Subsections


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

\begin{displaymath}
\psi_1({\skew0\vec r}_1,S_{z,1},{\skew0\vec r}_2,S_{z,2},\ldots,{\skew0\vec r}_A,S_{z,A})
\end{displaymath}

where 1 through $A$ 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

\begin{displaymath}
\psi_2({\skew0\vec r}_1,S_{z,1},{\skew0\vec r}_2,S_{z,2},\...
...kew0\vec r}_e,S_{z,e},{\skew0\vec r}_{\bar\nu},S_{z,\bar\nu})
\end{displaymath}

There is no way to describe how $\psi_1$ could evolve into $\psi_2$. 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 $p$ $\vphantom0\raisebox{1.5pt}{$\sim$}$ $\hbar$$\raisebox{.5pt}{$/$}$$\Delta{x}$, where the energy for relativistic particles is about $pc$, 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 $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $mc^2$. 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 ${\skew0\vec p}_e$, respectively ${\skew0\vec p}_{\bar\nu}$. The final state is written as

\begin{displaymath}
\psi_2 = \psi_{2,\rm nuc} \big\vert 1e,{\skew0\vec p}_e\big\rangle \big\vert 1\bar\nu,{\skew0\vec p}_{\bar\nu}\big\rangle
\end{displaymath} (A.274)

where $\psi_{2,\rm {nuc}}$ is the nuclear part of the final wave function. The electron ket $\big\vert 1e,{\skew0\vec p}_e\big\rangle $ is a “Fock-space ket,” and should be read as “one electron in the state with angular momentum ${\skew0\vec p}_e$.” The antineutrino ket should be read as “one antineutrino in the state with angular momentum ${\skew0\vec p}_{\bar\nu}$.”

Similarly, the state before the decay is written as

\begin{displaymath}
\psi_1 = \psi_{1,\rm nuc} \big\verte,{\skew0\vec p}_e\big\rangle \big\vert\bar\nu,{\skew0\vec p}_{\bar\nu}\big\rangle
\end{displaymath} (A.275)

where $\big\verte,{\skew0\vec p}_e\big\rangle $ means “zero electrons in the state with angular momentum ${\skew0\vec p}_e$,” and similar for the antineutrino ket. Written in this way, the initial and final wave functions are no longer inconsistent. What is different is not the form of the wave function, but merely how many electrons and antineutrinos are in the states with momentum ${\skew0\vec p}_e$, respectively ${\skew0\vec p}_{\bar\nu}$. Before the decay, the “occupation numbers” of these states are zero electrons and zero antineutrinos. After the decay, the occupation numbers are one electron and one neutrino. It is not that the initial state does not have occupation numbers for these states, (which would make $\psi_1$ and $\psi_2$ inconsistent), but merely that these occupation numbers have the value zero, (which does not).

(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 $T_3$ and is defined to be ${\textstyle\frac{1}{2}}$ if the nucleon is a proton and $-{\textstyle\frac{1}{2}}$ 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

\begin{displaymath}
\Psi_N({\skew0\vec r}_1,S_{z,1},T_{3,1},{\skew0\vec r}_2,S_{z,2},T_{3,2},
\ldots,{\skew0\vec r}_A,S_{z,A},T_{3,A};t)
\end{displaymath}

During the decay, the $T_3$ value of one nucleon will change from $-{\textstyle\frac{1}{2}}$ to ${\textstyle\frac{1}{2}}$.

Of course, the name nucleon type for $T_3$ 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 $\psi_1$ describes the nucleus before the decay, and a state $\psi_2$ 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:

\begin{displaymath}
E_1 \equiv H_{11} \equiv \langle\psi_1\vert H\psi_1\rangle
\end{displaymath}

where $H$ is the (relativistic) Hamiltonian. The value of $H_{11}$ is the expectation value of the energy $E_1$ when nature is in the state $\psi_1$. 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 $E_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m_{{\rm {N}}1}c^2$.

