Author: Ling fong Li
1
Weak Interaction
Classi…cation
Because the strong interaction e¤ects are di¢ cult to compute reliably, we distinguish processe involving
hadrons from leptons in weak interactions.
(a) Leptonic weak interactions
Here all particles are leptons and there is no complication from strong interactions. Once the interaction is known we can calculate these processes accurately since the interaction is weak and
perturbation theory is applicable.
Examples:
_
!e +
+ e
+e!
+e
_
_
!
+ +
!e + + e
(b) Semi-leptonic interactions
These processes involve both leptons and hadrons. Since we can calculate the leptonic parts quite
reliabley, these processes can be used to study the properties of hadrons very similar to the case of
ep scattering.
Examples :
!
+
K + ! 0 + e+ + e
_
_
n!p+e + e
+p! ++n
(c) Non-leptonic interactions
Here all particles are hadrons and they are the most di¢ cult reactions to study because of the strong
interaction e¤ects. This class of reactions di¤er from the normal strong interaction in the slower
decay rates and smaller crossections.
Example :
K+ ! + + 0 K0 ! + +
+ 0
+ !P + 0
!P +
1.1
Selection Rules in Weak Interaction
(a) Leptonic Interaction
Two neutrino experiments: from -decay and from decay are di¤erent
If they were the same then the following chain of reactions should be possible,
n !p+e+
+p !
+
+n
However, only e+ is observed in the …nal product and no + has been seen. A simple explanation
that is the neutrino from -decay called e is di¤erent from neutrino from -decay accompanied by
called
and there is also muon number and electron number conservation. In these conservation
laws we assign the electron number Le as
e ;
e+ ;
e
e
Le = 1
Le = 1
Similarly, for the muon number L
;
+;
L =1
L = 1
As a consequenc of these conservation laws, the reaction
! e + are forbidden and experimentally this is indeed the case. Lepton number conservations seem to hold up very well for many
years until neutrino oscillations have been observed recently.
(b) Semi-leptonic decays
(a) The strangeness changing reactions, S 6= 0 seem to be about a factor of 10 or so smaller than
those which conserve strangeness, S = 0:
(b) It has been observed that hadrons in the strangeness changing decays satisfy the selection rule
S=
Q
For examples,
K+ !
0 +
;
K+ 9
but
(c) Absence of S = 1 neutral currents
For example,
+
(KL !
(K + !
+
)
)
(d) No S = 2 transistion been obsverbed
For example,
9 ne
10
+ +
e
_
e
9
_
e
When quark model was developed, all these properties can be accormodated by writing the hadronic
weak current as
_
_
J had = u (1
(1
5 ) d cos c + u
5 ) s sin c
where
c
0:25 is the Cabbibo angle.
(c) Non-leptonic interactions
Here we havethe I = 1=2 rule,
K+ !
(Ks !
+ 0
+
)
' 1:5
10
3
This rule is very di¢ cult to explain because the strong interaction e¤ect is hard to study accurately.
1.2
Milestones of Weak Interaction
(a) Neutrino and Nuclear decay,
The emission of electron e from many di¤erent nuclei,
(A; Z) ! (A; Z + 1) + e
was observed at early study of radioactivity. The unusual feature is that the energy spectrum of e
is continuous rather than a sharp line as expected in a 2 body decay. If the basic mechanism for e
emission is
n!p+e
the energy momentum conservation will require e to hav a single energy while experimentally a
continuous distribution of energy was observed. Pauli (1930) postulated the presence of neutrino
which carries away energy and momentum in nuclear -decay,
n!p+e +
so that energy momentum conservation can be saved.
_
e
(b) Fermi Theory
Fermi (1934) proposed to explain the
interaction Lagrangian in the form,
GF
LF = p [p(x)
2
Fitting nuclear
decay by making analogy with QED to write down the weak
n(x)][e(x)
e (x)]
GF : Fermi coupling constant
decay reates give
GF '
10 5
;
Mp2
This theory works very well for
interaction was added
Mp is the proton mass
J = 0;
decays of many nuclei.
