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Accepted for the Libraries I Bate accepted S&asu~k
Higgs Flavor Multiplets
by
Ersen Bilgin
Advisor: David R. TuckerSmith
A thesis submitted in partial fulfillment
of the requirements for the
Degree of Bachelor of Arts with Honors
in Physics
WILLIAMS COLLEGE
Williamstown, Massachusetts
May 21,2006
Acknowledgements
There are many people without whose input this thesis would have been impos sible. First of all, I would like to thank my advisor, David TuckerSmith. Not having taken any field theory courses, I knew nothing about high energy physics when I started working on this thesis in the summer of 2005. He first taught me all the material I needed from scratch, and then guided me through every step of the project. His input on the writing of this thesis is invaluable. I would also like to thank Owen Simpson for his work in the summer of 2005. Owen worked on this project for the entire summer and had substantial contributions to many parts of this thesis. I also express my gratitude to Kevin Jones for his comments on the final draft of my thesis. His comments gave this thesis its final shape. Last, but not the least, I would like to thank Hang Song for her support and company.
Abstract
In this thesis, we propose an SU(3) flavor symmetric extension to the standard supersymmetric model. We replace the single pair of Higgs fields in the standard supersymmetric model with multiplets of Higgs fields that transform nontrivially under SU(3) flavor transformations to make the Lagrangian flavor symmetric. We break the flavor symmetry by introducing multiplets of scalar fields that get vac uum expectation values in the flavor symmetry scale, AF loi6GeV. By coupling these scalar fields to the Higgs, we are able to make all but one pair of Higgs fields heavy, thereby preventing excessive fermion flavor mixings. Our model accom modates all the experimental constraints from past collider experiments including bounds on the masses of the Higgs fields and known fermion masses and mixings. Lastly, we show that the new particles we introduce in our model are detectable by collecting more data than was done in previous ef ecolliders without the need to increase the collision energy.
Contents
1 Introduction 9
1.1 Standard Model ........................... 9
1.2 Supersymmetry ........................... 11
1.3 SU(3) Flavor Symmetry ....................... 11
1.4 Research Objectives and Motivations ................ 13
2 Restrictions on Higgs Masses 17
2.1 Mass Terms in the Lagrangian ................... 17
2.1.1 Diagonalizing Mass Matrices ............... 19
2.2 Higgs Masses ............................21
2.2.1 Flavor Changing Processes ................. 21
2.2.2 Unified Theory at High Energies .............. 24
2.3 Constructing models with a single pair of light Higgs fields .... 25
2.3.1 Simple Example: 333 ................... 26
2.3.2 A 6 Flavor Breaking Field, S ,i.e. W =axm~j mvlh .~ 27~~~~
2.3.3 ore combinations of 6 and 6 ............... 28
2.3.4 3's as the Symmetry Breaking Fields ........... 29
2.3.5 Other Flavor Breaking Field Structures .......... 29
2.3.6 The Adjoint Model ..................... 30
3 Fermion Masses and Mixings 33
3.1 333Model ............................. 33
3.2 Adjoint Model ............................ 34
3.2.1 One 6 as the flavor breaking sector. S ........... 34
3.2.2 The 6 and 3 as the flavor breaking fields .......... 36
3.2.3 Matching the masses of the Fermions ........... 37
3.2.4 Mixing Angles ....................... 40
4 Higgs Sector Masses 43
4.1 Thepterm ............................. 46
4.2 Supersymmetry Breaking Terms .................. 47
4.3 Only 3 as the symmetry breaking field ............... 48
4.4 Only 6 as the symmetry breaking field ............... 49
4.5 Both 3 and 6 as the symmetry breaking fields ........... 50
4.6 Goldstone's Theorem ........................ 51
4.6.1 Proof of Goldstone's Theorem ............... 51
4.6.2 Application of Goldstone's Theorem ............ 52
4.7 Including A terms .......................... 53
4.8 Coupling 3 and 6 to the Higgs ................... 54
4.8.1 When 6S has no VEV ................... 56
5 Experimental Signatures 65
5.1 S fields that give masses to light Higgs ............... 65
5.2 S fields that give masses to heavy Higgs .............. 66
5.2.1 The 333 Model ...................... 66
5.2.2 6+3 1 =8 Model with only 6 as the flavor breaking sector . 70
5.2.3 6 +3 1 =8 Model with 6 and 3 as the flavor breaking sector ............................74
5.2.4 6 +3 1 =8 Model with 6 and 3 as the flavor breaking sector ............................ 74
6 Conclusions 79
Chapter 1
Introduction
Standard Model
In midsixties, Glashow, Salam and Weinberg proposed a mathematical model that successfully describes the stong, weak and electromagnetic interactions. This model, which is now referred to as the standard model of elementary particle physics, has not been contradicted by any experimental result in the last forty years, unless one counts the discovery of neutrino masses. In fact, all but one of the elementary particles that the standard mode predicts have already been exper imentally observed. These particles are listed in Table 1.l.
Spin 1 Type I Flavor
e electron v, electron neutrino
Leptons p muon v, muon neutrino
Fermions (spin t) T d tau down quark v, u tau neutrino up quark
Quarks s strange quark c charm quark
b bottom auark t tor, auark
y photon
Wi,Z W*,ZBosons
G Gluons
Higgs (spin 0) H Higgs
Table 1.1: Standard Model Ingredients In the standard model, all the particles are mathematically represented as quan
tum fields. Instead of the Hamiltonian, as is used in nonrelativistic Quantum Me chanics, the Lagrangian is typically used to describe the interactions among these quantum fields. The Lagrangian is a functional that depends on functions of space and time and their derivatives (i.e. L (e(?, t)),,u(r', t), . . .)).
Each of these quantum fields (except, possibly, for the neutrinos) have both righthanded and lefthanded components. The notation used in the standard model to represent these fields is shown in Table 1.2, where the lower index rep resents the flavor of the particle.
Ll,2,3 = ( ::), ( ), ( :i )
"Lefthanded lepton doublets"
Q12.3 ( 2 ), ( zE ), ( ) "Lefthanded Quark doublets " ttt
Ei = eR, PR, IR "~i~hthin'ded
Lepton Singlets" "tt
0i,2,3= d;, bR
sR7
"Righthanded Down Quark Singlets"
tt't 'E,2,3 = 'R' 'R' t~ "Righthanded Top Quark Singlets"
Table 1.2: Notation used in the Standard Model
The standard model Lagrangian is constrained by gauge symmetry. A gauge symmetric Lagrangian is invariant under gauge transformations. There are three different kinds of gauge symmetries in the standard model: U(1), SU(2) and SU(3). Any term that respects all three of these symmetries is allowed and any thing that does not respect all of these symmetries is forbidden from being in the Lagrangian. The U(l) gauge transformations act nontrivially on all the fermionic fields and force us to introduce a gauge boson that couples to all the fermionsthat is, gauge symmetry gives rise to gauge interactions. For example, ef e annihila tion via virtual photon exchange is a kind of gauge interaction. SU(2) transforma tions only affect the lefthanded particles, and "mix" leptons with their neutrino counterparts, and up quarks with their down quark counterparts, e.g. under SU(2), we have: where V (?, t) is a 2 x 2 unitary matrix with unit determinant. The U(l) and SU(2) symmetries give rise to electromagnetism and weak interactions. The SU(3) sym metries only involve the quarks and require interactions between the quarks and the gluons, giving rise to the strong interactions.
1.2 Supersymmetry
The models that we study in the rest of the paper are all supersymmetric. Super symmetic theories have at least twice as many particles as the standard model. In these theories, each known particle in the standard model has a superpartner that obeys the opposite statistics (FermiDirac instead of BoseEinstein or viceversa).
The ingredient of supersymmetry that is most essential to this thesis is the superpotential. In all supersymmetric models, part of the Lagrangian is derived from a superpotential, which is a scalar function of superfields. Each superfield contains fermionic and bosonic components (e.g. the quark superfield contains spin; quarks and spin0 squarks, as they are called). More importantly, in the standard supersymmetric model, the the fermion mass terms are derived from:
where HU and HD are singlet Higgs fields, which will be discussed in detail in Chapter 2. The superpotential in Equation 1.2 and therefore the Lagrangian in the standard supersymmetric model is not flavor symmetric. That is, the superpoten tial above is not invariant under flavor transformations. When we transform all the flavor multiplets (e.g. Q + VIQ, Uc+ V2UC, DC + V3DC,EC+ V4Ec,L + V5L, where the V's are 3 x 3 transformation matrices), we don't get back the same superpotential, because of the arbitrariness of the R matrices.
1.3 SU(3) Flavor Symmetry
In the standard supersymmetric model, all but the Yukawa coupling terms in the superpotential have an SU(3) flavor symmetry. If we also impose an SU(3) flavor symmetry on the Yukawa coupling terms, we require these terms to be invariant under SU(3) flavor transformations. An SU(3) transformation is the multiplication of every flavor multiplet in the Lagrangian by the same 3 x 3 unitary matrix of determinant 1 (and hence the name, SU(3), Special Unitary 3 x 3 matrix). Unlike the gauge symmetries, this symmetry is a global symmetry; we are not allowed to make different choices for the transformation tensors at every point in space and time. A global transformation requires us to use the same transformation matrix on all the fields at every point in space and time.
To make the standard supersymmetric model flavor symmetric, we replace R"'Hu with a flavor multiplet of Higgs fields, that also transforms under flavor transformations. If we take the quark and leptons to transform as triplets under SU(3), there are only two possibilities for how the Higgs multiplets transform to make the superpotential flavor symmetric. The two possibilities for the Higgs multiplets to make the standard model superpotential flavorinvariant are the 3 and 6 representations of SU(3). 3 is a vector of size three (three independent fields), and 6 is a symmetric matrix with raised indices (with six independent fields)'. Each of these independent fields can be thought of as different Higgs flavors (just like e, p and r are different fermion flavors represented as a 3 in the Lagrangian). Using a 6 will introduce six Higgs flavors whereas a 3 will introduce only three flavors.
For example, if we replace RijHUwith a 3 (i.e. eijkHU k), we get the term:
where 3means that the term on the right hand side of the equation is a part of the
Lagrangian, and
1 1 0 if i,j,k is a cyclic permutation of 1,2,3 if ij,k is anticyclic permutation of 1,2,3 otherwise (1.4)
If we apply an SU(3) flavor transformation to this term where the transforma tion tensor is T, we get:
Thus, replacing RijHUwith a 3 in Equation 1.6 makes it flavor invariant. This is also true for the other Yukawa coupling terms in the superpotential. So the full
'when transforming the tensors with raised indices, we use the compex conjugate of the SU(3) transformation matrix.
flavor invariant Yukawa coupling terms are:
where A's are overall coupling constants multiplying the full sums. Similarly, we can create an invariant by replacing AijHo with a 6 (i.e. Hi):
If we transform this term with the SU(3) transformation tensor, T, we get:
where on the third line TT? was replaced with I, the identity matrix, since T is unitary. Thus, the full flavor invariant Yukawa coupling terms with the 6 Higgs fields are:
Wyuk= + ,I~L,E;H&' (1.7)
AIQi~;H:' + I~Q~D;H~ + h.c.
where A's are overall coupling constants multiplying the full sums.
Instead of using a symmetric matrix (H:) and an antisymmetric matrix (eijkHUk), we could have used a generic 3 x 3 matrix for the Higgs sector that transforms as H + THT~. We preferred to use the symmetric and antisymmetric matrices, because every generic 3 x 3 can be written in terms of its symmetric and anti symmetric components, i.e. H = aHSy,+pHanli. Furthermore, the Higgs sector may turn out to be symmetric or antisymmetric, rather than a mixture of the two. Because of this reason, it is better to explicitly write out the two components of the Higgs.
1.4 Research Objectives and Motivations
In this thesis, we investigate the idea that the entire Lagrangian is flavor symmet ric, and explore some phenomenological consequences. Some fields that make the entire Lagrangian flavor symmetries (such as new Higgs fields) could exist. If these fields were out of the detection range of the earlier colliders, we wouldn't
have observed them by now, and we wouldn't have any evidence for or against their existence. There are a number of motivations for considering the possibility of flavor symmetry.
First, it is a reasonable addition to already existing symmetries, since different flavors of particles are very similar to each other already. Their only distinguish ing feature is their masses. It is not too hard to imagine that these fields are indistinguishable from one another at high energies. Just like the gauge symmetry in the standard model, the flavor symmetry we propose is a spontaneously broken symmetry. This means that even though the Lagrangian is flavor symmetric, the symmetry breaks in the vacuum state resulting in the different flavors of particles that we observe.
Second, even though the standard model has been very successful in explain ing the experimental observations, it does not provide any predictions about the masses and flavor mixings of the particles. There are no bounds set by gauge symmetry on the masses of the elementary particles. The theory is fit to the ex perimental data by input parameters. One may hope that introducing new sym metries to the theory may reduce the number of input parameters, and result in more predictions about particle properties. As it turns out, this is not the case for the particular model we develop, but the framework described here may still be a good starting point.
