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