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Chapter 4 Introduction to Salam-Weinberg-Glashow model
4.1 Neutral weak currents
Weare now su ciently safe with the charged currents (and old version is quite good) but hypothesis about weak isotriplet of W leads to neutral currents in quark sector: 1 j neutr:ud = 2 (uO u ; dC O dC )= = 1 (uO u ; dO d(cos C )2 ; sO s(sin C )2; 2 dO scos C sin C ; sO dcos C sin C ): and, correspondingly, in lepton sector: j neutr:lept: = 1 ( e O e ; eO e)+ 2 + 1 ( O ; O ): 2 (We do not write here explicitly weak neutral operator O as it can diverge nally from the usual charged one (1 + 5):) Up to the moment when 70


neutral currents were discouvered experimentally presence of these currents in theory was neither very intriguing nor very disturbing. But when in 1973 one of the most important events in physics of weak interaction of the 2nd half of the XX century happened { neutral currents were discouvered in the interactions of neutrino beams of the CERN machine with the matter, it was become immediately clear the contradiction to solve: although neutral currents interacted with neutral weak boson (to be established yet in those years) with more or less the same coupling as charged currents did with the charged W bosons, there were no neutral strange weak currents which were not much suppressed by Cabibbo angle. Even more: neutral currents written aboveopened channel of decay of neutral K mesons into ; + pair with approximately the same coupling as that of the main decay mode of the charged K ; meson ( into lepton pair ; ). Experimentally it is suppressed by 7 orders of magnitude!!! ;(K
s
0

!

;

+

)=;(Ks0

!

all) < 3:2 10

;

7

Once more havewe obtained serious troubles with the model of weak interaction!? In what way, clear and understandable, it is possible to save it? It turns out to be su cient to remind of the J= particle and its interpretation as a state with the 'hidden' charm (cc): New quark with charm would save

situation!

Indeed now the number of quarks is 4 but in the weak isodoublet only 3 of them are in action. And if one (Glashow, Iliopoulos, Maiani) assumes that the 4th quark also forms a weak isodoublet, only with the combination of d and s quarks orthogonal to dC = dcos C + ssin C , namely, sC = scos C ; dsin C ? Then apart from charged currents

4.1.1 GIM model

j = c (1 + 5)s

C

neutral currents should exist of the form: j neutr:cs = 1 (cO c ; sC O sC )= 2 71


1 (uO u ; sO s(cos )2 ; dO d(sin )2+ C C 2 +dO scos C sin C + sO dcos C sin C ): The total neutral current yields: j neutr:ud = 1 (uO u+ 2 +cO c ; dO d ; sO s):

This is so called GIM mechanism proposed in 1970 by Glashow, Iliopoulos, Maiani in order to suppress theoretically decays of neutral kaons already suppressed experimentally. (For this mechanism Nobel price was given!) Now we should understand what is the form of the operator O . But this problem is already connected with the problem of a uni cation of weak and electromagnetic interactions into the electroweak interaction. Indeed the form of the currents in both interactions are remarkably similar to each other. Maybe it would be possible to attach to the neutral weak current the electromagnetic one? It turns out to be possible, and this is the main achievement of the Salam-Weinberg model. But we cannot add electromagnetic current promptly as it does not contain weak isospin. Instead we are free to introduce one more weak- interacting neutral boson Y ascribing to it properties of weak isosinglet. We shall consider only sector of u and d quarks and put for a moment even C = 0 to simplify discussion. L = g 1 (uL uL ; dL dL)W3 + 2 +g0(auL uL + buR uR + cdL dL + qdR dR)Y = e 2 (uL uL + uR uR) ; 1 (dL dL + dR dR)]A + 3 3 + J neutr:ud Z 0: 72

There are no strangeness-changing neutral currents at all!

4.1.2 Construction of the Salam-Weinberg model


Having two vector boson elds W3 Y we should transfer to two other boson elds A , Z 0 (one of them, namely, A we reserve for electromagnetic eld) and take into account that in fact we do know the right form of the electromagnetic current. It would be reasonable to choose orthogonal transformation from one pair of elds to another. Let it be gZ 0 + g0A g0 0 + pg2 + g02 Y = ;pZ2 + ggA : W3 = g 02 Substituting these relations into the formula for currents we obtain in the left-hand side (LHS) of the expression for the electromagnetic current the following formula pg2gg+0 g02 ( 1 + a)uL uL + buR uR+ 2 +(; 1 cdL dL + qdR dR)A ]= eJ emA 2 wherefrom 1 1 a=6 b= 2 c=6 3 gg0 1 q = ; 3 e = pg2 + g02 : Then for the neutral currentwe obtain (g pg22++gg020)2 1 (uL uL; 2 02 ;dL dL)W3 ; pg2g+ g02 J em = pg 2 + g 02 1 g g 2 (uL uL ; dL dL)W3 ; pg2 + g02 g02 ; g g g2 + g02 J em Let us nowintroduce notations 0 sin W = pg2g+ g02 cos W = pg2 g+ g02 : 73


