› ~eboli › aula1.pdf · Standard Model - USPPBSM-2006 Oscar Eboli´ The kinetic term for the...

Standard Model

Oscar EboliUniversidade de Sao Paulo

Departamento de Fısica [email protected]

December 4, 2006

PBSM-2006 Oscar Eboli

PLAN

➫ I. Elementary particles

➫ II. Basic tools

➫ III. Model bulding

➫ IV. Comments

1


I. Elementary Particles

➫ The known elementary spin-12 fermions are

(νe

e−

)L

e−R

(ud

)RBG

L

uRBGR dRBG

R


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

2


➫ 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

3


➫ There is a large amount of precise data to be understood

4


II. Basic Tools

➫ The basic element for the construction of the SM are gauge theories andthe phenomenon of spontaneous symmetry breakdown.

5

PBSM-2006 Oscar Eboli��

��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

6


➩ 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.


➩ 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.

7


��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

8


➪ 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

9


➪ 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!

10


➪ 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!

11


➫ 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

12


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.



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

13


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!

14


��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

15


➬ 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) =⇒


➬ 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

16


➬ 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.

17


➭ 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

)

18


➭ 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.

19


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.

20


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.

21


➮ 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

22


��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

23


��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.

24


��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

25


➱ 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]

26


➱ 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!

27


➱ 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

28


➱ 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.

29


� 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

30


➲ 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.

31


��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) ,

32


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.

33


��“Feynman rules”

➥ V ff vertices

➥ V V V and V V V V vertices

34


➥ hff and hV V vertices

35


��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

36


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.

37


➳ 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

38


� 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!

39


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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