An alternative theory of electroweak interaction without using any $SU(2)$ doublet

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by  Anirban Karan

An alternative theory of electroweak interaction without using any SU(2) doublet.
Anirban Karan∗
The Institute of Mathematical Sciences, HBNI, Taramani, Chennai 600113, India
(Dated: January 5, 2017)
In standard model electroweak interaction is described by SU(2)L × U(1)Y gauge symmetry
of Lagrangian. In this letter, we have derived an alternative way that can describe electroweak
interactions(asinstandardmodel)withoutusinganySU(2)doubletorSU(2)generator. Inaddition
to that, this theory can incorporate some new physics aspects which cannot be answered in the
framework of standard model. In this method, we only need local gauge invariance of Lagrangian
under a generalised U(1)Q gauge transformation, where Q is electric charge, and then everything
follows through.
Electroweakinteractionisone ofwidely studied topics independent real quantity and q is the charge of Aµ.
±
±
inhighenergyphysics,boththeoreticallyandexperimen- Now we demand this Lagrangian to be invariant under
tally. In theory, weak interactions are described using Local gauge transformation
7 non abelian gauge theory[1] whereas electromagnetic in-
1 teractions are explained by abelian U(1) gauge theory. Aµ± Aµ±e±iqθ(x) (3)
Q →
0
Standardmodel(SM)givesgoodtheoreticalexplanation
2 Under this transformation
ofelectroweakinteractionstakingSU(2) U(1) gauge
n symmetryofLagrangian[2–8]. OurapproLa×chdiffeYrsfrom ∂µAν (∂µAν)e±iqθ iq(∂µθ)Aνe±iqθ, (4)
a ± → ± ± ±
J the existing theory largely,but it can produce same the- Fµν Fµνe±iqθ iqe±iqθ[(∂µθ)Aν (∂νθ)Aµ], (5)
oretical results as SM; in addition, it can also incorpo- ± → ± ± ±− ±
3
rate different new physics aspects like flavour changing

] neutral current (FCNC) at tree level, non-unitary CKM (F+)µνF−µν (F+)µνF−µν
h →
p matrix,newelectroweakgaugebosons,fourthgeneration +2q2[A+µAµ−∂νθ∂νθ A+µAν−∂νθ∂µθ]
quarks or leptons etc. which cannot be addressed in the − (6)
p- framework of SM. In this approach, we take all charged +2iq[∂µθ{(∂µAν−)A+ν −(∂µAν+)A−ν}]
he fields including charged bosons to be transformed as −2iq[∂νθ{(∂µAν−)A+µ−(∂µAν+)A−µ}]
H Heiqhθ and all neutral gauge bosons to be trans-
[ → WeseefromEq.(6)thatlotsofextratermswillappear
formed as V V +Λ∂ θ under a generalised U(1)
1 µ → µ µ Q in the transformed free Lagrangian. As ∂µθ is a vector
gaugetransformation(Qdenotescharge,θ isthecharge
h
v quantity,weneedatleastonemorerealgaugebosonfield
of particle H and Λ is some constant) and claim that
9 Aµ which will transform as
7 demanding local gauge invariance of Lagrangian under 3
7 this gauge transformation will lead to electroweak inter- Aµ Aµ+λ∂µθ (λ is a constant) (7)
0 actions. Inthis context,ourtechniquecanbethoughtas 3 → 3
0 an extension of QED (Quantum Electrodynamics). We in order to cancel those extra terms in Lagrangian. As
.
