Post on 27-Feb-2020
Standard Model
Oscar EboliUniversidade de Sao Paulo
Departamento de Fısica Matematicaeboli@fma.if.usp.br
December 4, 2006
PBSM-2006 Oscar Eboli
PLAN
➫ I. Elementary particles
➫ II. Basic tools
➫ III. Model bulding
➫ IV. Comments
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I. Elementary Particles
➫ The known elementary spin-12 fermions are
(νe
e−
)L
e−R
(ud
)RBG
L
uRBGR dRBG
R
PBSM-2006 Oscar Eboli
I. Elementary Particles
➫ The known elementary spin-12 fermions are
(νe
e−
)L
e−R
(ud
)RBG
L
uRBGR dRBG
R
(νµ
µ−
)L
µ−R
(cs
)RBG
L
cRBGR sRBG
R
(ντ
τ−
)L
τ−R
(tb
)RBG
L
tRBGR bRBG
R
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➫ There are also 12 vector particles (forces): γ W± Z Gcc
particle mass (GeV)e± 5.11× 10−4
µ± 0.1057τ± 1.777νi ' 0u (1.5− 3.0)× 10−3
d (3− 7) × 10−3
c 1.25s 95× 10−3
t 174.2b 4.20− 4.70γ < 6× 10−26
W± 80.403Z 91.1876Gcc 0
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➫ There is a large amount of precise data to be understood
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II. Basic Tools
➫ The basic element for the construction of the SM are gauge theories andthe phenomenon of spontaneous symmetry breakdown.
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��Abelian Gauge Theories
➩ The free Dirac Lagrangian
L0 = Ψ(i 6 ∂ −m)Ψ is invariant under Ψ =⇒ Ψ′ = eiα Ψ
➩ For a local transformation α(x): ∂µ(eiαΨ) = eiα (Ψi∂µα+ ∂µΨ)
➩ We introduce a gauge field to make this transformation local
∂µ =⇒ Dµ = ∂µ − ieAµ with Aµ =⇒ A′µ = Aµ +1e∂µα(x)
in such way that
DµΨ =⇒ D′µΨ′ = eiα(x)DµΨ leading to L0 =⇒ L0 + eΨ 6 AΨ
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➩ The kinetic term for the gauge field Aµ is also fixed by its transformationproperties
Fµν = ∂µAν − ∂νAµ is such that Fµν =⇒ Fµν so Lk = −14FµνFµν
➩ The field Aµ is massless since a term m2AµAµ is not gauge invariant.
➩ Notice local gauge invariance =⇒ introduction of a new field with fixedinteractions with the fermions.
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➩ The kinetic term for the gauge field Aµ is also fixed by its transformationproperties
Fµν = ∂µAν − ∂νAµ is such that Fµν =⇒ Fµν so Lk = −14FµνFµν
➩ The field Aµ is massless since a term m2AµAµ is not gauge invariant.
➩ Notice local gauge invariance =⇒ introduction of a new field with fixedinteractions with the fermions.
➩ However this is not the only possibility respecting local invariance, e.g.
ΨσµνΨ Fµν
so we also require that the local gauge invariant interactions arerenormalizable.
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��Non-Abelian Gauge Theories Yang & Mills
➪ Consider the multiplet of a non-abelian group M . For instance, SU(2)
Mi =(M1
M2
)
➪ The free Dirac lagrangian is invariant under the global transformations
L0 = iM 6 ∂M ≡ i∑jk
M j 6 ∂δjkMk
M =⇒M ′ = exp(igtaαa)M ≡ GM
where ta are the representation of the algebra of the group for the multiplet M
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➪ The free lagrangian is not invariant under local transformation since
∂µM =⇒ ∂µ(GM) = G(∂µM) + (∂µG)M
➪ To impose the local invariance we introduce the covariant derivative
Dµ ≡ ∂µ + igtabaµ such that DµM =⇒ D′µM
′ = G(DµM)
which leads to
tabaµ =⇒ (tabaµ)′ = G(tabaµ)G−1 +i
g(∂µG)G−1
➪ The locally invariant lagrangian is
L0 −∑jk
gbaµM jγµ(ta)jkMk
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➪ In analogy with the electromagnetism (abelian case) we define the fieldstrength
taF aµν ≡ ∂µ(tabaν)− ∂ν(tabaµ) + ig[tabaµ, t
fbfν ] =1ig
[Dµ, Dν]
which transforms astaF a
µν =⇒ G(taF aµν)G
−1
➪ The kinetic term for the gauge fields is
Lg = −14F a
µνFa µν
➪ vector fields are massless: m2baµbaµ is incompatible with local gauge
invariance.
