Flavour Oscillations, and the States of the Neutral K

Two particles differing only in which quark is the antiquark can turn into each other, and everything in this chapter follows from that single possibility.

🎯 Why this matters

Oscillation is possible only because nothing conserved forbids it. That a particle can become its own antiparticle is therefore a measurement of which quantum numbers are exact and which are merely nearly so.

§7.11 found a complex phase in the CKM matrix that no choice of field conventions can remove, and left it there. This chapter is where it becomes visible.

Two phenomena run through it, and the book is careful to keep them apart because they are constantly confused. Mixing is the fact that a neutral flavoured meson is produced in one basis and propagates in another — the same weak-versus-mass-eigenstate mismatch as §7.9, except that now you watch it as a function of time instead of inferring it from rates. CP violation is the fact that matter and antimatter are not interchangeable.

They are not the same thing, and the cleanest proof is that CP violation has been observed in charged B decays (§8.10), where there is no mixing at all.

§8.1 Four systems, one equation

Because the top quark decays before it can hadronize (§7.11’s problem 7.20), there are exactly four neutral flavoured meson pairs in nature: three down-type — $K^0(d\bar s)$ , $B^0(d\bar b)$ , $B_s^0(s\bar b)$ — and one up-type, $D^0(u\bar c)$ .

Why exactly four — every neutral pair you can build, and the two that nature deletes

down-typed, s, bup-typeu, c, tK⁰ = d s̄1947 · T/τ = 13B⁰ = d b̄1987 · T/τ = 8.2B_s⁰ = s b̄2006 · T/τ = 0.24D⁰ = u c̄2007 · T/τ = 1584u t̄never formsc t̄never forms

Three quarks of each charge type give C(3,2) = 3 distinct pairs each, so six — and the top removes two.

Sixty years separate the first from the last, and the order is the T/τ column, not the mass.

Supplied. A neutral flavoured meson needs two quarks of the same charge type and different flavour, so the count is a choice of two from three, twice over: six candidates. The top quark’s lifetime is shorter than the time it takes to hadronize, so it never appears inside a bound state at all (Problem 7.20) and two of the six are deleted before the question of mixing arises.

The remaining four are the entire subject of this chapter, and their discovery dates — 1947, 1987, 2006, 2007 — track the difficulty column rather than the mass: the D⁰ is the lightest of the four after the kaon and was the last to be established, because 1584 periods per lifetime means nothing oscillates before everything has decayed.

The quantum mechanics of all four is identical. What differs is two numbers, and their ratio decides everything:

P_\pm(t) = \tfrac14\left[\htmlClass{t-d}{e^{-\Gamma_1 t} + e^{-\Gamma_2 t}} \;\pm\; \htmlClass{t-o}{2\,e^{-(\Gamma_1+\Gamma_2)t/2}\cos(\Delta m\, t)}\right]

(8.21)

Bettini p. 322. The upper sign is the survival probability of the flavour you made; the lower is the appearance of its antiparticle. Everything in this chapter's first half is this one line.

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.

A flavour oscillation therefore has two timescales competing: the period $T = 2\pi/\Delta m$ , and the shorter of the two lifetimes — because once the short-lived eigenstate is gone there is only one state left and nothing to beat against. Their ratio is the organising number of the whole chapter:

Survival and appearance — one equation, four systems

survival — still K⁰
appearance — now K⁰-bar
P₊ + P₋ — decay alone, no mixing

Δm ×1.00 — the measured value · drag to see what changing the period alone would do

Δm

5.29e-3 ps⁻¹

period T = 2π/Δm

1.19e+3 ps

cτ (short)

2.68 cm

T / τ — the whole story

13.3

K⁰. The only system whose two eigenstates have wildly different lifetimes — 89.5 ps against 51.2 ns, a factor 571. The oscillation is damped away long before one period is complete, so you see less than a quarter of a cycle.

The dashed curve is pure exponential decay — what would happen with no mixing at all. The two solid curves always sum to it: mixing redistributes the survivors between the two flavours, it does not change how many there are.

