The conflict between natural flavor conservation of Higgs couplings and cabibbo mixing in SU(2)L × U(1)

ANNALS

OF PHYSICS

The Conflict

124, 37-60 (1980)

between Natural and Cabibbo GINO

SEGI&*

Flavor Mixing

Conservation of Higgs Couplings in W(2), x U(1)

AND H. ARTHUR

WELDON+

Department of Physics, University of Pennsylvania, Philadelphia,

Pennsylvania 19104

The general problem of conservation of strangeness and other quark flavors by the exchange of several neutral Higgs mesons is investigated in Sum x U(1). We find that the horizontal symmetries necessary to enforce this conservation conflict with the known Cabibbo mixing. In particular, if the quarks form an irreducible representation of the horizontal symmetry, the mixing angles are all trivial (i.e., 0 or n/2); if they form a reducible representation, it is possible to have some nontrivial mixing angles, but only if there are several unmixed generations of quarks with exactly the same relative pattern of masses and mixings.

1. INTRODUCTION

It is well known that in the original Weinberg-Salam model [l] the direct and induced neutral currents conserve strangeness, charm, and other quark flavors to order olG via the GIM mechanism [2]. The experimentally observed suppression of KLo -+ p+t~- and K+ --f rr + vC, the extreme smallness of the JCL- K, mass difference, and the nonobservation of any charm changing neutral currents all point toward flavor conservation of neutral currents. In a fundamental theory one would like these currents to be conserved naturally, i.e., because of the symmetries built into the theory rather than from cleverly chosen parameters in a given model. Glashow and Weinberg [3] obtained three requirements for natural flavor conservation of neutral currents in any SU(2) x U(1) model: All quarks of the same charge and helicity must (1) transform according to the same irreducible representation of N(2), (2) have the same value of T3, and (3) acquire their mass from the vacuum expectation value of a single neutral Higgs meson. The last condition trivially guarantees that Higgs exchanges conserve flavor because, when there is only one neutral Higgs field coupled to each charge sector, diagonalizing the quark mass matrix also ‘diagonalizes the Yukawa couplings. As Glashow and Weinberg noted, condition (3) might be too strong in that it could be possible to have more than one Higgs field coupled to a particular charge sector and arrange discrete symmetries so judiciously that all the Yukawa couplings are simultaneously diagonalizable. Then neutral Higgs * Supported in part by the U. S. Department of Energy under Contract No. EY-76-C-02-3071. + Supported in part by the National Science Foundation.

37 OOO3-4916/80/010037-24$05.00/O Copyright 0 1980 by AcademicPress, Inc. All rights of reproduction in any form reserved.

38

SEGRti AND WELDON

exchanges would again conserve flavor regardless of the vacuum expectation values. We shall investigate whether this is actually possible. The motivation for introducing discrete symmetries and additional Higgs fields is that one would like some of the quark masses and/or mixing angles to be calculable, not just free parameters [4, 51. We are interested in the standard model with all lefthanded quarks in SU(2) doublets and all right-handed quarks in SU(2) singlets. The only Higgs fields that couple to fermions are SU(2) doublets. Generally the various Yukawa coupling constants are unconstrained and there is no restriction on the fermion masses or the mixing angles which emerge from diagonalizing the mass matrix. To relate these quantities it is necessary to impose some constraints on the Yukawa couplings. When there is only one Higgs field coupled to each charge sector one can only require invariance of the Lagrangian under some group of symmetries which transform the quarks among themselves and take 9 -+ eixq. However, Barbieri, Gatto, and Strocchi [6] have shown that such symmetries will never lead to relations for the mixing angles in SU(2), x U(1). One must therefore have more than one Higgs doublet coupled to quarks of the same charge. The problem that arises is whether one can allow several neutral Higgs fields to couple to quarks of the same charge and still have their exchange conserve all quark flavors naturally. A partial answer to this question was first provided by Gatto, Morchio, and Strocchi [7], who noted that with natural flavor conservation the generalized Cabibbo angles are uniquely determined by the symmetry group that is chosen and by the representations employed. They have recently extended their results, obtaining necessary and sufficient conditions on the quark and Higgs representations for natural flavor conservation and have shown that it may not be possible to obtain physically acceptable mixing angles [8,9]. Our results, which overlap with the above, are that either all the mixing angles are trivial (i.e., 0, n/2) or the quarks decouple into unmixed generations that are copies of each other. Having trivial mixing angles is physically unacceptable. The alternative possibility allows for arbitrary mixing angles but is rather unattractive. This can best be seen by displaying an example with, e.g., eight quarks. The charged current turns out to be

J-” = [ii, C, ii’, f’lL y“

1 cos 0 -sin 8 sine cos e o

o

0

0

0

0

0 0 cos e -sin e sin e cos e

d it s’ L,

with mass relations md/ms = rn,~lrn,~;

mu/m, = m,fl/m,~.

One may, of course, have more quarks in each generation and also more generations but the same features remain: All mixing angles between generations vanish; corresponding mass ratios are identical; mixing angles within each generation are identical and undetermined by any symmetries [IO].

39

FLAVOR CONSERVATION OF HIGGS COUPLINGS

We therefore conclude that in an X7(2), x U(1) model either one must abandon the hope of obtaining calculable mixing angles or accept some flavor nonconserving Higgs exchanges. We have dealt with the problem at some length because of the stringent limits imposed on strangeness nonconserving neutral currents by the smallness of the KL - K, mass difference and the relative agreement of its magnitude with calculations excluding Higgs exchanges [I 11. The case in which one imposes strangeness conservation on Higgs couplings, but not total flavor conservation, is dealt with in Appendix B. The paper is organized as follows: Section 2 is a general presentation of the problem in which we establish methods and notation. Sections 3 and 4 deal respectively with the proofs of our contentions if the quarks belong to irreducible and reducible representations of the horizontal symmetries. Examples are given in both sections. Appendix A includes some more technical details on reducible representations.