1 emphasisethatanykindofSU(2)doubletsorSU(2)gen- the extra terms in Eq. (6) contain products of Aµ, Aµ
+ −
0 erators are not used at all in our method. and ∂µθ, we can expect that Aµ and Aµ will interact
7 ± 3
Letusfirstassumethatthereexisttwocomplexgauge with each other. So, we need to redefine our Lagrangian
:1 boson fields (Aµ+ and Aµ−) which are charge conjugate to accordingly. Observing Eq. (6) we can infer that it must
v each other. Then we can write them as complex linear be in the following form
Xi combinations of two real gauge boson fields Aµ and Aµ
1 2 1
ar in the following manner Lgauge =Linv − 2(F+)µνF−µν +β1(A+·A−)A23
Aµ+ = Aµ1√−2iAµ2 and Aµ− = Aµ1√+2iAµ2 (1) e++βη2(AA+·(A∂3µ)A(Aν−)A·A3)+ρm(nwAit3hµ(m∂µ,Anνm=)A+n,ν ) (8)
mn 3ν m nµ −
The free fieldLagrangianforthese gaugebosonswith-
where ρ ,β and η are some arbitrary constants
out any kind of interaction looks like mn 1,2 mn
with the constrains ρ = ρ ,β = β and η =
mn nm 1 2 mn
Lgauge =−14(F1µνF1µν +F2µνF2µν)=−21(F+)µνF−µν −whηincmh aarnedinLvianrviandtenuontdeesrsluo−mcamlagtaiuogneosfy−malml ethtroys.e terms
(2)
Now we want to retain the Lagrangian in the same
where Fµν =∂µAν ∂νAµ with j =+, ,1,2. form as in Eq. (2) and write it as
j j − j −
traHnesrfoerwmeasteioentAhaµ±tL→gauAgµ±eeis±iinqθvawrihaenrteuθndiesraglsopbaacleg-atiumgee Lgauge =−14Fjµν(Fj)µν (9)
e
2
where Fµν are some operators, equivalent to Fµν. In As µ,ν are dummy variables, they have been inter-
j j
order to do that we define Fµν as changed in some terms while calculating to get the
j T2
above form.
Fµν =Fµν +α AµAν for (j,k,l)=(1,2,3)
j j jkl k l In order to respect the global gauge symmetry, every
(10)
and j =k =l terminLagrangianshouldbeinproductformofA and
+
6 6
This particular form of Fjµν is chosen keeping in mind Ata−in.sBountlythAe+reoarroensloymAe−te(rsmucshinasTA2µ+anAdν+To3rwAhν−icAh−cνoonr-
the structure of Lgauge, given by Eq. (8). As there is no (∂µAν+)Aµ+ etc.). These terms spoil the gauge invariance
higher dimensional operator in Eq. (8), so we keep the of Lagrangian. To maintain the global gauge symmetry,
terms, only quadreaticin Aµj, in definition of Fjµν. Again, the coefficients of these terms must go to zero .This de-
if we rewrite Eq.(8) in terms of Aµ1,Aµ2 and Aµ3, we shall mand gives us the following conditions
findthat thereis no termin likeAν(Aµ∂ A )or
A4 etc(j =1,2,3). ThesesteLrmgasugaerenotgkaugjeiνnvajµriant α123+α213 = α132+α231 (15)
j
also. So we take the conditioen j = k = l while defining α2 +α2 = α2 +α2 (16)
Fjµν. One can of course redo th6e an6alysis keeping all 21α3213α223311 = α112233α132132 (17)
termsinFµν. Butthisonlymakesthecalculationtedious
j α +α =0 = α = α =g (say) (18)
and finally shows that the coefficients of all those terms 312 321 ⇒ 321 − 312 c
mustvanishinordertorespectthelocalgaugeinvariance Combining the conditions in Eq. (15), Eq. (16) one can
and symmetry between Aµ1, Aµ2 and Aµ3. get two sets of solutions and each set satisfies Eq. (17).
Now, let us find (F ) Fµν using the definition of F ,
j µν j j
given by Eq. (10). Solution I: α = α =g (say)
213 123 b
− (19)
(Fj)µνFjµν =(Fj)µνFjµν +2αjkl(Fj)µνAµkAνl and α132 =−α231 =ga (say))
+α α AµAν(A ) (A ) (11) Solution II: α =α
jkl jmn k l m µ n ν 213 132 (20)
= 1+ 2+ 3 and α123 =α231)
T T T
Here, (Fj)µνFjµν is broken in three parts T1,T2 and Thoughthetwosetsofsolutionslookdifferent,eventually
3 to make the calculation easier. 1, 2 and 3 de- both of them produce same results. So, let us take the
T T T T
notequadratic,cubicandquarticinteractionsamongthe first one, Eq. (19) and proceed.
gauge bosons respectively.