Now neutral vector elds are related byformula

W3 = cos W Z 0 + sin W A

Y = ;sin W Z 0 + cos W A :

Finally weak neutral current in the sector of u and d quarks reads g 1 (u u ; d d ) ; sin2 J em]: W cos W 2 L L L L Nowwe repeat these reasonings for the sector of c and s quarks and restore Cabibbo angle arriving at the neutral weak currents in the model with 4 avors: neutr JW :GW S = cosg 1 (cL cL + uL uL ; dL dL ; sL sL); W2

J em ]: (4.1) Remember now that the charged currententers Lagrangian as W L = gp (c (1 + 5)sC W + + u (1 + 5)dC W ++ 22 +sC (1 + 5)cW ; + dC (1 + 5)uW ;) and in the 2nd order of perturbation theory in ud-sector one would have 2 1 L(2) = 8 (M 2gW q2) u (1 + 5)dC dC (1 + 5)u + H:C: W+ what should be compared to p2 Lef f = GF u (1 + 5)dC dC (1 + 5)u + H:C: Upon neglecting square of momentum transfer q2 in comparing to the W boson mass one has 2 GF = gW = p2 8MW 2 2 = 8M 2 e 2 W sin W
2

;sin

W

74


wherefrom that is

M

W

2

p2

e2 = 24 8GF 8GF

p

1200GeV

2

MW 35GeV !!! (Nothing similar happened earlier!) Measurements of the neutral weak currents give the value of Weinberg angle as sin2 W =0 2311 0 0003. But in this case the prediction becomes absolutely de nite: MW = 73GeV : As is known the vector intermediate boson W was discouvered at the mass 80 22 0 26g\w which agree with the prediction as one must increase it by 10% due to large radiative corrections.

4.1.3 Six quark model and CKM matrix
0 01 d B s0 C = @A b0 0 10 1 Vud Vus Vub d B Vcd Vcs Vcb C B s C @ A@ A Vtd Vts Vtb b

But nowadays wehave 6 and not 4 quark avors. So wehave to assume that there is a mix not of two avors (d and s) but of all 3 ones (d s b):

This very hypothesis has been proposed byKobayashi and Maskawa in 1973. The problem is to mix avors in such a way as to guarantee disappearence of neutral avor-changing currents. Diagonal character of neutral currentis achieved bychoosing of the orthogonal matrix VCKM of the avor transformations for the quarks of the charge -1/3. Even more it occurs that it is now possible to introduce a phase in order to describe violation of CP-invariance ( with number of avours less then 3 one can surely introduce an extra phase but it could be hidden into the irrelevant phase factor of one of the quark wave functions). Usually CabibboKobayashi-Maskawa matrix is chosen as

VCKM =

0 c12c13 s12c13 = B ;s12c23 ; c12s23s13ei 13 c12c23 ; s12s23s13ei @ s12s23 ; c12c23s13ei 13 ;c12s23 ; s12c23s13e
75

i

13 13

1 s13e;i 13 C s23c13 A : (4.2) c23c13


Here cij = cos ij sij = sin ij (i j =1 2 3) while ij - generalized Cabibbo angles. At 23 =0 13 = 0 one returns to the usual Cabibbo angle s = 12: Let us write matrix VCKM with the help of Eqs. (1,5,7) as

V

CKM

= R1( 23)D (ei 0 sin cos
23 23

13

=

2

)R2( 13)D(e
=
2

i 13 =

2

)R3( 12)=

i 13 =2 0 sin 13 CB e 0 0 0 C 1 0 A@ 1 0A 0 cos 13 0 0 e;i 13=2 0 1 cos 12 sin 12 0 B ;sin 12 cos 12 0 C : @ A 0 01 Matrix elements are obtained from experiments with more and more precision. Dynamics of experimental progress could be seen from these two matrices divided by10 years in time:
(1990) VCKM 0 :218 B 0:09747 to 0::9759 009734 to = @ :218 to 0 224 : to 0:03 to 0:019 0:029 to

0 0 B 1 cos 23 @0 0 ;sin 23 0 B cos 13 @0 ;sin 13

1 C A

0 ;i 13 Be 0 @ 10
0

00 1 0C A 0 ei 13=2

1
(4.3)

1

=

0 to 0 9757 B 0:09742 to 0::225 = @ :219 0:04 to 0:014
The charged weak current c

