Charm, Charged Mesons, and the Unitarity Triangle

Every measurement in this chapter ends as one point in one plane, and the test is not any single number but whether determinations sharing no apparatus intersect.

🎯 Why this matters

A test built that way fails usefully. If the regions missed each other the picture would say which pair disagrees and by how much, which is more than a discrepancy in any single number could tell you.

Three short sections close the chapter. §8.9 does the one quark sector left — charm, the only up-type system. §8.10 does CP violation without any mixing at all. And §8.11 collects everything into a single picture, which is the point the whole chapter has been building towards.

§8.9 Charm — the up-type sector, and a difference of asymmetries

The $D^0$ is the hardest of the four systems, for the reason §8.1 identified: its oscillation period is 1600 lifetimes, so nothing oscillates before everything has decayed. Charm oscillations have still not been directly observed. The mass difference is extracted indirectly:

\Delta m_c = (9.67 \pm 1.4)\times10^{-3}\;\text{ps}^{-1} = (6.37 \pm 0.92)\;\mu\text{eV}

CP violation in charm was the last to be found — 2019, LHCb — and it appears in singly Cabibbo-suppressed decays, where $c \to u d\bar d$ or $c \to u s\bar s$ (§7.12’s Example 7.1 classified exactly these).

Fig. 8.14 — the two SCS channels, and their conjugates

Click a vertex or an internal line.

Bettini Fig. 8.14. Both K⁺K⁻ and π⁺π⁻ are SCS, so both are suppressed by one power of tan θ_C — which is what makes them comparable, and what makes their difference meaningful.

The measured quantity is not an asymmetry but a difference of two asymmetries:

\Delta A_{CP} = A_{CP}(K^+K^-) - A_{CP}(\pi^+\pi^-) = (-15.7 \pm 2.9)\times10^{-4}

⚙️ Engineer’s bridge — subtracting two asymmetries kills the instrumental ones — provided the physics does not cancel too

The reason for the subtraction is the same instinct as §8.8’s double ratio, one level up, and it is worth seeing what it buys and what it risks.

What it buys. Two instrumental asymmetries contaminate any raw measurement here, and neither is small compared with $10^{-3}$ :

the $D^{*+}$ and $D^{*-}$ production cross-sections differ, because the LHC collides protons on protons and the initial state is not CP symmetric;
the detection efficiencies for the two tag charges differ — $\pi^+$ versus $\pi^-$ in one sample, $\mu^+$ versus $\mu^-$ in the other.

Both are properties of the tag, not of the final state. So they are identical for $K^+K^-$ and $\pi^+\pi^-$ events, and they cancel exactly in the difference.

What it risks. If the physics asymmetry were also the same in both channels, it would cancel too and the measurement would return zero by construction. This is the failure mode of every differential measurement: subtract away the noise and you may subtract the signal.

Why it does not happen here is a symmetry argument, and a pretty one. Under U-spin — the SU(2) subgroup of flavour SU(3) that rotates $d \leftrightarrow s$ , exactly as isospin rotates $u \leftrightarrow d$ — the $K^+K^-$ and $\pi^+\pi^-$ final states are the two members of a $U = 0$ combination:

|0,0\rangle = \tfrac{1}{\sqrt2}\left(\left|\tfrac12,\tfrac12\right\rangle\left|\tfrac12,-\tfrac12\right\rangle \htmlClass{t-minus}{-} \left|\tfrac12,-\tfrac12\right\rangle\left|\tfrac12,\tfrac12\right\rangle\right)

The U-spin singlet. The same Clebsch–Gordan combination that appears everywhere two spin-½ objects are coupled — and here its antisymmetry is what turns a subtraction into a doubling.

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.

and the antisymmetry means the two decay amplitudes are equal and opposite. So the two CP asymmetries have opposite signs, and subtracting them doubles the signal rather than cancelling it.

The engineering version of this is a differential pair: drive the two halves of a circuit in antiphase and take the difference. Common-mode disturbances — supply noise, temperature, the LHC’s proton-proton initial state — subtract away; the differential signal adds. It works only because the signal was arranged to be antisymmetric in the first place, which is exactly the condition U-spin supplies.

