1. Phenomenology of extended Electroweak Sector
C´edric Travelletti
Semester Thesis , Departement Physik, ETH Z¨urich
Advisor: Prof. Dr. Gino Isidori.
February 1, 2016
Abstract
We investigate the phenomenology of an electroweak model with
gauge group SU(2) × SU(2) × U(1). The symmetry is broken via two
Higgs-like fields: an SU(2) bidoublet first breaks SU(2)×SU(2) down
to the diagonal SU(2) subgroup and, in a second time, the remaining
gauge groups are broken to U(1)em in the standard way by a Higgs
doublet. We first compute the mass eigenstates of the gauge bosons. In
a second time we derive the Feynman rules for fermions-gauge bosons
interactions. Finally we compute the decay widths of the two neutral
massive gauge bosons predicted by our model.
1
3. 1 Introduction
Despite the experimental success of the standard model of particle physics,
theoretical considerations justify attempts to extend it. One of the possible
way of extending the standar model is to add new gauge groups. These will
then imply the existence of new gauge bosons and new interactions between
these and fermions. The additional gauge groups may either be postulated
for model-building purposes as in [7], or may be viewed as effective formu-
lations of higher dimensional physics as in [2]. In this thesis, we will simply
add an additional SU(2) group to the standard gauge structure. The mo-
tivation behind this is to explain the higher masses of the third generation
fermions by coupling them to the new gauge group, while leaving the cou-
pling of the first two generations as in the standard model. We introduce
two Higgs-like fields which break the gauge symmetry in two stage: first
down to the usual unbroken electroweak symmetry group and then down
to the electromagnetic U(1) group. The organisation of the thesis is the
following: we begin by describing the model we will use and its particle
content. We then compute the masses of the gauge bosons predicted by the
model. To this end, we consider the two symmetry breaking separately and
diagonalize the mass matrices perturbatively in the ratio of the symmetry
breaking scales. We then turn to the coupling of the bosons to fermions
and derive the Feynman rules of the model. Finally, we look for practical
implications of the model, compute the decay width of the massive neutral
gauge bosons and discuss experimental constraints.
3
4. 2 Model
We will study a model with gauge group SU(2)0 × SU(2)1 × U(1). The
corresponding gauge bosons (before symmetry breaking) will be denoted by
Wµ0 , Wµ1 , Bµ and the respective coupling constants by g0, g1, g2. The
gauge symmetry is broken by two Higgs-like fields which we denote Σ1 and
Σ2. They transform under the (2, ¯2, 0), respectively (1, 2, 1) representations.
Drawing on [11], we assume the first field to be a real bi-doublet, which
we parametrize as Σ2(x) = h(x)1 + iTaπa(x), where the h and πa fields are
real, Ta denote the usual SU(2) generators: Ta = σa
2 and σa are the Pauli
matrices; summation over a is intended. The second Σ-field is assumed to
be a complex doublet, which we also write in matrix form. For example
it may be viewed as the standard model Higgs using Σ2 = (iσ2φ , φ) =
φ0 φ+
−φ+ φ0 , with the S.M. Higgs doublet: φ =
φ+
φ0 .
For simplicity we assume that the first Higgs acquires a diagonal vacuum
expectation value Σ1 = α11, which breaks SU(20) × SU(2)1 down to the
diagonal SU(2) subgroup. The second Higgs is assumed to acquire a V.E.V.
Σ2 = α21, which will break the remaining gauge groups down to a U(1)
which we will identify as the usual electromagnetic U(1). We note that the
SU(2)×SU(2) gauge freedom is enough to gauge away the 6 real Goldstones,
thus, for the rest of this thesis, we will work in unitary gauge, ignore the
Goldstones and treat the gauge bosons as having 3 different polarizations.
Concerning the fermions, we assume that the first two generations trans-
form under the fundamental representation of the SU(2)1 gauge group and
acquire mass by interacting with Σ2, whereas third generation fermions are
assumed to transform under the fundamental representation of SU(2)0 and
acquire their mass via Σ1. This allows the top quark to get a large mass in
a natural way if we assume α1 α2.
We now turn to the mathematical formulation of our model.
The covariant derivatives for the Σ fields are given by:
DµΣ1 = ∂µΣ1 − ig0Wa
0µ
Ta
0 Σ1 + ig1Wa
1µ
Σ1Ta
1
DµΣ2 = ∂µΣ2 − ig1Wa
1µ
Ta
1 + ig2Y2BµT3
Σ2
The kinetic part of the Lagrangian reads:
Lkin =
1
4
Tr DµΣ1 (Dµ
Σ1)†
+
1
4
Tr DµΣ2 (Dµ
Σ2)†
4
5. This part of the Lagrangian is the only one we need in order to compute the
mass eigenstates of the gauge bosons.
3 Gauge Bosons Masses
We now compute the mass eigenstates of the gauge bosons after symmetry
breaking. To this end we will evaluate the kinetic term of the Σ-fields at
their VEV and diagonalize the result to extract the masses.
The VEV part of the kinetic lagrangian is:
LkinV EV
=
1
2
f1
4
−g0W3
0µ
+ g1W3
1µ
2
+ 2 −g0W+
0µ
+ g1W+
1µ
(c.c.)
+
1
2
f2
4
−g1W3
1µ
+ g2Y2Bµ
2
+ 2g1W+
1µ
W−
1µ
=W+
0µ
W−
0µ
f2
1
4
g2
0 + W+
1µ
W−
1µ
f2
1
4
g2
1 +
f2
2
4
g2
1 + W+
0µ
W−
1µ
−
f2
1
4
g0g1
+ W−
0µ
W+
1µ
−
f2
1
4
g0g1 + W3
0µ
2 1
2
f2
1
4
g2
0 + W3
1µ
2 1
2
f2
1
4
g2
1 +
f2
2
4
g2
1
+ B2
µ
1
2
f2
2
4
g2
2Y 2
2 + W3
0µ
W3
1µ
1
2
−
f2
1
4
g0g1 + (symm.)
+ W3
1µ
Bµ
1
2
−
f2
2
4
g1g2Y2 + (symm.) ,
where (c.c) denotes the complex conjugate of the preceding term and (symm.)
stands for the preceding term with the order of the fields exchanged.
For the charged gauge bosons we assume a mass term of the form
m2
ijWiµ Wµ
j .
