$SU(4)_L \otimes U(1)_N$ model for the electroweak interactions

11 Pages·0.12 MB·

by  F. Pisano

IFT-P.003/94
hep-ph/9401272
January 1994
model for the electroweak interactions
SU(4) U(1)
L N
⊗
F. Pisano and V. Pleitez
4
9
9 Instituto de F´ısica Teo´rica
1
Universidade Estadual Paulista
n
a
J Rua Pamplona, 145

7
1 01405-900– Sa˜o Paulo, SP

1
Brazil
v
2
7
2
1
0
Abstract
4
9
/
h
p Assumingtheexistenceofright-handedneutrinos,weconsideranelectroweak
-
p model based on the gauge symmetry SU(4) U(1) . We study the neutral
e L⊗ N
h
: currents coupled to all neutral vector bosons present in the theory. There
v
i
X are no flavor changing neutral currents at tree level, coupled with the lightest
r
a
neutral vector boson.
PACS numbers: 12.15.-y
Typeset using REVTEX
1
Symmetry principles have been used in elementary particle physics at least since the
discovery of the neutron. A symmetry is useful to both issues: the classification of particles
and the dynamics of the interactions among them. The point is that there must be a part
of the particle spectrum in which the symmetry manifests itself at least in an approximate
way. This is the case for quarks u and d and in the leptonic sector for the electron-neutrino
and electron. For instance, the SU(2) appears as an approximate symmetry in the doublets
(ν ,e)T. If one assumes this symmetry among these particles and in the sequential families
e
as well, almost all the model’s predictions are determined.
The fullsymmetry oftheso calledStandard Modelis thegaugegroupSU(3) SU(2)
c L
⊗ ⊗
U(1) . This model spectacularly explains all the available experimental data [1]. Usually
Y
it is considered that this symmetry emerges at low energies as a result of the breaking of
higher symmetries. Probably, these huge symmetries are an effect of grand unified scenarios
and/or their supersymmetric extensions.
Considering the lightest particles of the model as the sector in which a symmetry is
manifested, itisinteresting that theleptonsector couldbethepartofthemodeldetermining
new approximate symmetries. For instance, ν,e and ec could be in the same triplet of an
SU(3) U(1) symmetry. This sort of model has been proposed recently [2]. In this case
L N
⊗
neutrinos can remain massless in arbitrary order in perturbation theory, or they get a mass
in some modifications of the models [3]. If we admit that right-handed neutrinos do exist,
it is possible to build a model in which νc,ν and e are in the same multiplet of SU(3) [4].
In fact, if right-handed neutrinos are introduced it is a more interesting possibility to have
ν,e,νc and ec in the same multiplet of a SU(4) U(1) electroweak theory.
L N
⊗
Notice that using the lightest leptons as the particles which determine the approximate
symmetry, if each generation is treated separately, SU(4) is the highest symmetry group to
be considered in the electroweak sector. A model with the SU(4) U(1) symmetry in the
⊗
lepton sector was suggested some years ago in Ref. [5]. However, quarks were not considered
there. This symmetry in both, quarks and leptons, was pointed out recently [6] and here we
will consider the details of such a model.
2
Hence, our model has the full symmetry SU(3) SU(4) U(1) . These sort of models
c L N
⊗ ⊗
are anomaly free only if there are equal number of 4 and 4 (considering the color degrees
∗
of freedom), and furthermore requiring the sum of all fermion charges to vanish. Two of the
three quark generations transform identically and one generation, it does not matter which
one, transforms in a different representation of SU(4) U(1) [7]. This means that in
L N
⊗
these models as in the SU(3) SU(3) U(1) ones [2], in order to cancel anomalies, the
c L N
⊗ ⊗
number of families (N ) must be divisible by the number of color degrees of freedom (n).
f
Hence the simplest alternative is n = N = 3. On the other hand, at low energies these
f
models are indistinguishable from the Standard Model.
The electric charge operator is defined as
1 1 2
Q = (λ λ √6λ )+N, (1)
3 8 15
2 − √3 − 3
where the λ-matrices are a slightly modified version of the usual ones [8],
1 1
λ = diag(1, 1,0,0), λ = ( )diag(1,1, 2,0), λ = ( )diag(1,1,1, 3).
3 8 15
− √3 − √6 −
Leptons transform as (1,4,0), one generation, say Q , transforms as (3,4,+2/3) and
1L
the other two quark families, say Q , α = 2,3, transform as (3,4 , 1/3),
αL ∗
−
ν u j
a 1 i
     
l d d
 a   1   ′i 
faL =   , Q1L =   , QαL =   , (2)
     
