- A.44.1 Form of the wave function
- A.44.2 Source of the decay
- A.44.3 Allowed or forbidden
- A.44.4 The nuclear operator
- A.44.5 Fermi’s golden rule
- A.44.6 Mopping up

A.44 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 14.52. It also describes the Fermi integral plotted in figure 14.50, as well as Fermi’s (second) golden rule.

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
14.52.

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.44.1 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 number the nucleons. However, the decay creates an electron and a antineutrino out of nothing. Therefore, after the decay the classical wave function is of the form

There is no way to describe how could evolve into . You cannot just scribble in two more arguments into a function somewhere half way during the evolution. That would be voodoo mathematics. And there is also a problem with one nucleon turning from a neutron into a proton. You should really cross out the argument corresponding to the old neutron, and write in an argument for the new proton.

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 , where the energy for relativistic particles is about , so the kinetic energy of a light particle confined to a 1 fm range is about 200 MeV. What conceivable force could be strong enough to hold electrons and neutrinos that hot? And how come the effects of this mysterious force never show up in the atomic electrons that can be very accurately observed? How come that electrons come out in beta decays with only a few MeV, rather than 200 MeV?

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 . A relativistic quantum treatment is therefore necessary.

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 ,
respectively . The final state is written as

(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 and is defined to be if the nucleon is
a proton and if the nucleon is a neutron. (Some older
references may define it the other way around.) The general form of
the nuclear wave function therefore becomes

During the decay, the value of one nucleon will change from to .

Of course, the name nucleon type

for is not
really acceptable, because it is understandable. In the old days, the
names 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.44.2 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
describes the nucleus before the decay, and a state
describes the combination of nucleus, electron, and antineutrino after
the decay. That makes the system into a so-called “two state
system.” The unsteady evolution of such systems was discussed
in chapter 7.6 and {D.38}. The key to the
solution were the Hamiltonian coefficients.

The first
one is:

where is the (relativistic) Hamiltonian. The value of is the expectation value of the energy when nature is in the state . Assuming that the nucleus is initially at rest, the relativistic energy is just the rest mass energy of the nucleus. It is given in terms of its mass by Einstein’s famous relation .

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 is zero, no decay will occur. And most of the Hamiltonian does not produce a contribution to . But there is a small part of the Hamiltonian, call it , that does produce a nonzero interaction. That part is due to the weak force.

Unfortunately, Fermi had no clue what was. He assumed that beta decay would not be that much different from the better understood decay of excited atomic states in atoms. Gamma decay is the direct equivalent of atomic decay for excited nuclei. Beta decay is definitely different, but maybe not that different. In atomic decay an electromagnetic photon is created, rather than an electron and antineutrino. Still the general idea seemed similar.

In atomic decay is essentially proportional to the product of the
charge of the excited electron, times the spatial eigenstate of the
photon, times a 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 that involves an annihilation operator instead of a creation operator. An annihilation operator destroys photons. However, that does not produce a contribution to ; if you try to annihilate the nonexisting photon in the initial wave function, you get a zero wave function. On the other hand for the earlier term, the creation operator is essential. It turns the initial state with no photon into a state with one photon. States with different numbers of particles are orthogonal, so the Hamiltonian coefficient would be zero without the creation operator. Looked at the other way around, the presence of the creation operator in the Hamiltonian ensures that the final state must have one more photon for the decay to occur. (See addendum {A.15} for more details on electromagnetic interactions, including a more precise description of . See also {A.25}.)

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 is the nucleon number and is the remaining still unknown part of the Hamiltonian. In indicating this unknown part by , the assumption is that it will be possible to write it as some generic dimensional constant times some simple nondimensional operator acting on nucleon number .

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 is the volume of the periodic box. It is easy to check that this wave function is indeed normalized. Also, it is seen that it is indeed an eigenfunction of the -momentum operator with eigenvalue , and similar for the - and -momentum operators.

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 . However, near the nucleus the Coulomb field of the nucleus, and to some extent that of the atomic electrons, affects the kinetic energy of the electron, and hence its momentum. Therefore, the energy eigenfunction that has momentum far from the nucleus differs significantly from the above exponential closer to the nucleus. And this wave function must be evaluated at nucleon positions inside the nucleus! The problem is particularly large when the momentum is low, because then the electron has little kinetic energy and the Coulomb potential is relatively speaking more important. The problem gets even worse for low-energy positron emission, because a positively-charged positron is repelled by the positive nucleus and must tunnel through to reach it.

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 . The constant is simply defined so that the operator has a magnitude that is of order one. That means that should never be greater than about one, though it could be much less if and turn out to be almost orthogonal. It is found that has a rough value of about 100 eV fm, depending a bit on whether it is a Fermi or Gamow-Teller decay. Figure 14.52 simply used 100 MeV fm.