The Hamiltonian coefficient for the final state is similarly

\begin{displaymath}
E_2 \equiv H_{22} \equiv \langle\psi_2\vert H\psi_2\rangle
\end{displaymath}

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

\begin{displaymath}
E_2 =
\langle 1\bar\nu,{\skew0\vec p}_{\bar\nu}\vert\lan...
...vec p}_e\rangle\vert 1\bar\nu,{\skew0\vec p}_{\bar\nu}\rangle
\end{displaymath}

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

\begin{displaymath}
H_{21} \equiv \langle\psi_2\vert H\psi_1\rangle
\end{displaymath}

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

\begin{displaymath}
H_{21} =
\langle 1\bar\nu,{\skew0\vec p}_{\bar\nu}\vert\...
...0\vec p}_e\rangle\vert\bar\nu,{\skew0\vec p}_{\bar\nu}\rangle
\end{displaymath}

If $H_{21}$ is zero, no decay will occur. And most of the Hamiltonian does not produce a contribution to $H_{21}$. But there is a small part of the Hamiltonian, call it $H'$, that does produce a nonzero interaction. That part is due to the weak force.

Unfortunately, Fermi had no clue what $H'$ 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 $H'$ 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 $\widehat a^\dagger $:

\begin{displaymath}
H' \propto e \psi_{\rm photon}({\skew0\vec r}) \widehat a^\dagger
\end{displaymath}

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 $H'$ that involves an annihilation operator $\widehat a$ instead of a creation operator. An annihilation operator destroys photons. However, that does not produce a contribution to $H_{21}$; 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 $H_{12}$ 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 $H'$. 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

\begin{displaymath}
H_{21} = \langle \psi_{2,\rm nuc} \vert \sum_{i=1}^A g h_i...
...}_{\bar\nu}}({\skew0\vec r}_i)
\vert\psi_{1,\rm nuc}\rangle
\end{displaymath}

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

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

\begin{displaymath}
\psi_{\bar\nu,{\skew0\vec p}_{\bar\nu}}
= \frac{1}{\sqrt...
...m i}(p_{x,\bar\nu}x + p_{y,\bar\nu}y + p_{z,\bar\nu}z)/\hbar}
\end{displaymath}

where ${\cal V}$ 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 $x$-​momentum operator $\hbar\partial$$\raisebox{.5pt}{$/$}$${\rm i}\partial{x}$ with eigenvalue $p_x$, and similar for the $y$-​ and $z$-​momentum operators.

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

\begin{displaymath}
\psi_{e,{\skew0\vec p}_e}({\skew0\vec r})
= \frac{1}{\sq...
...\cal V}}}
e^{{\rm i}(p_{x,e}x + p_{y,e}y + p_{z,e}z)/\hbar}
\end{displaymath}

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 ${\skew0\vec p}_e$. 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 ${\skew0\vec p}$ 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 ${\skew0\vec p}_e$ 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

\begin{displaymath}
H_{21} = \frac{g}{\sqrt{{\cal V}}} \langle \psi_{2,\rm nuc...
...u}) \cdot {\skew0\vec r}_i/\hbar}
\psi_{1,\rm nuc}\rangle %
\end{displaymath} (A.276)

That leaves the still unknown operator $gh_i$. The constant $g$ is simply defined so that the operator $h_i$ has a magnitude that is of order one. That means that $\langle\psi_{2,\rm {nuc}}\vert h_i\psi_{1,\rm {nuc}}\rangle$ should never be greater than about one, though it could be much less if $\psi_{2,\rm {nuc}}$ and $h_i\psi_{1,\rm {nuc}}$ turn out to be almost orthogonal. It is found that $g$ has a rough value of about 100 eV fm$\POW9,{3}$, depending a bit on whether it is a Fermi or Gamow-Teller decay. Figure 14.52 simply used 100 MeV fm$\POW9,{3}$.


