§7.11The Quark Mixing Matrix

§7.10 left the mixing as a 2 × 2 rotation through one angle. Then the third family turned up, and the generalisation is not routine — it produces, for free and unavoidably, the only source of CP violation in the Standard Model.

Write the two-family case as a matrix, Eq. (7.85), and the shape of the question becomes clear:

Kobayashi and Maskawa’s 1973 paper does the obvious thing — three families, so a 3 × 3 matrix $V$, the Cabibbo–Kobayashi–Maskawa matrix CKM matrix the 3 × 3 unitary matrix V relating the weak and mass bases of the down-type quarks. Nine complex entries, reduced by unitarity and by rephasing to three angles and one phase; that phase is the Standard Model's only source of CP violation. defined in §7.11 — open in glossary — and then asks the non-obvious question: how many independent parameters does it actually have?

Counting, which is the whole argument

The matrix must be unitary, $V V^\dagger = 1$, because $(d', s', b')$ and $(d, s, b)$ are both orthonormal bases. That is a strong constraint, and counting what survives it is the entire content of the section.

A complex 3 × 3 matrix has 18 real parameters. Unitarity is 9 real conditions, so 9 survive. If the matrix happened to be real it would be orthogonal, with only 3 free parameters — the three Euler angles. So 6 of the 9 are phases.

But phases are partly fiction. Quark fields are defined only up to an arbitrary phase, and (7.89) is unchanged under the rephasing of Eq. (7.90),

Three such rephasings on the down-type fields and three on the up-type fields would remove six phases — except that one combination does nothing at all: rotating every quark by the same phase leaves $V$ untouched. So only 5 of the 6 rephasings are independent, and

one phase survives, irremovably.

print("N   params  angles  phases  removable  PHYSICAL PHASES")
print("-" * 56)
for N in (1, 2, 3, 4):
    params    = N*N                  # a unitary N x N matrix has N^2 real parameters
    angles    = N*(N-1)//2           # the rotations of a real orthogonal N x N
    phases    = params - angles
    removable = 2*N - 1              # one phase per field, minus the global one that does nothing
    print(f"{N:<3} {params:^7} {angles:^7} {phases:^7} {removable:^10} {phases-removable:^12}")

print("\nwith one family there is nothing to mix.")
print("with two, every phase can be rephased away -- the matrix is REAL,")
print("  the single parameter is the Cabibbo angle, and CP is safe.")
print("with three, one phase survives no matter what you do to the fields.")
print("\nthree families is the SMALLEST number that permits CP violation,")
print("and Kobayashi and Maskawa said so in 1973 -- when the third family")
print("did not exist.  The tau appeared in 1975, the b in 1977, the top")
print("in 1995.  The prediction preceded all three.")

N   params  angles  phases  removable  PHYSICAL PHASES
--------------------------------------------------------
1      1       0       1        1           0      
2      4       1       3        3           0      
3      9       3       6        5           1      
4     16       6      10        7           3      

with one family there is nothing to mix.
with two, every phase can be rephased away -- the matrix is REAL,
the single parameter is the Cabibbo angle, and CP is safe.
with three, one phase survives no matter what you do to the fields.

three families is the SMALLEST number that permits CP violation,
and Kobayashi and Maskawa said so in 1973 -- when the third family
did not exist.  The tau appeared in 1975, the b in 1977, the top
in 1995.  The prediction preceded all three.

So the CKM matrix has three angles and one phase. Choose a convention — the book uses the standard one, three rotations applied in order — and multiplying them out gives:

Nine measurements, not one matrix

The elements are not handed down. Each is a different experiment, and the third row is not measured from decays at all — there are no hadrons containing a top quark, so $V_{td}$ and $V_{ts}$ come from B-meson oscillation frequencies, an entirely different kind of observable that Chapter 8 will set up.

Testing unitarity, and Question 7.3

Unitarity of the mixing matrix unitarity of the mixing matrix the statement that the weak eigenstates form an orthonormal basis. It is testable row by row, and any failure would mean a fourth generation or something stranger. defined in §7.11 — open in glossary is not an assumption you are entitled to. If new physics exists — a fourth family, or anything that couples to quarks — the 3 × 3 matrix need not be unitary, and the parametrisation in (7.92) would simply be wrong. So each row and column is summed independently:

import numpy as np
V  = np.array([[0.97373, 0.2243, 3.82e-3],
               [0.221,   0.975,  40.8e-3],
               [8.6e-3,  41.5e-3, 1.014]])
dV = np.array([[0.00031, 0.0008, 0.20e-3],
               [0.004,   0.006,  1.4e-3],
               [0.2e-3,  0.9e-3, 0.029]])

def total(idx, n):
    v = np.array([V[i][j] for i, j in idx[:n]])
    e = np.array([dV[i][j] for i, j in idx[:n]])
    return (v**2).sum() - 1, np.sqrt((((2*v*e))**2).sum())

sets = {'1st row':    [(0,0),(0,1),(0,2)], '2nd row':    [(1,0),(1,1),(1,2)],
        '1st column': [(0,0),(1,0),(2,0)], '2nd column': [(0,1),(1,1),(2,1)]}
book = {'1st row': -0.0015, '2nd row': 0.0010, '1st column': -0.0028, '2nd column': 0.0040}

