lecture 11: weak interactions - university of oxfordbiller/particle_course/lectures/lecture_1… ·...

Post on 14-Jul-2020

7 Views

Category:

Documents

0 Downloads

Preview:

Click to see full reader

TRANSCRIPT

Lecture 11: Weak Interactions

•  Cross-Section and the W Coupling •  The Cabibbo Angle and the CKM Matrix •  Parity Violation •  Kaons and Mixing •  CP Violation

Sections 4.51, 8.1, Chapter 10

Useful Sections in Martin & Shaw:

(from ''Telephone Poles and Other Poems," 1963)

Neutrinos, they are very small. They have no charge, they have no mass And do not interact at all. The earth is just a silly ball To them, through which they simply pass, Like dustmaids down a drafty hall Or photons through a sheet of glass...

John Updyke

in fact, point-like in the Standard Model

and little (< 2eV) hardly

true

should not be taken to indicate a sensitive detection technique

interaction cross-section much higher than for typical neutrino energies

obvious foreshadowing of electroweak theory

Cosmic Gall

+ ν e (Pauli)

Beta Decay

n → p + e-

(Pauli)

''Inverse" Beta Decay

νe + p → n + e+

σ ~ λ2 × λ/c τ

''cross-sectional area" of ν wave packet

time spent by wave packet in presence of the proton

typical timescale for weak interaction to occur

νe + p → n + e+ Inverse β-decay:

(Pontecorvo)

From standard β-decay, the lifetime of the free neutron is τ ~ 1000 s and the energies of the e- and νe are ~ 1 MeV

⇒ λ = h/p ≃ 1200fm = 1.2x10-10cm

thus, σ ~ (1.2x10-10cm)3/[(3x1010 cm/s)(1000s)] ~ 10-43cm2

Note σ ∝ E-3t-1 and, from previous discussion, t-1 ∝ E5

⇒ σ ~ 10-43 (EMeV)2 cm2 Almost exactly right! (and very, very small!!!)

Interaction Length for a 1 MeV Neutrino in Lead

λ = 1/(σρ)

σ ~ 10-43 cm2 (per proton)

ρ = (11.4 g/cm3) x [1/(207 g/mole)] x (6.02x1023 atoms/mole) x (82 protons/atom)

= 2.7x1024 protons/cm3

λ = 1/(2.7x10-19) cm

= 3.7 x 1018 cm = 4 light-years !!

n → p + e- + νe

νe + p → n + e+

Reines and Cowan, 1956 (Nobel Prize – 1995 !!)

Parity Violation in Weak Interactions First suggested in 1956 by Lee & Yang based on review of kaon decay modes

60Co

e-

60Co

e-

P nuclear spins aligned by cooling to 0.01 oK in a magnetic field

Should be the same under parity transformation, but fewer electrons are actually seen going forward !

Directly observed by Wu et al. in 1957 from the decay 60Co → 60 Ni* + e- + νe

γ (1.173 MeV) + γ (1.332 MeV)

(degree of polarisation determined from the anisotropy of γ-rays)

Garwin, Lederman & Weinrich (1957)

e+ νµ

νe

νµ

µ+

π+

precess polarised muons (polarised)

Also, in 1958, Goldhaber et al. measured the helicity of the neutrino:

e- + 152Eu(J=0) → 152Sm*(J=1) + νe

152Sm(J=0) + γ

events were chosen with the final states collinear ⇒ γ and νe travel in opposite directions, so helicity of the neutrino is found from that of the gamma

⇒ all neutrinos are left-handed !

Leon Lederman, Melvin Schwartz and Jack Steinberger, 1962

Neutrinos of the ''Second Kind" (not as popular as the Spielberg sequel)

Assume some Yukawa-like exchange process is at work.

Weak interactions obey a simple symmetry :

So, for example, for the process π- → µ- + νµ (pion decay):

but, unfortunately, it is found experimentally that the couplings are not the same!

αWud ≃ 0.95 αW

d

u

W- νµ

µ- π-

⇒ W±

It can change u↔d (like β-decay) s↔c t↔b

and, for leptons, e↔νe µ↔νµ τ↔ντ

β-decay (n→p+e-+νe) tells us the exchange particle must be charged

s

u

W- νµ

µ- Κ-

Another hitch:

shouldn’t occur, but does ! (albeit infrequently)

We can explain all this (or, at least, parameterize our ignorance) by adopting the somewhat bizarre notion that the weak interaction actually couples to mixtures of quarks.

