Labelle P. Supersymmetry DeMYSTiFied
Подождите немного. Документ загружается.
376
Supersymmetry Demystified
It is more useful to restrict this symmetry to the value ϕ = π , in which case the
R-symmetry is simply referred to as R-parity (or sometimes, matter parity) because
the phase is equal to either −1or+1.
Now let’s impose that the MSSM lagrangian be invariant under R-parity. We
take the gauge superfields to be invariant under this symmetry. It is then clear that
all the kinetic energy terms of the left-chiral superfields are invariant because they
contain the product of the hermitian conjugate of each superfield with the superfield
itself. On the other hand, the superpotential is not automatically invariant.
Recall that extracting the F term is equivalent to computing the following inte-
gral:
W
F
=
d
2
θW
We see that in order for this term to be invariant under R-parity, the superpotential
must be invariant (because d
2
θ is invariant under R-parity).
Now we need to assign the R-parity of the MSSM left-chiral superfields. We
want to keep terms with two fermion superfields (quark or lepton) and one Higgs
or with two Higgs superfields, but we need to eliminate terms with three fermion
superfields or one Higgs and one fermion superfield. Obviously, we should take
the two Higgs superfields to be even under R-parity, whereas we take all quark and
lepton superfields to be odd.
As an aside, there is a simple formula that gives the R-parity of the component
fields:
R = (−1)
3B+L+2s
(15.44)
where B and L are the baryon and lepton quantum numbers, and s is the spin.
Because of the factor of (−1)
2s
, particles and their superpartners have opposite
R-parities. This formula is sometimes written as
R = (−1)
3(B−L)+2s
(15.45)
which is obviously equivalent to Eq. (15.44) because the difference in the exponent
is 6L, which is always even.
Consider, for example, the spinor component χ
¯
e
of the superfield E. In this case,
B = 0, L =−1, and s = 1/2, so this field has even parity. Since it multiplies a
factor θ which is odd under R-parity, the superfield E has odd parity, in agreement
with our previous assignment. It is easy to check that all the standard model particles
(and the Higgs scalars) have R = 1, whereas all their superpartners have R =−1.
We see that imposing invariance under R-parity eliminates all interactions vi-
olating conservation of the baryon and lepton quantum numbers. This turns out
to be true only because we are working with renormalizable interactions. When
CHAPTER 15 Introduction to the MSSM
377
nonrenormalizable interactions are included, in the spirit of effective field theo-
ries, this correspondence is no longer valid, and therefore one may write baryon
and lepton number violating terms that respect R-parity (but, of course, they are
suppressed by factors of some high energy scale).
One striking implication of invariance under R-parity is that there are no interac-
tions coupling a single superpartner to two standard particles. Therefore, the lightest
supersymmetric particle (LSP) is absolutely stable! It cannot decay to anything. It
means that the universe must be filled with these particles. Experiments carried
on the charge-to-mass ratio of matter rules out the possibility that such particles
may be charged. The LSP therefore must be electrically neutral. There is much
leeway in the masses of the superpartners, as we will discuss briefly in Chapter 16,
so different models have different particles playing the role of the LSP, but it is
usually taken to be the neutralino, the scalar superpartner to the electron neutrino.
This opens up the exciting possibility that the LSP could account for dark matter
(in supergravity theories, the gravitino is also a candidate of choice). More details
can be found in Refs. 8 and 47.
Other consequences of R-parity that are relevant for their search at particle
accelerators is that superpartners can be produced only in pairs and that a super-
partner may decay only into standard particles plus an odd number of superpartners.
More detailed discussion of the phenomenology of the MSSM can be found in the
two books cited above, as well as in Refs. 1 and 5.
15.8 The General MSSM Superpotential
Now let’s generalize to include all three generations. We get
W = y
ij
u
U
i
(Q
j
◦ H
u
) − y
ij
d
D
i
(Q
j
◦ H
d
) − y
ij
e
E
i
(L
j
◦ H
d
) + μ H
u
◦ H
d
(15.46)
where the indices i and j label the generations. The signs in the various terms have
been chosen in order for the mass terms for the quark and leptons come out with
the right sign if we take the Yukawa couplings to be positive, as we will briefly
discuss in the last chapter.
