Electroweak Interaction

Susumu Oda

Graduate School Lecture Experimental Particle Physics21 December 2018 (Fri.)11 January 2019 (Fri.)

1 / 65

Fermi Theory for Weak Interaction

In Fermi theory, the weak interaction is phenomenologicallydescribed as the interaction of four fermions.The Lagrangian is

L4−fermion = −GF√2J†µJ

µ,

where GF is called the Fermi coupling constant and isirrespectively of processes

GF = 1.166378(6)× 10−5 GeV−2

= 1.03× 10−5m−2proton. (1)

2 / 65

Diagram in Fermi Theory

+µJ µ

J

2FG

−

3 / 65

Charged Current

Jµ is called the charged current and is separated into one byleptons and one by quarks.

Jµ = J leptonµ + Jquark

µ

4 / 65

Charged Current, Chirality Operator

We will write Dirac fields ψe, ψνe as e, νe and so on.Since the weak interaction is the V − A type, the chargedcurrent of leptons is expressed as

J leptonµ = eγµ(1− γ5)νe + µγµ(1− γ5)νµ + τγµ(1− γ5)ντ ,

where γ5 is the chirality operator and is defined by

γ5 ≡ −iγ0γ1γ2γ3.

5 / 65

Projection Operators

The projecton operator PL, which extract the left-handedcomponent of spinors, is defined as

ψL ≡ PLψ =1− γ5

2ψ.

Similarly, the projecton operator PR, which extract theright-handed component of spinors, is defined as

ψR ≡ PRψ =1 + γ5

2ψ.

6 / 65

Left-Handed Doublet

The left-handed doublet of an electron and anelectron-neutrino is defined as

Le ≡(νeLeL

)= PL

(νee

).

7 / 65

Charged Current of Electrons and

Electron-Neutorinos

We concentrate on the charged current of electrons andelectron-neutrinos. Using Le, it is expressed as follows.

Jelectronµ = eγµ(1− γ5)νe

= 2eγµPLνe

= 2eLγµνeL

= 2(νeL eL

)( 0 01 0

)γµ

(νeLeL

)= 2Le

(0 01 0

)γµLe

8 / 65

Pauli Matrices

If we recall Pauli matrices are

σ1 =

(0 11 0

)σ2 =

(0 −ii 0

)σ3 =

(1 00 −1

),

we can find (0 01 0

)=

σ1 − iσ2

2.

9 / 65

SU(2) Structure of Charged Current

If we introduce

jaµ =∑

k=e,µ,τ

Lkγνσa

2Lk (a = 1, 2, 3), (2)

we can write the charged current using it as

J leptonµ = 2

(j1µ − ij2µ

)≡ 2j1−i2

µ ,

J leptonµ

†= 2

(j1µ

†+ ij2µ

†)= 2

(j1µ + ij2µ

)≡ 2j1+i2

µ .

This means that the charged current is the 1∓ i2 componentsof the SU(2) current.Jquarkµ also has the same property and Jµ satifies the SU(2)

algebra.

10 / 65

Weak Isospin

This SU(2) is called the weak isospin and the generators arewritten as T a(a = 1, 2, 3) and the magnitude is written as T .This weak isospin SU(2) is calledSU(2)L (the subscript of L represents the left handed) orSU(2)W (the subscript of W represents the weak isospin).

11 / 65

Electric Charge and Weak Isospin

Electric charge Q Weak isospin T 3

Le =

(νeLeL

)0−1

+12

−12

Although Le is a doublet of SU(2)L, Q and T 3 are notindependent. U(1)EM of the electromagnetic interaction andSU(2)L are not orthogonal.We need to consider the weak interaction and theelectromagnetic interaction at the same time and unify them.

12 / 65

Weak Hyper Charge

We introduce the U(1)Y group which is orthogonal to theSU(2)L group and call its generator (=charge) the weak hypercharge Y .If we assume doublets Le, Lµ and Lτ have Y = −1,

Q = T 3 +Y

2(3)

is satisfied.

