Labelle P. Supersymmetry DeMYSTiFied
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396
Supersymmetry Demystified
so that
∂
2
V
∂
R
+
u
2
vev
= a
1
+ 2c
v
2
u
− v
2
d
+
g
2
2
v
2
d
= b cot β +m
2
W
cos
2
β
∂
2
V
∂
R
−
d
2
vev
= a
2
− 2c
v
2
u
− v
2
d
+
g
2
2
v
2
u
= b tan β +m
2
W
sin
2
β
∂
2
V
∂ R
+
u
∂ R
−
d
vev
= b +
g
2
2
v
u
v
d
= b +
m
2
W
2
sin(2β) (16.29)
This gives the following mass matrix:
V =
⎛
⎜
⎝
b cot β +m
2
W
cos
2
β b + m
2
W
sin(2β)/2
b + m
2
W
sin(2β)/2 b tan β +m
2
W
sin
2
β
⎞
⎟
⎠
The eigenvalues are
b sec(2β) +
m
2
W
2
±
1
2
)
2 cot(2β)b + m
2
W
cos(2β)
2
+ 4
b + m
2
W
sin(2β)/2
2
Again, using b = m
2
A
0
sin(2β)/2 simplifies greatly the result, which turns out to be
0 and m
2
A
0
+ m
2
W
The state with zero mass gets eaten through the Higgs mechanism. The massive
state is a new scalar field. Because its mass contains m
A
0
, it does not have any upper
limit.
Finally, we have the pair I
+
u
, I
−
d
. It’s easy to see that we get all the same terms
as for the pair R
+
u
, R
−
d
except that the cross-terms in I
+
u
I
−
d
have the opposite signs
as the cross-terms in R
+
u
R
−
d
. This means that the off-diagonal elements of the mass
matrix will have the opposite sign to what we previously had, but this does not
change the eigenvalues, so we again get a massless state and a massive state with
mass m
2
A
0
+ m
2
W
.
The two massless states that we just found provide the longitudinal modes of the
W
±
. The two charged massive states are referred to as H
+
and H
−
with degenerate
masses:
m
2
H
±
= m
2
W
+ m
2
A
0
To summarize, we have found that the spectrum of the Higgs sector in the
MSSM consists into three massless and five massive states. The three massless
CHAPTER 16 Phenomenological Implications
397
states play the same role as the massless states in the standard model: They become
the longitudinal modes of the massive W
±
and Z. One massive state, the A
0
, has a
mass squared
m
2
A
0
=
2b
sin(2β)
Since there is no lower limit on β, there is no upper limit on how massive this
state could be. This is not very exciting from the point of view of phenomenology
because there is no way to “kill” the model: If the A
0
is not observed, it will simply
put an upper limit on β. The masses of the other four states all can be expressed in
terms of m
W
, m
Z
, m
A
0
, and β:
m
2
H
±
= m
2
W
+ m
2
A
0
m
H
0
=
m
2
A
0
+ m
2
Z
2
+
1
2
)
m
2
A
0
+ m
2
Z
2
− 4m
2
A
0
m
2
Z
cos
2
2β
m
2
h
0
=
m
2
A
0
+ m
2
Z
2
−
1
2
)
m
2
A
0
+ m
2
Z
2
− 4m
2
A
0
m
2
Z
cos
2
2β (16.30)
If we consider the limit β → 0 (and b = 0), we get the following limits for the
Higgs masses:
m
A
0
→∞
m
H
±
→∞
m
H
0
→∞
m
h
0
→ m
Z
|cos(2β)|
On the other hand, in the limit b → 0 (and β = 0) we get
m
A
0
→ 0
m
H
±
→ m
W
m
H
0
→ m
Z
m
h
0
→ 0
The important point to notice is that the mass of the h
0
does not tend to infinity
in any limit. Even better, it turns out that the value corresponding to the first limit,
398
Supersymmetry Demystified
m
Z
|cos(2β)|,istheupper limit to the mass of the h
0
! This can be seen more clearly
if we invert the last equation of Eq. (16.30) to get
m
2
A
0
= m
2
h
0
m
2
Z
− m
2
h
0
m
2
Z
cos
2
(2β) − m
2
h
0
which shows clearly that
m
h
0
< m
Z
|cos(2β)| (16.31)
This means that the MSSM predicts a Higgs particle lighter than the Z ! But this is
catastrophic because it is already experimentally excluded
4
:
m
H
≥ 114.4 GeV (confidence level of 95%)
However, it is important to keep in mind that all the results we have obtained in
this section are only tree-level relations. They are modified by loop corrections, and
sometimes the corrections are sizable. Calculations of loop diagrams in the MSSM
are beyond the scope of this book. Let’s just mention that the main correction
to the bound (16.31) arises from loop diagrams containing the top quark and its
supersymmetric partner, the stop (or top squark), and is given by
m
h
0
<
,
-
-
.
