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1. Basic Concepts + Appendices 2. Nuclear Phenomenology 3. Particle Phenomenology 4. Experimental Methods 5. Quark Dynamics: The Strong Interaction à 6. Weak Interactions And Electroweak Unification 7. Models And Theories Of Nuclear Physics 8. Applications Of Nuclear Physics 9. Outstanding Questions and Future Prospects F. Ould-Saada
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6.1 Charged and Neutral Currents. 6.2 Symmetries of the Weak Interaction. 6.3 Spin Structure of the Weak Interactions. 6.3.1 Neutrinos
6.3.2 Particles with mass: chirality ±
0
6.4 W and Z Bosons. 6.5 Weak Interactions of Hadrons: Charged Currents. 6.5.1 Semileptonic decays
6.5.2 Selection rules 6.5.3 Neutrino scattering 6.6 Meson Decays and CP Violation. 6.6.1 CP invariance 6.6.2 CP violation in K0L decay 6.6.3 CP violation in B decays 6.6.4 Flavour oscillations 6.6.5 CP violation and the Standard Model 6.7 Neutral Currents and the Unified Theory. 6.7.1 Electroweak unification 6.7.2 The Z0 vertices and electroweak reactions 14/05/14
F. Ould-Saada
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Wider aspects of weak interactions (WI) Unification with electromagnetism (EM) è Successful Electroweak theory (EW) ¡
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WI à spin-­‐1 gauge bosons – force carriers between quarks and leptons §  Charged bosons W+, W-­‐ à charged currents §  Neutral boson Z0 à neutral currents ▪  Observed in 1973 in a bubble chamber experiment νµ + N → νµ + X
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Successes of Unified theory of EW interactions §  Existence of neutral currents §  Prediction of existence of charm quark, prior to discovery in 1974 §  Prediction of masses of W+, W-­‐ and Z0 , prior to discovery in 1983 ¡
EW theory predicts existence of new spin-­‐0 boson – Higgs … confirmed at LHC … 14/05/14
F. Ould-Saada
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New unification becomes manifest at high energies §  At low energies weak and EM interactions clearly separated §  From general Amplitude M (q 2 ) =
M X = M W,Z ≈ 80 − 90 GeV
M X = Mγ = 0
g2!2
q 2 − M X2 c 2
gw ≈ gem
#
$ Unification ⇒ 2
q ≈ high
%
§  Start with low energies à deduce general properties valid at all energies §  Then consider ▪  How unification arises & Some consequences 14/05/14
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Parity – P Charge conjugation – C Time reversal – T §  ... Conserved in Strong and EM interactions §  What about Weak interactions? 14/05/14
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Soon after discovery of strange particles, a puzzle appeared the §  two spin-­‐0 particles (then referred to τ+ and θ+), of same mass and same lifetime, observed to ¡
decay into two pions and three pion: P(π)= -­‐1 è θ + → π +π 0 ; τ + → π +π +π −
P(θ + ) = P(π +π 0 ) = +1
P(τ + ) = P(π +π +π − ) = −1
€
¡
C, P and T symmetries thought to be conserved in all particle interactions. €
§  Puzzle! ¡
§  (θ+ and τ+ correspond charged K-­‐meson decaying weakly (in 10-­‐8s) according to: K + → π +π 0
T.D.Lee, C.N. Yang, 1956, §  Weak interactions do not conserve parity! §  measure some pseudo-­‐scalar quantity BR = 21.16 ± 0.14%
K + → π +π +π − BR = 5.59 ± 0.05%
▪  Helicity of some particle or scalar product of the spin of a particle and the € momentum of some other particle ¡
Parity violation: Wu et al. 1957 60
Co→60Ni * + e − + νe
§  In magnetic solenoid at 0.01K §  Nuclear spins align parallel to B field direction ¡
¡
If no asymmetry were detected in the emission of the decay electrons, the real world and the mirror world would be indistinguishable Experiment observed “forward-­‐backward decay asymmetry”, fewer electrons emitted in forward hemisphere then in the backward wrt spins of decaying nuclei 14/05/14
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60
60
*
Co→ Ni + e + v e
¡
Within months of Lee and Yang’s suggestion C.S. Wu and E. Ambler studied decay of cobalt-­‐60 ¡
Point of experiment: §
§
¡
−
€
Observe some spatial asymmetry in the emission of β-decay electrons from the cobalt which could then lead to a difference between β-decay and its mirror image Realisation: §
Necessity to establish some direction in space §
Put magnetic field B ~ 0.05 T across specimen of cobalt cooled to a very low temperature of T ~ 0.01 K (to avoid depolarisation) ▪  of which cobalt was aware ▪  with respect to which the emission of β electrons could be measured ▪  Spin of the nuclei align predominantly in direction of magnetic field ▪  In a radioactive source at room temperature, the spins are oriented randomly. By measuring emission of β-decay electrons along or against orientation of nuclear spin – i.e. orientation of B-­‐field -­‐ any asymmetry can be measured §  Through Parity: !
! !
! !
§
▪  Direction of magnetic field and spin do not change ▪  BUT direction of emission of electrons will change! Parity :
!
