104
CHAPTER 12. PROBLEM SET SOLUTIONS
12.4
Problem Set 4 Solutions
1. A nice derivation on the theoretical lower bound on Higgs mass (note exper­
imental lower bound is µH ≥ 120GeV by now) can be found in G. Attarelli,
G. Isideri, Phys. Lett. B337 p141 (’94) (available on the SPIRES). This uses
two–loop corrections and the result for (µt � 174GeV ) as a function of cut-off
scale Λ is as follows:
m H (GeV)
150
140
130
NOT ALLOWED
100
log
90
3
Λ
1 GEV
19
Figure 12.41: Lower Bound.
I will sketch the argument, see the paper for details.
Tree level Higgs potential is
Vtree = −
µ0 ∅2 λ0 ∅4
+
2
24
(12.164)
where we fix λ0 and µ0 at renormalization scale µ0 = µ = v � 245GeV (at weak
scale).
As µ → µ + δµ, Higgs field strength scales as ∅ → ∅(1 + δη) and λ → λ + δλ.
Accordingly the renormalized potential is
VRen =
1
λ(µ)((1 + δη(µ))∅)4
24
(12.165)
For one–loop corrections we can write
1 + δη � eδη = exp[−
�
log
O
Λ
µ0
γ[t� ]dt]
(12.166)
where
γ≡−
µ
δη
δη
δη = −
µ ≡ − �
δt
δ log µ0
δµ
(12.167)
12.4. PROBLEM SET 4 SOLUTIONS
105
These one–loop effects can turn the tail of Vtree (∅) in downwards for large ∅
hence VRen (∅) may become unstable.
Vtree
Ο
Figure 12.42: Vtree .
From Equation 12.165 the condition for stability is clearly
λ(µ) > 0
(12.168)
Thus we shall look at how λ evolves:
λ
∂
= β(t) =
(−δλ + 2δ∅ )
t
∂ ln µ
where δλ is the counterterm for
Figure 12.43: δλ Equation 12.169.
and δ∅ is Higgs self–renormalization
Figure 12.44: δ∅ Equation 12.169.
We just draw contributing diagrams and give the result:
take only top quark.
→ δλ (from log–divergent pieces)
(12.169)
106
CHAPTER 12. PROBLEM SET SOLUTIONS
+ 5 permutations
+ 5 permutations
+
+ 2 permutations
+
+ 2 permutations
+
+
+ 5 permutations
+
+ 5 permutations
Figure 12.45: Contributing Diagrams δλ .
+
Figure 12.46: Contributing Diagrams δ∅ .
take only top.
→ δ∅ (from log–divergent pieces)
Therefore,( see above mentioned paper) for the result. Actually I will quote
two–loop result:
dλ
1
9
27
=
[4λ2 + 12λgt2 − 36gt4 − 9λg12 − 3λg22 + g12 g22 + g14 ]
2
dt
16π
2
4
(12.170)
where gt is the Yukawa coupling for top quark, g1 and g2 are U (1)Y and SU (2)W
coupling.
We have
√
2mt
(a + δt (µ0 ))
V
3m2H
λt (µ0 ) =
(a + δλ (µ0 ))
V2
gt (µ0 ) =
(12.171)
(12.172)
12.4. PROBLEM SET 4 SOLUTIONS
107
at µ0 = weak scale � 100GeV . (note that the funny factors of 3 in
coming from my unusual Vtree definition).
λ
m2H
is
Therefore, we see that λ can become negative, hence VRen unstable for partic­
ular values of Λ, m2t and m2H . To find the range of m2H for which λ > 0 fix
m2t � 174GeV (ignore QCD corrections). Then solve the above equation for dλ
dt
numerically and the one gets the range as shown in Figure 12.41.
2. Consider
1
L[∅] = ∂µ ∅∂ µ ∅ − V (∅)
2
(12.173)
Suppose we expand around < ∅ >= v(x) where V (∅) looks like
V(O)
v
O
Figure 12.47: V (∅).
For simplicity lets take
V (∅) � ω∅2 + C
(12.174)
with ω < 0.
The ground state is a coherent state such that, for
d3 k 1
�
�
√ (ak eik·�x + ak+ e−ik·�x ) = ∅ˆ+ + ∅ˆ− (12.175)
3
(2π) 2k0
�
d3 k 1
�
√ ξ(k)eik·x� |ξ >
(12.176)
∅ˆ+ (�x)|ξ > =
(2π)3 2k0
∅(�x) =
�
Lets consider a 0 + 1 dimensional system for simplicity:
108
CHAPTER 12. PROBLEM SET SOLUTIONS
1
Ĥ = ω[at a + ]
2
< ξ |H |ξ > = ?