Where it breaks: the differential pair only doubles the signal if the two halves really are driven in antiphase, and here that is a symmetry argument rather than a wiring diagram. U-spin is broken by the strong interaction itself (unlike isospin, which is broken only electromagnetically), so the breaking is sizeable and not fully known. What survives the argument is that the two signs cannot be made equal — enough to guarantee the subtraction does not cancel the physics, not enough to predict the size.

charm, and how small its CP violation is

import numpy as np
hbar = 6.582119569e-4        # eV ps
dmc, tauD = 9.67e-3, 0.4103

print(f"(8.82) Delta m_c = {dmc:.3e} ps^-1 = {dmc*hbar*1e6:.2f} ueV")
T = 2*np.pi/dmc
print(f"   period T = 2 pi / Delta m_c = {T:.1f} ps = {T/tauD:.0f} lifetimes")
print( "   -- which is why charm oscillation has still not been seen directly")

print("\nthe measured difference of asymmetries (8.87):")
print(f"   Delta A_CP = -15.7e-04 +- 2.9e-04   ->  {15.7/2.9:.1f} sigma")

print("\nput the four CP violations of this chapter side by side:")
rows = [('K0','mixing','|eps|',2.232e-3), ('K0','decay',"|eps'|",3.71e-6),
        ('D0','decay','Delta A_CP',15.7e-4), ('B0','interference','sin 2 beta',0.675)]
for s, k, n, v in rows:
    print(f"   {s:4s} {k:13s} {n:15s} {v:.2e}")
print("\nthe B0 is 400x larger than anything else here.  that is not because")
print("the CKM phase is bigger there -- it is the same delta_13 -- but because")
print("interference with mixing reaches it WITHOUT the two small factors that")
print("throttle a decay asymmetry: a second amplitude and a strong phase.")

prints

(8.82) Delta m_c = 9.670e-03 ps^-1 = 6.36 ueV
 period T = 2 pi / Delta m_c = 649.8 ps = 1584 lifetimes
 -- which is why charm oscillation has still not been seen directly

the measured difference of asymmetries (8.87):
 Delta A_CP = -15.7e-04 +- 2.9e-04   ->  5.4 sigma

put the four CP violations of this chapter side by side:
 K0   mixing        |eps|           2.23e-03
 K0   decay         |eps'|          3.71e-06
 D0   decay         Delta A_CP      1.57e-03
 B0   interference  sin 2 beta      6.75e-01

the B0 is 400x larger than anything else here.  that is not because
the CKM phase is bigger there -- it is the same delta_13 -- but because
interference with mixing reaches it WITHOUT the two small factors that
throttle a decay asymmetry: a second amplitude and a strong phase.

§8.10 CP violation without any mixing

The charged B settles the question the chapter opened with. $B^\pm$ cannot mix — electric charge is exactly conserved — so any CP violation observed in its decay is violation in the decay, full stop. And it was observed, in 2008.

The payoff is the unitarity-triangle angle

\gamma = \htmlClass{t-arg}{\arg}\left(-\frac{\htmlClass{t-num}{V_{ud}V_{ub}^*}}{\htmlClass{t-den}{V_{cd}V_{cb}^*}}\right)

The one angle of the unitarity triangle that tree-level decays can measure — which is why it is the only one whose determination carries no lattice-QCD uncertainty.

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.

and the reason it matters is stated in one sentence in the book, which is worth unpacking: γ is the only angle measurable from decays described by tree-level diagrams alone. Every other angle in the triangle needs a box diagram — and a box diagram needs a lattice calculation of a hadronic matrix element, which is where the theoretical error lives.

Fig. 8.16 — two trees to the same final state

Click a vertex or an internal line.

Bettini Fig. 8.16. Two tree diagrams — no loops, no boxes, no lattice. They interfere only because the D⁰ and D̄⁰ can decay to the SAME final state, so the kaon's own trick from §8.6 is used again one level down.

The interference is arranged the same way as $J/\psi K_S$ was: the two amplitudes reach different charm states, $D^0$ and $\bar D^0$ , which would be distinguishable — until you insist that the $D$ decay into a channel both can reach. LHCb uses $D \to K_S^0K^+K^-$ .