Fig. 8.1's four panels, as one widget. The physics is Eq. (8.21) throughout; only Δm and the two widths change between presets. Switch between K⁰ and B_s⁰ and watch T/τ move by a factor of 55.

Δm and the shorter lifetime are the only inputs that differ. The last column is what you actually see.

system↕	$\Delta m$ (ps⁻¹)↕	shorter $\tau$ (ps)↕	$T = 2\pi/\Delta m$ ↕	$T/\tau$ ↕	what you see↕
$K^0$	$5.293\times10^{-3}$	89.5	1190 ps	13	a fraction of one cycle, heavily damped — but the two lifetimes differ by 571, which is a gift nothing else offers
$D^0$	$9.67\times10^{-3}$	0.41	650 ps	1600	essentially nothing. The appearance probability never exceeds $2\times10^{-6}$ , which is why mixing was not seen until 2007 and oscillation still has not been
$B^0$	0.5065	1.52	12.4 ps	8	a third of a cycle within three lifetimes; the measured range spans ±6 ps, one full period — so period and lifetime come from the same data set
$B_s^0$	17.7683	1.50	0.354 ps	0.24	four oscillations per lifetime. The frequency is known to five figures; the difficulty is resolving 0.35 ps at all

That column of ratios — 13, 1600, 8, 0.24 — is why the four discoveries are spread across 47 years. The kaon in 1960, the $B^0$ in 1987, the $B_s^0$ in 2006, $D^0$ mixing in 2007. Nothing about the theory changed; the experiments had to catch up with the numbers.

⚙️ Engineer’s bridge — you are measuring a beat, which is why the impossible precision is possible

Stop and look at what §8.2 is going to claim. The two neutral kaon mass eigenstates differ in mass by

\Delta m = 3.48\;\mu\text{eV}

out of a mass of 497.6 MeV. That is a fractional difference of $7\times10^{-15}$ — the equivalent of measuring the Earth–Moon distance to within three microns. No mass spectrometer has ever come within nine orders of magnitude of that.

It is measurable because nobody measures either mass. What is measured is the beat. Two states with slightly different masses are two monochromatic waves with slightly different frequencies, and their superposition beats at the difference frequency — which in natural units is exactly $\Delta m$ . You count oscillations of a 1.2 ns beat, and you are done.

This is the single most transferable idea in experimental physics, and an engineer meets it constantly:

a superheterodyne receiver mixes a 100 MHz signal against a 100.455 MHz local oscillator and works entirely at the 455 kHz difference, because everything is easier there;
an optical interferometer resolves displacements of $\lambda/1000$ by counting fringes, not by measuring two path lengths;
a phase-locked loop holds a frequency to parts in $10^{12}$ by driving the phase error to zero rather than measuring frequency at all;
and, in software, you compare two large trees by comparing hashes, or ship a diff instead of a state.

The general rule: a difference can be measured to a precision that would be absurd for either of the quantities being differenced, provided you can arrange for the two to interfere rather than measuring them separately and subtracting. Subtraction propagates both errors; interference does not.

Where it breaks: a beat measures $|\Delta m|$ and not its sign, because $\cos(\Delta m\,t)$ is even — so the technique that buys you fourteen digits of sensitivity throws away one bit, and §8.3 has to recover it with a separate refractive-index argument. The other limit is coherence: interference only outperforms subtraction while the two amplitudes remain coherent, and here the short-lived eigenstate decays away, so the beat is legible for a few $\tau_S$ and then simply is not there. An engineer chooses an integration time; this measurement is handed one.

§8.2 Telling K⁰ from K̄⁰

The neutral kaons differ in exactly one quantum number, strangeness, which the strong and electromagnetic interactions conserve and the weak interaction does not. Strong interactions therefore produce them as definite-strangeness states:

K^- + n \to K^0 + \pi^- \quad (S = +1), \qquad K^+ + p \to \bar K^0 + \pi^+ \quad (S = -1)

and — more usefully — strong interactions can also tell them apart afterwards, because strangeness conservation permits different reactions for each:

This is the property Pais and Piccioni's regeneration experiment (§8.4) exploits, and it is the only handle the strong interaction gives you on the flavour.

reaction↕	$K^0$ ( $S=+1$ )↕	$\bar K^0$ ( $S=-1$ )↕	why↕
charge exchange, $\to K^+ n$	yes	no	$K^0 \to K^+$ keeps $S = +1$ ; $\bar K^0 \to K^+$ would need $\Delta S = 2$
hyperon production, $\to \Lambda\pi^+$	no	yes	the $\Lambda$ carries $S = -1$ , available only from the $\bar K^0$

Aside — Question 8.1, and a sign convention worth pinning down

Does the $K^0$ charge-exchange with neutrons? Does the $\bar K^0$ ?