2. FORMULATION

OF THE PROBLEM

2.1. General notation

We work in SU(2), x U(1) with N quark doublets #nL = ($, general Yukawa coupling may be written

a = 1, 2,..., N. The

where all right-handed quarks are SU(2) singlets and );)&=: ia,x,+. The two multiplets of scalar fields, vpoland xa , may be entirely disjoint or may have fields in common. We can omit the indices on naL and think of nL as a column vector in flavor space; similarly for nR , pL , and pR . In order to constrain the Yukawa couplings it is necessary to impose a set of symmetries {Ki] under which, for example, nL+ KLinL,

nR---f KRinR,

9)01- %po -

These symmetries, labeled by i, may be either discrete or continuous. We shall refer to them as horizontal symmetries in contrast to the gauge group, which acts vertically. The invariance of JZyua means that (2.2)

for all transformations in the symmetry group. The positive quarks will also transform under the symmetry group. Since pL is in a doublet with nL , it must undergo the same transformation: PL -

XL
*

The symmetries of pR are not similarly constrained so that

40 Invariance of 9&

SEGRh

AND

WELDON

requires

Note that the sets (F,) and {FL> are in no way equivalent and generally do not contain even the same number of matrices. The sets {KRi} and {K:} are generally inequivalent, but represent the same symmetry group. If the Yukawa couplings are to conserve strangeness, charm, and other flavors, they must be simultaneously diagonalizable by unitary matrices U, , U, , UL , Vi which satisfy (2.4)

for every 01,where pW and pi are all diagonal. Obviously these are extremely strong demands on r, and will only be satisfied if the symmetries in (2.2) and (2.3) have been judiciously chosen. 2.2. The Cabibbo Matrix

Denote the physical (i.e., mass eigenstate) quarks by daL and uaL , a = 1,2,..., N. These are linear combinations of naL and paL given by dL = ULinL , UL

=

(2.5)

U&L.

The charged current coupled to W bosons is

in terms of the original fields. Rewriting this in terms of the physical, mass eigenstate, quarks gives J-u = iGLyuAab dbL ,

(2.6)

where the generalized Cabibbo matrix is given by A = l$U,.

(2.7)

It is this matrix A with which we will be most concerned. Let us transform the horizontal symmetries KLi to the physical quark basis (2.5) by defining s’ r ULtKLiuL ) Ti = ULtKLilJ; so that dL - SidL ,

uL ---, Tiur.

(2.8)

FLAVOR

CONSERVATION

for the Q = 2/3, Q = --l/3 Cabibbo matrix we have

OF HIGGS

41

COUPLINGS

quarks, respectively. From the definition AtpA

(2.7) of the

= si.

(2.9)

This has nothing to do with whether the Yukawa couplings conserve all flavors: It merely states that the Cabibbo matrix is a change of basis from the physical Q = 2/3 quarks to the physical Q = -l/3 quarks. 2.3. The Constraint

of Flavor

Conservation

When the Yukawa couplings conserve all flavors, si and Ti are such that the Cabibbo matrix A is severely constrained. To see this, first use (2.4) to write (2.2) in the physical quark basis: (CJL+K;+iJL) fm( UR+KRilJR)

Multiplying

= @,*p,,

(2.10)

.

this by the vacuum expectation values

of the neutral Higgs fields gives

where ii?i is the diagonal mass matrix. To eliminate equation by its own adjoint and obtain (ULtK;+UL)

i@i@+(UL+KLiU,)

U, and KR we may multiply

= [&5@,*9;,]

p,,l-“+.

this

(2.11)

This is of the form (P)+ @iI@(?P) = diagonal.

(2.12)

Now focus on a particular Si. Regardless of the form of ii?ii?P we know that each column of Si is an eigenvector of &?A@ because the right-hand side is diagonal. But since J?Zfi+ is already diagonal, its eigenvectors are just

u1=

1 0o

)

rg

=

0 1 o

,

0 0 z'3 =

1

3.'.

.

(2.13)

These, of course, represent the physical quarks d, s, b, etc., in some order. We shall assume that there are no mass degeneracies among quarks of the same charge. The eigenvectors (2.13) are then unique, i.e., no linear combination of them is an eigen-

42

SEGRi? AND WELDON

vector, and the columns of a particular matrix Si are just the column vectors (2.13) possibly arranged in a different order and multiplied by different phases. We have thus shown that the matrix Si has exactly one nonzero entry in each column and exactly one nonzero entry in each row. Such a matrix is called monomial. Because Si is unitary, each nonzero entry is a complex number of the form I?“. These matrices Si just permute the quark vectors (2.13): Sill, = eiVb , where the phase cy.= a(i, a) depends on i and a. These unitary, monomial matrices are slight generalizations of the ordinary permutation matrices, whose nonzero entries are always + 1 [12]. If we now refer back to (2. IO), the reason for this simplification is clear: All the Yukawa couplings fb, are diagonal in the physical quark basis and since the horizontal symmetries transform these diagonal couplings among themselves as in (2.10) they must be represented in this basis as generalized permutation matrices. We can apply the same argument to the horizontal symmetries (2.8) of the positive quarks and conclude that the Ti are also unitary, monomial. For the examples of Sections 3 and 4 one may check these results by direct computation. Let us now digress to the special case of one Higgs field. Then (2.11) reduces to (si)+ imi?f’(S~ = ii?lA+. Since @ is diagonal and, by assumption, nondegenerate, all the Si and similarly the Ti have to be diagonal. From (2.9) the Cabibbo matrix A therefore transforms diagonals into diagonals; this requires the mixing angles to be either 0,7~/2, or, if there is degeneracy among the diagonal entries of Si, undetermined. This is the argument of Ref. [6]. In the general case, (2.9) still applies and the question of what the mixing angles are is reduced to the following: Given two equivalent representations of a symmetry group by sets of unitary, monomial matrices (Si} and {P) what can we say about the matrix A which accomplishes the equivalence? This is an unusual question in that ordinarily one only asks if two representations are equivalent and not what is the matrix which accomplishes the equivalence. The answer to this question depends on whether the quarks are in irreducible or reducible representations of the horizontal symmetry and is provided in Sections 3 and 4, respectively.