13 / 65

Structure of the Weinberg-Salam Theory

(Electroweak Unification Theory) to Be Described

Spontaneoussymmetrybreaking

Gauge SU(2)L × U(1)Y −→ U(1)EMgroups

Generators T a Y Q = T 3 + Y2

14 / 65

Gauge FieldsThe gauge field of SU(2)L is

W⃗µ = (W 1µ ,W

2µ ,W

3µ) =

∑a=1,2,3

W aµ

and the coupling constant is g.The gauge field of U(1)Y is Bµ and the coupling constant is g′.The kinetic terms of the Lagrangian is

Lgauge = −1

4

(∂µW⃗µ − ∂νW⃗µ − gW⃗µ × W⃗ν

)2−1

4(∂µBν − ∂νBµ)

2 , (4)

where

W⃗µ × W⃗ν =∑a,b,c

εabcWbµW

cν

is caused by the fact that SU(2) is a non-Abelian group.

15 / 65

Higgs Field

The Higgs doublet is

Φ =

(ϕ1

ϕ2

),

where ϕ1 and ϕ2 are complex scalar fields and the number ofdegrees of freedom is four.The vacuum expectation value after the symmetry breaking is

⟨0|Φ|0⟩ =1√2

(0v

).

16 / 65

Electric Charges of Higgs Field

Since v is non-zero real scalar, the Higgs doublet has to haveY = +1 so that the electric charge Q = 0.Then, for ϕ1

Q = T 3 + Y/2 = +1/2 + 1/2 = +1

and for ϕ2

Q = T 3 + Y/2 = −1/2 + 1/2 = 0.

17 / 65

Covariant Derivative

The covariant derivative for the Higgs field Φ is defined as

DµΦ =

(∂µI + ig′

1

2IBµ + ig

1

2W⃗µ · σ⃗

)Φ

where I is the unit matrix, σi is Pauli matrices and

W⃗µ · σ⃗ = W 1µσ

1 +W 2µσ

2 +W 3µσ

3.

18 / 65

U(1)Y Transformation

Φ(x) → Φ′(x) = eiθ(x)Φ(x)

is transformed as

Bµ(x) → B′µ(x) = Bµ(x)−

2

g′∂µθ(x)

under the U(1)Y transformation.

19 / 65

SU(2)L Transformation

Φ(x) → Φ′′(x) = U(x)Φ(x) = exp(iσ⃗ · θ⃗(x)

)Φ(x)

is transformed as

W⃗µ(x) · σ⃗ → W⃗ ′′µ (x) · σ⃗

= U(x)W⃗µ(x) · σ⃗U †(x)− 2i

gU(x)∂µU

†(x)

under the SU(2)L transformation.

20 / 65

Kinetic Term and Potential of Higgs Field

The kinetic term and the potential of the Higgs field in theLagrangian is

LHiggs = (DµΦ)†DµΦ + µ2Φ†Φ− λ

(Φ†Φ

)2. (5)

21 / 65

Higgs Field after Symmetry Breaking

After the symmetries are broken, using v2 = µ2/λ, the Higgsfield is represented as

Φ(x) = exp

(iχ⃗(x)

v· σ⃗)

1√2

(0

v + ϕ(x)

).

If we set U−1 = exp(i χ⃗(x)

v· σ⃗), we have

Φ′ = UΦ =1√2

(0

v + ϕ(x)

)and we can redefine Φ′ as Φ to remove the freedom of χ⃗.

22 / 65

Kinetic Term and Potential of Higgs Field

after Symmetry Breaking

Equation (5) becomes

LHiggs

=1

2

∣∣∣∣{∂µI + i

2

(gW 3

µ + g′Bµ g(W 1

µ − iW 2µ

)g(W 1

µ + iW 2µ

)−gW 3

µ + g′Bµ

)}·(

0v + ϕ(x)

)∣∣∣∣2 + µ2

2(v + ϕ)2 − λ

4(v + ϕ)2

23 / 65

Kinetic Term and Potential of Higgs Field

after Symmetry Breaking

LHiggs

=1

2

∣∣∣∣( 0∂µϕ(x)

)+i

2

(g(W 1

µ − iW 2µ

)−gW 3

µ + g′Bµ

)(v + ϕ(x))

∣∣∣∣2+µ2

2(v + ϕ)2 − λ

4(v + ϕ)2

=1

2(∂µϕ)

2 +1

8g2[(W 1

µ

)2+(W 2

µ

)2](v + ϕ)2

+1

8

(gW 3

µ − g′Bµ

)2(v + ϕ)2 +

µ2

2(v + ϕ)2 − λ

4(v + ϕ)4 .