💡 What this really says — the strong phase varies across the Dalitz plot and γ does not — which is how you separate them

§8.8 established that a decay asymmetry needs a weak phase and a strong phase, and warned that the strong phase is usually incalculable — sitting in front of the answer as an unknown multiplier. The $B^\mp \to DK^\mp$ measurement gets around that in a way worth admiring.

The strong phase here, $\delta_{BDK}$ , is a function of the Dalitz-plot coordinates $m^2(K_S K^+)$ and $m^2(K_S K^-)$ — it varies from point to point across the plot, because the $D \to K_SK^+K^-$ amplitude runs through different resonances in different regions. The weak phase $\gamma$ does not: it is one number everywhere.

So instead of one measurement contaminated by one unknown, you have a two-dimensional map in which the two behave differently. Under CP conjugation $\gamma \to -\gamma$ while $\delta_{BDK}$ is unchanged except that the two Dalitz coordinates swap. Comparing the $B^+$ and $B^-$ Dalitz plots point by point therefore separates the constant from the varying — fitting for $\gamma$ once and $\delta_{BDK}$ at every point simultaneously.

The move is one an experimentalist reaches for constantly: when a nuisance parameter and a signal are degenerate in a single measurement, find a variable they depend on differently and measure along it. It is the same reason a spectrum beats a single-bin count, and why an angular distribution beats a total rate.

A rate asymmetry needs two phase differences, not one

strong phase difference δ_S1 − δ_S2 = 130°weak phase difference δ_W1 − δ_W2 = 66°|A₂| / |A₁| = 0.60

|A|²

0.2065

|Ā|²

1.8860

asymmetry (|A|² − |Ā|²) / (|A|² + |Ā|²)

-0.8026

Both present. Flipping the weak phase reflects A₂ about the direction of A₁, and because the two are not collinear the reflected chain closes somewhere else. The rate difference is −4|A₁||A₂| sin(δ_S1−δ_S2) sin(δ_W1−δ_W2) = -1.6796, matching |A|² − |Ā|² = -1.6796.

The same two-phasor picture as §8.8, now for B∓ → DK∓. The weak phase difference IS γ, and the strong phase δ_BDK changes as you move across the Dalitz plot — drag the strong slider to see the asymmetry sweep through zero and back while γ stays fixed.

Two independent LHCb determinations:

\gamma = 68.7^{+5.2}_{-5.1}\;^\circ, \qquad \gamma = 65.9^{+3.3}_{-3.5}\;^\circ

§8.11 One point, in one plane

Everything now collects. Unitarity applied to the first and third columns of the CKM matrix gives three complex numbers summing to zero — the unitarity triangle of §7.11 — and dividing through by $V_{cd}V_{cb}^*$ fixes two of its vertices at $(0,0)$ and $(1,0)$ . Only the apex is free, so every measurement in this chapter becomes a constraint on one point in one plane.

Every measurement in the chapter, as one point in one plane

ρ̄

0.159

η̄

0.348

R_b

0.383

R_t

0.910

22.5°

65.4°

α + β + γ = 180° by construction — the sum is 180° wherever you put the apex, so it is not a test. What is a test is whether all four bands can be satisfied at once. sin 2β here = 0.707. Click a band label for what measures it.

Fig. 8.17, made draggable. Two constraints are sides (circles, from the moduli |V_ij|) and two are angles (rays, from CP asymmetries). The three preset apexes are the story: the angle-based point sits close to the global fit, the side-based one does not.

The two kinds are genuinely different. Sides need moduli and lattice input; angles need CP asymmetries and, for γ, nothing beyond tree diagrams.

constraint↕	kind↕	from↕	theory input↕
$R_b = 0.413$	side, circle on (0,0)	$(4.406\pm0.079)\|V_{ub}/V_{cb}\|$ — semileptonic $b$ decays	form factors; \|V_ub\| is the matrix's hardest element, with a long-standing exclusive/inclusive tension
$R_t = 0.967$	side, circle on (1,0)	$\propto\sqrt{\Delta m_B/\Delta m_s}$ — §8.6 and §8.7	only the lattice ratio ξ = 1.20, known to ~2 %, because the hadronic factors largely cancel
$\sin2\beta = 0.675$	angle at (1,0)	$B^0\to J/\psi K_S$ interference — §8.6	none — every rate factor cancels in the asymmetry
$\gamma = 65.9°$	angle at (0,0)	$B^\mp\to DK^\mp$ — §8.10	none — tree diagrams only, no box, no loop

does the triangle close?

