Anomalous Flavor U(1) X for Everything

Transcription

1 Anomalous Flavor U(1) X for Everything Herbi K. Dreiner, 1 Hitoshi Murayama, Marc Thormeier,3 1 Physikalisches Institut der Universität Bonn, Nußallee 1, Bonn, Germany Institute for Advanced Studies, Einstein Dr., Princeton, NJ 0850, USA 3 Lawrence Berkeley Laboratory, University of California, Berkeley, CA 970, USA Abstract We present an ambitious model of flavor based on an anomalous U(1) X gauge symmetry with one flavon, only two right-handed neutrinos and only two mass scales: M grav and m 3/. In particular, there are no new scales introduced for right-handed neutrino masses. The X-charges of the matter fields are such that R-parity is conserved automatically, the higher dimensional operators are sufficiently suppressed to guarantee a proton lifetime in agreement with experiment, and the phenomenology is viable for quarks, charged leptons, as well as neutrinos. In our model one of the three light neutrinos is massless. The price we have to pay for this very successful model are highly-fractional X-charges which can likely be improved with less restrictive phenomenological ansätze for mass matrices. dreiner@th.physik.uni-bonn.de On leave of absence from Department of Physics, University of California, Berkeley, CA 970, USA. murayama@ias.edu thor@ias.edu 1

2 1 Introduction The fermionic mass spectrum of the standard model (SM) suggests that the entries of the quark and charged lepton Yukawa matrices exhibit hierarchical patterns. Froggatt and Nielsen (FN) in Ref. [?] presented an idea to explain these texture structures and traced them back to a flavor U(1) X symmetry beyond the SM, broken by an SM-singlet, called the flavon. For early models see e.g. Refs. [?,?]. Then string theory gave a theoretical motivation for the existence and the breaking scale of the U(1) X (in particular, it turned out that anomalousness of the U(1) X is a blessing), so the FN scenario was naturally embedded, see e.g. Ref. [?]. It is natural to assume that the origin of the aforementioned hierarchy also leaves its fingerprints on the other (Yukawa) coupling constants, which opens up many more applications. It is tempting to use the idea of FN for investigating why R-parity violating Yukawa coupling constants have not lead yet to the observation of exotic processes, see e.g. Ref. [?]. 1 Or to provide an explanation (see e.g. Ref. [?]) why (with or without grand unification) higher-dimensional genuinely supersymmetric operators like QQQL do not cause a short-lived proton; after all, these operators are suppressed by just a single power of the reduced Planck scale, so the dimensionless coefficient must be aedequately tiny. Furthermore, the idea of FN can easily be combined (see Ref. [?]) with the Giudice-Masiero (GM) mechanism (see Refs. [?,?]) to have a natural explanation for the µ-term of the minimal supersymmetric extension of the SM (MSSM). Last, but not least, neutrino data can also be interpreted in the light of the FN idea, see e.g. Ref. [?]. However, when dealing with neutrinos the question is posed what distinguishes the neutrino from the flavon. The aim of this note is to dovetail all of these different aspects of the FN scenario and at the same time be curmudgeonly about letting string theory introduce other beyond-mssm symmetries or particles. So our intention is to construct a minimalistic supersymmetric FN model with the following features: There is only one additional symmetry group beyond SU(3) C SU() W U(1) Y in the visible sector, namely a generation-dependent local U(1) X. For the cancellation of the U(1) X gauge anomalies we invoke the Green-Schwarz 1 Another interesting possibility is to suppress the R-parity violations down to phenomenologically acceptable levels using a flavor symmetry rather than forbidding them exactly [?]. However, it will render the Lightest Supersymmetric Particle not an attractive candidate for the Cold Dark Matter.

3 mechanism, see Ref. [?]. We employ the notation that a left-chiral superfield Φ carries the U(1) X charge X Φ. There is only one (left-chiral) flavon superfield A, which acquires a vacuum expectation value (VEV) A via the Dine-Seiberg-Wen-Witten mechanism, see Refs. [?,?,?,?], thus breaking U(1) X. In analogy to the three species (i.e. generations) of X- and SM-charged matter superfields {Q i, L i, U i, D i, E i } (i = 1,, 3), there are three species of superfields whose only gauge group is U(1) X : the flavon superfield and two right-handed neutrino superfields N I (I = 1, ). Thus the flavon superfield is so to speak a right-handed neutrino superfield without lepton number. The model produces a viable phenomenology: Quark masses and mixings and charged lepton masses agree with the data, see Refs. [?,?]; the neutrino masses and mixings are in accord with the recent measurements, see Ref. [?] and references therein; the lower bounds on proton longevity, see e.g. Ref. [?,?], and other rare processes are satisfied; broken R-parity (for a review see e.g. Ref. [?]) is forbidden by virtue of the X-charges. There are only two mass scales: the mass of the gravitino m 3/ 1 TeV (we assume gravity-mediation of supersymmetry breaking, see Refs. [?,?,?,?]) and the reduced Planck scale M grav GeV where gravity becomes strong. {Q i, L i, U i, D i, E i, H D, H U } are the only SM-charged superfields, and we made the (unsuccessful) attempt here that those fields together with {A, N I } possibly are the only U(1) X -charged superfields. Our paper is structured as follows: In Section we review the idea of Froggatt and Nielsen. In Section 3 we derive constraints on the X-charges such that conserved R-parity is guaranteed. In Section we review the conditions on the X-charges in order to have an anomaly-free theory. We are then able to finish the argument started in the previous Section. In Section 5 we first review the fermionic We simply assume that the supersymmetry breaking effects are sufficiently flavor blind, such as the dilaton-dominated breaking or the same modular weights to all three generations (see, e.g., Ref. [?,?]). 3

