III The Standard Model

6Weak decays

6.3 Muon decay

We now look at our first decay process, the muon decay:

−

→ e¯ν

This is in fact the only decay channel of the muon.

−

¯ν

We will make the simplifying assumption that neutrinos are massless.

The relevant bit of L

eff

−

√

α†

where the weak current is

= ¯ν

(1 − γ

)e + ¯ν

(1 − γ

)µ + ¯ν

(1 − γ

)τ.

We see it is the interaction of the first two terms that will render this decay

possible.

To make sure our weak field approximation is valid, we need to make sure

we live in sufficiently low energy scales. The most massive particle involved is

= 105.658 371 5(35) MeV.

On the other hand, the mass of the weak boson is

= 80.385(15) GeV,

which is much bigger. So the weak field approximation should be valid.

We can now compute

M =



−

(k)¯ν

(q)ν

)



eff



−

(p)



Note that we left out all the spin indices to avoid overwhelming notation. We

see that the first term in

is relevant to the electron bit, while the second is

relevant to the muon bit. We can then write this as

M = −

√



−

(k)¯ν

(q)



¯eγ

(1 − γ

)ν

|0ihν

)| ¯ν

(1 − γ

)µ



−

(p)



= −

√

¯u

(k)γ

(1 − γ

(q)¯u

)γ

(1 − γ

(p).

Before we plunge through more computations, we look at what we are interested

in and what we are not.

At this point, we are not interested in the final state spins. Therefore, we

want to sum over the final state spins. We also don’t know the initial spin of

−

So we average over the initial states. For reasons that will become clear later,

we will write the desired amplitude as

spins

|M|

spins

∗

spins



¯u

(k)γ

(1 − γ

(q)¯u

)γ

(1 − γ

(p)





¯u

(p)γ

(1 − γ

)¯v

(q)γ

(1 − γ

(k)



spins



¯u

(k)γ

(1 − γ

(q)¯v

(q)γ

(1 − γ

(k)





¯u

)γ

(1 − γ

(p)¯u

(p)γ

(1 − γ

)



We write this as

αβ

2αβ

where

αβ

spins

¯u

(k)γ

(1 − γ

(q)¯v

(q)γ

(1 − γ

(k)

2αβ

spins

¯u

)γ

(1 − γ

(p)¯u

(p)γ

(1 − γ

To actually compute this, we recall the spinor identities

spins

u(p)¯u(p) =

p + m,

spins

v(p)¯v(p) =

p − m

In our expression for, say,

, the

¯v

is already in the right form to apply

these identities, but

¯u

and

are not. Here we do a slightly sneaky thing. We

notice that for each fixed

α, β

, the quantity

αβ

is a scalar. So we trivially have

αβ

(

αβ

). We now use the cyclicity of trace, which says

(

) =

(

This applies even if

and

are not square, by the same proof. Then noting

further that the trace is linear, we get

αβ

= Tr(S

αβ

)

spins

¯u

(k)γ

(1 − γ

(q)¯v

(q)γ

(1 − γ

)

spins

(k)¯u

(k)γ

(1 − γ

(q)¯v

(q)γ

(1 − γ

)

= Tr

(

k + m

)γ

(1 − γ

)

qγ

(1 − γ

)

Similarly, we find

2αβ

= Tr

(1 − γ

)(

p + m

)γ

(1 − γ

)

To evaluate these traces, we use trace identities

Tr(γ

···γ

) = 0 if n is odd

Tr(γ

) = 4(g

µν

ρσ

− g

µρ

νσ

+ g

µσ

νρ

)

Tr(γ

) = 4iε

µνρσ

This gives the rather scary expressions

αβ

= 8



+ k

− (k · q)g

αβ

− iε

αβµρ



2αβ

= 8



+ q

− (q

· p)g

αβ

− iε

αβµρ

0µ



but once we actually contract these two objects, we get the really pleasant result

|M|

= 64G

(p · q)(k · q

It is instructive to study this expression in particular cases. Consider the case

where

−

and

go out along the +

direction, and

¯ν

along

−z

. Then we have

k · q

+ k

− k

As m

→ 0, we have → 0.

This is indeed something we should expect even without doing computations.