m
2
Z
+
3g
2
8π
2
m
2
t
M
2
W
ln
m
2
˜
t
m
2
t
Note that if SUSY were unbroken, the stop mass would be equal to the top mass,
and the logarithmic correction would be identically zero. Using a mass of 1 TeV as
an estimate of the stop mass, this brings up the bound on the h
0
to approximately
m
h
0
< 130 GeV
which is not yet excluded experimentally.
CHAPTER 16 Phenomenological Implications
399
16.6 Masses of the Leptons, Quarks,
and Their Superpartners
Let’s first have a look at the masses of the leptons and quarks in the MSSM.
Consider the term in the superpotential equal to
y
ij
u
U
i
Q
j
◦ H
u
We would like to extract from this the coupling of the standard model particles
to the Higgs bosons. The Higgs bosons are the scalar part of the superfield H
u
.
The standard model particles are contained in U
i
and Q
j
. The F terms containing
a scalar field and two fermions when we expand the product of the three chiral
superfields, is −φχ
A
· χ
B
/2. Therefore, we finally get
−
y
ij
u
2
¯
u
Li
· u
j
H
0
u
+
¯
u
Lj
· u
i
H
0
u
−
¯
u
Li
· d
j
H
+
u
−
¯
u
Lj
· d
i
H
+
u
The first two terms will generate masses for the fermions. Setting H
0
u
= v
u
, this
gives
−
y
ij
u
2
v
u
¯
u
Li
· u
j
+
¯
u
Lj
· u
i
=−y
ij
u
v
u
¯
u
Li
· u
j
using the symmetry of the Yukawa couplings.
If we add the hermitian conjugate, we get a Dirac mass term. After going to a
diagonal basis, this will lead to masses for the up, charm, and top quarks of the
form
m
u,c,t
= v
u
y
u,c,t
where the y
u,c,t
are linear combinations of the y
ij
u
.
Now consider another term in the superpotential that leads to a fermion mass:
−y
ij
d
D
i
Q
j
◦ H
d
Keeping only the terms that yield fermion masses, we get (we pick up an extra
minus sign due to the fact that the vev of iτ
2
H
d
is a doublet with components
400
Supersymmetry Demystified
(0, −v
d
) in contrast with the vev of iτ
2
H
u
which had a vev equal to (v
u
, 0))
−
y
ij
d
2
H
0
d
¯
d
i
· d
j
+
¯
d
j
· d
i
=−y
ij
d
v
d
¯
d
i
· d
j
If we add the hermitian conjugate, this generates Dirac mass terms. After diago-
nalization, this yields masses
m
d,s,b
= v
d
y
d,s,b
The third and final term will give the masses to the leptons:
−y
ij
e
E
i
L
j
◦ H
d
Again, keeping only the relevant terms (that generate fermion masses), we get
−
y
ij
e
2
H
0
d
(
¯
e
Li
· e
Lj
+
¯
e
Li
· e
Li
) =−y
ij
e
H
0
d
¯
e
i
· e
Lj
Again, after setting the Higgs field to its vev, this generates Dirac masses of the
form v
0
d
y
ij
e
. After diagonalizing, we get lepton masses
m
e,μ,τ
= v
d
y
e,μ,τ
The end result is that we get exactly the same type of fermion mass terms as in
the standard model, so in this respect, the MSSM can’t be differentiated from the
standard model.
On the other hand, once we introduce soft supersymmetry breaking terms, which
include mass terms for scalar fields, we lose all predictive power for the masses of
the sleptons and squarks.
16.7 Some Other Consequences of the MSSM
We have seen that a prediction of the minimal supersymmetric standard model
(MSSM) is a relatively light Higgs scalar field. The phenomenology of the MSSM
is extremely rich, and as mentioned earlier, this topic alone could fill an entire book.
Let us only mention again that if R-parity is conserved, the lightest supersym-
metric particle (LSP) is absolutely stable, and because of this, it should fill the
universe and represents therefore a promising dark matter candidate. The exact
nature of the LSP is model-dependent, with the two most likely candidates being
CHAPTER 16 Phenomenological Implications
401
the neutralino and the gravitino (which appears only in theories of supergravity,
i.e., when SUSY is gauged).