νc   u  u 
 a   ′   α 
     
     
 lc   J   d 
 a     α 
 L  L  L
     
where a = e,µ,τ; u and J are new quarks with charge +2/3 and +5/3 respectively; j and
′ i
d , i = 1,2 are new quarks with charge 4/3 and 1/3 respectively. We remind that in
′i − −
Eq. (2) all fields are still symmetry eigenstates. Right-handed quarks transform as singlets
under SU(4) U(1) .
L N
⊗
Quark masses aregenerated by introducing thefollowing HiggsSU(3) SU(4) U(1)
c L N
⊗ ⊗
multiplets: η (1,4,0), ρ (1,4,+1/3) and χ (1,4,+1).
∼ ∼ ∼
3
η0 ρ+ χ
1 1 −1
     
η ρ0 χ
 1−     −− 
η =  ρ =  χ =  . (3)
     
 η0   ρ+   χ 
 2   2   −2 
     
     
η+  ρ++   χ0 
 2     
     
     
In order to obtain massive charged leptons it is necessary to introduce a (1,10 ,0) Higgs
∗
multiplet, because the lepton mass term transforms as f¯cf (6 10 ). The 6 will leave
L L ∼ A⊕ S A
some leptons massless and some others degenerate. Therefore we will choose the H = 10 .
S
Explicitly
H0 H+ H0 H
1 1 2 2−
 
H+ H++ H+ H0
 1 1 3 3 
H =  . (4)
 H0 H+ H0 H 
 2 3 4 4− 
 
 
 H H0 H H 
 2− 3 4− 2−− 
 
 
If H0 = 0, H0 = 0 the charged leptons get a mass but neutrinos remain massless,
h 3i 6 h 1,2,4i
at least at tree level. In order to avoid mixing among primed and unprimed quarks we
can introduce another multiplet η transforming as η but with different vacuum expectation
′
value (VEV). The corresponding VEVs are the following η = (v,0,0,0), ρ = (0,u,0,0),
h i h i
η = (0,0,v ,0), χ = (0,0,0,w), and H = H0 = v for the decuplet. In this way we
′ ′ 24 3 ′′
h i h i h i h i
have that the symmetry breaking of the SU(4)L U(1)N group down to SU(3)L U(1)N′
⊗ ⊗
is induced by the χ Higgs. The SU(3)L U(1)N′ symmetry is broken down into U(1)em by
⊗
the ρ,η, η and H Higgs.
′
The Yukawa interactions are
1
= G fc f H +F Q¯ u η +F Q¯ u ρ
−LY 2 ab aL bL 1k 1L kR αk αL kR ∗
+F Q¯ d ρ+F Q¯ d η +h Q¯ u η +h Q¯ d η
1′k 1L kR α′k αL kR ∗ 1 1L ′R ′ αi αL ′iR ′∗
+Γ Q¯ J χ+Γ Q¯ j χ +H.c., (5)
1 1L R αi αL iL ∗
where a = e,µ,τ; k = 1,2,3; i = 1,2 and α,β = 2,3. We recall that up to now all fields are
weak eigenstates.
4
The electroweak gauge bosons of this theory consist of a 15 Wi, i = 1,...,15 associated
µ
with SU(4) and a singlet B associated with U(1) .
L µ N
The gauge bosons √2W+ = W1 iW2, √2V = W6 iW7, √2V = W9 iW10,
1− 2−
− − − − − −
√2V = W13 iW14, √2U = W11 iW12 and √2X0 = W4 +iW5 have masses
3− −−
− − − −
g2 g2
2 2 2 2 2 2 2 2
M = (v +u +2v ), M = (v +u +2v ), (6a)
W 4 ′′ V1 4 ′ ′′
g2 g2
2 2 2 2 2 2 2 2
M = (v +w +2v ), M = (v +w +2v ), (6b)
V2 4 ′′ V3 4 ′ ′′
g2 g2
2 2 2 2 2 2 2
M = (v +v ), M = (v +w +4v ). (6c)
X 8 ′ U 4 ′′
The mass matrix for the neutral vector bosons (up to a factor g2/4) in the W3,W8,W15,B
basis is
v2+u2+v 2 1 (v2 u2 v 2) 1 (v2 u2+2v 2) 2tu2
′′ √3 − − ′′ √6 − ′′ −
 
