Quark Mixing and the GIM Mechanism · Elementary Particle Physics

Universality never failed. What failed was the assumption that the quark you weigh and the quark the interaction couples to are the same object.

🎯 Why this matters

The repair costs a number. An angle nobody predicts enters as a measured constant, and the count of such constants only grows from here — three more angles and a phase before the chapter ends.

§7.8 ended with universality confirmed to 0.2 % across the three lepton families. Ask the same question of quarks and it fails immediately, and not subtly — by an order of magnitude.

The failure has a pattern. Weak decays of hadrons come in two kinds: those that conserve strangeness and those that change it by one unit. The $\Delta S = 0$ decays behave; the $|\Delta S| = 1$ decays are suppressed by roughly a factor 20. Compare, at quark level, two decays that differ in one leg and nothing else — the other two quarks are spectator quarks , carrying flavour through the decay without taking part in it:

Fig. 7.12 — the same decay twice, differing only in which quark decays

Click a vertex or an internal line.

Bettini Figs. 7.12 and 7.13 collapsed into one. The two light quarks are spectator quarks: they carry flavour through the decay and take no part in it, which is the approximation that lets a hadron decay be computed from a quark decay.

The cleanest version of the discrepancy is in the mesons, where there is no nuclear structure to argue about. The book compares $K^- \to \mu^-\bar\nu$ with $\pi^- \to \mu^-\bar\nu$ , which are the same decay with $s$ in place of $d$ :

the discrepancy, in the cleanest case there is

mK, mpi, mmu = 493.677, 139.57039, 105.6583755      # MeV
fKfpi = 1.1932                                       # lattice, Sec. 6.10

G_pi = 1.00/2.60e-8
G_K  = 0.64/1.24e-8
print("measured, from the book's own (7.71):")
print(f"  Gamma(pi -> mu nu) = 1.00 / 2.60e-08 s = {G_pi:.3e} /s")
print(f"  Gamma(K  -> mu nu) = 0.64 / 1.24e-08 s = {G_K:.3e} /s")
print(f"  (7.72) ratio = {G_K/G_pi:.3f}                          book: 1.34")

rate = lambda mP: mP * (1 - (mmu/mP)**2)**2
R = rate(mK)/rate(mpi)
print("\nexpected at equal couplings, from Eq. (7.73) as printed:")
print("  Gamma ~ f^2 |V|^2 m_mu^2 m_P (1 - m_mu^2/m_P^2)^2")
print(f"  m_K[1-(m_mu/m_K)^2]^2 / m_pi[1-(m_mu/m_pi)^2]^2 = {R:.3f}")
print(f"\nso the kaon channel is short by a factor {R/(G_K/G_pi):.1f} -- and the")
print( "strong-interaction correction makes it WORSE, because")
print(f"f_K/f_pi = {fKfpi:.3f} > 1, pushing the expectation to {R*fKfpi**2:.2f}.")
print(f"\nuniversality, taken literally, is wrong here by a factor of {R*fKfpi**2/(G_K/G_pi):.0f}.")

prints

measured, from the book's own (7.71):
Gamma(pi -> mu nu) = 1.00 / 2.60e-08 s = 3.846e+07 /s
Gamma(K  -> mu nu) = 0.64 / 1.24e-08 s = 5.161e+07 /s
(7.72) ratio = 1.342                          book: 1.34

expected at equal couplings, from Eq. (7.73) as printed:
Gamma ~ f^2 |V|^2 m_mu^2 m_P (1 - m_mu^2/m_P^2)^2
m_K[1-(m_mu/m_K)^2]^2 / m_pi[1-(m_mu/m_pi)^2]^2 = 17.670

so the kaon channel is short by a factor 13.2 -- and the
strong-interaction correction makes it WORSE, because
f_K/f_pi = 1.193 > 1, pushing the expectation to 25.16.

universality, taken literally, is wrong here by a factor of 19.

Erratum — Eq. (7.73) evaluates to 17.67, not 8.06

The formula in (7.73) is printed correctly:

\frac{\Gamma(K\to\mu\nu)}{\Gamma(\pi\to\mu\nu)} = \frac{m_K\left[1-(m_\mu/m_K)^2\right]^2}{m_\pi\left[1-(m_\mu/m_\pi)^2\right]^2}

Both outer squares are there. But it evaluates to 17.67, and the printed value is 8.06.