3. QUARKS IN IRREDUCIBLE

REPRESENTATIONS

In this section we will discuss the case in which both left- and right-handed quarks are in irreducible representations of the horizontal symmetry. Before we proceed to the proof that the mixing angles are trivial (0 or 77/2) if the neutral Higgs couplings conserve all quark flavors, we shall display some simple examples.

FLAVOR

3.1. Four-Quark

CONSERVATION

OF

HIGGS

43

COUPLINGS

Example

In the first example the horizontal group is the symmetric group S3. This is the group of all permutations of three objects, e.g., X, , xb , x, . There are three irreducible representations: a trivial one-dimensional representation 1 ,u, = 311” (s, + 3-b + x,)

(3.1)

which is invariant under all six permutations; a nontrivial one-dimensional representation under which the even permutations are represented by fl and the odd permutations by -- 1; and a two-dimensional representation 1 Xl

=

21/2

6%

-

%A

(3.2)

1 52 = 6112(x, + xt, - 2x,), which transforms as u cos a u sin u I -sin a: cos n

I I Xl

x,

(3.3)

’

where (\: = 0,2~/3, &r/3, and 0 = +l. We assign all the quarks to the two-dimensional representation, i.e., #L - 2, - 2. Then the discrete symmetries K Li, KRi, K: are all the same and equal nR "2>PR to the six matrices (3.3). The Higgs fields are (pl , C& - 2, x0 - 1, xh - 1’; by imposing an additional discrete symmetry such that #L couples to its via y fields and to pR via x fields, the most general SU(2)L X U( 1) X s3 invariant coupling is 2

Yuk

=

.f[6fbbR +d$lPlR

+

$'hR) +

$dhR)iO

ql

+

($,%R +

-

h($lp,R

ihR> -

d &PlR))7;,

.

(3.4)

We may write the Yukawa coupling of $L to nR in matrix notation as

The couplings of CJ+and ?Z may be simultaneously

diagonalized: (3.6)

where (3.7) Thus there are no strangeness changing neutral Higgs couplings.

44

SEGRk

AND

WELDON

This example was first noted by Pakvasa and Sugawara [13]. However, we have arranged different Yukawa couplings and S, assignments for thep, in order that there be no charm changing neutral currents (see Appendix B for the original version): (3.8) These couplings may also be diagonalized:

where U’=U’-1 L

R

21/”

I11i -i I*

(3.10)

The diagonal couplings (3.6) and (3.9) provide simple examples of the general requirement (2.4) for flavor conservation. One may easily verify that the matrices Si and Ti in (2.8) obtained by transforming the six matrices in (3.3) to the physical quark basis by (3.7) and (3.10) are indeed monomial as argued above. The problem with this model is that the Cabibbo matrix is trivial: A = U;‘U,

=

Io1 01 I.

(3.11)

3.2. Six-Quark Example The previous model was so simple that the vanishing of the Cabibbo angle might appear to be a fluke. We now present a six-quark model illustrating the same phenomena based on the symmetric group S, . The 1, l’, and 2 representations of S, are also representations of S, . There are, in addition, two representations that are three dimensional. One of these is the obvious generalization of (3.2) with basis Xl

=

$&a

-

x*)9

I

x3

=&ip

(3.12) (x, + .xb + x, - 3x,).

This is the 3, and the 4! = 24 matrices of the group may be straightforwardly determined by permutting a, b, c. There is also a 3’ representation. The explicit matrices for the 3’ are displayed in Hammermesh [14]. The following assignments lead to flavor conserving Higgs exchanges: $L N 3, nx - 3~PR N 3’. The Higgs coupled to nR are (yr , cpZ)N 2 and y0 N 1; those coupled 1’. (It makes no difference whether (xl , xZ) are the to PR are (xl , x2) 2 and x -

FLAVOR

CONSERVATION

OF

HIGGS

45

COUPLINGS

same or different than (vl, yZ)). The general Yukawa coupling of & to nR is, using the matrix notation (3.5),

r a%

=

7%

-2q

0

0

These couplings may all be diagonalized: 00 U,T&U,

0

-2f

0

0 I +9J2 I 0

f

0

= q+ 0 -3112f

0

3112f

0

0

by

0

3112 3112 (3.15)

(Of course, the inclusion of v0 is trivial but it allows one to fit the d, s, b masses.) The Yukawa coupling of $L to pR is

Note that the coupling of (xl, xZ) differs from that of (yr, p2) in (3.13) because ff 3 btltp, - 3’. Amazingly, we may diagonalize (3.16):

nR

l$r;f,

u;, =

with

Again the generalized Cabibbo matrix is trivial: A = U;tiJ,

=

1 0 0 1 0 . 0 0 1

I I 0

(3.18)

3.3. The General Case

The simplest possible transformation property of the N quark they form a single, irreducible representation of the horizontal in the previous examples. In such case the following applies:

doublets symmetry

$L is that group, as

46

SEGRI? AND WELDON LEMMA.

If {Si} and {Ti} are two sets of monomial matrices each of whichforms a unitarity, irreducible representation of a symmetry group and they are equivalent via a unitary matrix A, ASi = TiA, (3.19) then A is also monomial.

