the standard model in the lhc era i: histories and...
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The Standard Modelin the LHC era
I: Histories and Symmetries
Fawzi BOUDJEMA LAPTh-Annecy, France
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Introduction
EW theory is the combination of two fundamental principles
Gauge Symmetry Principle
Hidden symmetry or Spontaneous symmetry breaking
This allowsX a correct quantum descriptionX high degree of precision (LEP,SLC,Babar, ...LHC)
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Charge conservation
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Charge conservation
Charge can not just disappear like that!
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Charge conservation
Total net charge is conserved, global charge conservation
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Charge conservation
but for a flicker of a second, as if charge was not conserved
charge can not disappear at one point and reappear instantaneously at another point
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Charge conservation
Must have Local charge conservation
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Charge conservation
A change (in time) in the charge density is accompanied by a current flow
Local conservation of the electric charge
Charge must be conserved locally
continuity equation of the electric charge
∂ρ/∂t+ ∇.j = 0
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Introduction: In the beginning, there was light! Prior to Maxwell
div−→E = ∂iEi = ρ (Gauss)
div−→B = 0 (no magnetic charge)
Curl−→E = ǫijk∂jEk = −∂
−→B
∂t(Faraday)
Curl−→B =
−→j (Ampere)
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Introduction: In the beginning, there was light! Prior to Maxwell
div−→E = ∂iEi = ρ (Gauss)
div−→B = 0 (no magnetic charge)
Curl−→E = ǫijk∂jEk = −∂
−→B
∂t(Faraday)
Curl−→B =
−→j (Ampere)
Unfortunately the mathematics implies that
div(Curl−→B ) = ∇.(∇× B) = 0 ⇒ ∇.j = divj = 0
∇.j = divj = 0
in conflict with the continuity equation, local conservatio n.
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Introduction: Electromagnetism as a prototype
Maxwell equations: Unify−→E and
−→B
Local conservation of the electric charge∂j = 0 jµ = (ρ,
−→j )
div−→E = ρ div
−→B = 0
Curl−→E +
∂−→B
∂t= 0 Curl
−→B − ∂
−→E
∂t=
−→j
(Gauge Invariance) electrostatic field (force) depends only on difference of potential
The quantum (for photons ) is the vector potentialAµ(x) = (V,−→A ),
−→B = Curl
−→A
−→E = −GradV − ∂
−→A/∂t.
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Introduction: Electromagnetism as a prototype
Different Aµ lead to the same physical fields E,B.
Gauge invariance Aµ(x) → Aµ(x) + ∂µΛ(x) .
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Introduction: Electromagnetism as a prototype
Different Aµ lead to the same physical fields E,B.
Gauge invariance Aµ(x) → Aµ(x) + ∂µΛ(x) .
But Gauge invariance Fµν(x) → Fµν(x) .
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Introduction: Electromagnetism as a prototype
Different Aµ lead to the same physical fields E,B.
Gauge invariance Aµ(x) → Aµ(x) + ∂µΛ(x) .
With Aµ the equations can be written in a manifestly covariant form, with
Fµν = ∂µAν − ∂νAµ
Fµν =1
2εµνρσ∂
νFρσ
∂µFµν = jν ∂µF
µν = 0.
All of this can be derived from the Lagrangian
Lem = −1
4FµνF
µν ≡ 1
4
(
(−→E + i
−→B )2 + (
−→E − i
−→B )2
)
.
only two Transverse polarisations/helicity states no longitudinal polarisationEPAM, Taza, Maroc, March. 2011 F. BOUDJEMA, The Standard Model – p. 6/76
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Aside, Equation of motion:
The dynamics, the physics, is encoded in the action
S =
Zd 4x L [φi(x), ∂µφi(x)] .
The principle of least action: δS = 0 when varying δφi leads to the Euler–Lagrange equations
∂L∂φi
− ∂µ
„∂L
∂ (∂µφi)
«= 0
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Introduction: Electromagnetism as a prototype
Fµν ≡ ∂µAν − ∂νAµ =
0BBBBB@
0 −E1 −E2 −E3
E1 0 −B3 B2
E2 B3 0 −B1
E3 −B2 B1 0
1CCCCCA
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Introduction: Electromagnetism as a prototype
The equation of motion for the free field, j = 0, lead to
∂µFµν = 0 ⇒ �Aν − ∂ν
`∂.A
´= 0
The freedom from GI allows us to take the gauge fixing
∂A = 0
Out of the 4 degrees of freedom/components in Aµ, this condition freezes one of them
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Introduction: Electromagnetism as a prototype
The equation of motion for the free field, j = 0, lead to
∂µFµν = 0 ⇒ �Aν − ∂ν
`∂.A
´= 0
The freedom from GI allows us to take the gauge fixing
∂A = 0
Out of the 4 degrees of freedom/components in Aµ, this condition freezes one of themMoreover there is still a lot freedom in ∂A = 0, there is still an invariance, overcounting, dueto
Aµ(x) + ∂µΛ(x) �Λ = 0
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Introduction: Electromagnetism as a prototype
The equation of motion for the free field, j = 0, lead to
∂µFµν = 0 ⇒ �Aν − ∂ν
`∂.A
´= 0
The freedom from GI allows us to take the gauge fixing
∂A = 0
Out of the 4 degrees of freedom/components in Aµ, this condition freezes one of themTo remember: Working in a particular gauge, breaks gauge invariance, in fact it hides the
gauge symmetry, the physics is independent of the gauge fixing
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Introduction: Electromagnetism as a prototype
The equation of motion for the free field, j = 0, lead to
∂µFµν = 0 ⇒ �Aν − ∂ν
`∂.A
´= 0
The freedom from GI allows us to take the gauge fixing
∂A = 0
Out of the 4 degrees of freedom/components in Aµ, this condition freezes one of them
The free field is then
�Aν = 0, k2Aν = 0
Photon is massless, it has 2 transverse polarisations
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Introduction: Electromagnetism as a prototype
The equation of motion for the free field, j = 0, lead to
∂µFµν = 0 ⇒ �Aν − ∂ν
`∂.A
´= 0
The freedom from GI allows us to take the gauge fixing
∂A = 0
Out of the 4 degrees of freedom/components in Aµ, this condition freezes one of them
The free field is then
�Aν = 0, k2Aν = 0
Photon is massless, it has 2 transverse polarisations
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Introduction: Electromagnetism as a prototype
The equation of motion for the free field, j = 0, lead to
∂µFµν = 0 ⇒ �Aν − ∂ν
`∂.A
´= 0
The freedom from GI allows us to take the gauge fixing
∂A = 0
Out of the 4 degrees of freedom/components in Aµ, this condition freezes one of them
The free field is then
�Aν = 0, k2Aν = 0
Photon is massless, it has 2 transverse polarisations
EPAM, Taza, Maroc, March. 2011 F. BOUDJEMA, The Standard Model – p. 8/76
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Introduction: Electromagnetism as a prototype, Musings
Lpem = −1
4FµνFµν − jµAµ.