Finally, depending on how it is implemented, flavor symmetry may have im portant implications for collider physics. For instance, because we realize this symmetry by replacing the single Higgs pair in the standard supersymmetric model with flavor multiplets of Higgs fields that transform under flavor transformations, there are new particles in the theory that one might hope to be able to detect. As we will see, the extra Higgs particles themselves are out of reach, but other scalars required for the flavor symmetry breaking may be detectable.
Constructing a viable model in which flavor symmetry is realized by enlarging the Higgs sector has its challenges. The model we build must accommodate the known masses and mixing angles of the fermions. It should not lead to rates for rare processes in excess of their experimental bounds. Also, all new particles with substantial Higgs component should have masses greater than around 100 GeV to have evaded detection until now. The main purpose of this thesis is to develop a model that satisfies these requirements, and to study its phenomenology.
In Chapter 2, we construct a model in which all but one pair of Higgs fields are heavy in order to avoid excessive fermion flavor mixings. In Chapter 3, we show that this model can accommodate the experimental values for the fermion masses and the mixing angles. In Chapter 4, we show that the term in the standard supersymmetric model that give masses to the fermionic components Higgs fields is forbidden in our model due to flavor symmetry. However, the fermionic Higgs particles can still acquire masses through the expectation values of other fields in our model and be out of the detection range of past collider experiments. Finally, in Chapter 5, we propose methods by which the new particles in our model can be probed in future colliders. It turns our that the particles can be detected by collecting more data than that was done in previous e+ecolliders. No increase in the energy of collisions is necessary for this purpose.
Chapter 2
Restrictions on Higgs Masses
2.1 Mass Terms in the Lagrangian
The mass terms for a field + in the Lagrangian are terms of the form:
where rn is the mass of +. Terms of this form are not gauge invariant. For example, if + = u, then the mass term of the up quark contains both the left handed and right handed components of the up quark. As can be seen in Table 1.2, these components appear in different multiplets that transform differently under gauge
transformations. Under an SU(2) transformation u~ is unchanged whereas
(2)
is modified, causing an overall change in the up quark mass term.
Since these mass terms are not gauge invariant, they are not allowed in the standard model Lagrangian. Instead, the standard model uses a new field called the Higgs field, H. H is an SU(2) doublet, and unlike Liand Qiit has spin zero. The couplings of the Higgs field to fermion multiplets form gauge invariant terms
which in turn give masses to the fermions. Because Higgs field is a spinzero field, it is allowed to take a nonzero value in the vacuum (ground) state without spoiling Lorentz invariance. All other fields have to be zero in the vacuum state because they all have directions, being fields with spin.
To find the value that the Higgs fields gets in the vacuum state (the vacuum expectation value or VEV), we first need the potential of the Higgs field. The VEV of the field is the value that minimizes the potential. In the standard model, the Higgs potential is:
In the above equation, R has to be positive to bound the potential from below. If m2 > 0, then H = 0 minimizes the potential, and if m2 < 0, H~H= m2/2R v2 minimizes the potential. Then, the VEV of H is:
Any direction of the Higgs field minimizes the potential as long as ,/v; + vi =
v. To minimize the potential, the field has to choose an arbitrary direction in SU(2)xU(l) space even though the potential and the Lagrangian for the field are symmetric. The SU(2) direction is related to the relative magnitudes of vl and v2 whereas the U(l) direction is related to the phases of vl and v2. This is called the spontaneous breaking of the symmetry.
In the standard model, there are three gauge invariant terms that can be written by coupling the Higgs field to the fermions. These terms are called the Yukawa coupling terms:
where a sum over i and j is implied '. We can write any Higgs doublet as
where x is the space and time dependence, h, are the real components of the fields and hi are the complex components. Using our SU(2) freedom, we can rewrite this as
This is just like choosing the coordinates of the SU(2) space so that one of the axes aligns with the Higgs field. Note that we can do this because all gauge symmetries are local symmetries, meaning that the choice of transformations is
'1n the rest of this paper, a sum will be implied for every repeated index in a product
18
made independently at each point in space and time. We can also use our U(l) freedom to further simplify the field to make
This field will have some VEV in the vacuum state:
Expanding about the minimum we can write the field as
where h(x)is the real Higgs field with zero VEV.
If we plug in this form of the Higgs into Equation 2.4 and expand, we get terms like Rv(d;d~+ dLd~)which are clearly mass terms. We also get interaction terms with the Higgs boson such as Rh(x)(d;dL + dLdR).
2.1.1 Diagonalizing Mass Matrices
Because of the arbitrariness of the constants R in Equation 2.4, the mass terms are not diagonal in general. In addition to single particle mass terms like Rv(d;dL), we also get mass terms made up of two particles such as Rv(s:d~).These mixed terms affect the masses of both particles that appear in them. When we have terms like these, we know that we are not in a mass eigenstate basis for the quarks.
To switch to an eigenstate basis, we need to diagonalize the mass terms. To demonstrate how this can be done, let's first look at the lepton masses. Then we can also diagonalize the quark mass terms.
Lepton Masses
Lepton mass terms in the standard model are:
. . &epton mass = vh:LiEf
We can rewrite this in matrix form as:
Lepton mass = LTm~EC
where rn~= v~i
is called the lepton mass matrix. To switch to a basis of mass eigenstates for the leptons, we need to diagonalize rn~.We can rewrite rn~as rnE= V*~U+ (2.12)
where mE is diagonal with real and positive entries, and V and U are unitary matrices. Plugging this form of the mass matrix into Equation 2.1 1, we get:
where L' = and E'" =
V~L UtEC.Since iEjj is now diagonal, there are no cross mass terms, and the elements of L' and E'" are mass eigenstates. If we write every lepton field in the Lagrangian in terms of L' and E'", all the lepton fields become mass eigenstates, and it turns out that this doesn't change any of the interaction terms. Furthermore, the mass of each lepton can be read off as the elements of the new mass matrix 6.
Quark Masses
Diagonalizing the quark mass matrix is different, because both the up and down quarks appear in Q in the Lagrangian (see Table 1.2). When we expand the two quark terms in Equation 2.4, we get two separate mass matrices for up and down quarks:
Lquarkmass = uTrnuuC+ DTm~Dc (2.15)
where U and D are up and down quark flavor triplets respectively. To diagonalize these mass matrices, we need to apply different transformations to the downtype and the uptype quarks. Therefore, the upper components in the Q doublets are not transformed the same way the lower components are. Since Q terms also ap pear elsewhere in the Lagrangian, the interaction terms change in the quark mass eigenstate basis (the relevant terms turn out to be the chargedcurrent weak inter actions). Before the transformation, there was no mixing between the different flavors of quarks (Figure 2.1).
When the up quarks are transformed differently than the down quarks, the ver tex in Figure 2.1 becomes nonzero even when i # j. This means that there is some mixing between the flavors when we are in the basis of quark mass eigen states. There are experimental confirmations of these mixings, and any working model should allow for these experimentally observed mixings.
Figure 2.1 : This vertex exists only when i = j. Up and down quarks of different flavors cannot interact with each other
2.2 Higgs Masses
2.2.1 Flavor Changing Processes
By introducing new Higgs fields to the standard supersymmetric model, we intro duce the risk of violating certain experimental constraints on the flavor mixings of the fermions. Now that we have more than a single pair of Higgs fields, there will be new mass and interaction terms in the Lagrangian. We need to make sure that the new Higgs particles do not produce mixings that exceed the experimental bounds.
If more than a single pair of Higgs fields remains light after flavor symme try breaking, such mixings are typically introduced. For reasons to be discussed shortly, we will assume that the energy scale associated with the spontaneous breaking of flavor symmetry is enormous, AF 1016 GeV. This scale determines the masses of those Higgs fields that get heavy due to symmetry breaking. This means that the heavy Higgs fields have quadratic terms of order (AF)2 in their po tentia12, while the light Higgs have essentially flat potentials by comparison. The light fields get masses and potentially VEVs, through supersymmetry breaking effects which are of order As us* lo2 GeV. However, the quadratic terms with coefficients of order (A~)~
dominate the potentials of the heavy fields, and prevent them from acquiring nonzero VEVs due to the low energy effects of supersym metry breaking. As a result, in our model, only the light Higgs fields get nonzero VEVs.
In the standard supersymmetric model, there is a single pair of Higgs fields that give masses to the fermions through its VEV. The superpotential in the standard supersymmetric model is of the form:
2~o~
Higgs fields get masses will be discussed in Section 2.3
where y's are constants and h, is the time and space dependent up Higgs field. When we diagonalize the mass matrix of the up quarks, yij(H,), the Higgsfermion interaction term, yijh,QiUj, automatically gets diagonalized. This means that there are no interaction terms between the Higgs field and two different fields of fermions, and thus no flavor mixing of the fermions through the Higgs boson.
In the models we study, we have a sector of Higgs fields. When more than a single pair of Higgs fields couples to the quarks and leptons (i.e. if there is more than a single pair of light Higgs), the fermion mass terms in the Lagrangian are no longer proportional to the higgsfermionfermion interaction terms. For example, suppose two of the Higgs fields get VEVs in Equation 1.7. Then, the superpotential becomes:
where y's are matrices of constants that depend on the model. To switch to a mass eigenstate basis, we need to diagonalize the matrix (y;'(~~)' + y:(~~)'). When we do flavor transformations on Q and U to make the mass terms flavor diago nal, neither yyh, 1 nor y;'hU2 becomes diagonal. Thus, there are interaction terms between the Higgs field and two different flavors of fermions, causing mixings be tween these fields through the Higgs boson. Assuming that any of these particles have weakscale masses, this typically results in flavor mixings at a level much higher than what is experimentally observed.
For example, in the standard model, there is a very small mass splitting be tween the neutral K and K mesons. The neutral K meson is a composite particle that can be represented as 3;d where s and d are the strange and down quarks re spectively. Similarly, K is ds. In the standard model, the mass splitting between K and K comes from loop diagrams as in Figure 2.2
The vertices in Figure 2.2 correspond to terms involving the top, down, strange quarks and the W boson. Through diagrams like these, K can be converted into K and vice versa. This means that there are effective terms in the Lagrangian of the standard model that couple K to K.However, because this is a loop diagram (loop diagrams are like higher order terms in perturbation theory), the coupling is weak. So, the mass matrix of K and K contain small terms in the off diagonal elements, resulting in a small mass splitting, consistent with the observed value.
In our model, if we have sizeable couplings between the strange, down and the Higgs fields, we will have extra contributions to the offdiagonal elements of the KK mass matrix through diagrams like in Figure 2.3.
Figure 2.2: A loop diagram from the standard model that shows the mixing between K and K
Figure 2.3: A treelevel mechanism that couples k and & through the Higgs boson
The value of the diagram in Figure 2.3 is inversely proportional to the mass squared of the Higgs field that appears in the diagram. Also, this is not a looplevel contribution; it is referred to as a treelevel diagram, which is like a leading order term in perturbation theory. So, if the Higgs particle in the diagram is light, and the flavor changing vertices are unsuppressed, this contribution will overwhelm the standard model one. Therefore, we either need to avoid having these vertices in our models, or if they exist, they need to involve only the heavy Higgs fields.
A simple way to prevent light Higgs fields from appearing in diagrams like Figure 2.3 is to require all but one pair of Higgs fields to be heavy, with masses of order AF. When only a single pair of Higgs fields is light, instead of having a sum of two matrices as in the first line of Equation 2.17, we get a single matrix, which is also proportional to the interaction terms. Thus, when we apply transforma tions on the quarks to diagonalize this matrix, the interaction terms also become diagonal, avoiding the excessive rates for flavor changing processes.
It is not absolutely essential that all but one pair of Higgs doublets gets heavy. In fact, nonsupersymmetric [I] and supersymmetric [2] theories with Higgs flavor multiplets have been considered with multiple light Higgs pairs, but in this case satisfying bounds on flavor changing neutral currents becomes a complex model building challange, that we would like to avoid.
2.2.2 Unified Theory at High Energies
There is a second, independent motivation for making all but one pair of Higgs doublets heavy: it's the case that most naturally allows for a unified theory at high energies. When there is only a single pair of light Higgs, it can be mathemati cally shown that the electric, weak and strong interaction constants meet at high energies [3]. These calculations assume that the effective theory at low energies holds all the way up to the unification scale, 1016GeV (i.e. there are no new particles in that energy range etc.). However, this might still be an indication of the existence of a unified theory at such energies. This is one of the main rea sons supersymmetric theories are popular and we want a model that preserves this property. This is why we take AF1016GeV.
2.3 Constructing models with a single pair of light Higgs fields
To make all but one of the Higgs fields heavy, we introduce new symmetry break ing fields S that couple to the Higgs. These new fields are called symmetry break ing fields, because their VEVs break the flavor symmetry. For example, let's introduce a new flavor triplet ,which preserves the SU(3) flavor symmetry in our model and acquires a VEV
We know that the Lagrangian is invariant under SU(3) transformations, so we are allowed to transform every field with any SU(3) transformation matrix we choose. Without loss of generality, we can use an SU(3) transformation the VEV of the field
Clearly, this VEV is not symmetric under SU(3) transformations. If we multiply the vector, (S),by an SU(3) matrix, it will not be the same as before. However, we can still interchange the first two entries without changing the VEV. That is,
(S)is invariant under SU(2) transformations on its first two entries. In this case, we say that S spontaneously breaks the SU(3) symmetry to SU(2).
