Absolutely Stable Proton and Lowering the Gauge Unification Scale
aa r X i v : . [ h e p - ph ] A p r Absolutely Stable Proton and Lowering the Gauge Unification Scale
S.M. Barr
Department of Physics and AstronomyBartol Research Institute,University of Delaware, Newark, Delaware 19716, USA
X. Calmet
Physics & Astronomy,University of Sussex, Falmer, Brighton, BN1 9QH, UK
July 20, 2018
Abstract
A unified model is constructed, based on flipped SU (5) in which the proton is absolutely stable.The model requires the existence of new leptons with masses of order the weak scale. The possibilitythat the unification scale could be extremely low is discussed. It is an interesting question how low the unification scale can be. The most obvious issue is the protonlifetime. This is not necessarily a severe constraint on models of quark-lepton unification based on thePati-Salam group [1], since the gauge interactions in such models do not violate baryon number. Andrecently the possibility of a very low scale for Pati-Salam unification of quarks and leptons has beendiscussed [2].Proton decay is a much more serious constraint on models with gauge unification, by which wemean the unification of SU (3) of color and SU (2) of weak isospin in a single simple gauge group.(Gauge unification includes both “grand” unification and “flipped” unification [3].) A very low gaugeunification scale would be conceivable, however, if somehow the proton could be rendered absolutelystable. Of course, the lower the unification scale the less room for the Standard Model gauge couplingsto run. Nevertheless, if there were significant corrections to the gauge couplings coming from newphysics at or just above the unification scale, gauge coupling unification might nevertheless occur witha relatively low unification scale, perhaps even quite near the weak scale.In section 2 of the paper, we construct a simple model based on flipped SU (5) [3] in which theproton is absolutely stable due to an exact local symmetry. We prove the absolute stability of theproton in the model in section 3. In section 4, we show that the symmetry that forbids proton decayin the model is anomaly-free and can therefore be gauged, thus ensuring that quantum gravity effectsrespect it. We also show that the model can be embedded in SU (6) × SU (2).In section 5, we discuss the lepton sector, which has a rich low-energy phenomenology. The modelrequires the existence of several extra lepton doublets whose masses come from the Standard Model1iggs field and which therefore should be accessible to accelerator searches. Such extra lepton doubletswould have a large effect on the amplitude for Higgs decay to two photons. We show how agreementwith experiment is nonetheless possible. (The H → γγ decay rate can also be smaller or larger thanthe Standard Model prediction, depending on the particle content of the model.) In section 5, we alsoshow that small Majorana masses for the neutrinos can arise in a simple way despite the absence ofsuperheavy right-handed neutrinos in the model.In section 6, we briefly discuss the unification of gauge couplings, and consider the radical possibilitythat the unification scale could be near the weak scale, and even more radically that the gravity scalecould be near the weak scale as well, eliminating all large mass hierarchies. Proton decay can be suppressed if some of the baryon-number-violating gauge and Higgs couplingsto quarks and leptons have the effect of converting u or d quarks into fermions that are too heavy toappear in the final state of proton decay. One could call this “kinematic blocking” of proton decay.The most obvious way this might happen is that the unified interactions convert fermions of thelighter families into those of the heavier families, for example a d quark into a τ lepton. If the unifiedgroup is SU (5) and only the Standard Model quarks and leptons exist, then proton decay cannot becompletely suppressed by kinematic blocking even at tree-level, as it was shown already in [4]. Onthe other hand, it was shown in [5] that it can be completely suppressed at tree-level if the group isflipped SU (5), but only by tuning certain fermion mixing angle within SU (5) multiplets.If non-Standard Model quarks and leptons are introduced, more possibilities exist for kinematicblocking of proton decay. Already in 1980 an SU (5) model in which proton decay was kinematicallyblocked by heavy non-Standard Model fermions was proposed in [6]. That model, however, is no longerviable, as the new quarks and leptons it proposed should have been seen by experiments. In anotherinteresting paper [7], it was shown that proton decay can be forbidden to all orders in perturbationtheory in SU (5) by introducing into each family extra vectorlike fermions in + + 2 × ( + ).However, proton stability was due to global symmetries in that model, and thus not immune togravitational and other non-perturbative effects. Moreover, the pattern of vacuum expectation valuesassumed in that model cannot be exact without fine-tuning. (Another interesting model with kinematicblocking of proton decay in a supersymmetric SU (5) model with extra dimensions is [8].)Recently, in [9], it was shown that in flipped SU (5) with new non-Standard Model fermions com-plete kinematic blocking of proton decay can be achieved at tree level in a simple way without fine-tuning tuning of parameters. But even in that scheme, the proton decay was not absolutely forbidden,since loop-diagrams and non-perturbative effects could induce it.The model we present here, which like that of [9] is based on flipped SU (5), has an absolutelystable proton. Flipped SU (5) seems to be the smallest unification group that allows proton decay tobe completely forbidden by kinematic blocking in a realistic model without any fine tuning of mixingangles. In this model, as in [9], the new types of heavy fermions that block proton decay are quarksand leptons that are vectorlike under the Standard Model group. These vectorlike fermions will beassumed to have masses of at least hundreds of GeV. We will denote these new heavy fermions bycapital letters, and the known Standard Model fermions by lower-case letters.The mechanism for blocking proton decay does not depend on the number of families, so we willuppress family indices throughout this paper for notational simplicity; and the discussion will proceedjust as though there were only one family. It will be obvious that this does not matter.The gauge group of “flipped SU (5)” is SU (5) × U (1) X , with the weak hypercharge Y / U (1) X generator X and the SU (5) generator Y / ≡ diag ( , , − , − , − ), specifically Y / ( − Y / X ),with X normalized as in Table I,which gives the quark and lepton content of the model. Table I.
The fermion content of one family of the model. The left column gives the SU (5) × U (1) X quantum numbers. The other column shows how the different species of quark and lepton are containedin the SU (5) multiplets.10 (1) ψ αβ = " ψ , ψ a ψ a ! , ψ ab = " ν c , ud ! , D c ( − ψ α = " ψ ψ ! , ψ a = " νℓ − ! , u c (5) ψ = ψ = ℓ + ( − χ α = " χ χ ! , χ a = " N ′ L ′− ! , D (2) χ α = " χ χ ! , χ a = " N ′′ L ′′ + ! , d c ′ (5) σ ′ = L ′ + ′ (0) τ ′ = N ′ ′′ ( − σ ′′ = L ′′− ′′ (0) τ ′′ = N ′′ The left column in Table I gives the SU (5) × U (1) X representation, where the superscript is the valueof X . In the rest of the table, the Greek indices run from 1 to 5 and are the fundamental indicesof SU (5). The index values 1,2 correspond to the weak hypercharge group SU (2) L , while the values3,4,5 correspond to the QCD color group SU (3) c and are denoted by lower-case Latin letters, a, b, etc .(Note, however, that the superscript c when appearing alone denotes an antiparticle, for example d c denotes the anti-down quark.)One sees from Table I that a single family of fermions in this model consists of the usual 10 (1) +5 ( − + 1 (5) family of flipped SU (5) (denoted by the letter ψ ), plus a vectorlike pair 5 ( − + 5 (2) (denoted by χ ), and a vectorlike set of SU (5)-singlet fermions, denoted by σ and τ . This may looklike a somewhat arbitrary assortment, but, as will be seen in section 4, they fit exactly into a smallset of representations of SU (6) × SU (2), namely (15 ,
1) + (6 ,
2) + (1 , SU (6) × SU (2) would consist of (15 ,
1) + (6 , SU (5) family shown in the first three lines of TableI: it contains all the fermions of a Standard Model family with the exception of d c . Where the d c would normally be, one finds the new heavy field D c . And, conversely, d c occupies the place whereone would expect to find D c in the extra vectorlike pair 5 ( − + 5 (2) . This substitution is essentiallywhat leads to the kinematic blocking of proton decay. How it happens will be seen shortly.This explains the need for the extra vectorlike 5 ( − + 5 (2) of each family. What, however, explainshe need for the extra SU (5)-singlet fermions (the ones denoted by σ and τ )? The answer is thatthey are there to “mate” with the leptons in 5 ( − + 5 (2) to give them Dirac masses. One might askwhether these leptons could acquire mass more simply through a mass term M (2) ( − = M χ α χ α .Indeed, they could; but such a term would necessarily include the term M d c D , which would causemixing between d c and D c . That would mean that what we call D c in Table I would, through mixing,actually be partly the light Standard Model field. This would cause the kinematic blocking of protondecay to be imperfect; and proton decay would only be suppressed by a mixing angle. This is whathappens in the scheme proposed in [9]. This would not suppress proton decay sufficiently if theunification scale were very low. This is the reason why this scheme requires the existence of extraleptons whose masses come from electroweak breaking. These leptons have many phenomenologicalconsequences, as will be discussed in section 5. On the other hand, note that the extra quarks, D and D c , acquire mass from a Higgs field that is an electroweak singlet, i.e. Ω , and do not couple tothe Standard Model Higgs doublet. This difference between the extra quarks and extra leptons is apeculiar feature of the kind of model we are discussing.Breaking the gauge symmetries and giving quarks and leptons masses can be done with just thethree types of Higgs fields shown in Table II. Table II.