Charge exchange on a neutron means $K^0 n \to K^- p$ or $\bar K^0 n \to K^- p$ . Check strangeness: the $K^-$ is $s\bar u$ , so $S = -1$ .

$K^0 n \to K^- p$ would need $S: +1 \to -1$ , i.e. $\Delta S = 2$ . Forbidden — even the weak interaction cannot do that at first order (§7.10).
$\bar K^0 n \to K^- p$ has $S: -1 \to -1$ . Allowed, and it is the mirror of the $K^0 p \to K^+ n$ reaction in the table above.

So each neutral kaon charge-exchanges on exactly one of the two nucleons, and which one identifies its strangeness. The general rule: a $K^0$ ( $d\bar s$ ) has an $\bar s$ to give away and can make a $K^+$ ( $u\bar s$ ); a $\bar K^0$ ( $s\bar d$ ) has an $s$ and can make a $K^-$ ( $s\bar u$ ) or a hyperon.

The CP eigenstates, and why one kaon lives 571 times longer

Under CP the two flavour states exchange, so the CP eigenstates are their sum and difference:

\htmlClass{t-k1}{|K_1^0\rangle = \tfrac{1}{\sqrt2}\left(|K^0\rangle + |\bar K^0\rangle\right)} \;\; (CP = +1), \qquad \htmlClass{t-k2}{|K_2^0\rangle = \tfrac{1}{\sqrt2}\left(|K^0\rangle - |\bar K^0\rangle\right)} \;\; (CP = -1)

(8.6)

Two bases for one two-state system. The strong interaction produces the flavour states; CP — and therefore the decay — sorts by the sum and difference. Neither basis is more real than the other.

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.

Now work out the CP of the pion final states, because that is what decides which state can decay where. This is a genuinely nice piece of bookkeeping and the book does it properly:

🔢 Worked example — the CP eigenvalue of 2π and 3π

Two pions. For $\pi^0\pi^0$ , each $\pi^0$ has $CP = C \cdot P = (+1)(-1) = -1$ , so the pair has $(-1)^2 = +1$ .

For $\pi^+\pi^-$ the argument is different but the answer is the same: the pair is its own antiparticle-pair, so $C = (-1)^\ell$ , and parity gives $P = P_\pi^2(-1)^\ell = (-1)^\ell$ . Therefore

CP(\pi^+\pi^-) = (-1)^\ell \cdot (-1)^\ell = +1

for any $\ell$ — and the kaon is spinless so $\ell = 0$ anyway.

Three pions. $3\pi^0$ is immediate: $(-1)^3 = -1$ .

For $\pi^+\pi^-\pi^0$ , call $\ell$ the angular momentum of the $\pi^+\pi^-$ pair and $L$ that of the $\pi^0$ relative to it. The kaon has $J = 0$ , so $\ell + L = 0$ and hence $\ell = L$ . Then

P = P_\pi^3(-1)^{\ell+L} = (-1)(-1)^{2\ell} = -1, \qquad C = C(\pi^0)\,C(\pi^+\pi^-) = (+1)(-1)^\ell

CP(\pi^+\pi^-\pi^0) = (-1)^{\ell+1}

Both signs are formally available — but not in practice. The Q value is only $m_K - 3m_\pi \approx 79$ MeV, so the pions are barely above threshold and the centrifugal barrier crushes anything but $\ell = L = 0$ . With $\ell = 0$ , $CP = -1$ .

Conclusion. If CP is conserved: $K_1^0 \to 2\pi$ and $K_2^0 \to 3\pi$ , and neither can do the other’s job.