➪ There are triple and quartic interactions among the gauge bosons!
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➪ How can we describe the W± and Z that are massive?
��
��Realization of the Symmetry
➫ Fabri & Picasso showed that there are only two ways to realize acontinuous symmetry:
1. The symmetry is manifest, that is, the conserved charged operator Qannihilates the vacuum
Q|0〉 = 0
2. The symmetry is hidden:Q|0〉 6= 0
[in fact Q|0〉 is ill defined]
Exercise: Prove this theorem!
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➫ Therefore the vacuum |0〉 plays a big role in determining the effects of asymmetry!
➫ Goldstone Theorem: states that there are as many massless particles inthe spectrum as the number of generators of global continuous symmetriesthat do not annihilate the vacuum.
➫ Example. Consider a real scalar doublet φa under SO(2)
L =12∂νφ
a∂νφa − 12µ2φaφa − λ
4(φaφa)2
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For µ2 > 0, the vacuum satisfies 〈φa〉 = 0.the symmetry is manifest.
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For µ2 > 0, the vacuum satisfies 〈φa〉 = 0.the symmetry is manifest.
For µ2 < 0 we have that
〈φaφa〉 = − µ2
λ≡ v2
showing that there are manydegenarated vacua.
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For µ2 > 0, the vacuum satisfies 〈φa〉 = 0.the symmetry is manifest.
For µ2 < 0 we have that
〈φaφa〉 = − µ2
λ≡ v2
showing that there are manydegenarated vacua.
Choosing the vacuum state to be
〈φa〉 =(v0
)and writing φa = 〈φa〉+
(ηξ
)leads to
Lquad =12[∂νη∂
νη + 2µ2η2] +12∂νξ∂
νξ
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Exercise The above example worked at the classical level. Prove theGoldstone theorem in general, including quantum effects.
➫ Interesting facts:
• Rotations are generated by eiθσ2
• σ2〈φa〉 6= 0
• We could write the field as
φa = eiχσ2/v
(v + η
0
)
• The field χ turns out to be massless and η has mass −2µ2. Explain!
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��The Higgs–Kibble Mechanism
➬ When local gauge symmetries are hidden many of the massless statesdisappear from the spectrum!
➬ Example: Consider the scalar electrodynamics
L = |Dµϕ|2 − µ2|ϕ|2 − λ|ϕ|4 − 14FµνF
µν
➬ ϕ is a complex field and this model is invariant under a local U(1)
ϕ =⇒ eiqα(x)ϕ and Aµ =⇒ Aµ − ∂µα(x)
➬ For µ2 > 0 there is a unique minimum at ϕ = 0; the symmetry is manifestmA = 0 and m2
ϕ± = µ2
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➬ For µ2 < 0 there are many vacuasatisfying
|〈ϕ〉|2 = − µ2
2λ≡ v2
2
Choosing 〈ϕ〉 = v/2 and defining
ϕ =1√2eiξ/v(v + η) =⇒
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➬ For µ2 < 0 there are many vacuasatisfying
|〈ϕ〉|2 = − µ2
2λ≡ v2
2
Choosing 〈ϕ〉 = v/2 and defining
ϕ =1√2eiξ/v(v + η) =⇒
Lquad =12[∂νη∂
νη + 2λv2η2]+12∂νξ∂
νξ+qvAµ∂µξ +
q2v2
2AµA
µ − 14FµνF
µν
=12[∂νη∂
νη + 2λv2η2] +q2v2
2
(Aµ +
∂µξ
qv
) (Aµ +
∂µξ
qv
)− 1
4FµνF
µν
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➬ To extract the physical spectrum we make the gauge transformation
ϕ→ ϕ′ = e−iξ/v ϕ =1√2(v + η) and Aµ → A′µ = Aµ +
1qv∂µξ
➬ Then we have
Lq =12[∂νη∂
νη − 2λv2η2]− 14F ′
µνF′µν +
q2v2
2A′µA
′µ
➬ The spectrum includes a massive neutral particle and a massive vectorfield.
➬ The number of degrees of freedom is conserved, with the field ξ becomingthe longitudinal component of Aµ.
➬ Notice that the underlying symmetry related the masses mξ =√
2λv andmA = qv with λ.