B → B ; s → s ; x → −x
Any asymmetry of electron emission measured with respect to the direction of the magnetic field will appear to be reversed in the mirror image process §  Exactly what was measured! §
(L) Arrangement to measure emission from polarised nuclei. (R) Result of the earliest experiment showing parity non-­‐conservation. A normalised counting rate is shown for two directions of the external magnetic field. After adiabatic demagnetisation, the source warms up, the polarisation decreases, and the field disappears. Parity :
Result :
! !
⎛ σ ⋅ p ⎞
v
I (θ ) = 1 + α ⎜
=
1
+
α
cosθ
⎟
c
⎝ E! ⎠
! s
α = −1 and σ = !
s
! !
! !
s ⋅ p → −s ⋅ p
cosθ e → − cosθ e
! !
sco ⋅ pe
Measured : 〈 cos θ e 〉 = 〈 !
! 〉<0
sco pe
Expected if parity conserved: 〈 cos θ e 〉 = 0
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Soon after the observation of the WI’s violation of parity, it was discovered that it does not preserve charge conjugation symmetry, C, either. This was demonstrated by examining the spins of the electrons and positrons emitted in the decays of positively and negatively charged muons µ − → e−ν e vµ ≠ µ + → e+ν e vµ
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Angular distribution observed for µ+ and µ-­‐ decays of form ¡
C invariance would require that rates and angular distributions are same for µ+ and µ-­‐ ¡
P invariance would require 14/05/14
1
Γ µ ± (cosθ ) = Γ ± (1− a± cosθ )
2
−1
±
Γ± ≡ τ =
∫
+1
−1
Γ µ ± (cosθ )dcosθ
Γ+ = Γ−
a+ = a−
Γ µ ± (cosθ ) = Γ µ ± (−cosθ )
F. Ould-Saada
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C and P conservation would require ¡
Measured values were ¡
¡
But why do µ+ and µ-­‐ have the same lifetime if C is violated? … because the combined symmetry CP is conserved a+ = a− = 0
−a+ = a− = 1.00 ± 0.04
§  Violation of P being exactly compensated by that of P ¡
CP invariance requires ¡
… in agreement with experimental observations Γ µ + (cosθ ) = Γ µ − (−cosθ )
Γ+ = Γ−
τ+ = τ−
a+ = −a−
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How to survive in a strange world without symmetry as guidance! 14/05/14
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Helicity-­‐states: right-­‐handed R (a), left-­‐handed L (b) ¡
§  spin component along direction of motion of spin-­‐1/2 particle Non conservation of P and C has an important consequence on weak interactions. ¡
§  Massless neutrinos give us a good picture: §  “Almost” only left-­‐handed neutrinos and right-­‐
handed anti-­‐neutrinos exist in nature. ¡
As Abdu Salam said: “On reflecting a neutrino in a mirror, one sees nothing, §  the process being forbidden by P non-­‐
conservation. §  But if, in addition to P, the neutrino is turned into an anti-­‐neutrino by C operation, the “image" comes back! ” 14/05/14
F. Ould-Saada
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Relativistic fermions described by 2-­‐component spinor (and another 2-­‐component spinor for the antiparticle) §  Slow fermions: 2 spin states (up and down) §  Relativistic fermions: notion of spin no longer so useful ! 2 descriptions: ¡  helicity as component of spin in direction of motion of fermion §
§
§
§
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Spin can either be aligned with or against momentum Fermion referred to as being in h=+ or h=-­‐ state respect. Measurement of s and p à helicity However, helicity NOT invariant under proper Lorentz transformations (2 observers can measure opposite helicities) chirality or handedness to classify the 2 possible states of a relativistic fermion ▪  Chirality is invariant under proper Lorentz transformations ▪  Particle is left-­‐handed or right-­‐handed ¡
Both helicity and chirality states have important property that they are interchanged under parity operation LH ⎯Parity
⎯⎯→ RH ; h + ⎯Parity
⎯⎯→ h −
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¡
WIs couple uniquely to LH-­‐fermions and RH-­‐antifermions For massless fermions, helicity and chirality coincide §  A massless LH particle and a massless h-­‐ particle are one and the same thing ¡
Chirality: property of 4-­‐component spinor §  2 chiral states are eigenstates of γ5 with eigenvalues +1 (R) and -1 (L) §  chirality in general not conserved but good QN for massless particles ψ = ψL + ψR
1
(1− γ 5 ) ψ,
2
1
ψ R = (1+ γ 5 ) ψ,
2
ψL =
γ 5ψ L = −ψ L
γ 5ψ R = +ψ R
Lm = m(ψ Rψ L + ψ L ψ R )
¡
4-­‐fermion interaction constrains, very tightly, the spins which can couple through WIs §  Only LH fermions and RH anti-­‐fermions can take part in WIs ν L and ν R
§  Neutrinos and antineutrinos of opposite chiralities are believed to exist as mν>0 §  In this case they would not take part in any of the known interactions except gravity… ¡
¡
Measure of Neutrino helicity (àchirality for massless ν’s) Use β-­‐decay, deduce ν helicity from helicity of other decay products, using angular momentum conservation ¡
Electron capture §  Initial (Eu) and final (Sm) state nuclei are spin-­‐0 §  Experiment selected events in which photon was emitted in direction of Sm*, opposite to direction of neutrino ¡
Photon helicity measured §  helicity of neutrino inferred through angular momentum conservation (a) is observed!