A.44.3 Allowed or forbidden

The question of allowed and forbidden decays is directly related to the Hamiltonian coefficient , (A.276), derived in the previous subsection, that causes the decay.

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 , (A.276), can
be well approximated by replacing the exponential by 1, giving:

However, clearly this approximation does not work if the value of is zero for some reason:

If the decay is forbidden, higher order terms in the Taylor series will have to be used to come up with a nonzero value for . Since these higher order terms are much smaller, and drives the decay, a forbidden decay will proceed much slower than an allowed one.If the simplified coefficient is 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 and be exactly orthogonal? If you took two random wave functions for and , they definitely would not be. But and are not random wave functions. They satisfy a significant amount of symmetry constraints.

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 in the inner product in , the inner product would be zero unless the initial nucleus had zero spin. However, the final state is also in the inner product, and being at the other side of it, its angular variation acts to counteract that of the initial nucleus. Therefore, will be zero unless the initial angular momentum is exactly balanced by the net final angular momentum. And that is angular momentum conservation. The decay has to satisfy it.

Note that the linear momenta of the electron and antineutrino have become ignored in . Therefore, their orbital angular momentum is approximated to be zero too. Under these condition is zero unless the angular momentum of the final nucleus plus the spin angular momentum of electron and antineutrino equals the angular momentum of the original nucleus. Since the electron and antineutrino can have up to one unit of combined spin, the nuclear spin cannot change more than one unit. That is the first selection rule for allowed decays given in chapter 14.19.4.

Another important constraint is symmetry under the parity transformation . This transformation too does not affect inner products, so you can average the values before and after the transform. However, a wave function that has odd parity changes sign under the transform and averages to zero. So the inner product in is zero if the total integrand has odd parity. For a nonzero value, the integrand must have even parity, and that means that the parity of the initial nucleus must equal the combined parity of the final nucleus electron, and antineutrino.

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 14.19.6. 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 of orbital angular momentum are of the generic form, {D.14}

with the some constants. Therefore, the factor in brings in angular variation corresponding to one unit of angular momentum. That means that the total spin can now change by up to one unit, and therefore the nuclear spin by up to two units. That is indeed the selection rule for first forbidden decays.

And note that because changes sign when every is replaced by , the initial and final nuclear parities must now be opposite for not to be zero. That is indeed the parity selection rule for first-forbidden decays.

The higher order forbidden decays go the same way. For an
th-forbidden decay,

A.44.4 The nuclear operator

This subsection will have a closer look at the nuclear operator . While the discussion will be kept simple, having some idea about the nature of this operator can be useful. It can help to understand why some decays have relatively low decay rates that are not explained by just looking at the electron and antineutrino wave functions, and the nuclear spins and parities. The discussion will mainly focus on allowed decays.

Although Fermi did not know what was, Pauli had already established the possible generic forms for it allowed by relativity. It could take the form of a scalar (S), a vector (V), an axial vector (A, a vector like angular momentum, one that inverts when the physics is seen in a mirror), a pseudoscalar (P, a scalar like the scalar triple product of vectors that changes sign when the physics is seen in the mirror), or a tensor (T, a multiple-index object like a matrix that transforms in specific ways.) Fermi simply assumed the interaction was of the vector, V, type in analogy with the decay of excited atoms.

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), also should have terms with axial, A, and/or tensor, T, character. Work of Fierz combined with experimental evidence showed that the Hamiltonian could not have both S and V, nor both A and T terms. Additional evidence narrowed down to STP combinations or VA ones.

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 14.19.6, 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 is VA. Since so much of our knowledge about nuclei depends on experimental data, it may be worthwhile to keep this cautionary tale, taken from the Stanford Encyclopedia of Philosophy, in mind.

It may next be noted that will need to include a isospin creation operator to be able to turn a neutron into a proton. In Fermi decays, is assumed to be just that operator. The constant of proportionality , usually called the coupling constant , describes the strength of the weak interaction. That is much like the unit electric charge describes the strength of the electromagnetic interaction between charged particles and photons. In Fermi decays it is found that is about 88 eV fm. Note that this is quite small compared to the MeV scale of nuclear forces. If you ballpark relative strengths of forces, [30, p. 285] the nuclear force is strongest, the electromagnetic force about hundred times smaller, the weak force another thousand times smaller than that, and finally gravity is another 10 times smaller than that. The decay rates turn out to be proportional to the square of the interaction, magnifying the relative differences.

In Gamow-Teller decays, is assumed to consist of products of isospin creation operators times spin creation or annihilation operators. The latter operators allow the spin of the neutron that converts to the proton to flip over. Suitable spin creation and annihilation operators are given by the so-called “Pauli spin matrices,” chapter 12.10 When they act on a nucleon, they produce states with the spin in an orthogonal direction flipped over. That allows the net spin of the nucleus to change by one unit. The appropriate constant of proportionality is found to be a bit larger than the Fermi one.