import numpy as np
Vud, Vub, Vcd, Vcb, Vtd, Vts, Vtb = 0.97373, 3.82e-3, 0.221, 40.8e-3, 8.6e-3, 41.5e-3, 1.014
dmB, dms, mB, mBs = 0.5065, 17.7683, 5279.66, 5366.92

Rb = (Vud/Vcd)*(Vub/Vcb); Rt = (Vtd*Vtb)/(Vcd*Vcb)
print("the two sides, from Sec. 7.11's moduli:")
print(f"  R_b = 4.406 x |V_ub/V_cb| = {Rb:.4f}")
print(f"  R_t = |V_td V_tb| / |V_cd V_cb| = {Rt:.4f}")

xi = 1.20
vr = xi*np.sqrt(dmB*mBs/(dms*mB))
print("\nR_t the book's way, from the RATIO of the two oscillation frequencies:")
print(f"  |V_td/V_ts| = xi sqrt(dmB m_Bs / dms m_B) = {vr:.4f}")
print(f"  straight from (7.93):                      {Vtd/Vts:.4f}   -- agree to {abs(vr/(Vtd/Vts)-1)*100:.1f}%")
print( "  the hadronic factors enter only through xi = 1.20, known to ~2%,")
print( "  against ~10% for f_B alone.  that is what Delta m_s bought.")

rho = (1 + Rb**2 - Rt**2)/2; eta = np.sqrt(Rb**2 - rho**2)
b1 = np.degrees(np.arctan2(eta, 1-rho)); g1 = np.degrees(np.arctan2(eta, rho))
b = np.radians(np.degrees(np.arcsin(0.675))/2); g = np.radians(65.9)
r2 = np.tan(b)/(np.tan(g)+np.tan(b)); e2 = r2*np.tan(g)
print("\nwhere the apex lands, three ways:")
print(f"  from the two SIDES  (rho, eta) = ({rho:.3f}, {eta:.3f})  -> beta {b1:.1f}, gamma {g1:.1f} deg")
print(f"  from the two ANGLES (rho, eta) = ({r2:.3f}, {e2:.3f})  -> R_b {np.hypot(r2,e2):.3f}, R_t {np.hypot(1-r2,e2):.3f}")
print(f"  PDG global fit      (rho, eta) = (0.159, 0.348)  -> beta {np.degrees(np.arctan2(0.348,0.841)):.1f}, gamma {np.degrees(np.arctan2(0.348,0.159)):.1f} deg")
print("\nthe ANGLE-based apex sits much closer to the global fit than the")
print("SIDE-based one, and the gap is |V_ub| -- the element Sec. 7.11 already")
print("flagged as the hardest in the matrix.  the constraints DO overlap, but")
print("not perfectly, and that is a live tension rather than a rounding error.")
print("\nthe angle sum is not a test:")
print("  alpha + beta + gamma = 180 deg for ANY apex, by plane geometry.")
print('  book: 173 +- 6 deg, "compatible with 180".  what it really tests is')
print("  that the three angles were measured in three different experiments.")

prints

the two sides, from Sec. 7.11's moduli:
R_b = 4.406 x |V_ub/V_cb| = 0.4125
R_t = |V_td V_tb| / |V_cd V_cb| = 0.9671

R_t the book's way, from the RATIO of the two oscillation frequencies:
|V_td/V_ts| = xi sqrt(dmB m_Bs / dms m_B) = 0.2043
straight from (7.93):                      0.2072   -- agree to 1.4%
the hadronic factors enter only through xi = 1.20, known to ~2%,
against ~10% for f_B alone.  that is what Delta m_s bought.

where the apex lands, three ways:
from the two SIDES  (rho, eta) = (0.117, 0.395)  -> beta 24.1, gamma 73.5 deg
from the two ANGLES (rho, eta) = (0.148, 0.331)  -> R_b 0.363, R_t 0.914
PDG global fit      (rho, eta) = (0.159, 0.348)  -> beta 22.5, gamma 65.4 deg

the ANGLE-based apex sits much closer to the global fit than the
SIDE-based one, and the gap is |V_ub| -- the element Sec. 7.11 already
flagged as the hardest in the matrix.  the constraints DO overlap, but
not perfectly, and that is a live tension rather than a rounding error.

the angle sum is not a test:
alpha + beta + gamma = 180 deg for ANY apex, by plane geometry.
book: 173 +- 6 deg, "compatible with 180".  what it really tests is
that the three angles were measured in three different experiments.