This yields the mass matrix
λ2
1g2
0
1 −x1
−x1 x2
1 1 + λ2
For the neutral gauge bosons we assume a mass term of the form:
1
2
W3
0µ
W3
1µ
Bµ M2
Z
W3
0µ
W3
1µ
Bµ
5
6. We thus get the mass matrix:
M2
Zλ1g2
0
1 −x1 0
−x1 x2
1 1 + λ2 −x1x2λ2Y2
0 −x1x2λ2Y2 x2
2λ2Y 2
2
.
In order to simplify the previous expressions, we have used the following
parameters:
λ2
1 =
α2
1
4 λ2
2 =
alpha2
2
4 λ2 =
λ2
2
λ2
1
x1 = g1
g0
x2 = g2
g0
To find the mass eigenstates of the gauge bosons, we proceed as in [7] and
treat the two symmetry breaking successively: we first break SU(2)0 ×
SU(2)1 to SU(2) and diagonalize the resulting mass matrices. We then
break SU(2) × U(1) to U(1)em and compute the mass eigenstates by first
transforming to those of the first symmetry breaking. For the neutral gauge
bosons the process can be simplified by first identifying a zero eigenvector,
the photon.
3.1 First Symmetry Breaking
After the first symmetry breaking the mass matrices of the charged and
neutral gauge bosons are the same, for the W-bosons in W0, W1 basis:
m2
W = λ2
1g2
0
1 −x1
−x1 x2
1
This matrix is symmetric and can thus be diagonalized by a rotation so
we consider the change of basis:
W±
0µ
W±
1µ
=
cos ξ sin ξ
− sin ξ cos ξ
ˆW±
µ
˜W±
µ
.
The mass matrix in the new basis is then, ignoring the prefactor:
cos2 ξ + x1 cos ξ sin ξ + x1 sin ξ cos ξ + x2
1 sin2
ξ cos ξ sin ξ − x1 cos2 ξ − x1 − sin2
ξ + x1 sin ξ cos ξ
sin ξ cos ξ + x1 sin2
ξ − x1 cos2 ξ + x1 sin ξ cos ξ sin2
ξ − x1 sin ξ cos ξ + x1 − sin ξ cos ξ + x1 cos2 ξ
As done in [7], the off-diagonal terms can be made to vanish by setting
tan ξ = x1 = g1
g0
. The first eigenvalue is x2
1 + 1, while the second one
vanishes. Hence only the ˆW gets a mass, given by: m2
ˆW
= λ2
1 g2
0 + g2
1 . For
6
7. the neutral sector, after the fisrt symmetry breaking, the mass matrix is the
same as that of the charged boson (if we ignore Bµ). Hence we set:
W3
0µ
W3
1µ
=
cos ξ sin ξ
− sin ξ cos ξ
ˆZ0
µ
˜Z0
µ
.
Before we consider the second symmetry breaking, we first identify the pho-
ton. After the two SB, the neutral mass matrix is:
λ2
1g2
0
1 −x1 0
−x1 x2
1 1 + λ2 −x1x2λ2Y2
0 −x1x2λ2Y2 x2
2λ2Y 2
2
We get a normalized zero eigenvector by setting:
Aµ = e
1
g0
W3
0µ
+
1
g1
W3
1µ
+
1
g2Y2
Bµ ,
and identifying the electric charge:
1
e2
=
1
g2
0
+
1
g2
1
+
1
g2
2Y 2
2
.
We now try to express Aµ in terms of the fields we identified before and try
to get an analogue of the Weinberg angle.
Aµ ∝
g1
g0
W3
0µ
+ W3
1µ
+
g1
g2Y2
Bµ ∝ sin ξW3
0µ
+ cos ξW3
1µ
+ cos ξ
g1
g2Y2
Bµ
= ˜Z0
µ +
1
1 + tan2 ξ
g1
g2Y2
Bµ.
So by defining tan θ = g2Y2
1
g2
0
+ 1
g2
1
, we get an SM-like parametrization:
Aµ = sin θ ˜Z0
µ + cos θBµ.
We can also derive a relation between θ and the electric charge by noting
that: tan2 θ = g2
2Y 2
2
1
e2 − 1
g2
2Y 2
2
, and hence: e2 = g2
2Y 2
2 cos2 θ.
7
8. 3.2 Second Symmetry Breaking
We now consider: SU(2)×U(1) → U(1)em. The symmetry is broken by the
VEV of the second Σ-field: Σ2 = α21. The diagonalization procedure is
formally the same as in the last section.
3.2.1 Charged Sector
The charged mass matrix after the second symmetry breaking reads:
λ2
1g0
1 −x1
−x1 x2
1 1 + λ2
In ˆW, ˜W basis:
m2
ˆW
+ ∆m2
W δm2
W
δm2
W m2
˜W
where the masses and mass mixing terms are given by:
m ˜W = g2
0λ2
1x2
1λ2
cos2
ξ = g2
0λ2
2 sin2
ξ = g2
0λ2
2 1 −
1
1 + tan2 ξ
= λ2
2
1
1
g2
0
+ 1
g2
1
∆m2
W = g2
1λ2
2 sin2
ξ = x2
1m2
˜W
δm2
W = −g0g1λ2
2 sin2
ξ = −g0g1m2
˜W
.
3.2.2 Relations
Before studying the neutral part of the second symmetry breaking, it is
useful to derive some relations between the parameters of the model, which
will help us simplifying the mass matrices and also allow us to reduce the
number of redundant parameters in the final Feynman rules.