That single restriction explains the most striking number in the system — the gap between $K_S$ and $K_L$ :

why one kaon lives 571 times longer than the other

import numpy as np
hbar = 6.582119569e-4                      # eV ps
mK, mpi, mpi0 = 497.611, 139.57039, 134.9768   # MeV
tS, tL = 89.54, 51.16e3                    # ps
c = 0.299792458                            # mm/ps

print("the two lifetimes (8.8):")
print(f"  tau_S = {tS:8.2f} ps    tau_L = {tL:8.2f} ps")
print(f"  ratio tau_L/tau_S = {tL/tS:.0f}")
print("\nwidths (8.9):")
print(f"  Gamma_S = hbar/tau_S = {hbar/tS*1e6:.3f} ueV = {1000/tS:.2f} ns^-1     book: 7.4 ueV, 11.2 ns^-1")
print(f"  Gamma_L = hbar/tau_L = {hbar/tL*1e6:.4f} ueV                  book: 0.013 ueV")
print("\ndecay lengths (8.10):")
print(f"  c tau_S = {c*tS/10:6.2f} cm        book: 2.67 cm")
print(f"  c tau_L = {c*tL/1000:6.2f} m         book: 15.5 m")

print("\nand the reason: the K_L cannot use the 2-pion channel at all, so it")
print("is stuck with 3 pions, where the Q value is only")
Q3, Q2 = mK - 3*mpi, mK - 2*mpi
print(f"  m_K - 3 m_pi = {Q3:.1f} MeV   ({Q3/mK*100:.1f}% of the available mass-energy)")
print("against")
print(f"  m_K - 2 m_pi = {Q2:.1f} MeV  for the K_S")

g = 5000/mK
print(f"\npurifying a beam (a 5 GeV kaon has gamma = {g:.2f}):")
print(f"  gamma c tau_S = {g*c*tS/10:6.1f} cm      book: 27 cm")
print(f"  after 5 m the K_S component is down by exp(-500/{g*c*tS/10:.1f}) = {np.exp(-500/(g*c*tS/10)):.1e}")
print( "  -- so a few metres of flight leaves a PURE K_L beam, whatever")
print( "     you started with.  that is the whole experimental method.")

prints

the two lifetimes (8.8):
tau_S =    89.54 ps    tau_L = 51160.00 ps
ratio tau_L/tau_S = 571

widths (8.9):
Gamma_S = hbar/tau_S = 7.351 ueV = 11.17 ns^-1     book: 7.4 ueV, 11.2 ns^-1
Gamma_L = hbar/tau_L = 0.0129 ueV                  book: 0.013 ueV

decay lengths (8.10):
c tau_S =   2.68 cm        book: 2.67 cm
c tau_L =  15.34 m         book: 15.5 m

and the reason: the K_L cannot use the 2-pion channel at all, so it
is stuck with 3 pions, where the Q value is only
m_K - 3 m_pi = 78.9 MeV   (15.9% of the available mass-energy)
against
m_K - 2 m_pi = 218.5 MeV  for the K_S

purifying a beam (a 5 GeV kaon has gamma = 10.05):
gamma c tau_S =   27.0 cm      book: 27 cm
after 5 m the K_S component is down by exp(-500/27.0) = 8.9e-09
-- so a few metres of flight leaves a PURE K_L beam, whatever
   you started with.  that is the whole experimental method.

💡 What this really says — the long lifetime is evidence, not just a fact

It is worth pausing on what the 571 actually tells you, because the book slips it in as a remark and it is the chapter’s first real result.

$K_L \to 2\pi$ is forbidden by CP. $K_L \to 3\pi$ is allowed but strangled by phase space — 79 MeV of Q value against 218 for the two-pion channel, and the rate depends on that steeply. So the long-lived state is long-lived only because its fast channel is closed by a symmetry.

Turn that around: if CP were badly violated, the $K_L$ would decay to two pions at a competitive rate and there would be no factor 571 at all. The mere existence of a state that lives 51 nanoseconds — an eternity, on nuclear timescales — is direct evidence that CP violation is small, before anybody measures it.