The coupling terms between the Higgs and the S fields should obey all the symmetries in our model. For instance, to obey the U(l) gauge symmetry of the standard model, the sum of the hypercharges of the terms in a product must add up to zero. The hypercharges of H, and Hdfields are +$and $ respectively. The S fields we introduce will be neutral under all standard model gauge symmetries and so have hypercharge zero, only allowing for couplings where we have both H, and Hd and any number of S fields. To make the new terms flavor invariant, we can only use certain forms of S multiplets. For example, if we only have 3's as our Higgs fields, these can only couple to a 3 or a 6 of S fields, but we are not allowed to couple them to a flavorsinglet S field.
The calculation of the masses of the Higgs fields is different than the way we calculation of fermion masses described earlier. The Higgs fields are bosons, and their mass terms are in the scalar potential. In supersymmetric theories, the potential of a system is derived from the superpotential using the following:
where are all the fields in the superpotential. The mass terms in the potential are m2#+$, similar to the fermion mass terms. So, we need to follow a similar pro cedure when calculating the masses. We first need to find the mass matrix for the fields, and then do transformations on these fields to diagonalize the mass matrix. Then, we can read off the massessquared of the fields from the diagonalized mass matrices.
We considered several different possibilities for the Higgs and S multiplets. Below, we will list some of the models we tried, and explain why some of them were feasible models and some were not.
2.3.1 Simple Example: 333
Let's consider the case where we have 3's as our Higgs fields. The simplest sym metry breaking field that we can use to create flavor invariant terms is another 3. Using a single lowered index tensor representation for the 3 S field, we can write the following gauge and flavor invariant terms:
Because of the antisymmetry of E, the second term is just zero. So, we only have one term that gives masses to the Higgs. The only F terms that will result in mass terms in the potential will be and $. The derivatives with respect to S fields will cause only quartic terms in Higgs in the potential. Following the same argument that we had in the beginning of this section, without loss of generality, we can apply an SU(3) flavor transformation to all flavor multiplets and switch to a basis where only Sg gets an expectation value, and simplify the superpotential even further:
WHkg, = Nu Hd2(S 3) Hu2Hd1(S3) + (interaction terms) (2.23)
Using Equations 2.20 and 2.21, the above superpotential gives: But the mass matrices from this potential are already diagonalized. For Hu we have
and for Hd,we have:
Thus we get a single light pair of Higgs fields, and the other two Higgs are de generate. This automatically seems to satisfy the condition that all but one pair of Higgs fields is light. However, we have many more conditions that we need to satisfy in order to have a working model. As we will show in Chapter 3, this simple model does not accomodate the measured masses for the fermions. So, we need to find other models that have all but a single pair of Higgs heavy.
2.3.2 A 6 Flavor Breaking Field, S ,i.e. W mn
= tik,tjin~2~r~
The next simplest set of models that can create flavor invariant HiggsFermion couplings contain 6 as their Higgs flavor multiplets. A 6 is a 3x3 symmetric matrix of fields. One subtlety here is that if we assign fields to the components of this matrix in the naive way, e.g.
one finds that the kinetic terms for the fields are not canonically normalized. This issue is similar to wavefunction normalization in nonrelativistic quantum me chanics. To fix this, we need to multiply each field in the symmetric matrix by the appropriate constants to make the kinetic terms canonically normalized. A simple way of doing this is to check the sum H~~H~~
and make sure that there are no cross terms between flavors and that all coefficients are the same. When we do this we realize that the form:
makes the kinetic terms canonically normalized.
To make all but a single pair of Higgs heavy, we need to introduce the S fields. The simplest flavor multiplet that can couple to the 6 IIiggs and form flavor invariant terms is another 6. A 6 of S can couple to the Higgs field in the superpotential and form the invariant
Without loss of generality, we can pick a basis where in the vacuum state only the diagonal elements of Sm" get nonzero VEVs. We are allowed to do this, because every symmetric matrix can be diagonalized by a unitary transformation in the following way:
where U is a unitary matrix. In tensor notation, this corresponds to:
But any U(3) matrix can be written as U = VP, where V is an SU(3) matrix and P is some overall phase. Therefore, we can always find an SU(3) transformation which makes all but the diagonal elements of Sm" zero.
Using Equations 2.20 and 2.21 on the superpotential in Equation 2.29, we can find the potential and the mass matrix for both the up and down Higgs fields. When we diagonalize this mass matrix, we see that the diagonal elements (i.e. the eigenvalues of the mass matrix) are S:, , S;,, S:,and the three roots of the equation: ~s~,s;,s:,+ (. . .)x + (. . .)2+ (. . .)2= 0. This means that for any of the eigenvalues of the mass matrix to be zero, at least one of S 11, Szz and S33has to be zero. However, when either of these are zero, at least one root of the third degree equation will also be zero, making the number of zero eigenvalues at least two. We want only one pair of Higgs to be light and the others to be massive. So, the flavor breaking field cannot be a single 6.
2.3.3 More combinations of 6 and 6
We also tried the invariant in our superpotential. It gave us two or more zero eigenvalues (and thus two pairs of light Higgs) for the Higgs mass matrix.
If we ignore the terms without any Higgs fields in them, we see that the above invariant is very similar to the single 6 invariant we had in section 2.3.2. The only difference is that Sij in the single 6 case needs to be replaces with S ij+a~'j, where a is a constant. We can do the same substitution for the mass matrix from the 6 case to show that there has to be at least two zero eigenvalues.
2.3.4 3's as the Symmetry Breaking Fields
The 3 is a three dimensional single index tensors. Because they only have one index, we use several 3's to make invariants or combine them with other kinds of tensors before we can add them to the superpotential. Below is one of the combinations we tried:
The invariant with a single triplet at S, H~H:'S jSkS gives 5 zero eigen values. Note that, we are considering the case where the potential has only this term. We can always transform Si to leave only one nonzero VEV in the vac uum state. If the only nonzero element is $3, then the invariant simplifies to H23HFS3S3S3S3, leaving the mass terms (Hi3S3S3S3S3)2 and (H23S3S3~3~3)2 in the potential, V. So, the only entry in the Higgs mass matrices become leaving all the other elements, including the diagonals zero. This results in five zeros in the eigenvalues.
2.3.5 Other Flavor Breaking Field Structures
Although not exhaustively, we tried many other possibilities for the flavor break ing fields. It was not possible to give analytical solutions to many of the complex flavor breaking sectors due to our technical limitations. Therefore, in some cases, we picked a single term from the superpotential and focused on the effects of that particular term. For example, when we introduced two 6 S fields, Sik and Tjl, we used the superpotential W = H:'H:S~T~~, instead of including all the possible couplings such as H~H~S
,Tk,.
We used Mathematica to determine the number of massless fields for each superpotential in the table. First, using Equations 2.20 and 2.21, we derive the potential for the fields. Then, we find the mass matrix of the Higgs fields by as suming that only the S fields get VEVs. We are allowed to make this assumption, because in our models, we require the Higgs fields get no VEV. We are not in terested in cases where the Higgs fields acquire nonzero VEVs. We know that the minimum of the potential should be zero, since the potential is made up of only Fterms all of which are positive. So, the S VEVs must automatically make F = 131' = 0 for all 4'. This further simplifies our solutions.
In Table 2.1 is a list of some of the flavor breaking structures we tried. Since the analytical solutions we had were complex, there are no simple explanations about the number of massless fields for some of the superpotentials listed.
W Number of zeros
H;~H:~SisISyfsill' 5
Table 2.1 : List of different superpotential terms and the corresponding number of zero mass eigenstates. The A's referred in the table are the adjoint, 8, flavor breaking fields. (See Section
2.3.6 for the definition of this representation)
2.3.6 The Adjoint Model
We now describe a model which leaves just a single pair of Higgs fields light, and which will turn out to accommodate realistic fermion masses and mixings. In this model, we allow three different forms for the Higgs: 3, 6 and 8. The 8, or the adjoint, is also a two index tensors and can also be thought of as a 3 x 3 matrix. However, unlike the 6 or 6, the 8 does not have two raised or two lowered indices. Instead, it has one raised and one lowered index. It is not symmetric or asymmetric. The reason it is called 8 instead of 9 is that its trace is 0, which makes it have only eight independent entries. Because the 8 has one raised and one lowered index, we can multiply any invariant with it and the result would still be an invariant.
With the inclusion of the 8, we have a total of 17 Higgs fields. In addition to flavor and gauge symmetries, we also impose a U(l)x symmetry in which the charges of all the fields in a product should add up to zero. All the terms in the La grangian except for some of the HiggsS couplings already satisfy this symmetry, and are not affected. Only unwanted couplings between the Higgs and the S fields are forbidden. The charges of the fields under the U(l)x symmetry are shown in
Table 2.Z3.
Fermions Charge Higgs Sector Charge Q Uc DC L Ec N 1 13 3 1 5 3, 6, 3d 6d 8, 8d 6, 2 2 5 5 3 7 2 3, 2
Table 2.2: The U(l)xcharges of the flavor multiplets
With the introduction of the U(l)xsymmetry, every HS coupling term in the superpotential is forced to have one 8 Higgs that couple to a 3 or 6. The possible H S couplings in the superpotential are shown below in Table 2.3
Table 2.3: Possible H S couplings in the Superpotential
This model automatically makes all but one Higgs field heavy. 3, and 6, always couple to 8d, and 3d and 6dalways couple to 8,. Because there are 6 + 3 = 9 up Higgs fields coupling to 8 down Higgs fields, there is always one linear combination of 3, and 6, that doesn't couple to anything. The same is also true for one linear combination of 3d and 6d. These linear combinations that don't couple to any fields become the massless modes, and the rest become heavy. Solving for
3The bold numbers and subscripts correspond to the type of multiplet and field respectively.
31
the eigenvalues of the mass matrix confirms that all but one pair of Higgs fields get heavy in this model. In fact, we can simplify our superpotential further by only using a 6 for our flavor breaking sector. This still gives us only one light Higgs and 16 heavy ones. As we will show in Section 3.2.1, a 6 by itself cannot accommodate the experimental values for the fermion masses. Because of this, we are forced to include the 3, in our model.
This model does not give a massless pair of light Higgs fields if the U(l)x symmetry is not imposed. Without the U(l)x symmetry, higher order terms with four or more field multiplets such as 8,6d6,6, are allowed in the superpotential. These fourmultiplet terms in the superpotential must have the same dimensions as the threemultiplet terms. To obtain dimensional consistency, we have to divide the fourmultiplet terms by some energy scale which turns out to be Mpl. The new superpotential introduces couplings between the 6, terms to 6,, which are forbidden if U(l)x is imposed. Because of this, the linear combination of 6, of 3, that remained massless in the U(l)x symmetric model is allowed get massive. It
2
gets its mass through terms like 16,(6,)21 . If AFwas much less than Mpl, these mass terms would be negligible, but this is not the case and all Higgs fields gain masses of order hFg.Introducing the U(1)X symmetry avoids this problem
altogether by forbidding the higher order terms.
Chapter 3
Fermion Masses and Mixings
3.1 333Model
We showed in Section 2.3.1 that the simple model with triplets of the Higgs and flavor breaking fields automatically made one pair of Higgs light and the others heavy. Now, we need to check whether this simple model can fit the measured fermion masses. As we know, the fermions get masses through the vacuum ex pectation values of the Higgs fields in the Lagrangian. So, we need to look at the form of the Lagrangian in the vacuum state. In the vacuum state, the Higgs fields get VEVs, which are denoted by angle brackets:
where h' are the variable fields which correspond to the Higgs particles, and the ones in angle brackets are the expectation values. Using our SU(3) freedom, without loss of generality we can change to a basis
When we plug in this form of the Higgs field to the Yukawa coupling terms (Equation 1.7), we get the following lepton mass terms:
'We explained why we can do this in Section 2.3
We can rewrite the above in matrix form as:
From the form of this mass matrix, it is apparent that the third field remains massless and the first two have degenerate masses. However, we know experimen tally that all three leptons have masses and none of them are identical. Therefore, this simple model does not predict the correct masses and is not a working model.
3.2 Adjoint Model
The simplest model after the 333 is the model we introduced in Section 2.3.6. Both having a single 6 and having 6 and 3 as our symmetry breaking fields resulted in a single pair of light Higgs.
3.2.1 One 6 as the flavor breaking sector, S
The fermions can only couple to the 3 and the 6 Higgs tensors to make flavor invariant terms, so these will be the only two Higgs multiplets that appear in their mass terms. In fact, even if there were adjoint couplings in addition to 3 and 6, we could safely ignore them. We know from Section 2.3.6 that the fields in 8, are all heavy and don't get any expectation values. This can be confirmed by writing the Higgs mass matrix and diagonalizing it. Therefore, they cannot contribute to the Higgs mass terms. Thus, the relevant terms in the Lagrangian that form the mass terms of the fermions are:
To expand the above Lagrangian and find the mass matrices for the fermion triples, we first need to know the expectation values of the Higgs fields.
