribute virtually. If the SM is complete, the tree-determined apex and the loop-determined apex must coincide. Any daylight between them — a triangle that closes for trees but not for loops — is new physics. This is why LHCb and Belle II push every input to ever-higher precision: they are not confirming a known answer but stress-testing the single-phase hypothesis, hunting for the small crack that would reveal additional CP-violating interactions, the kind cosmology says must exist.