The Lagrangian is not invariant under the gauge transformat ion Aµ(x) → Aµ(x) + ∂µΛ(x) . But the
equation of motion requires ∂.j = 0.
The action is gauge invariant but what is important is the act ion
ZLpem =
Z − 1
4FµνFµν − jµAµ
!=⇒
ZLpem =
Z − 1
4FµνFµν − jµAµ
!−Z
Λ∂.j
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Introduction: Electromagnetism as a prototype, Musings
Lpem = −1
4FµνFµν +
1
2m2AµAµ − jµAµ.
�Aν +m2Aν − ∂ν`∂.A
´= jν =⇒
m2∂.A = ∂.j
The Lagrangian and the action are not invariant under the gau ge transformation
Aµ(x) → Aµ(x) + ∂µΛ(x) .
But the equation of motion requires ∂.A = 0.
This is the spin-1 condition, no longer a gauge fixing.
Then 3 degrees of freedom. An extra Longitudinal degree of freedom
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Quantization of the EM field
[P,X] = i
To quantize, find the conjugate momenta of each degree of freedom and impose equal time
commutation
Πµ =∂L
∂(∂0Aµ)= Fµ0 =⇒
Π0=0
Fix a gauge for Π0 6= 0
LGpem = −1
4FµνFµν − 1
2ξ
„∂.A
«2
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Gauge Invariance and minimal subtraction, introducing interactions
Recall the Lorentz force acting on a moving particle in a electromagnetic field
~F = q ~E + q~v × ~B
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Gauge Invariance and minimal subtraction, introducing interactions
this is derived from the Hamiltonian
H =1
2m(~p− q ~A)2 + qV ⇔ (H − qV ) =
1
2m(~p− q ~A)2
whereas for the corresponding free field
H =p2
2m
to introduce the interaction one has made minimum substitution
Pµ → Pµ − qAµ
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Gauge Invariance and minimal subtraction, introducing interactions
this is derived from the Hamiltonian
H =1
2m(~p− q ~A)2 + qV ⇔ (H − qV ) =
1
2m(~p− q ~A)2
whereas for the corresponding free field
H =p2
2m
to introduce the interaction one has made minimum substitution
Pµ → Pµ − qAµ
in QM
∂µ → Dµ = ∂µ + iqAµ = ∂µ + ieQAµ
Dµ is the covariant derivative.
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Introduction: Gauge Invariance in Quantum Mechanics
Take Schrdinger’s equation
(1/2m)(−i−→∇)2ψ = i∂ψ/∂t
invariant under a global phase transformation
ψ → exp(iλ)ψ
what about invariance under local phase transformation?
λ→ qΛ(x = (t,−→x ))
Possible only if one introduces a compensating vector field which transforms exactly like Aµ
◭.This prescription gives
(1/2m)“−i−→∇ + q
−→A”2ψ = (i∂/∂t+ qV )ψ .
This equation with interactions is invariant under gauge transformations
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Interaction of electrons
consider free Dirac particle whose equation of motion is
„iγµ∂
µ −m
«ψ =
„i∂/−m
«ψ = 0
easily derived from the matter Lagrangian
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Interaction of electrons
consider free Dirac particle whose equation of motion is
„iγµ∂
µ −m
«ψ =
„i∂/−m
«ψ = 0
easily derived from the matter Lagrangian
LM = ψ
„i∂/−m
«ψ
The interaction is obtained through the covariant derivative, leading to
L = ψ
„iD/−m
«ψ − 1
4FµνF
µν = LM + LG + LI
= ψ
„i∂/− eQA/−m
«ψ + LG
LI = −eJQµ A
µ ∂L∂Aµ
= −eJQµ ∂µJ
µ = 0
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Interaction of electrons
LM = ψ
„i∂/−m
«ψ
The interaction is obtained through the covariant derivative, leading to
L = ψ
„iD/−m
«ψ − 1
4FµνF
µν = LM + LG + LI
= ψ
„i∂/− eQA/−m
«ψ + LG
LI = −eJQµ A
µ ∂L∂Aµ
= −eJQµ ∂µJ
µ = 0
the electromagnetic spin-1/2 current
JQµ = Q ψγµψ
LI = − e JQµ A
µ UNIVERSAL COUPLING
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Interaction of electrons: Gauge Invariance
LM = ψ
„i∂/−m
«ψ
The free Dirac Lagrangian is invariant under a global symmetry : phase transformation
ψ → Uψ U = exp(iλ) U†U = 1 U is unitary
global means rigid, same for all x
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Interaction of electrons: Gauge Invariance
LM = ψ
„i∂/−m
«ψ
The free Dirac Lagrangian is invariant under a global symmetry : phase transformation
ψ → Uψ U = exp(iλ) U†U = 1 U is unitary
global means rigid, same for all x Promoting U to a local symmetry λ→ λ(x) only possible
by using covariant derivatives, compensating gauge field.
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Interaction of electrons: Gauge Invariance
LM → LQED = ψ
„iD/−m
«ψ − 1
4FµνF
µν
For λ = λ(x)
ψ∂/ψ → ψU†∂/
„Uψ
«6= ψ∂/ψ
must have
ψD/ψ → ψU†D/′„Uψ
«= ψD/ψ
D′ = UDU†
„Dψ
«′
= U
„Dψ
«
must require that
ψ → Uψ Aµ → Aµ + ∂µΛ
λ = −eQΛ(x) Universality
QED U(1) Abelian theory
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Fµν as a covariant derivative
[Dµ, Dν ]ψ =
((
∂µ + ieAµ
)(
∂ν + ieAν
)
− µ↔ ν
)
ψ
= ie
(
∂µAν − ∂νAµ + ie[Aµ, Aν ]
)
ψ
= ie
(
∂µAν − ∂νAµ
)
ψ Abelian
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Fµν as a covariant derivative
[Dµ, Dν ]ψ =
((
∂µ + ieAµ
)(
∂ν + ieAν
)
− µ↔ ν
)
ψ
= ie
(
∂µAν − ∂νAµ + ie[Aµ, Aν ]
)
ψ
= ie
(
∂µAν − ∂νAµ
)
ψ Abelian
[Dµ, Dν ] ≡ ieFµν
Fµν → UFµνU†
Tr
(
FµνFµν
)
→ Tr
(
UFµνFµνU†
)
= TrFµνFµν
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Interaction of electrons: Gauge Invariance, Recap
charges: ψ → Q, the antiparticle ψ → −Q the photon Aµ → Q = 0 Lagrangian
density has of course no charge whatsoever, is a true scalar in all respects.
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Interaction of electrons: Gauge Invariance, Recap
charges: ψ → Q, the antiparticle ψ → −Q the photon Aµ → Q = 0 Lagrangian
density has of course no charge whatsoever, is a true scalar in all respects.