GF
LGT = p [p(x)
2
to account for
+ h:c:
J = 1 nuclear
5 n(x)][e(x)
5 e (x)]
Later Gamow-Teller
+ h:c:
decays.
(c) Parity violation and V - A theory
puzzle
In 1950’s, it was observed that there are two decays
!
!
+
+
+
+
;
(even parity)
+
0
;
(odd parity)
while and have the same mass, charge and spin. It is very di¢ cult to understand these features
if the parity is a good symmetry.
1956 : Lee and Yang proposed that parity is not conserved in weak decays and suggested many
experiments to test this hyposethsis.
1957 : C. S. Wu showed that e in 60 Co decay has the property,
D! !E
! !
p 6= 0;
; p spin and momentum of e
This implies that the parity symmetry is violated in this decay.
(d) V-A theory (1958 Feynman and Gell-Mann, Sudarshan and Marshak, Sakurai)
As a result of parity violation, the e¤ective weak interaction was casted in the form with V
currents,
GF
Lef f = p J y J + h:c:
2
A
where
J (x) = Jl (x) + Jh (x)
Jl (x) =
and
_
e
_
Jh (x) = u
5 )e
(1
5 )(cos c d
c
+
_
(1
(1
5)
+ sin
;
c s)
leptonic current
(1)
hadronic current
: Cabibbo angle
Note that in V-A form the fermion …elds are all left-handed.
De…ne
1
(1
L
5)
2
Then we can simplify the form of weak currents,
Jl (x) = 2
eL
eL + 2
L
L
+ :::
Phenomenologically, the V-A theory has been quite successful in most of the weak interactions phenomena.
Di¢ culties:
(1) Not renormalizable
In the Fermi theory, 4 fermions interaction operator has dimension 6 and is not renormalizable. In
other words, the higher order graphs are more and more divergent. For example, in decay we have
(2) Violate unitarity
The tree amplitude for
+ e ! + e has only J = 1 partial wave at high energies and cross section
has the form,
( e) G2F S S = 2me E
On the other hand, unitarity for J=1 cross section is
(J = 1) <
Thus ( e) violates unitarity for E
blity, this violation is unacceptable.
1
S
300GeV: Since unitarity is a consequence of conservation of proba-
Intermidate Boson Theory(IVB)
In analogy with QED, we can introduce vector boson W to couple to the V-A current to mediate the weak
interaction
LW = g(J W + h:c:)
For example, the decay is now mediated by W-exchange.
Since weak interaction is short range, we need W-boson to be massive MW 6= 0. Use the massive W-boson
propagator in the form
g
k k
2
MW
g
! 2
2
MW
MW
+
k2
when jk j
MW
g2
GF
= p
2
MW
2
no longer violates unitarity. But the violation of unitarity
We see that this reproduces 4-fermion interaction with
In this theory, the scattering
shows up in other processes like
+e !
+
e
+
! W+ + W
and the theory is still non-renormalizable.
1.3
Construction of SU(2) U(1) model
Choice of group
In IVB theory, the interaction is described by
LW = g(J W + h:c)
For simplicity we neglect all other fermions excepts ; e and write the current as
J =
(1
5 )e
Recall that in the electromagnetic interaction, we have
Lem = eJ em A ;
where J em = e
e
De…ne the eletromagnetic and weak charges as the intergals of the time-component of the currents
Z
Z
1
1
3
T+ =
d xJ0 (x) =
d3 x y (1
T = (T+ )y
5 )e
2
2
Z
Z
Q = d3 xJ0em (x) =
d3 xey e
We can compute the commutator [T+ ; T ] = 2T3
Z
1
T3 =
d3 x[ y (1
4
5)
ey (1
5 )e]
6= Q
This means that these 3 charges, T+ ; T and Q don’t form a SU (2) algebra: The reason is that in order
for the eletric charge operator Q to be a generator of SU (2) it has to be traceless. In our case here it is
not the case. Futhre more the weak charges T have the V A form while the em charge Q is pure vector.