So, initially just considering the first two generations, the relevant quark doublets are:

u dʹ′

c sʹ′ ( ) and ( )

where dʹ′ ≡ d cosθC + s sinθC

sʹ′ ≡ -d sinθC + s cosθC

θC ⇒ ''Cabibbo angle"

or, alternatively d ≡ -sʹ′ sinθC + dʹ′ cosθC

s ≡ sʹ′ cosθC + dʹ′ sinθC

αWud = αW cos2θC

αWus = αW sin2θC

~ 1/20 ( θC = 12.7 + 0.1 degrees )

σ ( Κ- → µ- νµ )

σ ( π- → µ- νµ ) = tan2θC

(The factor of 1/20 delineates ''Cabibbo-suppressed" and ''Cabibbo-allowed" processes)

Generalizing to 3 generations and all possible mixings between quarks:

dʹ′ sʹ′ bʹ′

Vud Vus Vub Vud Vus Vub Vud Vus Vub

dʹ′ sʹ′ bʹ′ ( ) [ ] ( ) =

(Cabibbo, Kobayashi and Maskawa)

CKM matrix

αWus

αWud =

Kaons: Ko = ds Ko = sd (S = +1) (S = -1)

But S is not conserved in weak interactions so Ko-Ko mixing can occur:

u

u

d

s

s

d

W+ W- Ko Ko

We can thus define two orthogonal mixtures:

⎜K1o〉 = 1/√2 ( ⎜Ko〉 + ⎜Ko〉 )

⎜K2o〉 = 1/√2 ( ⎜Ko〉 - ⎜Ko〉 )

Note: C P ⎜K1o〉 = + ⎜K1

o〉 and C P ⎜K2o〉 = - ⎜K2

o〉

K1o → π+ π- ; πo πo

K2o → π+ π- πo ; πo πo πo

Allowed

K1o → π+ π- πo ; πo πo πo

K2o → π+ π- ; πo πo

Forbidden

Experimentally, 2 kaon states are observed with different lifetimes:

KSo → π+ π- ; πo πo τ ≃ 9x10-11s

So we associate KSo ⇔ K1

o and KLo ⇔ K2

o

However, in 1964, Christenson, Cronin, Fitch & Turlay discovered

KLo → π+ π-

(branching ratio ~ 2x10-3)

KLo → π+ π- πο ; πo πo πo ; π± lepton ν (ν) τ ≃ 5x10-8s

±

30 GeV protons

steel target

beam collimator

magnets sweeps out charged particles

lead-glass cuts out photons

KS+KL KL

18 m

KL beam direction

CM of π+π- pair

⎜KSo〉 = 1/√1+ ε 2 ( ⎜K1

o〉 - ε ⎜K2o〉 )

⎜KLo〉 = 1/√1+ ε 2 ( ε ⎜K1

o〉 + ⎜K2o〉 )

where ε ≡ small complex number parameterizing the size of the CP violation

(experimentally, ε ≃ 2.3x10-3 )

What does this mean??

Reason for antimatter assymmetry ??

Perhaps we can learn more from studying CP violation in other particle systems...

Basically compare the rates for

B0 = Ψ + KS0

(π+π- mode)

B0 = Ψ + KS0

versus

(π+π- mode)

dʹ′ sʹ′ bʹ′

Vud Vus Vub Vud Vus Vub Vud Vus Vub

dʹ′ sʹ′ bʹ′ ( ) [ ] ( ) = ??

CP violation could be parameterized as part of the mixing angles in the CKM matrix

Unitarity of the matrix is needed to allow for local gauge symmetry

Which imposes constraints on the angles:

γ

α

β

η

ρ

''Unitarity Triangle"

Matter-Antimatter Asymmetry Revisited:

Sakarov Conditions (1967)

1) Baryon Number Violation allows baryons and anti-baryons to appear and disappear independently of each other

2) CP Violation so the rate of appearance/disappearance of baryons is different from anti-baryons

Establishes Asymmetry

3) Non-Equilibrium Conditions since equilibrium would then tend to ''average-out" any asymmetry

Locks In Asymmetry

!!! (GUTs)

top related