Unitarity triangles and CP-violating invariants
Translate complex CKM orthogonality into triangle geometry and distinguish convention-dependent phases from the invariant Jarlskog area.
On this page
The question for today
What you’ll understand
- Construct a normalized unitarity triangle with explicit complex products.
- Relate its physical area to the Jarlskog invariant.
- Explain why an isolated matrix phase is not a CP observable.
Orthogonality of complex vectors can be drawn as a closed triangle. For three-family quark mixing, this elementary geometry becomes a precise consistency test linking several independent decay measurements. The drawing is meaningful only when we distinguish the triangle formed by actual CKM products from a rescaled plotting convention. Its area connects to a rephasing invariant that survives every arbitrary choice of quark field phases. We will use exact unitary identities, rather than a truncated expansion in a small mixing parameter.
Complex orthogonality closes a triangle
Take the inner product of the down and bottom columns of V. Unitarity gives . Each term is a complex number, hence a directed vector in a plane. Placing these three vectors head to tail closes a triangle. The construction requires their relative complex phases as well as magnitudes. Three positive lengths satisfying a geometric inequality are not by themselves proof that independently measured weak amplitudes satisfy the complex unitarity relation.
For plotting, divide by the nonzero middle product and choose an overall orientation so the vertices are zero, one, and . The closed vectors can then be written as z, one minus z, and minus one. This normalized triangle has base one and height the absolute imaginary part of z. Its area is convenient, but it is not the area of the original CKM-product triangle until the removed length scale is restored.
A quartet removes arbitrary field phases
A quark rephasing multiplies every entry of one row or column by a common phase. The quartet contains each relevant row and column phase once and its inverse once, so they cancel. Define J as the imaginary part of this quartet. Other distinct-row, distinct-column quartets give plus or minus J according to their index ordering. A sign therefore has meaning only after fixing the quartet convention. The unsigned triangle area is always half the absolute value of J.
Use the standard parameterization , where the real rotations have positive sine above and negative sine below the diagonal, and the 13 entry of the complex rotation is . Direct multiplication gives J as a product of three sines, appropriate cosines, and the sine of delta. A phase of ninety degrees does not guarantee a large CP invariant: a small mixing angle can suppress the whole product. Vanishing any necessary angle collapses every unitarity triangle.
Interpret geometry with its dynamical assumptions
Within the three-family Standard Model quark sector, a nonzero J with nondegenerate same-charge quark masses signals a physical CP phase. Exact degeneracy would permit additional rotations and can remove observable CP violation even if a chosen matrix representation looks complex. This qualification distinguishes a convenient matrix invariant from the full statement about the mass and interaction theory. The strong CP parameter is a separate issue; this discussion concerns the charged-current flavor sector and does not claim to classify every possible source of CP violation.
Experiments do not photograph this triangle. They infer side lengths and angles from decay rates, mixing frequencies, and interference asymmetries, with process-specific strong-interaction inputs. Comparing these constraints overdetermines a common geometry. A disagreement can test the assumed three-family unitary description or reveal underestimated inputs, but requires a complete uncertainty analysis. Conversely, a neatly drawn triangle built from one global fit is not an independent experimental verification of each side. Its diagnostic power comes from comparing genuinely distinct measurements under transparent dynamical assumptions.
Work it through
Restore the area scale
A hypothetical exactly unitary fit gives normalized apex and . Find the normalized area, original triangle area, and magnitude of J. These are illustrative values, not measured CKM inputs.
The normalized base has length one and the height is one half, so its area is . The real part locates the apex horizontally without changing this area.
Restoring the product magnitude rescales both lengths. Thus the original area is , rather than one quarter times 0.04.
Every original CKM unitarity triangle has area . Doubling the restored area therefore gives ; orientation would be needed to infer its sign.
The normalized area is 0.25, the physical CKM-product area is 0.0004, and the invariant magnitude is 0.0008.
The living notebook
Explore the model
Interactive lab · vectors
Conjugate directions and triangle height
Reset and set BOTH Angle = 30° and Magnitude = 1; predict x = √3/2 and y = 1/2. For a fresh comparison, Reset and set Angle = 150° and Magnitude = 1; predict x = −√3/2 and the same y = 1/2. Inspect the component readouts.
- x component
- 1.393
- y component
- 0.975
- Magnitude
- 1.7
Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forThe lab illustrates complex-plane projections and equal heights. It displays one vector, not a CKM triangle reconstruction, a Jarlskog calculation, or fitted flavor data.
Pause & reason
Check your understanding
Your turn
Try it yourself
Use , , and in the standard parameterization. Find J and the original triangle area. What happens when delta is replaced by zero at the same angles?
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
The first two cosine factors are , and .
Compare with the solution
The product is . The original unitarity-triangle area is therefore . These large toy angles are not a phenomenological quark fit. Setting delta to zero makes its sine vanish, so J and every original triangle area vanish while the mixing angles can remain nonzero. Thus mixing does not require CP violation, and geometric closure alone does not require a nonzero area.
Keep the idea
Take it with you
- Unitarity triangles arise from complex row or column orthogonality.
- Normalized plotting areas require a squared length rescaling before comparison with J.
- Rephasing-invariant quartets and nondegenerate masses underlie physical charged-current CP violation.
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