§8.6Oscillation and CP Violation in the B⁰

The kaon gave CP violation at the $10^{-3}$ level, buried in a wave function. The $B^0$ gives it at order one, as a clean sinusoid whose amplitude the Standard Model predicts. This section is where the CKM phase of §7.11 finally becomes something you can plot.

What is different about the B

Two structural differences from the kaon, and both simplify things:

That third row matters: with $|p/q| \approx 1$ and $|A_f| = |\bar A_f|$, all the CP violation lives in a phase, and a phase is what interference measures.

The box diagram, and why only the top matters

$B^0$ and $\bar B^0$ are connected at second order by a box diagram box diagram the second-order W-exchange loop connecting a neutral meson to its antiparticle. Its amplitude grows as the square of the internal quark mass, so the top dominates and Δm measures |V_tq V_tb|². defined in §8.6 — open in glossary — two W exchanges, with any up-type quark running in the loop:

The book says the internal-quark contribution “is proportional to the square of its mass”, so the diagrams with $u$ or $c$ are negligible. That is right, but it understates how the win is achieved:

mt, mc = 173.0, 1.27                              # GeV
Vtd, Vtb, Vcd, Vcb = 8.6e-3, 1.014, 0.221, 40.8e-3
hbar = 6.582119569e-4                             # eV ps

ckm_t, ckm_c = (Vtd*Vtb)**2, (Vcd*Vcb)**2
print("the two competing internal quarks, top and charm:\n")
print(f"  CKM weight |V_td V_tb|^2 = {ckm_t:.2e}     (top)")
print(f"             |V_cd V_cb|^2 = {ckm_c:.2e}     (charm)")
print(f"  ratio  {ckm_t/ckm_c:.2f}   -- essentially EQUAL.  the CKM factors do not choose.\n")
print(f"  mass weight (m_t/m_c)^2  = {(mt/mc)**2:.2e}\n")
print(f"  net top/charm = {ckm_t/ckm_c*(mt/mc)**2:.1e}")
print("\nso the top wins by four orders of magnitude entirely on its mass.")
print("the small CKM elements it must pay (|V_td| = 8.6e-03) are almost")
print("exactly cancelled by the charm's own smallness -- had the mass factor")
print("been absent, the two would have contributed equally.")
print("\nand this is why Delta m_B measures |V_td V_tb|:")
print(f"  Delta m_B = 0.5065 ps^-1 = {0.5065*hbar*1e3:.4f} meV     book (8.55): 0.3340 meV")
print( "  extracted:  |V_td||V_tb| = 8.4e-03        book (8.56): (8.4 +- 0.6)e-03")
print(f"  from Sec. 7.11's elements: {Vtd:.1e} x {Vtb} = {Vtd*Vtb:.2e}")

the two competing internal quarks, top and charm:

CKM weight |V_td V_tb|^2 = 7.60e-05     (top)
           |V_cd V_cb|^2 = 8.13e-05     (charm)
ratio  0.94   -- essentially EQUAL.  the CKM factors do not choose.

mass weight (m_t/m_c)^2  = 1.86e+04

net top/charm = 1.7e+04

so the top wins by four orders of magnitude entirely on its mass.
the small CKM elements it must pay (|V_td| = 8.6e-03) are almost
exactly cancelled by the charm's own smallness -- had the mass factor
been absent, the two would have contributed equally.

and this is why Delta m_B measures |V_td V_tb|:
Delta m_B = 0.5065 ps^-1 = 0.3334 meV     book (8.55): 0.3340 meV
extracted:  |V_td||V_tb| = 8.4e-03        book (8.56): (8.4 +- 0.6)e-03
from Sec. 7.11's elements: 8.6e-03 x 1.014 = 8.72e-03

The oscillation itself

With $|p/q| = 1$ the probabilities collapse to the simplest possible form — Eqs. (8.46) and (8.47):

and their normalised difference, the flavour asymmetry, is a bare cosine:

Measuring a time when you can only measure a length

Tagging, and a puzzle about time

The two neutral Bs from an $\Upsilon(4S)$ are one $B^0$ and one $\bar B^0$, but which is which is undetermined — a single wave function describes both, and their relative phase does not evolve. Then one of them decays semileptonically, and the sign of its lepton says what it was (§8.3’s $\Delta S = \Delta Q$ rule again, now $\Delta B = \Delta Q$):

That is flavour tagging flavour tagging inferring the flavour of one neutral meson from the decay of its coherently produced partner, or from the charge of the soft pion in D* → D⁰π. The technique that makes time-dependent CP measurements possible at all. defined in §8.6 — open in glossary . At the instant of the tag decay, the other B is known to be the opposite flavour, and its clock starts.