Assuming all the fields in 8, have zero VEVs, let's focus on the the F terms for the Higgs and Sfields containing 3, and 6,. If we only have a 6 as the fla vor breaking sector, these F terms are derived from the following superpotential terms:
w3a~:~;,~ki (3.6)
+~H,,H~~s~~c~~
where H: is 6., Huiis 3., H:~is Sd,Ski is 6,, and uand ,6 are constants. Since Hij is an 8, we can write it as the following sum:
where h:k are the eight 3 x 3 linearly independent traceless hermitian matrices. Now, we can find F by taking derivatives with respect to the ka's using Equation
Since the potential is entirely made up of Fsquaredterms which are positive, minimizing the potential is equivalent to setting all the Fterms to zero. Then, the VEVs of the fields will be the values that force the Fterms to be zero. To find the VEVs, we need to solve:
For notational convenience, we will drop the brackets until the end of this section. We can substitute
and write the Fa as a trace of matrices:
Any 3 x 3 matrix M can be written as a linear combination of the eight 3 x 3 linearly independent hermitian traceless matrices and the identity matrix, I. If we let
8
and plug this substitution into (3.1 I), we get Since the A's are linearly independent, AaRb = (~onstant)6,~,where dab is the kroneckerdelta function. Therefore, the equation above reduces to
(constant) 4, = 0 (3.14)
which means that for all a, 4, = 0. This implies from Equation 3.7 that if the value of any element in the adjoint Higgs is nonzero, the potential is not minimized. Therefore, the VEVs of the adjoint Higgs fields must be zero.
If all 4, are zero, then (aH, +PA)S = dl, which means that (aH, +PA) = dS'. Since S is a symmetric matrix, S' should also be a symmetric matrix. But, we know that A is antisymmetric, and H, is symmetric. The only way some linear combination of A and H, will be symmetric is if A is zero. Since A has HI, H2and H3 in its entries, all of these VEV's are forced to be zero. Thus, only the elements of the 6 H, can be nonzero. Moreover, H, has to be equal to the inverse of S.
When HIH3 become zero, then the Lagrangian with the Yukawa couplings, (3.3, simplifies. All the terms where H is a 3 disappear, and only the three terms with 6 H's remain. This implies that all the mass ratios, i.e. u : c : t, d : c : b and
e : p : T have to be the same. We know experimentally that this is not the case, disproving the model with only a single 6 as our symmetry breaking sector.
3.2.2 The 6 and 3 as the flavor breaking fields
The simplest symmetry breaking field after the single 6, is the flavor breaking field which is a combination of a 6 and a 3. The procedure for solving for the H's is the same as in the single 6 case. The only difference now is that the superpotential has two additional terms that generate F terms containing 6 and 3 Higgs fields. These terms are:
We can write H: as in Equation 3.7 and find F by taking derivatives with respect to the 4,'s:
Also, using the definitions from Equation 3.10 and
0 Sg S2
we can rewrite the equation in matrix trace form:
We can express (~H,S@) +PAS(@+ yHuB+ (AB) as a linear combination of the eight 3 x 3 linearly independent hermitian matrices and the identity matrix. So, the equation becomes:
Using the same argument from Section 3.2.1, for any choice of a the above equa tion has to be zero, which means:
(aHuS(6) +PAS@) + yHuB+ CAB) = (constant) I1 (3.20)
where 1is the 3 x 3 identity matrix.
Indeed, when we solve this equation we get a solution for the H's, confirming our previous prediction that we have a light Higgs in this model and that all the H's in the 6 and 3 Higgs matrices have light Higgs components. It turns out that the VEV each Higgs field gets is independent from the other Higgs fields allowing us to treat the Higgs VEVs as input parameters in Section 3.2.3.
3.2.3 Matching the masses of the Fermions
Now that we have a model with only a single pair of light Higgs fields, we need to compare our model with the experimental data. One experimental bound that our model has to accommodate is the fermion mass structure given in Table 3.1.
The up Higgs fields give masses to only the up quarks and the down Higgs fields give masses to both the down quarks and the leptons. Because of this,
Flavor Mass (GeV/c2) Flavor Mass (Gev/c2) Flavor Mass (~eV/c~) e 0.0005 1 1 u 0.003 d 0.006
0.106 c 1.3 s 0.1
r 1.777 1 t 175 b 4.3
P
Table 3.1 : Fermion Masses [4]. Note the hierarchy in the masses of the different flavors; me << m, << m,, etc.
setting the expectation values of the up Higgs to get the correct values for the up quark masses is easy. One can set the VEVs of the diagonal elements of the 6 Higgs to the masses of the up quarks2, and set the rest of the expectation values to zero. When the coupling constant that couples the 3 Higgs to the up quarks is also set to zero, the eigenvalues of the up quark mass matrix will be precisely the experimentally measured masses. That is, if we set R1 = 0 and
in Equation 3.5, the up quark mass terms in the Yukawa coupling terms will have the correct values. Setting the down quark and lepton masses is trickier. We first set the diagonal elements of the 6 Higgs down matrix to the masses of the down
quarks, i.e.
mdown 0
0 mstrange 0 (3.22)
0 0 mbottom
in Equation 3.5. Then, when we set the coupling constant that couples the 3 Higgs down to the down quarks to zero (i.e. R2 = 0 in Equation 3.5), the eigenvalues of the down quark mass matrix becomes the experimentally measured values for the masses5.
To set the lepton masses to the experimentally measured values, we use a perturbative approach by taking advantage of the hierarchy in the fermion masses.
2~heVEVs are not actually input parameters, but we assume they are in this section for sim plicity. Of course, these VEVs need to satisfy Equation 3.20. We did not explicitly show that 3.20 is satisfied, but, due to the high number of input parameters, we are convinced that it is.
3~ctually,the squares of the masses of the up quarks are the diagonal elements of the diago nalized form of M,M;, where Mu is the up quark mass matrix. The eigenvalues of MUM;give these diagonal elements. However, when Mu is hermitian (and in this case, it is diagonal, so it is also hermitian), the eigenvalues of only Mu directly give the masses of the up quarks.
We already set the expectation values in the 6 Higgs down matrix. If we make any changes to that, the masses for the down quarks will not be correct. However, we can still change the coupling constant that couples the 6 Higgs with the leptons (A6).We set this to m,/mb,,,,,. This way, the r mass will be correct, i.e. the mass matrix becomes:
Note that the second mass in the above mass matrix is zm,, which is less than m,. To raise the second mass to m,, we perturb this mass matrix with t<< 1:
Now, the eigenvalues of the down Higgs mass matrix becomes m, (1 t2)and m, (2+ t2)to the leading order in e. If we let
the second eigenvalue becomes m, and m, is not affected much since e2 << 1. The diagonalized mass matrix becomes:
We now apply the same procedure and perturb this mass matrix with 6 << 1, i.e.:
The perturbation 6 lowers the value of the first eigenvalue of this mass matrix. However, we need it to increase since Qmd > me. At this point, we realize that
mb
we can make zmdin the mass matrix negative. This doesn't affect anything since we only calculate the masssquared of the fields. When we do this, the eigenvalues of the matrix becomes m, (2 + 6') and m, (1 6'). Then, if we set
we get the eigenvalues to be me, m, and rn, to the first order in the perturbations.
To summarize, to get the correct lepton masses, we need to use Equation 3.22 with a negative sign for the H:' entry, set a6= 2and L3 = 1 in Equation 3.5, and let
(3.29)
The diagonalization we have in the first step naturally affects the 6 perturba tion. However, its effect is minimal since the diagonalization matrix is approxi mately the identity matrix. The matrix in Equation 3.24 is almost diagonal since E << 1. To exactly diagonalize it, we only need a diagonalization matrix with very small offdiagonal elements, which does not affect the 6 terms much.
In fact, if we do an exact calculation using the perturbations and starting with the masses in Table 3.1, we get me = 0.000525 GeV, m, = 0.107 GeV and m, = 1.7081 GeV. These are only off by less than 4%. To get the exact values, we can manually modify the perturbations.
3.2.4 Mixing Angles
Now that our model accomodates the correct masses for the fermions, we need to check whether it is able to accomodate the experimental values for the mixing angles of the quark mass eigenstates. These mixing angles are specified by the CKM matrix, which is a 3x3 matrix with complex entries. The probability of decay from one quark to another is proportional to the square of the value in the CKM matrix that corresponds to the two quarks in question.
Let MdM; and MUMA be diagonalized in the following way:
M,M; = V,M,M~U, (3.31)
where MdM; and MUM: are the diagonal, and Vd, Vu, Ud and U, are the diago nalizing matrices. Then, CKM = V~V~ (3.32)
where V,'s and Vd's columns are the eigenvectors of MUM: and M~M;respectively. Since the elements of the mass matrices are the VEVs of the Higgs fields, we can write down the relations between the Higgs fields and the CKM matrix, and set the expectation values of the Higgs fields to get the correct CKM matrix, while keeping the fermion masses within the experimental range.
In Section 3.2.3, we set the diagonal elements of H: to the masses of the down quarks for purposes of illustration. We would follow the same approach in this section too. However, to fit the CKM matrix, a different approach turns out to be more straightforward. In our calculations, instead of first getting the correct masses for the down quarks and using perturbations to get the correct lepton masses, we first set the masses of the leptons and use perturbations to get the correct down quark masses. That is, we first set:
This way, by setting Rg = 0 in Equation 3.5, the masses of the leptons are equal to their experimentally measured values. Then, by setting R5 = mdown/mT in Equation 3.5, the mass of the bottom quark is set to its experimental value. To set the masses of the down and strange quarks, we use a very similar perturbative approach as in Section 3.2.3 and introduce Hdk which perturbs the down quark mass matrix to set the eigenvalues to the experimentally measured masses.
By setting the masses of the down quarks and the leptons, we set all the VEVs in the down Higgs tensors. To accomodate the experimental values for the CKM matrix, we can only adjust the VEVs in the up Higgs mass matrices. We already have the values for the down quark mass matrix. Using Equation 3.30, we can solve for Vd. Now, we need to find a matrix V: which gives the correct CKM matrix when plugged in to Equation 3.32, and which can be derived from an up quark mass matrix that has the experimentally measured up quark masses as its eigenvalues.
As we mentioned earlier, the columns of V, are the eigenstates of MUM!. So, if we solve for the correct V, using Equation 3.32, we can solve for M,M; using the definition of eigenvectors and eigenvalues:
where VUi are the columns of V, and mi are the masses of the up quarks. Once we have all the entries in M,M:, it is not hard to find a mass matrix Mu from which MUM: can be derived. And using this mass matrix, we can find
the correct up Higgs expectation values that predict the correct masses for the up quarks and the correct entries in the CKM matrix.
Note that we did all of our numerical calculations in the real space even though all the fields are complex. However, if the calculations work in the real space, they should also work in the complex space. By switching to the complex space, we double our input parameters (the coupling constants, the VEVs, etc.), while only adding a single parameter to the list of experimental constraints. The masses were already real values, so they don't change when we switch to the complex space. The only addition is the phase information in the CKM matrix, which can be for mulated so that the complex phases of the entries are reduced to a single parameter. Since only one additional parameter needs to be satisfied and the input parameters are doubled in the complex space, we can be confident that all constraints can be satisfied in the complex space.
The Higgs expectation values that give the correct fermion masses and quark mixings are:
[
, 0.000~1 GeV 0 0 0.358 GeV
HV = 0.105 GeV 0
0 0 1.77 GeV
0.0973 GeV 0.599GeV 22.0 GeV
0.598 GeV 0.125 GeV 22.0 GeV
8.016 GeV 22.0 GeV 168 GeV
(3.35)
If the Higgs fields have the above expectation values and the constants in Equa tion 3.5 are
then, the all the entries in the CKM matrix, and the fermion masses are within the experimental bounds.
Chapter 4
Higgs Sector Masses
In Sections 3.2.3 and 3.2.4, we found expectation values of the Higgs fields that fit the experimental values for fermion masses and the CKM matrix. In the model we were working with, we only had a single pair of light Higgs fields which were the only fields that got nonzero expectation values in the vacuum state. These light fields were linear combinations of the fields in the 3 and 6 Higgs multiplets. Without loss of generality, we can change to the mass eigenstate basis and rewrite all the fields in the 3 and 6 Higgs in terms of the light and heavy mass eigenstatesl:
where a's are constants, and Hlightis the light eigenstate and Hheavyare the heavy eigenstates. If we take the expectation values of both sides in all equations and realize that heavy Higgs eigenstates have zero VEVs, we get:
for every Higgs field.
h he u and d subscripts are dropped as anindication that this argument applies to both up and down Higgs fields
Then, the ratios of light components among the fields are the same as the ratios of their expectation values. Combining this with the fact that the magnitude of the light field should be normalized (i.e. Ci=,lalkI2= I), we can calculate the values of the a's.