Two independent checks say 17.67 is what was meant:

The downstream physics needs it. §7.12 extracts the Cabibbo angle from exactly this comparison. With 17.67, the measured ratio 1.34 and the lattice $f_K/f_\pi = 1.193$ give $\theta_C = 13.00°$ and $\sin\theta_C = 0.2250$ — a 0.3 % match to the measured $|V_{us}| = 0.2243$ . With 8.06 you get $\theta_C = 18.9°$ and $\sin\theta_C = 0.324$ , wrong by 44 %, and (7.77)‘s own $\cos\theta_C = 0.97373$ becomes impossible.
The likely origin is visible. Drop both outer squares and the expression becomes the ratio of CM momenta, 7.91 — close to the printed 8.06. And the sentence introducing the equation says the ratio “is proportional to the ratio of the phase-space volumes”, which is exactly the one-power version. So the prose describes 7.91 while the formula computes 17.67, and the printed number belongs to the prose.

This is the same slip as §7.4’s, one section apart: two-body phase space carries one power of $p^*$ , the rate carries two, and the extra power lives in $|\mathcal{M}|^2$ . There the equation was right and the sentence was one power short; here the sentence and the number agree with each other and the equation is right. The §7.4 aside works through where the second power comes from.

Nothing in the argument changes — the conclusion is that $|\Delta S| = 1$ is suppressed by about an order of magnitude, and 17.67 makes that case more strongly than 8.06 does.

Cabibbo’s move

The repair is not to weaken universality but to change what it applies to. Quark mixing assumes the quark that couples to the W is not $d$ but a superposition, rotated by the Cabibbo angle :

d' = \htmlClass{t-d}{d\cos\theta_C} + \htmlClass{t-s}{s\sin\theta_C}

(7.74)

Bettini p. 296. One state, two components — and the coefficients are a sine and a cosine because they must square to one.

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.

There is exactly one matrix element, with exactly one coupling $G_F$ — universality is untouched. What used to look like two different couplings is one coupling resolved along two different axes, Eq. (7.76):

\mathcal{M} \propto \htmlClass{t-g}{G_F}\htmlClass{t-c}{\cos\theta_C}\,(\bar d_L\gamma^\alpha u_L) \;\; (\Delta S = 0), \qquad \mathcal{M} \propto \htmlClass{t-g}{G_F}\htmlClass{t-s}{\sin\theta_C}\,(\bar s_L\gamma^\alpha u_L) \;\; (|\Delta S| = 1)

(7.76)

Cabibbo's resolution. What looked like two different weak couplings — one for strangeness-conserving decays and a weaker one for strangeness-changing — is a single coupling resolved along two axes.

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.

The Cabibbo rotation, and what GIM does to it

ΔS = 0 rate ∝ cos²θ_C0.9482

|ΔS| = 1 rate ∝ sin²θ_C0.0518

strangeness-changing neutral current, d̄sexactly 0

The (c, s′) doublet contributes −cos θ_C sin θ_C, cancelling the (u, d′) term. Drag the slider: it stays at zero for every angle. GIM is not a tuning.

Cabibbo angle θ_C = 13.16° — the measured value

cos θ_C = |V_ud|

0.97374

sin θ_C = |V_us|

0.22767

suppression tan²θ_C

1 / 18.3

The cancellation is exact only if the two up-type quarks have equal mass. They do not, so a residue survives, of relative size (m_c² − m_u²)/M_W² × cos θ_C sin θ_C = 5.53e-5. That is the quantity GIM turned around: the observed SCNC rate is an upper bound on it, and therefore an upper bound on m_c — which is how a non-observation predicted a quark.

Fig. 7.14 made interactive. The slider is the Cabibbo angle; the toggle is the GIM mechanism, which the next section derives. At the measured 13.16° the strangeness-changing rate is 1/19 of the strangeness-conserving one — the 'about an order of magnitude' the book quotes.

⚙️ Engineer’s bridge — the interaction picks a basis, and it is not the one you labelled things in

This is one of the most transferable ideas in the book, and it is worth stating in general form before the physics buries it.