4 mass spectrum and its implications on the X-charges. We combine this with the results achieved before and arrive at Table??. Section 6 discusses how the flavon acquires a VEV (a key ingredient of the Froggatt-Nielsen scenario). We give a remark about tadpoles not endangering our model. Section 7 confronts the X-charges in Table?? with further constraints: there are only two mass scales in the game and attention must be paid to higher-dimensional operators which destabilize the proton. Section 8 is the heart of this paper, fixing the X-charge assignments by comparison with neutrino data. A preliminary result is Table 3, the main results are given in Tables??-??. Section?? finishes the paper. The Appendices??,??,?? complement Section 8: reviewing the seesaw mechanism, explaining how to extract masses from Froggatt-Nielsen textures and including supersymmetric zeros. The result of each section is summarized at the beginning. Hasty readers can skim through the paper by reading the beginning of each section together with tables. The Framework of Froggatt and Nielsen In this section, we review the framework to build models of flavor based on a U(1) X flavor symmetry, originally due to Froggatt and Nielsen (see Ref. [?]). For a review of the FN framework, possibly combined with the GM mechanism, see e.g. Ref. [?]. Here we shall give only a short sketch. Models of the same category as the one constructed in this text are found in Ref. [?]-Ref. [?]. Note that we do not introduce heavy vector-like fields unlike the original proposal (see Ref. [?]) but rather use a simple operator analysis. We also pay careful attention to supersymmetric zeros and how they are filled up by canonicalizing the Kähler potential. The idea of FN in its simplest form is as follows: One introduces the abovementioned U(1) X symmetry and the superfield A. Above the U(1) X breaking scale, e.g. the coupling constant G (U) ij in the MSSM superpotential Yukawa interaction term After A acquires a VEV one has, with

5 There are other contributions to Eqs. (??,??) produced by the breaking of supersymmetry from D-terms (GM mechanism) which are particularly important if the operators vanish due to supersymmetric zeros (e.g., X H D + X H U < 0 in the above example). Let us elaborate more on the last item because it is important. Assuming the gravity mediation of supersymmetry breaking, supersymmetry is broken by a leftchiral superfield Z in its F -component F Z m 3/ M grav. This allows us to write D-term operators of the form with a holormophic function F, [ d θ Z ( ) XF ( ) ] XF A A Θ[ X F ] Ω[ X F ] + Θ[X F ] Ω[X F ] F, (.1) M M M which give superpotential terms [ ] d θ m 3/ Ω[X F ] Θ[ X F ] ɛ X F + Θ[X F ] ɛ X F F. (.) Even if the U(1) X charge of the operator F is negative, the complex conjugate of ɛ would allow U(1) X invariants. They are very suppressed due to a factor of m 3/ /M grav, while they are relevant to the µ-parameter [?,?] or neutrino mass operators [?]. For example, one has 3 Ω [ X H D + X H U ] F Z M Ω [ X L i + X H U + X N j] g (µ) ɛ X H D +X H U H D H U, F Z M g(n) ij ɛ X Li+X HU +X N j L i H U N j. (.3) The total contribution from U(1) X breaking to the µ-term is then µ = M (µ) g (µ) ɛ X H D +X H U Θ [ ] [ ] X H D + X H U Ω XH D + X H U + m 3/ g(µ) ɛ X H D +X H U Ω [ X H D + X H U]. (.) It is most natural to have M (µ) = M grav as well, thus forcing one to have X H D +X H U to be ±1 (with ɛ 0., see below) or X H D +X H U = 1, < 0. In the latter case, the size of the µ-parameter is naturally of the same magnitude as the supersymmetry breaking effects, as desired phenomenologically. The supersymmetry-breaking 3 Using U(1) X gauge invariance, we can always take A real without a loss of generality. 5

6 contributions to the trilinear terms are usually safely neglected because of the strong suppression m 3/ /M grav, while they can be important in the case of neutrino masses. Since the Kähler potential from the outset does not have the canonical form, one must perform a transformation of the relevant superfields to the canonical basis, see Refs. [?,?,?,?,?] for more details. For example, for the quark doublets one obtains for the relevant Kähler potential term Q i H (Q) ij Q j = [ ] i [ ] j. DH (Q) U H (Q) Q δij DH (Q) U H (Q) Q (.5) D H (Q) is a diagonal matrix, its entries are the eigenvalues of the Hermitian matrix H (Q) ; the unitary matrix U H (Q) performs the diagonalization. We define the matrix C (Q) D H (Q) U H (Q). (.6) C (Q) is then absorbed into Q. This redefinition affects the superpotential, e.g. G (U) 1 H (H U ) C (Q) 1T G (U) C (U) 1. (.7) One has 3 Conserved R-Parity In this section, we show that it is possible to obtain conserved R-parity as an automatic consequence of the X-charge assignment. In general, it is desirable (if possible) to choose the X-charges such that superfieldoperators which give rise to exotic processes are either forbidden or strongly suppressed. For broken R-parity we shall follow the first path. In this and the next Section we shall for the purpose of generality we treat an arbitrary number of generations of {Q i, L i, U i, D i, E i } and {N I }, not restricting ourselves to i = 1,, 3 and I = 1,, Consider a general gauge invariant term of the R-parity violating MSSM with righanded neutrinos ( R p -MSSM +N), containing n Q 1 times the superfield Q 1, etc. The n... are non-negative integers if one deals with the superpotential, however they may be negative in case of the Kähler potential due to charge conjugation, e.g. the 6