The Higgs field content of the model.5 ( − H H α = " H H ! , H a = " H H − ! , H − / (3) H ˜ H α = " ˜ H ˜ H ! , ˜ H a = " ˜ H + ˜ H ! , ˜ H / (1) H Ω αβ = " Ω , Ω a Ω a ! , Ω ab = " Ω , Ω / Ω − / ! , Ω − / As is usual in flipped SU (5) models, the breaking of SU (5) × U (1) X down to the Standard Modelgroup is done by a Higgs field that transforms as 10 (1) , which we denote Ω αβ . The component Ω has Y / X = 1 and so has Y / SU (3) c × SU (2) L ,so that its vacuum expectation value (VEV) leaves the Standard Model group unbroken. This VEVis of order the unification scale. The Higgs fields that do the electroweak breaking are the SU (2) L doublets in the 5 ( − H and 5 (3) H , which we denote respectively by H and ˜ H .There are eight Yukawa terms that are needed to give quarks and leptons mass, which are listedin the first seven rows of Table III, where we write these terms using the various alternative notationsgiven in Table I. able III The Yukawa terms needed to give mass to the fermions. The ninth row is a term neededin the Higgs potential to align vacuum expectation values. dd c h ˜ H ∗ i = ψ a χ a h ˜ H ∗ i ⊂ (1) (2) h (5 (3) H ) ∗ i ( uu c , ν c ν ) h H ∗ i = ( ψ a ψ a , ψ ψ ) h H ∗ i ⊂ (1) ( − h (5 ( − H ) ∗ i ℓℓ c h H i = ψ ψ h H i ⊂ ( − (5) h ( − H i D c D h Ω i = ψ ab χ d h Ω i ǫ abd ⊂ (1) ( − h (1) H i L ′− L ′ + h ˜ H ∗ i = χ ψ h ˜ H ∗ i ⊂ ( − ′ (5) h (5 (3) H ) ∗ i N ′ N ′ h H ∗ i = χ τ ′ h H ∗ i ⊂ ( − ′ (0) h (5 ( − H ) ∗ i L ′′− L ′′ + h ˜ H i = σ ′′ χ h ˜ H i ⊂ ′′ ( − (2) h (3) H i N ′′ N ′′ h H i = τ ′′ χ h H i ⊂ ′′ (0) (2) h ( − H i Ω ˜ H ∗ H ∗ = Ω ˜ H ∗ H ∗ ⊂ (1) H (5 (3) H ) ∗ (5 ( − H ) ∗ νN ′′ h ˜ H i = ψ τ ′′ h ˜ H i ⊂ ( − ′′ (0) h (3) H i The ninth row of Table III gives a cubic term in the Higgs potential that is needed to tie the variousHiggs fields together, thereby aligning their VEVs and avoiding accidental global symmetries in theHiggs potential that would lead to goldstone bosons.The terms in the first nine rows of Table III are needed in the model. There are also terms thatmust be forbidden if proton decay is to be suppressed. We already mentioned one such term, namelyan explicit mass term of the form 5 (2) ( − , which would mix d c and D c . Let us suppose for a momentthat the terms in the first nine rows of Table III are the only non-trivial terms in the Yukawa sectorand Higgs potential. (We count as “trivial” any terms that always must be present no matter whatthe symmetries of the model are, such as the absolute square of any Higgs field.) With only thosenine terms, the model is easily found to have a U (1) × U (1) accidental symmetry, which we will call U (1) a × U (1) b .In Table IV we give the a and b charges for all the fermion and Higgs multiplets listed in TablesI and II. These charge assignments may look somewhat random, but it will be seen later that theyhave simple group theory interpretations if the flipped SU (5) group is embedded in SU (6) × SU (2).Moreover, note that the b values have a simple pattern: fermions that are odd-rank SU (5) tensorshave b = −
1, those that are even-rank tensors have b = +1, and Higgs bosons have b = 0. Table IV
The U (1) a and U (1) B charges of the fields of the model.field 10 (1) ( − (5) ( − (2) ′ (5) ′ (0) ′′ ( − ′′ (0) ( − H (3) H (1) H a − − − − b − − − U (1) a × U (1) b symmetry allows one more Yukawa interaction, 5 ( − ′′ (0) h (3) H i , which is shown inthe last row of Table III. This term, which couples ν to N ′′ , has the effect of mixing of ν and N ′′ .As we shall see, this is harmless and does not destabilize the proton. We will now prove that thesymmetry U (1) a forbids all operators of any dimension that would give proton decay. Proof of proton stability and gauging “baryon number”
