5Electroweak theory
III The Standard Model
5.3 Quarks
We now move on to study quarks. There are 6 flavours of quarks, coming in
three generations:
Charge First generation Second generation Third generation
+
2
3
Up (u) Charm (c) Top (t)
−
1
3
Down (d) Strange (s) Bottom (b)
each of which is a spinor.
The right handed fields have trivial
SU
(2) representations, and are thus
SU
(2)
singlets. We write them as
u
R
=
u
R
c
R
t
R
which have Y = Q = +
2
3
, and
d
R
=
d
R
s
R
b
R
which have Y = Q = −
1
3
.
The left-handed fields are in SU(2) doublets
Q
i
L
=
u
i
L
d
i
L
=
u
d
L
c
s
L
t
b
L
and these all have Y =
1
6
. Here i = 1, 2, 3 labels generations.
The electroweak part of the Lagrangian is again straightforward, given by
L
EW
quark
=
¯
Q
L
i
/
DQ
L
+ ¯u
R
i
/
Du
R
+
¯
d
R
i
/
Dd
R
.
It takes a bit more work to couple these things with
φ
. We want to do so in a
gauge invariant way. To match up the
SU
(2) part, we need a term that looks
like
¯
Q
i
L
φ,
as both
Q
L
and
φ
have the fundamental representation. Now to have an
invariant U(1) theory, the hypercharges have to add to zero, as under a U(1)
gauge transformation
β
(
x
), the term transforms as
e
i
P
Y
i
β(x)
. We see that the
term
¯
Q
i
L
φd
i
R
works well. However, coupling
¯
Q
i
L
with
u
R
is more problematic.
¯
Q
i
L
has
hypercharge
−
1
6
and
u
R
has hypercharge +
2
3
. So we need another term of
hypercharge −
1
2
. To do so, we introduce the charge-conjugated φ, defined by
(φ
c
)
α
= ε
αβ
φ
∗β
.
One can check that this transforms with the fundamental
SU
(2) representation,
and has
Y
=
−
1
2
. Inserting generic coupling coefficients
λ
ij
u,d
, we write the
Lagrangian as
L
quark,φ
= −
√
2(λ
ij
d
¯
Q
i
L
φd
i
R
+ λ
ij
u
¯
Q
i
L
φ
c
u
j
R
+ h.c.).
Here h.c. means “hermitian conjugate”.
Let’s try to diagonalize this in the same way we diagonalized leptons. We
again work in unitary gauge, so that
φ(x) =
1
√
2
0
v + h(x)
,
Then the above Lagrangian can be written as
L
quark,φ
= −(λ
ij
d
¯
d
i
L
(v + h)d
i
R
+ λ
ij
u
¯u
i
L
(v + h)u
j
R
+ h.c.).
We now write
λ
u
= U
u
Λ
u
S
†
u
λ
d
= U
d
Λ
d
S
†
d
,
where Λ
u,d
are diagonal and
S
u,d
, U
u,d
are unitary. We can then transform the
field in a similar way:
u
L
7→ U
u
u
L
, d
L
7→ U
d
d
L
, u
R
7→ S
u
u
R
, d
R
7→ S
d
d
R
.
and then it is a routine check that this diagonalizes
L
quark,φ
. In particular, the
mass term looks like
−
X
i
(m
i
d
¯
d
i
L
d
i
R
+ m
i
d
¯u
i
L
u
i
R
+ h.c.),
where
m
i
q
= vΛ
ii
q
.
How does this affect our electroweak interactions? The
¯u
R
i
/
Du
R
and
¯
d
R
i
/
Dd
R
are staying fine, but the different components of
¯
Q
L
“differently”. In particular,
the
W
±
µ
piece given by
¯
Q
L
i
/
DQ
L
is not invariant. That piece, can be explicitly
written out as
g
2
√
2
J
±µ
W
±
µ
,
where
J
µ+
= ¯u
i
L
γ
µ
d
i
L
.
Under the basis transformation, this becomes
¯u
i
L
γ
µ
(U
†
u
U
d
)
ij
d
j
L
.
This is not going to be diagonal in general. This leads to inter-generational
quark couplings. In other words, we have discovered that the mass eigenstates
are (in general) not equal to the weak eigenstates.
The mixing is dictated by the Cabibbo–Kabyashi–Maskawa matrix (CKM
matrix )
V
CKM
= U
†
u
U
d
.
Explicitly, we can write
V
CKM
=
V
ud
V
us
V
ub
V
cd
V
cs
V
cb
V
td
V
ts
V
tb
,
where the subscript indicate which two things it mixes. So far, these matrices
are not predicted by any theory, and is manually plugged into the model.
However, the entries aren’t completely arbitrary. Being the product of two
unitary matrices, we know
V
CKM
is a unitary matrix. Moreover, the entries are
only uniquely defined up to some choice of phases, and this further cuts down
the number of degrees of freedom.
If we only had two generations, then we get what is known as Cabibbo
mixing. A general 2
×
2 unitary matrix has 4 real parameters — one angle and
three phases. However, redefining each of the 4 quark fields with a global U(1)
transformation, we can remove three relative phases (if we change all of them by
the same phase, then nothing happens). We can then write this matrix as
V =
cos θ
c
sin θ
c
−sin θ
c
cos θ
c
,
where
θ
c
is the Cabibbo angle. Experimentally, we have
sin θ
c
≈
0
.
22. It turns
out the reality of this implies that CP is conserved, which is left as an exercise
on the example sheet.
In this case, we explicitly have
J
µ†
= cos θ
c
¯u
L
γ
µ
d
L
+ sin θ
c
¯u
l
γ
µ
s
L
− sin θ
c
¯c
L
γ
µ
d
L
+ cos θ
c
¯c
L
γ
µ
s
L
.
If the angle is 0, then we have no mixing between the generations.
With three generations, there are nine parameters. We can think of this as
3 (Euler) angles and 6 phases. As before, we can re-define some of the quark
fields to get rid of five relative phases. We can then write
V
CKM
in terms of
three angles and 1 phase. In general, this
V
CKM
is not real, and this gives us
CP violation. Of course, if it happens that this phase is real, then we do not
have CP violation. Unfortunately, experimentally, we figured this is not the case.
By the CPT theorem, we thus deduce that T violation happens as well.