You have a system with two natural bases. One is the basis in which the free system is diagonal — for quarks, the mass eigenstates $d$ and $s$ , the states that propagate with definite mass and that you use to label particles — the weak-versus-mass-eigenstate distinction. The other is the basis in which the interaction is diagonal — $d'$ and $s'$ , the states the W actually couples to. There is no law saying these coincide, and here they do not: they differ by a rotation of 13°.

Every apparent violation of universality is then a basis error. The coupling never varied. What varied is the overlap between the state you prepared (a $d$ , because you built it into a neutron) and the state the interaction acts on (a $d'$ ). Universality is a statement in the primed basis; you were testing it in the unprimed one.

An engineer meets this constantly. A filter is diagonal in frequency, a measurement is taken in time; a rotation is diagonal in its own principal axes, your sensors are bolted to the chassis; an error-correcting code is diagonal in its syndrome basis, the noise arrives in the physical one. In every case the transformation between bases is not a nuisance to be eliminated — it is the content, and it carries physical parameters. Here the rotation angle is a fundamental constant of nature that nothing predicts.

The pattern recurs immediately and gets more consequential each time. §7.11 makes the rotation 3 × 3 and finds it can carry a complex phase, which is the Standard Model’s only source of CP violation. Chapter 8 watches neutral mesons oscillate, which is the same mismatch observed over time. Chapter 10 does it for neutrinos, where the two bases differing is the entire phenomenon and requires mass the Standard Model does not supply.

Where it breaks: in signal processing you may work in whichever basis is convenient, because both are yours to choose and the transform is a bookkeeping step you could skip. Here you cannot skip it, and you do not get to choose: the two bases are selected by different physics and you have no apparatus that sees the weak one.

A quark propagates as a mass eigenstate — that is what having a definite mass and lifetime means — and interacts as a weak eigenstate, so no detector ever registers a $d'$ , and $d'$ has no mass to quote. The consequence is not aesthetic: it is why flavour is measured through interference and decay rates rather than by observing a state, and why ch08’s whole apparatus exists. A basis you can only ever infer is a different situation from a basis you merely chose not to use.

⚠️ The prime means “weak basis”, never “derivative”

$d'$ and $s'$ are states, not rates of change. The convention is universal in this subject and it collides badly with every other use of the prime you have met.

Two more traps in the same equations. $V$ is about to become the mixing matrix (§7.11) while it is still the vector bilinear covariant of §7.3 — both appear in this chapter. And $\theta_C$ is quoted through its sine: nobody reports the angle, they report $\sin\theta_C = |V_{us}| = 0.2243$ , because that is what a rate measures.

Note also that only the down-type quarks are written as mixed. That is a pure convention — the physics is in the relative rotation between the up-type and down-type bases, and putting all of it on the down-type side is a choice, not a fact about quarks.

Erratum — Eq. (7.77) prints 1 − cos θ_C where sin θ_C belongs

The values quoted are

\theta_C = 13.16°, \qquad \cos\theta_C = 0.97373 \pm 0.00031, \qquad \sin\theta_C = 0.02627

The first two are right. The third is not $\sin\theta_C$ — it is $1 - \cos\theta_C = 1 - 0.97373 = 0.02627$ , exactly.

The check is the book’s own. Two paragraphs earlier §7.9 says the coefficients “must satisfy the normalisation condition, namely the sum of their square must be one”. The printed pair gives

0.02627^2 + 0.97373^2 = 0.94884

which is not one. And directly: $\sin(13.16°) = 0.22767$ , which is also $\sqrt{1 - 0.97373^2} = 0.22771$ — consistent to the precision quoted, and equal to the measured $|V_{us}|$ to within the difference between a two-family rotation and the real three-family matrix.

Worth knowing because $\sin\theta_C \approx 0.225$ is one of the numbers you are expected to carry around; 0.026 would put the Cabibbo angle at 1.5°.