Any effective operator that leads to proton decay must involve only quark and lepton fields that arelighter than the proton, and thus not the new vectorlike fields denoted by capital letters in Table I.Consequently, it can be written in the general form( u, d ) m ( u c ) n ( d c ) p ( ν, ℓ − ) q ( ℓ + ) r ( N ′′ ) s ( h H i ) t ( h ˜ H i ) u ( h Ω i ) v ⊂ (10 (1) ) m (5 ( − ) n (5 (2) ) p (5 ( − ) q (1 (5) ) r (5 (2) ) s (5 ( − H ) t (5 (3) H ) u (10 (1) H ) v , (1)where the exponents m, n, p, q, r, s, t, u, v are integers. Note that here d stands for either d or s , and e stands for either e or µ , since we are not showing family indices. Also note that we have included N ′′ in this product. The reason is that this is not a purely heavy field, since (as we noted previously)there is mixing between the fields that are called ν and N ′′ in Table I, due to the last term in TableIV. If U (1) a is not explicitly broken, then the value of the generator a of the operator in Eq. (1) mustvanish, giving − n + p − q + 2 r + s − t + u = 0 . (2)By conservation of weak hypercharge, the value of Y / m − n + 13 p − q + r + 12 s − t + 12 u = 0 . (3)Multiplying Eq. (3) by 2 and subtracting Eq. (1) gives13 ( m − n − p ) = B = 0 , where B is baryon number. So no baryon-number-violating operators involving only quarks andleptons lighter than the proton exist to any order.Note that the above arithmetic shows that the linear combination of generators˜ B ≡ (cid:18) Y (cid:19) − a (4)is the same as baryon number for the Standard Model fermions. (On the heavy new fermions, however,˜ B has peculiar values: the heavy quarks D have ˜ B = , and the heavy leptons N ′ and L ′ − have˜ B = −
1, with the corresponding antiparticles having the opposite values.) As we shall see in the nextsection, U (1) a is anomaly free, and must be gauged in order to protect the stability of the proton fromquantum gravity effects. Thus, in this model we are gauging a quantum number that coincides withbaryon number on the Standard Model fields. The gauging of baryon number has been discussed inother recent papers [10]. Anomaly-freedom, and possible embedding in SU (6) × SU (2) The proof of proton stability just given assumed that the symmetry U (1) a is not explicitly broken. If itis a global symmetry, however, one would expect gravitational effects to break it explicitly. Therefore,to render the proton absolutely stable, it is necessary to gauge U (1) a , which would require that U (1) a be anomaly-free. And indeed, it turns out that it is. With the set of fermions given in Table I, and the U (1) a charges given in Table IV, both the anomalies of U (1) a alone and its mixed anomalies involving SU (5) × U (1) X vanish. That is, the following five conditions are satisfied: T r ( a ) = 0, T r ( a ) = 0, T r ( a X ) = 0, T r ( aX ) = 0, and T r ( a ( λ ) ) = 0 (where λ is an SU (5) generator), as can easily bechecked.The satisfying of all these conditions seems like an amazing coincidence, but actually it has a simpleexplanation based on the group SU (6) × SU (2). The explanation consists in these three facts: (1) thegroup SU (6) × SU (2) contains SU (5) × U (1) X × U (1) a ; (2) the set of SU (6) × SU (2) representations(15 ,
1) + (6 ,
2) + (1 ,
3) is anomaly-free; and (3) when decomposed under SU (5) × U (1) X × U (1) a thisset of representations contains exactly the set of fermions of one family of our model. We will nowdemonstrate each of these points.That the set of SU (6) × SU (2) representations (15 ,
1) + (6 ,
2) is anomaly-free is well-known andfollows simply from the possibility of embedding in E . That (1 ,