1 (v2 u2 v 2) 1(v2+4v2+u2+v 2) 1 (v2 2v2+u2 2v 2) 2 tu2
 √3 − − ′′ 3 ′ ′′ 3√2 − ′ − ′′ √3 
 
 
 1 (v2 u2+2v 2) 1 (v2 2v2+u2 2v 2) 1(v2 +v2 +u2 +9w2 +4v 2) 2 t(u2 +3w2) 
 √6 − ′′ 3√2 − ′ − ′′ 6 ′ ′′ √6 
 
 
 2tu2 2 tu2 2 t(u2+3w2) 4t2(u2+w2) 
− √3 √6 
 
 
(7)
where t g /g. The matrix in (7) has determinant equal to zero as it must be in order
′
≡
to have a massless photon. There are four neutral bosons: a massless γ and three massive
ones: Z,Z′,Z′′ such that MZ < MZ′ < MZ′′. The lightest one, say Z, corresponds to the
neutral boson of the Standard Model.
The photon field is
1 t 2√6
3 8 15
A = tW W tW +B , (8)
µ (1+4t2)21 µ − √3 µ − 3 µ µ!
with the electric charge defined as
gt g
′
e = = . (9)
| | (1+4t2)21 (1+4t2)12
5
In the following, we will use the approximation v = u = v v v = w v . In this
′′ 1 ′ 2
≡ ≪ ≡
approximation the three nonzero masses are given by [9]
g2
2 2
M (4λ )v , n = 0,1,2; (10)
n ≈ 4 n 2
where
1 1 2nπ +Θ
λ = A+2 A2 +3B 2 cos , (11a)
n
3 3
(cid:20) (cid:16) (cid:17) (cid:18) (cid:19)(cid:21)
3 5 1 1
2 2 2 2 2 2
A = +t + +t a , B = (1+3t ) (3+7t )a , (11b)
4 4 −8 − 4
(cid:18) (cid:19)
3 2A3 +9AB +27C
2 2
C = (1+4t )a , Θ = arccos , (11c)
32 " 2(A2 +3B)32 #
and we have defined a v /v . The respective eigenvectors are
1 2
≡
3 8 15
Z x W +y W +z W +w B, (12)
nµ ≈ n µ n µ n µ n
with
a2 4λ [2(a2 +3)t2 +a2 +1]
n
x = − w , (13a)
n −2t · (a2 1)(2λ a2) n
n
− −
√3
2 2 2 2
y = 32λ 4[8a +3+8(a +1)t ]λ
n 6t(a2 1)(2λ a2) n − n
n
− − h
2 2 2 2
+[5a +9+2(11a +17)t ]a w , (13b)
n
i
2
2 2 2
z = [2λ 2(a +1)t a ]w , (13c)
n n n
−√6t(a2 1) · − −
−
1
2
w = . (13d)
n 1+x2/w2 +y2/w2 +z2/w2
n n n n n n
The hierarchy of the masses is M > M > M . Hence we can identify the eigenvector with
0 2 1
n = 1 as being the neutral vector boson of the Standard Model.
However, we have checked numerically if the respective eigenvalue satisfies the relation
M2/M2 = 1/cos2 , with θ the weak mixing angle. Using a = 0.01 and t = 1.79 [10]
Z W W W
6
we obtain M2/M2 0.97 which does not agree with the value of the Standard Model
Z W ≈
M2/M2 1.30 with sin2θ = 0.2325. In fact, M /M 0.99. This suggests that, in
Z W ≈ W Z W ≈
this model, Z and W are mass degenerate at tree level. Hence, the right value for the
ratio M /M must arise only through radiative corrections. We recall that in models with
Z W
SU(3) U(1) symmetry M /M is bounded from above and the weak mixing angle has
L N Z W
⊗
an upper bound.
For these values of a and t the other two neutral bosons have MZ′/MW 60 and
≈
MZ′′/MW 189. If a < 0.1, all these results depend very weakly on the value chosen for a.
≈
The weak neutral currents have been, up to now, an important test of the Standard
Model. In particular it has been possible to determine the fermion couplings, so far all