We know weak interaction couples to left-handed particles and right-handed

anti-particles. If

= 0, then we saw that helicity are chirality the same. Thus,

the spin of ¯ν

must be opposite to that of ν

and e

−

¯ν

⇐

So they all point in the same direction, and the total spin would be

. But the

initial total angular momentum is just the spin

. So this violates conservation

of angular momentum.

But if

= 0, then the left-handed and right-handed components of the

electron are coupled, and helicity and chirality do not coincide. So it is possible

that we obtain a right-handed electron instead, and this gives conserved angular

momentum. We call this helicity suppression, and we will see many more

examples of this later on.

It is important to note that here we are only analyzing decays where the

final momenta point in these particular directions. If m

= 0, we can still have

decays in other directions.

There is another interesting thing we can consider. In this same set up,

= 0, but neutrinos are massless, then we can only possibly decay to

left-handed neutrinos. So the only possible assignment of spins is this:

¯ν

⇐

⇒

Under parity transform, the momenta reverse, but spins don’t. If we transform

under parity, we would expect to see the same behaviour.

¯ν

⇐

⇒

But (at least in the limit of massless neutrinos) this isn’t allowed, because weak

interactions don’t couple to right-handed neutrinos. So we know that weak

decays violate P.

We now now return to finish up our computations. The decay rate is given

Γ =

(2π)

× (2π)

(4)

(p − k − q − q

)

|M|

Using our expression for |M|, we find

Γ =

8π

k d

q d

(4)

(p − k − q − q

)(p · q)(k · q

To evaluate this integral, there is a useful trick.

For convenience, we write Q = p − k, and we consider

µν

(Q) =

(4)

(Q − q − q

By Lorentz symmetry arguments, this must be of the form

µν

(Q) = a(Q

+ b(Q

µν

where a, b : R → R are some scalar functions.

Now consider

µν

(4)

(Q − q − q

)q · q

= (a + 4b)Q

But we also know that

(q + q

)

= q

+ q

+ 2q · q

= 2q · q

because neutrinos are massless. On the other hand, by momentum conservation,

we know

q + q

= Q.

So we know

q · q

So we find that

a + 4b =

, (1)

where

I =

(4)

(Q − q − q

We can consider something else. We have

µν

= a(Q

+ b(Q

(4)

(Q − q − q

)(q · Q)(q

· Q).

Using the masslessness of neutrinos and momentum conservation again, we find

that

(q · Q)(q

· Q) = (q · q

)(q · q

So we find

a + b =

. (2)

It remains to evaluate

, and to do so, we can just evaluate it in the frame where

Q = (σ, 0) for some σ. Now note that since q

= q

= 0, we must have

= |q|.

So we have

I =

|q|

δ(σ − |q| − |q

i=1

δ(q

− q

)

|q|

δ(σ − 2|q|)

= 4π

∞

d|q| δ(σ − 2|q|)

= 2π.

So we find that

Γ =

(2π)



2p · (p − k)k · (p − k) + (p · k)(p − k)



Recall that we are working in the rest frame of µ. So we know that

p · k = m

E, p · p = m

, k · k = m

where E = k

. Note that we have

≈ 0.0048  1.

So to make our lives easier, it is reasonable to assume

= 0. In this case,

|k| = E, and then

Γ =

(2π)



E − 2(m

− 2(m

+ m



(2π)

k (3m

− 4E)

4πG

(2π)

dE E

(3m

− 4E)

We now need to figure out what we want to integrate over. When

is at rest,

then

min

= 0. The maximum energy is obtained when

, ¯ν

are in the same

direction and opposite to e

−

. In this case, we have

E + (E

¯ν

+ E

) = m

By energy conservation, we also have

E − (E

¯ν

+ E

) = 0.

So we find

max

Thus, we can put in our limits into the integral, and find that

Γ =

192π

As we mentioned at the beginning of the chapter, this is the only decay channel

off the muon. From experiments, we can measure the lifetime of the muon as

= 2.1870 × 10

−6

This tells us that

= 1.164 × 10

−5

GeV

Of course, this isn’t exactly right, because we ignored all loop corrections (and

approximated

= 0). But this is reasonably good, because those effects are

very small, at an order of 10

−6

as large. Of course, if we want to do more

accurate and possibly beyond standard model physics, we need to do better than

this.

Experimentally,

is consistent with what we find in the

τ → e¯ν

and

µ → e¯ν

decays. This is some good evidence for lepton universality, i.e. they

have different masses, but they couple in the same way.