At this point, there are 2 alternatives:
(a) Introduce another guage boson coupled to T3 : The generatros corresponding to these 4 gauge bosons
can then form the group SU (2) U (1) : This will be the choice we will adapt eventually.
(b) We can add new fermions to modify the currents such that T+ ; T and Q do form a SU(2) algebra
(Georgi and Glashow 1972). Here we introduce new fermions to extend the multiplet into triplets
0
1
E+
1
@ e cos + N sin A
(1
5)
2
e
0 + 1
E
1
@
N A
(1 + 5 )
2
e
and a singlet
1
(1 +
2
5 ) (N
cos
e sin
)
The weak charge is then
T+
Z
1
d3 x E + (1
=
5 ) ( e cos + N sin )
2
+
+ ( e cos + N sin ) (1
5 ) e + E (1 +
5) N
+ N y (1 +
5) e
It is then straightforward to verify that
[T+ ; T ] = 2Q
with
Q=
Z
3
h
y
d x E E
i
ee
y
Clearly, in this model only electromagnetic current is neutral and others all carry charges. When
neutral weak current reactions were discovered in 1973, this model is ruled out.
Unitarity argument
Equivalently we can argue from unitarity that it is necessary to introduce either new leptons or a new
_
guage boson. Consider in IVB theory the reaction +
! W + + W where both W 0 s are longitudinally
polarized. The lowest order amplitude corresponding to the following graph is
T
_
! W +W
=
=
i
( ig ")
/ (1
p/ k/ me
_
"/0 (p/ k)
/ "/(1
5)
2g 2 v p0
u (p)
2
2
(p k)
me
_
i v p0
ig "/0 (1
5)
The polarization vectors
"(i) (k)
with
"(i) "(j) =
ij
and
k "(i) = 0
5 ) u (p)
(2)
maybe chosen in the rest frame of W boson as
(i)
(i)
"0 = 0;
"j = ij
p
For a moving W boson with k = (E; 0; 0; k) with k = E 2 Mw2 ; we can make a Lorentz transformation
along the z axis: The transverse polarizations do not change while the longitudinal polarization becomes
2
MW
1
(3)
(k; 0; 0; E): In the high energy limit with k = E
" =
+
; we see that
MW
2E
"(3) =
k
+O
MW
MW
E
Then for longitudinally polarized W bosons, the scattering amplitude in Eq(2)
_
2g 2
k/0
0
v
p
(p/
k 2 2p k
MW
2g 2 _ 0 0
5 ) u (p)
2 v p k/ (1
MW
T
k/
(1
MW
k)
/
becomes,
5 ) u (p)
(3)
To show more explicitly this amplitude is a pure J = 1 partial wave, we take
p0 = (E; 0; 0; E)
p = (E; 0; 0; E) ;
!
!
Since
and
_
have opposite helicities, we have
1
0
1
p
u (p) = E @ ! !
p A
E
_
v p0 =
!
k 0 = E; k e ;
k = E; k e ;
p
E
y
1=2
! !0
p
E
with e = (sin ; 0; cos )
1=2
; 1
=
p
1
E
1=2
z
!
=
p
E
y
1=2 (
z;
1)
where
1=2
1
0
=
;
1=2
=
0
1
Then the combination in Eq(3) becomes,
_
0
0
v p k/ (1
5 ) u (p)
= E
=
y
1=2 (
1; 1)
1
1
1
1
4E
y
1=2
E
k
! !
1
1
E
k
k
e
! !
e
E
!