§7.10 The neutral current that should exist and does not

Cabibbo’s rotation fixes the charged current and immediately breaks something else. Write out the neutral-current term the theory now contains — the current of $d'$ with itself:

\bar d'_L\gamma_\alpha d'_L = \htmlClass{t-D}{\cos^2\!\theta_C\, \bar d_L\gamma_\alpha d_L + \sin^2\!\theta_C\, \bar s_L\gamma_\alpha s_L} + \htmlClass{t-X}{\cos\theta_C\sin\theta_C\left[\bar d_L\gamma_\alpha s_L + \bar s_L\gamma_\alpha d_L\right]}

(7.78)

Bettini p. 298. Rotating a basis and then squaring produces cross terms — and here the cross terms are a disaster.

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.

The prediction is testable in one line. $K^+ \to \pi^+\nu\bar\nu$ proceeds through that neutral current; $K^+ \to \pi^0 e^+\nu_e$ through the ordinary charged one. Similar diagrams, similar phase space, so similar rates:

Fig. 7.15 — the two kaon decays that should have been comparable

Click a vertex or an internal line.

Bettini Fig. 7.15. The spectator u is not drawn. Two decays of the same meson to similar final states, and the theory as it stood in 1963 predicted comparable rates.

Eight orders of magnitude is not a correction to be calculated. It is a mechanism that has not been found yet.

decay↕	vertex↕	expected↕	measured BR↕
$K^+ \to \pi^0 e^+\nu_e$	charged current, $\sin\theta_C$	ordinary	5.07 % — one of the kaon's six main modes
$K^+ \to \pi^+\nu\bar\nu$	neutral current, $\cos\theta_C\sin\theta_C$	comparable	$1.14^{+0.40}_{-0.33}\times10^{-10}$

Erratum — Eq. (7.80) quotes the wrong value, as a limit, for a well-measured decay

The book prints

\mathrm{BR}(K^+\to\pi^+\nu\bar\nu) = 1.14^{+0.40}_{-0.33}\times10^{-10}; \qquad \mathrm{BR}(K^+\to\pi^0e^+\nu_e) < 3\times10^{-3}

The first is right. The second is wrong twice over.

It is not a limit. $K^+\to\pi^0e^+\nu_e$ is the standard $K_{e3}$ decay, one of the kaon’s six principal modes — not a search, not a bound. And the argument being made requires a measured value: an upper limit on the unsuppressed decay cannot establish that the other one is suppressed, which is the entire point of the comparison.

The value is 17 times too small. $\mathrm{BR}(K^+\to\pi^0e^+\nu_e) = 5.07\%$ . That number is checkable against the book’s own arithmetic: §7.9’s Eq. (7.71) uses $\mathrm{BR}(K^+\to\mu^+\nu) = 0.64$ , and the kaon’s principal modes — $\mu^+\nu$ 63.6 %, $\pi^+\pi^0$ 20.7 %, $\pi^+\pi^+\pi^-$ 5.6 %, $\pi^0e^+\nu$ 5.07 %, $\pi^0\mu^+\nu$ 3.35 %, $\pi^+\pi^0\pi^0$ 1.76 % — sum to 100 %. There is no room for a mode at $3\times10^{-3}$ and no mode missing.

The suppression the section is establishing is therefore $1.14\times10^{-10}/5.07\times10^{-2} = 2.2\times10^{-9}$ , about nine orders of magnitude, not the eight the printed pair would give. The conclusion is strengthened, not weakened.

The mechanism

The GIM mechanism — Glashow, Iliopoulos and Maiani, 1970 — starts by noticing that $d'$ has an orthogonal partner nobody had used. If $(u, d')$ is a doublet, then $s'$ — the state orthogonal to $d'$ —

s' = \htmlClass{t-minus}{-}\,d\htmlClass{t-sin}{\sin\theta_C} + s\htmlClass{t-cos}{\cos\theta_C}

(7.80)

The state nobody had used. d′ is one row of a rotation; a rotation has two rows, and the second one had been sitting there unclaimed since 1963.

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.

should also sit in a doublet, and its partner does not exist. So postulate it. Call it charm.