Due to the relations among α , given by Eq. (18)
jkl
Atfirstwehaveevaluatedtheexpressionsof , and
1 2 and Eq. (19), the expressions of and in Eq. (13)
intermsofAµ, Aµ andAµ andthenusingtThe iTnverse T2 T3
T3 1 2 3 and Eq. (14) will now look like
transformation of Eq. (1) we rewrite them as
= 2ig F (AµAν AµAν)
T1 =(Fj)µνFjµν =2(F+)µνF−µν +(F3)µνF3µν (12) T−22i(−ga+cgb3)µ[Aν3ν{−(∂µ+A−ν−)A+µ+−−(∂µAν+)Aµ−}] (21)
T2 =2αjkl(Fj)µνAµkAνl +2i(ga+gb)[A3µ{(∂µAν−)A+ν −(∂µAν+)A−ν}]
=i[(α +α ) (α +α )][(∂ Aν)Aµ
213 123 − 132 231 µ + +
−+(i[∂(µαAν−)Aµ−α− )21+∂ν((αA+µAµ+α−A)]−[AµAµ−)(]∂Aµ3Aνν)A T+32=(g−a24+gaggb2b)(AA23+(A·A+3·)A(A−−) ·A3) (22)
213 123 132 231 3µ − +ν
− − { Nowletusfindhow , and transformunderthe
−(∂µAν+)A−ν}−A3ν{(∂µAν−)Aµ+−(∂µAν+)Aµ−}] localgaugetransformaTti1onT,2givenTb3yEq.(3)andEq.(7).
+iF [(α +α )(AµAν AµAν)
3µν 312 321 + +− − −
+(α312−α321)(Aµ−Aν+−Aµ+Aν−)] T1 →T1+4q2[A+µAµ−∂νθ∂νθ−A+µAν−∂νθ∂µθ]
(13) +4iq[∂µθ{(∂µAν−)A+ν −(∂µAν+)A−ν}] (23)
=α α AµAν(A ) (A ) −4iq[∂νθ{(∂µAν−)A+µ−(∂µAν+)A−µ}]
T3 jkl jmn k l m µ n ν
A2
= 23[(α2213+α2231−α2123−α2132)(A2++A2−) T2 →T2+2iλ(ga+gb)[∂µθ{(∂µAν−)A+ν −(∂µAν+)A−ν}]
++2(α(α212231α32+31α223α11+23αα2112332)+[(Aα21+32)A(A3)+2·+A(−A)−] A3)2] (14) +−42qiλ(g(gaa++ggb)bA)[∂ν+νAθ−{(ν∂[AµAµ3∂ν−µ)θA++µλ−∂µ(θ∂∂µµAθ]ν+)A−µ}]
− · · 2q(g +g )[∂ θAµAνA +∂ θAµAνA
+2(α213α231+α123α132)[(A+ A3)(A− A3) − a b µ + − 3ν µ − + 3ν
· · +2λAµAν∂ θ∂ θ]
1 + − µ ν
(α +α )2(A2 A2)2
− 4 312 321 +− − (24)
3
+2(g2+g2)(2λAµ∂ θ+λ2∂µθ∂ θ)A Aν
T3 →T3 a b 3 µ µ +ν − From QED , we know that its interaction with gauge
−4gagb[λ(Aµ+∂µθAν−A3ν +Aµ−∂µθAν+A3ν) boson Aµ is governed by the term qfψγµψAµ in the
+λ2Aµ+Aν−∂µθ∂νθ] Lagrangian where qf is charge of the fermion. Taking
the analogy from QED, one can take the interaction La-
(25)
grangianfor trial as
After adding Eq. (23),Eq. (24) and Eq. (25) one can
find that under local gauge transformation (Fj)µνFjµν Lint =−[gfψγµψAµ++(gf)∗ψγµψAµ−+gfψγµψAµ3] (31)
transforms as
whereg andg arecouplingsforthefermionψ withAµ
f f e +
(F ) Fµν (F ) Fµν and Aµ. As Lagrangian is hermitian, the fermion must
+j[2µ(νg2j+g→2)λ2j+µ4ν(gj +g )λq+4q2]A Aµ∂ θ∂νθ couple3with Aeµ− by (gf)∗, complex conjugate of gf and
a b a b +µ − ν g must be real.