This is the superpotential. Of course, to obtain the lagrangian, we must extract
the F term. The use of the symbol μ for the coefficient of the term with two Higgs
superfields (sometimes referred to as the μ-term) is conventional. This parameter
may be complex and should not be confused with the real parameter μ of the
standard model. Note that we haven’t included a Fayet-Illiopoulos term for the
U (1)
Y
symmetry. The reason is that D-type supersymmetry breaking is not possible
in the MSSM because it conflicts with the absence of light squarks and sleptons, as
we will discuss in Section 16.1 (and already mentioned in Chapter 14). Of course,
378
Supersymmetry Demystified
this does not give any theoretical justification for dropping that term. The real reason
is that we will have to include supersymmetry soft breaking terms which contain
all the possible interactions that would have been generated by a Fayet-Illiopoulos
term anyway, so there is no need to introduce one here.
The parameters y
ij
u
, y
ij
d
, and y
ij
e
correspond precisely to the Yukawa couplings
present in the standard model (more about this in Chapter 16). We therefore see
that, quite remarkably, the minimal supersymmetric extension of the standard model
contains the same number of parameters as the standard model since we have traded
the real parameters μ and λ (or, if you prefer, v and λ) for the MSSM complex
parameter μ. This is amazing since one could have expected that enlarging the
standard model to make it supersymmetric would have entailed the introduction
of a plethora of new constants. However, this is for the theory with unbroken
supersymmetry; when we will break SUSY, the number of new parameters will
unfortunately skyrocket.
We have succeeded in obtaining a supersymmetric version of the standard
model, but unfortunately, it is not phenomenologically viable. The problem is
that SUSY is unbroken (and it turns out, the SU(2)
L
⊗U (1)
Y
gauge symmetry is
unbroken as well). Indeed, it is easy to see that that if we write our superpotential
in terms of the scalar component fields, it is minimized—and equal to zero—when
all the scalar fields are equal to zero. Therefore, SUSY is unbroken, and each
particle of the standard model should have a superpartner of equal mass, which is
ruled out experimentally.
The question then becomes: How should we break SUSY? It turns out that the
simple mechanisms of SSB described in Chapter 14 are not phenomenologically
viable, and we are forced to include terms that explicitly break SUSY. Of course,
we shall include only terms that break SUSY softly because we don’t want to ruin
the most attractive aspect of SUSY as far as phenomenology is concerned: the
cancellation of quadratic divergences. We will discuss these issues in more details
in Chapter 16.
15.9 Quiz
1. Why are two Higgs superfields required in the MSSM?
2. How many new parameters (relative to the standard model) did we need to
introduce to build its supersymmetric extension?
3. What are the R-parity of the left-chiral superfields of the MSSM? Write down
a dimension 5 interaction that would respect R-parity.
4. What are the quantum numbers of the two Higgs superfields?
5. How do the Grassmann variables θ and
¯
θ transform under R-parity?
CHAPTER 16
Some Phenomenological
Implications of
the MSSM
In this chapter we will briefly discussed some phenomenological implications
of the minimal supersymmetric standard model (MSSM). This is a vast subject
that could easily fill an entire book just by itself. A choice therefore had to be
made between providing an overview of a large number of results and giving an
indepth coverage of only one or two aspects of particular interest. We have opted
for the second approach because it is more appropriate for a self-teaching book
whose purpose is to demystify a topic. Moreover, there are already many good
reviews available, several of which are mentioned in Section 16.7, whose purpose
is precisely to offer summaries of the MSSM phenomenology.
380
Supersymmetry Demystified
16.1 Supersymmetry Breaking in the MSSM
We learned about two mechanisms to spontaneously break supersymmetry (SUSY)
in Chapter 14. But, as we also discussed, both lead to the prediction that some super-
partners would be lighter than the known leptons, a prediction that is in conflict with
experimental observations. The prediction of the existence of light superpartners
was a consequence of the vanishing of the supertrace. In the case of F-type breaking,
the vanishing of the supertrace is automatic, whereas in the case of D-type breaking,
it follows from the fact that the hypercharges of the MSSM particles add up to zero,
making the right-hand side of Eq. (14.24) vanish. The fact that the hypercharges
add up to zero is bad news for spontaneous supersymmetry breaking (SSB), but it
is necessary for the cancellation of certain anomalies.
