Lecture 15: Hadronic Weak Decays, the Cabbibo …physics.lbl.gov/shapiro/Physics226/lecture15.pdfLecture 15: Hadronic Weak Decays, the Cabbibo Angle and the GIM Mechanism Oct 18, 2018

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