Supersymmetric GUTs
Michael Grefe Werkstattseminar – 4.11.2008
In unbroken SU(5) Grand Uni ed Theories (GUTs) we expect
0 gGUT = gs = g = 5/3 g (1) and p 02 2 2 g 3/5 gGUT 3 sin W = 2 02 = 2 2 = . (2) g + g gGUT + 3/5 gGUT 8 However, these parameters are not observed in low-energy experiments. Fur- thermore, we do not observe interactions through the 12 additional gauge bosons of SU(5). Thus the GUT symmetry must be broken at a high scale MGUT . In that case, the gauge couplings of the three Standard Model groups evolve independently below the GUT scale due to the renormalization of di- vergent radiative corrections to the gauge vertices.
1 Evolution of Gauge Couplings
0 The evolution of the gauge couplings gi (g3 = gs, g2 = g, g1 = 5/3 g ) is described by the beta function: p dgi dgi 1 3 5 i = = 2 bigi + (gi ), (3) d ln( / 0) d 16 O where the constants bi are determined from the one-loop corrections to the gauge vertices. The beta function can also be rewritten in terms of i = 2 gi /4 2 d i d(gi /4 ) gi dgi 1 2 3 = = = bi + ( ), (4) d d 2 d 2 i O i The rst order beta function for the inverse gauge couplings becomes very simple, 1 d i 1 d i 1 = 2 = bi , (5) d i d 2
1 and can easily be integrated using separation of variables: 0 1 0 1 d d ( ) = bi (6) i 2 0 MZ MZ leading to to the evolution equation
1 1 1 ( ) = (M) bi ln( /M). (7) i i 2 Under the assumption that there are no new particles (except for SM singlets) between the electroweak (or SUSY mass) scale and the GUT scale, MGUT and the uni ed gauge coupling GUT can be determined from the intersection of two gauge couplings according to equation (1):
1 1 bi 1 bj = (mZ ) ln(MGUT /mZ ) = (mZ ) ln(MGUT /mZ ) GUT i 2 j 2 (8)
2 1 1 MGUT = mZ exp i (mZ ) j (mZ ) (9) bi bj µ ¶ In order to solve these equations¡ we need the measured¢ input parameters at the electroweak scale: 1 em(mZ ) 128 , 2 ' sin W (mZ ) 0.231 , ' (10) s(mZ ) 0.118 , ' mZ 91.2 GeV. ' From these we can determine 1 and 2 at the electroweak scale: 2 1 3 cos W (mZ ) 1 (mZ ) = = 59.1 , 5 em(mZ ) 2 (11) 1 sin W (mZ ) 2 (mZ ) = = 29.6 . em(mZ )
In addition we need to know the constants bi for SU(3)c, SU(2)L and U(1)Y . These depend on the particle content of the theory.