7 term Q Q 1 has n Q = 1, n Q 1 = 1. The X-charge of this superfield-operator is, with X A = 1, X total = I (n N I X N I ) + i (n L i X L i + n E i X E i) + i (n Q i X Q i + n D i X D i + n U i X U i) + n H D X H D + n H U X H U. (3.1) The n... are not independent of each other due to SU(3) C SU() W U(1) Y invariance. They are subject to the conditions (with n Q 3 i=1 n Qi, etc.) gauge n Q n D n U = 3C, n H D + n H U + n Q + n L = W, Y H D n H D + Y H U n H U + Y Q n Q + Y D n D + Y U n U + Y L n L + Y E n E + Y N n N = 0. (3.) C is an integer, W is a non-negative integer, the Y... denote hypercharges. these three equations in terms of the numbers of quark superfields gives Solving n Q = W n H D n H U n L, n U = W C + n E n H D n L, n D = W C n E n H U. (3.3) We now state our central assumptions (using X A = 1): 1. All superfield-operators which conserve the ZZ -symmetry R p each have an overall integer X-charge. For scalar left-chiral superfields one has 5. All superfield-operators which do not conserve R p each have an overall fractional X-charge. To be more precise, all superfield-operators for which n Q n U n D + n L n N n E is odd each have an overall fractional X-charge. It follows that all R p superfield-operators are forbidden, R p is thus conserved exactly. Y Q = 1/3 Y L, Y U = /3 Y L, Y D = /3 Y L, Y E = Y L, Y H D = Y L, Y H U = Y L, Y N = 0. 5 Strictly speaking, R p we define here is a matter parity, because we are not flipping the sign of the fermionic coordinate θ. 7

8 One can draw several conclusions: The only texture zeros in the sense of Ref. [?] that one may have in e.g. G (U) are due to X Q i + X H U + X U j being negative. Since Q 1 has the same SM quantum numbers as Q etc., an R p -conserving superfield-operator Q 1 Φ 1 Φ...Φ n guarantees that Q Φ 1 Φ...Φ n is R p - conserving, too, etc. From Point 1. we thus find that it is necessary that X Q X Q 1 is integer, etc. Thus X Q = X Q 1 + Q 1, etc. (3.) }{{} integer For any SU(3) C SU() W U(1) Y invariant superfield-operator Φ 1 Φ...Φ n which violates R p, one has that Φ 1 Φ...Φ n Φ 1 Φ...Φ n conserves R p. From Point 1. we find that the X-charge of the latter operator, namely ( X Φ 1 + X Φ X Φ n), is integer. Point. demands that XΦ 1 + X Φ X Φ n is fractional. It follows that all R p superfield-operators have an overall half-odd-integer X-charge. 6 It follows immediately from the previous point that X N I is half-oddinteger; furthermore let L i Φ 1 Φ...Φ n be SU(3) C SU() W U(1) Y invariant and conserve R p, it follows that H D Φ 1 Φ...Φ n does not conserve R p, so that X L i X H D is half-odd-integer. So in summary For completeness sake it should be mentioned that the same reasoning to conserve R p as presented above can be applied to L p and B p instead (however, L p and B p are not free of discrete gauge anomalies and thus not viable, see Ref. [?,?]) but one cannot conserve any two of these three parities simultaneously this way. We find 7 6 This reasoning is not affected by e.g. on the one hand the superfield-operators L 1 L 1 E i being equal to zero due to SU() W gauge invariance, but on the other hand L 1 H D E i not vanishing. 7 Instead of Eq. (??) that X N 1, X L 1 X H D have to be half-odd-integer or integer (L p or B p, respectively). Furthermore we get n L n N n E = L + λ or n Q n U n D = B + β (L p or B p ) instead of Eq. (??). So X total integer has to be (3X Q 1 + X L 1 3 )C λ or (3X Q 1 + X L 1)C (L p or B p ; considering B p, one has that C = even B p, C = odd B p ). 8

9 3X Q 1 + X L 1 to be half-odd-integer, which unlike Eq. (??) cannot be combined with anomaly cancellation via the Green-Schwarz mechanism. Apart from R p, the only discrete anomaly-free gauge symmetry is B 3 in Ref. [?]. To have B 3 instead of R p is not very attractive in our case, as it allows (just like B p ) a tree-level tadpole term, namely the superpotential term which is linear in N. This requires the N to acquire VEVs, thus spoiling the idea that the flavon field alone breaks U(1) X. But is it possible to have B 3 together with R p by virtue of the X-charges? B 3 transformations act on superfields as {Q i, N I } {Q i, N I }, {D i, H U } e πi/3 {D i, H U }, {U i, L i, E i, H D } e πi/3 {U i, L i, E i, H D }, (3.5) to be compared with the result of R p transformations: {H D, H U } {H D, H U }, {Q i, U i, D i, L i, N I, E i } e iπ {Q i, U i, D i, L i, N I, E i } (3.6) With assumptions analogous to Point 1. and Point. one finds that all B 3 conserving operators have integer X-charges, while all B 3 operators have X-charges that are integer± 1: If an SM-invariant superfield-operator Φ 3 1Φ...Φ n violates B 3, then (Φ 1 Φ...Φ n ) 3 does not. This is incompatible with R p operators, which have half-odd-integer X-charges. Anomalies In this section, we work out requirements from the anomaly cancellation via the Green Schwarz mechanism on U(1) X charge assignments. The cancellation/absence of the mixed chiral anomalies of U(1) X group of the SM, itself and gravity demands, see e.g. Ref. [?], with the gauge A CCX k C = A W W X k W = A Y Y X k Y (relying on the Green-Schwarz mechanism) and = A XXX = A GGX 3 k X (.1) A Y XX = 0. (.) 9

