6Weak decays
III The Standard Model
6.4 Pion decay
We are now going to look at a slightly different example. We are going to study
decays of pions, which are made up of a pair of quark and anti-quark. We will
in particular look at the decay
π
−
(¯ud) → e
−
¯ν
e
.
π
−
e
−
¯ν
e
p
k
q
This is actually quite hard to do from first principles, because
π
−
is made up of
quarks, and quark dynamics is dictated by QCD, which we don’t know about
yet. However, these quarks are not free to move around, and are strongly bound
together inside the pion. So the trick is to hide all the things that happen in the
QCD side in a single constant
F
π
, without trying to figure out, from our theory,
what F
π
actually is.
The relevant currents are
J
α
lept
= ¯ν
e
γ
α
(1 − γ
5
)e
J
α
had
= ¯uγ
α
(1 − γ
5
)(V
ud
d + V
us
s + v
ub
b) ≡ V
α
had
− A
α
had
,
where the V
α
had
contains the γ
α
bit, while A
α
had
contains the γ
α
γ
5
bit.
Then the amplitude we want to compute is
M =
e
−
(k)¯ν
e
(q)
L
eff
W
π
−
(p)
= −
G
F
√
2
e
−
(k)¯ν
e
(q)
J
†
α,lept
|0ih0|J
α
had
π
−
(p)
= −
G
F
√
2
e
−
(k)¯ν
e
(q)
¯eγ
α
(1 − γ
5
)ν
e
|0ih0|J
α
had
π
−
(p)
=
G
F
√
2
¯u
e
(k)γ
α
(1 − γ
5
)v
ν
e
(q) h0|V
α
had
− A
α
had
π
−
(p)
.
We now note that
V
α
had
does not contribute. This requires knowing something
about QCD and
π
−
. It is known that QCD is a P-invariant theory, and experi-
mentally, we find that
π
−
has spin 0 and odd parity. In other words, under P, it
transforms as a pseudoscalar. Thus, the expression
h0| ¯uγ
α
d
π
−
(p)
transforms as an axial vector. But since the only physical variable involved is
p
α
, the only quantities we can construct are multiples of
p
α
, which are vectors.
So this must vanish. By a similar chain of arguments, the remaining part of the
QCD part must be of the form
h0| ¯uγ
α
γ
5
d
π
−
(p)
= i
√
2F
π
p
α
for some constant F
π
. Then we have
M = iG
F
F
π
V
ud
¯u
e
(k)
/
p(1 − γ
5
)v
ν
e
(q).
To simplify this, we use momentum conservation
p
=
k
+
q
, and some spinor
identities
¯u
e
(k)
/
k = ¯u
e
(k)m
e
,
/
qv
ν
e
(q) = 0.
Then we find that
M = iG
F
F
π
V
ud
m
e
¯u
e
(k)(1 − γ
5
)v
ν
e
(q).
Doing a manipulation similar to last time’s, and noting that
(1 − γ
5
)γ
µ
(1 + γ
5
) = 2(1 − γ
5
)γ
µ
Tr(
/
k
/
q) = 4k ·q
Tr(γ
5
/
k
/
q) = 0
we find
X
spins e,¯ν
e
|M|
2
=
X
spins
|G
F
F
π
m
e
V
ud
|
2
[¯u
e
(k)(1 − γ
5
)v
ν
e
(q)¯v
ν
e
(q)(1 + γ
5
)u
e
(k)]
= 2|G
F
F
π
m
e
V
ud
|
2
Tr
h
(
/
k + m
e
)(1 − γ
5
)
/
q
i
= 8|G
F
F
π
m
e
V
ud
|
2
(k ·q).
This again shows helicity suppression. The spin-0
π
−
decays to the positive he-
licity
¯ν
e
, and hence decays to a positive helicity electron by helicity conservation.
π
−
e
−
1
2
⇒
¯ν
e
⇐
1
2
But if
m
e
= 0, then this has right-handed chirality, and so this decay is forbidden.
We can now compute an actual decay rate. We note that since we are working
in the rest frame of
π
−
, we have
k
+
q
= 0; and since the neutrino is massless,
we have q
0
= |q| = |k|. Finally, writing E = k
0
for the energy of e
−
, we obtain
Γ
π→e¯ν
e
=
1
2m
π
Z
d
3
k
(2π)
3
2k
0
Z
d
3
q
(2π)
3
2q
0
(2π)
4
δ
(4)
(p − k −q)
8|G
F
F
π
m
e
V
ud
|
2
(k ·q)
=
|G
F
F
π
m
e
V
ud
|
2
4m
π
π
2
Z
d
3
k
E|k|
δ(m
π
− E − |k|)(E|k| + |k|
2
)
To simplify this further, we use the property
δ(f(k)) =
X
i
δ(k −k
i
0
)
|f
0
(k
i
0
)|
,
where k
i
0
runs over all roots of f. In our case, we have a unique root
k
0
=
m
2
π
− m
2
e
2m
π
,
and the corresponding derivative is
|f
0
(k
0
)| = 1 +
k
0
E
.
Then we get
Γ
π→e¯ν
e
=
|G
F
F
π
m
e
V
ud
|
2
4π
2
m
π
Z
4πk
2
dk
E
E + k
1 + k
0
/E
δ(k −k
0
)
=
|G
F
F
π
V
ud
|
2
4π
m
2
e
m
π
1 −
m
2
e
m
2
π
2
.
Note that if we set
m
e
→
0, then this vanishes. This time, it is the whole decay
rate, and not just some particular decay channel.
Let’s try to match this with experiment. Instead of looking at the actual
lifetime, we compare it with some other possible decay rates. The expression for
Γ
π→µ¯ν
µ
is exactly the same, with m
e
replaced with m
µ
. So the ratio
Γ
π→e¯ν
e
Γ
π→µ¯ν
µ
=
m
2
e
m
2
µ
m
2
π
− m
2
e
m
2
π
− m
2
µ
2
≈ 1.28 × 10
−4
.
Here all the decay constants cancel out. So we can actually compare this with
experiment just by knowing the electron and muon masses. When we actually
do experiments, we find 1
.
230(4)
×
10
−4
. This is pretty good agreement, but
not agreement to within experimental error. Of course, this is not unexpected,
because we didn’t include the quantum loop effects in our calculations.
Another thing we can see is that the ratio is very small, on the order of 10
−4
.
This we can understand from helicity suppression, because m
µ
m
e
.
Note that we were able to get away without knowing how QCD works!