f
−[4gagbλ2+4(ga+gb)λq+4q2]A+µAν−∂νθ∂µθ Now,accordingto our convention,under U(1)Q gauge
+[4q(ga+gb)+4λ(ga2+gb2)]Aν+A−νAµ3∂µθ etransformation ψ changes as ψ → ψeiqfθ. So, Lint will
[2q(g +g )+4g g λ][Aµ∂ θAνA +Aµ∂ θAνA ] transform as
− a b a b + µ − 3ν − µ + 3ν
+i[2λ(ga+gb)+4q][∂µθ{(∂µAν−)A+ν −(∂µAν+)A−ν}] Lint →−[gfψγµψAµ+eiqθ+(gf)∗ψγµψAµ−e−iqθ] (32)
−i[2λ(ga+gb)+4q][∂νθ{(∂µAν−)Aµ+−(∂µAν+)Aµ−}] −gfψγµψAµ3
(26)
Clearly, this trial-Lagrangianin Eq. (31) is not invari-
e
In order to make (Fj)µνFjµν as well as Lagrangian in- antunder globalgaugetransformation. So Eq.(31)does
variant under the gauge transformation, the coefficients notgivethecorrectformofinteractionbetweenfermions
ofallextratermsintheaboveexpression,Eq.(26),must andAµ. Actually,itisimpossibletomake invariant
± int
L
go to zero. Then we must have with only one fermionic field as ψγ ψ remains invariant.
µ
But ψ γ ψ can transform under U(1) gauge transfor-
q 2 µ 1 Q
ga =gb and λ=−g (27) mation if ψ1 and ψ2 have different electric charges. So,
b
we need at least two fermionic fields to describe the in-
Now, there exists some kind of symmetry among Aµ, teractionofAµ± withfermions. Butincaseofinteraction
AµandAµ. InsteadofstartingwithAµandAµ,wecoul1d with Aµ, two fermions are not needed necessarily as Aµ
2 3 1 2 3 3
have started with Aµ and Aµ and made the complex does not change under global gauge transformation.
2 3
fields as linear combinations of them. Then, of course, Now, we take two fermions as ψ1 and ψ2 with charges
wewouldhavegotAµ1 astheextragaugeboson,required q1 and q2 respectively and take the interaction La-
to make the Lagrangian gauge invariant. In that case, grangianas
everything would go in the same way and we would end
up with Lint =−[g21ψ2γµψ1Aµ++(g21)∗ψ1γµψ2Aµ−] (33)
[g ψ γ ψ Aµ+g ψ γ ψ Aµ]
g =g and λ= q (28) e − 1 1 µ 1 3 2 2 µ 2 3
c b
−g
b
Then int willetransformundeerglobalgaugetransforma-
L
CombiningEq.(27),Eq.(28),Eq.(18)andEq.(19),we tion as
infer that e
g ψ γ ψ Aµei(q+q1−q2)θ
q Lint →− 21 2 µ 1 +
ga =gb =gc =g, λ=−g and αjkl =−gǫjkl (29) e −(g21)∗ψ1γµψ2Aµ−e−i(q+q1−q2)θ (34)
[g ψ γ ψ Aµ+g ψ γ ψ Aµ]
Thus we have arrived at same Fµν as it should be − 1 1 µ 1 3 2 2 µ 2 3
j
iWn±µc,aWse3µofwSitUh(2g) gaasuwgeeaskymcomueptlriyngwhcoenrestAanµ±t,.Aµ3Aatctthaiss Tsyhmenm,etthreycisonditionethat mustholdeto respectthe global
point one can ask why Aµ is identified as Wµ, not as
3 3
photon. We know from QED that under U(1) gauge q+q q =0 (35)
Q 1 2
−
transformationphoton changes as Aµ Aµ ∂µθ when
ψ ψeiqfθ[5–7]. Solookingatthe valu→eofλ−inEq.(27) This Eq. (35) signifies electric charge conservation at
on→e can infer that Aµ is not photon. fermion-Aµ± interactionvertex. InSM(StandardModel),
3
Now let us look at the interactions of these gauge there are total 24 fermions including particles and anti-
bosons with fermions. Let, there is a fermionic field ψ particlesboth. Eq.(35)indicateswhichtwoofthem will