10 The A... are the coefficients of the SU(3) C -SU(3) C -U(1) X, SU() W -SU() W -U(1) X, U(1) Y -U(1) Y -U(1) X, U(1) X -U(1) X -U(1) X, grav.-grav.-u(1) X, U(1) Y -U(1) X -U(1) X anomalies, respectively. The factor of 3 in one of the denominators in Eq. (.1) is of combinatorial nature: One deals with a pure rather than mixed anomaly. The affine/kač-moody levels k... of non-abelian gauge groups have to be positive integers. One has [ ( X Q i + X U i + X D i) ], (.3) A CCX = 1 A W W X = 1 A Y Y X = 1 i=1 [ X H U + X H D + i=1 [ X H U + X H D ( 3 X Q i + X L i) ], (.) + 1 ( X Q i + 3X 3 U i + 8X D i + 1X L i + X E i) ] Y L, (.5) i=1 [ A Y XX = X H U X H D + i=1 A XXX = X H U 3 + X H D 3 + i=1 ( ) ] X Q i X U i + X i D X L i + X E i Y L, (.6) +X A 3 + I=1 X N I A GGX = X H U + X H D + i=1 +X A + I=1 ( ) 6X 3 Q i + 3X i3 U + 3X i3 D + X 3 3 L i + X E i 3 hidden sector + AXXX, (.7) ( ) 6X Q i + 3X U i + 3X D i + X L i + X E i hidden sector X N I + AGGX. (.8) We have not fixed the normalization of the hypercharges, and we used the standard GUT normalization for the generators of the non-abelian gauge groups: From Eqs. (3.,.) one finds (with Q,L 11 = 0, N g is the number of generations) 10

11 5 Phenomenological Constraints from Quarks and Charged Leptons In this section, we use phenomenologically acceptable forms of mass matrices for up-quarks, down-quarks, charged leptons, and the CKM matrix, and determine the U(1) X charge assignments consistent with them. We make the full use of anomaly cancellation conditions derived in the previous section. There are five viable patterns for quark mass matrices Eqs. ( ), and we will be left with three real parameters (X L 1, L 1, L 31 ) for each pattern, as shown in Table 1. At this point, the U(1) X charges for two right-handed neutrinos are left free. Combining further with the requirement of automatic R-parity conservation, we arrive at Table?? where parameters L 31, ζ, H, ν, N 1 are constrained to be integers. Even though each of the five patterns is phenomenologically viable, we pick the patterns Eqs. (5.9) and (5.11) because the CKM matrix comes out most successfully (the middle one in Eq. (5.7)). To identify phenomenologically acceptable mass matrices, we follow Ref. [?]. The mass eigenvalues are given at the GUT-scale [?], 8 m e : m µ : m τ or 5 : : 1, (5.1) m τ : m b 1, (5.) m d : m s : m b : : 1, (5.3) m b : m t 0,1, or 3 H D / H U, (5.) m u : m c : m t 8 : : 1, (5.5) m t H U, (5.6) and in addition V CKM 1 1 or 1 or 1. (5.7) The first 9 and the last choice for the CKM matrix require accidental cancellations of O(ɛ) among unknown O(1)-coefficients. The second choice is slightly preferred, 8 For fields except the top-quark the fermion masses renormalize practically only according to the anomalous dimensions due to gauge interactions, and hence their inter-generational ratios do not renormalize. 9 This shape of V CKM was only recently suggested in Ref. [?]. 11

12 which is why we will eventually discard the first and the third choice. H U and H D denote the VEVs of the two neutral Higgs scalars, V CKM is the Cabibbo- Kobayashi-Maskawa matrix and 0. is the Wolfenstein parameter, i.e. the (sine of the) Cabibbo angle. If one is dealing with U(1) X and one flavon superfield, the only pairs of u- and d-type quark mass matrices (after the Kähler potential has been diagonalized and thus textures have been filled up) which can be generated á la FN and which simultaneously reproduce the quark masses and mixings as displayed above are (see Refs. [?,?,?,?]; the textures in Eq. (5.10) are presented for the first time) G (U) G (U) G (U) G (U) G (U) , G (D) x, G (D) x, G (D) x, G (D) x, G (D) x , (5.8), (5.9), (5.10), (5.11). (5.1) Here, x = 0, 1,, 3 except for Eq. (5.9) where the choice is limited to x = 0, 1,. The first and the last of these pairs of matrices lead to the third choice for the CKM matrix in Eq. (5.7), the third pair corresponds to the first choice in Eq. (5.7). The second pair does not give m b : m t 3 H D / H U, see below Eq. (5.17). As a spot-check, we investigated the validity of G (U) in Eq. (5.11) with an ensemble of 3000 Mathematica c -randomly generated sets of O(1) and complex g (U) ij. In Figure 1 the logarithm to base (log λc ) of the positive square roots of the eigenvalues of G (U) G (U) is plotted against the 3000 trials. The result agrees very well with Eq. (5.5), apart from a largish scatter for 8. In order to reproduce the patterns Eq. ( ), the X-charges have to fulfill, see 1

13 Figure 1: The powers in of the positive square roots of the three eigenvalues of G (U) G (U) for an ensemble of 3000 Mathematica c -randomly generated sets of g (U) ij which are complex and O(1). 13