The mass term does not break charge (gauge) symmetry
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Interaction of electrons: Gauge Invariance, Recap
ψ has 2 chirality states
ψ = ψL + ψR = PLψ + PRψ PL,R =1
2(1 ∓ γ5)
ψL = ψPR ψR = ψPL
The em current conserves chirality and from the point of view of the gauge interaction eachcomponent ψL,R does not talk to each other
Je.mµ = QLψLγµψL +QRψLγµψL QL = QR = Q
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Interaction of electrons: Gauge Invariance, Recap
ψ has 2 chirality states
ψ = ψL + ψR = PLψ + PRψ PL,R =1
2(1 ∓ γ5)
ψL = ψPR ψR = ψPL
The em current conserves chirality and from the point of view of the gauge interaction eachcomponent ψL,R does not talk to each other
Je.mµ = QLψLγµψL +QRψLγµψL QL = QR = Q
ψL ⇋ ψR through the mass
m
ψRψL + ψLψR
!= mψRψL + h.c m = m⋆
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Interaction of electrons: Gauge Invariance, Recap
ψ has 2 chirality states
ψ = ψL + ψR = PLψ + PRψ PL,R =1
2(1 ∓ γ5)
ψL = ψPR ψR = ψPL
The em current conserves chirality and from the point of view of the gauge interaction eachcomponent ψL,R does not talk to each other
Je.mµ = QLψLγµψL +QRψLγµψL QL = QR = Q
ψL ⇋ ψR through the mass
m
ψRψL + ψLψR
!= mψRψL + h.c m = m⋆
in general ψL and ψR have different transformation properties. When they transform similarly, one has a
vector theory, like in QED
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Chiral limit
in the limit m→ 0 there is another (global) symmetry
ψ → exp(iλ5γ5)
ψLγµψL → ψLγµψL ψRγµψR → ψRγµψR
but
mψψ →6= mψψ
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Current-Current
Jµ Jν
Gµν
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Current-Current
Jµ Jν
GµνGµν is the Green’s function or propagator
.
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Current-Current
Jµ Jν
GµνGµν is the Green’s function or propagator
. FromLG = −1
4FµνF
µν − j.A
∂µFµν = Jν ⇒ Aµ = GµνJ
ν or
G−1µνAµ = Jν ⇒ G−1
µν = �gµν − ∂µ∂ν
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Current-Current
Jµ Jν
GµνGµν is the Green’s function or propagator
. FromLG = −1
4FµνF
µν − j.A
∂µFµν = Jν ⇒ Aµ = GµνJ
ν or
G−1µνAµ = Jν ⇒ G−1
µν = �gµν − ∂µ∂ν
what is the inverse, G?
G−1µνG
νρ = gρµ Gµρdoes not exist !
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Current-Current
Jµ Jν
GµνGµν is the Green’s function or propagator
. FromLG = −1
4FµνF
µν − j.A
∂µFµν = Jν ⇒ Aµ = GµνJ
ν or
G−1µνAµ = Jν ⇒ G−1
µν = �gµν − ∂µ∂ν
what is the inverse, G?
G−1µνG
νρ = gρµ Gµρdoes not exist !
must include gauge fixing
LG = −1
4FµνF
µν − 1
2ξ(∂A)2
Gµν =−ik2
gµν − (1 − ξ)
kµkν
k2| {z }
unphysical longitudinal part
!k.J = 0 ξ = 1 Feynman Gauge
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Current-Current
Jµ Jν
Gµν
The amplitude is
M = e2Jemµ GµνJem
ν = −i e2
k2Jem.Jem
#
Leff. = −e2
2
Jemµ Jem
µ
k2
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Current-Current
Jµ Jν
Gµν
The amplitude is
M = e2Jemµ GµνJem
ν = −i e2
k2Jem.Jem
#
Leff. = −e2
2
Jemµ Jem
µ
k2
Contact interaction
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Current-Current
Jµ Jν
Gµν
The amplitude is
M = e2Jemµ GµνJem
ν = −i e2
k2Jem.Jem
#
Leff. = −e2
2
Jemµ Jem
µ
k2
Fundamental interaction
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Massive case
LG = −1
4FµνF
µν +1
2M2A2
Gµν =−i
k2 −M2
(
gµν − kµkν
M2︸ ︷︷ ︸
longitudinal part
)
Leff. = −e2
2
Jemµ Jem
µ
k2 −M2
Leff. =e2
2
Jemµ Jem
µ
M2≡ GmJ
emµ Jem
µ k2 ≪M2
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Weak Interactions and non Abelian theories
First time we became aware of a new type of interaction was at play was through the
discovery of
β-decay: n→ p+ e− + νe (Fermi 1933)
Difficult to accept and set up theoretically.
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Weak Interactions and non Abelian theories
First time we became aware of a new type of interaction was at play was through the
discovery of
β-decay: n→ p+ e− + νe (Fermi 1933)
Difficult to accept and set up theoretically.
Heisenberg: Thought that n = (p+ e−)
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Weak Interactions and non Abelian theories
First time we became aware of a new type of interaction was at play was through the
discovery of
β-decay: n→ p+ e− + νe (Fermi 1933)
Difficult to accept and set up theoretically.
Heisenberg: Thought that n = (p+ e−)Bohr: Energy not conserved (letter from Gamow to Goudsmit: Well you know that he absolutely does not
like this chargeless, massless little things! Thinks that the continuous beta spectrum is compensated by the
“emittion” of gravitational waves which play the role of neutrinos but are much more physical things!)
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Weak Interactions and non Abelian theories
First time we became aware of a new type of interaction was at play was through the
discovery of
β-decay: n→ p+ e− + νe (Fermi 1933)
Difficult to accept and set up theoretically.
Heisenberg: Thought that n = (p+ e−)Bohr: Energy not conserved (letter from Gamow to Goudsmit: Well you know that he absolutely does not
like this chargeless, massless little things! Thinks that the continuous beta spectrum is compensated by the
“emittion” of gravitational waves which play the role of neutrinos but are much more physical things!)
Fermi’s article to Nature rejected: ”contains speculations too remote from reality to be of interest to the
reader”...
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Weak Interactions
It was also found that
β-decay: n→ p+ e− + νe semi-leptonic decay,
muon decay: µ− → e− + νe + νµ leptonic decay
muon capture: µ− + p→ n+ νµ
were of the same nature and have the same strength.
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Weak Interactions
It was also found that
β-decay: n→ p+ e− + νe semi-leptonic decay,
muon decay: µ− → e− + νe + νµ leptonic decay
muon capture: µ− + p→ n+ νµ
were of the same nature and have the same strength.
Universality, Gauge Interaction?
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Weak Interactions
It was also found that
β-decay: n→ p+ e− + νe semi-leptonic decay,
muon decay: µ− → e− + νe + νµ leptonic decay
muon capture: µ− + p→ n+ νµ
were of the same nature and have the same strength.
Universality, Gauge Interaction?
But important differences with QED: they involve
a change in the identity of the fermion
only left-handed field/component were found to interact.