(6)

24 / 65

W± Particles, Z0 Particle, Photon, Higgs ParticleHere, we introduce

Wµ ≡ 1√2

(W 1

µ − iW 2µ

),

W †µ =

1√2

(W 1

µ + iW 2µ

),(

Zµ

Aµ

)≡ 1√

g2 + g′2

(g −g′g′ g

)(W 3

µ

Bµ

)=

(cos θW − sin θWsin θW cos θW

)(W 3

µ

Bµ

).

θW is called the weak mixing angle or the Weinberg angle.Wµ represents the W+ particle, W †

µ the W− particle, Zµ theZ0 particle, Aµ the photon (massless), and ϕ the Higgsparticle.

25 / 65

Masses of W±, Z0 and Higgs ParticlesHere, we introduce

cos θW =g√

g2 + g′2,

sin θW =g′√

g2 + g′2,

MW =1

2gv,

MZ =1

2

√g2 + g′2v =

MW

cos θW,

MH =√

2µ2 =√2λv.

Coupling constants g, g′, the parameter λ of the potential andthe vacuum expectation value v/

√2 determine the mass of W

particles MW , the mass of the Z particle MZ and the mass ofthe Higgs particle MH .

26 / 65

Kinetic Term and Potential of Higgs Field

after Symmetry Breaking

Then, Equation (6) becomes

LHiggs =1

2(∂µϕ)

2

+M2WW

†µW

µ

(1 +

ϕ

v

)2

+1

2M2

ZZµZµ

(1 +

ϕ

v

)2

+µ2

2(v + ϕ)2 − λ

4(v + ϕ)4

27 / 65

Kinetic Term and Potential of Higgs Field

after Symmetry Breaking

LHiggs = M2WW

†µW

µ +1

2M2

ZZµZµ

+

(gMWϕ+

g2

4ϕ2

)(W †

µWµ +

1

2 cos2 θWZµZ

µ

)+1

2

((∂µϕ)

2 −M2Hϕ

2)− 1√

2MH

√λϕ3 − λ

4ϕ4

+const.

28 / 65

Lagrangian of Gauge Fields

Using Wµ, Zµ, Aµ, the Lagrangian of the gauge fieldsEquation (4) is represented as

Lgauge = −1

4FAµνF

Aµν − 1

4FZµνF

Zµν

−1

2(DµWν −DνWµ)

† (DµW ν −DνW µ)

+i(eFA

µν + g cos θWFZµν

)W µ†W ν

+g2

2

(|WµW

µ|2 −(Wµ

†W µ)2)

.

29 / 65

Lagrangian of Gauge Fields

Here,

FXµν ≡ ∂µXν − ∂νXµ,

e = g sin θW =gg′√g2 + g′2

,

gA3µ = g sin θWAµ + g cos θWZµ

= eAµ + g cos θWZµ,

DµWν =(∂µ + igA3

µ

)Wν

= (∂µ + ieAµ + ig cos θWZµ)Wν . (7)

Equation (7) means that the electric charge of Wµ is Q = +1.

30 / 65

Lepton Doublets and Singlets

Left-handed leptons constitute doublets of SU(2)L (T = 1/2)and have Y = −1, andright-handed leptons constitute singlets of SU(2)L (T = 0)and have Y = −2.

Weak Weak First Second Thirdisospin hyper charge generation generation generation

T = 12

Y = −1 Le ≡(

νee

)L

Lµ ≡(

νµµ

)L

Lτ ≡(

νττ

)L

T = 0 Y = −2 Re ≡ eR Rµ ≡ µR Rτ ≡ τR

31 / 65

Lepton Doublets and SingletsUsing Equation (3) Q = T 3 + Y/2, the electric charge Q canbe confirmed as follows.

νeL Q = +1

2+

1

2(−1) = 0,

eL Q = −1

2+

1

2(−1) = −1,

eR Q = 0 +1

2(−2) = −1.

Right-handed neutrinos νeR, νµR, ντR are not experimentallyobserved and are not included in the Standard Model.Even if those exist, their electric charges Q = 0. If those aresinglets of SU(2)L (T = 0), their hyper charges should beY = 0. Since they do not interact electromagnetically norweakly, it does not contradict the experimental facts.