The relevant operators then become, [5],

for Fermi and Gamow-Teller decays respectively. Here the three are the Pauli spin matrices of chapter 12.10. The are the equivalents of the Pauli spin matrices for isospin; in the combinations shown above they turn neutrons into protons, or vice-versa. Please excuse: using the clarity now made possible by modern physical terminology, they create, respectively annihilate, isospin. The upper sign is relevant for beta-minus decay and the lower for beta-plus decay. The Gamow-Teller operator absorbs the spin part of the electron and antineutrino wave functions, in particular the averaging over the directions of their spin.

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 will usually be pretty small, if the decay is allowed at all. There is one big exception, though: mirror nuclei. In a decay between mirror nuclei, a nucleus with a neutron number decays into one with neutron number . In that case, the nucleon that changes type remains in the same spatial orbit. Therefore, the Fermi inner product equals one, and the Gamow Teller one is maximal too. Allowed decays of this type are called “superallowed.” The simplest example is the beta decay of a free neutron.

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 . (The reason that it is instead of 1 like for mirror nuclei can be seen from thinking of isospin as if it is just normal spin. Mirror nuclei have an odd number of nucleons, so the net nuclear isospin is half integer. In particular the net isospin will be in the ground state. However, nuclei with zero spin have an even number of nucleons, hence integer net isospin. The isospin of the twenty decays is one; it cannot be zero because at least one nucleus must have a nonzero net nucleon type . The net nucleon type is only zero if the number of protons is the same as the number of neutrons. It is then seen from (12.9) and (12.10) in chapter 12 that the isospin creation or annihilation operators will produce a factor .)

These decays therefore allow the value of the Fermi coupling constant to be determined from the decay rates. It turns out to be about 88 eV fm, regardless of the particular decay used to compute it. That seems to suggest that the interaction with neighboring nucleons in a nucleus does not affect the Fermi decay process. Indeed, if the value of is used to analyze the decay rates of the mirror nuclei, including the free neutron that has no neighbors, the data show no such effect. The hypothesis that neighboring nucleons do not affect the Fermi decay process is known as the “conserved vector current hypothesis.” What name could be clearer than that? Unlike Fermi decays, Gamow-Teller decays are somewhat affected by the presence of neighboring nuclei.

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.44.5 Fermi’s golden rule

The previous four subsections have focussed on finding the Hamiltonian coefficients of the decay from a state to a state . Most of the attention was on the coefficient that drives the decay. The next step is solution of the Schrödinger equation to find the evolution of the decay process.

The quantum amplitude of the pre-decay state will be
indicated by and the quantum amplitude of the final decayed
state by . The Schrödinger equation implies that
increases from zero according to

(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 1. All these nuclei
then evolve according to the Schrödinger equation, above, over a
time interval that is short enough that stays close to
one. (Because the perturbation of the nucleus by the weak force is
small, the magnitudes of the coefficients only change slowly on the
relevant time scale.) In that case, can be dropped from the
equation and its solution is then seen to be

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

Then at the final time , assume that the state of all the
nuclei is 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 of the amplitude
of the decayed state. Therefore, a fraction of the
nuclei will be found to have decayed and will be found
to be still in the pre-decay state . After this
measurement,

the entire process then repeats for the
remaining nuclei that did not decay.

The bottom line is however that a fraction did.
Therefore, the ratio gives the specific decay rate,
the relative amount of nuclei that decay per unit time. Plugging in
the above expression for gives:

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 for the decay will not be the same as the one for : you average over the initial magnetic quantum number, but sum over the final one.) If there are different excitation levels of the final nucleus that can be decayed to, you also need to sum over these. And if there is more than one type of decay process going on at the same time, they too need to be added together.

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 of the nucleus, electron and
antineutrino after the decay is exactly equal to the energy of
the nucleus before the decay. However, the summation above shows that
that is not quite true. For a final state that has exactly
equal to , the last fraction in the summation is seen to be
unity, using l’Hospital. For a final state with an energy
of, for example, , the ratio is quite comparable.
Therefore decay to such a state proceeds at a comparable rate as to a
state that conserves energy exactly. There is 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 assures that, chapter 5.3, and there may be more decay processes adding to the uncertainty in energy. If there is some uncertainty in energy, then is not an exact relationship.

To narrow this effect down more precisely, the fraction is plotted in
figure 7.7. The spikes in the figure indicate the energies
of the possible final states. Now the energy states are almost
infinitely densely spaced, if the periodic box in which the decay is
assumed to occur is big enough. And the box must be assumed very big
anyway, to simulate decay in infinite space. Therefore, the summation
can be replaced by integration, as follows:

where is the number of final states per unit energy range, often called the density of states .