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Weak Interactions a
Fermi postulated a current-current interaction
LFermi = −4GF√2J+
µ Jµ − GF√
2= 1.03510−5M−2
P
Jµ = Lleptonsµ +Hhadrons
µ
Structure of the current was purely (V −A): Parity violation
restrict myself to leptonic current
J−µ = eγµ
(1 − γ5)
2νe + µγµ
(1 − γ5)
2νµ + τγµ
(1 − γ5)
2ντ
= eγµνeL+ · · ·
J+µ = (J+
µ )† = νe LγµeL + · · ·analogy Jem
µ = eLγµeL + eRγ
µeR
for em same entity e ⇆ e here eL ⇆ νe looks like it is not the same entity, some charge notconserved. Nope.Make it EL ⇆EL
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Weak Current
Jµ = eγµνeL + · · ·
Jµ = EL γµ ? EL
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Doublets and Isospin:
Define the doublet
EL =
(
νeL
eL
)
EL =
(
νeLeL
)
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Doublets and Isospin:
Define the doublet
EL =
(
νeL
eL
)
EL =
(
νeLeL
)
J+µ = νeLγµeL =
(
νeLeL
)
γµ
(
0 1
0 0
) (
νeL
eL
)
=1√2EL γµ
︸︷︷︸
spin
τ+︸︷︷︸
weak isospin
EL
=√
2EL γµT+L EL
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Doublets and Isospin:
This is the same maths are your spin/ 2-level system in QM
(
↑↓
)
QM of rotations:
O(3) ∼ SU(2)
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Doublets and Isospin:
Pauli matrices, fundamental representationPauli matrices, 3 generators
τ1 =
0 1
1 0
!τ2 =
0 −ii 0
!τ3 =
1 0
0 −1
!
τ± =1√2
„τ1 ± iτ2
«
"τi
2,τj
2
#= ifijk
τk
2= iǫijk
τk
2fijkstructure constants
X
i
(titi)ab = CF δab =N2 − 1
2Nδab =
3
4δab
X
ij
fijkfijl = CAδkl = Nδkl = 2δkl
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Neutral weak current
where is τ3??
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Neutral weak current
We are forced to consider the group SU(2)w:
there is no group of continuous transformation that has only 2 generators
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Neutral weak current
We are forced to consider the group SU(2)w:
there is no group of continuous transformation that has only 2 generators
J+µ = EL γµ
τ+
√2EL →
J3µ = EL γµ
τ3
2EL =
1
2
νeL
γµνeL− eLγµeL
!
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Neutral weak current
We are forced to consider the group SU(2)w:
there is no group of continuous transformation that has only 2 generators
J+µ = EL γµ
τ+
√2EL →
J3µ = EL γµ
τ3
2EL =
1
2
νeL
γµνeL− eLγµeL
!
This is a neutral current but it is not the em current, this would have been too good
neutrinos have no em charge
no eR, P violation
strength G! This current is much weaker at lower energies than the em current and istherefore very difficult to detect at those earlier energies
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Neutral weak current
We are forced to consider the group SU(2)w:
there is no group of continuous transformation that has only 2 generators
J+µ = EL γµ
τ+
√2EL →
J3µ = EL γµ
τ3
2EL =
1
2
νeL
γµνeL− eLγµeL
!
This is a neutral current but it is not the em current, this would have been too good
neutrinos have no em charge
no eR, P violation
strength G! This current is much weaker at lower energies than the em current and istherefore very difficult to detect at those earlier energies
However part of Jemµ is contained in J3
µ
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The rest of the current and the hypercharge current
J3µ = EL γµ
τ3
2EL =
1
2
νeL
γµνeL− eLγµeL
!
Jemµ (Q) = Q
eLγµeL + eRγµeR
!= −
eLγµeL + eRγµeR
!
= − eRγµeR
!+ J3
µ − 1
2
νeL
γµνeL+ eLγµeL
!
= J3µ − 1
2ELγµIEL −
eRγµeR
!= J3
µ + Yµ
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The rest of the current and the hypercharge current
J3µ = EL γµ
τ3
2EL =
1
2
νeL
γµνeL− eLγµeL
!
Jemµ (Q) = Q
eLγµeL + eRγµeR
!= −
eLγµeL + eRγµeR
!
= − eRγµeR
!+ J3
µ − 1
2
νeL
γµνeL+ eLγµeL
!
= J3µ − 1
2ELγµIEL −
eRγµeR
!= J3
µ + Yµ
Yµ is the hypercharge current
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The rest of the current and the hypercharge current
J3µ = EL γµ
τ3
2EL =
1
2
νeL
γµνeL− eLγµeL
!
Jemµ (Q) = Q
eLγµeL + eRγµeR
!= −
eLγµeL + eRγµeR
!
= − eRγµeR
!+ J3
µ − 1
2
νeL
γµνeL+ eLγµeL
!
= J3µ − 1
2ELγµIEL −
eRγµeR
!= J3
µ + Yµ
Yµ = yeR(eRγµeR) + yEL
(ELγµIEL)
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The rest of the current and the hypercharge current
J3µ = EL γµ
τ3
2EL =
1
2
νeL
γµνeL− eLγµeL
!
Jemµ (Q) = Q
eLγµeL + eRγµeR
!= −
eLγµeL + eRγµeR
!
= − eRγµeR
!+ J3
µ − 1
2
νeL
γµνeL+ eLγµeL
!
= J3µ − 1
2ELγµIEL −
eRγµeR
!= J3
µ + Yµ
Yµ = yeR(eRγµeR) + yEL
(ELγµIEL)
This is indeed a U(1) currentEL is a doublet and there is a separate entity eR which is a singlet (under SU(2)) each
entity has its own hyperchargeyeR
= −1 yeL= −1/2
Q = T3 +Y
2=τ3
2+ y
QeR= 0 − 1 = −1 QeL
= −1
2− 1
2= −1
QνL= +
1
2− 1
2= −1
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Unitarity
For the effective QED operator and with α = e2/4π
Leff. = − e2
2
Jemµ (e+, e−)Jem
µ (µ+, µ−)
k2
The cross section e+e− → µ+µ− behaves as
σ ∝ α2/s
decreases as the energy decreases. Less probability, P, to produce muons.
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Unitarity
For the effective QED operator and with α = e2/4π
Leff. = − e2
2
Jemµ (e+, e−)Jem
µ (µ+, µ−)
k2
The cross section e+e− → µ+µ− behaves as
σ ∝ α2/s
decreases as the energy decreases. Less probability, P, to produce muons.For the Fermi interaction
LFermi = −4GF√2J+
µ (eνe)Jµ −(µ, νµ)
GF√2
= 1.03510−5M−2P
The cross section e−νe → µ−νµ behaves as
σ ∝ G2F × s
The probability P increases indefinitely.
But P < 1, unitarity must be preserved.
This means something must happen at some energy to restore P < 1
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Unitarity
Fermi Contact interaction
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Unitarity
Fermi Contact interaction
Where is the underlying fundamental interaction?
