- A.45.1 Form of the wave function
- A.45.2 Source of the decay
- A.45.3 Allowed or forbidden
- A.45.4 The nuclear operator
- A.45.5 Fermi’s golden rule
- A.45.6 Mopping up

A.45 Fermi theory

This note needs more work, but as far as I know is basically OK. Unfortunately, a derivation of electron capture for zero spin transitions is not included.

This note derives the Fermi theory of beta decay. In particular, it gives the ballparks that were used to create figure 14.54. It also describes the Fermi integral plotted in figure 14.52, as well as Fermi’s (second) golden rule. There is also a final subsection on electron capture, A.45.7.

When beta decay was first observed, it was believed that the nucleus
simply ejected an electron. However, problems quickly arose with
energy and momentum conservation. To solve them, Pauli proposed in
1931 that in addition to the electron, also a neutral particle was
emitted. Fermi called it the neutrino,

for
small neutral one.

Following ideas of Pauli in 1933,
Fermi in 1934 developed a comprehensive theory of beta decay. The
theory justifies the various claims made about allowed and forbidden
beta decays. It also allows predictions of the decay rate and the
probability that the electron and antineutrino will come out with
given kinetic energies. This note gives a summary. The ballparks as
described in this note are the ones used to produce figure
14.54.

A large amount of work has gone into improving the accuracy of the Fermi theory, but it is outside the scope of this note. To get an idea of what has been done, you might start with [23] and work backwards. One point to keep in mind is that the derivations below are based on expanding the electron and neutrino wave functions into plane waves, waves of definite linear momentum. For a more thorough treatment, it may be a better idea to expand into spherical waves, because nuclear states have definite angular momentum. That idea is worked out in more detail in the note on gamma decay, {A.25}. That is news to the author. But it was supposed to be there, I think.

A.45.1 Form of the wave function

A classical quantum treatment will not do for beta decay. To see why,
note that in a classical treatment the wave function state before the
decay is taken to be of the form

where 1 through

There is no way to describe how

You might think that maybe the electron and antineutrino were always
there to begin with. But that has some major problems. A lone
neutron falls apart into a proton, an electron and an antineutrino.
So supposedly the neutron would consist of a proton, an electron, and
an antineutrino. But to confine light particles like electrons and
neutrinos to the size of a nucleon would produce huge kinetic
energies. According to the Heisenberg uncertainty relation

Further, a high-energy antineutrino can react with a proton to create a neutron and a positron. That neutron is supposed to consist of a proton, an electron, and an antineutrino. So, following the same reasoning as before, the original proton before the reaction would consist of a positron, an electron, and a proton. That proton in turn would supposedly also consist of a positron, an electron, and an proton. So the original proton consists of a positron, an electron, a positron, an electron, and a proton. And so on until a proton consists of a proton and infinitely many electron / positron pairs. Not just one electron with very high kinetic energy would need to be confined inside a nucleon, but an infinite number of them, and positrons to boot. And all these electrons and positrons would somehow have to be prevented from annihilating each other.

It just does not work. There is plenty of solid evidence that
neutrons and protons each contain three quarks, not other
nucleons along with electrons, positrons, and neutrinos. The electron
and antineutrino are created out of pure energy during beta decay, as
allowed by Einstein’s famous relativistic expression

In particular, it is necessary to deal mathematically with the
appearance of the electron and an antineutrino out of nothing. To do
so, a more general, more abstract way must be used to describe the
states that nature can be in. Consider a decay that produces an
electron and an antineutrino of specific momenta

(A.274) |

Similarly, the state before the decay is written as

(A.275) |

(You could also add kets for different momentum states that the final electron and antineutrino are not in after the decay. But states that have zero electrons and neutrinos both before and after the considered decay are physically irrelevant and can be left away.)

That leaves the nuclear part of the wave function. You could use
Fock-space kets to deal with the disappearance of a neutron and
appearance of a proton during the decay. However, there is a neater
way. The total number of nucleons remains the same during the decay.
The only thing that happens is that a nucleon changes type from a
neutron into a proton. The mathematical trick is therefore to take the
particles to be nucleons, instead of protons and neutrons. If you
give each nucleon a nucleon type

property, then the
only thing that happens during the decay is that the nucleon type of
one of the nucleons flips over from neutron to proton. No nucleons
are created or destroyed. Nucleon type is typically indicated by the
symbol

During the decay, the

Of course, the name nucleon type

for isobaric spin” or “isotopic spin

were used, because nucleon type has absolutely nothing to do with
spin. However, it was felt that these nonsensical names could cause
some smart outsiders to suspect that the quantity being talked about
was not really spin. Therefore the modern term isospin

was introduced. This term contains nothing to give the secret away
that it is not spin at all.