Since A is the 0 or 42. It is essential example, take Si choose Ti = si;

Cabibbo matrix this means that all the mixing angles are necessarily that the representations be irreducible for this result to hold. For to be the six 3 x 3 permutation matrices (i.e., all entries 0 or 1) and then Ap = pA

is satisfied by (3.20)

which is certainly not monomial. The loophole is, of course, that the permutation matrices are reducible since they all leave invariant the vector v=

1 1. 1

II

To prove the lemma we will rely on the irreducibility to show that A transforms an arbitrary diagonal matrix into a diagonal matrix. As a first step let us take the (a, d) matrix element of (3.19): 5 A&& b=l

= 2 T;,A,,

.

C==l

For a fixed a, d, and i the monomial one c contribute:

property of Si and Ti implies that only one b and

A&&

= T&A,, .

(Of course, b depends on d and i, c depends on a and i.) We therefore have that 1f&b 1 . 1& 1 = 1 Tic 1 * 1& If we define new matrices with all phases removed by 6%9mn = I sim I>

G%m = I Ttzn1, &)m, = I 4nn I,

1.

(3.21)

FLAVOR

CONSERVATION

OF HIGGS

47

COUPLINGS

then S,,i and Toi are just permutation matrices. For example, a particular Si is diagonal if and only if the corresponding Soi = I. (More generally the number of nonzero diagonal entries of a particular Si is given by Tr(S,,i).) Because of (3.21) these matrices satisfy [ 151 A,S,i = T,iA, .

(3.22)

Thus if Soi = I, Tt = I. This means that every diagonal Si is transformed into a Ti which is also diagonal. (More generally, Tr(S,i) = Tr(Toi).) From this result we can immediately show that A is monomial in the special cases N = 2 or 3 (i.e., four quarks or six quarks). Here we use the fact that if the Si form an irreducible representation, then for any a # b there must exist some matrix S with Sh # Stb but neither equal to zero [16]. Any 2 x 2 monomial representation that is irreducible must therefore have at least one matrix Si that is diagonal and nondegenerate. Because of (3.22) the corresponding Ti is diagonal and hence A must be monomial. A similar argument applies to 3 x 3 representations because any matrix which has two nonzero diagonal entries must be completely diagonal because of the monomial property. (To obtain a diagonal matrix with no degeneracies one need only add two diagonal matrices Si and Sj with S$ # St, and Si, # Si, .) The same argument does not apply for N 3 4 because a matrix with two nonzero entries satisfying S& # Si, need have no other diagonal entries at all. (In fact, there are irreducible monomial representations whose only diagonal matrices are multiples of the identity matrix [17].) For the general case it is very convenient to use the fact that there always exists a permutation matrix P satisfying the same relation (3.22) as A, does [18], viz., PSoi = TtP.

(3.23)

Restoring the phases to Ti gives a new monomial s'i

=

matrix (3.24)

PtTip 2

which has exactly the same pattern of nonzero entries as Si and is related to Si by (PtA) Si = S’i(PtA)

(3.25)

because of (3.19). For example, if Si, # 0 then S;i # 0 and the (1,2) matrix element of (3.25) gives (P+A)ll S;, = S;;(P+A)z,

.

(3.26)

Next we use the fact that any N x N matrix can be expressed as a linear combination of the irreducible set {S}. In particular, we can choose coefficients ci such that

595/124/I-4

48

SEGRi? AND WELDON

is a diagonal matrix with nondegenerate eigenvalues. The vanishing of an off-diagonal element, e.g., D,, = 0, means that

This is a nontrivial constraint only for those matrices among {P} for which Si, # 0. But for these, S;i # 0 and (3.26) gives

Applying the same argument to the other off-diagonal elements shows that D’ = c +yi 2 is completely diagonal. Therefore the relation (P+A) D = D’(P+A)

requires that P+A be monomial since it transforms an arbitrary diagonal matrix into a diagonal matrix. Because P is just a permutation matrix, this already shows that A must be monomial. In fact we can be a bit more precise about the determination of A by returning to (3.25). There Si and Sfi have the same pattern of nonzero entries and since, P+A is monomial we may conclude that P+A is actually diagonal. This determines A completely.

4. QUARKS IN REDUCIBLE REPRESENTATIONS

It is not surprising that when quarks are assigned to reducible representations of the horizontal symmetry in such a way that the Higgs couplings conserve all flavors, the previous result will frequently apply so that all mixing angles will be trivial. We first present an example in which the mixing angles are not trivial, however, and then show that this is the only real exception. 4.1. Eight-Quark Example

We now present an example in which not all the mixing angles vanish. We again use the symmetric group S, and essentially double the example of Section 3.1 by letting #L transform as the sum of two irreducible representations, #L N 2 @ 2. Similarly, nR -2@2andp,w 2 @ 2. As in Section 3.1 we let $,n, couple to a pair of H&s, CR , ~2) N 2. This just doubles (3.5) to give 0 a 0 b r&h

=

Tl

0” 8;

I:+%

c 0 d 0

0 Z -a 0 E -c

b 0 0 -b d 0’ 0 -d

(4.1)

FLAVOR

CONSERVATION

OF HIGGS

49

COUPLINGS

Obviously by using the same U, and ZJ, as (3.7) we may obtain

where Z and T3 are 2 x 2 diagonal. We can easily diagonalize these matrices by cos w -sin w

2x

x2+

::$I= I: :I

(4.3)

for some angles w and 4. Thus (4.4)

is completely diagonal with ULcos w -UU,sin w UL sin w UL cos w

Y/L =

(4.5)

and similarly for gR . Note that if we label the quark states as (a, d’, s, s’), then the mass matrix resulting from (4.4) gives rnd/rndr = m,lm,p .

(4.6)

The obvious choice of Higgs coupled to $LpR is x0 - 1, x - 1’ as in Section 3.1. We first show that this will not conserve all positive quark flavors and will then show what choice does succeed. A straightforward doubling of (3.8) gives

ci,

a’ 0 b’ 0 a’ 0 = fll c’Od’O+i 0 c’ 0

0 -a” 0 - (.”