with massm. The correspondingfree fieldLagrangianis interactwitheachotherthroughAµ±. Forexample,inSM
we write the charged current interactions (not consider-
=ψ(i∂/ m)ψ (30) ing left handedness of weak interaction) between e− and
fermion
L −
e
4
ν as u(ν )γ u(e−)Wµ and u(e−)γ u(ν )Wµ [5, 8]and of demanding local gauge invariance. From whole La-
e e µ + µ e −
see that the condition in Eq. (35) holds. grangian, the portion which is not invariant under local
At low energy scale, charged weak interaction is ob- gauge transformation is sorted out as
served to be left handed only. It can be incorporated
in theory assuming that Aµ1 and Aµ2 interact with left L′ =ψ(i∂/)ψ−gχfψLγµψLAµ3 (37)
handed fermions and right handed fermions differently.
where χ is the coupling of ψ with Aµ in terms of g.
Itisevidentfromthestructureof ,givenbyEq.(33), f 3
int
L Under the local gauge transformation, given by Eq. (3)
that one right handed fermion cannot interact with left
handed one through Aµ± as ψL e= 1−2γ5ψ and ψR = and Eq. (7), L′ changes as
w1+i2tγh5ψth.eTwakeainkggtahuegceobuopsloinngsstoofbaellzreirgohtwheacnadnedmfaetrcmhioonusr L′ →L′−qfψLγµψL∂µθ−qfψRγµψR∂µθ (38)
λgχ ψ γ ψ ∂µθ
theory with observation. Another method of explaining − f L µ L
it is to take different vertex factor[8] for charged weak
From the above Eq. (38) it is evident that we need
interaction. In QED,vertexfactorfor any Feynmandia-
one more gauge boson (B ) to cancel the extra piece in
gram is taken to be proportional to γ . In case of weak µ
µ thetransformedLagrangianforψ . WeknowfromQED
interaction if vertex factor is chosen to be proportional R
that photon also cancels exactly this extra term. Then
to γ (1 γ ) then, only left handed fermions will couple
to wµeak−gau5ge bosons. Bµ would act as a photon for right handed fermions.
But it will not act as photon for left handed fermions
It is alwayspossible to write the couplings in terms of
as λ and χ have some non zero value in Eq. (38). Now
weak coupling constant g. Then, charged weak interac- f
the question arises whether B will couple to ψ at all.
tion can be written as, µ L
The answer is yes, otherwise it would be impossible to
g describe QED in this scenario as photon couples to ψ
= [V (ψ ) γ (ψ ) Aµ R
Lcharged −√2 jk j L µ k L + (36) and ψL equally. One important point to note that Bµ
e +(Vjk)∗(ψk)Lγµ(ψj)LAµ−] should not couple to Aµ± or Aµ3, otherwise the gauge in-
variance achieved in will be gone and we have
gauge
L
wherej,kdenotetwodifferentflavoursoffermionswhich to start from scratch again. Taking into account these
must satisfy the chargeconservationrelation q+q =q considerations, we adde two interaction terms in L′ as
k j
as shown in Eq. (35). As in SM W±µ have unit charge so tfψRγµψRBµ and tfψLγµψLBµ and a kinetic term in
charged weak interaction only happens between up type like 1(∂µBν ∂νBµ)(∂ B ∂ B ).