14 Ref. [?], X Q i + X H U + X U j = X Q i + X H D + X D j = X L i + X H D + X E i = y 3 + y 7 y 5 y 0 + x 3 + y + x 3 + y + x 3 y + x + x + x 1 y + x x x + z + x + x x i ij, (5.13) ij, (5.1), (5.15) X L 3 + X E 3 = X Q 3 + X D 3, (5.16) with x = 0, 1,, 3 if y = 7, 1, 0, 1 or x = 0, 1, if y = 6 and z = 0, 1. (5.17) Eq. ( ) are in order of increasing y. The cases y = 6, 7 necessarily need supersymmetric zeros in the (1,)- and (1,3)-entries of G (U), G (D), which is why y = 6, x = 3 has to be excluded. x = 3 is our preferred value, since it requires a small value of tan β, i.e. H U is of the same order of magnitude as H D, which is more natural than H U H D. In the rest of this text we shall not deal with y = 7, 1, 1 anymore, because y = 6, 0 produces the best fit to the CKM matrix, not requiring any (mild) fine-tuning [as has already been stated below Eq. (5.7)]. The set of X-charges which is constrained by the anomaly conditions of Section and gives rise to the phenomenology explained above is displayed in Table 1, see Ref. [?]. Should one wishes to impose SU(5) invariance on the U(1) X charge assignments, one has to work with y = 1, z = 0, and L 1 = L 31. We will not be able to do so, however, as we will see later on. This is consistent with our philosophy of not having an additional mass scale for grand unification. We now check whether Table 1 can be combined with R p being conserved by virtue of the X-charges. Eq. (3.) is fulfilled if the Ḷ.. are integer. Eq. (??) is automatically fulfilled, as seen from Eqs. (5.13,5.1,5.15). With Table 1 we see 1

15 X H D = [ 1 5+9x+6z x + 18y +6x + z + 5xz X L 1 (36 + 6x + 9z) (1 + x + z) L 1 (6 + x + z) L 31 ] X H U = X H D z X Q 1 = [ XL 1 L 1 L x + 6y + z] X Q = X Q 1 1 y X Q 3 = X Q 1 3 y X D 1 = X H D X Q x X D = X D y X D 3 = X D y X U 1 = X H D X Q z X U = X U y X U 3 = X U y X L = X L 1 + L 1 X L 3 = X L 1 + L 31 X E 1 = X H D+ X L 1 + x + z X E = X H D+ X L 1 + x L 1 X E 3 = X H D X L 1 + x L 31 Table 1: The constrained X-charges to reproduce phenomenologically acceptable mass matrices Eqs. ( ), with the normalization X A = 1. X L 1, L 1, L 31 are real numbers, for x, y, z see Eq. (5.17). SU(5) invariance would require y = 1, z = 0, L 1 = L

16 For the upcoming calculations it is useful to know that X L i + X H U + X E j + z 0 = x + + z 0 + z 0 + z 0 = x + + z 0 + z 0 + ij ij + 0 L 1 L 31 L 1 0 L 1 L 31 L 31 L 31 L z + L 31 3ζ L 31 z L ζ 0 z L ζ L 31 z + L 31 3ζ 0 ij ij. (5.18) It is worth pointing out that there already exists a model in the literature which fulfills all the necessary constraints for R p being conserved due to the X-charges, namely Ref. [?]. This model is in the tradition of the papers Ref. [?] and Ref. [?]. In the former, a general analysis of D-flat directions and the seesaw mechanism leads the authors to conserved R p, in the latter, the authors worked out a concrete model. They considered three beyond-sm U(1)s, two of them being generation-dependent and non-anomalous, one being generation-independent and anomalous. In Ref. [?] the before-mentioned symmetries were not gauged separately but together, so that the model falls into the category of models considered here. In our notation, the authors work with x = 3, y = 0, z = 0, L 31 = 3, ζ =, H = 1. 6 The VEV of the Flavon; Tadpoles Because we would like to construct a complete theory of flavor out of only two mass scales, M grav and m 3/, the mass scale of the U(1) X breaking must be a derived scale. Indeed, the vacuum expectation value of the flavon is determined dynamically thanks to the anomalous nature of U(1) X. We show explicitly that our X-charge assignments can successfully lead to an expansion parameter ɛ = A /M grav = as desired phenomenologically. We, however, point out an important caveat in a class of string-derived models. We also show that tadpoles are of no concern. 16

17 In the string-embedded FN framework the expansion parameter ɛ (which will be identified with ) has its origin solely in the Dine-Seiberg-Wen-Witten mechanism, due to which the coefficient of the Fayet-Iliopoulos term is radiatively generated. One has, see Ref. [?], However, there is a very important caveat which one should keep in mind: Eq. (??) together with the assumption that the dimensionless prefactors like g (...) ij are of O(1) might well not be justified by superstring theory. In Ref. [?] it is nicely and clearly demonstrated how a prototype string theory (Ref. [?]) produces an effective supersymmetric theory with a superpotential, including the coupling constants. Translating their result to our notation we get e.g. instead of Eq. (??) Θ [ X Q i + X H U + X U j] Ω [ XQ i + X H U + X U j ] g C kc C XQ i +X H U +X U j I XQ i +X H U +X U j ( A π M grav ) XQ i +X H U +X U j Q i H U U j. (6.1) C... is a O(1) Clebsch-Gordan coefficient and I... a world sheet integral. For large X Q i + X H U + X U j naively one would expect I XQ i +X H U +X U j I 1 X Q i +X H U +X U j (with I 1 70), but due to destructive interference effects of the integrands actually 10 Below the U(1) X breaking scale ɛ M grav there are three singlets {A, N I }, with A = A A. One must thus wonder whether these lead to tadpoles causing quadratic divergences and thus possibly destabilizing the hierarchy between the weak scale and M grav, see Refs. [?,?,?,?]. First of all, in our model R p is conserved before and after the breaking of U(1) X. This prevents any N I -tadpole term. Possible A -tadpoles are however harmless due to the high mass of A, given by ɛ M grav. 7 µ-parameter and Proton Decay So far, we are left with two patterns Eqs. (5.9) and (5.11) with y = 6, 0, respectively, each with possible choices x = 0, 1,, 3 and z = 0, 1. In this section, 10 We thank Mirjam Cvetič for pointing this out. 17