1=2
! !
e
1=2
= 4Ek sin
We have then
T
GF E 2 sin
as
E!1
(4)
The partial wave expansin for the helicity amplitude is of the form,
T
3 4; 1
2 (E; ) =
1
X
(2J + 1) T J3
4; 1 2
(E) dJ ( )
J=M
where
=
1 =
2 = 1=2 and 3 = 4 = 0 are the helicities of the initial and …nal particles with
J ( ) is the usual rotation matrix with d1 ( ) = sin :
=
1;
=
and
M
=
max
(
;
)
=
1:
d
1
2
3
4
10
It is clear that T in Eq(4) corresponds to a pure J = 1 partial wave and violates the unitarity bound of
T J=1 (E)
constant at high energies. To cancel this bad high energy behavior we need other diagrams
for this reaction. There are 2 possibilities: s-channel or u-channel exchange diagrams.
(a) The heavy lepton alternative; the u-channel exchange in the following diagram(a) yields the amplitude,
Tu
_
! W +W
_
0
2g 2 v p0
=
"/(p/ k)
/ "/0 (1
5)
u (p)
2
(p k 0 )
m2E
2g 02 _ 0 0
2 v p k/ (1
MW
=
5 ) u (p)
If g 2 = g 02 ; this will cancel the bad behavior given in Eq(3)
(b) The neutral vector boson alternative; the s-channel exchange in the diagram (b) given above gives
the amplitude
Ts
_
! W +W
=
_
i v p0
"
i
if
g
+
(1
5 ) u (p) L
(k + k) (k + k)
MZ2
Choose the ZW W coupling to have Yang-Mills structure
h
L
= if 0 k 0 k g
2k 0 + k
g
"0
k 0 " (k) =
and
Ts '
if 0
k0
k
if 0
2
MW
k0
k
" "0
k 0 " (k)
#
1
(k + k)2 MZ2
+ k 0 + 2k
we get
L
#"
"0
g
i
2k 0 " "0 + 2k "0 "
k k0
ff0 _ 0 0
2 v p k/ (1
MW
5 ) u (p)
Thus if we choose f f 0 = 2g 2 ; this will also canel the amplitude in Eq(3). This corresponds to the
case of adding an additional U (1) symmetry discussed before.
In fact if one demands that all the amplitudes which violate unitarity be cancelled out, one ends up
with a renormalizable Lagrangian which is the same as the one derived from the algebraic approach.
We now choose the gauge group to be SU (2)
L=
U (1). The Lagrangian for the gauge …elds is then
1 i
F Fi
4
1
G G
4
where
F i = @ Ai
G
@ Ai + g
=@ B
ijk
Aj Ak
SU (2)
@ B
U (1)
gauge …elds
gauge …eld
Fermions
Clearly, from the stuctrue of the weak charge current given in Eq(1) ; e form a doublet under SU(2),
lL =
1
(1
2
5)
e
For convenience, we introduced left-handed and right-handed …elds
Then
T+ =
Note that
L
1
(1
2
Z
+
3
L eL )d x;
(
Q
5)
;
T =
Z
T3 =
It is straightforward to show that
[Q
Z
1
(
2
[
L
eL
5)
;
=
3
(e+
L L )d x;
+
L L
Q=
Z
L
+
R
+
(e+
L eL + eR eR )
3
e+
R eR ]d x
+ e+
L eL )
T3 ; Ti ] = 0 ;
Thus we can take Q T3 to be U (1) charge Y
charges for fermions are
lL =
1
(1 +
2
R
i = 1; 2; 3
2(Q
T3 ); sometime called weak hypercharge. The Y
Y =
1;
eR Y =
2
With these quantum numbers, the Lagrangian for gauge coupling is
L2 = lL i
D ll + lR i
where
D lR
= (@
ig
~
~ A
2
ig
D lL = (@
ig
~
~ A
2
ig
D
0
(5)
Y
B )
2
For example,
Spontaneous Symmetry Breaking
The symmetry braking pattern we want is SU (2)
doublet with hypercharge Y = 1;
=
The Lagrangian contaning
y
0
0
Y
B )lL
2
U (1) ! U (1)em . Choose scalar …elds in SU(2)
; Y =1
is of the form,
L3 = (D
)y (D
)
V( )
where
D
= (@
ig
~ A~
2
0
ig
B )
2
and
V( )=
2 y
+ (
y
)2
In addition there is a coupling between leptons and scalar …eld ;
_
L4 = f LL eR + h:c:
As we have seen before, the spontaneous symmetry breaking is generated by the vaccum expectation value
r
2
1
0
< >0 = h0j j0i = p
with v =
2 v
Write the scalar …eld in the form
(x) = U
1
(~ )
0
v+ (x)
p
2
!