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Gauging, the vector bosons
To each current associate a vector particle, spin-1, a gauge particle
QED recap
Jemµ −→ eJµ(Q)Aµ L = −eJµ(Q)Aµ
Turn the derivatives of the free Lagrangian into covariant derivatives to get the interactionwith the gauge field
Lint,QED = iψeγµ
∂µ + ieQAµ
!ψe
Universality
e↔ Aµ
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Gauging, the vector bosons
To each current associate a vector particle, spin-1, a gauge particle
Ji=±,3µ −→ W i=±,3
µ −→ g
Yµ −→ Bµ −→ g′
g 6= g′ (partial unification)
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Gauging, the vector bosons
To each current associate a vector particle, spin-1, a gauge particle
L = iELγµ
∂µI + ig
„τ i
2
«W i
µ + ig′(yEL)IBµ
!EL
+ ieRγµ
∂µ + ig′(yeR
)Bµ
!eR
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Gauging, the vector bosons
To each current associate a vector particle, spin-1, a gauge particle
L = iELγµ
∂µI + ig
„τ i
2
«W i
µ + ig′(yEL)IBµ
!EL
+ ieRγµ
∂µ + ig′(yeR
)Bµ
!eR
τ iW i = τ3W 3 + τ1W 1 + τ2W 2 = τ3W 3 + τ+W+ + τ−W−
W±µ =
W 1µ ∓ iW 2
µ√2
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Gauging, the vector bosons
To each current associate a vector particle, spin-1, a gauge particle
L = iELγµ
∂µI + ig
„τ i
2
«W i
µ + ig′(yEL)IBµ
!EL
+ ieRγµ
∂µ + ig′(yeR
)Bµ
!eR
τ iW i = τ3W 3 + τ1W 1 + τ2W 2 = τ3W 3 + τ+W+ + τ−W−
W±µ =
W 1µ ∓ iW 2
µ√2
Lcc = − g√2
J+
µ W+ + J−
µ W−,µ
!
W± have electric charge, they should couple to the photon
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The weak mixing angle (unorthodox way)
Counting the gauge fields one has 4 3 of the triplet W and B.
The photon must emerge as the physical field Aµ of W 3 −B. The other orthogonal physicalfield is the Zµ boson. Two requirements, i) Aµ couples to the em current ii) with strength e.
LNC = − gJ3
µW3 + g′YµB
µ
!= −eJµ(Q)Aµ − gZJ
Zµ Z
µ
0@ W 3
µ
Bµ
1A =
0@ cos θW sin θW
− sin θW cos θW
1A0@ Zµ
Aµ
1A .
using JQµ = J3
µ + Yµ
g sin θW = g′ cos θW = e
JZµ = J3
µ − sin2 θW JQµ ; gZ =
g
cos θW=
g
cos θW sin θW
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Feynman Rules, coupling g, sW unspecified as yet
γ
f fe Q
f2
Z
f f
θ θs ce
f f(v − a )γ5 23/2
W
l νl−
5(1− γ )
g
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Cross sections
f−e
e+
f
θ− e+
f f
eγ , Z
e–
e+
µ–
µ+
γ, Z e–
e+
ν
ν
Z
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Weak Interactions
It was also found that
β-decay: n→ p+ e− + νe semi-leptonic decay,
muon decay: µ− → e− + νe + νµ leptonic decay
muon capture: µ− + p→ n+ νµ
were of the same nature and have the same strength.
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Weak Interactions
It was also found that
β-decay: n→ p+ e− + νe semi-leptonic decay,
muon decay: µ− → e− + νe + νµ leptonic decay
muon capture: µ− + p→ n+ νµ
were of the same nature and have the same strength.
Universality, Gauge Interaction?
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Weak Interactions
It was also found that
β-decay: n→ p+ e− + νe semi-leptonic decay,
muon decay: µ− → e− + νe + νµ leptonic decay
muon capture: µ− + p→ n+ νµ
were of the same nature and have the same strength.
Universality, Gauge Interaction?
But important differences with QED: they involve
a change in the identity of the fermion
only left-handed field/component were found to interact.
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Weak Interactions
It was also found that
β-decay: n→ p + e− + νe semi-leptonic decay,
muon decay: µ− → νµ + e− + νe leptonic decay
were of the same nature and have the same strength.
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Weak Interactions
It was also found that
β-decay: n→ p + e− + νe semi-leptonic decay,
muon decay: µ− → νµ + e− + νe leptonic decay
Old writing of the (isospin) doublet
p
n
!
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Weak Interactions
It was also found that
β-decay: n→ p + e− + νe semi-leptonic decay,
muon decay: µ− → νµ + e− + νe leptonic decay
Old writing of the (isospin) doublet
p
n
!
Modern writing of the (isospin) doublet
p = (uud)
n = (udd)
!
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Weak Interactions
It was also found that
β-decay: n→ p + e− + νe semi-leptonic decay,
muon decay: µ− → νµ + e− + νe leptonic decay
Old writing of the (isospin) doublet
p
n
!
Modern writing of the (isospin) doublet
p = (uud)
n = (udd)
!EW=⇒
u
d
!
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Weak Interactions
It was also found that
β-decay: n→ p + e− + νe semi-leptonic decay,
muon decay: µ− → νµ + e− + νe leptonic decay
Old writing of the (isospin) doublet
p
n
!
Modern writing of the (isospin) doublet
p = (uud)
n = (udd)
!EW=⇒
u
d
!
Qu = 2/3 Qd = −1/3
u
d
!
L
and singlets uR; dR (both R have hypercharge)
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Feynman Rules, coupling g, sW unspecified as yet
γ
f fe Q
f2
Z
f f
θ θs ce
f f(v − a )γ5 23/2
W
quqd g(1− γ )
5
LZNC = − e
2 sin θW cos θWZµ
X
f
fγµ(vf−afγ5) f , af = T f3 vf = T f
3
`1 − 4|Qf | sin2 θW
´
u d νe e
2 vf 1 − 83
sin2 θW −1 + 43
sin2 θW 1 −1 + 4 sin2 θW
2 af 1 −1 1 −1
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Cross sections
f−e
e+
f
θ− e+
f f
eγ , Z
e–
e+
q
q
γ, Z
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Inclusion of hadrons
When the precision got better (1960) it was in fact found that the Fermi constant in βdecay was 3% smaller than that measured in muon decay!
β-decay: d→ u + e− + νe semi-leptonic decay,
muon decay: µ− → νµ + e− + νe leptonic decay
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Inclusion of hadrons
When the precision got better (1960) it was in fact found that the Fermi constant in βdecay was 3% smaller than that measured in muon decay!
β-decay: d→ u + e− + νe semi-leptonic decay,
muon decay: µ− → νµ + e− + νe leptonic decay
Moreover at that time we knew of strange particles, Λ ≡ (uds) for example.Λ seemed to decay like the neutron but the associated effective coupling was measuredmuch smaller: strangeness suppression!
Λ → p+ e− + νe ≡ s → u+ e− + νe
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Inclusion of hadrons
When the precision got better (1960) it was in fact found that the Fermi constant in βdecay was 3% smaller than that measured in muon decay!
β-decay: d→ u + e− + νe semi-leptonic decay,
muon decay: µ− → νµ + e− + νe leptonic decay
Moreover at that time we knew of strange particles, Λ ≡ (uds) for example.Λ seemed to decay like the neutron but the associated effective coupling was measuredmuch smaller: strangeness suppression!
