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Lecture 15: Hadronic Weak Decays, the CabbiboAngle and the GIM Mechanism
Oct 18, 2018
Reminder: Charged Current Weak Interactions with ParityViolation
• Wu et al showed that polarized Co60 decays had angular distribution
I(θ) = 1 + α(σ · pE
)= 1 + α
v
ccos θ
with later experiments verifying that α = −1
• Charged current weak interactions are left handed, aside from mass terms
• Garwin et al confirmed using µ-decay
• Goldhaber et showed that ν is left handed (and ν is right handed)
Reminder: Four Fermi Theory with V-A
• OK as long as q2 << MW
• Matrix element
M =GF√
2Jµ1 J2 µ
where J1,J2 are the two currents and GF ∼ 10−5 GeV−2
NB: The√
2 is convention and we missing in Tues lecture
• We know now that currents exchange a W±
• Leptonic current:J` µ = ψνell (1− γ5)ψ`
where ψ is a spinor
γ5 = iγ0γ1γ2γ3 =
(0 II 0
), γ0 =
(I 00 −I
), γi =
(0 σi
−σi 0
)• Note: γ5 is a pseudoscalar, so leptonic current is V −A• If momentum transfer is large, replace 4-point interaction withW -propagator
• This is called a “charged current” interaction since a W± is exchanged.
We’ll get to “neutral currents” in Lecture 19
An aside on numerical factors
• I was sloppy Tuesday on numerical factors. Let’s be more careful this time.• Before 1956 matrix element was defined
Mfi = GF gµν
[ψ3γ
µψ1
] [ψ4γ
νψ2
]
• Modification to add V-A interactions
Mfi =1√
2GF gµν
[ψ3γ
µ(1− γ5
)ψ1
] [ψ4γ
ν(1− γ5
)ψ2
]
where 1/√
2 is to keep value of GF the same• Replacing 4-Fermi interaction with propagator
Mfi =
[gW√
2ψ3γ
µ(1− γ5
)ψ1
](−gµν + qµqν/M2W
q2 −M2W
)[gW√
2ψ4γ
ν(1− γ5
)ψ2
]
• In limit q2 << MW
Mfi =g2W
8M2W
gµν
[ψ3γ
µψ1
] [ψ4γ
νψ2
]
⇒GF√
2=
g2W8m2
W
Where we finished last time: Tau Decay
• mτ = 1.777 GeV
• Several possible decays:
τ− → e−νeντ
τ− → µ−νµντ
τ− → du ντ
In last case, the du turns intohadrons with 100% probability
• All diagrams look like µ-decay
• If GµF = GeF = GF , predict:
Γτ−→e− = Γτ−→µ−
= (mτ/mµ)2 Γ(µ)
(difference in available phase space)
• Using the measured τ -lifetime and BR,check consistency of GF
GτF /GµF = 1.0023± 0.0033
GeF /GµF = 1.000± 0.004
Lepton universality for GF
• For quark decays, need a factor of 3 forcolor. Predict
BR(τ → hadrons) =3
3 + 1 + 1= 60%
• Experimental result:
BR(τ → hadrons) = (64.76± 0.06)%
Difference from 60% understood (QCDcorrections; as for R)
Pion Decay (I) π−(q)→ µ−(p) + νµ(k)
τπ+ = 2.6× 10−8 s
• ud annihilation into virtual W+
• Depends on π+ wave function at
origin
I Need phenomenological parameter that
characterizes unknown wave function
• Write matrix element
M =GF√
2Jµπu(p)γµ (1− γ5) v(k)
where Jµπ = f(q2)qµ since qµ is theonly available 4-vector
• But q2 = m2π so Jπ = fπqµ. fπ has
units of mass (matrix element mustbe dimensionless)
• After spinor calculation, result for
decay width:
Γ =G2F
8πf2πmπm
2µ
(1−
m2µ
m2π
)2
• This came from:
|M|2 ∼ G2Fm
2µ
(m
2π −m
2µ
)f2π
Phase Space ∼|p|
8πm2π
• We’ll examine what this means onthe next page
Pion Decay (II)
• From previous page
Γ =G2F
8πf2πmπm
2µ
(1−
m2µ
m2π
)2
• Result for electron same with mµ → me
• Thus
ΓeΓµ
=m2e
m2µ
(m2π −m2
e
m2π −m2
µ
)2
• Since me = 0.51 MeV,
mµ = 105.65 MeV and
mpi+ = 139.57 MeV
ΓeΓµ∼ 1.2× 10−4
This agrees with measurements
• Physically, result comes from helicitysuppression
• Spin 0 pion, right-handedantineutrino forces µ− to beright-handed
• But µ− wants to be left-handed
I rh component ∼ (v/c)2 ∼ mµ• The less relativistic the decay product
is, the larger the decay rate
Charged Kaon Decays
• K± mass larger than π±