As discussed in Section 14.9, one way out is to introduce explicit SUSY-breaking
terms that we take to break SUSY softly, in order to retain the nice ultraviolet
behavior that cures the hierarchy problem. As also mentioned in that section, the
presence of these terms may be justified in the context of models where there
is a hidden sector interacting with the MSSM particles through some messenger
interaction. This topic is beyond our scope (see Refs. 43 and 47 for more details);
we will simply introduce the soft SUSY-breaking terms by hand without justifying
their existence any further.
Recall the three types of possible soft SUSY-breaking terms, as described in Sec-
tion 14.9. These are mass terms for the scalar particles (by mass terms, we include
terms that are bilinear in two scalar fields), interaction terms between scalar fields
that are holomorphic and superrenormalizable (in other words, terms of the form
φ
i
φ
j
φ
k
, where the indices don’t have to be different), and finally, mass terms for
the spin 1/2 superpartners of the gauge fields, the gauginos. Because the mass terms
and the cubic interactions may, in principle, mix different families, this amounts
to a very large number of new parameters, 105 to be precise! More specifically,
there are
•
5 real parameters and 3 phases that violate charge conjugation C and parity P
(CP violation) in the fermion sector of the gauge and Higgs supermultiplets
(i.e., in the higgsinos and gauginos sector)
•
21 masses, 36 mixing angles, and 40 CP-violating phases in the slepton and
squark sector
These parameters are, of course, constrained by experimental results on CP
violation, flavor changing neutral currents, etc. But there is still quite a bit of
leeway. A popular model is minimal supergravity or minimal SUGRA, in which
CHAPTER 16 Phenomenological Implications
381
there are only five additional parameters relative to the MSSM. More detail can be
found in Section 9.2 of Ref. 1 and in Section 8.1.1 of Ref. 5, as well as in Ref. 10.
Now that SUSY breaking is “taken care of,” we need to make sure that the
electroweak gauge symmetry is also broken and that the predicted masses for the
gauge bosons and Higgs fields do not conflict with experiments. We will therefore
focus on this issue in the next section.
16.2 The Scalar Potential, Electroweak
Symmetry Breaking, and All That
We have many fields that contribute to the scalar potential: the sleptons, the squarks,
and the Higgs fields, which means that the scalar potential is fairly long. However,
we are interested only in the value of the potential when the fields are set equal to
their vacuum expectation values (vevs) to see if the electroweak gauge symmetry
is broken spontaneously.
One thing is clear: We do not want to spontaneously break down charge or color
invariance. Therefore, we must take all the scalar fields that have a nonzero charge
or that carry color to have vanishing vevs. This includes all sleptons (except the
sneutrino) as well as the squarks and the charged Higgs. In addition, we do not
want R-parity to be broken spontaneously, to avoid the reappearance of baryon and
lepton number violating terms, so we take the vev of the sneutrino, which is odd
under R-parity, to be zero as well. We therefore only need to include the Higgs field
terms when we write down the scalar potential. Note that the only fields that may
have a nonzero vev are the neutral Higgs scalar fields H
0
u
and H
0
d
with hypercharges
1 and −1, respectively.
Four types of terms contribute to the scalar potential after elimination of the
auxiliary fields:
V =
i
∂W
∂φ
i
2
+
1
2
D
2
+
1
2
D
a
D
a
+ V
SSB
=
i
∂W
∂φ
i
2
+
g
2
2
i
Y
i
2
H
†
i
H
i
2
+
g
2
2
H
†
u
τ
i
2
H
u
+ H
†
d
τ
i
2
H
d
H
†
u
τ
i
2
H
u
+ H
†
d
τ
i
2
H
d
+ V
SSB
(16.1)
where V
SSB
is the soft SUSY-breaking potential, to which we will come back shortly.
