On the Flavor Problem in Strongly Coupled Theories - THEP Mainz

20 Chapter 1. Introduction: Problems beyond the Standard Model
Here {κ = (κ − , κ 0 ), �κ} and {h = (h − , h 0 ), η} denote each five Goldstone bosons from
the breaking G/H. Now assume that the diagonal subgroup F = SU(3)diag of G is
gauged, so that the kinetic part of (1.35) becomes
Lkin = (Dµφ1) † (D µ φ1) + (Dµφ2) † (D µ φ2) (1.37)
with Dµ = ∂µ − ig T a A a µ, a = 1, . . . , 8. This is an explicit breaking of the subgroup H,
so that only I = H ∩ F = SU(2)diag (generated by the T a s with the Pauli matrices
in the upper left corner) remains unbroken. The interaction terms from (1.37) can be
written as
�
i
�
|Dµφi| ∋ Tr g 2 (A a µT a ) 2 (φ1φ † †
1 + φ2φ 2 )
�
, (1.38)
where g is the weak coupling constant in this model and (ignoring the singlet η)
φ1φ † †
1 + φ2φ 2 =
�
h † h
0
0
f 2 − h † �
.
h
(1.39)
And, if we denote the new gauge bosons by X and the SM ones as usual, we find in
the mass eigenbasis
so that
A a µT a = 1
√ 2
�
Tr g 2 (A a µT a ) 2 (φ1φ † †
1 + φ2φ 2 )
�
⎛
B
⎜
⎝
0 µ + X8 µ
√3
W − µ
X 0
µ
W + µ
X 8 µ
√3 − B 0 µ
X + µ
X 0 µ
X − µ
−2
√ 3 X 8 µ
= g2
�
2 8
f
2 3 X8 µX µ8 + X 0
µX µ0 + X + µ X µ−
�
⎞
⎟
⎠ , (1.40)
+ g2
2 h2 � W + µ W µ− + 2B 0 µB µ0 − X + µ X µ− − 2X 8 µX µ8�
+ 2g2
√ 3 h 2 X 8 µB µ0 . (1.41)
Two things become evident from this result. First, the five gauge bosons corresponding
to the broken T a s, a = 1, . . . , 5 have eaten the κ fields from (1.36) and gain masses
proportional to the scale f, while the remaining three gauge bosons will become
massive if the Higgs field h acquires a vev. Second, the couplings to the Higgs in
the second line of (1.41) are of the same magnitude but of opposite sign between the
new gauge fields and the weak gauge fields. This leads to a cancellation (between
equal spin fields), which will assure that no quadratic divergent loop diagrams appear
in the theory. The reason is, that if one covariant derivative in (1.37) was replaced
by Dµ → ∂µ, all GBs of the corresponding scalar would be exact. Thus, only gauge
interactions including both scalars φ1 and φ2 break the shift symmetry and can give