First of all, the definition of ξ yields:
cos ξ =
sin ξ
x1
(i)
From the definiton of θ we also get:
cot θ = cos ξ
g1
g2Y2
= sin ξ
g0
g2Y2
(iii)
Relating this to the mass of the ˜W:
cot2
ξ =
m ˜W
λ2
2g2
2Y 2
2
(ii)
8
9. 3.2.3 Neutral Sector
For the neutral case we first go in ˆZ, ˜Z, B basis:
m2
ˆZ
+ x2
1m2
˜W
−x1m2
˜W
λ2
2g1g2Y2 sin ξ
−x1m2
˜W
m2
˜W
−λ2
2g1g2Y2 cos ξ
λ2
2g1g2Y2 sin ξ −λ2
2g1g2Y2 cos ξ λ2
2g2
2Y 2
2
We then go to the basis given by the photon and its orthogonal comple-
ment:
ˇZ0
µ
Aµ
=
cos θ −sinθ
sin θ cos θ
˜Z0
µ
Bµ
In ˆZ, ˇZ, A basis, the neutral mass matrix reads:
m2
ˆZ
+ x2
1m2
˜W
−x1m2
˜W
λ2
2g1g2Y2 sin ξ
−x1m2
˜W
cos ξ − λ2
2g1g2Y2 sin ξ sin θ m2
˜W
cos θ + λ2
2g1g2Y2 cos ξ sin θ −λ2
2g1g2Y2 cos ξ cos θ − λ2
2g2
2Y 2
2 sin θ
−x1m2
˜W
sin θ + λ2
2g1g2Y2 sin ξ cos θ m2
˜W
sin θ − λ2
2g1g2Y2 cos ξ cos θ −λ2
2g1g2Y2 cos ξ sin θ + λ2
2g2
2Y 2
2 cos θ
1 0 0
0 cos θ sin θ
0 − sin θ cos θ
Using the relations ii) and iii), we see that the last row of the first matrix
vanishes. We simplify further entry by entry:
For the (2,2) entry:
m2
˜W
+ cos2
θ + 2λ2
2g1g2Y2 cos ξ sin θ cos θ + λ2
2g2
2Y 2
2 sin2
θ
vi)
= λ2
2g2Y2 cos2
θ g2Y2 cot2
θ + 2g1 cos ξ tan θ + g2Y2 tan2
θ
iii)
= λ2
2g2
2Y 2
2 cos2
θ cot2
θ + 2 cot θ tan θ + tan2
θ
(cot θ+tan θ)2
1
sin2 θ
=1+cot2 θ
ii)
= m2
˜W
+ λ2
2g2
2Y 2
2 .
For the (1,2) entry:
− x1m2
˜W
cos θ − λ2
2g1g2Y2 sin ξ sin θ = cos θ −x1m2
˜W
− λ2
2g1g2Y2 sin ξ tan θ
vi)
= cos θ −x1m2
˜W
− λ2
2x1g2
2Y 2
2
vii)
= − cos θx1m2
˜W
1 + tan2
θ
= −x1m2
˜W
1
cos2 θ
= −
1
e
x1g2Y2m2
˜W
iv)
= −
1
e
λ2
2g0g1g2Y2 sin2
ξ
After these simplifications, we get the following mass matrix:
m2
ˆZ
δm2
Z 0
δm2
Z m2
ˇZ
0
0 0 0
9
10. With masses and mixings:
m2
ˆZ
= m2
ˆW
m2
ˇZ
= m2
˜W
+ λ2
2g2
2Y 2
2
∆m2
Z = −g2
1λ2
2 sin2
ξ δm2
Z = −
1
e
λ2
2g0g1g2Y2 sin2
ξ.
3.3 Perturbative Diagonalization
In order to diagonalize the final mass matrices, we will identify the leading
entry and then expand the results in inverse powers of the leading entry.
We first consider the diagonalization of a general two by two real symmetric
matrix. Such matrices can be diagonalized by rotations so we compute:
cos θ sin θ
− sin θ cos θ
a b
b c
cos θ − sin θ
sin θ cos θ
=
a cos2
θ + 2b cos θ sin θ + c cos θ sin θ −a cos θ sin θ + b cos2
θ − b sin2
θ + c cos θ sin θ
−a cos θ sin θ − b sin2
θ + b cos2
θ + c cos θ sin θ a sin2
θ − 2b sin θ cos θ + c cos2
θ
.
Using trigonometric relations we see that the matrix is diagonal if
tan 2θ =
2b
a − c
.
Before computing the eigenvalues, we should find a perturbative parameter
to expand in. In our case, we will assume that the first symmetry breaking
happens at much higher energy than the second one, hence λ1 λ2. This
assumption is justified by the fact that current results do not hint at new
physics at the present energy scale. Noting that m2
ˆW
, m2
ˆZ
∝ λ2
1, whereas all
other entries are ∝ λ2
2, we can write the mass matrices in the general form:
a
1 + α β
β γ
,
where a = m2
ˆW
, respectively m2
ˆZ
and the other terms are of order
λ2
2
λ2
1
.
Expanding the eigenvalues, we get:
κ1 = α 1 +
γ2
4
+ β2
+ O
λ6
2
λ6
1
κ2 = α γ −
γ2
4
− β2
+ O
λ6
2
λ6
1
10
11. 3.4 Results of Diagonalization
Applying the above method to the charged and neutral mass matrices, we
get the mass eigenstates of the gauge bosons. We here summarize our results.
The eigenstates and mixing angles are given by:
W±
µ = cos ηW
ˆW±
µ + sin ηW
˜W±
µ Z0
µ = cos ηZ
ˆZ0
µ + sin ηZ
ˇZ0
µ
W±
µ = − sin ηW
ˆW±
µ + cos ηW
˜W±
µ Zo
µ = − sin ηZ
ˆZ0
µ + cos ηZ
ˇZ0
µ
tan 2ηW =
2δmW
m ˆW − m ˜W
tan 2ηZ =
2δmZ
m ˆZ − m ˇZ
The masses are given by:
mW = m ˆW + ∆m2
ˆW
+
1
m2
ˆW
1
4
m4
˜W
+ δm2
W m2
Z
= m2
ˆZ
+ ∆m2
ˆZ
+
1
m2
ˆZ
1
4
m2
ˇZ
+ δm2
Z
= m2
ˆW
+ m2
˜W
x2
1 +
m ˜W
4m2
ˆW
−
gog1m2
˜W
m2
ˆW
m2
W = m2
˜W
−
m4
˜W
m2
ˆW
1
4
− g0g1 m2
Z = m2
ˇZ
−
1
m2
ˆZ
1
4
m2
ˇZ
+ δm2
Z .