Now compute the neutral current of $s'$ with itself. Every term is the same as (7.78) with $\cos$ and $\sin$ exchanged — and, crucially, the cross term picks up a minus sign, because (7.81) has one:

\bar s'_L\gamma_\alpha s'_L = \htmlClass{t-diag}{\sin^2\!\theta_C\,\bar d_L\gamma_\alpha d_L + \cos^2\!\theta_C\,\bar s_L\gamma_\alpha s_L} \htmlClass{t-cross}{- \cos\theta_C\sin\theta_C\left[\bar d_L\gamma_\alpha s_L + \bar s_L\gamma_\alpha d_L\right]}

(7.83)

The neutral current of the charm partner with itself. Every term matches the d′ case with cos and sin exchanged — except the last one, which arrives with the opposite sign.

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.

Add them and the strangeness-changing terms are gone:

\htmlClass{t-w}{\bar d'_L\gamma_\alpha d'_L + \bar s'_L\gamma_\alpha s'_L} = \htmlClass{t-m}{\bar d_L\gamma_\alpha d_L + \bar s_L\gamma_\alpha s_L}

(7.84)

Bettini p. 299. The Cabibbo angle has vanished from both sides — that absence is the GIM mechanism.

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.

The right-hand side has no $\theta_C$ in it at all. That is worth dwelling on:

the cancellation is a property of the rotation, not of its size

import numpy as np

print("(7.78) + (7.83), coefficients of [dbar_d, sbar_s, dbar_s, sbar_d]")
print("over 1000 Cabibbo angles drawn uniformly from 0 to 180 degrees:\n")
worst = np.zeros(4)
for th in np.random.default_rng(1).uniform(0, np.pi, 1000):
    c, s = np.cos(th), np.sin(th)
    d_p = np.array([c*c, s*s,  c*s,  c*s])       # (7.78)
    s_p = np.array([s*s, c*c, -c*s, -c*s])       # (7.83)
    worst = np.maximum(worst, np.abs(d_p + s_p - np.array([1, 1, 0, 0])))
print(f"  max |coefficient - (1, 1, 0, 0)| = {worst.max():.1e}\n")
print("exactly diagonal, at EVERY angle.  GIM is not a tuning that happens")
print("to work at 13 degrees -- it works because the rotation is orthogonal,")
print("so the neutral current is the same in both bases.  The Cabibbo angle")
print("is INVISIBLE to the neutral current.")

mu_, mc, MW = 2.16e-3, 1.27, 80.377
th = np.radians(13.16)
res = (mc**2 - mu_**2)/MW**2 * np.cos(th)*np.sin(th)
print("\nthe residue, since the cancellation needs m_u = m_c and they differ:")
print(f"  (m_c^2 - m_u^2)/M_W^2 x cos.sin = {res:.2e}   in amplitude")
print(f"                                    {res**2:.2e}   in rate")
obs = 1.14e-10/5.07e-2
print(f"\nobserved BR(K->pi nu nu)/BR(K->pi e nu) = {obs:.2e}")
print(f"  the GIM estimate lands within a factor {max(obs,res**2)/min(obs,res**2):.1f} of it -- close enough")
print( "  to say the mechanism accounts for the size, which is what let GIM")
print( "  turn the NON-observation into an upper bound on the charm mass.")

prints

(7.78) + (7.83), coefficients of [dbar_d, sbar_s, dbar_s, sbar_d]
over 1000 Cabibbo angles drawn uniformly from 0 to 180 degrees:

max |coefficient - (1, 1, 0, 0)| = 2.2e-16

exactly diagonal, at EVERY angle.  GIM is not a tuning that happens
to work at 13 degrees -- it works because the rotation is orthogonal,
so the neutral current is the same in both bases.  The Cabibbo angle
is INVISIBLE to the neutral current.

the residue, since the cancellation needs m_u = m_c and they differ:
(m_c^2 - m_u^2)/M_W^2 x cos.sin = 5.53e-05   in amplitude
                                  3.06e-09   in rate

observed BR(K->pi nu nu)/BR(K->pi e nu) = 2.25e-09
the GIM estimate lands within a factor 1.4 of it -- close enough
to say the mechanism accounts for the size, which is what let GIM
turn the NON-observation into an upper bound on the charm mass.

💡 What this really says — a non-observation that predicted a particle

It is worth being precise about what GIM actually did, because “they predicted charm” undersells it.