A.44.3 Allowed or forbidden

The question of allowed and forbidden decays is directly related to the Hamiltonian coefficient $H_{21}$, (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

\begin{displaymath}
\frac{1}{{\cal V}} e^{{\rm i}({\skew0\vec p}_e + {\skew0\vec p}_{\bar\nu}) \cdot {\skew0\vec r}_i/\hbar}
\end{displaymath}

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:

\begin{displaymath}
e^{{\rm i}({\skew0\vec p}_e + {\skew0\vec p}_{\bar\nu}) \c...
...}_{\bar\nu})\cdot{\skew0\vec r}_i}{\hbar}\right)^2
+ \ldots
\end{displaymath}

Since the first term in the Taylor series is by far the largest, you would expect that the value of $H_{21}$, (A.276), can be well approximated by replacing the exponential by 1, giving:

\begin{displaymath}
H_{21}^0 = \frac{g}{\sqrt{{\cal V}}} \langle \psi_{2,\rm nuc} \vert \sum_{i=1}^A h_i
\psi_{1,\rm nuc}\rangle
\end{displaymath}

However, clearly this approximation does not work if the value of $H_{21}^0$ is zero for some reason:

If the simplified coefficient $H_{21}^0$ is nonzero, the decay is allowed. If it is zero, the decay is forbidden.
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 $H_{21}$. Since these higher order terms are much smaller, and $H_{21}$ drives the decay, a forbidden decay will proceed much slower than an allowed one.

Why would a decay not be allowed? In other words why would $\psi_{2,\rm {nuc}}$ and $h_i\psi_{1,\rm {nuc}}$ be exactly orthogonal? If you took two random wave functions for $\psi_{2,\rm {nuc}}$ and $\psi_{1,\rm {nuc}}$, they definitely would not be. But $\psi_{2,\rm {nuc}}$ and $\psi_{1,\rm {nuc}}$ 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 $\psi_{1,\rm {nuc}}$ in the inner product in $H_{21}^0$, 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, $H_{21}^0$ 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 $H_{21}^0$. Therefore, their orbital angular momentum is approximated to be zero too. Under these condition $H_{21}^0$ 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 ${\skew0\vec r}\to-{\skew0\vec r}$. 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 $H_{21}^0$ 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,

\begin{displaymath}
H_{21}^1 = \frac{g}{\sqrt{{\cal V}}} \langle \psi_{2,\rm n...
...r\nu})\cdot{\skew0\vec r}_i}{\hbar}
\psi_{1,\rm nuc}\rangle
\end{displaymath}

has to be nonzero. If it is, the decay is a first-forbidden one. Now the spherical harmonics $Y_1^m$ of orbital angular momentum are of the generic form, {D.14}

\begin{displaymath}
r Y_1^m = \sum_j c_j r_j
\end{displaymath}

with the $c_j$ some constants. Therefore, the factor ${\skew0\vec r}_i$ in $H_{21}^1$ 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 ${\skew0\vec r}_i$ changes sign when every ${\skew0\vec r}$ is replaced by $\vphantom0\raisebox{1.5pt}{$-$}$${\skew0\vec r}$, the initial and final nuclear parities must now be opposite for $H_{21}^1$ 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 $\ell$th-forbidden decay,

\begin{displaymath}
H_{21}^\ell =
\frac{g}{\ell!\sqrt{{\cal V}}} \langle \ps...
...kew0\vec r}_i}{\hbar}\right)^\ell
\psi_{1,\rm nuc}\rangle %
\end{displaymath} (A.277)

must be the first nonzero inner product. Note that an $\ell$th-forbidden decay has a coefficient $H_{21}$ proportional to a factor of order $(pR/\hbar)^\ell$, with $R$ the nuclear radius. Since the decay rate turns out to be proportional to $\vert H_{21}\vert^2$, an $\ell$th-forbidden decay is slowed down by a factor of order $(pR/\hbar)^{2\ell}$, making highly forbidden decays extremely slow.