32 / 65

Lagrangian of Leptons

The Lagrangian of leptons are separated into interaction andmass terms.

Llepton = Lint.lepton + Lmass

lepton

33 / 65

Lepton Interaction Term

The lepton interaction term is

Lint.lepton =

∑j=e,µ,τ

Ljiγµ

(∂µ −

i

2g′Bµ + ig

σ⃗

2· W⃗µ

)Lj

+∑

j=e,µ,τ

Rjiγµ (∂µ − ig′Bµ)Rj

=g

2√2

(W †

µJµ +W µJ†

µ

)+ eAµJ

µem +

g

cos θWZµJ

µZ .

34 / 65

Lepton Interaction Term

Here, the electromagnetic current Jµem, the charged current

Jµ, the neutral current JµZ are

Jµem = −

∑k=e,µ,τ

kγµk,

Jµ =∑

k=e,µ,τ

kγµ(1− γ5)νk,

JµZ = jµ3 − sin2 θWJ

µem,

respectively.jµa is defined in Equation (2).By unifying the weak interaction and the electromagneticinteraction, the unknown neutral current Jµ

Z is appeared.

35 / 65

Diagram of Charged Current

in the Standard Model

+µJ µ

J22

g

22

g2W-M2p

νµg

36 / 65

Charged Current

The propagator of the W particle is

gµνp2 −M2

W

.

If the momentum p is much smaller than the mass of the Wparticle MW , we approximately have

−g2

8

1

M2W

≃ −GF√2.

Then, we have

GF√2

=g2

8M2W

=g2

8 ·(12gv)2 =

1

2v2.

37 / 65

Charged Current

Using Equation (1), we have

v ≃ 246 GeV,

MW =

(g2

8

√2

GF

) 12

=

( √2

8GF

e2

sin2 θW

) 12

=37.3

sin θWGeV,

MZ =MW

cos θW=

37.3

cos θW sin θWGeV =

74.6

sin 2θWGeV.

38 / 65

Diagram of Neutral Current

in the Standard Model

µν , eν µν , eν

, u, d-e , u, d-e

0Z

By measuring neutral current processes involving the Zparticle, θW can be determined.

39 / 65

Neutral Current

In 1972–1973, cross section measurement determinedsin2 θW ≃ 0.23.From this value, MW ≃ 78 GeV and MZ ≃ 87 GeV weredetermined.In 1983, the W and Z particles were discovered.As of 2018,

MW = 80.379± 0.012 GeV,

MZ = 91.1876± 0.0021 GeV

are experimentally determined and are in excellent agreementwith theoretical values including higher order corrections.

40 / 65

Lepton Mass Term

We will next look at the lepton mass term Lmasslepton.

The Dirac mass term has the form of −mψψ. By using thefollowing properties

PR =1 + γ5

2,

PL =1− γ5

2,

P †R = PR,

P 2R = PR,

ψ = ψ†γ0,

γ5γ0 = −γ0γ5,

41 / 65

Lepton Mass Term

we obtain

ψψ = ψ (PR + PL)ψ

= ψ (PRPR + PLPL)ψ

=(ψPR

)(PRψ) +

(ψPL

)(PLψ)

= (PLψ) (PRψ) + (PRψ) (PLψ)

= ψLψR + ψRψL.

In SU(2)L, since left-handed leptons are doublets andright-handed leptons are singlet, ψψ is a doublet and is notunchanged under the SU(2)L transformation.

42 / 65

Lepton Mass Term

We focus on electron-neutrinos νe and electrons e. Using theHiggs doublet Φ, we consider

−fe(LeΦ

)Re + h.c. (8)

Since the doublets and singlet are

Le =(νeL eL

),

Φ =

(ϕ1

ϕ2

)=

(ϕ+

ϕ0

),

Re = eR,

LeΦ is a singlet and (LeΦ)Re is a singlet, which is uncangedunder the SU(2)L transformation.

