Role of Kaons in Development of SM
• Kaons played essential role in understanding particle physics
• Have already seen two examples:
I K+ → 2π, 3π: Parity violation
I K0 → µ+µ+: Very low BR due to GIM mechanism
• Today and Thursday, explore special role of neutral K system
I Mixing
I CP Violation
• Will also include some discussion of neutral B mesons, which exhibitsimilar phenomena
Reminder: Strange Particle Phenomenology
• Stange pseudoscale mesons: 2 isodoublets(K+
K0
) (K
0
K−
)where K+ = su, K− = us, K0 = sd, K
0= ds
• Because strangeness conserved in SI, K0 and K0
are distinct particles
• Strange particles are pair produced via SI
π−p → ΛK0
π+p → pK+K0
• First reaction has much lower threshold than second
I Can produce a pure K0 beam
• Today, neutal K beams produced using high intensity proton beams
hitting targets
I Background a big issue for K0 experiments, most notably from neutrons
Neutral Kaon Decays
• m(K0) = 497 MeV. Not many decay modes open
I Fully leptonic decays highly suppressed (GIM)I K0 → π−`+ν` (and charge conjugate) occurs
I Since parity not conserved in WI, both 2π and 3π decays possible
• Since both K0 and K0
can decay to same states, they can mix throughvirtual decays
K0 ←→⟨
πππππ
⟩←→ K
0
• These are 2nd order WI with ∆S = 2
• If we start with a pure K0 state at t = 0, it will at some later time have a
combination of K0 and K0
|K(t)〉 = α(t)∣∣K0⟩+ β(t)
∣∣∣K0⟩
with√α2 + β2 = 1
Neutral Kaon Mixing (I)
• Can describe this 2nd order WI in quark terms:
• If there were no WI, K0 and K0
would be degenerate in mass
• WI break the degeneracy
• Because physical observables (eg mass, lifetime) are eigenstates ofcomplete Hamiltonian (SI+WI), must select correct linear combination of
K0 and K0
define the states that propagate and decay
The mass and lifetime eigenstates are not the flavor eigenstates!
Neutral Kaon Mixing (I)
• We can almost guess the correct basis to use
I We know WI don’t conserve P since ν are LH and ν are RHI Parity would turn a LH ν into a RH νI But Charge Conjugation turns a ν into a ν
I Hence, CP turns a a LH ν into a RH ν
• WI Lagrangian appears to be CP invariant
• In fact, CP is violated in CKM matrix (∼ 10−3 effect)
• But the CP basis is close to the correct one and that’s what we’ll use
today
I We’ll add the small CP violating piece on Thurs
Neutral Kaon Mixing (II)
• Neutral Kaons transform under CP (not unique definition)
CP∣∣K0⟩ =
∣∣∣K0⟩
CP∣∣∣K0
⟩=
∣∣K0⟩• Therefore, we can write
|K1〉 =1√2
(∣∣K0⟩+∣∣∣K0
⟩)CP |K1〉 = |K1〉
|K2〉 =1√2
(∣∣K0⟩− ∣∣∣K0⟩)
CP |K2〉 = − |K2〉
• |K1〉 and |K2〉 are CP eigenstates and almost the physical basis
CP of Possible Hadronic Decays
π0π0:
I Bose symmetry: WF symmetric
P (π0π0) → π0π0
Cπ0 → π0
CP (π0π0) → +1(π0π0)
π+π−:
I Bose symmetry: WF symmetric
P (π+(~p)π−(−~p)) → π+(−~p)π−(~p)
Cπ± → π∓
CP (π+π−) → +1(π+π−)
π0π0π0:
I Any two π0 combo must haveeven ` (Bose stats)
I J = 0 so ` of 3rd π0 also evenwrt other two
I But π has intrinsic parity P = −1
P (π0π0π0) → (−1)3π0π0π0
Cπ0 → π0
CP (π0π0π0) → −1(π0π0π0)
π+π−π0:
I Small Q suggests ` = 0. If so,same argument as above
I Both CP states allowed but
CP (π+π−π0) = −(π+π−π0)
state highly dominant
2π states have CP = +1 and 3π states have CP = −1
Hadronic Decays of the |K1〉 and |K2〉
• Associating the CP states with the decays:
|K1〉 → 2π
|K2〉 → 3π
• However, very little phase space for 3π decay: Lifetime of |K2〉 muchlonger than of |K1〉