• More options for decay
Leptonic
Same calculation as for π±
Helicity suppression make decay rateto muons larger that to electrons
BR(K− → µ
−νµ) = (63.56± 0.11)× 10
−2
BR(K− → e
−νe) = (1.582± 0.007)× 10
−5
Semi-Leptonic
3-body decay: No helicity suppression
BR(K− → π
0µ−νµ) = (3.352± 0.033) %
BR(K− → π
0e−νe) = (5.07± 0.04) %
More phase space for decay to e
• Hadronic (several diagrams possible)
BR(K− → π
−π0) = (20.67± 0.08) %
BR(K− → π
−π0π0) = (1.760± 0.023) %
BR(K− → π
−π−π+) = (5.583± 0.024) %
Some Observations
• Leptonic decays of µ and τ demonstrate that GF the same for alllepton species
• Leptonic decay of charged pion and kaon tell us nothing about GFsince fπ and fK (which depend on wf at origin) are unknown
• If we want to ask whether GF is the same for hadronic currents asleptonic ones, we need to look at semileptonic decays
I Analog of β-decay
• But, well have to make sure that we are not affected by stonginteraction corrections!
(V-A) and the Hadronic Current
• At low q2 we measure WI using hadrons and not quarks
• Need to worrry about whether binding of quarks in hadron affects thecoupling
1− γ5 → CV − CAγ5
• Experimentally, for the neutron
CV = 1.00± 0.003; CA = 1.26± 0.02
I Vector coupling unaffected: protected by charge conservation (CVC)
I Axial vector coupling modified (PCAC)
• Experimental implication: for precision tests of hadronic weak
interactions, study decays that can only occur through CV term
I This means decays between states of the same parityI Best option is “superallowed” β-decay with 0+ → 0+ transition
I In addition to no axial vector component, such transitions cannot occur
via γ decay
Is GF Really Universal?
• Muon decay rate is
Γ =G2Fmµ
5
192π3
in approximation where me ignored
• Same formula holds for nuclear β-decay
• A good choice of decay: O14 → N14∗ e+νe (0+ → 0+)
• Correcting for available phase space we find
Gµ = 1.166× 10−5
Gβ = 1.136× 10−5
Close but not the same!
• What’s going on?
Extend Our Study to More Hadron Decays
• Compare the following:
Decay Quark Level Decay
014: p→ ne+νe u→ de+νeπ− → π0e−νe d→ ue−νeK− → π0e−νe s→ ue−νeµ+ → νµe
+νe
• After correcting for phase space factors, GF obtained from p and π−
agree with each other, but are slightly less than obtained from µ.
• GF obtained from K− decay appears much smaller
• Either GF is not universal, or something else is going on!
An Explanation: Choice of weak eigenstates
• Suppose strong and weak eigenstates of quarks not the same
• Weak coupling:
• Here d′ is an admixture of down-type quarks
• Normalization of w.f. for quarks means if d′ = αd+ βs, then√α2 + β2 = 1
• Can force this normalization by writing α and β in terms of an angle
d′ = d cos θC + s sin θc
or (d′
s′
)=
(cos θc sin θC− sin θc cos θc
)(ds
)
The Cabbibo Angle
• Usingd′ = d cos θC + s sin θc
we predict
p & π decay ∝ G2F cos2 θC
K decay ∝ G2F sin2 θC
µ decay ∝ G2F
• Using experimental measurements, find
cos θc = 0.97420± 0.00021
sin θc = 0.2243± 0.0005
• However, in addition to the d′ there is an orthoginal down-typecombination
s′ = s cos θc − d sin θc
Does it iteract weakly?
A New Quark (Discussion from the Early 1970’s)
• It’s odd to have one charge 2/3 quark and two charge -1/3 quarks
• Suppose there is a heavy 4th quark
• We could then have two families of quarks. In strong basis:(ud
),
(cs
)Call this new quark “charm”
• Then, the weak basis is(u
d′ = d cos θC + s sin θc
),
(c
s′ = s cos θc − d sin θc
)
• There is a good argument for this charm quark in addition to GF ...