(12.177)
(12.178)
t
Solution to equation a|ξ >= ξ |ξ > is |ξ >= eξa |o > in this case
1
1 2
t
< ξ|H |ξ >= ω < o|eξa (at a + )eξa |o >= ω(ξ 2 + )eξ
2
2
(12.179)
Consider an excitation:
< ξ |aHat |ξ >=
d d
d d
1
�
< ξ � |H |ξ > |ξ=ξ� =
(ξξ � + )eξξ ω >< ξ|H |ξ >
�
�
dξ dξ
dξ dξ
2
(12.180)
Excitation get lower and lower energy (straightforward) since ω < 0 (for classical
unstable potential).
This reasoning is easily generalized to more non-trivial potentials which exhibit
non-stable behavior classically and to 4D QFT, in which case
�
|ξ >= exp[
d3 k 1
�
√ ξ(k)eik·x� akt ]|0 >
3
(2π) 2k0
(12.181)
3. (Thanks to Guide Festuccier) Observed value of
mb
|µ � 1.62
mτ µz
(12.182)
Both in supersymmetric and non-supersymmetric GUTs b and τ are in the same
multiplet and get the same mass through Higgs coupling, hence
mb
=1
|µ
mτ µGU T
(12.183)
To obtain the value at µ = µz we should run the Yukawa couplings down to
weak scale. One can ignore Yukawa couplings of other matter except top-quark:
MSSM:
12.4. PROBLEM SET 4 SOLUTIONS
d
ln λt
dt
d
ln λb
dt
d
ln λt
dτ
t
109
�
1
[6λ2t + λ
2b −
ci gi2 ]
2
16π
�
1
2
2
=
[−6λ
+
λ
−
c�i gi2 ]
τ
b
16π 2
�
1
2
2
=
[6λ
−
3λ
−
c�i� gi2 ]
τ
b
2
16π
µ
= ln( )
µz
=
(12.184)
(12.185)
(12.186)
(12.187)
where gi are ((U (1), SU (2), and SU (3)) coupling for i = 1, 2, 3 respectively and
the coefficients are:
MSSM:
13
16
, 3, )
5
3
7
16
= ( , 3, )
15
3
9
= ( , 3, 0)
5
ci = (
(12.188)
c�i
(12.189)
c��i
(12.190)
SM:
17 9
, , 8)
20 4
1 9
= ( , , 8)
4 4
9 9
= ( , , 0)
4 4
ci = (
(12.191)
c�i
(12.192)
c��i
(12.193)
Subtract equations for λb and λτ and neglect λb with respect to λt to get
MSSM:
2π
1
16
20
d
λb
(λt2 − g32 +
g12 )
ln( )2 �
dt
λτ
4π
3
15
(12.194)
d
λb
1
ln( )2 �
(λt2 − 8g32 + 2g
12 )
dt
λτ
4π
(12.195)
SM:
2π
110
CHAPTER 12. PROBLEM SET SOLUTIONS
One straightforwardly gets
MSSM:
� t(µGU T ) 2 �
mb
α3 (µz ) 8
− 12
λt dt
0
16π
(µz ) = (e
)9
)(
α3 (µGU T )
mτ
(12.196)
(ignoring O(g12 ) with respect to O(g22 )).
SM:
� t(µGU T ) 2 �
1
α3 (µz ) 4
mb
λt dt
)(
(µz ) = (e− 16π2 0
)7
α3 (µGU T )
mτ
(12.197)
We have to obtain α3 = α2 = α1 |GU T in MSSM vs. SM.
This is easily done by looking at one–loop running of MSSM and SM couplings
and their unification at µGU T , see for instance R. Mohapatra hep-th 9801235
r2.
1 2
g
(4π)2
1
�
24
1
�
42
α ≡
M SSM
αGU
T
SM
αGU
T
(12.198)
(12.199)
(12.200)
One obtains
mb
1 � 2
λt )|M SSM × 2.56
(µz )|M SSM = exp(−
mz
16π 2
mb
1 � 2
(µz )|SM = exp(−
λt )|SM × 2.51
mz
16π 2
(12.201)
(12.202)
The difference mostly depends of the running of top-Yakawa coupling in MSSM
vs. SM. According to R. H. Mohapatra, (“Suppersymmetry and Unification,”
mb
mb
� 3 in SM which is bad as m
� 1.62 in reality and,
Springer-Verlag, 2003) m
τ
τ
mb
�
2.3
in
MSSM.
But
the
latter
result
is
model-dependent.
Especially on the
mτ
particular breaking mechanism.