Lgauge −4 − µ ν − ν µ
quarkanddowntypequarkandsimilarlyforleptonsand e As localgauge symmetry is restoredin ψ after intro-
R
the anti- particles. Interaction between quarks and lep- deuction of Bµ, we leave the right handed part and shift
tonsorparticleandanti-particlethroughW±µ isimpossi- our focus to left handed part.
ble. But,theinteractingfermionsneednottobeofsame
generation. Thus mixing of different generations comes L′left =ψL(i∂/)ψL−gχfψLγµψLAµ3−tfψLγµψLBµ (39)
automatically. Though there exists an ambiguity in lep-
tonsectorasneutrinoandanti-neutrinobothareneutral Heeretf isthecouplingofψL withBµ. Ifweassumethat
Bµ, interacting with ψ , changes as B B +λ ∂ θ
incharge,butthiscanbesolvedeitherbydemandingto- L µ → µ b µ
under the gauge transformation, then demanding local
talleptonnumberconservationorbyassumingneutrinos
gauge invariance of L′ will lead to
to be Majorana particle.
V inthe aboveEq.(36)givesthe strengthofinterac-
tionjkfor twodifferentfermionswith Aµ in termsofg. At qf e+gλχf +λbtf =0 (40)
+
this point there is no restriction on Vjk. Now, one can Now, it is always possible to write λb and tf as
make a column matrix Ψ with only downtype quarksor
neutrinos (accordingto SMconvention)anddefine Ψ′ as q ′
λ = and t =g ξ (41)
Ψ′ = VΨ where V is a matrix made with V . The ki- b −g′ f f
jk
netic termforthe downtype fermions(orneutrinos)can
whereg′issomeconstantandξ issomeparameterwhich
be written as Ψ(i∂/)Ψ. If it is demanded that the kinetic f
term can be written as Ψ′(i∂/)Ψ′ also, then V has to be dependsonflavourandchiralitybothofafermion. Then
combiningtheaboveEq.(40)withEq.(29)andEq.(41)
unitaryandV canbeidentifiedwithCKMmatrixofSM
one would get
[5, 8, 9]. But in general, there is no reason to demand
this extra condition in our approach. So we can take V q =q(χ +ξ ) (42)
f f f
to be completely arbitrary.
As initially Aµ,Aµ and Aµ were equivalent, it is legit- In case of SM, one can identify χ with the value of T
1 2 3 f 3
imate to assume that Aµ also interact with left handed (third component of weak isospin), ξ with Y (half of
3 f 2
fermions only. Now let us look at the consequences weak hypercharge) and q with unit charge e. Then the
5
aboveEq.(42)indicatestoGell-Mann-Nishijimaformula as λ = 0, Eq. (48). With these constrains mass terms
z
[5, 8, 10]. for gauge boson can be written in Lagrangianas,
Now,onecanmixAµ andBµ throughWienbergangle 1
(θW) to get photon (A3µ) and Z-boson(Zµ) [2, 3, 5, 8]as Lmass = 2m2ZZ2+m2WAµ+A−µ (50)
Now let us see whether there is any scope for
Zµ =Aµcosθ Bµsinθ (43) FCNC(flavouer changing neutral current) at tree level.
Aµ =Aµ3 sinθW+−BµcosθW (44) Let us write down the neutral current interaction La-
3 W W grangian along with the kinetic term and two extra
FCNC terms as
Solving for Aµ and Bµ from Eq. (43) and Eq. (44) one
can write the3interaction part of L′left, in Eq. (39) in L′ =ψj,τ(i∂/)ψj,τ −ψj,τγµψj,τ(qjAµ+gzCj,τZµ) (51)
terms of Aµ and Zµ as ψ γ ψ (ha Aµ+hz Zµ) h.c.
e − j,τ µ k,τ j,k,τ j,k,τ −
e
= [(gχ sinθ +g′ξ cosθ )Aµ where τ(= L,R), h.c. stands for hermitian conjugate.ψj
Lnuet,L − f W f W (45) and ψk are two fermions with different flavour but same
e +(gχfcosθW −g′ξfsinθW)Zµ]ψLγµψL charge; haj,k and hzj,k are their couplings with Aµ and
Zµ. The coefficient C is coming because interaction
j,τ
Identifying the coupling of photon in the above Eq. (45) of a fermion with Z can be written as C ψ γ ψ Zµ +
L L µ L
to be qf, one have CRψRγµψRZµ and gz is the coupling constant for that
interaction.