A.44.4 The nuclear operator

This subsection will have a closer look at the nuclear operator $h_i$. 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 $h_i$ 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), $gh_i$ 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 $h_i$ 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 $gh_i$ 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 $gh_i$ will need to include a isospin creation operator to be able to turn a neutron into a proton. In Fermi decays, $h_i$ is assumed to be just that operator. The constant of proportionality $g$, usually called the coupling constant $g_{\rm {F}}$, describes the strength of the weak interaction. That is much like the unit electric charge $e$ describes the strength of the electromagnetic interaction between charged particles and photons. In Fermi decays it is found that $g_{\rm {F}}$ is about 88 eV fm$\POW9,{3}$. 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$\POW9,{34}$ 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, $h_i$ 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 $g_{GT}$ is found to be a bit larger than the Fermi one.

The relevant operators then become, [5],

\begin{displaymath}
h_i = \tau_1\pm{\rm i}\tau_2
\qquad
h_i = (\tau_1\pm{\rm i}\tau_2) \sum_{j=1}^3 \sigma_j
\end{displaymath}

for Fermi and Gamow-Teller decays respectively. Here the three $\sigma_j$ are the Pauli spin matrices of chapter 12.10. The $\tau_i$ 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 $\langle\psi_{2,\rm {nuc}}\vert h_i\psi_{1,\rm {nuc}}\rangle$ 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 $N_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $Z_1\pm1$ decays into one with neutron number $N_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $Z_2\mp1$. 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 $\sqrt{2}$. (The reason that it is $\sqrt{2}$ 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 ${\textstyle\frac{1}{2}}$ 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 $T_{3,\rm {net}}$. 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 $\sqrt{2}$.)

These decays therefore allow the value of the Fermi coupling constant $g_{\rm {F}}$ to be determined from the decay rates. It turns out to be about 88 eV fm$\POW9,{3}$, 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 $g_{\rm {F}}$ 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 $\psi_1$ to a state $\psi_2$. Most of the attention was on the coefficient $H_{21}^\ell$ 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 $\psi_1$ will be indicated by $\bar{a}$ and the quantum amplitude of the final decayed state $\psi_2$ by $\bar{b}$. The Schrö­din­ger equation implies that $\bar{b}$ increases from zero according to

\begin{displaymath}
{\rm i}\hbar \dot{\bar b} = H_{21}^\ell e^{{\rm i}(E_2-E_1)t/\hbar} \bar a
\end{displaymath}

(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 $\bar{a}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. All these nuclei then evolve according to the Schrö­din­ger equation, above, over a time interval $t_c$ that is short enough that $\bar{a}$ 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, $\bar{a}$ can be dropped from the equation and its solution is then seen to be

\begin{displaymath}
\bar b = - H_{21}^\ell \frac{e^{{\rm i}(E_2-E_1)t_c/\hbar}-1}{(E_2-E_1)}
\end{displaymath}

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

\begin{displaymath}
\bar b = - H_{21}^\ell e^{{\rm i}\frac12(E_2-E_1)t_c/\hbar...
...ac12(E_2-E_1)t_c/\hbar\Big)}{\frac12(E_2-E_1)t_c/\hbar}
t_c
\end{displaymath}

Then at the final time $t_c$, 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 $\vert\bar{b}\vert^2$ of the amplitude of the decayed state. Therefore, a fraction $\vert\bar{b}\vert^2$ of the nuclei will be found to have decayed and $1-\vert\bar{b}\vert^2$ will be found to be still in the pre-decay state $\psi_1$. After this measurement, the entire process then repeats for the remaining $1-\vert\bar{b}\vert^2$ nuclei that did not decay.

The bottom line is however that a fraction $\vert\bar{b}\vert^2$ did. Therefore, the ratio $\vert\bar{b}\vert^2$$\raisebox{.5pt}{$/$}$$t_c$ gives the specific decay rate, the relative amount of nuclei that decay per unit time. Plugging in the above expression for $\bar{b}$ gives:

\begin{displaymath}
\lambda_{\rm single\ final\ state} = \frac{\vert H_{21}^\e...
...c/\hbar\Big)}
{\Big(\frac12(E_2-E_1)t_c/\hbar\Big)^2} t_c %
\end{displaymath} (A.278)

To get the total decay rate, you must still sum over all possible final states. Most importantly, you need to sum the specific decay rates together for all possible electron and antineutrino momenta.