43 / 65

Lepton Mass Term

Here, fe is a Yukawa coupling (coupling between spin-1/2fermion and spin-0 boson) constant, h.c. stands for Hermitianconjugate.On weak hyper charge Y , since Le is the Dirac conjugate ofLe, it has Y = −(−1). Φ has Y = +1 and Re has Y = −2.Therefore, we have

−(−1) + (+1) + (−2) = 0

and the mass term is unchanged under the U(1)Ytransformation.

44 / 65

Lepton Mass Term

After the symmetry breaking, Equation (8) becomes

−fe(νeL eL

)( 0v√2

)eR + h.c.

= −fev√2eLeR + h.c.

= −fev√2(eLeR + eReL)

= −meee.

45 / 65

Lepton Mass Term

Here,

me =fev√2.

The Yukawa coupling constant fe and the vacuum expectationvalue v/

√2 determine the electron mass me.By breaking the

electroweak symmetry, the electron mass arises.

fe =

√2me

v=

me√2MW

g.

46 / 65

Diagram of Lepton Yukawa Coupling

R-τ ,

R-µ , R

-e L-τ ,

L-µ , L

-e

ef

2v

47 / 65

Mass Term of Leptons of Three Generations

and Neutrino Masses

The mass term of leptons of three generations Lmasslepton is

Lmasslepton = −

∑j=e,µ,τ

fj[(LjΦ

)Rj +Rj

(Φ†Lj

)]. (9)

Since there is no neutrino mass term, the masses of neutrinosare zero in the Standard Model.

48 / 65

Yukawa Coupling Constants

In Equation (9), fj is real.This can be generalized using complex Gij as

Lmass,gen.lepton = −

∑i,j=e,µ,τ

[Gij

(LiΦ

)Rj +G∗

ijRj

(Φ†Lj

)].

Any complex matrix can be transformed to a real diagonalmatrix using two unitary matrices UL and UR.

G = U †LFUR,

where F is a real diagonal matrix and its component Fij isFij = 0 for i ̸= j.

49 / 65

Yukawa Coupling Constants

We can define

R′i =

∑j

URijRj,

L′i =

∑j

ULijLj.

And, we can substitute R′i → Ri and L

′i → Li, and define

Fjj = fj. Then, we can obtain the original Lmasslepton.

Therefore, the generalization does not make any difference inthe case of leptons.However, as we will see, there is a difference, which is theorigin of the Cabibbo-Kobayashi-Maskawa (CKM) matrix, inthe case of quarks.

50 / 65

Three Generations of Quarks

There are three generations for quarks as well as leptons.

Electric First Second Thirdcharge Q generation generation generation

Up uA + 23 u c t

type (A = 1, 2, 3) up charm topDown dA − 1

3 d s btype down strange bottom

51 / 65

Doublets and Singlets of Quarks

If we assume left-handed quarks are doublets of SU(2)L andright-handed quarks are singlets of SU(2)L, those are classifiedas follows.

Weak Weak First Second Thirdisospin hyper charge generation generation generation

T = 12

Y = + 13

qL1 =

(ud

)L

qL2 =

(cs

)L

qL3 =

(tb

)L

T = 0 Y = + 43

uR cR tRY = − 2

3dR sR bR

52 / 65

Electric Charges of Quarks

We can confirm that quarks satisify Q = T 3 + Y/2 by

uL Q = +1

2+

1

2

(+1

3

)= +

2

3,

dL Q = −1

2+

1

2

(+1

3

)= −1

3,

uR Q = 0 +1

2

(+4

3

)= +

2

3,

dR Q = 0 +1

2

(−2

3

)= −1

3.

53 / 65

Lagrangian of QuarksThe Lagrangian of quarks Lquark is

Lquark = Lint.quark + Lmass

quark.

The quark interaction term Lint.quark is

Lint.quark =

∑A=1,2,3

{q(W )RA iγ

µ

(∂µ + ig′

1

6Bµ + ig

σ⃗

2· W⃗µ

)q(W )LA

+u(W )LA iγµ

(∂µ + ig′

2

3Bµ

)u(W )RA

+d(W )

LA iγµ(∂µ − ig′

1

3Bµ

)d(W )RA

},

where (W ) stands for the eigenstates of the weak interaction,

ψ(W )RA = ψ

(W )

LA and ψ(W )LA = ψ

(W )

RA .