A.45.2 Source of the decay

Next the source of the decay must be identified. Ultimately that must be the Hamiltonian, because the Hamiltonian describes the time evolution of systems according to the Schrödinger equation.

In a specific beta decay process, two states are involved. A state
Hamiltonian coefficients.

The first
one is:

where

The Hamiltonian coefficient for the final state is similarly

Using the form given in the previous section for the final wave function, that becomes

It is the expectation value of energy after the decay. It consists of the sum of the rest mass energies of final nucleus, electron, and antineutrino, as well as their kinetic energies.

The Hamiltonian coefficient that describes the interaction between the
two states is crucial, because it is the one that causes the decay.
It is

Using the form for the wave functions given in the previous section:

If

Unfortunately, Fermi had no clue what

In atomic decay photon creation

operator

In words, it says that the interaction of the electron with the electromagnetic field can create photons. The magnitude of that effect is proportional to the amplitude of the photon at the location of the electron, and also to the electric charge of the electron. The electric charge acts as a “coupling constant” that links electrons and photons together. If the electron was uncharged, it would not be able to create photons. So it would not be able to create an electric field. Further, the fact that the coupling between the electron and the photon occurs at the location of the electron eliminates some problems that relativity has with action at a distance.

There is another term in

Fermi assumed that the general ideas of atomic decay would also hold
for beta decay of nuclei. Electron and antineutrino creation
operators in the Hamiltonian would turn the zero-electron and
zero-antineutrino kets into one-electron and one-antineutrino ones.
Then the inner products of the kets are equal to one pairwise.
Therefore both the creation operators and the kets drop out of the
final expression. In that way the Hamiltonian coefficient simplifies
to

where the index

To write expressions for the wave functions of the electron and
antineutrino, you face the complication that unbound states in
infinite space are not normalizable. That produced mathematical
complications for momentum eigenstates in chapter 7.9.2,
and similar difficulties resurface here. To simplify things, the
mathematical trick is to assume that the decaying nucleus is not in
infinite space, but in an extremely large “periodic
box.” The assumption is that nature repeats itself spatially; a
particle that exits the box through one side reenters it through the
opposite side. Space wraps around

if you want, and
opposite sides of the box are assumed to be physically the same
location. It is like on the surface of the earth: if you move along
the straightest-possible path on the surface of the earth, you travel
around the earth along a big circle and return to the same point that
you started out at. Still, on a local scale, the surface on the earth
looks flat. The idea is that the empty space around the decaying
nucleus has a similar property, in each of the three Cartesian
dimensions. This trick is also commonly used in solid mechanics,
chapter 10.

In a periodic box, the wave function of the antineutrino is

where

The wave function of the electron will be written in a similar way:

This however has an additional problem. It works fine far from the nucleus, where the momentum of the electron is by definition the constant vector

The usual way to deal with the problem is to stick with the
exponential electron wave function for now, and fix up the problem
later in the final results. The fix-up will be achieved by throwing
in an additional fudge factor. While “Fermi fudge
factor” alliterates nicely, it does not sound very respectful,
so physicists call the factor the Fermi function.

The bottom line is that for now

That leaves the still unknown operator

A.45.3 Allowed or forbidden

The question of allowed and forbidden decays is directly related to
the Hamiltonian coefficient

First note that the emitted electron and antineutrino have quite
small momentum values, on a nuclear scale. In particular,
in their combined wave function

the argument of the exponential is small. Typically, its magnitude is only a few percent, It is therefore possible to approximate the exponential by one, or more generally by a Taylor series:

Since the first term in the Taylor series is by far the largest, you
would expect that the value of

However, clearly this approximation does not work if the value of

If the decay is forbidden, higher order terms in the Taylor series will have to be used to come up with a nonzero value forIf the simplified coefficientis nonzero, the decay is allowed. If it is zero, the decay is forbidden.

Why would a decay not be allowed? In other words why would

One very important one is symmetry with respect to coordinate system
orientation. An inner product of two wave functions is independent of
the angular orientation of the coordinate system in which you evaluate
it. Therefore, you can average the inner product over all directions
of the coordinate system. However, the angular variation of a wave
function is related to its angular momentum; see chapter
7.3 and its note. In particular, if you average a wave
function of definite angular momentum over all coordinate system
orientations, you get zero unless the angular momentum is zero. So,
if it was just

Note that the linear momenta of the electron and antineutrino have
become ignored in

Another important constraint is symmetry under the parity
transformation

Since the electron and antineutrino come out without orbital angular momentum, they have even parity. So the nuclear parity must remain unchanged under the transition. (To be sure, this is not absolutely justified. Nuclear wave functions actually have a tiny uncertainty in parity because the weak force does not conserve parity, chapter 14.19.8. This effect is usually too small to be observed and will be ignored here.)