3) is anomaly-free follows simply fromthe fact that it is a real representation. Therefore, obviously, the combined set (15 ,
1) + (6 ,
2) + (1 ,
3) isanomaly-free under SU (6) × SU (2). Moreover, SU (6) × SU (2) obviously contains the subgroup SU (5) × U (1) × U (1) , where U (1) is the subgroup of SU (6) corresponding to the diagonal generator T ≡ diag ( , , , , , − ), and U (1) is the subgroup of SU (2) corresponding to the diagonal generator T ≡ diag ( , − ). If one defines X ≡ T + 5 T , then the (15 ,
1) + (6 ,
2) + (1 ,
3) is easily seen todecompose under SU (5) × U (1) X into exactly the set of fermions in Table I. And if one identifies a with 2 T , one immediately finds that those fermions have exactly the values of a given in Table IV.It should be mentioned that if the model is embedded in SU (6) × SU (2), then all the Yukawacouplings shown in Table III can arise from just a few types of terms, namely terms of the form(15 , , h (15 , H i , (15 , , h (6 , H i , and (6 , H (1 , h (6 , H i ∗ .The symmetry U (1) b is not contained in SU (6) × SU (2). yet it turns out, quite remarkably, thatall its anomalies vanish over the set of fermions shown in Table I. This includes both the anomaliesof U (1) b alone and its mixed anomalies with SU (5) × U (1) X × U (1) a . This involves altogether eight trace conditions: T r ( b ) = 0, T r ( b ) = 0, T r ( b X ) = 0, T r ( bX ) = 0, T r ( b a ) = 0, T r ( ba ) = 0, T r ( baX ) = 0, and T r ( b ( λ ) ) = 0. Because of this it is possible to gauge the full group SU (5) × U (1) X × U (1) a × U (1) b . However, it is not necessary to gauge U (1) b to insure proton stability. If U (1) b is not gauged, and therefore presumably broken explicitly by gravity effects, then several moreYukawa terms would be allowed besides those shown in Table III. (In particular, it would allow thecoupling 5 ( − (2) h (1) H i .) Those additional terms would cause Standard Model leptons to mix withheavy, vectorlike leptons, but it can easily be shown that it would not cause protons to decay.Though it is not important for proton stability that U (1) b be anomaly free, it is quite interestingthat it is, since it involves eight independent non-trivial conditions, as we saw. The question iswhether there is also some underlying group-theoretical explanation for these cancellations based onembedding in SU (6) × SU (2), as there was for the anomaly-freedom of U (1) a . It happens there is apartial explanation, as we will now see.Consider the group SU (6) × SU (2) × U (1) T , where (15 ,
1) has T = 1, (6 ,
2) has T = −
2, and (1 , T = 3. One can easily easily check that U (1) T is anomaly-free. There are four conditions: T r ( T ) = 15 · (1) + 12 · ( − + 3 · (3) = 15 −
96 + 81 = 0 ,T r ( T ) = 15 · (1) + 12 · ( −
2) + 3 · (3) = 15 −
24 + 9 = 0 ,T r (( T ) T ) = 1 · · (1) + 2 · · ( −
2) = 0 ,T r (( T ) T ) = 6 · · ( −
2) + 1 · · (3) = 0 . (5)Though this is a surprising coincidence, it is far less surprising than the satisfying of eight anomaly-cancellation conditions for U (1) b at the SU (5) × U (1) X level. If one now defines b ≡ T + T , onediscovers that the representations in Table I have exactly the b values given in Table IV. All of this isdisplayed in Table V. Table V
How the generators of U (1) X , U (1) a , and U (1) b are related to those of SU (6) × SU (2) × U (1) T X = a = b =[ SU (6) × SU (2)] T SU (5) T T T T + 5 T T T + T (15 ,
10 1 0 1 1 0 1” 5 − − − , − −
12 12 − −
1” 1
52 12 − − − − − − −