experimental data are in agreement with the model. In the present model, the neutral
currents couple to the Z neutral boson as follows
n
g
NC = [ψ¯ γµψ L +ψ¯ γµψ R ]Z (14)
Ln −2c L L ψ R R ψ nµ
W
where c cosθ and
W W
≡
1 1 4
Ln = c x + y + z + w t , (15a)
u1 − W n √3 n √6 n 3 n !
1 1 2
Ln = c x y z w t , (15b)
uα − W n − √3 n − √6 n − 3 n !
2 1 4
Ln = c y + z + w t , (15c)
u′ − W −√3 n √6 n 3 n !
4
Rn = Rn = Rn = c w t (16)
u1 uα u′ −3 W n
for the charge 3/2 quarks, and
1 1 4
Ln = c x + y + z + w t , (17a)
d1 − W − n √3 n √6 n 3 n !
1 1 2
Ln = c x y z w t , (17b)
dα − W − n − √3 n − √6 n − 3 n !
7
2 1 2
Ln = c y z w t , (17c)
d′i − W √3 n − √6 n − 3 n !
2
Rn = Rn = Rn = c w t, (18)
d1 dα d′i 3 W n
for the charge 1/3 quarks. We have checked numerically that only for n = 1 we have (for
−
a given value of t and a < 0.1)
1 1 1 1
L = L = L = L , (19a)
u1 u2 u3 6 u′
and
1 1 1 1 1
L = L = L = L = L . (19b)
d1 d2 d3 6 d′1 d′2
Hence, we can introduce a discrete symmetry, as in Model I of Ref. [4], in order to obtain a
mass matrix which does not mix u with u and d with d , k = 1,2,3 and i = 1,2. That is,
k ′ k ′i
the mass matrices have a tensor product form, next they can be diagonalized with unitary
matrices which are themselves tensor products of unitary matrices. We see that in this
case the GIM mechanism [11] is implemented, at tree level, in the Z ( Z0) couplings. We
1
≡
must stress that, if the new quarks u and d are very heavy, the requirements for natural
′ ′i
(independent of mixing angles) flavor conservation in the neutral currents to order αG [12]
F
break down, and it should be necessary to impose the restriction that the mixing angles
between ordinary and the new heavy quarks must be very small [13].
It is useful to define the coefficient V = (L+R)/2 and A = (L R)/2. In the Standard
−
Model, at tree level, we have VSM = t 2Q sin2θ and ASM = t , where t is the
ψ 3Lψ − ψ W ψ 3Lψ 3Lψ
weak isospin of the fermion ψ. Hence, we have VSM 0.19 and ASM = 0.5 for the charge
U ≈ U
2/3 sector, and VSM 0.345 and ASM = 0.5 for the charge 1/3 sector. In our model,
D ≈ − D − −
also at tree level, using a = 0.01 and t = 1.79 we obtain V 0.19, A 0.5 for u ; and
U U k
≈ ≈
V 0.345, A 0.5 for d . We see that the values are in agreement with the values
D D k
≈ − ≈ −
of the Standard Model. On the other hand Vu′ 0.310, Au′ 0 and Vd′ 0.15, Ad′ 0.
≈ − ≈ i ≈ i ≈
For leptons we have
8
4 4 4 3
Ln = Ln + c w t,Rn = Ln c w t, Ln = L + c w t, Rn = z . (20)
ν u1 3 W n ν − u′ − 3 W n l d1 3 W n l −√6 n
For a = 0.01 and t = 1.79 we obtain V 0.5, A 0.5 and V 0.036, A 0.5
ν ν l l
≈ ≈ ≈ − ≈ −
which are also in agreement with the values of the Standard Model, VSM = ASM = 0.5
ν ν