§  Helicity of neutrino found to be negative §  Neutrino is LH Pion decay at rest ¡
§  Momentum and spin conservation π + →l +v l
! ! !
pl + pν = 0
" " !
sl + sν = 0
!
!
vl − LH ⇒ pν & sν antiparallel ⇒ l + − LH ?
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2 cases: §  But relativistic l+ is RH! €
§  l+=µ+ : muon non-­‐relativistic , both helicities allowed §
2
2 &
#
m
m
−5
e
e
l+=e+ : electron relativistic , helicity suppression à ∝
≈
2
=
2.6
×10
%
(
2
2Ee2
$ mπ '
Spin-dependence: V-A interaction
-  V: proper vector à sign chge under P
-  A: axial-vector à no sign change
14/05/14
F. Ould-Saada
Electromagnetic current (γ): transforms as V
Weak current (W): as V-A because of P-violation
18
¡
The measured ratio is ¡
π+àµ+νµ
Γ(π + → e+ν e )
−4
=
(1.230
±
0.004)
×10
Γ(π + → µ +ν µ )
§  Aother decay where both P and C are violated ¡
Combined CP is conserved 14/05/14
F. Ould-Saada
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¡
¡
Muons emitted in pion decays are polarised Highest energy electrons emitted in muon decay satisfy (show this) §  The electron is relativistic, contrary to the muon µ − → e − veν µ
mµ c 2 ⎛ me2 ⎞
⎜1 + 2 ⎟ >> mec 2
Ee =
2 ⎜⎝ mµ ⎟⎠
Helicity arguments + fact that e-­‐R is helicity suppressed à Electrons emitted opposite the muon spin à The angular distribution of high energy electrons leads to a forward-­‐
backward asymmetry 14/05/14
F. Ould-Saada
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¡
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Dirac (particle = antiparticle) or Majorana (particle=antiparticle) §  à neutrino-­‐less double β-­‐decay υe = υe ⇒ 76Ge→76Se + e − + e −
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F. Ould-Saada
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p + p →W ± + X ∓
W ± → l ± + νl or νl ⎫⎪ MW = 80.4GeV / c 2
−25
τ
≈
3
×
10
s
⎬
2
0
+ −
Z → l l , νl νl ⎪⎭ M Z = 91.2GeV / c
p + p →Z 0 + X 0
2α W
Γ(W →eν e ) =
MW c 2
3
gW2
⇒ αW =
≈ (4.2 ± 0.2) × 10 −3 ≈ 0.6α
4 π!c
¡
¡
Weak interactions are not that weak!! Apparent weakness solely a consequence of its short range because Mw heavy 2 2
g
!
2
M (q ) = 2
q − M X2 c 2
Weak interactions: point, zero-­‐range α eff
M X = M W,Z ≈ 80 − 90 GeV
2
¡
2
λ = h / p large ⇒F.qOuld-Saada
<< M X
14/05/14
# E &2
2
= αW %
;
E
<<
M
c
(
W
2
$ MW c '
22
¡
Converting a conventional ”fixed target” proton accelerator into a proton-­‐antiproton collider in order to provide the earliest possible opportunity for discovering massive gauge bosons §  C. Rubbia et al., 1976 §  S. van der Meer, stochastic cooling of antiprotons ¡
Better chance than e+e-­‐ , …that time p + p →W ± + X ∓ →l ± + ν l or ν l + X ∓
p + p →Z 0 + X 0 →l + l − + X 0
¡
¡
Manifestation of heavy W and Z bosons §
§
distribution of transverse momentum of emerging leptons Invariant mass distribution of di-­‐leptons §
W discovery, January 1983 ; UA1 and UA2 experiments 109 p-­‐pbar collisions of which 106 recorded ▪
§
UA1: 5 Wà eν candidates; UA2: 4 Wàeν candidates Z discovery, June 1983 p + p → W ± + X ∓ → e± + νl or νl + X ∓
p + p → Z 0 + X 0 → e+e− + X 0
http://cerncourier.com/cws/article/cern/29053
http://nobelprize.org/nobel_prizes/physics/laureates/1984/press.html
¡
Lepton-­‐quark symmetry and quark mixing ⎛ u ⎞ ⎛ c ⎞
⎜⎜ ⎟⎟, ⎜⎜ ⎟⎟
⎝ d ⎠ ⎝ s ⎠
⎛ν e ⎞ ⎛ν µ ⎞
⎜⎜ − ⎟⎟, ⎜⎜ − ⎟⎟
⎝ e ⎠ ⎝ µ ⎠
d → u + e− + ve ; π −(du ) → µ− + vµ ; K −(su ) → µ− + vµ
¡
Symmetry à gud=gcs=gW Allowed : d + u → W − ; s + c → W −
Forbidden : s + u → W − ; d + c → W −
But : K − decay observed!