1” 1 − − − , − − − U (1) a must be local to prevent gravity-induced proton decay. If U (1) a is gauged, however, its gauge boson could create difficulties. For if U (1) a is not spontaneously brokenthere is a new long-range force, while if it is spontaneously broken the proof of proton stability couldbe invalidated, since it depended on conservation of a .First, consider the case that SU (5) × U (1) X × U (1) a is not embedded in SU (6) × SU (2). Then onecan simply introduce a scalar field η which is neutral under SU (5) × U (1) X but has charge a ( η ) = 0under U (1) a . This field can obtain a vacuum expectation value that makes the mass of the U (1) a gauge boson large enough to avoid conflict with experiment. (There is also nothing to prevent thegauge coupling of U (1) a being small.) The existence of such a field would modify Eqs. (1) and (2).One must put into the operator of Eq. (1) a factor ( h η i ) w , where w is some integer. Then an additionalterm wa ( η ) would appear on the left-hand side of Eq. (2). This changes Eq. (4) to B = wa ( η ). Aslong as a ( η ) is not of the form 1 /w for some integral value of w , proton decay cannot happen.Another possibility is that U (1) a is not spontaneously broken, but has such a tiny gauge couplingconstant that the resulting long-range force has not been seen. This seems highly implausible, but iscertainly possible.If SU (5) × U (1) X × U (1) a is embedded in SU (6) × SU (2), then the possibilities are more limited.The gauge coupling of U (1) a cannot then be arbitrarily small, and the possible values of a ( η ) arerestricted. Moreover, if h η i is large compared to the electroweak scale, one requires that it break U (1) a without breaking the electroweak gauge group. The smallest SU (6) × SU (2) multiplet that hasa component that can do this is (6 , η , (where the 6 is the SU (6) index and the 2 is the SU (2) index) has Y / I L = 0, and a = 1. But this value of a ( η ) allows the proton to decay.The smallest multiplet that can break U (1) a above the electroweak without allowing proton decay is (21 ,
3) of SU (6) × SU (2). This has a component that has I L = Y / a = 2. This allowsoperators that give ∆ B = ±
2, and thus possibly neutron-antineutron oscillations, but not protondecay.
The model presented above has six doublets of extra leptons (two for each family). This raisesseveral possible phenomenological problems, including consistency with the measured value of the ρ parameter, the rates for H → γγ and H → Z γ , and the stability of the Higgs potential. We shalldiscuss these in turn.The effect of the new fermions on the ρ parameter can be made small if the extra lepton doubletsare not “split”, i.e. if the neutral and charged components have the same or nearly the same mass.This seems somewhat artificial, but perhaps could be enforced by some symmetry, though we havenot investigated this possibility.The extra lepton doublets will definitely contribute very significantly to the amplitude for H → γγ .This process comes, as is well known, from one-loop triangle graphs, where in the Standard Model theamplitude is dominated by the W boson loop and t quark loop [11, 12]. In the model presented here,one must include the diagrams with the extra charged leptons running around the loop. The matrixelement squared for H → γγ is given by | M | = g m H π m W (cid:12)(cid:12)P i αN c e i F i (cid:12)(cid:12) , where i stands for the typeof particle in the loop, and F i is given (for i being a gauge boson, fermion, or scalar, respectively) by F gauge = 2 + 3 τ + 3 τ (2 − τ ) f ( τ ), F fermion = − τ (1 + [1 − τ ] f ( τ )), F scalar = τ (1 − τ f ( τ )), for τ > τ ≡ (2 m i /m H ) , and f ( τ ) = (sin − p /τ ) . For the Standard Model contributions, one has F SM ∼ = F W + F t ∼ = +8 . − . .