In the earlier chapters, we were focusing on the dynamics at the flavor break ing scale& 1016 GeV. Because of this, we assumed that the masses of the light Higgs fields were zero. This cannot be correct. There is an experimental lower bound of around 100 GeV on the light Higgs masses as can be seen in Fig ure 4.1. In the figure, t2corresponds to the square of the Higgs component of mass eigenstates, because reducing the coupling constant in the Standard Model and lowering the Higgs content have the same implications on the probability of production. Anything above the curve in the plot is ruled out. For example, the standard model Higgs field, which is a pure mass eigenstate, has 5 = 1, and its mass would be bounded below by around 115 GeV. If the Higgs masses were less than 100 GeV, they would have been observed in earlier collider experiments. We will assume that the masses of the light Higgs fields are generated by super symmetry breaking effects, and hence are of order supersymmetry breaking scale, As us Y 1 TeV. Because As us is much smaller than AF, it is then appropriate to ignore their masses at the flavor breaking scale.
The lower bound on the masses are relaxed due to the loop corrections. Through out this thesis, we are dealing with the treelevel masses. A mass of 80 85 GeV at treelevel corresponds to a physical mass of 100 GeV after loop cor rections (see e.g. [7]). In fact, if it weren't for this effect, the MSSM (Minimal supersymmetric modelthe most popular extension of the standard model) would be ruled out.
If a mass eigenstate is not made of only Higgs fields, the lower bound is re laxed further. The bound we get from collider experiments relaxes as the Higgs component in the mass eigenstate decreases. The collider experiment with the most stringent lower bound on the Higgs mass, LEP 11, aimed to produce the Higgs particle through processes as in Figure 4.2. The Higgs field is an S U(2) x U(1) doublet, and it couples to the Z boson through terms like Z Z h in the Lagrangian. If this process occured often enough, the Z and the Higgs could be identified by their decay products. The S field on the other hand, is a scalar field and has no coupling to the Z boson. If some mass eigenstate field is a linear combination of Higgs and S fields, the probability that we would observe it through a process like Figure 4.2 is proportional to the Higgs component of the field squared. We can
PI
1I
G LEP
0 ;(a)
4s = 91210 GeV
Figure 4.1: Figure 10a in [5].LEP bounds on Higgs properties. More precisely, this plot applies to a particle with the properties of the Standard Model Higgs, except couplings to the gauge bosons is 5times that in the standard model
always write this field in terms of its Higgs and scalar components:
Then, instead of writing the Lagrangian in terms of H and S, we can write it in terms of mass eigenstates. Equation 4.3 implies that in the eigenstate basis, H =aY+. . ..Then, the HiggsZ couplings in the Lagrangian becomes
ZZh+aZZY (4.4)
The value of a diagram like Figure 4.2 is proportional to the square of the coupling constants. Thus, if a mass eigenstate has a Higgs component, the probability that the Zboson will decay into the eigenstate will be a2times that of a pure Higgs mass eigenstate2. As the Higgs component of the field decreases, so does our chance of observing the field through elecronpositron collisions. Because of this, if the Higgs component of any field is less than lo%, we wouldn't have observed these fields in earlier collider experiments and even a mass of 1 GeV would be allowed.
Figure 4.2: The electron and the positron collide in linear accelerators to create the Higgs particle with the mediation of the Z boson
4.1 The pterm
As we mentioned earlier, in supersymmetric models, every particle has a super partner. For example, fermionic quarks (spin112) have scalar superpartners called
2~incea mixture of electroweak singlet S fields with Higgs would decay through the same channels as the Higgs, if it were produced, there would be no trouble detecting it.
squarks. If supersymmetry were exact, we would expect squarks to be degenerate in mass with the quarks, and would have already detected them. This is why in realistic supersymmetric models, supersymmetry is assumed to be spontaneously broken. Supersymmetry breaking can give acceptable masses to scalars like the squarks, and, as we will see, to the components of the Higgs.
In this chapter, we will show that supersymmetry breaking can give exper imentally acceptable masses to the scalar components of the Higgs. However, supersymmetry breaking has no effect on the masses of the fermions, including the fermionic Higgs particles. In the minimal supersymmetric standard model (MSSM), the fermionic Higgs fields get masses through the term
in the superpotential. Flavor symmetry prevents us from including this term in our model, which is necessary to satisfy experimental constraints (91. In next to minimal supersymmetric model (NMSSM), instead of the above pterm, a singlet scalar field S couples to the Higgs
The VEV of this S field then creates an effective pterm, ((S)HuHd) which gives masses to the fermionic Higgs.
To make our model realistic, we also need scalar S fields coupling to the Higgs. However, instead of having a singlet S, we have S flavor multiplets in order to conserve the flavor symmetry of our model. Similar to the NMSSM, these S fields generate effective pterms through their VEVs and give masses to the fermionic components of the Higgs. Note that the S fields we had in ear lier chapters have VEVs of order AF. The S fields we introduce in this chapter will turn out to have VEVs of order AsusY in order to generate pterms of order
ns.sy3.
4.2 Supersymmetry Breaking Terms
The potential of a supersymmetric model with spontaneously broken supersym metry contains Aterms and soft mass terms in addition to the Fterms derived from the superpotential using 2.21.
3Couplings between these two types of S fields are forbidden by the U(l)x symmetry
The soft mass terms are of the form &rn2q52,where m is an input parameter and is the scalar component of a field. There are no soft mass terms for the fermionic components of any fields, thus breaking the supersymmetry between the fermionic and scalar compoents of fields. For example, a supersymmetric model with a triplet S field would have the term
in addition to the Fterms derived from the superpotential. The typical expected scale for these terms is TeV. All of S fields in the triplet have the same coefficient because they are all part of the same flavor triplet. If they had different coefficients, some flavors would be explicitly preferred over others and we would no longer have flavor symmetry.
The Aterms look just like superpotential terms, except that they are made only out of the scalar components of the superfields. The coupling constants are input parameters with units of energy. For example, given the superpotential in Equation 4.14 we would expect the following Aterms, which are added to the scalar potential:
V 3A~E~~~c~~~skn + A~Sjs ii (4.8)
il~Jm~
4.3 Only 3 as the symmetry breaking field
For now, we will assume that the S fields do not couple to the Higgs fields for simplicity. We will study the dynamics of the S fields first, and then add the couplings to the Higgs fields in the later sections.
The simplest choices for the S fields allowed by flavor and gauge symmetries are 3 and 6. Let's first start with the simple case where we only introduce a single 3 as our symmetry breaking field.
The superpotential of a theory with only a 3 S (without the Higgs for simplic ity) would be: W = Etik~iS.S 0
J k(4.9)
The above superpotential equals zero, because all the terms cancel when the sum is written out explicitly. Since the superpotential is zero, the potential will have no Fterms. The only term in the potential will be the soft mass term.
All of these fields have the same coefficient because they are all part of the same flavor triplet. If they had different coefficients, some flavors would be explicitly preferred over others and we would no longer have flavor symmetry.
If m2is positive in the above equation, then the minimum of the potential will be at (S 1, S2,S 3) = (0,0,0) and the S fields will not get any VEVs. If m2is nega tive, there will be no stable minimum. Even when we include the Higgs couplings to the superpotentials, the VEV of the S fields will either be zero, or there will be no stable minimum. The potential terms derived from HiggsS couplings will be at most quadratic in S4. These quadratic terms in S cannot stabilize the potential with nonzero (S) since the highest order terms already in the potential are the quadratic m2 terms. Thus, a model with only single 3 as the symmetry breaking field cannot generate an effective pterm.
4.4 Only 6 as the symmetry breaking field
The next simplest choice for the new symmetry breaking fields is a 6. In this case, we are allowed to have a quartic term in the potential which comes from the superpotential term:
W 2eijkelmnS jmSkn
il~ (4.1 1)
Then the superpotential becomes (ignoring the Higgs couplings for simplicity5, and setting the Aterms to zero)
(4.12) Even though we have quartic terms in the potential there are some directions where these quartic terms cannot stabilize the potential. For example, in the di rection
where a is a variable, the quartic term disappears. If m2 is negative, the potential goes to m in this direction. If m2 is positive, then the minimum is at zero, and
4~11terms in the superpotential are of the form H,HdS. Deriving the potential using Equations
2.20 and 2.21 results in terms like (Higgs14 or (Higgs)'s2
5~hecouplings with the Higgs can at most contribute to the quadratic term, which would be equivalent to modifying the soft mass
the S fields do not get VEVs. Thus, a model with a single 6 multiplet as the flavor breaking field is not suitable for generating a pterm either.
4.5 Both 3 and 6 as the symmetry breaking fields
If we introduce both 3 and 6, for certain values of the soft masses, the potential is stable and forces some fields to get VEVs. In this case, the superpotential becomes (without Higgs):
W > E~~~E~~~S~~S~~S
kn + S jSiJ (4.14)
This superpotential gives rise to the following potential (neglecting Aterms for now):
where the softmasssquared terms can be positive or negative. In this form of the potential both the 3 and 6 have stabilizing quartic terms. However, we still do not have a stable potential for every value of rn: and m:. For example, if mz is negative, there is a direction in which the quartic terms for the 6 disappears (as in Section 4.4). In this direction, the potential is not stable, and thus mt should be positive. If mi is positive, mt should be negative in order for the S fields to get VEVs. However, the potential is stable about the point where the 3 gets a VEV and 6 does not, only if mi is less than mt. If we write the mass matrix for the 6 Higgs assuming that it has no VEV, we realize that the diagonal elements of the mass matrix are of order m: (from mi Ci,,j(Si,i)2) and the offdiagonals are of order (SJ2 (from the cross terms of (E~~~E~~~S~~S~~
+ SiS1I2). For the potential minimum to be stable, all the eigenvalues of the mass matrix have to be positive(i.e. there are no directions in the potential where the quadratic term is negative). This will happen if mz > (S). But (S)2 1rn3l2. So, we need mi to be less than m:.
Given the above potential, it is straightforward to numerically minimize it us ing Mathematica. Using the values that the fields get at the minumum as their vacuum expectation values, we can rewrite the potential as a series expansion about the minimum. At the minimum, we can calculate the mass matrix of the fields and find its eigenvalues which correspond to the masses of the mass eigen states. When we do this with a relatively small negative mi and a positive m:, all the fields in 3 get nonzero VEVs and five linear combinations of the fields in 3 are massless. Goldstone's theorem confirms this finding.
4.6 Goldstone's Theorem
Goldstone's theorem [8] states that for every spontaneously broken continuous symmetry transformation, one massless particle will appear in the system. These massless particles are called the Goldstone bosons.
4.6.1 Proof of Goldstone's Theorem
Let us have a potential V(#"), where a is an index and 4" are the fields that V is a function of. If $",e the values of the fields that minimize this potential, the first derivatives of the potential with respect to 4" evaluated at 4; will be zero. That is,
If we Taylor expand the potential about $0, the coefficient of the quadratic terms, which are also entries in the mass matrix become:
A general symmetry in the Lagrangian is defined by the invariance of the La grangian under transformations of the form:
where 'Z is an infinitesimal constant, T is the transformation tensor, and a sum over a' is implicit. At the vacuum state, i.e. when the fields have the constant values of their VEVs with no additional fluctuations, the only terms left in the Lagrangian are the potential terms. Then, the potential should also be invariant under these transformations, which means
Since 'Z is an infinitesimal constant, we can rewrite this as
If we differentiate this with respect to 4band take the value at 4 = #o, we get:
By Equation 4.16, the first term in the equation above is zero. If the symmetry is not broken, the VEVs of the fields should be invariant under flavor transfor mations with T,4, i.e. 4; = 4; + a~,",@$. Then, T:,$$ = 0. In this case, the mass matrix, ("m&,)mocan be anything, and all mass eigenstates can be massive states. However, if the potential is not invariant under T,", transformations, then T,",,df # 0, and we have ()mo (T;,&) = 0. This means that (Imo has a zero eigenvalue, giving rise to a massless state. Furthermore, the mass eigenstate corresponding to the zero eigenvalue is (T:,,d$). So, just by looking at the VEVs of the fields, and the transformation matrix, we can tell where the massless Higgs lives.
4.6.2 Application of Goldstone's Theorem
In the example in Section 4.5, we had an SU(3) symmetric Lagrangian to begin with. SU(3) transformation tensors are unitary 3x3 matrices with determinant 1. Every SU(3) transformation matrix can be written as the following:
where 13's are real constants and T's are hermitian traceless matrices. In this form, M is unitary since
Every M in this form also has determinant 1. Peskin and Shroeder shows in
[I I] that for any diagonalizable square matrix det(M) = Exp [Tr(ln(M))], which is equivalent to In(det(M)) = Tr(ln(M)) is true. If we apply this to the M from above, we get:
In (det (M)) = Tr (ln (M))
In (det (M)) = Tr (ln (eibTa))
In (det (M)) = Tr (id, T,)
In (det (M)) = iQ,Tr (T,)
ln(det (M)) = 0
since Tais traceless. This means that det(M) = 1.