The GIM Mechanism (I)
• Glashow, Iliopoulous, Maiani (GIM) proposed existence of this 4th quark(charm)
• Charm couples to the s′ in same way u couples to the d′
• Reason for introducing charm: to explain why flavor changing neutralcurrents (FCNC) are highly suppressed
• Two examples of FCNC suppression:
1. BR(K0L → µ+µ−) = 6.84× 10−9
2. BR(K+ → π+νν)/BR(K+ → π0µν) < 10−7
• Why are these decay rates so small?
• It turns out that there is also a Z that couples to ff pairs, but it doesnot change flavor (same as γ)
• If only vector boson was the W±, would require two bosons to be
exchanged
I Need second order charged weak interactions, but even this would give a
bigger rate than seen unless there is a cancellation
The GIM Mechanism (II)
• Consider the “box” diagram
• M term with u quark ∝ cos θC sin θC
• M term c quark ∝ − cos θC sin θC
• Same final state, so we add M’s
• Terms cancel in limit where we ignore quark masses
This Cancellation is Not a Accident!
• Matrix relating strong basis to weak basis is unitary
d′i =∑j
Uijdj
• Therefore is we sum over down-type quark pairs∑i
d′id′i = ∼ijk djU†jiUikdk
=∑j
djdj
• If an interaction is diagonal in the weak basis, it stays diagonal in thestrong basis
• Independent of basis, there are no d←→ s transitions
No flavor changing neutral current weak interactions(up to terms that depend on the quark masses)
Some Questions and Answers about the GIM Mechanism
• Why is mixing in the down sector?
I This is convention.I Charged current interactions always involve an up-type and a down-type
quark
I Can always define basis to move all mixing into either up or down sector
• Why is there no Cabbibo angle in the lepton sector?
I Actually, there is!I Before people observed neutrino oscillations, they thought ν’s were
massless.I If all ν were massless, or had same mass, then free to redefine flavor basis
to remove the mixing
I We now need to define mixing angles for neutrinos as well as quarks
More Than Two Generations
• Generalize to N families of quark (N = 3 as far as we know)
• U is a unitary N ×N matrix and d′i is an N -column vector
d′i =
N∑j=1
Yijdj
• How many independent parameters do we need to describe U?
I N ×N matrix: N2 elementsI But each quark has an unphysical phase: can remove 2N − 1 phases
(leaving one for the overall phase of U)
I So, U has N2 − (2N − 1) independent elements
• However, an orthogonal N ×N matrix has 12N(N − 1) real parameters
I So U has 12N(N − 1) real parameters
I N2 − (2N − 1)− 12N(N − 1) imaginary phases (= 1
2(N − 1)(N − 2))
• N = 2 1 real parameter, 0 imaginary
• N = 3 3 real parameters, 1 imaginary
• Three generations requires an imaginary phase: CP Violation inherent
The CKM Matrix
• Write hadronic current
Jµ = − g√2
(u c t
)γµ
(1− γ5)
2VCKM
dsb
• VCKM gives mixing between strong (mass) and (charged) weak basis
• Often write as
VCKM =
Vud V us VubVcd V cs VcbVtd V ts Vtb
• Wolfenstein parameterization:
VCKM =
1− λ2/2 λ Aλ3(ρ− iη)−λ 1− λ2/2 Aλ2
Aλ3(1− ρ− iη) −Aλ2 1
+O(λ4)
Here λ is the ≈ sin θC .
Best Fit for CKM Matrix from PDG
• From previous page
VCKM =
1− λ2/2 λ Aλ3(ρ− iη)−λ 1− λ2/2 Aλ2
Aλ3(1− ρ− iη) −Aλ2 1
+O(λ4)
• Impose Unitary and use all experimental measurements
λ = 0.22453± 0.00044 A = 0.836± 0.015
ρ = 0.122+0.018−0.17 η = 0.355
+0.12−0.11
• Result for the magnitudes of the elements is: 0.97446± 0.00010 0.22452± 0.00044 0.00365± 0.000120.22438± 0.00044 0.97359± 0.00011 0.04214± 0.000760.00896± 00024 0.04133± 0.00074 0.999105± 000032
• We’ll come back to this in Lecture 18
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