0 b’ d’

a” 0 0 -b” C” 0 0 -d”

b” 0 d”’ 0

(4.7)

Using the UL and lJA of (3.10) yields

But now we cannot simultaneously a’ c’

I

b’ d’

diagonalize

I

and

a”

Ic”

b”

I

d” *

(4.9)

(Note that since x0 and x are different irreducible representations of S, , no symmetry can relate the singly primed couplings to the doubly primed.)

50

SEGRk

AND

WELDGN

To conserve flavor the Higgs coupled to $LpR must be in a single representation of S, . Let us abandon (4.7) and instead use (x1 , xZ) - 2 so that 0 CR

=

a’

0

b’

’

0

b’

0

(4.10)

21

c’

0

d’

0

0 -c’

0 -d’

Note that (x1, xZ) may be the same or different than (vl , qz) in (4.1). Obviously

We may now diagonalize this, in contrast to (4.8), by ~, =

UL cos w’ -UL sin w’ UL sin w’ UL cos w’

L

and similarly for gR . With quarks labeled as (u, u’, c, c’) we obviously obtain m,lm u’ = m,/m,n in analogy to (4.6). The general Cabibbo matrix is ICOS e -Zsin Isine k0se

A = 9;f?ii!!, =

e

(4.12)

’

where B = w - w’. To recognize the implications of this we may write out the charged current (2.6) and choose quark names appropriately to obtain cog e 0 -sin e 0 0 ~0~ e 0 -sin J-” = [II, ii’,‘, T, L-‘IL ye sin e o cos e 0 0 sin e 0 c0se

Thus the primed and unprimed explicit by reordering:

e

sin =

[i&

T,

U’,

E’]L

d d’ s

.

d L

generations do not mix. We may make this more 1cos e -sin e

J-”

0

y”

e

cos

o

o

0

0

0

0

0

0

case sin

d j,

-sine

e

Note that the Cabibbo angle is nonzero but undetermined.

cos

e

s’ L

.

(4.13)

FLAVOR

CONSERVATION

OF

HIGGS

51

COUPLINGS

4.2. Genernl Results

Among all the reducible representations of the horizontal symmetry group, we are interested only in those which allow the Yukawa couplings to be simultaneously diagonalizable as in (2.4). In Appendix A we investigate the various possible decompositions of n, and nR into irreducible representations and shall quote the results here. We reject as unrealistic any assignment in which some of the quarks have no Yukawa couplings at all, for there is then an unbroken chiral symmetry which keeps these quarks massless to all orders. The first result is that if all r, are diagonalizable, then nL. and IZ~ must decompose into the same number of representations: for example, into two each so that its = #’ @ ni2’ and nR = PZ~)@ @. The next result shown in Appendix A is that the representations II~‘), ni2), nf), $’ must be either all different or all the same for the r, to be diagonalizable. This gives us only two cases to consider. Case (i). The representations np’, ni2), ng), L$!’ are all different. Then, as shown in Appendix A, (2.4) forces I’, to be block diagonal:

(4.14) For there to be no massless quarks we require y(l) and yc2) to be square. Since the left-hand quarks are in SU(2) doublets, p;” must be the same representation as @) and pi2’ must be the same representation as ni2’. Thus again pF’, pi2’, pf’, pg’ must all be different representations and therefore I’: has exactly the same diagonal form as (4.14) with blocks of the same size:

ixdh

f(l) = W, &?I a’r,o

0 ctly;(*)

Pi;

/I pR I.

(4.15)

The first generation of quarks, n(l) and p(l), does not mix with the second generation, PZ@)and ~(~1. The case is thus reduced to that of Section 3 because the discrete symmetries Si and Sfi must be block-diagonal monomial matrices. The Cabibbo matrix must be block diagonal and hence, by our lemma, each block on the diagonal must be monomial. The mixing angles are again 0 or r/2. Case (ii).

If all the representations are the same, K$’ = PZ~’ = ny’ = ng’, then

where yoi is &N x 4N if I’, is N x N. Because the left-handed quarks are all in SU(2) doublets this means $) =pL” = pL2’ = pg’ = pjp2). Of course, the positives and negatives may couple to Higgs in different representations so the Clebsch-Gordon matrices yII may differ: (4.17)

52

SEGRiS AND WELDON

For neutral Higgs exchanges to conserve flavor it is necessary that there exist unitary matrices U, and U, such that (4.18)

UL+YJJR = ?a

where yo are all diagonal. We may separately diagonalize the coupling constants as in (4.3) and then find that

is’ entirely diagonal with 4YL =

UL cos w I -UULsinw

UL sin w ULcoso .

The Yukawa couplings of the positive quarks, I’:,

are similarly diagonalized by

u;cos WI Ui sin w’ - UL sin 01’ Ui cos w’ ’

f@!;=

(4.20)

so that the Cabibbo matrix is A = W$&

=

u$Y, cos 0 - U;fUL sin e U;tUL sin 0 U;tUL cos 0 I’

(4.21)

where 8 = o - w’. Since U, and Ui each diagonalize the Yukawa couplings of irreducible representations, we may again use Lemma (3.19) to conclude that their product ULtUL is some monomial matrix. To recognize the implication of the N x N Cabibbo matrix (4.21) it is convenient to put N = 2k. Then ULtUL is a k x k monomial matrix. By reordering columns and absorbing phases we can arrange that U;fUL = IkXk.

Thus the 4k quarks separate into k generations of four each (like u, c, d, s) whose mixing among themselves is always given by sin 8. This mixing angle depends on the values of a, b, c, etc., but not on the symmetry. (This is the argument of Ref. [6] again.) Quarks of different generations do not mix. Effectively all that has happened is that a single combination of Higgs fields couples to each generation of quarks so that flavor conservation is trivial. This is just the Glashow-Weinberg result [3] applied k times. The only way to have just one generation (i.e., k = 1) is to have just one Higgs field to begin with.