If every SU(3) transformation tensor M can be written in terms of hermitian traceless matrix T,, the maximum number of linearly independent SU(3) transfor mations will be the same as the number of independent entries in T,. In a 3x3 hermitian matrix, there are three real diagonal entries. T, is traceless only two of these are linearly independent. Since Tais hermitian, there are only six real independent entries in the offdiagonal elements, making the total of independent entries eight.
The spontaneous breaking of the SU(3) symmetry in the example in Section
4.5 still leaves some freedom in the Lagrangian. In fact, after the breaking we are left with an SU(2) symmetry. After the breaking of the symmetry, we can switch to a basis where the VEV of 3 is (0, 0, v). As we argued earlier, none of the fields in 6 get VEVs either. This means that we can apply SU(2) transformations to the first two terms without changing the overall VEV. The number of linearly independent SU(2) transformations are three. So, 8 3 = 5 continuous symmetries are broken which results in 5 massless particles according to Goldstone's Theorem.
4.7 Including A terms
In Sections 4.4 and 4.5, the 6 S fields could not get any VEVs, because the quadratic term for the VEVs had to be positive. The positive quadratic and quar tic terms forced the minimum of the potential to be at zero for all the 6 S fields. The addition of Aterms in Equation 4.8 to the potential generates effective linear terms, which in combination with the positive quadratic terms create a nonzero minimum for some of the fields in 6 S.
The linear term comes from the term A2SiS jSij when both S i and S j have nonzero VEVs. When we switch into a basis where in the vacuum state (Si)= (0,O, v), a linear term for only S is generated. This means that only S gets a VEV, and the other fields in the 6 are zero in the ground state. That is,
si= [ :::]
oov
when (Si)= (0,0, v).
In this form, the vacuum still has an SU(2) symmetry, and according to Gold stone's Theorem, there should be five massless fields. When we write out the potential and minimize it, we confirm that there are five massless fields. The only difference is that, the inclusion of the Aterms to the potential partially move the massless fields into the 6. Recall that, in Section 4.5, the fields in 6 did not have any massless components. This has a simple explanation using the Goldstone the orem. We showed in the proof of the Goldstone theorem that the massless field lives in the fields whose VEVs are shifted by the spontaneously broken transfor mation. In Section 4.5, the 6 S field+s did not acquire VEVs. If the VEVs are zero to start with, they will remain zero after any transformation, and thus won't be shifted. Because of this, the Goldstone boson did not live in any of the 6 S fields. In the model that we are considering in this section, the 6 S fields acquire VEVs, which shift after being transformed by the spontaneously broken SU(3) transformation. Therefore, the Goldstone boson must have some 6 S fields.
4.8 Coupling 3 and 6 to the Higgs
In this section, we will study the dynamics our model when the S fields couple to the Higgs. These couplings explicitly break flavor symmetry if we only include light Higgs, so we should expect some Goldstone fields to get masses. In addition to the restrictions on the masses of the new fields we introduce, there are bounds on some of the parameters in our superpotential. For the fermionic partners of the Higgs scalars to gain masses greater than 100 GeV, the effective p term in the superpotential, i.e., the coefficient of HUHdterms, should be greater than 100 GeV. Note that this is the coefficient of the superpotential when the superpotential is expanded about the VEVs of the fields. That is, if the superpotential is of the form:
W 3 a SHUHd (4.24)
where S is some field and a is a coupling constant, then p = a(S). Furthermore, \l(~,)~ 174 GeV. This is due to the fact that
+ (Hd)2should be the Higgs VEVs are connected to the known ZO boson mass and the electroweak gauge couplings in the following way:
When we write out all the flavor and gauge symmetric terms with the Higgs and the S fields, the full superpotential is:
Note that the coupling constants for every term in the above equation are omitted for simplicity. The full superpotential in Equation 4.26 reduces to the following effective superpotential when every Higgs field is replaced by its light component:
where the E'S have been omitted in the last line, ai, bi and ci have been absorbed into ai,and dij and eij have been absorbed into yij.
Even though it is not explicit in the equation above, replacing every Higgs field with its light component creates big suppressions for every term containing Higgs. These suppressions are due to the hierarchy in the fermion masses (i.e. mdown<< mstrange<< mbotlom)that were used in Section 3.2.3 and 3.2.4. In these sections, we force the Higgs expectation values to follow the same hierarchy in order to obtain the experimentally known CKM matrix and fermion masses.
When we replace the Higgs multiplets with their light components as above, we use the solution to Equation 4.2, which follows directly from the hierarchical Higgs expectation values. This relation causes the light fields to live in similar places in both up and down Higgs multiplets in Equation 4.26. From Equation 3.35, it is clear that Hi3 term will dominate over any other term in the sums that contain H:. The same is also true for H,?~.Unfortunately, there are no terms in
4.26 where Hj3 and H,33are multiplied together. Due to the specific features of our model's tensour structure, the form of 4.26 suggests that the next biggest term would be terms containing ~j~ . ~These terms would carry a suppression of 2~
,
order lo'. We had several degrees of freedom when choosing the numbers in Equation
3.35. However, the freedom we had is not enough to prevent the suppressions.
We only used MdMl and MUM,', in our calculations for the CKM matrix and the fermion masses. Instead of diagonalizing M~M;all at once, if we diagonalize in dividual mass matrices, we see that any right handed rotation on the mass matrices do not affect the GKM matrix, i.e.:
M~M; = v~LM~v~R(v~LM~v~R)' (4.28)
In the last line, VdRv;, cancels as long as it is a unitary matrix. Thus, any right handed unitary rotation of the mass matrices do not affect the CKM matrix or the fermion masses.
A righthanded rotation of the mass matrix corresponds to the right handed rotation of the Higgs tensors. This doesn't remove the suppressions in Equation
4.26 since righthanded rotations can only move the numbers in Equation 3.35 horizontally. We need H:2 to be greater than any of the other terms in the third row of H,.
To overcome the problem due to these suppressions in the superpotential, we will aim to generate big S VEVs so thatp = a (S)is 100 GeV. We cannot simply make the coefficient a larger because this will cause the perturbative approach to quantum field theory to eventually break down. To be consistent, we require that the coefficient is no larger than order unity.
Differentiating the superpotential with respect to each field, we get the poten tial:
V = + 6?6? +
laiuD + 6y3sj12+ ~Y,~UD + 3si3sj12
lai3~j12(1~12 + Vsoft mass + (A terms) (4.31)
+ 1~1~)
where VsOpmass includes all the soft mass terms and the quartic Higgs term, (U28 + D~)~.
Also, note that every field appearing in this potential is a com plex field. Thus, we need to write out all the real and complex components of the fields before manipulating the potential.
To find the VEVs of the scalar fields, we need to minimize the potential. Even though it is possible to minimize the above potential with all its 22 fields (includ ing real and complex parts), it is more illustrative to analyze it term by term.
4.8.1 When& has noVEV
One way to simplify the potential in Equation 4.31 is to make the soft mass term for 6s very big. This way, any VEV that gS gets will be very costly (i.e. will increase the potential). Thus, the minimum of the potential has 0 VEV for all the fields in 6s and the potential becomes:
V = laiUD12+ lyijUD + 3si3sj12+
1a~3s~1~(1~1~+ 101') + V,,,, + (A terms) (4.32)
Let's look at the effects of each term in the potential above. First, we will consider the case where yij = 0 and there are no Aterms. From Equation 4.27, we see that setting yij = 0 is the same as removing the 6s UD couplings from the superpotential.
Potential without Aterms or yij
In this case, the potential reduces to:
Clearly, when we make all the softmassessquared positive, no field in the above potential will acquire a VEV. The minimum of the potential will be 0. To generate VEVs for the fields, we want to make m: negative. With the help of the quartic term, 13si3sj/2, the magnitude of the VEV of 3s will be stable. If either U or D gets a VEV, the direction of 3s will be determined by the term jai3s112(1 ul2+1Dl2), which is the only 3s term that couples to U and D. To minimize this term, 3s will point in an orthogonal direction to a'. The VEVs of U and D would be independent of the VEV of 3~, since lai3s,l2 = 0.
In this case, 3s and the Higgs fields are decoupled. We can independently adjust m;, mi and mi to get any VEV we want for the fields. This freedom results in 5 symmetries in the potential. These symmetries are:
All these symmetries are spontaneously broken, giving rise to 5 massless fields, two of which live entirely in the Higgs.
Since we can adjust the Higgs and the S VEV independently, we can easily make (S) >> (U),but we still cannot get a big p term. In fact, for this form of the potential, p = ai(3si)= 0.
This potential can neither generate a big p term, nor can it give masses to all of the Higgs fields. So, we need to look at the other terms in the potential.
Potential without Aterms
If we include the y term to the potential in the previous section we get:
The y term in the potential prevents S and the Higgs fields from decoupling. When y is nonzero, the two terms, lyijUD + 3si3sj12and jmi3si12([U12+ +Dl2) dictate the direction of (3si). If 3si points orthogonally to ai as in the previous section, lyijUD+ 3si3s,12becomes positive. If 3sipoints in a direction as to make lyCjUD+ 3si3sj12zero, lai3si/2becomes nonzero. Thus, to minimize to potential, 3sineeds to point somewhere in between the two directions, where the exact di rection depends on the relative size of y and a. Since, 3s, is no longer orthogonal to ai,p = ai(3s,)f 0.
Although, this form of the potential is able to generate a nonzero pterm, it is still far from fulfilling all the experimental constraints. The potential has two symmetries that are simultaneously broken to give two massless fields, both of which have Higgs components. The two symmetries are:
Both these symmetries are spontaneously broken when we minimize the poten tial (i.e. the VEVs of the fields change under these transformations). Therefore, there should be two Goldstone bosons. The Goldstone boson we get from the sec ond symmetry is expected and harmless. The second symmetry is nothing more than the U(l)y symmetry of the standard model. The massless field due to this symmetry corresponds to longitudinal polarization of the Z gauge boson, which is experimentally known to exist. The first symmetry on the other hand, gives rise to a massless field which contradicts experiments.
Potential with the Aterms
Since we are assuming that the 6s fields do not get any VEVs, the only Aterm that we need to consider is Ai3s, UD. Then the potential becomes:
V = + lyijUD+ 3si3s,~2
18~~1~ + lct.i3~i12(~~~2+ 1~1 +~ Vsoft mass + Ai3siuD
) (4.41)
Here, we have three different A constants because the Aterms are:
A ~"~~j~3~,3~~3~,+ ~"3~~3~~6~terms = + ~'~3~,6;3~, (4.42)
We can absorb the constants due to the light components into the A's and write the potential as above.
The addition of the Aterm explicitly breaks the first symmetry in 4.40,leaving only one spontaneously broken symmetry in the potential. This spontaneously broken symmetry gives rise to the only Goldstone boson, which is "eaten" by the Zboson, as mentioned in the previous section. The rest of the eigenstates, on the other hand, now have light masses. This is due to the fact that the y term is suppressed because of the hierarchical arrangement of the light Higgs fields. The y term explicitly breaks the symmetry:
That is, if the y term did not exist at all, we would get a Goldstone boson when this symmetry is spontaneously broken. This Goldstone boson gets a mass when the y term is introduced. However, its mass is proportional to the size of y. If y is very small, as it is the case in our model, the symmetry is not very strongly broken and the Goldstone boson cannot get a big mass. The S and H components of this Goldstone boson can be read off the symmetry above. For very small 4,we can approximate the symmetry as:
So, the change in the fields after a small transformation is:
63s,, 6U and 6Dcorrespond to ct.~;,4" in Equation 4.18.
Using the argument from Section 4.6.1, we realize that the ratio of the S, U and D components of the Goldstone boson is (3Si) : (U) : (D). So, unless (3si) is much greater than the VEV of the Higgs fields, the Goldstone will have a substantial Higgs component. This violates the experimental bounds, since any mass eigenstate with more than 0.1 Higgs component has to be heavy. So, if we can make (S) >> (H),all the masses will be within the experimental bounds. We already had this requirement to get a large y term. So, without any additional constraints on our model, the light mass eigenstates have less than 10% Higgs components, and are within the experimental range.
Increasing the ratio of (U) to (D)
When rn; > rn;, increasing the VEV of D becomes much more costly than increas ing (U). Therefore, when we minimize the potential, we always get (U) > (D). Experimentally we are required to have (D) <: (U) 5 50(D) [7]. If we let tanp = (U)/(D),we can rewrite the potential as:
U2
+ Vsannloss + ~'3~;(4.46)
tanp
If we make tanp large, the magnitude of 3s is determined to a large extent by rn: because of the suppressions due to the fermion mass hierarchy. If we also have a relatively large A, the Aterm determines the direction of 3si. By pointing
u2
in the opposite direction of A', 3sican make A13siZ get a large negative value, which lowers the potential. Thus, to minimize the potential, 3Sipoints along Ai, assuming that the Aterm dominates over the other terms involving 3si. Since we want a big yterm, it is to our advantage to make Ai and ai point in the same direction. This way, 3s; points in the same line as a' and p = ai3sj is maximized. To have less than 10% Higgs component in the light fields, we need
This can be done by raising the value of rn;. This requires some finetuning because the VEV of both Higgs fields become zero when the soft masses are big. The amount of required finetuning increases with Ai and decreases with tanp.