382
Supersymmetry Demystified
The first D term is for the U (1)
Y
gauge symmetry, and the second D term is for
the SU(2)
L
gauge symmetry (we will be more explicit shortly). There is no corre-
sponding SU(3)
C
term because the Higgs fields are color singlets. In this equation,
φ
i
stands for the four Higgs fields H
+
u
, H
0
u
, H
−
d
, H
0
d
and the SU(2) doublets H
u
and
H
d
are defined as
H
u
≡
⎛
⎝
H
+
u
H
0
u
⎞
⎠
and H
d
≡
⎛
⎝
H
0
d
H
−
d
⎞
⎠
(16.2)
Now let’s consider each term in Eq. (16.2) in turn. To calculate the first term, we
simply express the part of the superpotential that contains the Higgs superfields in
terms of the corresponding scalar fields. The term in the superpotential that contains
only the Higgs superfields is
μ H
u
◦ H
d
F
+ h.c. =
i
μ H
T
u
iτ
2
H
d
+ h.c. (16.3)
Now we replace the superfield doublets by their scalar field components H
u
and
H
d
given in Eq. (16.2) and expand:
W =
i
μ
H
+
u
H
−
d
− H
0
u
H
0
d
+ h.c. (16.4)
The first term of Eq. (16.2) is therefore quite simple:
∂W
∂φ
i
2
=|μ|
2
|H
0
u
|
2
+|H
+
u
|
2
+|H
0
d
|
2
+|H
−
d
|
2
The U(1)
Y
contribution in Eq. (16.1) is
g
2
8
|H
+
u
|
2
+|H
0
u
|
2
−|H
0
d
|
2
−|H
−
d
|
2
2
where we used the fact that the hypercharge assignments of H
u
and H
d
are 1 and
−1, respectively.
The SU(2) contribution is
g
2
2
H
†
u
τ
i
2
H
u
+ H
†
d
τ
i
2
H
d
H
†
u
τ
i
2
H
u
+ H
†
d
τ
i
2
H
d
CHAPTER 16 Phenomenological Implications
383
Expanding this out, we get
g
2
8
|H
+
u
|
2
+|H
0
u
|
2
−|H
0
d
|
2
−|H
−
d
|
2
2
+
g
2
2
H
+
u
H
0
†
d
+ H
0
u
H
−†
d
H
+†
u
H
0
d
+ H
0†
u
H
−
d
The SUSY-breaking terms are
V
SSB
= δm
2
H
u
H
†
u
H
u
+ δm
2
H
d
H
†
d
H
d
+ (bH
u
◦ H
d
+ h.c.) (16.5)
Watch out here! The invariant SU(2) dot product between two SU(2) doublets
involved the usual iτ
2
, but the SU(2)-invariant dot product between a doublet and
an antidoublet involves the identity 2 ×2 matrix, which is why there is no dot-
product symbol or matrices between the doublets in the first two terms of Eq. (16.5).
Expanding, we have
V
SSB
= δm
2
H
u
H
+†
u
H
0†
u
1
⎛
⎝
H
+
u
H
0
u
⎞
⎠
+ δm
2
H
d
H
0†
d
H
−†
d
1
⎛
⎝
H
0
d
H
−
d
⎞
⎠
+b
H
+
u
H
0
u
iτ
2
⎛
⎝
H
0
d
H
−
d
⎞
⎠
− b
∗
H
0†
d
H
−†
d
iτ
2
⎛
⎝
H
+†
u
H
0†
u
⎞
⎠
(16.6)
where the extra minus sign in the last term comes from the fact that the dagger of
iτ
2
is −iτ
2
. So we finally get
V
SSB
= δm
2
H
u
|H
+
u
|
2
+ δm
2
H
u
|H
0
u
|
2
+δm
2
H
d
|H
0
d
|
2
+ δm
2
H
d
|H
−
d
|
2
+ b(H
+
u
H
−
d
− H
0
u
H
0
d
+ h.c.)
Combining all terms together, we find that the full scalar potential is
V =
|μ|
2
+ δm
2
H
u
|H
+
u
|
2
+|H
0
u
|
2
+
|μ|
2
+ δm
2
H
d
|H
0
d
|
2
+|H
−
d
|
2
+
b
H
+
u
H
−
d
− H
0
u
H
0
d
+ h.c.
