6Weak decays
III The Standard Model
6.5 K
0
-
¯
K
0
mixing
We now move on to consider our final example of weak decays, and look at
K
0
-
¯
K
0
mixing. We will only do this rather qualitatively, and look at the effects
of CP violation.
Kaons contain a strange quark/antiquark. There are four “flavour eigenstates”
given by
K
0
(¯sd),
¯
K
0
(
¯
ds), K
+
(¯su), K
−
(¯us).
These are the lightest kaons, and they have spin
J
= 0 and parity
p
=
−ve
. We
can concisely write these information as J
p
= 0
−
. These are pseudoscalars.
We want to understand how these things transform under CP. Parity trans-
formation doesn’t change the particle contents, and charge conjugation swaps
particles and anti-particles. Thus, we would expect CP to send K
0
to
¯
K
0
, and
vice versa. For kaons at rest, we can pick the relative phases such that
ˆ
C
ˆ
P
K
0
= −
¯
K
0
ˆ
C
ˆ
P
¯
K
0
= −
K
0
.
So the CP eigenstates are just
K
0
±
=
1
√
2
(
K
0
∓
¯
K
0
).
Then we have
ˆ
C
ˆ
P
K
0
±
= ±
K
0
±
.
Let’s consider the two possible decays
K
0
→ π
0
π
0
and
K
0
→ π
+
π
−
. This
requires converting one of the strange quarks into an up or down quark.
d
¯s ¯u
W
u
¯
d
K
0
π
−
π
+
d
¯s ¯u
W
¯
d
u
K
0
π
0
π
0
From the conservation of angular momentum, the total angular momentum of
ππ is zero. Since they are also spinless, the orbital angular momentum L = 0.
When we apply CP to the final states, we note that the relative phases of
π
+
and π
−
(or π
0
and π
0
) cancel out each other. So we have
ˆ
C
ˆ
P
π
+
π
−
= (−1)
L
π
+
π
−
=
π
+
π
−
,
where the relative phases of π
+
and π
−
cancel out. Similarly, we have
ˆ
C
ˆ
P
π
0
π
0
=
π
0
π
0
.
Therefore ππ is always a CP eigenstate with eigenvalue +1.
What does this tell us about the possible decays? We know that CP is
conserved by the strong and electromagnetic interactions. If it were conserved
by the weak interaction as well, then there is a restriction on what can happen.
We know that
K
0
+
→ ππ
is allowed, because both sides have CP eigenvalue +1, but
K
0
−
→ ππ
is not. So
K
0
+
is “short-lived”, and
K
0
−
is “long-lived”. Of course,
K
0
−
will still
decay, but must do so via more elaborate channels, e.g. K
0
−
→ πππ.
Does this agree with experiments? When we actually look at Kaons, we find
two neutral Kaons, which we shall call
K
0
S
and
K
0
L
. As the subscripts suggest,
K
0
S
has a “short” lifetime of
τ ≈ 9 × 10
−11
s
, while
K
0
L
has a “long” lifetime of
τ ≈ 5 × 10
−8
s.
But does this actually CP is not violated? Not necessarily. For us to be
correct, we want to make sure
K
0
L
never decays to
ππ
. We define the quantities
η
+−
=
|hπ
+
π
−
|H
K
0
L
|
|hπ
+
π
−
|H |K
0
S
i|
, η
00
=
|
π
0
π
0
H
K
0
L
|
|hπ
0
π
0
|H |K
0
S
i|
Experimentally, when we measure these things, we have
η
±
≈ η
00
≈ 2.2 × 10
−3
6= 0.
So K
0
L
does decay into ππ.
If we think about what is going on here, there are two ways CP can be
violated:
– Direct CP violation of s → u due to a phase in V
CKM
.
– Indirect CP violation due to K
0
→
¯
K
0
or vice-versa, then decaying.
Of course, ultimately, the “indirect violation” is still due to phases in the CKM
matrix, but the second is more “higher level”.
It turns out in this particular process, it is the indirect CP violation that is
mainly responsible, and the dominant contributions are “box diagrams”, where
the change in strangeness ∆S = 2.
d
u, c, t
s
¯s
¯u, ¯c,
¯
t
¯
d
W W
K
0
¯
K
0
d
¯s
s
¯
d
W
W
K
0
¯
K
0
Given our experimental results, we know that
K
0
S
and
K
0
L
aren’t quite
K
0
+
and
K
0
−
themselves, but have some corrections. We can write them as
K
0
S
=
1
p
1 + |ε
1
|
2
(
K
0
+
+ ε
1
K
0
−
) ≈
K
0
+
K
0
L
=
1
p
1 + |ε
2
|
2
(
K
0
−
+ ε
2
K
0
+
) ≈
K
0
−
,
where
ε
1
, ε
2
∈ C
are some small complex numbers. This way, very occasionally,
K
0
L
can decay as K
0
+
.