i~ (x) ~
where U (~ ) = exp[
]
v
(6)
Gauge Transformation
The scalar …eld in Eq(6) is in the form of gauge transformation. We can then simplify the form by a gauge
transformation
1
0
0
= U (~ ) = p
v
+
(x)
2
~0
~
~ A
~ A
= U (~ )
U
2
2
1
(~ )
i
(@ U )U
g
1
Now the …eld ~ (x) disappears from the Lagrangian as a consequence of gauge invariance. From L4 (Yukawa
coupling), VEV of the scalar …eld gives
1
L4 = f p (lL <
2
(x) _
> eR + h:c:) + f p ( e L eR + h:c:)
2
as consequence the electron is now massive with electron mass
f
me = p v
2
Mass spectrum
We now list the mass spectrum of the theory after the spontaneous symmetry breaking:
(a) Fermion mass
fv
me = p
2
(b) Scalar mass(Higgs)
0
V( )=
2 2
+ v
3
+
4
4
! m =
p
2
(c) Gauge boson masses
From the covariant derivative in L3
v2
L3 =
2
y
0
0
0
0
~0
~0
~ A
gB
~ A
gB
(g
+
)(g
+
) +
2
2
2
2
;
we get the mass terms for the gauge bosons,
v2 2 1 2
fg [(A ) + (A2 )2 ] + (gA3
8
1
2
W + W + MZ2 Z Z +
= MW
2
L3 =
0
g B )2 g +
=
0
1
where
1
W + = p (A1
2
1
The …eldA = p
0
Z =p
(g A
g2 + g02
0
1
(g A3
g 2 +g 0 2
+ gB )
2
MW
=
iA2 );
3
MZ2
gB );
g2v2
4
0
g2 + g 2 2
=
v
4
massless photon
does not appear in L3 and is massless. This clearly corresponds to photon …eld.
For convience we de…ne
0
tan
W
g
=
g
: Weinberg angle or weak mixing angle
W
Then we can write
Z = cos
WA
3
sin
A = sin
3
WA
Note that there is a relation between MW ; MZ and
=
MZ2 =
WB
cos
W
WB
of the form,
W
2
MW
MZ2 cos2
g2v2
sec2
4
=1
W
which is a consequence of the doublet nature of the scalar …elds.
The weak interactions mediated by W and Z bosons can be read out from Eq(5)
(a) Charged current
g
1
Lcc = p (J y W y + h:c:) J y = J 1 + iJ 2 =
2
2
(1
Again to get 4-fermion interaction as low energy limit, we require
g2
GF
= p
2
8MW
2
which implies that
sp
2
GF
v=
246Gev
This is usually referred to as the weak scale.
(b) Neutral Current
The Largrangian for the neutral currents is
3
3
LN C = gJ A
0
g
g
+ J Y B = eJ em A +
2
cos
where
e = g sin
W;
and
JZ = J3
sin2
WJ
em
JZZ
W
5 )e
is the weak neutral current. We can de…ne the weak neutral charge as
Z
Z
Q = J0Z d3 x = (T3 sin2 W Q)
This means that the coupling strength of fermions to Z-boson is proportional to the quantum number
T3 sin2 W Q.
In particluar, Z boson can contribute to the scattering
e
+e!
e
+e
The measurement of this corss section in the 1970’s give sin2
and MZ 90 GeV.