(1)
11
12. 4 Coupling to Fermions
We now consider the coupling of the gauge bosons mass eigenstates to
fermions and derive the corresponding Feynman rules. In order to get the
Feynman rules for fermion-gauge boson interactions, we compute the covari-
ant derivatives corresponding to the various representations. For left-handed
fermions in SU(2)0 :
DL
µ = ∂µ − i g0Wa
0µ
Ta
0 − i
g2
2
Y Bµ
= ∂µ +
i
2
−g0W3
0µ
− g2Y Bµ −g0W1
0µ
+ i g0W2
0µ
−g0W1
0µ
− i g0W2
0µ
g0W3
0µ
− g2Y Bµ
= ∂µ −
i
√
2
g0W+
0µ
0 1
0 0
−
i
√
2
g0W−
0µ
0 0
1 0
−
i
2
g0W3
0µ
+ g2Y Bµ
1 0
0 −1
−
i
2
g2Y Bµ
= ∂µ −
i
√
2
g0 cos ξ ˆW+
µ + sin ξ ˜W+
µ
0 1
0 0
−
i
√
2
g0 (+ ↔ −)
0 0
1 0
−
i
2
g0 cos ξ ˆZ0
µ + sin ξ ˜Z0
µ
1 0
0 −1
−
i
2
g2Y cos θ Aµ − sin θ ˇZµ
= ∂µ −
i
√
2
g0 cos ξ cos ηW W+
µ − sin ηW W+
µ + sin ξ sin ηW W+
µ + cos ηW Wµ
0 1
0 0
−
i
√
2
g0 (+ ↔ −) −
i
2
cos ξ cos ηZ Z0
µ − sin ηZ Z0
µ + sin ξ sin θAµ + cos θ ˇZµ
1 0
0 −1
−
i
2
g2Y cos θAµ − sin θ sin ηZZ0
µ + cos ηZZ0
µ ,
12
13. where (+ ↔ −) stands for the preceding term, with W+ fields replaced by
W− ones. Grouping terms we get:
DL
µ = ∂µ −
i
√
2
g0 (cos ξ cos ηW + sin ξ sin ηW ) W+
µ
0 1
0 0
−
i
√
2
g0 (− cos ξ sin ηW + sin ξ cos ηW ) W+
µ
0 1
0 0
−
i
√
2
g0 (+ ↔ −)
0 0
1 0
−
i
2
g0 (cos ξ cos ηZ + sin ξ cos θ sin ηZ) Z0
µ
1 0
0 −1
+
i
2
g2Y sin θ sin ηZZ0
µ
1 0
0 1
−
i
2
g0 (cos ξ sin ηZ + sin ξ cos θ cos ηz) Z0
µ
1 0
0 −1
+
i
2
g2Y sin θ cos ηZZ0
µ
1 0
0 1
−
i
2
(g0 sin ξ sin θ + g2Y cos θ) Aµ
1 0
0 0
−
i
2
(g0 sin ξ sin θ − g2Y cos θ) Aµ
0 0
0 −1
(2)
We now proceed to simplify this expression, beginning with the leptonic
case. We note that the first term of the last line must vanish in order for
the neutrino to be neutral. Using relation iii), we can achieve this by setting
Y = −Y2 for leptons.
The second term of the last line can also be simplified:
g0 sin ξ sin θ − g2Y cos θ = sin θ (g0 sin ξ + g2Y2 cot θ)
iii)
= sin θ g0 sin ξ + g2Y2 sin ξ
g0
g2Y2
= 2g0 sin θ sin ξ
iii)
= 2g0 cos θ
g2Y2
g0
= 2e
We now consider the Z boson term. Beginning with the up part:
− g0 sin ξ cos θ sin ηZ + g2Y sin θ sin ηZ = sinηZ sin θg2Y2 −
g0
g2Y2
sin ξ cot θ − 1
iii)
= − sin ηZ sin θg2Y2 cot2
θ + 1 = − sin ηZ
g2Y2
sinθ
iii)
= − sin ηZg0 sin ξ
1
cosθ
= − sin ηZg0 sin ξ
g2Y2
e
= − sin ηz
m ˜W
λ2
g2Y2
e
.
13
14. And for the down part:
g0 sin ξ cos θ sin ηZ + g2Y sin θ sin ηZ = sin ηZ cos θ (g0 sin ξ − g2Y2 tan θ)
ii)
= sin ηZ cos θ g0 sin ξ −
λ2g2
2Y 2
2
m2
˜W
= sin ηZ cos θ
m ˜W
λ2
−
λ2g2
2Y 2
2
m2
˜W
= − sin ηZe
λ2g2Y2
m ˜W
1 −
m ˜W
λ2g2Y2
2
.
The Z boson term can be simplified in the same way, replacing sin ηZ by
cos ηZ.
For right-handed fermions in SU(2)0 :
DR
µ = ∂µ −
i
2
g2Y2Bµ
= ∂µ +
i
2
g2Y2 sin θ sin ηZZ0
µ + cos ηZZ0
µ −
i
2
e
YR
Y2
Aµ, (3)
and hence we set YR = −2Y2 in order to get the same electromagnetic
couling for left- and right-handed fermions.
14
15. So finally, for SU(2)0 leptons, the kinetic part of the lagrangian is:
Lkin = (¯ντ , ¯τ)L iγµ
Dµ
ντ
τ L
+ ¯τRiγmu
DµτR
= (¯ντ , ¯τ)L iγµ
∂µ
ντ
τ L
+ ¯τRiγmu
∂µτR
+ g0
γµ
√
2
γµ
W+
µ ¯ντL τL (cos ξ cos ηW + sin ξ sin ηW ) + g0
γµ
√
2
γµ
W−
µ ¯τLντL (...same...)
+ g0
γµ
√
2
γµ
W+
µ ¯ντL τL (− cos ξ sin ηW + sin ξ cos ηW ) + g0
γµ
√
2
γµ
W−
µ ¯τLντL (...same...)
+
γµ
2
sin ηZ
m ˜W
λ2
g2Y2
e
+
1
2
g0 cos ξ cos ηZ Z 0
µ ¯ντL ντL
+
1
2
sin ηZ
m ˜W
λ2
g2Y2
e
1 −
m ˜W
λ2g2Y2
2
Z0
µ ¯τLτL −
1
2
g0 cos ξ cos ηZZ 0
µ ¯τLτL
+
γµ
2
cos ηZ
m ˜W
λ2
g2Y2
e
Z0
µ¯ντL ντL −
γµ
2
g0 cos ξ sin ηZ Z0
µ¯ντL ¯ντL
+
1
2
cos ηZ
m ˜W
λ2
g2Y2
e
1 −
m ˜W
λ2g2Y2
2
Z0
µ¯τLτL +
γµ
2
g0 cos ξ sin ηZZ0
µ¯τLτL
− eAµ¯τLτL + g2Y2 sin θ sin ηZγµ
Z0
µ ¯τRτR + g2Y2 sin θ cos ηZZ0
µ¯τRτR − eAµ¯τRτR.