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14/05/14
à quark mixing! F. Ould-Saada
25
¡
Quarks are mixed and Lepton-­‐quark symmetry applies to doublets ⎛ u ⎞ ⎛ c ⎞
⎜⎜ ⎟⎟, ⎜⎜ ⎟⎟
⎝ d ' ⎠ ⎝ s' ⎠
d ' = d cosθC + s sin θC
s' = s cosθC − d sin θC
θC = 13!
¡
¡
14/05/14
With additional couplings, theory agrees with experiment Cabbibo angle extracted from measurements F. Ould-Saada
26
¡
¡
¡
Allowed processes ~ Suppressed processes ~ cos2 θ C ≈ 0.95
sin 2 θ C ≈ 0.05
Charm quark introduced in order to explain the suppression of Flavour Changing Neutral Currents (FCNC) see section 6.7 14/05/14
F. Ould-Saada
GIM mechanism – 1971
Charm discovery – 1974
27
¡
Weak interaction does not produce “bound states”. This is consequence of: §  its weakness compared to EM and strong interactions. §  very small range compared to gravitation ¡
3 vector bosons mediate WIs: §  W± (charged currents) , Z˚ (neutral currents) ¡
3 types of processes §  Leptonic, semi-­‐leptonic, hadronic ¡
Leptonic ¡
semi-­‐leptonic CC: µ − → e−ν µν e ; ν µe− → ν e µ− ; NC :
ν µe− → ν µe−
CC : n → p + e − + ν e d →u + e − + ν e
" p"→n + e + + ν e "u"→d + e + + ν e
€
¡
hadronic Λ0 → p + π −
s→u+u +d
Characteristics: Long decay times and flavour violation
¡
2 à 3 quark generations §  more complicated Quark mixing scheme ! d ' $ ! Vud
#
& = ##
" s' % " Vcd
¡
Vus $! d $ ! cosθC
&#
& = ##
&
Vcs %" s % " – sin θC
sin θC $! d $
&#
&
cosθC &%" s %
⎛ u ⎞ ⎛ c ⎞ ⎛ t ⎞
⎜⎜ ⎟⎟ ,⎜⎜ ⎟⎟ ,⎜⎜ ⎟⎟
⎝ d ⎠ ⎝ s ⎠ ⎝ b ⎠
⎛ νe ⎞ ⎛ ν µ ⎞ ⎛ τ µ ⎞
⎜⎜ − ⎟⎟ ,⎜⎜ − ⎟⎟ ,⎜⎜ − ⎟⎟
⎝ e ⎠ ⎝ µ ⎠ ⎝ τ ⎠
GIM mechanism generalised by Kobayashi and Maskawa (1972) §  Cabibbo-­‐Kobaiashi-­‐Maskawa (CKM) 3X3 matrix §  Unitary: VV+= 1 ¡
Vij: transition probability ià j by W emission §  3 mixing angles and an imaginary phase to account for CP-­‐violation à section 6.6.5 §  CKM prediction of quarks b,t … before c was discovered! ⎛ d '⎞ ⎛ Vud
⎜ s ' ⎟ = ⎜ V
⎜ ⎟ ⎜ cd
⎝ b ' ⎠ ⎝ Vtd
Vus Vub ⎞ ⎛ d ⎞
Vcs Vcb ⎟ ⎜ s ⎟
⎟ ⎜ ⎟
Vts Vtb ⎠ ⎝ b ⎠
Vij (i = u, c, t; j = d , s, b)
http://nobelprize.org/nobel_prizes/physics/laureates/2008/index.html ¡  Continue first at slide 60 to further introduce CKM matrix and quark mixing … ¡
Why is Σ+ decay suppressed/forbidden? §  Selection rules Σ − →ne −v e
Σ − (1197) ≡ dds
sdd → udd + e−ve
Σ + →ne +v e
Σ + (1189) ≡ uus
uus → udd + e+ ve
ΔS = − ΔQ = ±1
→ forbidden decays
Γ(Σ + →ne +v e )
< 5 × 10 −3
−
−
Σ →ne v e
ΔS = 0, ΔQ = ±1
+
−
u → d + W ; W → d + u , ...
ΔS = ΔQ = ±1
+
−
u → s + W ; W → s + c , ...