and VSM 0.035 ASM = 0.5. Notice also that at tree level neutrinos are still massless
l ≈ − l −
but they will get a calculable mass through radiative corrections. In this kind of model
it is possible to implement the Voloshin’s mechanism i.e., in the limit of exact symmetry,
a magnetic moment for the neutrino is allowed, and a mass is forbidden [5]. We will not
discuss this issue here.
Finally, we write down the charged current interactions in terms of the symmetry eigen-
states. In the leptonic sector they are
g
CC = ν¯ γµl W+ +νcγµl V+ +lcγµνcV+ +lcγµl U++ +H.c. (21)
Ll −2 L L µ L L 1µ L L 2µ L L µ
h i
We have also the interaction (g/2)ν¯cγµν X0.
L L
In the quark sector we have
g
CC = u¯ γµd W+ +(u¯ γµd +u¯ γµd )V+
LQ −√2 kL kL µ ′L 1L αL ′iL 1µ
h
+ J¯ γµu +d¯ γµj V+ + J¯ γµu
L 1L αL iL 2µ L ′L
(cid:16) (cid:17) (cid:16)
+ d¯ γµj V+ + J¯ γµd u¯ γµj U++ +H.c., (22)
′iL iL 3µ L 1L − iL iL µ
(cid:17) (cid:16) (cid:17) i
where k = 1,2,3; α = 2.3; i = 1,2. We have also interaction via X0 among primed and
unprimed quarks of the same charge as u¯ γµu and so on.
′L L
ACKNOWLEDGMENTS
We would like to thank the Conselho Nacional de Desenvolvimento Cient´ıfico e Tecno-
l´ogico (CNPq) full (FP) and partial (VP) financial support. We also would like to thank
M.D. Tonasse for useful discussions.
9
REFERENCES
[1] Particle Data Group, Phys. Rev. D45, Part II, (1992).
[2] F. Pisano and V. Pleitez, Phys. Rev. D46, 410(1992); P. H. Frampton, Phys. Rev. Lett.
69, 2889(1992); R. Foot, O. Hernandez, F. Pisano and V. Pleitez, Phys. Rev. D47,
4158(1993).
[3] F. Pisano, V. Pleitez and M.D. Tonasse, “Radiatively induced electron and electron-
neutrino masses” preprint IFT-P.059/93.
[4] J.C. Montero, F. Pisano and V. Pleitez, Phys. Rev. D 47, 2918(1993).
[5] M.B. Voloshin, Sov. J. Nucl. Phys. 48, 512(1988).
[6] V. Pleitez, “ Possibilities beyond 3-3-1 models” preprint IFT-P.010/93 (extended ver-
sion).
[7] The possibility that flavor changing neutral currents processes can discriminate which
generation transforms differently in 3-3-1 models it was recently considered by D.
Dumm, F. Pisano and V. Pleitez, Flavor changing neutral currents in SU(3) U(1)
⊗
models, Preprint IFT-P.026/93 (extended version), and J.T. Liu, Generation non-
universality and flavor changing neutral currents in the 331 model, hep-ph/9312312.
[8] D. Amati et al., Nuovo Cimento. 34, 1732(1964). See the Appendix of Ref. [4].
[9] See for example F. Cajori, An Introduction to the Theory of Equations, Dover Publi-
cations, 1969; p.68-72.
[10] This value for t is chosen in order to obtain the right value for the neutral coupling for
quarks in (15)-(18).
[11] S.L. Glashow, J. Iliopoulos and L. Maiani, Phys. Rev. D2, 1285(1970).
[12] S.L. Glashow and S. Weinberg, Phys. Rev. D 15, 1958(1977).
10