The book puts it exactly right: “this very fact shows that the CP violation by weak interactions is small, if it occurs at all.” In 1964 it turned out to be $2\times10^{-3}$ , and §8.5 is that story.

The mass difference

The masses are defined for $K_S$ and $K_L$ , not for $K^0$ and $\bar K^0$ , and their difference is too small to measure by any direct means:

m(K^0) = 497.611 \pm 0.013\;\text{MeV}, \qquad \Delta m = m_L - m_S = 3.481 \pm 0.006\;\mu\text{eV} = 5.293 \pm 0.009\;\text{ns}^{-1}

how small 7 x 10^-15 is

import numpy as np
hbar_ns = 6.582119569e-7        # eV ns
mK_eV, dm_eV = 497.611e6, 3.481e-6
c = 0.299792458                 # m/ns

r = dm_eV/mK_eV
print(f"Delta m / m(K0) = {dm_eV:.3e} eV / {mK_eV:.3e} eV = {r:.1e}")
print( "                                                     book: 7 x 10^-15")
print("\nwhat a fractional precision of {:.1e} means elsewhere:".format(r))
print(f"  Earth-Moon distance  384400 km  ->  {384400e3*r*1e6:.1f} um")
print(f"  the Earth's diameter  12742 km  ->  {12742e3*r*1e6:.3f} um  (about one wavelength of UV)")
print(f"  one second in         {1/r/3.156e7/1e6:8.1f} million years")

T = 2*np.pi*hbar_ns/dm_eV
print("\nnobody measures either mass to anything like this.  the beat does it:")
print(f"  T = 2 pi / Delta m = {T:.2f} ns          book: 1.2 ns")
g = 10000/497.611
print( "  first oscillation maximum, 10 GeV beam:")
print(f"     gamma = {g:.1f}, gamma c T/2 = {g*c*T/2:.2f} m    book: 3.6 m")
print("\nso the apparatus is a few metres of empty space, and the measurement")
print("is counting.  that is what the beat buys you.")

prints

Delta m / m(K0) = 3.481e-06 eV / 4.976e+08 eV = 7.0e-15
                                                   book: 7 x 10^-15

what a fractional precision of 7.0e-15 means elsewhere:
Earth-Moon distance  384400 km  ->  2.7 um
the Earth's diameter  12742 km  ->  0.089 um  (about one wavelength of UV)
one second in              4.5 million years

nobody measures either mass to anything like this.  the beat does it:
T = 2 pi / Delta m = 1.19 ns          book: 1.2 ns
first oscillation maximum, 10 GeV beam:
   gamma = 20.1, gamma c T/2 = 3.58 m    book: 3.6 m

so the apparatus is a few metres of empty space, and the measurement
is counting.  that is what the beat buys you.

One detail the book flags and it is easy to miss: $\Delta m > 0$ is a measured fact, not a convention. Nothing in the definitions forces $m_L > m_S$ , and a beat measurement gives only $|\Delta m|$ . The sign comes from a completely different experiment — a kaon beam crossing matter has a refractive index that depends on $\Delta m$ with sign, exactly as light does. The answer is that the heavier neutral kaon is the longer-lived one.

⚠️ K₁⁰ and K₂⁰ are not K_S and K_L, and the book uses both within a page

Four states, two bases, and they are nearly but not exactly the same — which is the most dangerous kind of nearly.

	definition	what it is good for
$K^0$ , $\bar K^0$	definite strangeness	what strong interactions make and detect
$K_1^0$ , $K_2^0$	definite CP	what decides which pion channel is open
$K_S$ , $K_L$	definite mass and lifetime	what propagates

If CP were exact the last two rows would coincide. They do not, and the difference is the parameter $\varepsilon = 2.2\times10^{-3}$ of §8.5. Until then the site follows the book in treating $K_1^0 \approx K_S$ and $K_2^0 \approx K_L$ , and says so wherever it matters.