The interference asymmetry

Now the observable. The quantity to measure is λ_f λ_f the product (p/q)(Ā_f/A_f), whose imaginary part is the CP-violating observable in interference. |λ_f| = 1 when CP is violated in neither the mixing nor the decay. defined in §8.6 — open in glossary . Take a final state $f$ that is a CP eigenstate and that both flavours can reach — then a $B^0$ can get there directly, or oscillate into a $\bar B^0$ first and get there from the other side, and the two paths interfere:

For $f = J/\psi K^0$ the Standard Model prediction is unusually clean, and the argument is worth following. Look at Fig. 8.10: the decay diagram contains $V_{cb}$ and $V_{cs}$, both real to excellent approximation, so the decay contributes no phase. The mixing box contains $V_{td}$ twice, and $V_{td}$ is the one element carrying the CKM phase. Hence

The result, and how to read the plot

Do not read $\sin2\beta$ off the plot. The published curves have an amplitude of roughly 0.45, not 0.68, and the book lists the three reasons: background, time resolution of 1–1.5 ps against a 12.4 ps period, and mis-tags — events where a $B^0$ was called a $\bar B^0$. All three dilute the amplitude towards zero and none of them shifts the phase, so the fitted value is corrected upward for each:

import numpy as np
meas = [('BaBar', 0.687, 0.028, 0.012), ('Belle', 0.667, 0.023, 0.012)]
for nm, v, st, sy in meas:
    print(f"{nm} {v} +- {st} (stat) +- {sy} (syst)   combined {np.hypot(st, sy):.4f}")

w = [(v, np.hypot(st, sy)) for _, v, st, sy in meas]
avg = sum(v/s**2 for v, s in w) / sum(1/s**2 for v, s in w)
err = 1/np.sqrt(sum(1/s**2 for v, s in w))
print(f"\nweighted mean sin 2 beta = {avg:.4f} +- {err:.4f}")
print(f"  -> 2 beta = {np.degrees(np.arcsin(avg)):.1f} deg,  beta = {np.degrees(np.arcsin(avg))/2:.1f} deg     (PDG global fit: 22.2 deg)")

print("\nwhat makes this measurement remarkable is what is NOT in it:")
print("  no hadronic matrix element, no lattice input, no form factor.")
print("  every rate factor cancels in the ratio (8.53), and the amplitude")
print("  is a pure CKM phase.")
print("\ncontrast with Delta m_B, on the same page:")
print("  |V_td||V_tb| = 8.4e-03 +- 0.6e-03, i.e. +-7%, and the error is")
print("  dominated by the LATTICE calculation of the box's hadronic matrix")
print("  element -- not by the frequency, which is known to 0.4%.")

BaBar 0.687 +- 0.028 (stat) +- 0.012 (syst)   combined 0.0305
Belle 0.667 +- 0.023 (stat) +- 0.012 (syst)   combined 0.0259

weighted mean sin 2 beta = 0.6754 +- 0.0198
-> 2 beta = 42.5 deg,  beta = 21.2 deg     (PDG global fit: 22.2 deg)

what makes this measurement remarkable is what is NOT in it:
no hadronic matrix element, no lattice input, no form factor.
every rate factor cancels in the ratio (8.53), and the amplitude
is a pure CKM phase.

contrast with Delta m_B, on the same page:
|V_td||V_tb| = 8.4e-03 +- 0.6e-03, i.e. +-7%, and the error is
dominated by the LATTICE calculation of the box's hadronic matrix
element -- not by the frequency, which is known to 0.4%.

Where this goes next

$\sin2\beta \approx 0.68$ is one angle of the unitarity triangle, measured to 3 % with essentially no theoretical input. §8.11 will demand that it agree with every other side and angle simultaneously.

First, §8.7 runs the same programme on the $B_s^0$, which oscillates 35 times faster — a 0.35 ps period against 12.4 ps — and therefore needs the LHC rather than a beauty factory. Then §8.8 returns to the hardest case: CP violation in the decay amplitudes themselves, where the observable is a double ratio and the answer is $\varepsilon'$, a thousand times smaller than $\varepsilon$.