1.1 Standard Model In the Standard Model the constants in the one-loop beta functions are given by 2 1 11 bi = T (Ri)d(Rj)d(Rk) + T (Si)d(Sj)d(Sk) C (Gi). (12) 3 3 3 2 2 The fermion multiplets transform according to the representation Ri with respect to Gi and the bosons according to Si. T (R) is the normalization of the generators T a in the representation R: Tr T aT b = T (R) ab. (13) The generators of the fundamen£ tal represen¤ tation of SU(N) are usually nor- malized to T (R) = 1/2. d(R) is the dimension of the representation R and C2(R) is the quadratic Casimir operator for the representation R: a a TikTkj = C2(R) ij. (14)
It is easy to see that T (R) and C2(R) are related by the identity
C2(R)d(R) = T (R)d(G), (15) where d(G) is the dimension of the adjoint representation (i.e. the number of generators of the group). For the adjoint representation of SU(N) we have
T (G) = C2(G) = N. (16)
2 For a representation of U(1) one has C2(G) = 0 and T (R) = Y , where Y is the appropriately normalized hypercharge. For SU(3) we only have contributions from the SU(3) quark triplets 2 1 11 4 b = (2 1 + 1 1 + 1 1) NG 3 = NG 11. (17) 3 3 2 3 3 In the SU(2) case we have contributions from one SU(2) quark doublet, one SU(2) lepton doublet and SU(2) Higgs doublets 2 1 1 1 11 b = (3 1 + 1 1) NG + (1 1) NH 2 2 3 2 3 2 3 4 1 22 = NG + NH (18) 3 6 3 and in the U(1) case we have contributions from all particles according to their hypercharge (with a normalization factor of 3/5)
2 3 1 2 2 2 1 2 1 2 b = 2 3 + 1 3 + 1 3 + 2 1 1 3 5 6 3 3 2 " µ ¶ µ ¶ µ ¶ µ ¶ 2 2 1 3 1 + 1 1 1 NG + 2 1 NH (19) 3 5 2 # "µ ¶ # 4 1 = N + N . 3 G 10 H
3 i i Name Scalars Fermions L (SU(3)c, SU(2)L)Y i i ˜i ˜L i L Sleptons, leptons L = i L = i (1, 2) 1 e˜L eL 2 ˜ i µ ¶i c i µ c¶i E = e˜R E = eR (1, 1)+1 i i ˜i u˜L, h i uL, h Squarks, quarks Qh = ˜i Qh = i (3, 2)+ 1 dL, h dL, h 6 ˜ i µ i ¶ c i µ c i ¶ Uh = u˜R, h Uh = uR, h (3, 1) 2 3 ˜ i ˜ i c i c i Dh = dR, h Dh = dR, h (3, 1) 1 + 3 0 ˜ 0 Hd ˜ Hd Higgs, higgsinos Hd = Hd = ˜ (1, 2) 1 Hd Hd 2 µ +¶ µ ˜ +¶ Hu ˜ Hu Hu = 0 Hu = 0 (1, 2)+ 1 H H˜ 2 µ u ¶ µ u ¶ Table 1: Chiral supermultiplets of the MSSM.
We can summarize these results and insert the number of fermion gener- ations NG = 3 and Higgs doublets NH = 1:
4 1 41 b1 3 10 0 10 b = N 4 + N 1 22 = 19 . (20) 2 G 3 H 6 3 6 b 4 0 11 7 3 3 Using these parameters the gauge couplings do not unify at the high scale. 13 The three intersection points range from MGUT 10 GeV to MGUT 17 ' 1 ' 10 GeV and correspond to a uni ed coupling in the range GUT 40 to 1 ' GUT 47. Th'us new physics beyond the Standard Model is needed to make GUTs viable. One possibility is to introduce additional particles in the SM. For instance six Higgs doublets at (mZ ) lead to a uni cation of gauge couplings 13 O at MGUT 4 10 GeV. Another possibility is to change the particle content due to the'intro duction of supersymmetry.
1.2 Minimal Supersymmetric Standard Model In the supersymmetric case there is a scalar partner for each fermion that transforms according to the same representations with respect to the gauge groups. Thus the contributions of fermions and bosons to bi can be combined in one term. The fermionic partners of the gauge bosons, the gauginos, also transform according to the adjoint representation and therefore alter the
4 Name Gauge bosons Gauginos (SU(3)c, SU(2)L)Y (1) (1) ˜ B boson, bino A = B = B (1, 1)0 (2) a a (2) a ˜ a W bosons, winos A = W = W (1, 3)0 (3) a a (3) a a gluons, gluinos A = G = g˜ (8, 1)0
Table 2: Gauge supermultiplets of the MSSM.