14/05/14 F. Ould-Saada
€
32
§  Pure hadronic decays ΔS = 0, ± 1
Ω− (1672) = sss ( S = −3)
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33
v e e − → ve e −
¡
¡
v e e − → ve e −
14/05/14
F. Ould-Saada
¡
W couples only to LH fermions Neutrinos have negative helicity, idem for relativistic electrons à cm spin/momentum configuration à isotropy 34
¡
Matrix element: GF 4π(!c)3αW
( 1.54 ) ⇒
=
2
(M W c2 )2
2
Qmax
<< MW2 c 2 ⇒
f (ve e− → ve e− ) = −2 2GF
dσ
1 GF2 2
−
( 1.74 ) ⇒
(νe e ) = 2
ECM
4
dΩ
4π (!c)
2
CM
HE: E
Cross section increases linearly with Eν
¡
14/05/14
F. Ould-Saada
2me c 2GF2
≈ 2me c Eν ⇒ σ tot(νe e ) =
Eν
4
π(!c)
2
−
35
¡
cm spin/
momentum configuration v e e − → ve e −
§  Initial state: Jz=+1 ; Final state: Jz=-­‐1
§  Scattering through 180o forbidden §  amplitude with factor (1+cosθ) 14/05/14
F. Ould-Saada
dσ
1 GF2
−
2
2
(ν e e ) =
E
(1+
cos
θ
)
dΩ
16π 2 (!c) 4 CM
2
HE: E CM
≈ 2me c 2 Eν
1
⇒ σ tot (ν e e − ) = σ tot (ν e e − )
3
36
¡
¡
Quark densities u(x), d(x), ... Ignore strangeness and consider isoscalar nucleus 2
2
F
4
m cG
1
σtot (ν e N) = N
Eν (H + H )
π(!c)
3
mN c 2GF2
1
σtot (ν e N) =
E
(
H + H)
ν
4
π(!c)
3
¡
¡
¡
14/05/14
1
∫ x[u(x) + d(x)]dx
H ≡ ∫ x[u (x) + d (x)]dx
H≡
0
1
0
y=
H
σ (ν N) 1+ 3y
⇒ R ≡ tot e =
H
σtot (ν e N) 3 + y
R~constant~0.51 €of anti-­‐quarks in nucleon à y~0.2 à level of existence Other experiments à y~15-­‐20% F. Ould-Saada
37
σ ∝E
σ (ν N )
≈2
σ (ν N )
14/05/14
F. Ould-Saada
38
¡
Meson decays §  allow sensitive tests of CP violation ¡
Application of basic QM §  leads to surprising effects: test of symmetry between particles and antiparticles ¡
Particle mixing: neutral K-­‐, B-­‐ and D-­‐ mesons ¡
CP invariance §  What is a neutral kaon? 14/05/14
F. Ould-Saada
K 0 (498) ≡ ds ; K 0 ≡ sd
S = +1
S = −1
39
Charged kaons §  interpreted the two particles “θ+(θ-)”and “τ+(τ-)” as different decay channels of K+ (K-­‐). ▪  The θ-τ puzzle led to parity non-­‐conservation. ¡
θ + → π +π 0 ; τ + → π +π +π −
+
+
0
P(θ ) = P(π π ) = +1
K + → π +π 0
BR = 21.16 ± 0.14%
K + → π +π +π − BR = 5.59 ± 0.05%
P(τ + ) = P(π +π +π − ) = −1
θ 0 → π 0π 0
; π +π −
¡
What are the analogue neutral decays? ¡
The question is how to assign particles θ0 and τ0 to states K0 and K0bar?
§  Production (through strong interactions) and decay (through WIs) of neutral
kaons:
K − + p →K 0 + n €
€
€
K + + n →K 0 + p
π − + p = Λ0 + K 0
τ 0 → π 0π 0π 0 ; π +π −π 0
¡
¡
What is the lifetime of the neutral kaon? Proper decays length? ▪  A well defined eigenstate should have one characteristic decay time (one exponential). ▪  Instead, the data show two different lifetimes! Interpretation possible if we assume K0 and K0 bar as superposition of two states with different lifetimes 0
0
0 0
+ −
−10
K1 ≡ θ → π π
;
π π
K 20 ≡ τ 0 → π 0π 0π 0 ; π +π −π 0
τ1 ≅ 0.9 × 10
τ 2 ≅ 5 × 10 −8 s
s
K10 ≡ θ 0 → π 0π 0
;
π +π −
K 20 ≡ τ 0 → π 0π 0π 0 ; π +π −π 0
The neutral kaons K0 and K0 are 2 distinct particles
(as€
they have different strangeness QNs), that can
both decay into two pions (or three pions)
è Possibilities of oscillations!
τ1 ≅ 0.9 × 10 −10 s
τ 2 ≅ 5 × 10 −8 s
¡
Parity and Charge conjugation π 0 π 0 and π +π - ⇒ P = Pπ2 (−1)L = +1#%
$
2
⇒ Cπ 0 = +1
%&
⇒ CP(ππ ) = +1
! !
! !