At tree level the cancellation is exact and says nothing about masses. The content is at one loop, in the box diagrams of Fig. 7.16. Diagram (a) has a $u$ in the loop with vertex factors $\cos\theta_C\sin\theta_C$ ; diagram (b) has a $c$ with $-\cos\theta_C\sin\theta_C$ . Identical apart from the internal mass. If $m_u = m_c$ they cancel exactly; since they do not, what survives is proportional to $m_c^2 - m_u^2$ .

So the experimental bound on strangeness-changing neutral currents becomes an upper bound on the charm mass. That is the whole trick, and it is running the usual logic backwards: instead of a theory predicting a rate, a non-observed rate constrains a particle nobody had seen. A charm quark much heavier than a couple of GeV would have left a visible SCNC signal; there wasn’t one, so it isn’t.

Four years later the $J/\psi$ turned up at 3.1 GeV (§4.9) — a $c\bar c$ bound state, so $m_c \approx 1.5$ GeV, inside the bound.

The same logic is still in use and is the reason FCNC searches are worth doing at all. Any process that the Standard Model forbids at tree level and suppresses at loop level is a place where a small new contribution is not buried under a large known one. The sensitivity comes from the suppression, exactly as it did in §7.4, where the value of $\pi \to e\nu$ came from being nearly forbidden.

One more observation the book makes in passing, and it matters for §7.13: the Cabibbo rotation is invisible to the neutral current. Eq. (7.84) has no $\theta_C$ , so the flavour-conserving neutral current is the same object in either basis. That is why the Z⁰ discovered at Gargamelle in 1973 conserves flavour exactly, and why flavour-changing neutral currents remain, fifty years later, one of the sharpest places to look for physics beyond the Standard Model.

🔢 Worked example — the Cabibbo angle, from two lifetimes

Everything needed is in Eq. (7.71) plus one lattice number.

Measured: $\Gamma(K\to\mu\nu)/\Gamma(\pi\to\mu\nu) = (0.64/1.24\times10^{-8}) \div(1/2.6\times10^{-8}) = 1.34$ .

Expected at equal couplings: the corrected Eq. (7.73) gives 17.67, and the strong interaction contributes $(f_K/f_\pi)^2 = 1.1932^2 = 1.4237$ , so the expectation is $17.67\times1.4237 = 25.15$ .

The shortfall is the mixing:

\tan^2\theta_C = \frac{1.34}{25.15} = 0.0533 \;\Longrightarrow\; \theta_C = \mathbf{13.00°}, \quad \sin\theta_C = \mathbf{0.2250}

against a measured $|V_{us}| = 0.2243$ — agreement to 0.3 %, from two lifetimes, two branching ratios and one lattice calculation. This is the number that would have come out 44 % wrong had Eq. (7.73)‘s printed 8.06 been used.

🔑 If you remember only three things

The discrepancy was an order of magnitude, not a subtlety. Nothing delicate needed explaining; the rate was simply wrong by a large factor.
A rotation redistributes strength without losing any. The two pieces are a unit vector resolved onto two axes, so nothing has been suppressed.
The mechanism came from taking the rotation seriously. Computing a decay the rotation demands, and finding it absent, is what forced a fourth quark.

Where this goes next

§7.11 promotes the 2 × 2 rotation to a 3 × 3 matrix once the third family exists. That is not a routine generalisation: a real 3 × 3 orthogonal matrix has three angles, but a unitary one has three angles and a phase, and no amount of rephasing the quark fields can remove it. That leftover phase is the Standard Model’s only source of CP violation and the reason Chapter 8 exists.

§7.12 then measures $\sin\theta_C$ properly, in two independent ways — including the $K/\pi$ comparison this page has been using, done carefully.

✅ Check yourself — quark mixing and GIM

0/6 answered · 0 correct

1.Strangeness-changing decays are suppressed by about a factor 20 relative to strangeness-conserving ones. How does Cabibbo's proposal fix this without abandoning universality?
2.Eq. (7.73) prints the value 8.06 for the K/π rate ratio. What is wrong, and how do you know?
3.Eq. (7.77) quotes θ_C = 13.16°, cos θ_C = 0.97373, sin θ_C = 0.02627. Which check catches the error fastest?
4.Why is the GIM cancellation not a fine-tuning?
5.In what sense did GIM predict the charm quark?
6.Which of these are consequences of the weak and mass bases not coinciding? (Select all that apply.)