6. Since the six new charged leptons in our model must be heavyenough not to have been seen, τ for them is large and F L ± ∼ = 6( − / ∼ = −
8. This is larger than theStandard Model contribution and of opposite sign. If there are additional fermions that contribute to H → γγ , the total amplitude can be close to − H → Z γ [13, 12] is not a difficulty for the model. The present limits on this decay arevery loose, and the contribution of charged leptons to it are highly suppressed, since the Z coupling tothe charged leptons is proportional to the well-known factor I L − Q sin θ W = − + 2(0 . ∼ = − . O (1) Yukawa couplings will give large radiative contribu-tions to the Higgs quartic self-coupling. However, we envision the unification scale being much lowerthan it is in typical unified models, and the unified theory may be effective theory valid only belowsome cutoff. If that cutoff scale is relatively low, the Higgs quartic coupling can remain positive belowthat scale.We conclude that the existence of the extra leptons is compatible with present limits. One mightworry, on the other hand, that the new quarks D = D c would present phenomenological problems,for instance by substantially affecting the H → D + D c quarks (unlike the extra leptons) do not couple to the Standard ModelHiggs doublet, but get their mass from a Standard Model singlet Higgs field (Ω ). Thus they donot contribute to the Higgs decay amplitudes, the ρ parameter, or the running of the Higgs quarticcoupling. Moreover, their mass could be much higher than the weak scale.eturning to the leptons, there remains the question of neutrino mass. Realistic neutrino massesseem at first sight to be a problem for the model. The first question is how those masses can befractions of an eV, since this would not emerge from the usual see-saw mechanisms if the unificationscale is very low. The second question is how to avoid neutrino masses that are of the same order asthe quark and charged lepton masses.If one looks at the Yukawa couplings allowed by SU (5) × U (1) X × U (1) a × U (1) b , all of which areshown in Table III, one finds several mass terms for neutral fermions. Specifically, the second, eighth,and tenth lines of Table III have operators that give, respectively, operators of the form ν c ν h H ∗ i , N ′′ N ′′ h H i , and νN ′′ h ˜ H i . Ignoring family indices, this gives a mass matrix of the following form: (cid:16) ν, ν c , N ′′ , N ′′ (cid:17) h H ∗ i h ˜ H i h H ∗ i h ˜ H i h H i h H i νν c N ′′ N ′′ . (6)This matrix has non-zero determinant, and the VEVs that appear in it are of order the electroweakscale. Thus, one would not expect neutrino masses of order a fraction of an eV unless some Yukawacouplings were extremely small. For example, if the Yukawa coupling in the terms ν c ν h H ∗ i , and N ′′ N ′′ h H i were of order ǫ and those in νN ′′ h ˜ H i were of order 1, then there would be (for each family)one pseudo-Dirac neutrino with mass of order the electroweak scale (composed approximately of ν and N ′′ ) and one pseudo-Dirac neutrino with mass of order ǫ times the electroweak scale (composedapproximately of ν c and N ′′ ), as can be seen from the form of Eq. (5). One would therefore need tohave ǫ be of order 10 − to 10 − even for the third family. This seems contrived.A more attractive possibility arises if there is an additional type of neutral fermion introducedfor each family. Let is call it S and say that it is neutral under SU (5) × U (1) X × U (1) a , but has b = −
1. Then one can have a coupling of the type ψ S h (Ω ) ∗ i = ν c S h (Ω ) ∗ i , which is contained in10 (1) (0) h (10 (1) H ) ∗ i . This term is invariant under SU (5) × U (1) X × U (1) a × U (1) b . The mass matrixthen has the form (cid:16) S, ν, ν c , N ′′ , N ′′ (cid:17) h Ω i h H ∗ i h ˜ H i h Ω i h H ∗ i h ˜ H i h H i h H i Sνν c N ′′ N ′′ . (7)This matrix has one zero eigenvalue. So (for each family) there is a massless neutral fermion. Thesecan be given tiny masses in various ways, one of which we will describe shortly. These light neutrinosare linear combinations of S , ν and N ′′ , as can be seen from Eq. (6). Of course, to be consistent withbounds on lepton universality, these linear combinations would have to be mostly ν ; but this onlyrequires certain ratios of Yukawa couplings in Eq. (6) to be of order 10 − .One can give a small mass to the neutrinos by a higher-dimension operator of the form S S h ζ i n ,where ζ is a 1 (0) of SU (5) × U (1) X and has a = 0 and b = 1 /n . Such an operator might be inducedby gravity. The value of n needed to get realistic neutrino masses would depend on the gravity scale. Low scale grand unification?