coe cient of the C2(Gi) term. One obtains:
bi = T (Ri)d(Rj)d(Rk) 3 C (Gi). (21) 2 Inserting the particle content of the MSSM one has 1 b = (2 1 + 1 1 + 1 1) NG 3 3 = 2 NG 9, (22) 3 2 1 1 b = (3 1 + 1 1) NG + (1 1 + 1 1) N H 3 2 2 2 2 2 = 2 NG + N H 6 (23) 2 and
3 1 2 2 2 1 2 1 2 b = 2 3 + 1 3 + 1 3 + 2 1 1 5 6 3 3 2 " µ ¶ µ ¶ µ ¶ µ ¶ 2 2 2 3 1 1 + 1 1 1 NG + 2 1 + 2 1 N H (24) 5 2 2 2 # "µ ¶ µ ¶ # 3 = 2 N + N . G 5 2H
Inserting the number of fermion generations NG = 3 and the number of pairs of Higgs doublets N2H = 1 the results can be summarized as: 3 33 b1 2 5 0 5 b = N 2 + N 1 6 = 1 (25) 2 G 2H b 2 0 9 3 3 With these altered coe cients the one-loop calculation results in a uni cation 16 of gauge couplings at MGUT 2 10 GeV, corresponding to a uni ed gauge 1 ' coupling GUT 24. In fact, using' the precision input parameters from LEP, one has to con- sider the two-loop beta function for the evolution of the gauge couplings including also one-loop threshold e ects due to particle masses (at the elec- troweak scale and also at the GUT scale).
5 60
50 α −1 1
40
α−1 30 α −1 2 20
10 α −1 3 0 2 4 6 8 10 12 14 16 18
Log10(Q/1 GeV)
Figure 1: Renormalization group evolution of the inverse gauge couplings in the Standard Model (dashed) and the MSSM (solid) including two-loop e ects. The sparticle mass thresholds are varied between 250 GeV and 1 TeV. Figure taken from [4].
2 Higgs Sector of Minimal SU(5) SUSY GUT
In order to break the SU(5) GUT group down to SU(3)c SU(2)L U(1)Y one introduces a new Higgs multiplet in the adjoint represen tation (24) of SU(5). To maintain the electroweak symmetry breaking to SU(3)c U(1)em and to generate masses for the fermions two additional Higgs multiplets H (5) and H (5 ) that contain the usual SU(2)-doublet Higgs multiplets as well as color-triplet partners of these have to be included. The superpotential of the minimal SUSY SU(5) model is given by
1 2 1 3 W = fV Tr + f Tr + H + 3 V H 2 3 (26) 1 + √2Y ij H + Y ijε ¡ H ,¢ d i j 4 u i j where i, j = 1, 2, 3 are generation indices and Greek indices are SU(5) indices. The chiral super elds (10) and (5 ) are left-handed matter supermulti- plets. The adjoint Higgs multiplet is given by
a a (8,1) (3,2) 1 2 13 3 0 = T = + (1,1) . (27) (3 ,2) (1,3) 2√15 0 3 12 2 µ ¶ µ ¶
6 We neglect the Yukawa part for this discussion and rewrite the superpo- tential as 1 1 W = fV a b Tr T aT b + f a b c Tr T aT bT c. (28) 2 3 Now we can substitute Tr T aT b = 1/2 ab and 1 Tr T aT bT c = Tr T a T b, T c + T b, T c 2 (29) 1 1 = Tr T a ¡£f bcdT d +¤ dbcd©T d =ª¢ f abc + dabc 2 4 ¡ ¢ ¡ ¢ with f abc = 2 Tr T a T b, T c and dabc = 2 Tr T a T b, T c . The superpotential becomes £ 1¤ 1 © ª W = fV ( a)2 + fdabc a b c. (30) 4 12 The conditions for a SUSY conserving minimum of the superpotential are ∂W ∂W ∂W = 0, = 0, = 0 (31) ∂ a ∂H ∂H and they are at the same time approximate conditions for the minimum of the scalar potential. We nd ∂W 1 1 = fV a + fdabc b c = 0 (32) ∂ a 2 4 and see that a = 0 is a solution that does not break SU(5). Therefore we h i choose a VEV for (1,1) and use
3 3 12 12 12 12 12 12 3 2 2 3 1 d = 4 Tr T T T = 4 3 = (33) 23√15 √15 giving the condition 1 1 f (1,1) V (1,1) = 0 (34) 2 2√15 µ ¶ ® ® that is solved by (1,1) = 2√15 V , so that