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 $\vert H_{21}^\ell\vert$ for the decay $1\to2$ will not be the same as the one for $2\to1$: 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

\begin{displaymath}
\lambda = \sum_{{\rm all\ }{\skew0\vec p}_e,{\skew0\vec p}...
...t_c/\hbar\Big)}
{\Big(\frac12(E_2-E_1)t_c/\hbar\Big)^2} t_c
\end{displaymath}

Based on energy conservation, you would expect that decays should only occur when the total energy $E_2$ of the nucleus, electron and antineutrino after the decay is exactly equal to the energy $E_1$ of the nucleus before the decay. However, the summation above shows that that is not quite true. For a final state that has $E_2$ exactly equal to $E_1$, the last fraction in the summation is seen to be unity, using l’Hospital. For a final state with an energy $E_2$ of, for example, $E_1+\hbar/t_c$, 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 $H_{21}^\ell$ 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 $E_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E_1$ 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 $E_2$ 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:

\begin{displaymath}
\lambda = \int_{{\rm all\ }E_2}
\frac{\vert H_{21}^\ell\...
.../\hbar\Big)^2} t_c
\frac{{\rm d}N}{{\rm d}E_2} {\,\rm d}E_2
\end{displaymath}

where ${\rm d}{N}$$\raisebox{.5pt}{$/$}$${\rm d}{E_2}$ is the number of final states per unit energy range, often called the density of states $\rho(E_2)$.

Now assume that the complete problem is cut into bite-size pieces for each of which $\vert H_{21}^\ell\vert$ 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 ${\rm d}{N}$$\raisebox{.5pt}{$/$}$${\rm d}{E_2}$ 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:

\begin{displaymath}
\lambda =
\frac{\vert H_{21}^\ell\vert^2}{\hbar^2} t_c \frac{{\rm d}N}{{\rm d}E} \frac{2\pi\hbar}{t_c}
\end{displaymath}

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 $t_c$ 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

\begin{displaymath}
\Delta E_{\rm slop} = \frac{2\pi\hbar}{t_c}
\end{displaymath}

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

\begin{displaymath}
\fbox{$\displaystyle
\lambda_{\rm single\ final\ state} ...
... \varepsilon
\qquad
\varepsilon t_c \sim 2\pi\hbar
$} %
\end{displaymath} (A.279)

Here $\varepsilon$ is the amount that energy conservation seems to be violated, and is related to a typical time $t_c$ between collisions by the energy-time uncertainty relationship shown.

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 $\varepsilon$ is so large that ${\rm d}{N}$$\raisebox{.5pt}{$/$}$${\rm d}{E}$ 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 $\bar{b}$ becomes appreciable, $\bar{a}$ can no longer be close to one since the probabilities $\vert\bar{a}\vert^2$ and $\vert\bar{b}\vert^2$ 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

\begin{displaymath}
\vert H_{21}\vert \ll \varepsilon \ll E
\qquad \varepsilon \equiv \frac{2\pi\hbar}{t_c}
\end{displaymath} (A.280)

The first inequality means that the perturbation causing the decay must be weak enough that there is only a small chance of decay before a collision occurs. The second inequality means that there must be enough time between collisions that an apparent energy conservation from initial to final state applies. Roughly speaking, collisions must be sufficiently frequent on the time scale of the decay process, but rare on the quantum time scale $\hbar$$\raisebox{.5pt}{$/$}$$E$.