15
16. 4.1 Feynman Rules
The Feynman rules for fermion-gauge boson interactions may be directly
read off the lagrangian.
For SU(2)0 leptons:
e−
L W +
ντ
i
g0
√
2
γµ (cos ξ cos ηW + sin ξ sin ηW ) PL
ντ W −
eL
e−
L
W+
ντ
i
g0
√
2
γµ (− cos ξ sin ηW + sin ξ cos ηW ) PL
ντ W−
eL
ντ Z 0
ντ
i
γµ
2
m ˜W
λ2
g2Y2
e
sin ηZ +
1
x1
cos ηZ PL
ντ Z0
ντ
i
γµ
2
m ˜W
λ2
g2Y2
e
cos ηZ −
1
x1
sin ηZ PL
τ−
Z 0
τ−
i
γµ
2
sin ηZe
λ2g2Y2
m ˜W
1 −
m ˜W
λ2g2Y2
2
PL + 2PR − i
γµ
2
cos ηZ
1
x1
m ˜W
λ2
PL
τ−
Z0
τ−
i
γµ
2
cos ηZe
λ2g2Y2
m ˜W
1 −
m ˜W
λ2g2Y2
2
PL + 2PR + i
γµ
2
sin ηZ
1
x1
m ˜W
λ2
PL
e− A
e−
ieγµ
Where we have used: g0 cos ξ = 1
x1
m ˜W
λ2
.
16
17. For the SU(2)1 leptons, the results are the same, up to the replacement
cos ξ → − sin ξ and sin ξ → cos ξ, we also recall that g0 sin ξ = g1 cos ξ.
We thus get the following Feynman rules:
e−
L W +
νe
i
g1
√
2
γµ (− sin ξ cos ηW + cos ξ sin ηW ) PL
νe W −
eL
e−
L
W+
νe
i
g1
√
2
γµ (sin ξ sin ηW + cos ξ cos ηW ) PL
νe W−
eL
νe Z 0
νe
i
γµ
2
m ˜W
λ2
g2Y2
e
sin ηZ − x1 cos ηZ PL
νe Z0
νe
i
γµ
2
m ˜W
λ2
g2Y2
e
cos ηZ + x1 sin ηZ PL
e−
Z 0
e−
i
γµ
2
sin ηZe
λ2g2Y2
m ˜W
1 −
m ˜W
λ2g2Y2
2
PL + 2PR + i
γµ
2
cos ηZx1
m ˜W
λ2
PL
e−
Z0
e−
i
γµ
2
cos ηZe
λ2g2Y2
m ˜W
1 −
m ˜W
λ2g2Y2
2
PL + 2PR − i
γµ
2
sin ηZx1
m ˜W
λ2
PL
e− A
e−
ieγµ
17
18. 4.1.1 Quark Sector
In order to compute the kinetic part of the quark lagrangian, we first need
to determine which weak hypercharge we should assing to them. For left-
handed quarks, considering the electromagnetic part of the leptonic covari-
ant derivative, eq.2:
−
i
2
(g0 sin ξ sin θ + g2Y cos θ) Aµ
1 0
0 0
−
i
2
(g0 sin ξ sin θ − g2Y cos θ) Aµ
0 0
0 −1
,
we see that, assuming from now on Y2 = 1, the assignment Y = 1
3 produces
the desired result: electric charge +2
3 for up quarks and −1
3 for down quarks.
For the right-handed quarks, eq.3 sugessts Y = 4
3 for up quarks and Y = −2
3
for down quarks.
We get the Feynman rules for the 3rd generation:
t Z 0
t
− i
γµ
2
sin ηZe
λ2g2Y2
m ˜W
1
3
−
m ˜W
λ2g2Y2
2
PL +
4
3
PR +i
γµ
2
cos ηZ
1
x1
m ˜W
λ2
PL
t Z0
t
− i
γµ
2
cos ηZe
λ2g2Y2
m ˜W
1
3
−
m ˜W
λ2g2Y2
2
PL +
4
3
PR −i
γµ
2
sin ηZ
1
x1
m ˜W
λ2
PL
The rules for the bottom quark can be obtained from the above ones, re-
placing the underlined terms by: − → + and 4
3 → −2
3.
18
19. 5 Phenomenological Implications
We now turn to the predictions of the model for actual particle processes.
Using the Feynman rules we derived, we will compute the decay widths of
the two neutral gauge bosons.
5.1 S.M. Z Decay Width
Before proceeding to the extended model, it will be useful to compute the
decay width of the Z-boson in the standard model. The calculation is easier
and allows us to identify techniques which will prove useful in the extended
calculation.
We consider the tree-level diagram:
Z0
µ (p, λ)
¯u (p1, s)
v (p2, s )
The matrix element for the diagram is:
iM = µ (p, λ) ¯u (p1, s) Aµ
v p2, s ,
where the standard Z vertex Feynman rule is given by(ref):
Aµ
= −
i
2
γµ g
cos θW
−2Qf sin2
θW (PL + PR) + σ3
PL
= −
i
2
γµ g
cos θW
−2Qf sin2
θW 1 + σ3 1 − γ5
2
.
In order to average the squared matrix element over possible fermion
spins we will have to compute spinor sums. We proceed as in [8], using the
identities: s us(p)¯us(p) = /p + m and s vs(p)¯vs(p) = /p − m.
s,s
¯us
(p1) γµ
vs
(p2) ¯vs
(p2) γν
us
(p1) =
s,s
¯us
αγµ
αβ /p2
− m
βγ
γν
γδuδ(i)
= /p1
+ m
δα
γµ
αβ /p2
− m
βγ
γν
γδ
= Tr /p1
+ m γµ
/p2
− m γν
.
19
20. s,s
¯us
(p1) γµ
vs
(p2) ¯vs
(p2) γν
γ5
us
(p1) = Tr /p1
+ m γµ
/p2
− m γν
γ5
.
(ii)
s,s
¯us
(p1) γµ
γ5
vs
(p2) ¯vs
(p2) γν
us
(p1) = Tr /p1
+ m γµ
γ5
/p2
− m γν
.