1.3.1
W
0:22. This yields MW
80 GeV
Generalization to more than one family.
From 4-fermion and IVB theory, the form of weak currents of leptons and hadrons gives the following
multiplets structure,
u
e
uR ; dR ; sR
;
eR ; R
d
e L
L
L
where
d = cos
Cd
+ sin
Cs
The neutral current in the down quark sector is of the form
LN C
1
1
1
+ sin2 C )d L sin2 W (dR dR + sR sR )]Z
2
3
3
1
1
2
+ sin W )[(dL dL + sL sL ) + sin W cos W (dL sL + sL
= (
2
3
= [d
(
dL ) + :::
The term dL sL + sL dL gives rise to S = 1 neutral current processes, e.g. KL !
same order of magnitude as charged current interaction. But experimentally,
R=
(KL ! + + )
(K + ! + )
10
+
+
with
8
Thus we can not have S = 1 neutral curent process at the same order of magnitude as the charged
current process.
GIM mechanism
Glashow, Iliopoulos and Maiani (1970) suggested that there is a 4-th quark, the charm quark c, which
couples to the orthogonal combination s = sin c d + cos c s so that the multiplets look like
u
d
;
L
c
s
L
As a result , the
d (
S = 1, neutral current is canceled out. The new current is of the form
1 1
+ sin2
2 3
W)
1 1
+ sin2
2 3
d +s (
W)
s =(
1 1
+ sin2
2 3
W )(d
d+s
s)
which conserves the strangeness.
Quark mixing
Before the spontaneous symmetry braking, fermions are all massless because L and R have di¤erent quantum numbers under SU (2) U (1). So the mass term L R + h:c: is not invariant under SU (2) U (1)
groups and can not appear in the Lagrangian. When we have more than one doublets, iR ; iL all have
the same quantum numbers with respect to SU (2) U (1) group we call them "weak eigenstates". Here
the index i labels the di¤erent families of weak eigenstates. When spontaneous symmetry breaking takes
place, fermions obtain their masses through Yukawa coupling.
0
LY = (fij giL uRj + fij giL dRj ) + h:c:
Note that from the requirement of renormalizability we need to write down all possible terms consistent
with SU (2) U (1) symmetry. Since Yukawa coupling constants fij; , fij0 are arbitrary, the fermion mass
matrices are in general not diagonal. When mass matrices are diagonalized the mass matrix we obtain the
mass eigenstates which are not the same as the weak eigenstates. The mass matrices in the up and down
sectors are given by
v
v
0
(u)
(d)
mij = fij p
mij = fij p
2
2
These matrices which are sandwiched between left and right handed …elds can be diagonalized by bi-unitary
transformations, i.e. given a mass matrix mij ;there exits unitary matrices S and T such that
S y mT = md
is diagonal. Basically, S is the unitary matrix which diagnoalizes the hermitian combination mm+ , i. e.
S + (mm+ )S = m2d
Biunitary transformation
Write
0
m2d = @
De…ne
0
md = @
and
m21
m22
1
A
m23
1
m1
A
m2
m3
H = Smd S y
hermitian
De…ne a matrix V by
V
H
1
m
Then
VVy =H
1
mmy H
1
=H
1
Sm2d S y H
1
=H
1
H 2H
So V is unitary and we have
S y HS = md ;
=)
S y mV y S = md
Or
S y mT = md ;
with
T = V yS
1
=1
If we write the left-handed doublets, (weak eigenstates) as
0
0
u
0
d
q1L =
c
0
s
q2L =
L
L
These weak eigenstates are related to mass eigenstates by unitary transformations,
0
u
0
c
= Su
0
u
c
d
0
s
;
= Sd
d
s
Note that in the coupling to charged gauge boson W , we have
_
g
LW = p W [ q 1L
2
y
q1L + q2L
y
q2L ] + h:c:
and is invariant under unitary transformation in q1L ; q2L space, i.e.