Aside — α + β + γ = 180° is geometry, not physics

The book checks that $\alpha+\beta+\gamma = (173\pm6)°$ is “compatible with 180°”, and it is easy to read that as a test of the Standard Model. It is not — or rather, not in the way it looks.

The three angles are the interior angles of a plane triangle. Their sum is 180° for any apex whatsoever, unitary matrix or not. Drag the apex in the widget above anywhere you like and the readout stays at 180.

What the check actually tests is that the three angles were measured in three different experiments, on three different final states, with three different sets of systematics — and that the three answers are mutually consistent with belonging to one triangle. That is a real and non-trivial statement. But it is a statement about the measurements agreeing, not about the geometry closing.

The genuine geometric test is the one the widget makes visible: whether the two sides and the two angles can be satisfied simultaneously by a single point. Four constraints, two degrees of freedom — two conditions’ worth of redundancy, and it is that redundancy that could have failed.

Table 8.1 — the chapter in one table

Bettini Table 8.1, p. 348, with four corrections in bold — see the erratum below. Γ is computed as ħ/τ throughout.

↕	τ (ps)↕	cτ (μm)↕	Γ (ps⁻¹)↕	Δm (ps⁻¹)↕	Δm (meV)↕
$K_L$	$(51.16\pm0.21)\times10^3$	$15.3\times10^6$	$2.0\times10^{-5}$	$(5.292\pm0.009)\times10^{-3}$	$(3.483\pm0.006)\times10^{-3}$
$K_S$	89.54 ± 0.04	$2.67\times10^4$	0.011	—	—
$D_1, D_2$	0.4103 ± 0.0010	123	2.4	$(9.67\pm\mathbf{1.4})\times10^{-3}$	$(6.13\pm0.86)\times10^{-3}$
$B_H, B_L$	1.520 ± 0.005	459	0.65	0.5065 ± 0.0019	0.3337 ± 0.0033
$B_{sH}, B_{sL}$	1.497 ± 0.015	439	0.67	17.7683 ± 0.0060	11.5 ± 0.5

Erratum — three of Table 8.1’s entries disagree with the text, all by a factor of ten or by arithmetic

Table 8.1 is the chapter’s reference table, and it does not close against itself.

1. The $B_s$ width contradicts its own lifetime. The table gives $\tau = 1.497$ ps and $\Gamma = 0.86$ ps⁻¹, but $1/1.497 = \mathbf{0.668}$ . Every other row checks exactly — $1/51160 = 2.0\times10^{-5}$ ✓, $1/89.54 = 0.011$ ✓, $1/0.4103 = 2.4$ ✓, $1/1.520 = 0.65$ ✓ — so the $B_s$ row is the only failure, and it is off by 29 %. The error propagates: the table’s $\Gamma$ (meV) column gives 0.57, which is $0.86\,\text{ps}^{-1}\times\hbar$ ; the value consistent with the lifetime is 0.44 meV.

2. Three uncertainties are a factor of ten from the text.

quantity	in the text	in Table 8.1
$\tau_S$	$89.54 \pm \mathbf{0.04}$ ps, Eq. (8.8)	$\pm\mathbf{0.004}$
$\Delta m_D$	$\pm\mathbf{1.4}\times10^{-3}$ ps⁻¹, Eq. (8.82)	$\pm\mathbf{0.14}\times10^{-3}$
$\Delta m_s$	$\pm0.0051 \oplus 0.0032 = \pm\mathbf{0.006}$ ps⁻¹, Eq. (8.62)	$\pm\mathbf{0.06}$

Three of the same slip in one table — and in both directions, the table being too tight twice and too loose once — which suggests it was assembled from a different source rather than from the chapter. The PDG values agree with the text in all three cases.

The table above uses the text’s values, and $\Gamma = \hbar/\tau$ throughout.