18 we narrow down the choices further. The first one is the µ-term which we need phenomenologically comparable to m 3/. This requirement picks z = 1. Another requirement is the adequate stability of the proton against Planck-scale D = 5 operators, which eliminates y = 6 and prefers larger x. The resulting U(1) X charge assignments are shown in Table. In order to get a satisfactory µ-term we have to rely on the GM mechanism to have µ m 3/, since X H U + X H D = is not possible, see Table 1. This requires X H U + X H D < 0, see Eq. (.), hence we need z = 1. Thus, see Eq. (??), Next, we consider the proton decay constraints. We might be forced not to work with small x and/or y = 6. This is because of the R p -conserved proton-destabilizing operators χ ijkl M grav Q i Q j Q k L l (i, j, k must not all be the same), where For both equations above one sees that suppressing one of the three operators by choosing an appropriate ζ and/or L 31 makes one or both of the others less suppressed. The average X-charges of Q 1 Q 1 Q L i ( i X Q 1 Q 1 Q L i/3) and Q Q 1 Q L i are x + y + z and x, which are not very high (note that already in Ref. [?] 3 3 it was anticipated that e.g. z = 1 gives a more stable proton than z = 0). Thus already now we can see that the model could get into trouble due to proton decay if we work with the wrong choices for x, y, ζ, L 31. For a first crude estimate we use For a more quantitative investigation we will in the next Section rely on Ref. [?] s treatment of the so-called best-fit scenario. Translated to our notation, they state that the X-charges have to fulfill (with m squark =1 TeV, y = 0, z = 1 ) O(1) ɛ x+10+ L 31 ζ < 10 8, O(1) ɛ x+10 L 31 +ζ HU H D < (7.1) in order not to be in conflict with experiment. With m t /m b at high energies being 100 (see Ref. [?]) we get from m b H D ɛ x, m t H U that 100 ɛ x H U / H D. 18

19 Thus O(1) ɛ x+10+ L 31 ζ < 10 8, O(1) ɛ x+10 L 31 +ζ < (7.) Note that our model with y = 0, z = 1, L 1 = L 31 = 1, and ζ = 1 is a special case of the best fit model in Ref. [?] with m = 1 in their notation, while they took X L 3 a free parameter because they do not impose the anomaly cancellation conditions nor conserved R p as a consequence of the U(1) X symmetry. 8 Neutrino Phenomenology Our study of the neutrino sector is far more constrained than most models in the literature. This is because there is no GUT scale, which is a factor of 100 lower than M grav, to suppress the mass scale of the Majorana mass terms GeV, or equivalently, boost the light neutrino masses to the required orders of magnitude. In typical seesaw models (see Refs. [?,?,?,?]), it is achieved using an extra symmetry, such as gauged U(1) B L. However, in our scenario there is no additional symmetries beyond the MSSM gauge groups and U(1) X nor additional mass scales beyond M grav and m 3/. Therefore, in our scenario, the mass scales of right-handed neutrinos originate from M grav, suppressed by powers of ɛ. The successful neutrino phenomenology together with proton decay constraints determine the U(1) X charge assignments down to four discrete choices. 11 Because this discussion is rather long, we have divided this section into the following subsections. In Section 8.1, we review our phenomenological understanding of neutrino mixings and discuss their implications on U(1) X charge assignments. The phenomenology requires ζ = L 31 = 1 as well as z = 1, thus justifying the GM mechanism for the µ-parameter from a completely different reasoning. The resulting charge assignments are shown in Table 3. Section 8. is the corresponding discussion of neutrino mass eigenvalues. Here we encounter different possibilities depending on if LL, LR and RR entries of the neutrino mass matrices are induced by the GM 11 We cannot, however, uniquely determine the U(1) X charges of right-handed neutrinos, because they practically drop out from the light neutrino masses and mixings. This fact is well-known, see, e.g., [?]. 19

20 ( 1 X H D = x (x + 1 H ) 10 (7+x) ) ( H ) (6 + 6 H L 31 ) (7 + x)ζ X H U = 1 X H D ( ) X Q 1 = 1 1 X 3 H D + x H ζ X Q = X Q 1 1 X Q 3 = X Q 1 3 X D 1 = X H D X Q x X D = X D 1 1 X D 3 = X D 1 1 X U 1 = X H D X Q X U = X U 1 3 X U 3 = X U 1 5 X L 1 = 1 + X H D + H X L = X L L ζ X L 3 = X L 1 + L 31 X E 1 = X H D+ 5 X L 1 + x X E = X H D+ 1 X L 1 + x + L 31 3ζ X E 3 = X H D X L 1 + x L 31 X N 1 = 1 + ν X N = 1 + ν + N 1 Table : In addition to Table??, the constraints from the µ-parameter z = 1 and proton decay y = 0 are imposed. Furthermore, the proton decay prefers x =, 3 over x = 0, 1. L 31, ζ, H, ν, N 1 are integers. 0