0
q1L
0
q2L
q1L
q2L
=V
V V y = 1 = V yV
We can use this feature to put all mixing in the down quark sector,
0
qiL =
Here U is a 2
u
00
d
;
L
c
00
s
00
;
d
00
s
where
L
=U
d
s
2 unitary matrix. Clearly, we can extend this to 3 generations with result
0 1
0 00 1
d
d
u
c
t
00
A
@
@
s A
=U
qiL :
;
;
s
00
00
00
d
s
b
00
b
b
Now U is a 3 3 unitary matrix, usually called the Cabibbo-Kobayahsi-Maskawa (CKM) matrix.
CP violation Phase
CP violetion can come form complex coupling to gauge bosons. The gauge coupling of W to quarks
is governed by the 3 3 unitary matrix U discussed above. This unitary matrix U can have many complex
entries. However, in diagonalzing the mass matrices, S y (mmy )S = m2d There is ambiguity in the matrix S,
0
in the form of diagonal phases. In other words, if S diagonalizes the mass matrix, so does S
0 i
1
e 1 ::: :::
B ..
.. C
0
..
.
. C
S =SB
@ .
A
..
n
.
::: e
We can use this property to rede…ne the quark …elds to reduce the phase in U . It turns out that for n n
unitary matrix, number of independent physical phases left over is
(n
1)(n 2)
2
Thus to get CP violetion we need to go to 3 generations or more (Kobayashi Mskawa). Here we give a
constructive proof of this statement. Let us start with a …rst doublet written in the form,
q1L =
u
U11 d + U12 s + U13 b
If U11 has phase ;
U11 = R11 ei ;
R11
real
then this phase
can be absorbed in the rede…niton of the u quark …eld
u ! u0 = ue
and we can write
i
u0
0 s + U0 b
R11 d + U12
13
q1L = ei
Similarly, we can factor out the complex phases of U21 and U31 by rede…nition of c and t quark …elds.
These overall phases are immaterial because there are no gauge couplings between doublets with di¤erent
family indices. Finally we can absorb two more phases of U12 and U13 by a rede…nition of the s and b
…elds. The doublets now take the form
u0
R11 d + R12 s + R13 b
0
c
;
R21 d + R22 ei 1 s + R23 ei 2 b
L
t0
R31 d + R32 ei 3 s + R33 ei 4 b
;
L
;
L
Now we have reduced the number of parameters to 13. The normalization conditons of each down-like
state gives 3 real conditions aen orthogonality conditions among di¤erent states give 6 real conditions on
the parameters, Now we are down to 4 parameters. Since we need 3 parameters for the real orthogonal
matrix, we end up with one independent phase.
Flavor consevation in neutral current interaction
It turns out that the coupling of neutral Z boson to the fermions conserve ‡avors. This can be illustrated
as follows. We …rst write the neutral currents in terms of quark …elds which are weak eigenste,
X_
T3 ( i ) sin2 W Q ( i ) i
JZ =
i
i
Separate into left- and right-handed …elds and distingush the up and down components,
JZ
=
X
_0
(uLi
i
_0
+uRi
1
2
sin2
_0
sin2
W
2
3
W
2
3
u0Ri + d Ri
u0Li + d Li
_0
sin2
1
+ sin2
2
W
1
3
W
1
3
d0Li
d0Ri
0 and mass eigen states q 0 are related by unitary matries,
Since weak eigen states qiL
iL
u0Li = U (uL )ij uLj ;
_0
We see that these unitary matrices cancel out in the combination, uLi u0Li so that the neutral current in
terms of mass eigenstates has the same form as the one in terms of weak eigenstates. Thus it conserves
all quark ‡avor. Note this feature is due to the fact that all quarks with same helicity and electric charge
have the same quantum number with repect to SU (2) U (1) gauge group. In other words, if there are
quarks in representations other than SU (2) doublets, then there will be ‡avor changing neutral current if
the new quarks have the same eletric charge as either u or d quark.