But the table is not uniformly the culprit, and the one place it wins is worth recording. The $K_L$ row gives $\Delta m = 5.292\times10^{-3}$ ps⁻¹ and $3.483\times10^{-3}$ meV, where p. 319 gives $5.293$ and $3.481$ . Convert each pair with $\hbar = 6.5821\times10^{-4}$ eV ps:

5.292\times10^{-3}\,\hbar = 3.4832\;\mu\text{eV} \;\checkmark \qquad\text{but}\qquad 5.293\times10^{-3}\,\hbar = 3.4839\;\mu\text{eV} \ne 3.481

The table’s pair converts into itself to four digits; the text’s does not. Both agree inside the quoted $\pm0.006$ , so nothing is wrong with either number and no result on this site changes — §8.1 and §8.3 quote the text’s 5.293 and the table above quotes the table’s 5.292, deliberately. It is recorded because the natural reading of the three rows above — “the table is unreliable” — is too strong.

What the chapter established

Every entry traces back to δ₁₃ = 65.6° of §7.11. The spread of six orders of magnitude is about how each observable reaches that phase, not about the phase itself.

year↕	system↕	mechanism↕	observable↕
1964	$K^0$	mixing	$\|\varepsilon\| = 2.23\times10^{-3}$ — Christenson, Cronin, Fitch, Turlay
1999–2002	$K^0$	decay	$\mathrm{Re}(\varepsilon'/\varepsilon) = 1.66\times10^{-3}$ — NA48, KTeV, after 30 years
2001	$B^0$	interference	$\sin2\beta = 0.675$ — BaBar, Belle. Order one, and theoretically clean
2008	$B^\pm$	decay	CP violation with no mixing at all — and later, γ from trees alone
2019	$D^0$	decay	$\Delta A_{CP} = -1.57\times10^{-3}$ — LHCb, the up-type sector at last
2021	$B_s^0$	interference	$a_{CP} = 0.236$ — LHCb, the largest asymmetry of all

Read that column of dates. Fifty-seven years, four mesons, three mechanisms — and every one of them is the same single phase, $\delta_{13} = 65.6°$ , reached by a different route. The spread from $10^{-6}$ to $10^{-1}$ is not a spread in the physics; it is a spread in how much each observable is throttled on its way to the phase.

🔑 If you remember only three things

Charm’s asymmetry is a difference of asymmetries. Either one measured alone would be dominated by the apparatus rather than by the physics.
Angles summing to 180° is not a test. Any three angles of a triangle do that; the test is whether independently measured sides and angles describe the same triangle.
A result with no mixing in it is the cleanest of the three kinds. Nothing has to propagate, so nothing about the states can enter the answer.

Where this goes next

Two threads leave this chapter.

Chapter 9 explains where the W and Z came from, and supplies the electroweak theory that has been assumed throughout — including the $g_L$ and $g_R$ that §7.13 refused to name.

Chapter 10 repeats the whole of this chapter for neutrinos. The formalism is identical — OscillationLab will be mounted again — but with two differences that change everything: the mixing angles are large rather than small, and the phenomenon requires neutrino mass, which nothing in Chapters 7–9 provides.

And one thread leaves the Standard Model altogether. All of the CP violation established here, combined, is far too small to explain why the universe contains matter and not antimatter. Sakharov’s conditions require CP violation, and the CKM phase supplies some — but short of what is needed by many orders of magnitude. That gap is one of the clearest pieces of evidence that the Standard Model is incomplete, and Chapter 12 returns to it.

✅ Check yourself — charm, charged B, and the triangle

0/6 answered · 0 correct

1.LHCb measures ΔA_CP = A_CP(K⁺K⁻) − A_CP(π⁺π⁻) rather than either asymmetry alone. What does the subtraction buy, and why doesn't it also cancel the signal?
2.Why is U-spin a weaker tool than isospin?
3.Why is γ the only unitarity-triangle angle whose measurement is free of theoretical uncertainty?
4.In B∓ → DK∓, the strong phase δ_BDK is unknown. How is γ extracted anyway?
5.The book checks that α + β + γ = (173 ± 6)°, compatible with 180°. What does that test?
6.R_t is obtained from √(Δm_B/Δm_s) rather than from Δm_B alone. Why does that matter?