Another consequence of R-parity is that the superpartners of the known particles
can only be created in pairs. If they are weakly interacting, their existence thus could
show up in the form of missing energy in scattering events, missing energy that
would be equal to at least twice the mass of the LSP.
Unfortunately, we are running out of space (and I am running out of energy!),
so we won’t be able to discuss any of these fascinating topics more. To learn more
about the phenomenology of the MSSM, you are invited to consult References 1,
5, 8, 10, 12, 29, 32, 35 and 47.
16.8 Coupling Unification in
Supersymmetric GUT
We will cover one last topic. If you have read about SUSY in particle physics
before, you have probably heard about a very exciting feature of the theory: a near-
perfect unification of the coupling constants at the so-called unification scale. To
be precise, this is not a feature of the MSSM itself but of supersymmetric grand
unified theories (GUT). Let’s see how this works.
First, let’s review the situation in nonsupersymmetric theories and why things
don’t look so good for them. We need to consider the “running” of coupling con-
stants with energy. This is given by the renormalization group (RG) equation
1
α(E)
=
1
α(E
0
)
+
β
2π
ln
E
E
0
where E
0
and E are two energy scales, and β is the “beta function” of the theory
calculated in perturbation theory (an explicit expression will be provided shortly).
α is generically defined as the square of the coupling constant over 4π . We therefore
define
α
g
≡
g
2
4π
α
g
≡
g
2
4π
α
s
≡
g
2
s
4π
with g
, g, and g
s
being the coupling constants of the U (1)
Y
, SU(2)
L
, and SU(3)
C
groups, as introduced in Section 15.2. The term coupling constants is misleading
because these parameters vary with energy. Note that the α’s themselves are also
often referred to as the coupling constants of the three forces.
402
Supersymmetry Demystified
The coupling constants g
, g, and g
s
are not the parameters extracted directly
from experiments. Instead, the following quantities are usually quoted (see Ref. 4):
α
−1
em
(M
Z
) ≈ 128
α
−1
s
(M
Z
) ≈ 8.47
sin
2
θ
W
(M
Z
) ≈ 0.2312
where, as indicated, all the values are obtained at the energy scale E = M
Z
. Our
goal is not to offer a detailed numerical analysis of the data, so we will not bother
with the uncertainties. The errors are in the last digit of each result quoted, α
s
having
the largest percentage uncertainty at ±1.7%.
The fine structure constant α
em
is, of course, defined as
α
em
=
e
2
4π
=
g
2
4π
sin
2
θ
W
where we have used e = g sin θ
W
. Therefore, we have
1
α
g
(M
Z
)
=
sin
2
θ
W
α
em
(M
Z
)
≈ 29.6 (16.32)
On the other hand, g
= g tan θ
W
= e/ cos θ
W
,so
1
α
g
(M
Z
)
=
cos
2
θ
W
α
em
(M
Z
)
≈ 98.4 (16.33)
Now that we have the values of all the inverse constants at the Z mass, the next
step is to calculate the beta functions β
1
,β
2
, and β
3
corresponding to each group
and to “run” the coupling constants to see if they converge to a single value at some
energy scale. This scale then would be interpreted as a unification scale,orGUT
scale, at which the three forces would merge into a single, unified gauge theory.
However, it does not make sense to do this directly with the coupling constants
of the standard model. To understand why, consider the gauge-covariant derivative
CHAPTER 16 Phenomenological Implications
403
of a weak doublet of quarks, for example. It is of the form
D
μ
= ∂
μ
+ig
s
λ
a
2
A
a
μ
+ig
τ
i
2
W
i
μ
+ig
Y
2
B
μ
We chose the example of a weak doublet of quarks simply to have all three gauge
fields present in the covariant derivative.
We see right away that the value of g
is totally arbitrary because any change
in it can be compensated by a change in the hypercharge assignments of all the
particles. We could divide the value of g
by an arbitrary factor c
, and all physical
results would remain identical if we would in addition multiply by the same factor,
the hypercharge of all particles. In other words, we may make the replacement
g
Y →
g
c
c
Y
and define g
/c
as a new coupling constant without affecting any physical result!
It would simply rescale the hypercharge assignments of all the particles without
changing the physics.
The same freedom is present in the nonabelian gauge groups. We may rescale
the SU(2)
L
and SU(3)
C
generators by factors of c and c
s
, respectively, if we also
rescale the corresponding coupling constants by the same factors. In other words,
we could make the replacements
gτ
i
→
&
g
c
'
cτ
i
g
s
λ
a
→
g
s
c
s
c
s
λ
a
without affecting any physical result.