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 ${\skew0\vec p}_e$ and ${\skew0\vec p}_{\bar\nu}$ of the electron and antineutrino. Their wave functions were approximately of the form

\begin{displaymath}
\psi_{{\skew0\vec p}} = \frac{1}{\sqrt{{\cal V}}} e^{{\rm i}(p_x x + p_y y + p_z z)/\hbar}
\end{displaymath}

where ${\cal V}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell^3$ 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 $e^{{\rm i}{p}_xx/\hbar}$ is 1 at $x$=0, and it must be 1 again at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell$. That requires ${p}_x\ell$$\raisebox{.5pt}{$/$}$$\hbar$ to be a whole multiple of $2\pi$. Therefore $p_x$ must be a whole multiple of $2\pi\hbar$$\raisebox{.5pt}{$/$}$$\ell$. Successive possible $p_x$ values are therefore spaced the finite amount $2\pi\hbar$$\raisebox{.5pt}{$/$}$$\ell$ apart. And so are successive $p_y$ and $p_z$ values.

Figure A.27: Possible momentum states for a particle confined to a periodic box. The states are shown as points in momentum space. States that have momentum less than some example maximum value are in red.
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...y$}}
\put(-45,104.5){\makebox(0,0)[b]{$p_z$}}
\end{picture}
\end{figure}

Graphically this can by visualized by plotting the possible momentum values as points in a three-di­men­sion­al $p_x,p_y,p_z$ 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 $2\pi\hbar$$\raisebox{.5pt}{$/$}$$\ell$. The volume (in this three-di­men­sion­al momentum plot, not physical volume) of that little cube is $(2\pi\hbar/\ell)^3$. Since $\ell^3$ is the physical volume ${\cal V}$ of the periodic box, the volume in momentum space taken up by each momentum state is $(2\pi\hbar)^3$$\raisebox{.5pt}{$/$}$${\cal V}$

That allows the number of different momentum states to be computed. In particular, consider how many states have magnitude of momentum $\vert{\skew0\vec p}\,'\vert$ less than some maximum value $p$. For some example value of $p$, these are the red states in figure A.27. Note that they form a sphere of radius $p$. That sphere has a volume equal to $\frac43\pi{p}^3$. Since each state takes up a volume $(2\pi\hbar)^2$$\raisebox{.5pt}{$/$}$${\cal V}$, the number of states $N$ is given by the number of such volumes in the sphere:

\begin{displaymath}
N_{\vert{\skew0\vec p}\,'\vert\mathrel{\raisebox{-.5pt}{$\...
...e\leqslant$}}p}=\frac{\frac43\pi p^3}{(2\pi\hbar)^3/{\cal V}}
\end{displaymath}

The number of electron states that have momentum in a range from $p_e$ to $p_e+{\rm d}{p_e}$ can be found by taking a differential of the expression above:

\begin{displaymath}
{\rm d}N_e = \frac{{\cal V}p_e^2}{2 \pi^2\hbar^3} {\rm d}p_e
\end{displaymath}

(Here the range ${\rm d}{p}_e$ 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 $T_{\bar\nu}$ is fixed in terms of that of the electron and the energy release of the decay $Q$ by:

\begin{displaymath}
T_{\bar\nu} = Q - T_e
\end{displaymath}

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:

\begin{displaymath}
T_e = \sqrt{(m_{\rm e}c^2)^2 + p_e^2 c^2} - m_{\rm e}c^2
\qquad
T_{\bar\nu} = p_{\bar\nu} c %
\end{displaymath} (A.281)

where $c$ is the speed of light and the extremely small rest mass of the neutrino was ignored. With the neutrino energy fixed, so is the magnitude of the neutrino momentum:

\begin{displaymath}
p_{\bar\nu} = \frac{1}{c}(Q - T_e)
\end{displaymath}

These result shows that the neutrino momentum is fixed for given electron momentum $p_e$. Therefore there should not be a neutrino momentum range ${\rm d}{p}_{\bar\nu}$ 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

\begin{displaymath}
\Delta E_{\rm slop} = \Delta T_{\bar\nu} = \Delta p_{\bar\nu} c
\equiv \varepsilon
\end{displaymath}