21 mechanism, schematically shown in Table. Section 8..1 discusses cases 1.),.), and 3.), while Section 8.. discusses cases.), 5.), and 6.). We find successful solutions to cases.) and 6.). The former case is a similar to standard seesaw scenario, and U(1) X charge assignments are shown in Tables?? and??. The latter case has the right-handed neutrino masses from the GM mechanism and hence they are present below the electroweak scale. The U(1) X charge assignments are shown in Tables?? and??. 8.1 Neutrino Mixing If there are no filled-up supersymmetric zeros in G (U) and G (D), then, see Ref. [?] From Eq. (??) one can observe furthermore that there are no supersymmetric zeros for the superfield operators QQQL, so that the canonicalization of the Kähler potential does not affect the order of ɛ-supression. With the results of Ref. [?], also translated to the mass matrices of charged leptons and light neutrinos we find that the powers of ɛ of Q 1 Q 1 Q L 1, Q 1 Q 1 Q L, Q 1 Q 1 Q L 3, Q Q 1 Q L 1, Q Q 1 Q L, Q Q 1 Q L 3 are again not changed when going to the mass basis. 1 With Eq. (??) and , , , , , , , (8.1) together with Eq. (7.), one finds that x = 0, 1 cases are ruled out, only x =, 3 are viable, while x = 3 is allowed even for m squark = 100 GeV. 13 Will we be able to have an X-charge assignment such that no hidden sector fields are needed in order to cancel the anomalies of U(1) X with itself and gravity? Table 1 and Eq. (.8) give 1 This was assumed to be true in Ref. [?], while we explicitly verified it. 13 In the language of Ref. [?], the model with L 31 = ζ = 1, z = 1, and y = 0 is classified as (no) h.o.. 1

22 It should be mentioned that phenomenology might also suggest the so-called anarchical scenario, see Refs. [?,?,?], i.e. instead of Eq. (??) one has U MNS (8.) However, for z = 1 and due to Eq. (??) this is not an option in our case. 8. Neutrino Masses After looking at the mixing, let us now investigate the relationship between the neutrino mass spectrum and the X-charges. For future reference we here state the experimental status, allowing for three possible neutrino mass solutions, see e.g. Ref. [?] and references therein: hierarchical (m ν 3 is much larger than m ν, which is much larger than m ν 1 which in our case is zero), m ν m ν ev, m ν 3 m ν ev, (8.3) inverse hierarchical (m ν is minutely larger than m ν 1, which is much larger than m ν 3 which in our case is zero; this is not possible in our scenario), m ν m ν ev, m ν 3 m ν ev, (8.) quasi-degenerate (all m ν are almost identical; this is not possible in our scenario). The scenario sketched so far (with x =, 3, y = 0, z = 1, ζ = 1, L 31 = 1) generates a superpotential (neglecting the tiny contributions of the GM mechanism

23 X H D = 5 39 x(6+x H )+30 H 10 (7+x) X H U = 1 X H D ( ) X Q 1 = 1 3 X 3 H + x D H X Q = X Q 1 1 X Q 3 = X Q 1 3 X D 1 = X H D X Q x X D = X D 1 1 X D 3 = X D 1 1 X U 1 = X H D X Q X U = X U 1 3 X U 3 = X U 1 5 X L 1 = 1 + X H D + H X L = X L 1 1 X L 3 = X L 1 1 X E 1 = X H D+ 5 X L 1 + x X E = X H D+ 3 X L 1 + x X E 3 = X H D+ 1 X L 1 + x X N 1 = 1 + ν X N = 1 + ν + N 1 Table 3: In addition to Table, we required successful neutrino mixings, i.e., L 31 = ζ = 1. The remaining parameters H, ν, N 1 are integers, while x = or 3 to satisfy proton decay constraints. 3

24 to G (U,D,E) ) with [for G (E) see Eq. (5.18)] W MSSM = g (U) ij ɛ 8 ɛ 5 ɛ 3 ɛ 7 ɛ ɛ Q i H U U j ɛ 5 ɛ 1 ij ɛ ɛ 3 ɛ 3 + m 3/ g(µ) ɛ H D H U + g (D) ij ɛ or 3 ɛ 3 ɛ ɛ Q i H D D j ɛ 1 1 ij ɛ 5 ɛ 3 ɛ + g (E) ij ɛ or 3 ɛ ɛ 1 L i H D E j (8.5) ɛ ɛ 1 and W ν = ( Θ[X N I + X N J ] M grav γ IJ ɛ X N I +X N J ij + + m 3/ γ IJ ɛ X N I +X ) N J N I N J ( Θ[X L i + X H U + X N J ] g (N) ij ɛ X L i +X H U +X N J + + m 3/ g M (N) ij ɛ XLi +X HU +X ) N J L i H U N J grav ( ψ ij Θ[X L i + X H U + X L j + X H U ] M grav + m 3/ ψ ) ij ɛ X L M grav M i +X H U +X Lj +X H U grav ɛ X L i +X H U +X Lj +X H U L i H U L j H U ; (8.6) summation over repeated indices is implied. The two equations above do not contain any factors of Ω[...], because by construction all R p -conserving terms have integer X-charge. We will in turn investigate the different possibilities for generating the mass terms, as given in Table. With y = 0 all exponents in G (U,D,E) are positive. From this one may feel inspired to assume either that all exponents in the mass terms of the neutrinos are positive (the case in the lower right-hand corner of Table ), or (less restrictive) simply that for a given array of neutrino coupling constants all exponents are either negative or positive. Are the cases sketched in Table Majorana or pseudo Dirac neutrinos? One has pseudo Dirac neutrinos if M Dirac LR M Maj LL, M Maj RR. We will investigate case by case (starting in the lower right-hand corner, proceeding anti-clockwise) whether this condition can be met when M Dirac LR is generated via the FN mechanism, since