To put it another way, for a nonabelian gauge group,
*
we impose a normalization
condition on the generators of the form
Tr(G
i
G
j
) = N δ
ij
(16.34)
where the G
i
are the generators. The normalization constant is arbitrary; it is usually
chosen to be 1/2, but it could be changed at the condition of rescaling appropri-
ately the corresponding coupling constant (here, the trace is understood as being
*
More specifically, for a simple group.
404
Supersymmetry Demystified
calculated with respect to the fermions present in the theory, as we will demonstrate
explicitly shortly).
With this in mind, we can rewrite the covariant derivative as
D
μ
= ∂
μ
+i
g
s
c
s
c
s
λ
a
2
A
a
μ
+i
&
g
c
'
cτ
i
2
W
i
μ
+i
g
c
c
Y
2
B
μ
The problem is now clear. Since the standard model is based on a direct product
of three groups, all three coupling constants can be independently rescaled. We can
use the beta functions of the three gauge groups to run the coupling constants, but
if we can choose arbitrarily which value to take for each coupling constant at the
Z mass, it is meaningless to ask if they meet at some well-defined GUT scale.
There is, however, a way to set the relative scale of the coupling constants. The
key observation is that within each nonabelian group, there is only a single coupling
constant even though there are several generators. Consider the SU(2)
L
interaction,
for example. We could not write the covariant derivative as
D
μ
= ∂
μ
+ig
1
τ
1
2
W
i
μ
+ig
2
τ
2
2
W
i
μ
+ig
3
τ
3
2
W
i
μ
with the g
i
being unequal. The reason is that the group structure of the interaction,
through Eq. (16.34), imposes a common normalization to all generators, implying
that g
1
= g
2
= g
3
.
Going back to the issue of the three independent coupling constants of the stan-
dard model, the solution is now obvious: If the three forces of the standard model
are unified into a single theory at some high energy scale, their gauge groups must
be subgroups of a larger, grand unified group. This implies that there is a common
normalization condition that must be imposed on all the generators of the standard
model, including the hypercharge!
The overall normalization chosen does not matter, only the fact that we must use
the same one for all the generators. For the SU(2)
L
and SU(3)
C
, we will pick τ
3
and
λ
3
as the representative generators because they are diagonal in the basis used for
the particles of the standard model, which will make our life easier (any generator
can be used in principle, but if they are not diagonal, more work is needed to find
their eigenvalues; note that λ
8
is also diagonal and could be used as easily as λ
3
).
Imposing that all the generators of the standard model have the same normalization
then leads to the relation
Tr
c
Y
2
2
= Tr
c
τ
3
2
2
= Tr
c
s
λ
3
2
2
(16.35)
CHAPTER 16 Phenomenological Implications
405
Recall that T
3
is defined as the operator τ
3
/2, so the second term of Eq. (16.35)
may be written as
Tr
c
τ
3
2
2
= Tr(cT
3
)
2
Since the overall scale is irrelevant, let’s set c
s
= 1 [in other words, we choose to
set the normalization for the generators of the unified group equal to the normaliza-
tion of the SU(3)
C
group of the standard model]. With this choice, the normalization
of the other generators is uniquely fixed:
c
2
=
1
4
Tr(λ
3
)
2
Tr(T
3
)
2
c
2
=
Tr(λ
3
)
2
Tr(Y )
2
(16.36)
The values of these coefficients determine the coupling constants we must evolve
with the renormalization group. These are g
s
, g/c, and g
/c
. In other words, we
will evolve α
s
, α
g
/c
2
and α
g
/c
2
.
By taking the trace, we mean summing over all the eigenvalues corresponding
to the particles of the standard model. The hypercharge assignments of the standard
model particles are given in Chapter 12. The eigenvalues of T
3
and the hypercharge
are related by Y = 2(Q − T
3
), where, by a slight abuse of the language, we use the
symbol T
3
to represent the eigenvalue corresponding to the operator with the same
name, and where Q is the electric charge. The eigenvalues of λ
3
are, of course, zero
for the leptons. Each generation contributes equally to the traces in Eq. (16.35), so
we can simply impose the relation for the first-generation. The quantum numbers
of the first-generation particles are listed in Table 16.1.
Keeping in mind that each quark comes in three colors, we get, for the first
generation,
Tr(Y
2
) = 3
1
3
2
+ 3
4
3
2
+ 3
1
3
2
+ 3
−
2
3
2
+ (−1)
2
+ (−2)
2
+ (−1)
2
=
40
3
(16.37)
It is also straightforward to calculate
Tr(T
3
)
2
= 3
1
2
2
+ 3
−
1
2
2
+
−
1
2
2
+
1
2
2
= 2 (16.38)