As Ai increases, the vacuum expectation value of Ai3si increases. This term appears in the offdiagonal elements of the Higgs mass matrix while the Higgs soft masses appear in the diagonal elements. That is, if the Higgs fields have zero VEVs (or the Higgs VEVs are very small compared to the Higgs soft masses):
If both eigenvalues of the matrix are positive for a given (3s,), we can be confi dent that the vacuum state that we are considering is stable6. These eigenvalues correspond to the second derivatives in the mass eigenstate direction. If we are at the minimum of a potential (i.e. in the vacuum state), the second derivative of the potential in any direction has to be positive (or zero). Otherwise, other states with lower energy would be accessible and the system would spontaneously shift to the lower energy state.
The determinant of a matrix is equal to the product of its eigenvalues. There fore, if the determinant of the above mass matrix is negative, then the mass matrix has one negative eigenvalue resulting in an unstable potential about zero Higgs VEV. To make this point stable, we can slowly increase the masssquared terms for the Higgs until the determinant is positive.
To make the VEVs of the Higgs field arbitrarily small (relative to (3s,)), we need to make the potential only slightly unstable at the point where the Higgs fields are zero. In this case, the minimum of the potential will not be far from (U, D) = (0,0),giving small VEVs to the Higgs. Slightly unstable potential means that the determinant of the above mass matrix is only slightly negative. This will happen if we balance the offdiagonal elements and the diagonal elements of the mass matrix. As the values of the offdiagonal elements increase, it becomes harder to get a very small negative value for the determinant of the mass matrix; a delicate finetuning is required7.
When tanp is big, the quartic Higgs term, V(U2 D')', becomes more significant due to the fact that (U2) > (0'). SO, the Higgs potential is more strongly restored by the quartic terms and U and D acquire smaller VEVs relative to (3s,). This makes it much easier to get (3s,) > ~((U))Z+ ((D))2, and reduces finetuning.
6~echnically,we also need to check that the first derivatives also vanish in that direction
7~hisis like trying to get a b = 1 when a is a very large number. If a = 2, we only need a single digit accuracy for b, since 2 3 = 1. If a = lo5, we need b to be exactly 100001, which requires six digits of accuracy
The greater value we have for tanp, the less severe is the finetuning to get a big VEV for 3si,due to the effects of the quartic Higgs term. However, as we mentioned earlier, the value of tanp needs to be less than around 50. We can also make A' small to make finetuning less severe. However, if we make Ai arbitrarily
2
low, laL3sil term will dominate the quadratic terms for the Higgs fields and 3Si will point in a direction orthogonal to ai. This causes the pterm to be zero.
In fact, it is not hard to derive an expression relating the Ai to p. First, let A' = Aai to make it explicit that we want A' to point in the same direction as ai. Then, let's define 8 to be the angle between ai and 3si. This angle is well defined, since both ahnd 3si are threevectors (let's say with magnitudes a and S respectively). With these definitions, we can rewrite the Aterm as:
1 Ul2 . (uI2 A~scos8u2
Ai3Si= Aaz3Si

tanp tanp tanp where I am using a negative sign to explicitly show that the Aterm is negative.
2
We can also rewrite the lai3s,1 term using the same definitions (when tanp >> 1):
2
lac3sillu12= a2s2cos28 (4.50)
If we assume y is small8, then these two terms in the potential are the only terms that are functions of 8. At the minimum of the potential, $ = 0. So,
dV AaS sin 0lUl2
2a2s21u12cos8sin8 = 0dQ tanp
Solving for cos 0, we get:
A
cos 8 =
2aS tanp Then, p is
A
p = ai3si = aS cos 8 = 
2 tanp which tells us that we need A when tanp 100 GeV in order to get
4 T~V~ =
p 100 GeV.
8~e
can make this assumption since y has a suppression due to the hierarchy in the fermion masses. If this suppression is not small enough, we can always make the coupling constants in the superpotential that are absorbed into y smaller.
'~ecallthat A is defined by the relation A' = Aa'
If we rewrite the mass matrix from Equation 4.54 in terms of 9, and substitute in A = 2,~tanp and simplify, we get:
m: 2,~~
tan/? 2,~~ m:
tanp
This form of the mass matrix clearly shows that increasing tanp increases the fine tuning necessary to get (3~~)
>> (H), which contradicts our previous prediction that increasing tanp reduces fine tuning by making the quartic Higgs term larger. From this, we can conclude that tanp has two competing affects on the fine tuning, and a more detailed analysis is necessary to see whether increasing tanp increases or reduces fine tuning necessary to get (3s,) >> (H).
Furthermore, when tanp 15, the offdiagonal elements of the mass matrix become (600 GeV)2. This means that the softmassessquared need to be of order (T~V)~
scale to get a small determinant.
We now list the results for a sample point in the parameter spacelo. When Ai = (2.66 GeV, 90.8 GeV, 501 GeV) (as expressed in Equation 4.46), all the coupling constants in the full superpotential are unity except for the 6sHiggs coupling constants which are all 0.1, the soft masses are m: = (2.09 TeV)2, mi =
(0.017 T~v)~and m: = (1.74 T~v)~,
we get
p = 127 GeV (4.55) (Si)= (329GeV,77.8GeV,1190GeV) (4.56)
The masses and the Higgs contents of the mass eigenstates are given in Table 4.1. All of these results are well within the experimental bounds, and the light fields in Table 4.1 may be detectable in the future, as we will discuss in Section 5.1.
'O~ote that, instead of the numbers given in Section 3.2.4, a different solution for the Higgs expectation values were used in our calculations
Mass Higgs component
2.47 TeV 0.149
2.10 TeV
2.10 TeV
94.7 GeV
14.3 GeV
12.5 GeV
8.16 GeV
6.58 GeV
2.87 GeV Table 4.1 :The masses and the Higgs components of mass eigenstates in our model
Chapter 5
Experimental Signatures
In previous chapters, we found models that accommodate the experimental data. Now, we need to find ways to test our model experimentally. In this chapter, we will first suggest ways to detect the light scalar fields from Chapter 4. Then, we will show that the S fields from Chapter 3 do not strongly couple to the light Higgs, which prevents us from observing them.
S fields that give masses to light Higgs
Table 4.1 shows that there are many mass eigenstates in our model with less than
0.1 Higgs component. Five of these fields have masses of less than 30 GeV. If these light particles were ever created, they would be within the probing range of existing colliders. Unfortunately, these particles were not observed in earlier collider experiments [5][6J, because the probability of their production through diagrams like Figure 4.2 is proportional to the square of their Higgs components as we mentioned in Chapter 4. For example, the probability of the production of the 2.87 GeV mass eigenstate in Table 4.1 is 130 times lower than that of a pure Higgs mass eigenstate. As the probability of detection decreases, the collider experiments have to collect much more data to increase the chance of producing these particles.
Previous collider experiments, LEP I and LEP 11, in particular, collected only enough data to rule out the existence of any light mass eigenstates with less than
0.1 Higgs component. To confirm or rule out the existence of particles with smaller Higgs components, such as the 2.87 GeV mass eigenstate from Table 4.1 a future e+ecollider would need to produce 2 times more Z's than that were collected at LEP I. There are plans for an experiment that would produce a number of 2's well in excess of this [lo].
If our model with the parameters given in Chapter 4 reflects reality, the five light particles in our model with small Higgs components might be detected sim ply by collecting much more data. Of course, the masses most likely will not be the exact masses given in Table 4.1, because of the arbitrariness of the input pa rameters chosen for the model in Chapter 4. The detected masses of these fields (or the bounds on such masses) would fix (or put bounds on) the input parameters of our model.
Another distinct feature of our model would be the decay products of the light mass eigenstates. The lightest mass eigenstate in our model (which is also the most easily detectable because of its 0.088 Higgs component) has mass of only
2.87 GeV. So, it cannot decay into the bottom quark (which the heavier Higgs par ticles would decay to) because of conservation of energy. So, with the parameters given in Chapter 4, we would expect to detect particles less than GeV, such as p+ppairs
5.2 S fields that give masses to heavy Higgs
We might also hope to see the S fields that give masses to heavy Higgs in future colliders, because their masses are of order As even though their VEVs are of order AF. At the flavor breaking scale, AF, we know that the heavy Higgs fields do not acquire VEVs. The VEV of the light Higgs is also negligible at this energy scale, since it acquires a VEV of order Asusr through supersymmetry breaking effects. So, the superpotential at this energy scale results in a flat potential for the S fields when we expand it about the VEVs of the Higgs fields. A flat potential means that the S fields are massless at AF Their potential is only modified at Asusy by supersymmetry breaking effects, giving them masses of order Asusy. The masses at this scale are within the reach of future colliders.
5.2.1 The 333Model
We know that the model where we have only a single 3 Higgs up, a 3 Higgs down and a 3 flavor breaking sector doesn't work with the experimental data. However, we will still try to find the experimental signatures of this model to compare with the experimental signatures we get from more complex models.
In the 333 Model ( 3 Higgs up, 3 Higgs down and 3 S model), we have a single term in the superpotential which is:
From this superpotential, we can calculate the potential, V, using equations (2.20) and (2.21). The potential gives us the mass terms for the Higgs and we can cal culate the mass matrix for the Higgs particles in terms of the expectation values of the S 's. Then, we need to diagonalize the mass matrix to figure out what linear combination of the H's gives the light Higgs and which two linear combinations give the heavy ones. This is the same procedure as Chapters 2 and 3. The nor malized eigenvector corresponding to the zero eigenvalue is the light Higgs in the basis where the components of H, i.e. HI, Hz and H3 are the basis vectors. The other two eigenvectors are the heavy Higgs fields in the same basis. Since we want to figure out the couplings of the light Higgs to the symmetry breaking fields, we need to change to a new basis where the light Higgs and the two heavy Higgs' are the basis vectors themselves.
After changing basis, V will be a function of Hul, Hdl, Huhl, Huh2,Hdhl and Hdh2, where the subscript h represents a heavy Higgs and 1 represents a light Higgs. If we expand V right away, it looks like the light Higgs fields couple to the S fields. However, we cannot do this, because we are looking for couplings at the low energy scales and this potential is the potential at AF.We first need to derive an effective potential for low energies that only includes light particles. This is done by solving for the heavy Higgs in terms of the light ones using the following equations. This procedure is also called "integrating out" the heavy fields.
After we substitute the solutions to the heavy Higgs' into the potential, we write the flavor breaking sector as a sum of its vacuum expectation values, vl, v2, v3,and the time and space dependent fields sl(x), s2(x) and s3(x). When we do these substitutions, we see that all the terms in V cancel out and we get V = 0. This means that the light Higgs don't couple to any of the symmetry breaking fields and thus in a model like this, there are no observable experimental signatures for these S fields.
Instead of doing all these calculations, we could have realized that the light Higgs don't couple to the flavor breaking fields by doing some simplifying trans
formations. Every 3 field S = , where v's are the vacuum expecta
tion values and the s's are the time and space dependent fields, can be rewritten as follows:
(5.4)
where V(x) is a S U(3) matrix and
If we substitute the new definition of S, equation (5.4), into the superpotential defined in (5.1), we get W = eiikH~ Hdj Vkk' S (5.6) Then, we let
Ajk = €ijk H . =
U1
Hu1 Ha1 0 We can always rewrite Ajk and Hdjin terms of V which is an S U(3) transforma tion. We do this as follows:
If we plug equations (5.7), (5.8) and (5.9) into (5.6), and rewrite in matrix trace form, we get: Furthermore, equation (5.8)implies that Hu = Vi' H;, . If we let AYk' in (5.8)be
~fj'k'
Ei'j'k'~
ui,,we get:
1
Also, since V is an S U(3)matrix, we know that V+V= Vil.,V."' = I. Because identity matrix doesn't affect anything, we can rewrite this in tensor form and add it to the equation above by coupling it with HL and E.
But we know that the definition of the determinant of V is
Because of the definition of E,any cyclic permutation of 1,2and 3, will also give the determinant, and any anticyclic permutation of 1,2 and 3 gives the negative of the determinant. Sums like Vi,,V*'~V~~,
where any two of i, j or k are equal equal to zero because all terms cancel out. This is precisely the definition of E. so, ~i , V*J,,vkk,Ei'j'k' = €ijk and
J
Thus, we have just shown (5.8)implies Hui= Vi'HL,.This means, if we change into a new basis where Huiand Hd are the basis fields, the superpotential remains the same and can be expressed as1
where Sf(x)= . If we need to, we can always use the definitions of
the unprimed fields in terms of the primed fields to change back into the original basis. If we plug in the definition of Sf,we get:
Using (2.20) and (2.21), we get
If we plug in the expectation values for S', we see that the mass matrices of the Higgs fields are diagonal, and one of the three diagonal elements in each mass matrix is zero. Since the mass matrix is diagonal, it's eigenvalues are the diagonal
elements and the eigenvectors corresponding to these eigenvalues are
B],[!]
and ( 1. These are already orthogonal and one of the basis elements already
.. z
corresponds to the light higgs fields, so, we don't need to change basis. We just let Hu3and Hd3be HuLand Hdlrespectively and the other fields be the heavy Higgs fields. Solving for the heavy Higgs fields in the equations (5.2) and (5.3) gives zero for all the heavy Higgs fields. Since all the terms in the potential have heavy Higgs fields, the potential reduces to zero, resulting in no coupling between the light Higgs field and the flavor breaking fields.