Λ → p+ e− + νe ≡ s → u+ e− + νe
breakdown of universality? gauge principle ???
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Inclusion of hadrons
When the precision got better (1960) it was in fact found that the Fermi constant in βdecay was 3% smaller than that measured in muon decay!
β-decay: d→ u + e− + νe semi-leptonic decay,
muon decay: µ− → νµ + e− + νe leptonic decay
Moreover at that time we knew of strange particles, Λ ≡ (uds) for example.Λ seemed to decay like the neutron but the associated effective coupling was measuredmuch smaller: strangeness suppression!
Λ → p+ e− + νe ≡ s → u+ e− + νe
breakdown of universality? gauge principle ???
Universal coupling but apparent non universality due to mixing, cf Z,W
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Inclusion of hadrons
J−µ = cos θcuγµPLd+ sin θc
∆s6=0z }| {uγµPLs
= uγµPL
„cos θcd+ sin θcs
«
| {z }d′
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Inclusion of hadrons
J−µ = cos θcuγµPLd+ sin θc
∆s6=0z }| {uγµPLs
= uγµPL
„cos θcd+ sin θcs
«
| {z }d′
cos θc = Vud = 0.97
sin θc = Vus = 0.24
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Inclusion of hadrons
J−µ = cos θcuγµPLd+ sin θc
∆s6=0z }| {uγµPLs
= uγµPL
„cos θcd+ sin θcs
«
| {z }d′
with d′ universality reinstated
use weak current eigenstate
UL =
u
d′
!
L
what to do with the orthogonal state to d′?
s′L = − sin θcd+ cos θcs
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Flavour changing neutral currents
J±µ = ULγµτ
+UL
⇓
J3µ =
1
2
„uLγµuL − d′Lγµd
′L
«
=1
2
uLγµuL − cos2 θcdLγµdL − sin2 θcsLγµsL − sin θc cos θc
„dLγµsLsLγµdL| {z }∆S=1 , FCNC
«!
of course one requires the em current to be diagonal
Jµ(Q) =2
3uγµu− 1
3
„dγµd+ sγµs
«
= J3µ + Yµ
ex. find all quantum numbers of s including ...sR
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GIM mechanism
∆S = 1 lead to FCNC K0(ds) → µ+µ− occurs at tree-level and leads to a large rate for thisdecay! But experimentally
BKµ =Γ(K0 → µ+µ−)
K+ → µ+νµ∼ 10−8
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GIM mechanism
∆S = 1 lead to FCNC K0(ds) → µ+µ− occurs at tree-level and leads to a large rate for thisdecay! But experimentally
BKµ =Γ(K0 → µ+µ−)
K+ → µ+νµ∼ 10−8
The sdZ coupling must be eliminated!
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GIM mechanism
∆S = 1 lead to FCNC K0(ds) → µ+µ− occurs at tree-level and leads to a large rate for thisdecay! But experimentally
BKµ =Γ(K0 → µ+µ−)
K+ → µ+νµ∼ 10−8
GIM=Glashow Illiopoulos Maiani postulate that a cousin of the u exists, the c quark thatshould form a doublet with s′L
CL =
u
s′
!
L
with this new entry J3µ is diagonal and no FCNC occur!
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GIM mechanism
∆S = 1 lead to FCNC K0(ds) → µ+µ− occurs at tree-level and leads to a large rate for thisdecay! But experimentally
BKµ =Γ(K0 → µ+µ−)
K+ → µ+νµ∼ 10−8
GIM=Glashow Illiopoulos Maiani postulate that a cousin of the u exists, the c quark thatshould form a doublet with s′L
CL =
u
s′
!
L
with this new entry J3µ is diagonal and no FCNC occur!
J+µ =
„u, c
«
L| {z }family space
γµ τ+VCabbibo
d
s
!)family space
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GIM mechanism
∆S = 1 lead to FCNC K0(ds) → µ+µ− occurs at tree-level and leads to a large rate for thisdecay! But experimentally
BKµ =Γ(K0 → µ+µ−)
K+ → µ+νµ∼ 10−8
GIM=Glashow Illiopoulos Maiani postulate that a cousin of the u exists, the c quark thatshould form a doublet with s′L
CL =
u
s′
!
L
with this new entry J3µ is diagonal and no FCNC occur!
J+µ =
„u, c
«
L| {z }family space
γµ τ+VCabbibo
d
s
!)family space
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With a bonus!
d
s
W−
W+
νµ
µ−
µ+
u
g cos θc
+g sin θc
K0
+
d
s
W−
W+
νµ
µ−
µ+
c
−g sin θc
g cos θc
K0
if mu = mc total cancellation, GIM suppression even at one-loop.if mc ≫ muu rate still too large. J. Ellis and M.K. Gaillard, predicted mc in the range1 − 3GeV to account for the experimental rate.
Loop calculations and masses!
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Generalisation, third family
J+µ =
(
u c t
)
L︸ ︷︷ ︸
gauge eigenstates
γµ τ+VCKM
( d
s
b
)}
family space
uidj
i jV
W
u c t
d s b
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Gauge invariance, non-Abelian theory
G = SU(2)L ⊗ U(1)Y theory
ψ1(x) =
0@ u
d
1A
L
, ψ2(x) = uR , ψ3(x) = dR .
ψ1(x) =
0@ νe
e−
1A
L
, ψ2(x) = νeR , ψ3(x) = e−R .
As in the QED start from the free Lagrangian (no masses)
L0 =3X
j=1
i ψj(x) γµ ∂µψj(x)
L0 is invariant under global G transformations in flavour space:
ψ1(x)G−→ ψ′
1(x) ≡ exp {iy1β} UL ψ1(x), UL ≡ expniτi
2αio
(i = 1, 2, 3)
ψ2(x)G−→ ψ′
2(x) ≡ exp {iy2β} ψ2(x)
ψ3(x)G−→ ψ′
3(x) ≡ exp {iy3β} ψ3(x) ,EPAM, Taza, Maroc, March. 2011 F. BOUDJEMA, The Standard Model – p. 47/76
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Requiring local gauge transformation, αi = αi(x) and β = β(x) we must make ∂µ → Dµ
Dµψ1(x) ≡h∂µ − i gfWµ(x) − i g ′ y1Bµ(x)
iψ1(x) , fWµ(x) ≡ τi
2W i
µ(x)
Dµψ2(x) ≡ [∂µ − i g ′ y2Bµ(x)] ψ2(x) ,
Dµψ3(x) ≡ [∂µ − i g ′ y3Bµ(x)] ψ3(x) ,
Dµψj(x) transforms (covariantly) like ψj(x) ⇒
Bµ(x)G−→ B′
µ(x) ≡ Bµ(x) +1
g ′∂µβ(x),
fWµG−→ fW ′
µ ≡ UL(x)fWµ U†L(x) − i
gUL(x) ∂µU
†L(x).