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

\begin{displaymath}
\Delta N_{\bar\nu} = \frac{{\cal V}p_{\bar\nu}^2}{2 \pi^2\...
...{1}{c} \varepsilon
\qquad p_{\bar\nu} = \frac{1}{c} (Q-T_e)
\end{displaymath}

The number of complete system states in an electron momentum range ${\rm d}{p}_e$ is the product of the number of electron states times the number of antineutrino states:

\begin{displaymath}
{\rm d}N = {\rm d}N_e \Delta N_{\bar\nu}
= \frac{{\cal V...
...^4\hbar^6 c}\, p_e^2 p_{\bar\nu}^2\, \varepsilon {\,\rm d}p_e
\end{displaymath}

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

\begin{displaymath}
\lambda_{\rm single\ final\ state} =
\frac{2\pi}{\hbar\varepsilon} \vert H_{21}^\ell\vert^2
\end{displaymath}

Therefore the total specific decay rate is

\begin{displaymath}
\lambda = \int_{p_e=0}^{p_{e,\rm max}}
\frac{{\cal V}^2\...
...ll\vert^2}{2 \pi^3\hbar^7 c} p_e^2 p_{\bar\nu}^2 {\,\rm d}p_e
\end{displaymath}

where the maximum electron momentum $p_{e,max}$ can be computed from the $Q$-​value of the decay using (A.281). (For simplicity it will be assumed that $\vert H_{21}^\ell\vert^2$ has already been averaged over all directions of the electron and antineutrino momentum.)

The derived expression (A.277) for the Hamiltonian coefficient $H_{21}^\ell$ can be written in the form

\begin{displaymath}
\vert H_{21}^\ell\vert^2 = \frac{g^2}{{\cal V}^2} \frac{1}...
...{\sqrt{p_e^2+p_{\bar\nu}^2} R}{\hbar}\right)^{2\ell} C_N^\ell
\end{displaymath}


\begin{displaymath}
C_N^\ell \equiv
\overline{
\bigg\vert\bigg\langle \psi...
...
\bigg)^\ell
\psi_{1,\rm nuc}\bigg\rangle\bigg\vert^2
}
\end{displaymath}

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

It is not easy to say much about $C_N^\ell$ in general, beyond the fact that its magnitude should not be much more than one. This book will essentially ignore $C_N^\ell$ 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$\POW9,{17}$ year.

The decay rate becomes after clean up

\begin{displaymath}
\lambda =
\frac{1}{2\pi^3}
\frac{g^2m_{\rm e}^4 c^2}{\...
...tilde p_e^2 \tilde p_{\bar\nu}^2
F^\ell {\rm d}\tilde p_e %
\end{displaymath} (A.282)

where the $\tilde{p}$ indicate the electron and antineutrino momenta nondimensionalized with ${m_{\rm e}}c$. Also,
\begin{displaymath}
\tilde Q \equiv \frac{Q}{m_{\rm e}c^2}
\quad
\tilde p_...
...2} + 1
\quad
\tilde R \equiv \frac{m_{\rm e}c R}{\hbar} %
\end{displaymath} (A.283)

Here $\tilde{Q}$ is the $Q$-​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 $\tilde{R}$ a suitably nondimensionalized nuclear radius, and is small.

The factor $F^\ell$ 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 $F^0$ for allowed decays is called the “Fermi function” and is given by

\begin{displaymath}
F(\tilde p_e,Z_2,A) =
\frac{2(1+\xi)}{\Gamma^2(1+2\xi)} ...
...^{2-2\xi}}
e^{\pi\eta}\vert\Gamma(\xi+{\rm i}\eta)\vert^2 %
\end{displaymath} (A.284)


\begin{displaymath}
\xi \equiv \sqrt{1-(\alpha Z_2)^2}
\quad
\eta \equiv \...
...alpha = \frac{e^2}{4\pi\epsilon_0\hbar c} \approx \frac1{137}
\end{displaymath}

where $\alpha$ is the fine structure constant and $\Gamma$ the gamma function. The nonrelativistic version follows for letting the speed of light go to infinity, while keeping $p_e$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${m_{\rm e}}c\tilde{p}_e$ finite. That gives $\xi$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and

\begin{displaymath}
F(p_e,Z_2) = \frac{2\pi\eta}{1-e^{-2\pi\eta}}
\qquad
\eta = \frac{e^2}{4\pi\epsilon_0\hbar} \frac{m_{\rm e}}{p_e}
\end{displaymath}