q =gχ sinθ +g′ξ cosθ (46) Using Eq. (48) and demanding local gauge invariance
f f W f W
of this Lagrangian, one can show ha = 0. But, it is
j,k,τ
As all particles satisfy Eq. (42) and Eq. (46) both, then impossible to infer anything about hzj,k,τ as λz = 0. So,
one must have left handed or right handed FCNC at tree level might
exist, but it must be mediated by Z-boson only.
q q
g = and g′ = (47) Inconclusion,wehaveshownthatageneralizedU(1)Q
sinθW cosθW gaugesymmetrycanproduceelectroweakinteractions(as
in SM) without using any SU(2) gauge invariance. In-
Using Eq. (45) and Eq. (47) one can write the cou- teractions among gauge bosons, chargedcurrentinterac-
pling of left handed fermion with Z boson to be gz[χf tions, neutralcurrentinteractions everythingcomes nat-
(qf/q)sin2θW] where gz = q/(sinθW cosθW). One ca−n urally in correct form (like SM) once generalized U(1)Q
use this coupling of Z boson with right handed fermion
gauge symmetry is demanded. In addition, scopes for
also keeping in mind that χf = 0 for ψR. These results non-unitary CKM matrix and FCNC at tree level are
are already known [5, 7, 8]. alsodiscussed. Infuture,ifsomeotherelectroweakgauge
Now, if we assume that Aµ and Zµ transform as bosonorfourthgenerationquarkorleptonisexperimen-
Zµ Zµ +λz∂µθ and Aµ Aµ +λa∂µθ, then using tally discovered, that might put SM in trouble due to
→ →
Eq. (43),Eq. (44), Eq. (29),Eq. (41) and Eq. (47) it can SU(2) nature of weak interaction and unitarity of CKM
be easily shown that matrix, but they can be handled in our theory without
assuming any new model.
λ = 1 and λ =0 (48) IthankRahulSinhaandG.Rajasekaranfortheirvalu-
a z
−
able suggestions.
The above conditions in Eq. (48) hold for right handed
case also as photon couples to ψ and ψ equally. Thus
L R
wegettheknownresultofQEDforλ being 1andthis
a
−
value of λ can be used as a signature of photon.
a ∗
[email protected]
Themasstermsforgaugebosonscanbe introducedas
[1] C.N. Yang,R.L. Mills, Phys.Rev.96 (1954) 191.
1 [2] S.L. Glashow, Nucl. Phys.,22 (1961), p. 579.
= [m2 A2+m2A2+m2A2+m2Z2] (49)
Lmass 2 ph 1 1 2 2 Z [3] A. Salam, J.C. Ward,Phys. Lett. 13 (1964) 168
[4] S. Weinberg, Phys.Rev. Letters 19 (1967), 1264
wherem ,m ,m andm arethemassesofphoton,A , [5] Collider Physics by Barger and Phillips
ph 1 2 z 1
[6] An introduction to quantum field theory by Peskin and
A and Z bosons respectively. If we rewrite this La-
gr2angian in terms of Aµ and Aµ and demand global Schroeder
+ − [7] Gauge theory of elementary particle physics by Cheng
gaugeinvarianceof thenwemusthavem =m =
mass 1 2 and Li.
L
mW(say). Moreover,inordertomake mass localgauge [8] Introduction to elementary particles by D.Griffiths
L
invariantweneedm =0asunderlocalgaugetransfor- [9] M. Kobayashi, T. Maskawa, Progr. Theor. Phys., 49
ph
mation Aµ Aµ ∂µθ. But Z boson might be massive (1973), p.652
→ − [10] T. Nakano, K. Nishijima, Prog. Theor. Phys. Vol. 10
(1953) 581.