(iii)
s,s
¯us
(p1) γµ
γ5
vs
(p2) ¯vs
(p2) γν
γ5
us
(p1) = Tr /p1
+ m γµ
γ5
/p2
− m γν
γ5
.
(iv)
To proceed further we use the famous trace technology from [8]
Tr /p1
+ m γµ
/p2
− m γν
= Tr /p1
γµ
/p2
γν
− /p1
γµ
mγν
+ mγµ /p2γν
− m2
γµ
γν
(i)
= −4m2
ηµν
+ 4pµ
1 pν
2 + 4pν
1pµ
2 − 4p1 · p2ηµν
.
Tr /p1
+ m γµ
/p2
− m γν
γ5
= Tr /p1
γµ
/p2
− /p1
γµ
m + mγµ /p2 − m2
γµ
γν
γ5
(ii)
= −4i αµβν
p1α p2β
.
Here denotes the totally antisymmetric tensor. At the end of the calcula-
tion, this expression will have to be contracted with the polarization vectors
of the boson: µ(p, λ) and ν(p, λ). These being symmetric under µ ↔ ν,
their contraction with the antisymmetric tensor vanishes.
In the same way, the next trace also vanishes:
Tr /p1
+ m γµ
γ5
/p2
− m γν
= 0.(iii)
For the last trace we use γ5 2
= 1 and γµ, γ5 = 0.
Tr /p1
+ m γµ
γ5
/p2
− m γν
γ5
= Tr /p1
γµ
γ5
/p2
γν
γ5
− /p1
γµ
γ5
mγν
γ5
+ mγµγ5
/p2γν
γ5
− m2
γµ
γ5
γν
γ5
(iv)
= 4m2
ηµν
+ 4pµ
1 pν
2 + 4pν
1pµ
2 − 4p1 · p2ηµν
.
20
21. The spin averaged squared matrix element is then:
s,s
|M|2
=
g2
4 cos θW
µ
(p, λ) ν
(p, λ) −2Qf sin2
θW ¯uγµ
v +
σ3
2
¯uγµ
u − ¯uγµ
γ5
v
2
=
g2
4 cos θW
µ ν
4Q2
f sin4
θW (i) − 2Qf sin2
θW σ3
(i) + Qf sin2
θW σ3
(ii) + Qf sin2
θW σ3
(iii)
+
1
4
((i) − (ii) − (iii) + (iv))
=
g2
4 cos θW
µ ν
4Q2
f sin4
θW − 2Qf sin2
θW 2T3
+
1
2
4 T3 2
2(T3−Qf sin2 θW )
2
+2(Qf sin2 θW )
2
(i) + 2m2
f ηµν
Before summing over the Z polarizations, we compute some identities:
m2
f = p2
2 = (p − p1)2
= p2
− 2p · p1 + p2
1 → p1 · p =
1
2
m2
Z
p2
= (p1 + p2)2
= 2m2
f + 2p1 · p2 → p1 · p2 =
1
2
m2
Z − m2
f
Now for the polarization sum, using the identity:
λ
µ (p, λ) ν (p, λ) = −ηµν + pµpν
m2
Z
, we get:
λ
µ (p, λ) ν (p, λ) (i) = 4 −2p1 × p2 + 4p1 × p2 + 4m2
f +
1
m2
Z
2p × p1 p × p2 − p1 × p2p2
− m2
f p2
= 4 m2
Z + 2m2
f
λ
µ (p1, λ) ν (p2, λ) 2m2
f ηµν
= −6m2
f
We are now ready to compute the decay width. From ([9] eq.(46.17)), the
width is given by:
dΓ =
1
32π2
| ¯M|2 |p1|
m2
Z
dΩ,
where ¯M|2 denotes the squared matrix element averaged over incoming po-
larizations and summed over outgoing spins. Using momentum conservation,
we get: |p| =
m2
Z
4 − m2
f , the decay width is thus:
Γ =
1
3
1
16π
1
mZ
β
g2
4 cos2 θW
8 m2
Z + 2m2
f T3
− Qf sin2
θW
2
+ Qf sin2
θW
2
−
6
8
m2
f
m2
Z
,
21
22. where β = 1 −
m2
f
4m2
Z
and the result should be multiplied by 3 for decay into
quarks to take different colors into account. If we neglect fermion masses
and use the Fermi constant GF =
√
2
8
g2
cos2 θW m2
Z
, we get:
Γ ∼=
√
2Gf m3
Z
6π
T3
− Qf sin2
θW
2
+ Qf sin2
θW
2
,
which agrees exactly with the decay width computed by the Particle Data
Group (see eq. (48.24) in [9]).
5.2 Z Decay Width in extended Model
Now that the tools are set up, we are ready to compute the width in our
model. First of all, it will be useful to summarize all Z vertex Feynman rules
into one formula. Using the rules listed in section 4.1 and the identities from
3.2.2, we get, for quarks:
Z0
f
¯f
− 2T3
γµ
i
2
α sin ηZPL −
i
2
γµe cos ηZ
γ2g2Y2
m ˜W
J − 2T3 m ˜W
λ2g2Y2
2
PL + 2Qf PR ,
with the parameters α =
m ˜W
x1λ2
for 3rd generation quarks and α = −
x1m ˜W
λ2
for 1st and 2nd generations; J = 1
3. The same formula holds for leptons,
with J = −1. From now on, we will abbreviate: Θ :=
m ˜W
λ2g2Y2
.
To see why this formula also holds for leptons, we rearrange the neutrino-Z
boson interaction term in the kinetic lagrangian derived in chapter 4:
−g0 sin ξ cos θ cos ηZ − g2Y2 sin θ sin ηZ = −g0 sin ξ
e
g2Y2
cos ηZ − g2Y2 sin θ cos ηZ
= − cos ηZ
m ˜W
λ2
e
g2Y2
+ g2Y2
λ2g2Y2
m ˜W
cos θ
= − cos ηZ
m ˜W
λ2
e
g2Y2
+ e
λ2g2Y2
m ˜W
= cos ηZ
λ2g2Y2
m ˜W
1 +
m ˜W
λ2g2Y2
2
.