W iµ
G−→ Wi′µ ≡ W iµ − 1
gαi − (~α× ~W )i (infinitesimal )
This fixes the gauge-matter interaction
L =3X
j=1
i ψj(x) γµ Dµψj(x)
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Pure Radiation, Gauge kinetic Lagrangian
• for the U(1)Y field do the same as with the electromagnetic field.The field strength
Bµν ≡ ∂µBν − ∂νBµ BµνG−→ Bµν ,
LKin,B = −1
4Bµν B
µν
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Pure Radiation, Gauge kinetic Lagrangian
• for the U(1)Y field do the same as with the electromagnetic field.The field strength
Bµν ≡ ∂µBν − ∂νBµ BµνG−→ Bµν ,
LKin,B = −1
4Bµν B
µν
We can not just use W iµν = ∂µW i
ν − ∂νW iµ Note the right gauge transformation.
W i is isospin charged and the normal derivative should be turned into a covariant derivative
W iµν = ∂µW
iν − ∂νW
iµ − g ǫijk W j
µ Wkν
fWµνG−→ UL
fWµν U†L
LKin = −1
4Bµν B
µν − 1
2TrhfWµν
fWµνi
= −1
4Bµν B
µν − 1
4W i
µν Wµνi .
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Non-Abelian Couplings
Trilinear and Quartic Gauge Boson Couplings
at the LHC
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Gauge Invariance: gffV = gV V V
√s
[GeV]
σ(e
+e
−→
W+W
−(γ
)) [
pb
] LEP
only νe exchange
no ZWW vertex
GENTLE
YFSWW3
RACOONWW
Data
√s
≥ 189 GeV: preliminary
0
10
20
160 170 180 190 200
+
e-W+
W-W+e
Z γ,
W-e-
e+
• LEP legacy: We know that WWV can not deviate too much (10%) from SM gauge value.
• But slightest deviations are revealed at higher energies (LHC?)
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Origin of VVVV and VVV: Gauge Invariance
LGauge = −1
2[Tr(WµνW
µν) + Tr(BµνBµν)] GI kinetic term
Wµν =1
2
„∂µWν − ∂νWµ +
i
2g[Wµ,Wν ]
«=τ i
2
“∂µW
iν − ∂νW
iµ − gǫijkW j
µWkν
”
Bµν =1
2(∂µBν − ∂νBµ) Bµ = Bµ
LWWV = −ie(h
Aµ
“W−µνW+
ν −W+µνW−ν
”+ FµνW+µW−ν
i
+ cWsW
hZµ
“W−µνW+
ν −W+µνW−ν
”+ ZµνW+µW−ν
i)No ZZZ,ZZγ, Zγγ
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tri-linear couplings
LWWV = −ie
8><>:
264Aµ
“W−µνW+
ν −W+µνW−ν
”+
κγz }| {(1 + ∆κγ)FµνW+µW−ν
375
+ cWsW
26664
gZ1z }| {
(1 + ∆gZ1 )Zµ
“W−µνW+
ν −W+µνW−ν
”+
κZz }| {(1 + ∆κZ)ZµνW+µW−ν
37775
+ 1
M2W
“λγ F νλ + λZ
cWsW
Zνλ”W+
λµW−µ
ν
ffNo ZZZ,ZZγ, Zγγ
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LWWV1V2 =
− e2
(`AµA
µW+ν W
−ν −AµAνW+µ W
−ν
´
+ 2cW
sW
„AµZ
µW+ν W
−ν − 1
2AµZν(W+
µ W−ν +W+
ν W−µ )
«
+c2Ws2W
`ZµZ
µW+ν W
−ν − ZµZνW+µ W
−ν
´
+1
2s2W
`W+µW−
µ W+νW−ν −W+µW+
µ W−νW−
ν
´)
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Aspects of ColourQCD
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Make sense of the hadron spectrum economically
The Particle Data Book lists a plethora, that might seem as a zoo, of particles in the
hadron family, between the mesons and the baryons. But !
everything fits neatly with much fewer fundamental constitutents of spin-1/2
Quarks that come in 6 flavours
The whole spectrum of hadrons can be constructed out of
Mesons M = qq
Baryons B = qqq
This picture faces a major problem
∆++JZ=+3/2,J=3/2
= u↑u↑u↑
The wave function is symmetric for a fermion! and therefore does not obey spin-statistics.
For such a state to exist, the three u can not be the same, they must carry (at least) 3different quantum numbers.
∆++ = u↑u↑u↑
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Colour
Simplicity and minimality suggests number of colours is
Nc = 3
then each quark is qα, α = 1, 2, 3 =red,green blue.Then ∆++ ∼ ǫαβγ |u↑αu↑βu
↑γ〉.
Baryons and mesons are described by colour singlet construct
B =1√6ǫαβγ |qαqβqγ〉 , M =
1√3δαβ |qαqβ〉
We observe no free quarks and moreover we do not see any combination of states/particles
that carry colour: confinement
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Colour, more evidence
Re+e− ≡ σ(e+e− → hadrons)
σ(e+e− → µ+µ−).
Probability for quarks to hadronize is one and for s≪M2Z
e–
e+
q
q
γ
Re+e− ≈ NC
NfX
f=1
Q2f =
8>><>>:
23NC = 2 , (Nf = 3 : u, d, s)
109NC = 10
3, (Nf = 4 : u, d, s, c)
119NC = 11
3, (Nf = 5 : u, d, s, c, b)
.
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Colour, more evidence
Re+e− ≈ NC
NfX
f=1
Q2f =
8>><>>:
23NC = 2 , (Nf = 3 : u, d, s)
109NC = 10
3, (Nf = 4 : u, d, s, c)
119NC = 11
3, (Nf = 5 : u, d, s, c, b)
.
��������������������������������������������������������������������������������
0
1
2
3
4
5
6
0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
exclusive data
e+e– → hadrons
QCD
γγ2
Crystal Ball
PLUTO
BES
ω Φ J/ψ1S ψ2S
ψ3770
√s (GeV)
R
0
1
2
3
4
5
6
5 6 7 8 9 10 11 12 13 14
e+e– → hadrons
QCD
PLUTO
LENA
Crystal Ball
MD1
JADE
MARK J
ϒ1S ϒ2S 3S 4S
ϒ10860
ϒ11020
√s (GeV)
R
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Colour, more evidence
Re+e− ≈ NC
NfX
f=1
Q2f =
8>><>>:
23NC = 2 , (Nf = 3 : u, d, s)
109NC = 10
3, (Nf = 4 : u, d, s, c)
119NC = 11
3, (Nf = 5 : u, d, s, c, b)
.
RZhad = Nc
Pq v
2q + a2
q
v2µ + a2µ
= 20.095.
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Exact colour symmetry Symm3
Colour is what binds hadrons together.
NC = 3. Quarks belong to the triplet representation 3 of Symm3.
Quarks and antiquarks are different states. Therefore, 3∗ 6= 3.
Confinement hypothesis: hadronic states are colour singlets.