54 / 65

Quark Mass TermThe mass term Lmass

quark is written as

Lmassquark = −

∑A,B

(f(d)ABq

(W )RA Φd

(W )RB + f

(u)ABq

(W )RA Φ̃u

(W )RB + h.c.

),

where

Φ̃ ≡ iσ2Φ∗,

σ2 =

(0 −ii 0

),

⟨0|Φ̃|0⟩ =

(0 1−1 0

)(0v√2

)=

( v√2

0

)to give masses to up-type quarks uA. f

(X)AB are Yukawa

coupling constants and complex numbers.

55 / 65

Quark Mass Term

After the symmetry breaking, the mass term becomes

Lmassquark

= −∑A,B

(1√2f(d)ABvd

(W )

RA d(W )RB +

1√2f(u)ABvu

(W )RA u

(W )RB + h.c.

).

56 / 65

Quark Mass Term

Yukawa coupling constants f(X)AB can be transformd to a real

diagonal matrices using two unitary matrices SX and TX(X = d, u). Thier eigenvalues m

(X)A are interpreted as quark

masses.∑C,D

(S†X

)AC

(1√2f(X)CD v

)(TX)DB = m

(X)A δAB

This represents the product of matrices

S†XFTX = M.

57 / 65

Weak Interaction Eigenstates and Mass Eigenstates

Weak interaction eigenstates ψ(W ) and mass eigenstates ψ(M)

are related by SX and TX .

d(W )LA =

∑B

(Sd)AB d(M)LB

u(W )LA =

∑B

(Su)AB u(M)LB

d(W )RA =

∑B

(Td)AB d(M)RB

u(W )RA =

∑B

(Tu)AB u(M)RB

58 / 65

Electromagnetic Current of Quarks

The interaction terms with gauge fields are given similarly asleptons. The electromagnetic current Jµ

em is

Jµem = +

2

3

∑A

u(W )A γµu

(W )A − 1

3

∑A

d(W )

A γµd(W )A

= +2

3

∑A

u(M)A γµu

(M)A − 1

3

∑A

d(M)

A γµd(M)A .

Since this current does not mix the up type and the downtype, the form is the same for both interaction eigenstates(W ) and mass eigenstates (M).

59 / 65

Neutral Current of Quarks

Similarly, the neutral current JµZ is

JµZ = jµ3 − sin2 θWJ

µem,

jµ3 =∑A

q(W )RA γ

µσ3

2q(W )LA

=1

4

∑A

{u(W )A γµ(1− γ5)u

(W )A − d

(W )

A γµ(1− γ5)d(W )A

}=

1

4

∑A

{u(M)A γµ(1− γ5)u

(M)A − d

(M)

A γµ(1− γ5)d(M)A

}.

60 / 65

Charged Current of QuarksCharged current Jµ is

Jµ =∑A

u(W )A γµ(1− γ5)d

(W )A

= 2∑A

u(W )RA γµd

(W )LA

= 2∑A,B

u(M)RA γµ

(S†uSd

)AB

d(M)LB ,

where

VCKM ≡ S†uSd

is the Cabibbo-Kobayashi-Maskawa (CKM) matrix.This matrix orignates in the differences between theinteraction eigenstates and mass eigenstates and is due toYukawa couplings.

61 / 65

CKM Matrix

The components of the CKM matrix is written as

VCKM =

Vud Vus VubVcd Vcs VcbVtd Vts Vtb

.

It has physical freedom of three rotation angles and onecomplex phase.

62 / 65

CKM Matrix

Since the CKM matrix is the products of unitary matrices, it isalso a unitary matrix and satisfies

V †CKMVCKM = VCKMV

†CKM = I.

Using the components, the relations are∑i=u,c,t

V ∗imVin = δmn (m,n = d, s, b),∑

m=d,s,b

VimV∗jm = δij (i, j = u, c, t). (10)

63 / 65

Unitarity Triangles

For example, if we choose m = b and n = d in Equation (10),we have

V ∗ubVud + V ∗

cbVcd + V ∗tbVtd = 0.

This means that V ∗ubVud, V

∗cbVcd and V ∗

tbVtd form a triangle inthe complex plane.This triangle is called a unitarity triangle.

64 / 65

Unitarity Triangle

cdV*cbV

udV*ubV tdV*

tbV

65 / 65