22
23. The spin- and polarization-averaged squared amplitude for the vertex
diagram is then:
| ¯M|2
=
1
3
1
4
α2
sin2
θZ (¯1) + 2T3
α sin ηZ
e
2
cos ηZ
1
Θ
J − 2T3
Θ2
(¯1) + 2Qf (¯2)
+
e2
4
cos2
ηZ
1
Θ2
J2
− 4JT3
Θ2
+ 4 T3 2
Θ4
(¯1) + 2 J − 2T3
Θ2
2Qf (¯2) + 4Q2
f (¯1) ,
where the polarization sums are:
(¯1) =
λ
µ
(p, λ) ν∗
(p, λ)
s,s
¯us
γµ
PLvs
¯vs
γν
PLus
= 2 (pµ
1 pν
2 + pν
1pµ
2 − p1p2ηµν
)
(¯2) =
λ
µ
(p, λ) ν∗
(p, λ)
s,s
¯us
γµ
PLvs
¯vs
γν
PRus
= 2 (pµ
1 pν
2 + pν
1pµ
2 − p1p2ηµν
)
(¯3) =
λ
µ
(p, λ) ν∗
(p, λ)
s,s
¯us
γµ
PRvs
¯vs
γν
PLus
= (¯2)
(¯4) =
λ
µ
(p, λ) ν∗
(p, λ)
s,s
¯us
γµ
PRvs
¯vs
γν
PRus
= (¯1).
Which gives the decay width:
Γ =
1
48π
βmZ
1
2
α2
sin2
ηZ 1 −
m2
f
m2
Z
+ α sin ηZ cos ηZe
λ2g2Y2
m ˜W
σ3
J + Θ2
1 −
m2
f
m2
Z
+ σ3
Qf 6
m2
f
m2
Z
+ 2e2
cos2
ηZ
1
Θ2
Q2
f 1 −
m2
f
m2
Z
+ e2
cos2
ηZ
1
Θ2
Qf J + Qf σ3
Θ2
6
m2
f
m2
Z
+
J2
4
+ JL
σ3
2
Θ2
+
Θ4
4
1 −
m2
f
m2
Z
.
(4)
Using an approximation for the ratio mZ
m ˜W
, which we will derive next, we
can rewrite this as:
Γ =
1
48π
m
3
Z
e2
g2
2Y 2
2
β
1
2
sin
2
ηZ
1
λ2
2
x−2
1
x2
1
+ 2 cos
2
ηzQ
2
f e
2 1
Θ2m2
˜W
+ cos
2
ηZ e
2
J2
4
1
Θ2m2
˜W
+ J
σ3
2
1
m2
˜W
+
1
4
Θ2
m2
˜W
1 −
m2
f
m2
Z
(*)
+ sin ηz cos ηZ e
x−1
1
−x1
σ
3
J
g2Y2
m3
˜W
+
1
g2Y2λ2
2m ˜W
1 −
m2
f
m2
Z
+ cos
2
ηZ e
2
Qf J
1
Θ2m2
˜W
+ Qf σ
3 1
m ˜W
2
m2
f
m2
Z
+ sin ηz cos ηZ e
x−1
1
−x1
6Qf σ
3 g2Y2
λ2m3
˜W
m2
f
m2
Z
.
23
24. Here the upper terms in the braces apply to the 3rd generations fermions
and the lower terms to 1st and 2nd generation. In order to identify the
dominant terms, we note that the mass m ˜W is mainly governed by the
symmetry breaking scale λ2 so we assume O m ˜W = O (λ2) and group the
terms according to their depencende on these parameters. Here the first
line has a second order dependence and the second line a third order. The
last two lines depend on order 2 and 4 respectively and give corrections
proportional to the ratio of the masses of the involved fermion to that of the
Z boson.
5.2.1 Intermezzo: m ˜W vs mZ
In order to simplify the results, we compute the ratio between the Z boson
mass and m ˜W , which is the leading term in the W boson mass (see eq.1).
We recall that:
m2
Z = m2
ˇZ
−
1
m2
ˆZ
1
4
m2
ˇZ
+ δm2
Z
= m2
˜W
+ λ2
2g2
2Y 2
2 +
1
m2
ˆW
1
4
m2
˜W
+ λ2
2g2
2Y 2
2 −
1
e
λ2
2g0g1g2Y2 sin2
ξ .
Now
m2
˜W
m ˆW
= O
λ2
1
λ2
2
and assuming that the first symmetry breaking happens
at much higher energy than the second one, we may neglect such terms and
get: m2
Z ˜=m2
˜W
+ λ2
2g2
2Y 2
2 . Hence
m2
Z
m2
˜W
˜=1 + g2
2Y 2
2
1
g2
0
+
1
g2
1
tan2 θ
=
g2
2Y 2
2
e2 .
5.2.2 Decay to Light Fermions
Since decay to top quarks are forbidden at tree level (even given small mass
corrections coming from the extended model), we will ignore all fermion
masses. We note that the terms involving σ3 in eq. (*) will cancel between
up and down fermions (as long as the electric charge is not involved). Hence
for decays into leptons:
Γl¯l =
gens. up,down
Γf ¯f
∼=
1
48π
m3
Z
e2
g2
2Y 2
2
1
λ2
2
sin2
ηZ
2x4
1 + 1
x2
1
+ cos2
ηZ
e2
Θ2m2
˜W
6 +
3
2
+ cos2
ηZe2 3
2
Θ2
m2
˜W
+ 2 sin ηZ cos ηZ
e
g2Y2λ2
2m ˜W
1 − 2x2
1
x1
.
24
25. For first and second generation quarks we also neglect masses. Summing
over the first two generations we get:
Γ
(1,2)
q¯q =
colors gens.1,2 up,down
Γf ¯f
∼=
3
48π
m3
Z
e2
g2
2Y 2
2
1
λ2
2
sin2
ηZ2x2
1 + cos2
ηZ
e2
Θ2m2
˜W
20
9
+
1
9
+ cos2
ηZe2 Θ2
m2
˜W
− 4x1 sin ηZ cos ηZ
e
g2Y2λ2
2m ˜W
.
Finally for decay to a bottom quark we get:
Γ
(1,2)
q¯q
∼=
3
48π
m3
Z
e2
g2
2Y 2
2
1
λ2
2x2
1
sin2
ηZ + cos2
ηZ
e2
Θ2m2
˜W
2
9
+
1
4
1
9
+ cos2
ηZ
e2
m2
˜W
Θ2
4
−
1
2
1
3
+ sin ηZ cos ηZ
e
x1
−
g2Y2
3m3
˜W
+
1
g2Y2λ2
2m ˜W
.