25 X L i + X H U < 0 X L i + X H U > 0 5.) X L i + X H U < X N I M Maj LL : FN.) M Dirac LR : GM X N I M Maj LL : GM Maj MRR GM < 0 M Dirac LR : GM 6.) X L i + X H U X N I M Maj RR : GM M Maj LL : FN 3.) X N I < X L i X H U M Maj LL : GM M Dirac LR : GM 1.) M Dirac LR : FN M Maj RR : GM X N I M Maj RR : FN Maj MLL FN > 0.) X N I X L i X H U M Dirac LR : FN M Maj Maj LL : GM MRR FN M Dirac LR : FN M Maj RR : FN Table : The Majorana mass of the left-handed neutrinos is denoted by M Maj LL, the one for the right-handed neutrinos is given by M Maj RR, and the Dirac mass is. We work with two right-handed neutrinos. M Dirac LR 5

26 M Dirac LR generated by the GM mechanism produces neutrino mases which are too ] ij ɛ X L i +X N j +X H U H U m 3/ /M grav 10 5 ev. small, one has that [M Dirac LR 8..1 Positive X N First we will take the X-charges of all right-handed neutrino superfields to be positive. This ensures that the scalar component of A acquires a VEV, since its X-charge is negative and ξ X is positive. This way the D X -term does not acquire a VEV and supersymmetry is not broken by the DSWW-mechanism, see Ref. [?,?,?,?], at a much too high energy scale. Of course, guaranteeing a VEV for A does not automatically guarantee that the scalar components of the right-handed neutrino superfields do not get VEVs this we simply have to postulate, in order to conserve R p and to have only one flavon field. 1.) We now consider the case in the lower right corner of Table. All Θ[...] can be dropped: W ν = M grav γ IJ ɛ X N I +X N J N I N J + g (N) ij ɛ X L i +X H U +X N J L i H U N J + ψ ij M grav ɛ X L i +X H U +X L j +X H U L i H U L j H U, (8.7) for short (dropping all generational indices) W ν = M grav N T Γ N + L T G (N) H U N 1 + L T Ψ L H U H U. (8.8) M grav We get a Lagrangian with (n L/R are left/right-handed neutrinos in the interaction basis) Pseudo Dirac neutrinos are also not possible. One needs to fulfill two conditions which contradict each other: ( ) H U ɛ XLi +X Lj +X HU HU ɛ X M Li +X HU +X NJ, grav M grav ɛ X N I +X N J HU M grav ɛ X Li +X HU +X N J. (8.9) 6

27 .) Now we consider the lower case in the lower left-hand corner in Table, first the Majorana case. M Maj LL is suppressed by a factor of m 3/ /M grav, compared to the one in 1.), so that we can neglect it. We arrive at From the equation above it follows that det [ G (ν)] = det [ g (N) γ 1 g (N)T ] ɛ 6X H U + i X L i. (8.10) Since γ is a matrix and g (N) is a 3 matrix, the determinant of g (N) γ 1 g (N)T is zero, regardless of which values one has for the entries of γ, g (N) (this constrained seesaw mechanism was first proposed in Ref. [?]; for a model embedding it into a family-symmetry see Ref. [?]). Thus one of the three eigenvalues of G (ν) is definitely zero, so that the lightest neutrino is massless; we can use m ν to determine the absolute masses of the two other neutrinos. So, with Eq. (8.3), 1 with λ denoting the eigenvalues of the matrix in Eq. (??). For the hierarchical case one has ev HU λ middle, M grav ev HU M grav λ max. (8.11) Hence with x = 3, tan β is smallish (say, 3) and thus H U 16 GeV, one finds i.e. (see Appendix??) λ middle , λ max , (8.1) That we started with L 1 = L 31 = 1 was of course an inspired guess based on comparing Eq. (??) with the more-or-less handwaving Eq. (??). So we have to check 1 We easily obey the limit on neutrino masses from WMAP, see Refs. [?,?], namely mν 0.71 ev. 7

28 ɛ 10 that the X-charges given above indeed lead to the MNS-matrix we used as a starting point, to justify our guess in hindsight: 15 We get from Eq. (8.5) and Eq. (??) that ɛ ɛ ɛ ɛ ɛ ɛ G (E) G (E) ɛ x ɛ 1 1, G (ν) G (ν) ɛ 1 1, (8.13) ɛ 1 1 ɛ 1 1 so the two matrices which make up U MNS both have a structure as in Eq. (??). To schematically see this, consider the mass matrix in Eq. (??), dropping Ψ, Γ, G (N). It is diagonalized by the matrix given in Eq. (??), with its off-diagonal blocks approximated in Eq. (??). Now replace all η by ɛ, one finds that From the Tables??,?? we get furthermore that 3 H + I X does not allow for N I Green-Schwarz anomaly cancellation, see Eq. (??) and the text below it. So we are forced to require the existence of at least one X-charged superfield in the hidden sector. Pseudo Dirac neutrinos are possible, but require very large X N I. As a toy-model, consider the one-generational case. One of the two conditions not to have Majorana masses is 3.) Now we discuss the upper case in the lower left-hand corner in Table. The Majorana case gives that the mass matrix of the light neutrinos is to lowest order 8.. Negative X N Now we consider the case with X N < 0, which is less appealing than the previous one, because a VEV of A is no longer guaranteed. 15 An example that the starting rule-of-the-thumb guess [to apply Eq. (??) to Eq. (??)] does not automatically lead to the correct U MNS in the end: One might be willing to allow for a choice of the O(1)-coefficents to be such that z = is a possible. With Eqs. (??,??) this gives L 31 = 1, ζ = 0. Using Eq. (5.18) we get a G (E) in which the (1,3)-entry dominates, producing a U MNS which is not in accord with Eq. (??). 8