So what if either one of these selection rules is violated? In that
case, maybe the second term in the Taylor series for the electron and
antineutrino wave functions produces something nonzero that can drive
the decay. For that to be true,

has to be nonzero. If it is, the decay is a first-forbidden one. Now the spherical harmonics

with the

And note that because

The higher order forbidden decays go the same way. For an

A.45.4 The nuclear operator

This subsection will have a closer look at the nuclear operator

Although Fermi did not know what

Fermi ignored the spin of the electron and antineutrino. However,
Gamow & Teller soon established that to allow for decays where
the two come out with spin, (Gamow-Teller decays),

Finally, it was established in 1953 that the correct one was the STP combination, because experimental evidence on RaE, (some physicists cannot spell bismuth-210), showed that P was present. Unfortunately, it did not. For one, the conclusion depended to an insane degree on the accuracy of a correction term.

However, in 1955 it was established that it was STP anyway, because experimental evidence on helium-6 clearly showed that the Gamow-Teller part of the decay was tensor. The question was therefore solved satisfactorily. It was STP, or maybe just ST. Experimental evidence had redeemed itself.

However, in 1958, a quarter century after Fermi, it was found that beta decay violated parity conservation, chapter 14.19.8, and theoretically that was not really consistent with STP. So experimentalists had another look at their evidence and quickly came back with good news: “The helium-6 evidence does not show Gamow-Teller is tensor after all.”

The final answer is that

It may next be noted that

In Gamow-Teller decays,

The relevant operators then become, [5],

for Fermi and Gamow-Teller decays respectively. Here the three

So how do these nuclear operators affect the decay rate? That is best
understood by going back to the more physical shell-model picture. In
beta minus decay, a neutron is turned into a proton. That proton
usually occupies a different spatial state in the proton shells than
the original neutron in the neutron shells. And different spatial
states are supposedly orthogonal, so the inner product

If you allow for beta decay to excited states, more superallowed
decays are possible. States that differ merely in nucleon type are
called isobaric analog states, or isospin multiplets, chapter
14.18. There are about twenty such superallowed decays in
which the initial and final nuclei both have spin zero and positive
parity. These twenty are particularly interesting theoretically,
because only Fermi decays are possible for them. And the Fermi inner
product is

These decays therefore allow the value of the Fermi coupling constant

Besides the spin and parity rules already mentioned, Fermi decays must satisfy the approximate selection rule that the magnitude of isospin must be unchanged. They can be slowed down by several orders of magnitude if that rule is violated.

Gamow-Teller decays are much less confined than Fermi ones because of the presence of the electron spin operator. As the shell model shows, nucleon spins are uncertain in energy eigenstates. Therefore, the nuclear symmetry constraints are a lot less restrictive.

A.45.5 Fermi’s golden rule

The previous four subsections have focussed on finding the Hamiltonian
coefficients of the decay from a state

The quantum amplitude of the pre-decay state

(To use this expression, the quantum amplitudes must include an additional phase factor, but it is of no consequence for the probability of the states. See chapter 7.6 and {D.38} for details.)

Now picture the following. At the initial time there are a large
number of pre-decay nuclei, all with

Half of the exponential can be factored out to produce a real ratio:

Then at the final time measured.

The macroscopic surroundings of
the nuclei establishes whether or not electron and antineutrino pairs
have come out. The probability that a give nucleus has emitted such a
pair is given by the square magnitude measurement,

the entire process then repeats for the
remaining

The bottom line is however that a fraction

And there may be more. If the final nuclear state has spin you also
need to sum over all values of the magnetic quantum number of the
final state. (The amount of nuclear decay should not depend on the
angular orientation of the initial nucleus in empty space. However,
if you expand the electron and neutrino wave functions into spherical
waves, you need to average over the possible initial magnetic quantum
numbers. It may also be noted that the total coefficient

However, all these details are of little importance in finding a
ballpark for the dominant decay process. The real remaining problem
is summing over the electron and antineutrino momentum states. The
total ballparked decay rate must be found from

Based on energy conservation, you would expect that decays should only
occur when the total energy slop

in
energy conservation.