print("(7.94), recomputed from (7.93)'s measured values:")
for k, idx in sets.items():
    d, e = total(idx, 3)
    print(f"  {k:11s} {d:+.4f} +- {e:.4f}  ({abs(d)/e:.1f} sigma)   book: {book[k]:+.4f}")

print("\nQuestion 7.3 -- the same four with the third family dropped:")
for k, idx in sets.items():
    d3, _ = total(idx, 3)
    d2, e2 = total(idx, 2)
    print(f"  {k:11s} {d2:+.4f} +- {e2:.4f}   third-family term was worth {abs(d3-d2):.5f}")

print("\nso the answer to Question 7.3 is: you cannot tell.  every third-family")
print("term is smaller than the uncertainty on its own row, and in the first")
print("row it is 100x smaller than the DEFICIT.  this test has no power to")
print("distinguish two families from three.")
print("\nwhat it does show is that the first row does not reach 1, by 2.2 sigma,")
print("and that has nothing to do with the third family.")

(7.94), recomputed from (7.93)'s measured values:
1st row     -0.0015 +- 0.0007  (2.2 sigma)   book: -0.0015
2nd row     +0.0011 +- 0.0118  (0.1 sigma)   book: +0.0010
1st column  -0.0029 +- 0.0019  (1.6 sigma)   book: -0.0028
2nd column  +0.0027 +- 0.0117  (0.2 sigma)   book: +0.0040

Question 7.3 -- the same four with the third family dropped:
1st row     -0.0015 +- 0.0007   third-family term was worth 0.00001
2nd row     -0.0005 +- 0.0118   third-family term was worth 0.00166
1st column  -0.0030 +- 0.0019   third-family term was worth 0.00007
2nd column  +0.0009 +- 0.0117   third-family term was worth 0.00172

so the answer to Question 7.3 is: you cannot tell.  every third-family
term is smaller than the uncertainty on its own row, and in the first
row it is 100x smaller than the DEFICIT.  this test has no power to
distinguish two families from three.

what it does show is that the first row does not reach 1, by 2.2 sigma,
and that has nothing to do with the third family.

The book says the conditions “are satisfied within the statistics”, and for three of the four they comfortably are. The first row is the exception, and it is worth being straight about it: $-0.0015 \pm 0.0007$ is a 2.2σ deficit in the best-measured row of the matrix, driven by the two most precisely known elements in the whole table. It is not a rounding artefact and it does not go away when $|V_{ub}|$ is included, because $|V_{ub}|^2 = 1.5\times10^{-5}$ is a hundred times smaller than the discrepancy.

This is a live tension — usually called the Cabibbo angle anomaly — and it is exactly the kind of thing unitarity tests exist to find. It may be a lattice-QCD or radiative-correction issue in extracting $|V_{ud}|$ and $|V_{us}|$; it may not be. The honest statement is that the test’s most sensitive row is the one that does not pass cleanly.

That last row is the one to carry forward. It is a common misreading that CP violation is small because the CP-violating parameter is small. It is not: $\delta_{13} = 66°$ is an order-one angle, about as far from CP-conserving ($0$ or $180°$) as it can get. Every CP-violating effect in Chapter 8 is small because the phase can only reach observables through $s_{13} = 0.0037$, the smallest mixing angle there is.

Why a phase means CP violation

The book’s argument is compact: the phase appears in a wave function as $e^{i(\omega t + \delta)}$, which is not invariant under $t \to -t$ unless $\delta = 0$ or $\pi$. So T is violated; CPT is a theorem; therefore CP is violated.

A cleaner way to see the same thing, and the one Chapter 8 uses: CP conjugation takes $V_{ik} \to V_{ik}^*$. If every element can be made real by some choice of field phases, then CP is a symmetry of the mixing sector. The counting above says that is possible for two families and impossible for three. CP violation and an irremovable complex phase are the same statement.

Unitarity as geometry

Unitarity does not only constrain the moduli. Off-diagonal, it says any two different columns are orthogonal — six conditions, each a sum of three complex numbers vanishing:

Three complex numbers summing to zero form a triangle in the complex plane. Five of the six conditions give very flat triangles, because the three terms have wildly different sizes; this one — the first and third columns — has all three terms of comparable magnitude, so its triangle is fat and its angles measurable. Those angles, $\alpha, \beta, \gamma$ (or $\phi_2, \phi_1, \phi_3$ depending on whether you learned them at SLAC or KEK), are functions of $\delta_{13}$.

The point of drawing it is that every side and every angle can be measured separately, by different experiments with different systematics. If they all land on one consistent triangle, the three-family Standard Model survives; if the triangle fails to close, something else exists. That programme is §8.11, where the triangle gets its own widget and its constraint bands.

Where this goes next

§7.12 comes back down to earth and measures $|V_{ud}|$ and $|V_{us}|$ in two concrete cases, including the $K/\pi$ comparison §7.9 used informally. §7.13 then closes the chapter with the discovery of the neutral current at Gargamelle — a discovery made by the absence of a muon.

Chapter 8 is where the phase found here does its work.