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➭ Non-abelian case: Let’s consider a complex multiplet M (doublet) of anon-abelian group G (SU(2))
L = −14F a
µνFa µν + [(DµM)a]∗ (DµM)a − µ2Ma∗Ma − λ(Ma∗Ma)2
withDµ ≡ ∂µ + igtabaµ
➭ The vacuum also has 〈Ma∗Ma〉 = −µ2/2λ = v2/2. Choosing (in the SU(2)case)
〈M〉 =(
0v√2
)we can write M = exp
(iξata
v
) (0
v+η√2
)
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➭ The quadratic part of the Lagrangian still contains terms mixing ξa and baµ,which disappear in the unitary gauge
M →M ′ = exp(−iξ
ata
v
)M =
(0
v+η√2
)
and we are left with
Lq =12[∂νη∂
νη − 2λv2η2]− 14F a
µνFaµν +
g2v2
8(|b1µ − ib2µ|2 + b3µb
3µ)
➭ So we are left with 3 massive vector particles and a neutral scalar. Again,the symmetry relates their masses and couplings.
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III. Model Building
➮ Old style “practical” recipe for model building:
1. Choose the local gauge symmetry;
2. choose the matter content and its representation;
3. choose the symmetry breaking pattern;
4. verify whether the model satisfies the theoretical and experimentalconstraints; if not, return to 1.
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Standard Model: electroweak interactions
➮ The SM summarizes all our knowledge on electromagnetic, weak, andstrong interactions.
➮ The strong interactions are described by a local gauge theory based on theSU(3)C that remains unbroken. Let’s postpone dealing with this part of themodel. [Rogerio’s lecture]
➮ Let’s use the data to obtain the minimal model that fits the spectrum.
➮ We know three massive vector bosons and one massless =⇒ thesymmetry group must have 4 generators, with 3 symmetries being hidden.
➮ In the SM the gauge group is
SU(2)L × U(1)Y
that has 4 generators.
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➮ For a given choice of representation M for the matter
M =⇒M ′ = eigαataeig′Y βM
➮ Lorentz transformation of a multiplet member must be the same.
➮ Left-handed fermions are doublets of SU(2)L while right-handed fermionsare singlets.
➮ The hypercharge Y is fixed using the Gell-Mann–Nishijima relation
Q = t3 +Y
2
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��Gauge Sector
➯ Vector bosonsW 1
µ ; W 2µ ; W 3
µ for SU(2)L
Bµ for U(1)Y
with couplings g and g′. The kinetic lagrangian is
Lkg = −14BµνB
µν − 14W a
µνWaµν
whereBµν = ∂µBν − ∂νBµ
W aµν = ∂µW
aν − ∂νW
aµ − gεabcW b
µWcν
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��Leptons
➞ The multiplets and respective hypercharge are (without νR)
L =(νe
e
)L
⇐⇒ YL = −1 and R = eR ⇐⇒ YL = −2
The covariant derivative for the left and right multiplets are
DµL ≡(∂µ + i
g′
2Y Bµ + i
g
2σaW a
µ
)L and DµR ≡
(∂µ + i
g′
2Y Bµ
)R
The leptonic lagrangian is L` = iRDµγµR+ iLDµγ
µL
Important: the mass term meeReL is not SU(2)L × U(1)Y invariant =⇒ theleptons must be massless in the explicit realization of the symmetry.
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��Symmetry Breaking Sector
➱ We must break the symmetry to generated mass for the leptons, W± and Z.However, there is no direct indication so far on how this is realized in nature!
➱ In the minimal model we introduce a complex scalar doublet
ϕ =(ϕ+
ϕ0
)with Yϕ = +1
➱ This case is very similar to a previous example, but it has one extra gaugefield =⇒ massless vector.
➱ So we choose
Ls = |Dµϕ|2 − µ2|ϕ|2 − λ|ϕ|4 with Dµϕ =(∂µ + i
g′
2Y Bµ + i
g
2σaW a
µ
)ϕ
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➱ For the vacuum 〈ϕ〉 =(
0v/√
2
)we have
ta〈ϕ〉 6= 0
Y 〈ϕ〉 6= 0
Q〈ϕ〉 =12(σ3 + Y )〈ϕ〉 = 0
➱ Only the gauge boson associated to the electric charge is massless.
➱ We now use the unitary gauge for which ϕ =(
0v+h√
2
)to get
Ls =12
[∂νh∂
νh− 2λv2h2]+v2
8[g2|W 1
µ − iW 2µ|2 + (g′Bµ − gW 3
µ)2]
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➱ The charged W±µ = (W 1
µ ∓ iW 2µ)/
√2 have a mass MW = gv/2.
➱ We must diagonalize the mass matrix of the neutral vectors, obtaining
? A massive vector particle
Zµ =gW 3
µ − g′Bµ√g2 + g′2
= cos θWW 3µ − sin θWBµ
with MW/ cos θW . We wrote g′ = g tan θW .
? A massless gauge boson
Aµ =gBµ + g′W 3
µ√g2 + g′2
= cos θWBµ + sin θWW 3µ
➱ We must verify whether Aµ is really the photon!