® 2 2 = T 12 = V 2 . (35) h i (1,1) 3 ® 3 7 This VEV breaks SU(5) to SU(3) SU(2) U(1). In a similar way one can nd a VEV that breaks SU(5) to SU(4) U(1): 1 1 = V 1 . (36) h i 1 4 These three minima are degenerate and the minimalSU(5) SUSY GUT does not explain why the breaking to SU(3) SU(2) U(1) should be chosen. Now we want to check the other tw o conditions in equation (31): ∂W = H ( + 3 V 1) = 0 , ∂H (37) ∂W = ( + 3 V 1) H = 0 . ∂H These conditions require that the color-triplet components do not acquire a VEV. On the other hand, a VEV for the SU(2)-doublet Higgs bosons does not destroy the picture and allows for electroweak symmetry breaking at low energy. Inserting the VEV into the superpotential we see that it is ne-tuned such that the SU(2) Higgs doublets are massless while the color-triplet Higgs bosons obtain a mass parameter
MHc = MH c = 5 V. (38) The Higgs VEV generates masses for the SU(5) gauge bosons X and Y . This can be seenhfromi the kinetic Lagrangian of the 24 Higgs bosons: † Tr(D ) (D ) (39) L with the covariant derivative of the adjoint Higgs matrix
D = ∂ + igGUT [A , ] . (40) The mass terms for the gauge bosons are then given by 2 2 g Tr [A , ] . (41) GUT h i The Standard Model gauge bosons commute with and therefore remain massless. The masses for the X and Y bosons becomeh i 2 X1 Y1 2 X Y 2 2 2 g2 V 2 Tr X Y , 2 GUT 3 3 (42) X X X 3 1 2 3 Y Y Y 3 1 2 3 = 50 g2 V 2 X2 + X2 + X2 + Y2 + Y 2 + Y 2 GUT 1 2 3 1 2 3 ¡ 8 ¢ MX = MY = 5√2 gGUT V . (43) ⇒ 1 We will call this mass the GUT scale and from GUT 24 we nd gGUT 0.7 and therefore V 4 1015 GeV. ' ' In addition one' can nd the masses for the 24 Higgs bosons from the rst two terms in the superpotential. The results are: 5 M = M = fV, (44) (8,1) (1,3) 2 √ M (3,2) = M (3 ,2) = 5 2 gGUT V , (45) 1 M = fV. (46) (1,1) 2
Open Questions In the minimal SU(5) SUSY GUT model the mass of the 5 and 5 Higgs bosons has to be ne-tuned such that the SU(2) doublet Higgs bosons have vanishing mass and the color-triplet Higgs bosons have masses of (MGUT ) in order to suppress proton decay. This is known as the doublet-tripletO splitting problem. There are several models in the literature to obtain massless Higgs dou- blets without explicit ne-tuning of the parameters in the model: In the sliding singlet model one introduces a singlet super eld Z and adds a term H ZH to the superpotential. This would alter equation (37) to ( + Z) H + MH = 0 (47) and for a nonvanishing VEV of the Higgs doublet the new singlet ac- quires a VEV such that 3 V + Z + M = 0. h i The missing doublet model uses a superpotential without a mass term MH H and gives masses to the color-triplet Higgs bosons through mix- ing with particles in another representation. To achieve this one has to use larger representations of SU(5): Instead of the 24 one uses a 75 and one introduces the additional 50 and 50 . The Higgs doublet has no partner in 50 and therefore remains massless. Apart from the problem to generate the doublet-triplet splitting it is not clear that the color-triplet Higgs bosons are heavier than the GUT scale: In a perturbative theory the dimensionless couplings have to be small. Thus we have . 1 and the mass of the color-triplet Higgs bosons becomes
MHc = MH c = 5 V . MGUT . (48) Thus the Higgs triplet mass might be too low to suppress rapid proton decay.
9 References
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