The interaction must be quite strong ρ→ 2π τ ∼ 10−22s compared toµ→ eνeνµ 10−6s
Symm3 = SU(3)C
qq : 3 ⊗ 3∗ = 1 ⊕ 8 ,
qqq : 3 ⊗ 3 ⊗ 3 = 1 ⊕ 8 ⊕ 8 ⊕ 10 ,
qq : 3 ⊗ 3 = 3∗ ⊕ 6 ,
qqqq : 3 ⊗ 3 ⊗ 3 ⊗ 3 = 3 ⊕ 3 ⊕ 3 ⊕ 6∗ ⊕ 6∗ ⊕ 15 ⊕ 15 ⊕ 15 ⊕ 15′ ,
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RecapSU(2) set upSU(3), SU(N)
SU(2)weak isospin
• Matter fields in doublets, La, a=1,2• Generators of the group, τA Paulimatrices, A = 1...(N2 − 1) = 3
• N2 − 1 = 3 gauge bosons• Invariance under U = exp(iτAαA)
.
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RecapSU(2) set upSU(3), SU(N)
SU(2)weak isospin
• Matter fields in doublets, La, a=1,2• Generators of the group, τA Paulimatrices, A = 1...(N2 − 1) = 3
• N2 − 1 = 3 gauge bosons• Invariance under U = exp(iτAαA)
.
SU(3)C
• Matter fields, qa, a=1,2,3• Generators of the group, tA, A =
1...(N2 − 1) = 8
• N2 − 1 = 8 gauge bosons• U = exp(iτAθA)
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RecapSU(2) set upSU(3), SU(N)
SU(2)weak isospin
• Matter fields in doublets, La, a=1,2• Generators of the group, τA Paulimatrices, A = 1...(N2 − 1) = 3
• N2 − 1 = 3 gauge bosons• Invariance under U = exp(iτAαA)
.
SU(3)C
• Matter fields, qa, a=1,2,3• Generators of the group, tA, A =
1...(N2 − 1) = 8
• N2 − 1 = 8 gauge bosons• U = exp(iτAθA)
TR =1
2,
CF =4
3,
CA = 3,
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Gell-Mann Matrices
The fundamental representation T a = λa/2 is N –dimensional. For N = 2, λaare
the usual Pauli matrices, while for N = 3, they correspond to the eight Gell-Mann
matrices:
λ1 =
0BB@
0 1 0
1 0 0
0 0 0
1CCA , λ2 =
0BB@
0 −i 0
i 0 0
0 0 0
1CCA , λ3 =
0BB@
1 0 0
0 −1 0
0 0 0
1CCA
λ4 =
0BB@
0 0 1
0 0 0
1 0 0
1CCA , λ5 =
0BB@
0 0 −i0 0 0
i 0 0
1CCA, λ6 =
0BB@
0 0 0
0 0 1
0 1 0
1CCA,
λ7 =
0BB@
0 0 0
0 0 −i0 i 0
1CCA, λ8 =
1√3
0BB@
1 0 0
0 1 0
0 0 −2
1CCA.
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QCD Lagrangian
From the free Lagrangian
Lquarks =
nfX
i
qai (i6∂ −mi)abq
bi ,
turn to the covariant derivative
Dµ,ab = ∂µ 1ab + igs (t·Aµ)ab,
Aaµ are coloured vector fields, gluons
Lkin = −1
4FA
µνFµνA ,
FAµν = ∂µA
Aν − ∂νA
Aµ − gsf
ABCABµA
Cν ,
LQCD = − 14FA
µνFµνA +
Pni q
ai (i6D −mi)abq
bi − 1
2λ
`∂µAA
µ
´2+ Lghost.
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QCD Feynman rules
abcfGσ
c
Gνb
Gµa
Gνc
adefabcfgs2
G bµ Gσ
d
G eρ
qα
Gµa
q
gs 2γµ
αβa
λ gs
β
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QCD Feynman rules
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QCD Feynman rules
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Quantum numbers
◮
SU(3) SU(2)L U(1)Y Q = T3 + Y
Q = (uL, dL) 3 2 16
(23,−1
3)
uR 3 1 23
23
dR 3 1 −13
−13
L = (νL, eL) 1 2 −12
(0,−1)
eR 1 1 −1 −1
νR 1 1 0 0
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De quoi sommes nous faits ◮ 1
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De quoi sommes nous faits ◮ 1
•
•
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De quoi sommes nous faits ◮ 1
•
•
interaction lectromagnetique (QED)
interaction forte (QCD)
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Interaction faible
vers 1880, D’ou vient l’energie du soleil?
Kelvin, Helmholtz, etc...: ”contraction gravitationnelle
du nuage solaire...”
age de l’Univers (soleil): qq millions d’annees
Darwin (evolution, erosion,...): milliards d’annees pour
la terre!
.
.
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Interaction faible
vers 1880, D’ou vient l’energie du soleil?
Kelvin, Helmholtz, etc...: ”contraction gravitationnelle
du nuage solaire...”
age de l’Univers (soleil): qq millions d’annees
Darwin (evolution, erosion,...): milliards d’annees pour
la terre!
.
.
desintegration βn→ pe−νe
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Interaction faible
vers 1880, D’ou vient l’energie du soleil?
Kelvin, Helmholtz, etc...: ”contraction gravitationnelle
du nuage solaire...”
age de l’Univers (soleil): qq millions d’annees
Darwin (evolution, erosion,...): milliards d’annees pour
la terre!
.
.
reaction nucleaire: fusion
bruler de l’hydrogene:
41H + 2e→4 He+ 2ν + 6γ + Energie(CH4 + 2O2 → 2H2O + chaleur)
desintegration βn→ pe−νe
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La premiere famille
Spin
-1/2
Quarks
Leptons
u
d
νe
e
Mat
iere
Les Forces: Spin-1γ Photon Lumiere, QEDW±, Z Interaction faibleTheorie electrofaible
Gluons, gInteraction forte
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Le ciel encore...
Rayons Cosmiques
µ: muon a part la masse, en tout point comme l’electron
accelerateurs pour produire ces nouvelles particules ou d’autres
pour mieux les etudier
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Et!
Spin
-1/2
Quarks
Leptons
u
d
νe
e
Mat
iere
Spin
-1/2
Quark
sLep
tons
u c td s bνe νµ ντe µ τ1 2 3Generations
Mat
iere
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Spin-0???
Higgs, H ??
Spin-0
Masse
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Le Modèle Standard
Spin-1
γ Photon
Lumier
e,QED
W±, Z
Intera
ction
faible
Theorie
electro
faible
Gluons
Intera
ction
Spin
-1/2
Quark
sLepto
ns
u c td s bνe νµ ντe µ τ1 2 3Generations
Mat
iere
Higgs, H
??
Spin-
0
Masse
Spin
-1
γ
Photon
L
um
iere,
Q
E
D
W
±,Z
Interaction
faible
Theorie
electrofa
ible
Glu
ons
Interaction
Spin-1/2
Quarks
Lepton
su
c
t
d
s
b
νe
νµ
ντ
e
µ
τ
1
2
3
G
en
era
tio
n
s
Matiere
Higgs,
H??
Spin-0
Masse
EPAM, Taza, Maroc, March. 2011 F. BOUDJEMA, The Standard Model – p. 76/76