π π π and π π π : L ≡ L12 + L3 = 0 ⇒ L12 = L3 $
&
L3
3
L12
⇒ P = Pπ (−1) (−1) = −1
% ⇒ CP(πππ ) = −1
&
⇒ Cπ3 0 = +1
'
0
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0
0
+
-
0
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43
K10 =
1 "
$
2#
K 0 + K 0 '&
K 20 =
1 "
$
2#
K 0 − K 0 '&
%
K 0 and K 0
%
PK
¡
=−K
¡
; PK
0
=−K
0
K10 and K 20
are eigenstates of
CP :
CP K10 = + K10 ; CP K 20 = − K 20
K01 and K02 are like real particles to the weak interaction as are K0 and K0bar to the strong €
0
CP :
CP K 0 = K 0 ; CP K 0 = K 0
C K0 = − K 0 ; C K 0 = − K0
0
are not eigenstates of
CP π 0 π 0 = + π 0 π 0 ; CP π +π − =€
+ π +π −
CP π 0 π 0 π 0 = − π 0 π 0 π 0 ; CP π +π −π 0 = − π +π −π 0
K01 and K02 have different masses! K10 → ππ
K 20 → πππ
+& K S0 ≡ K10
mK − 2mπ ≈ 220MeV $&
% ⇒ Γ1 > Γ 2 → τ 1 < τ 2 ⇒ , 0
&- K L ≡ K 20
mK − 3mπ ≈ 90MeV &'
¡
When neutral K meson first produced by SI, it is definitely either K0 or K0 bar, depending on the strangeness of the reaction π − + p → Λ0 + K 0
¡
Immediately , at the point of creation, K0 is an equal mixture of K01 and K02 §  As K01 has shorter lifetime than K02, after a relatively long time, K02 only survive, which is a mixture of K0 and K0 bar §  We started with K0 and finished with 50% K0 bar! K 0 + N →Λ0 + π
K0 – K0-­‐bar oscillation frequency depends on mass difference between K01 & K02 €
¡  Experiments à ΔM~3.5 x 10-­‐6 eV ~1 part in 1014 , as M(K)~498 MeV ! ¡
¡
K-­‐Kbar oscillations ¡
B-­‐Bbar oscillations ¡
Exercise: Draw a similar diagram for D0 14/05/14
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¡  If CP were conserved, K01 decays into 3 pions, and K02 decays into 2 pions would be forbidden. §  It was such a forbidden decay that was observed in 1964, showing that CP is not conserved! ¡
In the experiment, K0 beam was allowed to travel ~18m to ensure as few K01 present as possible §  The products of the particle decays of K02 beam were then observed in detectors §  Observation: 50 K02 à π+π- out of 23 000 decays! 0
0&
−8
( τ ~ 5 × 10 s ,
K
+
ε
K
2
2
1'
L
.
1+ ε
−3
ε
≈
2
×
10
# 0
1
0&
−11
% K − ε K ( τ ~ 9 × 10
K S0 =
s.
2 $
1
2'
S
1+ ε
/
K L0 =
1
#
%
$
CP violation can occur in 2ways ¡
(a) CP violation by Mixing §  CP-­‐forbidden K01 component in K0L decays via CP-­‐
allowed process K01 à ππ, giving contribution proportional to probability |ε|2 of finding K01 component in K0L
¡
K L0 =
K S0 =
1
1+ ε
2
1
1+ ε
2
2
⎛
⎜
⎝
K 20 + ε K10 ⎞⎟⎠
⎛
⎜
⎝
K10 − ε K 20 ⎞⎟⎠
2
(a) ε (1+ ε ) −1 ≈ ε
2
(b) Direct CP violation §  CP-­‐allowed K02 component in K0L decays via CP-­‐
violating reaction K02 à ππ if (b) neglected:
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1
−
2 2
η± = η00 = ε (1+ ε )
48
¡
What was measured experimentally: M ( K L0 → π +π − )
η± ≡
M ( K S0 → π +π − )
M ( K L0 → π 0π 0 )
η00 ≡
M ( K S0 → π 0π 0 )
η± = (2.236 ± 0.007) × 10 −3
η00 = (2.225 ± 0.007) × 10 −3
ε = (2.232 ± 0.007) × 10 −3
¡
CP violation in K decays occurs dominantly by mixing of the CP eigen states in the physical states §  Direct CP violation also occurs and has been measured 14/05/14
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K 0 →π − + e + + νe
2
K 0 →π + + e − + νe
2
N ± ∝ 1 ± ε (1 + ε ) −1
N+ − N−
A≡ +
= 2ℜe(ε )
−
N +N
2
for pure K L0 beam, neglecting ε terms
2ℜe(ε ) ≈ 3.3 × 10 −3
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¡
The Low Energy Anti-­‐
proton Ring (LEAR) §  Very active in testing discrete symmetries ¡
¡
CPLEAR experiment proved in 1998 that time reversal symmetry, T, is not conserved in weak processes involving K mesons T-­‐invariance would require same probability for the inverse transformations K 0 →K 0
&
K 0 →K 0
¡
What is a neutral B-­‐meson? §  Short lifetime à no beams of B mesons B 0 (5279) ≡ db ; B 0 ≡ bd
§  B-­‐factories Υ(4s): M=10.58GeV, Γ=20MeV B + (5279) ≡ ub ; B − ≡ bu
B˜ = +1 B˜ = −1
e + e − → Υ (4 s ) → Bd0 Bd0 ; B + B −
J
PC
τ B ~ 1.5 ps
=1
−−
Analogof K S0 ,K L0 →BL0 ,B H0
€
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¡
BaBar at PEP-­‐II, SLAC, US ¡
BELLE at KEK-­‐B, Japan AKπ
Γ( B 0 → K −π + ) − Γ( B 0 → K +π − )
≡
Γ( B 0 → K −π + ) + Γ( B 0 → K +π − )
AKπ = −0.095 ± 0.013
B 0 / B 0 → J / ΨK S
Other decays studied where CP violation measured
→ K + K − KS
→ η ' KS
!
t =0 ⇒ K ,p =
¡
!