0:224 0:001 to 0:007 0:9752 0:030 to 0:058 C : A 0:058 0:9983 to 0:9996 (4.4) (2000) VCKM = 1 0:219 to 0:226 0:002 to 0:005 0:9734 to 0:9749 0:037 to 0:043 C : A 0:035 to 0:043 0:9990 to 0:9993 (4.5) ould be written as
CKM

1

; JW =(u c t) (1 + 5)V

0 B @

1 dC sA b

76


Neutral current would be the following in the standard 6-quark model of Salam-Weinberg: g NE pcos W JW JTR:6 = g1 pcos W 2 (tL tL + SL SL + uL uL; dL dL ; sL sL ; bL bL) ; sin2 W J em]:

4.2 Vector bosons

W

and Y as gauge elds

Bosons W and Y could be introduced as gauge elds to assure renormalization of the theory of electroweak interactions. We are acquainted with the method of construction of Lagrangians invariant under local gauge transformations on examples of electromagnetic eld and isotriplet of the massless -meson elds. We have introduced also the notion of weak isospin, so now we require local gauge invariance of the Lagrangian of the left-handed and right-handed quark (and lepton) elds under transformations in the weak isotopic space with the group SU (2)L SU (1). But as we consider left- and right- components of quarks (and leptons) apart we put for a moment all the quark (and lepton) masses equal to zero. For our purpose it is su cient to write an expression for one left-handed isodoublet and corresponding right-handed weak isosinglets uR dR:

L0 = qL(x)@

qL(x)+ uR(x)@ uR(x)+ dR (x)@ dR(x)
0 qL(x)= ei~ ~ qL(x)

This Lagrangian is invariant under a global gauge transformation

u0RL(x)= ei uRL(x) d0RL(x)= ei dRL(x) where matrices ~ act in weak isotopic space and ~ =( are arbitrary real phases.
RL 0 RL

1

2

3

),

RL

,

0 RL

77


Let us require nowinvariance of this Lagrangian under similar but local 0 gauge transformations when ~ and RL RL are functions of x. As it has been previously L0 is not invariant under such local gauge transformations: L00 = L0 + iqL(x) @~ ~ (x) qL(x)+ @x 0 R R +iuR @ @x(x) uR + idR @ @x(x) dR 0 +iuL @ L(x) uL + idL @ L(x) dL : @x @x In order to cancel terms violating local gauge invariance let us introduce ~ weak isotriplet of vector elds W and also weak isosinglet Y with the gauge transformations @U ~ ~ ~ W 0 = U y~ W U ; g1 @x U y W i~ (x)~ GDE U = e 0 0 Y 0 = Y ; g1 @ ( R + R + L + L) @x Y Interactions of these elds with quarks could be de ned by the Lagrangian constructed above 1 1 L = p g(uL dLW + + dL uLW ;)+ g 2 (uL uL ; dL dL)W3 + 2 g0(auL uL + buR uR + cdL dL + qdR dR)Y = Thus the requirementof invariance of the Lagrangian under local gauge transformations along the group SU (2)L SU (1) yields appearence of four mass~ less vector elds W Y . Earlier it has already been demonstrated in what way neutral elds W3 , Y by an orthogonal transformation can be transformed into the elds Z , A . After that one needs some mechanism (called mechanism of spontaneous breaking of gauge symmetry) in order to give masses to W Z and to leave the eld A massless. Usually it is achieved by so called Higgs mechanisn. Finally repeating discussion for all other avours we come to the already obtained formulae for the charged and neutral weak currents but already in the gauge-invariant symmetry with spontaneous breaking of gauge symmetry. 78


4.3 About Higgs mechanism
Because of short of time we could not show Higgs mechanism in detail as an accepted way of introducing of massive vector intermediate bosons W Z 0 into the theory of Glashow-Salam-Weinberg. We give only short introduction into the sub ject. Let us introduce rst scalar elds with the Lagrangian

L =T ;V =@ @ + dV ( ) = ;2 d

22

+

4

:

(4.6)

Let us look at V just as at ordinary function of a parameter and searchfor the minimum of the potential V ( ).
2

;
=0

4

3

=0:

Wehave 3 solutions:
23

1

min min

That is, with 2 h0 we would have two minima (or vacuum states) not at the zero point! How to understand this fact? Let us nd some telegraph mast and cut all the cords which help to maintain it in vertical state. For a while it happens nothing. But suddenly wewould see that it is falling. And maybe directly to us. What should be our last thought? That this is indeed aspontaneous breaking of symmetry! This example shows to us not only a sort of vanity of our existence but also the way to follow in searching for non-zero masses of the weak vector bosons W Z . By introducing some scalar eld with nonzero vacuum expectation value (v.e.v.) < >= v it is possible to construct in a gauge-invariant way the interaction of this scalar eld with the vector bosons W W 3 Y having at that moment zero masses. This interaction is bilinear in scalar eld and bilinear in elds W W 3 Y that is it contains terms of the kind 2jW j2. At this moment the whole Lagrangian is locally gauge invariantwhich assures its renormalization. Changing scalar eld to scalar eld with the v.e.v. equal to zero = ; < > < >= 0 we break spontaneously local 79

