III The Standard Model

6Weak decays

6.2 Decay rates and cross sections

We now have an effective Lagrangian. What can we do with it? In this section,

we will consider two kinds of experiments:

–

We leave a particle alone, and see how frequently it decays, and into what.

–

We crash a lot of particles into each other, and count the number of

scattering events produced.

The relevant quantities are decay rates and cross sections respectively. We will

look at both of these in turn, but we will mostly be interested in computing

decay rates only.

Decay rate

Definition

(Decay rate)

Let

be a particle. The decay rate Γ

is rate of

decay of

in its rest frame. In other words, if we have a sample of

, then this

is the number of decays of

per unit time divided by the number of

present.

The lifetime of X is

τ ≡

We can write

X→f

where Γ

X→f

is the partial decay rate to the final state f

Note that the

is usually a complicated mixture of genuine sums and

integrals.

Often, we are only interested in how frequently it decays into, a particular

particle, instead of just the total decay ray. Thus, for each final state

, we

want to compute Γ

X→f

, and then sum over all final states we are interested in.

As before, we can compute the S-matrix elements, defined by

hf|S |ii.

We will take

. As before, recall that there is a term 1 of

that corresponds

to nothing happening, and we are not interested in that. If

|fi 6

|ii

, then this

term does not contribute.

It turns out the interesting quantity to extract out of the

-matrix is given

by the invariant amplitude:

Definition (Invariant amplitude). We define the invariant amplitude M by

hf|S − 1 |ii = (2π)

(4)

− p

)iM

When actually computing this, we make use of the following convenient fact:

Proposition. Up to tree order, and a phase, we have

= hf|L(0) |ii,

where L is the Lagrangian. We usually omit the (0).

Proof sketch. Up to tree order, we have

hf|S − 1 |ii = i

x hf|L(x) |ii

We write

in momentum space. Then the only

-dependence in the

hf|L

(

)

|ii

factor is a factor of

i(p

−p

)·x

. Integrating over

introduces a factor of

(2π)

(4)

− p

). Thus, we must have had, up to tree order,

hf|L(x) |ii = M

i(p

−p

)·x

So evaluating this at x = 0 gives the desired result.

How does this quantity enter the picture? If we were to do this naively, we

would expect the probability of a transition i → f is

P (i → f) =

|hf|S − 1 |ii|

hf|fihi|ii

It is not hard to see that we will very soon have a lot of

functions appearing

in this expression, and these are in general bad. As we saw in QFT, these

functions came from the fact that the universe is infinite. So what we do is that

we work with a finite spacetime, and then later take appropriate limits.

We suppose the universe has volume

, and we also work over a finite

temporal extent T . Then we replace

(2π)

(4)

(0) 7→ V T, (2π)

(3)

(0) 7→ V.

Recall that with infinite volume, we found the normalization of our states to be

hi|ii = (2π)

(3)

(0).

In finite volume, we can replace this with

hi|ii = 2p

Similarly, we have

hf|fi =

(2p

V ),

where r runs through all final state labels. In the S-matrix, we have

|hf|S − 1 |ii| =



(2π)

(4)

− p

)



We don’t really have a δ(0) here, but we note that for any x, we have

(δ(x))

= δ(x)δ(0).

The trick here is to replace only one of the

(0) with

V T

, and leave the other

one alone. Of course, we don’t really have a

(

−p

) function in finite volume,

but when we later take the limit, it will become a δ function again.

Noting that p

= m

since we are in the rest frame, we find

P (i → f) =

(2π)

(4)

−

V T





We see that two of our V ’s cancel, but there are a lot left.

The next thing to realize is that it is absurd to ask what is the probability of

decaying into exactly

. Instead, we have a range of final states. We will abuse

notation and use

to denote both the set of all final states we are interested in,

and members of this set. We claim that the right expression is

i→f

P (i → f)



(2π)



The obvious question to ask is why we are integrating against

(2π)

, and

not, say, simply d

. The answer is that both options are not quite right. Since

we have finite volume, as we are familiar from introductory QM, the momentum

eigenstates should be discretized. Thus, what we really want to do is to do an

honest sum over all possible values of p

But of course, we don’t like sums, and since we are going to take the limit

V → ∞

anyway, we replace the sum with an integral, and take into account of

the density of the momentum eigenstates, which is exactly

(2π)

We now introduce a measure

dρ

= (2π)

(4)

−

(2π)

and then we can concisely write

i→f

dρ

Note that when we actually do computations, we need to manually pick what

range of momenta we want to integrate over.

Cross sections

We now quickly look at another way we can do experiments. We imagine we set

two beams running towards each other with velocities

and

respectively.

We let the particle densities be ρ

and ρ

The idea is to count the number of scattering events,

. We will compute

this relative to the incident flux

F = |v

− v

|ρ

which is the number of incoming particles per unit area per unit time.

Definition (Cross section). The cross section is defined by

n = F σ.

Given these, the total number of scattering events is

N = nρ

V = F σρ

V = |v

− v

|ρ

V σ.

This is now a more symmetric-looking expression.

Note that in this case, we genuinely have a finite volume, because we have

to pack all our particles together in order to make them collide. Since we are

boring, we suppose we actually only have one particle of each type. Then we

have

= ρ

In this case, we have

σ =

− v

We can do computations similar to what we did last time. This time there are a

few differences. Last time, the initial state only had one particle, but now we

have two. Thus, if we go back and look at our computations, we see that we will

gain an extra factor of

in the frequency of interactions. Also, since we are

no longer in rest frame, we replace the masses of the particles with the energies.

Then the number of interactions is

N =

(2E

)(2E

dρ

We are often interested in knowing the number of interactions sending us to each

particular momentum (range) individually, instead of just knowing about how

many particles we get. For example, we might be interested in which directions

the final particles are moving in. So we are interested in

dσ =

− v

dN =

− v

|4E

dρ

Experimentalists will find it useful to know that cross sections are usually

measured in units called “barns”. This is defined by

1 barn = 10

−28