342
21. CP Violation and Oscillation Phenomena
Suppose first that the penguin contributions could be neglected. Then the
asymmetry A π + π − would measure
(
)
(
)
¯
−2iβ A +−
V ub V
∗
−2iβ
ud
Imλ π + π − = Im e
= Im e
V ∗
A +−
ub V ud
= Ime
−2i(γ+β) = sin 2α
(21.54)
where α is defined as π − β − γ. Unfortunately, this simple result is spoiled by
the penguin contributions, which there is no good reason to ignore. However,
Gronau and London (1990) showed how an isospin analysis could disentangle the tree and penguin parts. The method involves the three amplitudes
A +− , A 00 (B
0
→ π
0 π
0 ), and A +0 (B
+
→ π
+ π
0 ).
First of all, note that Bose statistics for the 2π states requires them to have
only the symmetric isospin states I = 0 or 2, since the angular momentum is
zero. Next, the effective non-leptonic weak Hamiltonian H ˆ nl acting in the tree
diagram transition contains both ΔI = 1/2 and ΔI = 3/2 pieces; combining
with the initial I = 1/2 of the B meson, the first piece will lead only to the
I = 0 final state, while the second contributes to both I = 0 and I = 2 final
states. However, since the gluon in the penguin diagrams carries no isospin,
these diagrams can only change the isospin by ΔI = 1/2, which connects only
to the I = 0 final state. The conclusion is that the I = 2 final state is free of
penguins, and carries the pure tree phase.
This information can be exploited as follows (Gronau and London 1990).
First, the action of H ˆ nl on the B
0 state can be written as
1 1
1
1
ˆ
− › = √ A 3/2 |20› + √ A 1/2 |00›
(21.55)
H nl | 2 2
2
2
where as usual |II 3 › is the state with isospin I and third component I 3 . Expanding the states π
+ π
− , π
+ π
0 and π
0 π
0 in terms of definite isospin states,
one finds (problem 21.4)
A +− =
A +0 =
1
1
A 00 = √ A 3/2 − √ A 1/2
(21.56)
3
6
where A ij is the amplitude ‹π
i π
j
|H ˆ nl |B
i+j
›. The π
+ π
0 state can have only
I = 2, and arises solely from the tree diagram. Furthermore, the three complex
amplitudes A +− , A +0 and A 00 are expressed in terms of only two reduced
amplitudes A 3/2 and A 1/2 , leading to one relation between them:
1
√ A +− + A 00 = A +0 ,
(21.57)
2
1
√
6
A 3/2 +
1
√
3
A 1/2
√
3
2
A 3/2
21. CP Violation and Oscillation Phenomena
Suppose first that the penguin contributions could be neglected. Then the
asymmetry A π + π − would measure
(
)
(
)
¯
−2iβ A +−
V ub V
∗
−2iβ
ud
Imλ π + π − = Im e
= Im e
V ∗
A +−
ub V ud
= Ime
−2i(γ+β) = sin 2α
(21.54)
where α is defined as π − β − γ. Unfortunately, this simple result is spoiled by
the penguin contributions, which there is no good reason to ignore. However,
Gronau and London (1990) showed how an isospin analysis could disentangle the tree and penguin parts. The method involves the three amplitudes
A +− , A 00 (B
0
→ π
0 π
0 ), and A +0 (B
+
→ π
+ π
0 ).
First of all, note that Bose statistics for the 2π states requires them to have
only the symmetric isospin states I = 0 or 2, since the angular momentum is
zero. Next, the effective non-leptonic weak Hamiltonian H ˆ nl acting in the tree
diagram transition contains both ΔI = 1/2 and ΔI = 3/2 pieces; combining
with the initial I = 1/2 of the B meson, the first piece will lead only to the
I = 0 final state, while the second contributes to both I = 0 and I = 2 final
states. However, since the gluon in the penguin diagrams carries no isospin,
these diagrams can only change the isospin by ΔI = 1/2, which connects only
to the I = 0 final state. The conclusion is that the I = 2 final state is free of
penguins, and carries the pure tree phase.
This information can be exploited as follows (Gronau and London 1990).
First, the action of H ˆ nl on the B
0 state can be written as
1 1
1
1
ˆ
− › = √ A 3/2 |20› + √ A 1/2 |00›
(21.55)
H nl | 2 2
2
2
where as usual |II 3 › is the state with isospin I and third component I 3 . Expanding the states π
+ π
− , π
+ π
0 and π
0 π
0 in terms of definite isospin states,
one finds (problem 21.4)
A +− =
A +0 =
1
1
A 00 = √ A 3/2 − √ A 1/2
(21.56)
3
6
where A ij is the amplitude ‹π
i π
j
|H ˆ nl |B
i+j
›. The π
+ π
0 state can have only
I = 2, and arises solely from the tree diagram. Furthermore, the three complex
amplitudes A +− , A +0 and A 00 are expressed in terms of only two reduced
amplitudes A 3/2 and A 1/2 , leading to one relation between them:
1
√ A +− + A 00 = A +0 ,
(21.57)
2
1
√
6
A 3/2 +
1
√
3
A 1/2
√
3
2
A 3/2