=

s

;

2

:


gauge invariance of the whole Lagrangian but instead we obtain terms of the kind v2jW j2 which are immediately associated with the mass terms of the vector bosons W . A similar discussion is valid for Z boson. Nowwe shall show this "miracle" step by step. Firs let us introduce weak isodoublet of complex scalar elds

L =@ @ +

2

+(

)2:

where is a doublet, T =( + 0) with non-zero vacuum value, 0. It is invariant under global gauge transformations
0

hi

= v 6=

(4.7)

=U =e

i~ ~

:

Now let us as usual require invariance of the Lagrangian under the local gauge transformations. The Lagrangian Eq.(4.7) however is invariant only in the part without derivatives. So let us study "kinetic" part of it. With ~ (x) dependenton x wehave

@
it becomes

0

= U@ + @ U

@

0

@

0

=( @ U y +@

U y)( @ U + @

U )=

(U yU )@ + (@ (U y@ U ) + (@ U y)U@ + U y(@ U ) : Let us introduce as already known remedy massless isotriplet of vector mesons ~ W with the gauge transformation already proposed ~ ~ ~ W 0 = U~ WU y ; g1 (@ U )U y W butwithtwointeraction Lagrangians transforming under local gauge transformations as: 0 W y0 W 0 0 = ~~ ~~ Wy W + ~ ~ U yU W yU y(@ U ) + U y@ W U yU + =@ 80


~ ~ ~ W 0 @ 0 = W @ + W U y(@ U ) : The sum of all the terms results in invariance of the new Lagrangian with twointroduced interaction terms under the local gauge transformations. We can do it in a shorter wayby stating that ~ (@ + ig W 0 ) 0 = ~ (U@ +(@ U )+ igU W U yU +(@ U )U y ) = ~ U (@ + ig W ) : Then it is obvious that the Lagrangian ~ ~ (@ ; ig W y) (@ + ig W ) is invariant under the local gauge transformations. In the same way but with less di culties we can obtain the Lagrangian invariant under the local gauge transformation of the kind 0 = ei that is under Abelian transformations of the type use for photon previously: (@ ; ig0Y ) (@ + ig0Y ): But our aim is to obtain masses of the vector bosons. It is in fact already achieved with terms of the kind 2jW j2 and 2Y 2. Now we should also assure that our e orts are not in vain. That is searching for weak boson masses we should maintain zero for that of the photon. It is su cient to propose the Lagrangian ~ ~ (@ ; ig W y + ig0Y ) (@ + ig W ; ig0Y ) where g2jW j2 terms would yield with = + v masses of W bosons MW = v g, while term jgW 3 ; g0Y j2 would yield mass of the Z 0 boson pg2 + g02 MW q 2 + g 02 = v g = cos : MZ = v g g W
0

and

@ U y(@ U )

81


There is a net prediction that the ratio MW =MZ is equal to cos W . Experimentally this ratio is (omitting errors) 80=91 whichgives the value of Weinberg angle as sin2 W 0:23 in agreement with experiments on neutrino scattering on protons. Due to the construction there is no term jg0W 3 + gY j2 that is photon does not acquire the mass! Talking of weak bosons and scalar Higgs mesons we omit one important point that is in the previous Lagrangians dealing with fermions we should put all the fermion masses equal to zero! Why? It is because we use di erent left-hand-helicity and right-hand-helicity gauge transformations under which the mass terms are not invariantas

mqqq = mq qLqR + mqqRqL 1 qL = 2 (1 + 5)q qR = 1 (1 ; 5)q: 2 What is the remedy for fermion masses? Again we could use Higgs bosons. In fact, interaction Lagrangian of quarks (similar for leptons) with the same scalar eld , T =( + 0) with non-zero vacuum value, h i = v 6=0, can be written as Ldm = dqL dR + HC = d (u
d (uL
+

L

+

d+d

L

0

dR)=

with md = dv. (In a similar way lepton masses are introduced:
l l( L
+

d+d

L

0

dR + mddLdR )

l + lL 0lR + mllLlR)

with ml = l v.) So we could obtain now within Higgs mechanism all the masses of weak bosons and of all fermions either quarks or leptons. By this note we nish our introduction into the Salam-Weinberg model in quark sector and begin a discussion on colour.

82