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➱ Writing L` in terms of A and Z
Lnc = −eQ``γµ`Aµ − g
2 cos θWνLγ
µνLZµ −g
2 cos θW`γµ(gV − gAγ
5)`Zµ
with
gV = t3` − 2 sin2 θW Q`
gA = t3`
➱ The charged current interactions are given by
Lcc = − g√2νLγ
µeLW+µ + hc
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➱ Up to this point the leptons are massless. So we introduce the term
LY u = −Ge
[R(ϕ†L) + (LϕR)
]with Ge being the Yukawa coupling of the electron.
➱ Notice:∑Y = 2− 1− 1 = 0 so it is U(1)Y invariant.
➱ In the unitary gauge reads
LY u = −Ge
(v + h√
2
)(`R`L + `L`R) ;
so me = Gev/√
2.
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� Quark sector
➲ The quark multiplets are(ud
)L
uR dR with YL =13
YuR=
43
YdR= −2
3
➲ The interaction of the quarks with the gauge bosons are
Lcc = − g√2uLγ
µdLW+µ + hc
andLnc = −eQqqγ
µqAµ −g
2 cos θWqγµ(gq
V − gqAγ
5)qZµ
withgq
V = t3q − 2 sin2 θWQq and gqA = t3q
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➲ The quark masses are generated by
LY u = −Gu
[uR(χ†Lq) + (LqχuR)
]−Gd
[dR(ϕ†Lq) + (LqϕdR)
]with χ = iσ2ϕ∗ =
(ϕ∗0−ϕ−
)also being an SU(2)L doublet with Y = −1.
➲ In the unitary gauge
LY u = −Gu
(v + h√
2
)uLuR −Gd
(v + h√
2
)dLdR + h.c.
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��Quark mixing
➲ In general, for three generations qj =(uj
dj
)L
uRj dRj with j = 1, 2, 3
LY u = −∑jk
{qj
[G
(d)jk ϕdRk +G
(u)jk χuRk
]}+ h.c.
after SSB the mass matrices for the up (down) quarks are
∑jk
uLjMujkuRk with Mu
jk = G(u)jk
v√2
(Md
jk = G(d)jk
v√2
)
➲ After diagonalization dL = SLd d′L and dR = SR
d d′R we have
Mddiag = diag(md,ms,mb) , Mu
diag = diag(mu,mc,mt) ,
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and
LY u = −(
1 +h
v
) [d′Md
diagd′ + u′Md
diagu′]
➲ There is an important effect in the quark CC interactions since
∑j
uLjγµdLj =
∑jkn
u′Liγµ
(SL†
d
)ij
(SL
u
)jkd′Lk leading to
LCC =g
2√
2
W †µ
∑ij
u′i γµ(1− γ5)Vij d
′j + h.c.
➲ V = SL†
d SLu is unitary. It is the Cabibo-Kobayashi-Maskawa (CKM) matrix.
➲ The NC interactions are the same in new mass basis.
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��“Feynman rules”
➥ V ff vertices
➥ V V V and V V V V vertices
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➥ hff and hV V vertices
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��Number of free parameters
➥ 3 coupling constants
➥ 6 quark masses
➥ 3 CKM angles and 1 CP-violating phase
➥ 2 parameters for the Higgs potential
➥ 3 charged lepton masses
➥ for massive Dirac neutrinos: 3 masses, 3 mixing angles and 1 CP-violatingphase
➥ Total: 18 + 7 = 25
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IV. Comments
➳ SM is a consistent theory at the quantum level!
➳ Spontaneously broken gauge theories are renormalizable (t’Hooft) despitethe bad behaviour of the propagators seen in the unitary gauge
−iq2 −M2
V + iε
(gµν −
qµqνM2
V
)There are delicate cancellations already at tree level.
➳ an extensive use of conservations laws is needed to proof renormalizability=⇒ the model must be anomaly free.
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➳ In general chiral currents jµ5 = ψγµγ5ψ are
anomalous due to the triangular fermion loop➳ One way out is to cancel out the contributionsof different fermions.
➳ In the SM the only anomaly is proportional to
tr({ta, tb}Y
)∝
∑ferm. doub.
Y = 3× 13
+ (−1) = 0
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� A nice theorem
➴ Cornwall et al. showed (Phys.Rev.D10:1145,1974) that the most generalscalar, spinor and vector lagrangian respecting unitarity in perturbation theoryis equivalent to a spontaneously broken gauge theory!
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References
➺ Chris Quigg, Gauge Theories of the Strong, Weak, and ElectromagneticInteractions
➺ Ian Aitchison, An Informal Introduction to Gauge Field Theories
➺ Michael Peskin, An Introduction to Quantum Field Theory
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