5. CONCLUSIONS We set out to examine if we could construct an SU(2), x U(1) gauge model of weak and electromagnetic interactions in which the Higgs meson couplings, which

FLAVOR CONSERVATION OF HIGGS COUPLINGS

53

lead to quark masses and mixings, are both constrained by discrete symmetries so as to give us nontrivial mixing angles and be flavor diagonal in the neutral sector. The two requirements clash, as we have seen, and we must now face the alternatives. The obvious possibility is the unattractive one that the various mixing angles are arbitrary parameters, which we cannot determine. A second possibility is to embed SU(2), x U(1) in a larger gauge group which restricts the parameters: much work has been done along this line, particularly with regard to unified models of weak, strong, and electromagnetic interactions. A third possibility is to limit the restrictions on our models to conservation of strangeness and .other n-quark-type flavors, but allow violation of charm and p-quark flavors in the neutral Higgs couplings. In this case we are no longer able to obtain any general results and are back to proceeding by trial and error. Some examples of models of this type are given in Appendix B. Finally, rather than forbid flavor violation in neutral Higgs couplings, we might consider being satisfied with suppressing the violation by making the Higgs mesons very heavy. Typically the range of masses required by the limits on strangemess conservation is 100 GeV to I TeV, depending on the couplings. This is thought to invalidate the simple perturbative features of the weak interactions [19], though a recent proposal by Georgi and Nanopolous may avoid these objections. [20]

APPENDIX

A: FLAVOR

CONSERVATION

FOR REDUCIBLE

REPRESENTATIONS

Let us now investigate when it is possible to relax the condition that all the quarks of the same charge and helicity be in a single irreducible representation of the horizontal symmetry. We shall find that if the quarks are in reducible representations the Yukawa couplings can be simultaneously diagonalized only in some rather trivial cases. In several of these cases, some of the quarks have no Yukawa couplings at all. Such quarks respect an exact chiral symmetry and remain massless to all orders so that we do not consider them further. There are two cases in which the Yukawa couplings can be diagonalized without some of the quarks decoupling; the resulting pattern of mixing angles is discussed in Section 4.2. For definiteness we focus on the Yukawa couplings of the nL and nR quarks. Suppose that the left-handed quarks transform as the sum of two irreducible representations nil) and @’ . Let nR decompose into two representations &?’ and @‘. (Our argument may be easily generalized to the case of more representations or of unequal numbers of representations.) The general Yukawa coupling to qa is iLronR

=

[i$',

ii?']

(1) aA, bB, nR CC, dD, II .,) I3

(AlI

where a, b, c, d are arbitrary coupling constants and the matrices A, B, C. D are Clebsch-Gordon coefficients that couple nil’ x @ to v.’ np’ x $” to qa, etc. We shall suppose that the scalar fields vcI form a single irreducible representation since adding more representations will always make it more difficult to conserve flavor.

54

SEGRi?

AND

WELDON

We must now distinguish whether any of the quark representations are the same. There are four possibilities: (0 (ii) (iii) (iv>

n;“, n;” n;” nil’

np’, nr’, ng’ are all drtferent; = n;“’ = ng’ = ng), i.e., A, = B, = C, = D, ; * = ni,“’ but ng’ # nf’, i.e., A, = C, , B, = D 013 # np’ but ng) = nf), i.e., A, = B, , C, = De.

Because the ensuing argument is rather tedious, we shall first quote the results, viz,, when it is possible to diagonalize I’, . For case (i) this is only possible if b = c = 0 so that r, is block diagonal (or equivalently a = d = 0). For case (ii) there is no restriction. For cases (iii) and (iv) it is not possible to diagonalize r, at all. We also examine the possibility that nL forms two representations but nR forms only one and find that the only such r, which might be diagonalizable have the form

where A has twice as many columns as rows. Obviously if r, is N x N then only +N of the rows are linearly independent. Therefore +N of the quarks have no Yukawa couplings and remain massless. A similar objection results if nR forms two representations but nL forms only one. We will now demonstrate that it is impossible to diagonalize the reducible r, in (Al) except in the above trivial cases. Let U, and U, be unitary matrices such that U,+rJJ, is diagonal for all (Y.Then U,+F,I’,+U, must be diagonal for all 01and /3. All the following commutators must therefore vanish: [vi+,

cc+1

= 0,

(44

where r r += 01 B

I a I2 A&+

+ I b I2 B&Y+

ciiC,A,+ + dtiD,B,+

aEA&+ t biiB,D,+ 1c I2 C,C,+ + j d I2 D,D,+

’

(A3)

Note that the matrix element (A,),, is the Clebsch-Gordon coefficient for coupling the rth quark in n;” and the 8th quark in n 2’ to r The indices r, s, a are analogous to J, for the rotation group in that they label the* states within each multiplet. The orthogonality relation for the Clebsch-Gordon coefficients of an irreducible representation states that

or in matrix notation c A,A,+ = I,

and similarly for B, C, D.

(A4)

FLAVOR

CONSERVATION

OF HIGGS

COUPLINGS

55

Let us now specialize to case (i) in which all the representations are different. If in (A3) we set /3 = (Yand sum then the off-diagonal blocks vanish by orthogonality of the representations: c A&‘: a

= c B&D,+ = 0. m

We thus obtain (I a I2 + I b I”V o

(I c I2 -P I d l”V .

The requirement that w-9

then leads to (I a I2 + 1b I2 - I c I2 -

( d j2)(aEA,C,,+ + bilB,D,,+) = 0.

(A7)

Because all four representations are different it is not possible to impose any discrete symmetries that relate the coupling constants. Thus aZA,C,+ + baBuD,+ = 0.

Because of the orthogonality

(A@

relations c A,+B, = 1 ButA, = 0 CL LI

we may separate (AS) into aE = 0,

bZ = 0.