5.3 Decay Width of the new Neutral Boson
We now consider the additonal neutral boson predicted by the extended
model. We note that the the Z Feynman rules are the same as those of
the Z, except for the replacement: sin ηZ → − cos ηZ, cos ηZ → sin ηZ.
The only other difference with the Z boson lies in eq. 4, we get the correct
width for the Z by multiplying this equation by
mZ
mZ
and replacing the ratio
m2
f
m2
Z
→
m2
f
m2
Z
.
In order to compute the ratio of the masses, we again neglect terms of order
O
λ2
1
λ2
2
, use the previous approximation m2
Z
∼= m2
˜W
+ λ2
2g2
2Y 2
2 , and in the
same fashion approximate: m2
Z
∼= m2
ˆW
− g2
1λ2
2 sin2
ξ, where we recall that
m2
ˆW
= O λ2
1 . By Taylor expanding, we get the rough estimate:
m2
Z
m2
Z
∼=
m2
˜W
+ λ2
2g2
2Y 2
2
m2
ˆW
.
Using these replacement rules, we can get the decay widths of the Z to
light fermions by using the ones for the Z computed in the last section.
Nevertheless, if the mass of the new neutral boson happens to be large
enough, then decay to top quarks will also be allowed. We now compute the
decay width for this additional process. We will use eq.4, together with the
replacement rules mentioned above, but without neglecting the mass of the
25
26. top quark. The decay width is then given by:
Γ =
3
48π
mZ m
2
Z
e2
g2
2Y 2
2
β
1
2
sin
2
ηZ
1
λ2
2
1
x2
1
+
8
9
cos
2
ηze
2 1
Θ2m2
˜W
+ cos
2
ηZ e
2
1
36
1
Θ2m2
˜W
−
1
6m2
˜W
+
1
4
Θ2
m2
˜W
1 −
m2
t
m2
Z
+ sin ηz cos ηZ e
1
x1
−
g2Y2
3m3
˜W
+
1
g2Y2λ2
2m ˜W
1 −
m2
t
m2
Z
+ cos
2
ηZ e
2
−
2
9Θ2m2
˜W
+
2
3m ˜W
2
m2
f
m2
Z
+ sin ηz cos ηZ e
4
x1
g2Y2
λ2m3
˜W
m2
t
m2
Z
.
With this last process we now have all the partial decay widths of the mas-
sive neutral bosons and are hence also able to compute their total decay
width. These informations also allow us to compute the cross-sections of
other processes. For example τ pair production via e+e− → τ+τ− is, at
tree level, governed by the diagrams:
γ
e−
e+
τ+
τ−
+ Z0
e−
e+
τ+
τ−
+ Z 0
e−
e+
τ+
τ−
.
The photon diagram may be neglected and the other ones may be approxi-
mated by a Breit-Wigner form:
σZ ∝
ΓZe¯eΓZτ ¯τ
E2
cm − m2
Z + m2
ZΓ2
Z
σZ ∝
ΓZ e¯eΓZ τ ¯τ
E2
cm − m2
Z + m2
Z Γ2
Z
,
where ΓZ denotes the total decay width. We have analytical expressions
for each terms involved in these calculations and could in principle provide
a closed formula for the cross-section of the process. Nevertheless such a
formula would be highly complicated and of little practical use. This shows
one of the main shortcomings of the type of extensions of the standard
model studied in this thesis: adding more gauge groups makes the theory
analytically intractable.
26
27. 6 Conclusion
This thesis allowed us to compute the main phenomenological implications
which result from a simple extension of the gauge structure of the standard
model. Now that one has analytical expressions for the masses of the pre-
dicted particles, as well as for the decay widths of the neutral bosons, the
next logical step would be to fit these predictions to the actual data and
work out which region in the parameter space of the extended model are
allowed by the current experimental knowledge. According to the values
given by [9], there is some room for small corrections to the SM predic-
tions, for example the uncertainty on the mass of the Z boson is 0.002%
and the one on the decay width to leptons is 0.01%. Using the expressions
we computed for these quantities, we may put bounds on the parameters of
the model. Our approach to extensions of the standard model may also be
applied to more complicated choices of gauge structure, for example as in
[2]. The main limitation on the kind of extensions of the SM considered in
this thesis is that they soon become analytically intractable. We were only
able to compute the boson masses perturbatively and the expressions for the
Feynman rules or the decay widths are a lot more involved than the standard
ones. A quick argument allows to see how the complexity of such models
scales with the number of additional gauge groups: we start with the model
we considered: SU(2) × SU(2) × U(1) and iteratively add SU(2) groups,
coupling each group to the preceding via a bidoublet. We then compute the
masses of the bosons using the framework from this thesis. Each new group
yields a charged and a neutral two by two mass matrix. When considering
the symmetry breaking stage associated to the new group, these matrices
will be equal and only one of them has to be diagonalized. Then one has to
consider the symmetry breakings associated to the remaining groups. Each
time both matrices grow one rank larger and have to be diagonalized. Since
real symmetric matrices of rank n may be diagonalized by rotations, such on
operation requires the computation of dim(O(n)) = n(n−1)
2 different angles.
Thus if one adds n different SU(2) groups to the original SU(2) × U(1),
then the number of angles one as to compute in addition to the ones in
our model is: 1 + n−1
k=1 2dim(O(2 + k)) = 1 + 1
3(n3 + 3n2 + 2n − 6). This
clearly demonstrates how quickly these types of model become intractable
and shows the limitations of such frameworks to describe physics beyond
the standard model.
27
28. References
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J. He, M. Kurachi, M. Tanabashi: A Three Site Higgsless Model,
arXiv:hep-ph/0607124v5
[2] R. Foadi, S. Gopalakrishna, C. Schmidt: Higgsless Electroweak Sym-
metry Breaking from Theory Space, arXiv:hep-ph/0312324v2.
[3] R. Casalbuoni, S. De Curtis, D. Dominici and R. Gatto: Effective Weak
Interaction Theory with a possible new vector Resonance from a strong
Higgs Sector, Phys. Lett. 155B, 95 (1985).
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28