Two more traps arriving shortly. $B_H$ and $B_L$ are labelled by mass (heavy/light), not by lifetime, because the two B lifetimes are equal — the opposite convention to the kaon’s. And $\eta$ is about to acquire its third meaning in this book: the meson (§4.6), the amplitude ratios $\eta_{+-}$ and $\eta_{00}$ (§8.8), and the CP eigenvalue $\eta_f$ of a final state (§8.6).

Erratum — Eq. (8.8) and Table 8.1 disagree on the K_S uncertainty

Eq. (8.8) gives $\tau_S = 89.54 \pm \mathbf{0.04}$ ps. Table 8.1 on p. 348 gives the same central value with $\pm \mathbf{0.004}$ — an order of magnitude tighter.

The PDG value is $\pm 0.04$ , so the table is the one at fault. It is a small thing, but it is the second of three places where Table 8.1’s uncertainties disagree with the text by exactly a factor of ten (§8.9–8.11 has the others, on $\Delta m_D$ and $\Delta m_s$ ). Three of the same slip in one table suggests it was assembled from a source with a different convention rather than from the chapter.

Nothing here depends on it: a 0.04 % versus 0.004 % uncertainty on $\tau_S$ changes none of this page’s conclusions.

Erratum — Table 8.1’s B_s width contradicts its own lifetime

Looking ahead to the chapter’s reference table on p. 348: the $B_s^0$ row gives

\tau = 1.497 \pm 0.015\;\text{ps}, \qquad \Gamma = 0.86\;\text{ps}^{-1}

But $1/1.497 = \mathbf{0.668}$ ps⁻¹, not 0.86. Every other row in the table is internally consistent — $1/51160 = 2.0\times10^{-5}$ ✓, $1/89.54 = 0.011$ ✓, $1/0.4103 = 2.4$ ✓, $1/1.520 = 0.65$ ✓ — so the $B_s$ row is the only one that fails, and it fails by 29 %.

The error propagates: the table’s $\Gamma$ (meV) column gives 0.57 meV for the $B_s$ , and $0.86\;\text{ps}^{-1}\times\hbar = 0.566$ meV, so that column was computed from the wrong 0.86 rather than from the lifetime. The value consistent with $\tau = 1.497$ ps is 0.44 meV.

Nothing in this page depends on it — the site’s widget uses $\Gamma = 1/\tau$ throughout, which is why the $B_s$ preset shows $T/\tau = 0.24$ rather than the 0.31 the printed width would give. Flagged here because Table 8.1 is the chapter’s reference table and will be surfaced again in §8.11.

🔑 If you remember only three things

You tell them apart by what they do to matter. A K⁰ and a K̄⁰ scatter differently on nuclei, which is the only handle available on particles whose decays cannot separate them.
The factor of 571 is a phase-space consequence. One state can reach two pions and the other cannot, and the rest is the energy available.
The mass difference is the smallest number quoted anywhere in this book. A few microelectronvolts against masses of GeV is one part in 10¹⁵.

Where this goes next

§8.3 turns Eq. (8.21) into an experiment: the observable is the charge asymmetry between $K \to \pi^-\ell^+\nu$ and its conjugate, because semileptonic decays obey the $\Delta S = \Delta Q$ rule and so tag the strangeness that the pion channels cannot. §8.4 is the experiment that made the whole picture credible — Pais and Piccioni’s regeneration, where a beam that has become pure $K_L$ is passed through an absorber and the short-lived component comes back, proving the superposition was coherent all along.

Then §8.5 breaks it: the $K_L$ decays to two pions after all, two times in a thousand.

✅ Check yourself — oscillation and the neutral K states

0/6 answered · 0 correct

1.Why does the same equation produce such different behaviour in the K⁰, D⁰, B⁰ and B_s⁰ systems?
2.Δm for the neutral kaons is 3.5 μeV out of a mass of 497.6 MeV — a fraction of 7 × 10⁻¹⁵. How is that measurable?
3.If CP is conserved, why can K₂⁰ not decay to two pions?
4.The K_L lives 571 times longer than the K_S. What does that fact establish, before anyone measures a CP-violating rate?
5.A 5 GeV neutral kaon beam is produced at a target. What does it look like after five metres of vacuum?
6.In Eq. (8.21), what do the survival and appearance probabilities sum to?