How can energy not be conserved? The reason is that neither the
initial state nor the final state is an energy eigenstate, strictly
speaking. Energy eigenstates are stationary states. The very fact
that decay occurs assures that these states are not really energy
eigenstates. They have a small amount of uncertainty in energy. The
nonzero value of the Hamiltonian coefficient

To narrow this effect down more precisely, the fraction is plotted in
figure 7.7. The spikes in the figure indicate the energies

where

Now assume that the complete problem is cut into bite-size pieces for
each of which

That is

Fermi’s (second) golden rule.It describes how energy slop increases the total decay rate. It is not specific to beta decay but also applies to other forms of decay to a continuum of states. Note that it no longer depends on the artificial length

measurement.That is good news, since that time interval was obviously poorly defined.

Because of the assumptions involved, like dividing the problem into
bite-size pieces, the above expression is not very intuitive to apply.
It can be rephrased into a more intuitive form that does not depend on
such an assumption. The obtained decay rate is exactly the same as if
in an energy slop range

all states contribute just as much to the decay as one that satisfies energy conservation exactly, while no states contribute outside of that range.

(Note that if you ballpark

The good news is that phrased this way, it indicates the relevant
physics much more clearly than the earlier purely mathematical
expression for Fermi’s golden rule. The bad news is that it
suffers esthetically from still involving the poorly defined time

It may be noted that the golden rule does not apply if the evolution
is not to a continuum of states. It also does not apply if the slop
range measured.

(If
Measurements,

or rather interactions with the larger
environment, are called collisions.

Fermi’s golden
rule applies to so-called collision-dominated

conditions. Typically examples where the conditions are not collision
dominated are in NMR and atomic decays under intense laser light.

Mathematically, the conditions for Fermi’s golden rule can be
written as

(A.280) |

It should also be noted that the rule was derived by Dirac, not Fermi.
The way Fermi got his name on it was that he was the one who named it
a golden rule.

Fermi had a flair for finding memorable
names. God knows how he ended up being a physicist.

A.45.6 Mopping up

The previous subsections derived the basics for the rate of beta decay. The purpose of this section is to pull it all together and get some actual ballpark estimates for beta decay.

First consider the possible values for the momenta

where

In a periodic box the wave function must be the same at opposite sides
of the box. For example, the exponential factor

Graphically this can by visualized by plotting the possible momentum
values as points in a three-dimensional volume

(in this three-dimensional momentum plot, not physical volume) of that
little cube is volume

in momentum space taken up by each momentum state is

That allows the number of different momentum states to be computed.
In particular, consider how many states have magnitude of momentum
volume

equal to
volume

volumes

in
the sphere:

The number of electron states that have momentum in a range from

(Here the range

Now the kinetic energy of the antineutrino

Here the kinetic energy of the final nucleus is ignored. The heavy final nucleus is unselfish enough to assure that momentum conservation is satisfied for whatever the momenta of the electron and antineutrino are, without demanding a noticeable share of the energy for itself. That is much like Mother Earth does not take any of the kinetic energy away if you shoot rockets out to space from different locations. You might write equations down for it, but the only thing they are going to tell you is that it is true as long as the speed of the nucleus does not get close to the speed of light. Beta decays do not release that much energy by far.

The electron and antineutrino kinetic energies are related to their
momenta by Einstein’s relativistic expression,
chapter 1.1.2:

These result shows that the neutrino momentum is fixed for given
electron momentum

Therefore the total amount of neutrino states for a given electron momentum is not zero, but

The number of complete system states in an electron momentum range

Each of these states adds a contribution to the specific decay rate
given by

Therefore the total specific decay rate is

where the maximum electron momentum

The derived expression (A.277) for the Hamiltonian coefficient

where the overline indicates some suitable average over the directions of the electron and antineutrino momenta.

It is not easy to say much about

The decay rate becomes after clean up

Here

The factor

where

For beta-plus decay, just replace

To ballpark the effect of the nuclear charge on the electron wave function, this book will use the relativistic Fermi function above whether it is an allowed decay or not.

For allowed decays, the factor in the decay rate that is governed by
the

Note that

(A.286) |

(A.287) |

A.45.7 Electron capture

Electron capture is much more simply to analyze than beta decay, because the captured electron is in a known initial state.

It will be assumed that a 1s, or K-shell, electron is captured, though
L-shell capture may also contribute to the decay rate for heavy
nuclei. The Hamiltonian coefficient that drives the decay is

In this case, it is an electron annihilation term in the Hamiltonian that will produce a nonzero term. However, the result will be the pretty much same; the Hamiltonian coefficient simplifies to

Here

where

The square Hamiltonian coefficient for

The decay rate for electron capture is

where the first ratio is the decay rate of a single state, with

Put it all together, including the fact that there are two K
electrons, and the electron-capture decay rate becomes