If a unified model has an absolutely stable proton due to an exact symmetry, then obviously protonlifetime limits would not constrain the unification scale at all. The question would then arise how lowthe unification scale could be. Could it be near the electroweak scale? If the unification scale is low,one has to explain how the gauge couplings are able to unify. One possibility is to exploit an idea firstproposed by Shafi and Wetterich many years ago [14] (see also [15, 16]).Let us consider an effective operator for physics below some cutoff scale Λ given by c Λ Tr ( G µν G µν A ) , (8)where G µν is the Grand Unified Theory field strength and A is a scalar multiplet in the adjointrepresentation of SU (5). The scale Λ is kept as a free parameter for the time being. Upon sym-metry breaking at the unification scale M U , the Higgs field gets a vacuum expectation value hAi = M U (2 , , , − , − / √ πα G , where α G is the value of the SU (5) gauge coupling at M U .The dimension 5 operator modifies the gauge kinetic terms of SU (3) × SU (2) × U (1) below thescale M u to −
14 (1 + ǫ ) F µν F µν U(1) −
12 (1 + ǫ ) Tr (cid:16) F µν F µν SU(2) (cid:17) −
12 (1 + ǫ ) Tr (cid:16) F µν F µν SU(3) (cid:17) (9)with ǫ = ǫ − ǫ √ √ π c √ α G M u Λ . (10)If we were to take Λ = M u , then ǫ = ǫ − ǫ √ √ π c √ α G . (11)We can now perform a finite field redefinition A iµ → (1 + ǫ i ) / A iµ to canonically normalize thekinetic terms of the gauge bosons. Then the corresponding redefined coupling constants are g i → (1 + ǫ i ) − / g i . We get the unification condition: α G = (1 + ǫ ) α ( M u ) = (1 + ǫ ) α ( M u ) = (1 + ǫ ) α ( M u ) . (12)We now wish to consider low scale unification. Direct observational bounds on the heavy gaugebosons of SU (5) /G SM as well as on the color octet scalars lead us to consider a unification scale inthe few TeV region. We thus take M U a few TeV which implies that there is very little runningfor the gauge couplings and we can use the LEP values, at least to first approximation. We take α ( M Z ) = 0 . α ( M Z ) = 0 . α is a free parameter, we will use c to obtain thenumerical unification of α and α . We need c = 5 r π (cid:18) α − α α + 2 α (cid:19) √ α G . (13)We then find ǫ = 5 (cid:18) α − α α + 2 α (cid:19) α G (14)nd thus α = α G (cid:16) α − α α +2 α (cid:17) α G (15)Numerically we have c = − . / √ α G . If I take α G = 0 .
05 for illustration, We get c = −
7. In asense we see that if the grand unified theory is strongly coupled, the Wilsonian expansion works bestas the Wilson coefficients get smaller: for α G = 1, we get c = − . U (1) of hypercharge is not purely a subgroup of SU (5), but lies partly in the U (1) X , whosegauge coupling is an independent, free parameter. This coupling can be chosen to give the observedvalue of the hypercharge (and electromagnetic) gauge coupling.Interestingly, the Planck scale could also be lowered to the TeV region to remove all hierarchies.There are two known mechanisms for that. One is to assume that large extra dimensions open up atan energy scale of a few TeV [17, 18]. The other one relies on a large hidden sector of particles whichlead to a running of the Planck mass [19]. Note that the running can also be obtained by a scalarfield with a large non-minimal coupling to the Ricci scalar [20]. Acknowledgements
We thank K.S. Babu, Ilia Gogoladze, and Qaisar Shafi for useful conversations. This work is supportedin part by DOE grant No. DE-FG02-12ER41808.
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