5.2.2 6+31=8 Model with only 6 as the flavor breaking sector
In this model, when we only have a 6 as the flavor breaking sector, the superpo tential is:
We can rewrite the adjoint as a linear combination of eight linearly independent, unitary 3 x 3 hermitian traceless matrices. This way, the kinetic terms in the Lagrangian would be canonically normalized. If Ra are these matrices and 8" are the coefficients in the linear combinations then,
Using equations (2.20) and (2.21), we get the terms in the potential involving the up Higgs as follows:
To find the light Higgs couplings with the light symmetry breaking fields, we first need to change to a new basis where the mass eigenstates are the basis vectors. Then, we need solve for the heavy Higgs in terms of the light one using the equation
for every heavy Higgs. Before doing this, we first simplify the potential by series expanding to the powers of the light fields, and concentrating on the lowest order term.
In order to do this kind of expansion, we first need to figure out the lowest order light Higgs term in the solution for the heavy higgs. We know that when expanded the potential will have the form:
where mplis a mass scale that we need to have the same units for every term. This mass scale, mpl turns out to be very big compared to the Higgs fields and therefore the first term in each row dominates.
We don't have any terms of the form (heavyj(1ight)because the mass matrix is diagonalized which means there are no cross terms between any of the mass eigenstates. When we write down equation (5.22)for the potential above, we get an equation of the form:
dV 1
= + (lightj4+
2m2(heavyj+(mpl(light)2+(lightj3 ...)
dheavy PI +2(heavy)(mpl(lightj+ (light)' + ...) + ... = 0 (5.24)
Solving for (heavy), we get
(mpl(light)2 + (light)3 + &(light)4 + . ..)
(heavy) = (5.25)
2m2 + 2(mpl(light) + (light)2 + . . .) + . . .
By expanding this in powers of &,we see that the heavy fields are O(light2).
With this information, we see that the lowest order term in (light) series ex pansion in equation (5.21) is 0(light4). One might think that we can get lower order terms from the first row of (5.21) by choosing the light components of HLJ and HUi, and the vacuum expectation values of the S fields. However, there should not be any (construrt)~~,,,~,
terms in the Lagrangian at this energy scale. If these terrns existed, the light Higgs field would have masses of order lot6GeV. So, we know that Hlighi can only couple to the time and space dependent field components
of the S,not their expectation values.
We also realize that the second and the third lines of (5.21) don't have any light Higgs light S couplings of order 0(light4). We know that the elements of the adjoint Higgs matrix don't have any light Higgs components. Since these elements are made up of only heavy Higgs fields, each element of adjoint Higgs matrix is O(light2). If these fields couple to the S field, the coupling terms be come light^)^. Because of this, we can ignore the last two rows of (5.21) and concentrate on the first row.
As we mentioned earlier, when we rewrite the Higgs fields in terms of their light and heavy components, and expand the first row in (5.21), many terms can cel. The series expansion of the potential to the lowest order in light fields gives:
In this form of the potential, we can factor the square in two parts and only check one part to see whether the terms in there cancel. If they do, they will also cancel in the second part and they won't appear in the overall expansion. For example, we know that all the light Higgs terms that are multiplied by the vacuum expectation value cancel on the first side of the equation. This is due to the fact that the light Higgs field cannot have quadratic terms. The only O(light4) terms that do not cancel are the terms with the heavy Higgs fields and the vacuum expectation values and also the terms with light Higgs and the s particles. Using this fact, we
2~e
also have 0(light4)terms like HH(S)~ in the expansion. However, we are not interested in terms that do not involve the S field itself. So, we are ignoring these terms
can simplify the entire expression as follows:
where cq are constants, Hi are the heavy Higgs fields, dz are some other constants, h is the light Higgs field, and 1, are the time and space dependent components of the S fields. The vacuum expectation value components of the S fields are incorporated into the constants cq and d:.
This simple representation makes it very easy for us to solve for the heavy higgs fields using (5.22). When we apply these equations to the potential above, we get
If we write this in matrix form, we get
where the indices a and i in cg are the row and the column indices of C respectively, H is a column vector where the rows are Hi, D's row and column indices are a and a in d:, and L is another column vector with rows 1,.
Furthermore, from the simplified version of the potential, we see that CC? is the mass matrix of the heavy higgs fields. Because we already diagonalized the mass matrix for the up Higgs fields by changing our basis, we know that CCt should also be diagonal. Furthermore, none of the masses in the diagonal are zero since all elements in the diagonal correspond to heavy Higgs masses. The determinant of a diagonal matrix is simply the product of its diagonal elements. Because there are no zeros in the diagonal, we see that det(CC1)# 0. This also means det(C) f 0. A matrix with nonzero determinant is invertible. So, we can multiply both sides of equation (5.29)with (CT)',which gives us:
(CH + hDL)*= 0 (5.30)
Since V = (CH + ~DL)~(CH = 0 + O(1ighg). Since
+ hDL)*,we realize that V ~(lighg)terms are very small because of our mass scale, we conclude that there are no observable light Higgs light S field couplings in this model.
5.2.3 6 + 3 1 = 8 Model with 6 and 3 as the flavor breaking sector
We can generalize the argument in Section (5.2.2) to the 6 + 3 1 = 8 model with 6 and 3 as the flavor breaking sector. Adding 3 S field to our model doubles the number of terms in the superpotential, and causes V, in equation (5.21) to be more complicated. However, all the arguments in Section (5.2.2) still apply. Two of the sums in the potential are forced to be 0(ligh9), and we can still rewrite the potential as in equation (5.27). This means that there are no observable S light Higgs coupling in this model either.
5.2.4 6 + 3 1 = 8 Model with 6 and 3 as the flavor breaking sector
There were no O(light4) light S light Higgs couplings in the previous two mod els, because in both the potential could be reduced to a single sum. When we change the 3 in the previous model to a 3, the potential can no longer be reduced to a single sum. In this case, the superpotential becomes
From here, we can write the equations for F using equation (2.21). We know that for each F should go to zero, if we substitute the Higgs fields with their light components and the S fields with their vacuum expectation values. Using this fact, we can calculate the light components of each Higgs field like we did earlier in Section 3.2.1.
The first set of F's are:
But this is the same equation as in Section 3.2.1. This equation forces Hujto have no light components and the 6 S field to have light components only in the diagonal elements, that are inversely proportional to the diagonal elements of Ski.
The next set of F's are: Since Hui doesn't have any light components, and the 6 Higgs has light compo nents only on the diagonal, we are left with the following three equations:
The above equations force the offdiagonal elements of the adjoint Higgs HLi to have nonzero light components.
The final set of F's are obtained by differentiating the superpotential with re spect to the 6 fields. Because 6 is symmetric, we can write the F terms as follows after simplifying using what we know about the light components of 6 and 3:
With these final set of equations for F, we see that the diagonal elements of the adjoint Higgs don't get any light components and the offdiagonal elements do.
Now that we know where the light fields appear, we can look at the potential to see whether they couple to light S fields. Using the F's calculated above and the equation (2.20), we calculate the potential to be:
Since all of the sums in the potential have both light and heavy components, we can write 0(light4) terms of the potential as follows:
where everthing is defined in the same way as in equation (5.27). To find whether there are light Higgs light S couplings in this potential that are observable at low energies, we first need to integrate out the heavy Higgs fields. We do this by solving for the heavy Higgs fields from equation (5.22). Then we plug in the solutions to the heavy Higgs' back to the above potential and see whether any terms remain. In fact the calculation will be exactly the same as the calculations in Section 5.2.2 after which we rewrite the potential as a single sum as follows:
where C and D are redefined constants. If we imagine Hi and 1, to be column vectors and C and D as matrices, then the potential becomes:
V = (CH + dlh)'(CH + dlh) = (H~c~
+ h*lidt)(cH + dlh)
Integrating out the heavy Higgs, i.e. the H vector, we get:
We also know that the mass matrix of the heavy Higgs is the CtC in the H~C~CHterm in the potential. So, we know that C~Cis a diagonal mass ma trix with no zeros in the diagonal. This makes the determinant of CtC nonzero, which means that C also has a nonzero determinant. All matrices with nonzero determinants are invertible, so C' exists. When we multiply both sides of the above equation with Cl, we get:
When we plug in this result to the potential, we see that the potential goes to zero and that there are no light S light Higgs couplings. These calculations show us that in order to have O(light4) light Higgs light S couplings, we need to have terms of the form ~dj~hl,))~
in the superpotential. When we integrate out the heavy Higgs, the potential terms derived from these superpotential terms will remain in the potential.
We might have guessed this result from the beginning. We know that the S VEV gives masses to the heavy Higgs fields at AF.At this energy scale, the light Higgs field remains massless which should mean that it has no coupligs to the S fields. Although this sounds like a good explanation, it is only an intuitive one, and in this chapter we showed that this is indeed correct.
Chapter 6
Conclusions
In this thesis, we have shown that a flavor symmetric supersymmetric model is feasible. The flavor symmetric model we developed accommodates the experi mental values for the fermion masses and mixing angles. The light pair of Higgs fields behaves the same way as the corresponding field in the standard supersym metric model. The ratio of vacuum expectation values (VEVs) of the up and down Higgs fields is within the allowed range, and with some fine tuning, the masses of all particles are within the experimental bounds.
We constructed our model by assuming that flavor symmetry is broken at an energy scale AF loi6GeV by some scalar fields that we referred to as the Sfields. These new fields also give masses to Higgs fields by coupling to them in the superpotential. The way they couple to the Higgs is similar to how the single S field in NMSSM (next to minimal supersymmetric model) couples to the Higgs. In our model, we investigated what happens if instead of a singlet S field, we have a sector of S fields that preserve flavor symmetry.
To avoid predicting processes known from experiments to be rare, we chose the S and Higgs multiplets so that all but one pair of Higgs fields acquire masses at the flavor symmetry breaking scale. Unfortunately, in Section 5.2, we showed that it is not possible to observe these S fields in the near future since they did not have significant couplings to the light Higgs fields.
We also investigated the implications of this model at low energies by work ing with an effective theory after integrating out the heavy Higgs fields. We were able to show that after supersymmetry breaking, all of the fields with substantial Higgs components had high enough masses to place them out of the detection range of past collider experiments. The hierarchy of the masses of fermion flavors caused major suppressions in terms involving the Higgs fields. These suppres sions, along with the requirement for the size of the effective pterm (the term that gives masses to the fermionic components of the Higgs) forced the S fields to acquire VEVs several times the Higgs VEV. This resulted in some finetuning of the soft masses to achieve the observed scale of electroweak symmetry break ing. A more thorough study of the large parameter space of the model would be necessary to determine whether this finetuning is unavoidable.
In spite of the finetuning, we can conclude that we may be living in a flavor symmetric universe where the symmetry is spontaneously broken at high energies. To stay well within the experimental bounds, we were not forced to have unnatural coupling constants. All of the coupling constants were order unity and the Aterms were not unnaturally large.
We also described in Section 5.1 an experimental signature of our model that can be probed at a future e+ecollider'. We showed that the light states in our model which are a mixture of Higgs and scalar fields would not have been de tected in earlier collider experiments. For the parameter space that place us within the experimental bounds, the Higgs content of the lightest mass eigenstates in our model is too small. However, the lightest mass eigenstate in our model with the parameters given in Chapter 4 would be detectable at an e+ecollider that pro duced 2 times more Z bosons than LEP I. There are plans for a future collider whose performance would easily exceed this [lo]
Even though we have done some work on building a flavor symmetric model in this thesis, the model is not yet complete. The next step in building the model would be checking to make sure that the fermionic partners of the scalar Higgs fields have the expected properties. We have only investigated the properties of the scalar components of the Higgs and the Sfields and shown that they lie well within the experimental range. In our model, the effective pterm was also forced to be around 100 GeV, which should be sufficient to give high enough masses to the fermionic Higgs fields. However, a detailed analysis is necessary to rigorously prove that this is the case.
Moreover, other (possibly more complex) flavor symmetric models can be studied. We only considered the model with a 3, a 6 and an 8 as our Higgs flavor multiplets. Certainly, other models with different Higgs sector structure exist. It is possible that these models do not have the Yukawa suppressions we had in our model and fit the experimental data without substantial finetuning (see [12] for an example with a 15 as the flavor breaking sector).
he signature arises due to the mixing of the Higgs scalars with neutral fields required for giving masses to the fermionic components of the Higgs fields.
Lastly, other flavor symmetries such as discrete or SU(2) symmetries can be studied. The model we developed was unable to predict any fermion masses. Instead, we had to fit the experimental data into our model using the model's input parameters. These new symmetries may reduce the number of required input parameters and also lead to predictions about the fermion masses.
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