• Physical states called “K-long” and “K-short”:
τ(KS) = 0.9× 10−10 sec
τ(KL) = 0.5× 10−7 sec
• We’ll use distinction that |K1〉, |K2〉 are the CP eigenstates and |KS〉,|KL〉 are true mass eigenstates (including CP violation)
A More Formal Treatment of Mixing
• Write our state ψ as linear combination of K0 and K0:
ψ = α∣∣K0⟩+ β
∣∣∣K0⟩
=
(αβ
)• Schrodinger eq tells us
idψ
dt= Hψ
where H is Hermitian matrix: “generalized mass matrix”
• In matrix form:
H =
(M − i
2Γ M12 − i
2Γ12
M∗12 − i2Γ∗12 M − i
2Γ
)• Diagonal elements equal from CPT
• If CP is a good symmetry, M12 and Γ12 are real
• Find eigenstates by diagonalizing the matrix
M = (m1 +m2)/2 ∆m ≡M12 = (m1 −m2)/2Γ ≡ Γ12 = (Γ1 + Γ2)/2 ∆Γ = (Γ1 − Γ2)/2
Time Dependence (I)
• Write wave functions
|K1(t)〉 = e−im1t−Γ1t/2 |K1〉
|K2(t)〉 = e−im2t−Γ2t/2 |K2〉
• Writing this in terms of strong eigenstates∣∣K0⟩
=1√
2
[e−im1t−Γ1t/2 |K1〉+ e−im2t−Γ2t/2 |K2〉
]∣∣∣K0
⟩=
1√
2
[e−im1t−Γ1t/2 |K1〉 − e−im2t−Γ2t/2 |K2〉
]• If a state ψ that is purely
∣∣K0⟩
is produced at t = 0, at a later time it
will be a combination of∣∣K0
⟩and
∣∣∣K0⟩
:
⟨K0∣∣ ∣∣K0
⟩=
1
2
[e−im1t−Γ1t/2 + e−im2t−Γ2t/2
]⟨K
0∣∣∣ ∣∣∣K0
⟩=
1
2
[e−im1t−Γ1t/2 − e−im2t−Γ2t/2
]
Time Dependence (II)
• Square to get probability:∣∣⟨K0∣∣ ∣∣K0
⟩∣∣2 =1
4
[e−Γ1t + e−Γ2t + 2e−(Γ1+Γ2)t/2 cos(∆mt)
]∣∣∣⟨K0
∣∣∣ ∣∣K0⟩∣∣∣2 =
1
4
[e−Γ1t + e−Γ2t − 2e−(Γ1+Γ2)t/2 cos(∆mt)
]
• The∣∣K0
⟩and
∣∣∣K0⟩
oscillate with frequency ∆m and at the same time
they decay
Observing the Oscillation
• Oscillation provides a way to measure ∆M
• Also demonstrates that this QM phenomenon is happening
• How do we see it?
1. Start with pure K0 beam (low energy)
Look at time dependence of hyperon yield in interactions (K0p→ Λπ)
∆mτ1 = 0.477± 0.2
2. Look for decays that tag the flavor: semileptonic
Observe time dependence in `+ vs `− rate
• Note: Phenomenology of K mixing depends on two things
I Large lifetime difference: time to mix before decaying
I Small mass difference: short oscillation frequency
• In B system things look somewhat different (we’ll discuss later)
What we expect to see as a function of ∆M
∣∣⟨K0∣∣ ∣∣K0
⟩∣∣2 =1
4
[e−Γ1t + e−Γ2t + 2e−(Γ1+Γ2)t/2 cos(∆mt)
]∣∣∣⟨K0
∣∣∣ ∣∣K0⟩∣∣∣2 =
1
4
[e−Γ1t + e−Γ2t − 2e−(Γ1+Γ2)t/2 cos(∆mt)
]
For the measured ∆M
• Start with a K0 beam
• After many KS lifetimes, have a pure KL beamI In absence of CP violation, equal parts K0 and K0
Observation of K0 −K0 Oscillation using semileptonic decays
• Use lepton flavor to distinguish K0 and K0
• Plot shows asymmetry N(`+)−N(`−)N(`+)+N(`−)
• Removes (trivial) lifetime dependence• We’ll come back to the non-zero value at large times
Thursday (CP violation)
Regeneration
• Suppose we put our K beam into matter
• K0 and K0
(SI) interaction rates very different (can make Λ from protons butnot Λ)
• Different absorption cross sections and different forward scattering amplitudes
• Define a complex index of refraction (see Jackson’s E&M book)
n = 1 +2πN
k2f(0)
(similar for n)
• Optical theorm
σtot =4π
kIm (f(0))
• Can include this complex index of refraction into Schrodinger Eq as a phase
∆(phase) =2πNvf
k(1− v2)12
• This changes the eigenstates to∣∣K′1,2⟩ = |K1,2〉 ± r |K2,1〉
• We’ll see an important application of this next time
How About the B system?
• Again, second order in weak interactions
• Different CKM matrix elements for B0 and BsI Larger ∆M for Bs than Bd
• Many possible final states for the decayI Difference in lifetime of the BL and BS states small