(A9)

Obviously at least two of the coupling constants must be zero, but we may go a bit farther by observing that [r,+r,

because U,+r,+I’,U, (A2). We find

, r,+r,l

= 0

must also be diagonal. Note that this is not justithe adjoint of

c r,+r, = / a

(I a I2 + I c I”)Z o

(I b I2i” I d I”)1 ’

which leads to

(I a I2- Ibla+lc12-

I d 12)(ifbA,+B, + ZdC,+D,) = 0.

From this we obtain iib = 0,

Ed==.

(AI01

56

SEGRB

AND

WELDON

There are two solutions to (A9) and (Al 1): Either b = c = 0 with a and d unrestricted or a = d = 0 with b and c unrestrcted. This is the result for case (i). We can also use this machinery to demonstrate what happens if the left- and righthanded quarks are grouped into an unequal number of representations. Suppose, for example, that nR forms a single irreducible representation but nL still forms two: np # IZL (2). Then 6412)

where the matrices A and C have more columns than rows to account for r being square. (This is very different than putting b = d = 0 in (Al).) If we repeat the same calculations then (A7) becomes (1 a (2 - / c I”) &A$“+

= 0.

It is not possible to impose 1a j = / c 1 because @ # nl”‘. Thus a = 0 or c = 0. In either case an entire multiplet of right-handed quarks has no Yukawa couplings at all and must therefore remain massless. For case (ii) we have already said that it may be possible to diagonalize r, . We therefore proceed to case (iii) and ask whether this is possible when #’ = ni2’ but ny’ # ng’, i.e., aA, CA,

r, =

bB, dBm ’

W3)

Setting A = C and B = D in (A3) gives

(I a I2+ I b l”Y ca raru+ = /

@ + @I

(cc + di;)l

(1 c I2 + 1d I”)1 1.

(-414)

On the other hand, 0 c c+c = ) (I a I2 +o I c 12Y (I b I2 + I d 12V’ a

which is identical to (AlO) and leads to (I a I2 -

1b I2 + I c I2 -

I d j2)(iibA,+Bv + CdA,+B,) = 0.

Discrete symmetries may relate a to c and b to d but nothing more. Hence iib + Zd = 0.

The commutator

of (A14) is more complicated (la I2 -

but leads to

1c I”) bil = (I b I2 (ab)(ciE) = @@(cd).

I d Is) at?,

(A13

FLAVOR CONSERVATION

57

OF HIGGS COUPLINGS

The only solutions are b = c = 0 with a and d unrestricted or a = d = 0 with b and c unrestricted. (It is not possible, for example, to arrange that a = c but b = -d.) These two possibilities are the same as resulted from case (i). For completeness we should consider the subcase in which @ = @’ still but nR forms a single irreducible representation. Then

which is analogous to (A12). The commutators here do not lead to any constraints on a and c. However, as already noted, such a Yukawa coupling yields $N massless quarks.

APPENDIX

B: PARTIAL

FLAVORCONSERVATION

The experimental limits on strangeness changing neutral couplings are much more stringent than those on charm changing neutral couplings. This prompts us to investigate relaxing our previous restriction that all neutral Higgs meson couplings preserve flavor. The obvious manner is to require flavor conservation only for the neutral Higgs couplings of Q = -6 quarks. We are then unfortunately no longer able to prove any general theorems regarding the Cabibbo matrix A and are reduced to the usual groping for an acceptable model. We present two examples in which the Cabibbo angle 0, is expressed in terms of quark mass ratios; to be more specific tan ec = md/ms. The first of these models was proposed by Pakvasa and Sugawara [13], while the second is new [21]. Model I

Following the discussion of Section 3.1, we have as horizontal symmetry group S, . The Higgs fields are {dl, I&} - 2, #L - 2, nR - 2, but rather than pR - 2 as in Section 3.1, we have pR - 1 @ 1 and also no x fields. The SU(2) x U(1) x S, invariant Yukawa coupling is now

where uRo and cRo are the two right-handed, charge Q, S, singlet fields. Only one linear combination of uRo and cRo appears in LZyuk and we associate it with the field cR . Diagonalization of the charge 6 quark mass matrix is then immediate with CR

=

~;“++gf-;;:

,

c

L

=

t’ZP2L

w%+ (Y12

+

Y22)l/2

*

58

SEGRh

AND

WELDON

The orthogonal combination of uRo and cRodoes not appear in gyUk, so the u quark remains massless in this model. The couplings of I,!& to nR via &2 are the same as in Section 3.1 and hence are diagonalized by the matrix U, given in (3.7): “=@

1

1 I -i

I i I*

The analogous matrix for the p quarks depends on the vacuum expectation value Q!

of

41.2

:

ui=(12’1 12 :, v2jy /-:f :;1. We have, with 1v1 i iu, j/21j2 = P~,~, after absorbing phases into the definitions of the quark fields

A=

(p12

/;; -;1”1.

:p22)‘i2

So tan tIC = p2/pI . From (3.4) and (3.6) one sees immediately mJm, . The Higgs couplings are not, however, simultaneously are charm changing Higgs mediated neutral currents.

that this ratio equals diagonalized so there

Model 2 This model has six quarks placed in three left-handed doublets #1,2,3L and six righthanded singlets. There are three Higgs fields $1,2,3 and we restrict the Yukawa couplings by imposing S, permutation symmetry and an additional discrete symmetry of the form (with 01~= e@b) 4 kj”Ik

?Jk,

P)k +

akvk,

nkR

+

cEkj2

nkR

>

PkR

-+PkR

,

where the ok are arbitrary, but different, phases. The invariant coupling is

This type of model, though not this specific example, was discussed in [4], so we do not repeat the details of the Higgs potential, etc. The n quark Higgs coupling is obviously diagonal so there are no strangeness changing neutral currents. The n quark masses are m&&b =S I u1.2.8 1so the ratios 1aI/v3 1 , ] v2/v3 I are fixed:

FLAVOR CONSERVATION OF HIGGS COUPLINGS

59

Since U, is the unit matrix (up to phases), the Cabibbo matrix A is equal to Ul’, the matrix which transforms the p quarks to the physical basis. The p quark mass matrix is

with x = f l/g’. We see that the three p quark masses and the three mixing angles are given in terms of the two known quantities 6,,, and two new parameters x and g’ (or f’ and g’). We proceed as usual by diagonalizing M,M,+ which determines the (mass)a eigenvalues and the unitary matrix UL . For arbitrary x, we find m, and m, comparable; for their ratio to be small we need x close to -4. If we expand x about this point, i.e.. x = -4 + E, we find as E + 0

and, keeping only the leading entries in the Cabibbo matrix,

Our predictions are a large value for m, N 40 GeV and small mixing angles with sin 8, = 6,/S, = md/ms . It is perhaps worth mentioning that if the n-quark symmetry is extended to the lepton sector, we obtain the SU(5) relations me/m, = m,/m, = mTlmb . The value mdlms for sin Bc does not agree well with experiment, so we are not portraying these as realistic models, but rather as examples of how one might proceed if one was willing to accept only partial flavor conservation for the neutral Higgs couplings. ACKNOWLEDGMENTS It is a pleasure to thank S. M. Barr and 5. Weyers for their assistance and interest in this work.

Norm ad&in proof. Recently R. Gatto, G. Morchio, G. Sartori, and F. Strocchi, Univ. of Geneva preprint UGA-DTP-1979-07-206, have presented a detailed exposition and extension of their earlier work on flavor conservation. Though our methods are somewhat different, we are in substantial agreement. For the case of quarks belonging to an irreducible representation of the horizontal symmetry group, they present additional solutions. We have implicitly assumed that the matrix A, , obtained from the Cabibbo matrix A by removing all phases, can be inverted. This is certainly true if all the diagonal entries in A are close to one in magnitude, the physically interesting case. If det A,, equals zero, there is, e.g., a four-quark solution with 8, = 71/4,in agreement with the work of Gatto, rt al. For the case of quarks belonging to a reducible representation of the horizontal symmetry

60

SEGR&

AND

WELDON

group we both conclude that there are no physically acceptable solutions for the mixing angles, but the discussion by Gatto et a/. is more complete than ours. Finally, we note that we have recently shown that if the mixing angles are zero at the tree level, they remain zero after all radiative corrections are included (G. Segr& and H. A. Weldon Phys. Lett. 86B (1979), 291). REFERENCES &JS. Rev. Lett. 19 (1967), 1264; A. SALAM, in “Elementary Particle Physics: Relativistic Groups and Analyticity,” Nobel Symposium No. 8, p. 367 (N. Svartholm, Ed.), Almquist and Wiksell, Stockholm, 1968. S. L. GLASHOW, J. IL~OPOULOS,AND L. MAIANI, Phys. Reu. D 2 (1970), 1285. S. L. GLASHOW AND S. WEINBERG, Phys. Rec. D 15 (1977), 1958; see also E. A. PASCHOS,Phys. Rev. D 15 (1977), 1966. See, for example, G. SEGR~, H. A. WELDON, AND J. WEYERS,Phys. Lett. 83B (1979), 351. Additional Higgs fields are also required for CP to be a spontaneously broken symmetry. R. BARBIERI, R. GATTO, AND F. STROCCHI,Phys. Lett. B 74 (1978), 344. R. GATTO, G. MORCHIO, AND F. STROCCHI, Phys. Lett. B 80 (1979), 265. R. GATTO, G. MORCHIO, AND F. STROCCHI, Phys. Lett. 83B (1979), 348. G. SARTORI, Phys. Lett. 82B (1979), 255. We have discussed discrete symmetries only, but these considerations apply as qell to gauged horizontal symmetries, as proposed by F. WILCZEK AND A. ZEE [Phys. Rev. Lett. 42 (1979), 4211. See, e.g., M. K. GAILLARD AND B. W. LEE, Phys. Rev. D 10 (1974), 897:For KI. - pLp- limits see M. J. Shochet et al., Phys. Rev. Lett. 39 (1977), 59. There are, in fact, many groups with the special property that all their irreducible representations are equivalent to monomial representations, e.g., all groups of order p3. pq. p”q, or pyr where p, q, Y are primes. S. PAKVASA AND H. SUGAWARA, Phys. Lett. B 73 (1978), 61. M. HAMERMESH, “Group Theory,” p. 224, Addison-Wesley, Reading, Mass.. 1962. Note that A, may not be unitary. This is an immediate consequence of the usual orthogonality relations for matrix elements of any irreducible representation as noted by N. ITO [Osaka Math. 6 (1954), 1191. Amazingly, for monomial representations this condition was proved sufficient to guarantee irreducibility by K. Shoda [Proc. Phys.-Math. Sot. Japan 15 (1933), 2491. For an example see N. ITO, Ref. [16]. This follows from the fact that permutation representations are obtained by inducing from a unique representation of a subgroup (viz., the trivial, 1 x 1 representation) and the only arbitrariness is in ordering the basis vectors. See, for example, M. Hall, “The Theory of Groups,” Chapt. 5, Chelsea Publishing Co., New York, 1959. C. E. VAYONAKIS, Lett. Nuovo Cirnento 17 (1976), 383; M. VELTMAN, Acta Phys. Polon. B 8 (1977), 475; B. W. LEE, C. QUIGG, AND H. THACKER, Phys. Rev. Lett. 38 (1977). 883. H. GEORGI AND D. V. NANOPOULOS, Phys. Lett. 82B (1979), 95. This model was developed in collaboration with J. Weyers.

1. S. WEINBERG,

2. 3.

4. 5. 6.

7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

17. 18.

19. 20. 21.