Now assume that the complete problem is cut into bite-size pieces for
each of which is about constant. It can then be taken out
of the integral. Also, the range of energy in figure 7.7
over which the fraction is appreciable, the energy slop, is very small
on a normal nuclear energy scale: beta decay is a slow process, so the
initial and final states do remain energy eigenstates to a very good
approximation. Energy conservation is almost exactly satisfied.
Because of that, the density of states will be
almost constant over the range where the integrand is nonzero. It can
therefore be taken out of the integral too. What is left can be
integrated analytically, [40, 18.36]. That gives:

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 of the time interval over which the system was supposed to evolve without

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.

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
, instead of already having shoved under the mat.
Therefore, it is more appealing to write things in terms of the energy
slop altogether:

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 is so large that is not constant
over it. And it does not apply for systems that evolve without being
perturbed over times long enough that the decay probability
becomes significant before the system is measured.

(If
becomes appreciable, can no longer be close to one
since the probabilities and must add to
one.) 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.44.6 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 and
of the electron and antineutrino. Their wave
functions were approximately of the form

where is the volume of the periodic box in which the decay is assumed to occur.

In a periodic box the wave function must be the same at opposite sides of the box. For example, the exponential factor is 1 at =0, and it must be 1 again at . That requires to be a whole multiple of . Therefore must be a whole multiple of . Successive possible values are therefore spaced the finite amount apart. And so are successive and values.

Graphically this can by visualized by plotting the possible momentum
values as points in a three-dimensional axis system. That is
done in figure A.27. Each point correspond to one
possible momentum state. Each point is the center of its own little
cube with sides . The volume

(in this three-dimensional momentum plot, not physical volume) of that
little cube is . Since is the
physical volume of the periodic box, the 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
less than some maximum value . For some example
value of , these are the red states in figure
A.27. Note that they form a sphere of radius .
That sphere has a volume

equal to
. Since each state takes up a
volume

, the number of
states is given by the number of such volumes

in
the sphere:

The number of electron states that have momentum in a range from
to can be found by taking a differential of the
expression above:

(Here the range is assumed small, but not so small that the fact that the number of states is discrete would show up.) Each momentum state still needs to be multiplied by the number of corresponding antineutrino states to find the number of states of the complete system.

Now the kinetic energy of the antineutrino is fixed in
terms of that of the electron and the energy release of the decay
by:

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 there should not be a
neutrino momentum range and so no neutrino states.
However, Fermi’s golden rule says that the theoretical energy
after the decay does not need to be exactly the same as the one before
it, because both energies have a bit of uncertainty. This slop in the
energy conservation equation allows a range of energies

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 is the product of the number of electron states times the number of antineutrino states:

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 can be computed from the -value of the decay using (A.281). (For simplicity it will be assumed that has already been averaged over all directions of the electron and antineutrino momentum.)

The derived expression (A.277) for the Hamiltonian coefficient
can be written in the form

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

It is not easy to say much about in general, beyond the fact that its magnitude should not be much more than one. This book will essentially ignore to ballpark the decay rate, assuming that its variation will surely be much less than that of the beta decay lifetimes, which vary from milliseconds to 10 year.

The decay rate becomes after clean up

Here is the -value or kinetic energy release of the decay in units of the electron rest mass, and the next two relations follow from the expression for the relativistic kinetic energy. The variable a suitably nondimensionalized nuclear radius, and is small.

The factor that popped up out of nothing in the decay rate is
thrown in to correct for the fact that the wave function of the
electron is not really just an exponential. The nucleus pulls on the
electron with its charge, and so changes its wave function locally
significantly. The correction factor for allowed decays is
called the “Fermi function” and is given by

where is the fine structure constant and the gamma function. The nonrelativistic version follows for letting the speed of light go to infinity, while keeping finite. That gives 1 and

For beta-plus decay, just replace by , because an electron is just as much repelled by a negatively charged nucleus as a positron is by a positively charged one.

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 -value and nuclear charge is

Note that also depends a bit on the mass number through the
nuclear radius in . The figure used

(A.286) |

(A.287) |

A.44.7 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 is the hydrogen ground state wave function, but
rescaled for a nucleus of charge instead of . It does
not contribute to making forbidden decays possible, because
is spherically symmetric. In other words, the 1s
electron has no orbital angular momentum and so cannot contribute to
conservation of angular momentum and parity. Therefore,
can safely be approximated by its value at the origin, from chapter
4.3,

where is the fine-structure constant.

The square Hamiltonian coefficient for th-forbidden decays then
becomes

The decay rate for electron capture is

where the first ratio is the decay rate of a single state, with the energy slop implied by Fermi’s golden rule.

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