0 ! &
K , p + K L , p ('
!
! &
1 #
At t ⇒ 2 %$ aS (t) K S0 , p + aL (t) K L0 , p ('
0
1 #
%
2$
−i mα c 2 t / "
aα (t) =
e#
%$%
&
oscillating time factor
0
S
−Γα t / 2"
e#
$&
Probability
§  K0 production in strong interactions à π − + p → K 0 + Λ0
S=
(α = S,L)
1 −Γα t / "
1
a
(t)
=
e
2 α
2
0
0
1
−1
§  After propagation: no particle decays
2
Strangeness oscillations: −1
decreases exponentially with τ = Γα
definite strangeness – mixture of S=1 and S=-­‐1, with oscillating intensities τ S << τ L , forτ S << t < τ L , only K L0 component survives ⇒ equal intensity of K 0 and K 0
!
!
K 0 ,K 0 at shorter times : A0 (t) K 0 , p + A0 (t) K 0 , p
1
1
A0 (t) = (aS (t) + aL (t)) ; A0 (t) = (aS (t) − aL (t))
2
2
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2
1
4
2
1
4
I(K 0 ) ≡ A0 (t) =
I(K 0 ) ≡ A0 (t) =
Δm ≡ mS − mL
[e
[e
]
cos( Δmc t /!)]
−ΓS t / !
+ e −ΓL t / ! + 2e −(ΓS +ΓL ) t / 2! cos( Δmc 2 t /!)
−ΓS t / !
+ e −ΓL t / ! − 2e −(ΓS +ΓL ) t / 2!
2
Δm⋅ τ S = 0.5 ⇒
¡
Intensity of the 2 components! ¡
Experimentally I(K 0 ) = f (distance from K 0source)
K 0 + p → π + + Λ0 , π 0Σ +
⇒ ΔmK = (3.483 ± 0.006) × 10 −12 MeV /c 2
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¡
BaBar §  B0-­‐B0bar oscillations §  ΔmB=mH-­‐mL BaBar :
N (unmixed ) − N ( mixed )
Am =
N (unmixed ) + N ( mixed )
ΔmB = (3.337 ± 0.033) × 10 −10 MeV / c 2
mixed: both mesons decay as B0 or B0bar (because of mixing)
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à Neutral flavoured mesons:
à Mix quantum-mechanically
_
- The states M0 and M0 in which
the mesons are produced by SI
are not the same as the
propagating states M1 and M2
( )
CP M 0 = M 0
CP − conjugate states
M 1 = M 0 + M 0 and M 2 = M 0 − M 0
CP M 0 + M 0 = M 0 + M 0
CP M 0 − M 0 = − M 0 − M 0
(
(
)
) (
)
http://en.wikipedia.org/wiki/Cabibbo–Kobayashi–Maskawa_matrix
¡
CKM quark-­‐mixing matrix Vαi=probability (α!i+W transition)
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¡
CKM standard parameterisation §  1. θ12 around z §  2. θ13 around y §  3. θ23 around x ¡
s12 >> s23 >> s13
Experimental values θ12 = 13.04±0.05° ; θ13 = 0.201±0.011° ; θ23 = 2.38±0.06° ; δ13 = 1.20±0.08 rad. ¡
¡
Wolfenstein parameterisation §  λ = s12 §  Aλ2 = s23 §  Aλ3(ρ − iη) = s13e−iδ Experimental values +0.021
+0.031
+0.015
λ = 0.2257 +0.0009
;
A
=
0.814
;
ρ
=
0.135
;
η
=
0.349
−0.0010
−0.022
−0.016
−0.017
§  CP violation most significant in B-­‐decays: Vub plays important role §  K and D decays: Vcd à CP violation much smaller §  CP observed for K not yet for D §  SM accounts for CP violation with 1 phase δ … However, not enough to account for matter-­‐antimatter asymmetry observed in Universe ¡
Various measurements to determine CKM parameters ¡
Unitarity conditions 2
2
2
1st row
2
2
2
2 nd row
2
2
2
1st column
Vud + Vus + Vub = (0.9992 ± 0.0011)
Vcd + Vcs + Vcb = (0.968 ± 0.181)
Vud + Vcd + Vtd = (1.001 ± 0.005)
(
Γ (D
Γ D0 → K +K –
0
→ π +K –
),
)
D0 = cu
(
(
(
Γ (D
Γ (D
)
)
)
→π π ) V
∝
→π K ) V
2
(
Γ (D
Γ D 0 → π +π –
0
→ π +K –
K + = us
0
+
–
2
cd
+
–
cs
(
Γ (D
Γ D 0 → K +π –
0
→ π +K –
)
)
s12 >> s23 >> s13
π + = ud π − = du
K − = su
2
Γ D 0 → π + K – ∝ Vcs Vud
2
2
Γ D0 → K +K –
Vcs Vus
2
∝
≈
tan
θC ≈ 0.05
2
2
0
+ –
Γ D →π K
Vcs Vud
0
),
)
2
Vud
2
Vud
2
2
≈ tan θC ≈ 0.05
(
Γ (D
Γ D 0 → K +π –
0
+
→π K
–
)∝ V
) V
2
cd
cs
2
Vus
2
Vud
2
≈ tan 4 θC ≈ 0.0025
Complete calculation taking into account phase space (easy) and colour field effects (difficult)
à
(
Γ (D
Γ D0 → K +K –
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0
→ π +K –
) ≈ 0.10; Γ ( D
Γ (D
)
0
→ π +π –
0
→ π +K –
) ≈ 0.04; Γ ( D
Γ (D
)
0
→ K +π –
0
→ π +K –
) < 0.02
)
64
(
(
)
)
Γ B − → D 0 K *−
Ratio :
Γ B − → D0 ρ −
2
2
B − = bu ; D 0 = cu ; ρ − = du ; K *− = su
2
Vus
Γ(B − → D0 K *− ) Vus Vcb
2
∝
=
∝
tan
θc ≈ 0.05
2
2
2
Γ(B − → D0 ρ − ) Vud Vcb
Vud
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Experimental value ~0.05
65
¡
Neutral currents predicted by unified theory of EM and weak forces before discovery in 1973 §  Mediated by neutral Z boson ¡
Glashow, Salam Weinberg à Nobel price 1979 – before discovery of Electroweak bosons W, Z in 1983 14/05/14
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ν µ + e− →ν µ + e−
ν µ + p →ν µ + p + π 0
CERN, 1973: First observation
of “neutral currents" in the
bubble chamber Gargamelle:
an incident neutrino interacts
with an electron in the liquid
and disappears.