+
g
2
+ g
2
8
|H
+
u
|
2
+|H
0
u
|
2
−|H
0
d
|
2
−|H
−
d
|
2
2
+
g
2
2
|H
0†
d
H
+
u
+ H
−†
d
H
0
u
|
2
384
Supersymmetry Demystified
Let us define
a
1
≡|μ|
2
+ δm
2
H
u
a
2
≡|μ|
2
+ δm
2
H
d
c ≡
g
2
+ g
2
8
(16.7)
Using this notation, the full scalar potential is
V = a
1
|H
+
u
|
2
+|H
0
u
|
2
+ a
2
|H
0
d
|
2
+|H
−
d
|
2
+b
H
+
u
H
−
d
− H
0
u
H
0
d
+ h.c.
+ c
|H
+
u
|
2
+|H
0
u
|
2
−|H
0
d
|
2
−|H
−
d
|
2
2
+
g
2
2
|H
0†
d
H
+
u
+ H
−†
d
H
0
u
|
2
(16.8)
Note that the full scalar potential contains five parameters. Of those, c and g
have known values; c ≈ 0.069 and g ≈ 0.651. The other three are free parameters
and could be either positive or negative.
16.3 Finding a Minimum
The situation is obviously quite a bit more complex than in the standard model
because we have twice as many Higgs fields.
As in the standard model, the equations are simplified if we use the SU(2)
invariance to set a component of either H
u
or H
d
equal to zero at the minimum (in
other words, we can set the vev of one component to be zero without any loss of
generality). Let’s choose to set H
+
u
= 0, which, of course, also implies H
+†
u
= 0at
the minimum.
In addition, we will impose that H
−
d
has a zero vev because otherwise electro-
magnetism would be broken spontaneously, which is not observed experimentally.
Of course, if we were to find out that it is impossible to have spontaneous symmetry
breaking of SU(2)
L
×U (1)
Y
to U (1)
em
if we impose H
−
d
=0, the MSSM would
be ruled out experimentally from the very start. As you might expect, this won’t be
the case, and we will indeed find that a zero value of the vev of H
−
d
does not clash
with SSB of the gauge symmetry.
A first consistency check is to verify that the full potential has indeed an extremum
at H
+
u
= H
−
d
= 0 (from now on, we drop the expectation value symbols around
CHAPTER 16 Phenomenological Implications
385
the vevs). In other words, we must verify that the partial derivatives
∂V
∂ H
+
u
∂V
∂ H
−
d
∂V
∂ H
+†
u
and
∂V
∂ H
−†
d
vanish when we set H
+
u
= H
−
d
= 0 (and, of course, H
+†
u
= H
−†
d
= 0 as well). That
this is actually the case is easy to see just by looking at the potential because there
are no terms containing only one of these four fields.
Therefore, at the minimum, the Higgs potential reduces to
V
n
= a
1
|H
0
u
|
2
+ a
2
|H
0
d
|
2
− b
H
0
u
H
0
d
+ h.c.
+ c
|H
0
u
|
2
−|H
0
d
|
2
2
where the label n is to indicate that this potential contains only the neutral fields
H
0
u
and H
0
d
.
Note that the only term that depends on the phase of the fields is the term pro-
portional to b. Without loss of generality, we then can take b to be real and positive
because its phase always can be absorbed into the phase of the fields. Actually,
b = 0 is also possible (nothing says that this term must be present), so we take
b ≥ 0
We will make one last simplification before finally working out the pattern of
SSB: We will assume that CP (the symmetry under the combined operations of
parity and charge conjugation) is not broken spontaneously, as is the case in the
standard model. The consequence is that any nonzero vev must be real, in which
case the potential at the minimum is simply
V
n
= a
1
H
0
u
2
+ a
2
H
0
d
2
− 2bH
0
u
H
0
d
+ c
H
0
u
2
−
H
0
d
2
2
(16.9)
Now we want to investigate if there are stable minima for nonzero values of H
0
u
and H
0
d
. To simplify the notation, let’s write the potential as
V
n
= a
1
x
2
+ a
2
y
2
− 2bx y + c(x
2
− y
2
)
2
(16.10)
Note that b and c are positive, whereas a
1
and a
2
could be either positive or negative
[see Eq. (16.7)].
One thing we must ensure is that the potential goes to plus infinity as we move
away from the origin in any direction. If we fix, say, x to any value and take y
to infinity (or vice versa), the last term, c(x
2
− y
2
)
2
, will always dominate all the
other terms, and since c is positive, we are safe. However, if we move away from