We assume that we just have two state mixing, and ignore details of the
strong interaction. Then as time progresses, we can write have
|K
S
(t)i = a
S
(t)
K
0
+ b
S
(t)
¯
K
0
|K
L
(t)i = a
L
(t)
K
0
+ b
L
(t)
¯
K
0
for some (complex) functions
a
S
, b
S
, a
L
, b
L
. Recall that Schr¨odinger’s equation
says
i
d
dt
|ψ(t)i = H |ψ(t)i.
Thus, we can write
i
d
dt
a
b
= R
a
b
,
where
R =
K
0
H
0
K
0
K
0
H
0
¯
K
0
¯
K
0
H
0
K
0
¯
K
0
H
0
¯
K
0
and
H
0
is the next-to-leading order weak Hamiltonian. Because Kaons decay
in finite time, we know
R
is not Hermitian. By general linear algebra, we can
always write it in the form
R = M −
i
2
Γ,
where
M
and Γ are Hermitian. We call
M
the mass matrix , and Γ the decay
matrix .
We are not going to actually compute
R
, but we are going to use the known
symmetries to put some constraint on
R
. We will consider the action of Θ =
ˆ
C
ˆ
P
ˆ
T
.
The CPT theorem says observables should be invariant under conjugation by Θ.
So if
A
is Hermitian, then Θ
A
Θ
−1
=
A
. Now our
H
0
is not actually Hermitian,
but as above, we can write it as
H
0
= A + iB,
where A and B are Hermitian. Now noting that Θ is anti-unitary, we have
ΘH
0
Θ
−1
= A − iB = H
0†
.
In the rest frame of a particle
¯
K
0
, we know
ˆ
T
K
0
=
K
0
, and similarly for
¯
K
0
. So we have
Θ
¯
K
0
= −
K
0
, Θ
K
0
= −
¯
K
0
,
Since we are going to involve time reversal, we stop using bra-ket notation for
the moment. We have
R
11
= (K
0
, H
0
K
0
) = (Θ
−1
ΘK
0
, H
0
Θ
−1
ΘK
0
) = (
¯
K
0
, H
0†
¯
K
0
)
∗
= (H
0
¯
K
0
,
¯
K
0
)
∗
= (
¯
K
0
, H
0
¯
K
0
) = R
22
Now if
ˆ
T
was a good symmetry (i.e.
ˆ
C
ˆ
P
is good as well), then a similar
computation shows that
R
12
= R
21
.
We can show that we in fact have
ε
1
= ε
2
= ε =
√
R
21
−
√
R
21
√
R
12
+
√
R
21
.
So if CP is conserved, then R
12
= R
21
, and therefore ε
1
= ε
2
= ε = 0.
Thus, we see that if we want to have mixing, then we must have
ε
1
, ε
2
6
= 0.
So we need R
12
6= R
21
. In other words, we must have CP violation!
One can also show that
η
+−
= ε + ε
0
, η
00
= ε − 2ε
0
,
where
ε
0
measures the direct source of CP violation. By looking at these two
modes of decay, we can figure out the values of
ε
and
ε
0
. Experimentally, we find
|ε| = (2.228 ± 0.011) × 10
−3
,
and
ε
ε
0
= (1.66 ± 0.23) × 10
−3
.
As claimed, it is the indirect CP violation that is dominant, and the direct one
is suppressed by a factor of 10
−3
.
Other decays can be used to probe
K
0
L,S
. For example, we can look at
semi-leptonic decays. We can check that
K
0
→ π
−
e
+
ν
e
is possible, while
K
0
→ π
+
e
−
¯ν
e
is not.
¯
K
0
has the opposite phenomenon. To show these, we just have to try to
write down diagrams for these events.
Now if CP is conserved, we’d expect the decay rates
Γ(K
0
L,S
→ π
−
e
+
ν
e
) = Γ(K
0
L,S
→ π
+
e
−
¯ν
e
),
since we expect K
L,S
to consist of the same amount of K
0
and
¯
K
0
.
We define
A
L
=
Γ(K
0
L
→ π
−
e
+
ν
e
) − Γ(K
0
L
→ π
+
e
−
¯ν
e
)
Γ(K
0
L
→ π
−
e
+
ν
e
) + Γ(K
0
L
→ π
+
e
−
¯ν
e
)
.
If this is non-zero, then we have evidence for CP violation. Experimentally, we
find
A
L
= (3.32 ± 0.06) × 10
−3
≈ 2 Re(ε).
This is small, but certainly significantly non-zero.