¡
Electroweak theory originally introduced to solve problems related to higher order diagrams §  Required by experiments §  Lead to infinite contributions (divergence) ¡
Problems solved when contribution due to Z0 and γ taken into account in unified theory §  Cancellation of divergences: T’Hooft, Veltman: 1999 Nobel prize ¡
Renormalisability is a consequence of a fundamental symmetry – Gauge invariance §  Cancellation follows if 2 fundamental relations hold ▪  Unification condition ▪  anomaly condition 14/05/14
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¡
Unification condition §  Involves Weinberg angle θW §  Coupling constant gZ à strength of neutral current vertices §  Weak and EM coupling constants related ¡
Anomaly condition §  Relates electric charges of leptons and quarks e
= gW sin θW = g Z cosθW
2 2ε 0
MW
cosθW =
, 0 < θW < π2
MZ
∑ Q + 3∑ Q
l
l
q
=0
q
q = u, d , s, c, b, t
3= number of colours
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¡
Standard Model predictions 2(!c) 2 gW2
πα (!c) 3
M =
=
GF
2GF sin 2 θ W
2
W
πα (!c) 3
M =
2GF sin 2 θ W cos 2 θ W
2
Z
€
Neutral current coupling à
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GZ
g Z2
=
2 ( M Z c 2 )2
GZ g Z2 M W2
= 2 ⋅ 2 = sin 2 θW
GF gW M Z
70
¡
Weinberg angle θW §  From comparison of neutral and charged current processes at low energies à prediction of W and Z masses sin 2 θW = 0.227 ± 0.014
M W = 78.3 ± 2.4 GeV / c 2
M Z = 89.0 ± 2.0 GeV / c 2
§  Nowadays: from various EW measurements à 3
2
⎫
πα
(!
c
)
M
2
2
W
MW =
; MZ =
⎬ ⇒
2
2
cos θW ⎭
2GF sin θW
§  Not in agreement with direct measurements of sin 2 θW = 0.2315 ± 0.0001
M W = 77.50 ± 0.03 GeV / c 2
M Z = 88.41 ± 0.04 GeV / c 2
gauge boson masses ▪  GF relation used valid at low energies ▪  If higher orders (loops with W,Z and top) taken into account à agreement between theory and experiment 14/05/14
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¡
Neutral current interactions conserve individual quark numbers §  Strangeness-­‐changing weak NC reactions forbidden to 1st order, §  such as K0àµ+µ- ν lν l Z 0 , l −l − Z 0 , l +l + Z 0
qqZ 0 , qqZ 0 ; q = u, d, c, s, t, b
uuZ 0 , ccZ 0 , d 'd ' Z 0 , s's' Z 0
d 'd ' Z 0 = (d cosθ c + ssin θ c )(d cosθ c + ssin θ c )Z 0
= ddZ 0 cos2 θ c + ssZ 0 sin 2 θ c + (dsZ 0 + sdZ 0 )sin θ c cosθ c
s's' Z 0 = ... ⇒ cancellation ⇒ ddZ 0 , ssZ 0
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σ QED
4πα2 (!c)2
=
3s
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σZ ≈
F. Ould-Saada
α Z2 s(!c)2
2( M Z c2 )4
73
1/ α s (Λ2QCD ) = 0
αs (Q 2 ) = [12π /(33 − 2n f )]/ ln(Q 2 / ΛQCD 2 )]
Λ QCD ~ 0.2 GeV ~ 1 fm
±
¡  Solve exercises pp 213-­‐216 §  6.1, 6.2, 6.3, 6.5, 6.6 §  6.7, 6.8, 6.9, 6.11 ¡  14/05/14
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