For beta-plus decay, just replace $Z_2$ by $-Z_2$, 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 $Q$-​value and nuclear charge is

\begin{displaymath}
f = \int_{\tilde p_e=0}^{\tilde p_{e,\rm max}}
\tilde p_e^2 \tilde p_{\bar\nu}^2 F {\rm d}\tilde p_e %
\end{displaymath} (A.285)

This quantity is known as the “Fermi integral.” Typical values are shown in figure 14.50.

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

\begin{displaymath}
A = 1.82+1.9\,Z_2 + 0.012{,}71\,Z_2^2- 0.000{,}06\,Z_2^3
\end{displaymath} (A.286)

for beta-minus decay and
\begin{displaymath}
A = -1.9+1.96\,Z_2 + 0.007{,}9\,Z_2^2- 0.000{,}02\,Z_2^3
\end{displaymath} (A.287)

for beta-plus decay, [23].


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

\begin{displaymath}
H_{21} =
\langle 1\bar\nu,{\skew0\vec p}_{\bar\nu}\vert\...
...e,1{\rm s}\rangle\vert\bar\nu,{\skew0\vec p}_{\bar\nu}\rangle
\end{displaymath}

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

\begin{displaymath}
H_{21} = \frac{g}{\sqrt{{\cal V}}}\langle \psi_{2,\rm nuc}...
...bar\nu}\cdot{\skew0\vec r}_i/\hbar}
\psi_{1,\rm nuc}\rangle
\end{displaymath}

Here $\psi_{100}$ is the hydrogen ground state wave function, but rescaled for a nucleus of charge $Ze$ instead of $e$. It does not contribute to making forbidden decays possible, because $\psi_{100}$ 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, $\psi_{100}$ can safely be approximated by its value at the origin, from chapter 4.3,

\begin{displaymath}
\psi_{100}({\skew0\vec r}_i) \approx \frac{1}{\sqrt{\pi a_...
...hbar^2}{m_{\rm e}e^2Z_1} = \frac{\hbar}{m_{\rm e}c\alpha Z_1}
\end{displaymath}

where $\alpha$ is the fine-structure constant.

The square Hamiltonian coefficient for $\ell$th-forbidden decays then becomes

\begin{displaymath}
\vert H_{21}^\ell\vert^2 =
\frac{g^2}{{\cal V}}
\frac{...
...2}
\left(\frac{p_{\bar\nu}R}{\hbar}\right)^{2\ell} C_N^\ell
\end{displaymath}


\begin{displaymath}
C_N^\ell = \equiv
\overline{
\bigg\vert\bigg\langle \p...
...R} \bigg)^\ell
\psi_{1,\rm nuc}\bigg\rangle\bigg\vert^2
}
\end{displaymath}

The decay rate for electron capture is

\begin{displaymath}
\lambda =
\frac{2\pi}{\hbar\varepsilon}\vert H_{21}^\ell...
...p_{\bar\nu} \qquad \Delta p_{\bar\nu} = \frac{\varepsilon}{c}
\end{displaymath}

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

\begin{displaymath}
\lambda =
\frac{2}{\pi^2}
\frac{g^2m_{\rm e}^4c^2}{\hb...
..._N^\ell
\tilde p_{\bar\nu}^{2\ell}
\tilde p_{\bar\nu}^2 %
\end{displaymath} (A.288)

where the $\tilde{p}_{\bar\nu}$ indicate the neutrino momentum nondimensionalized with ${m_{\rm e}}c$. Also,
\begin{displaymath}
\tilde Q \equiv \frac{Q}{m_{\rm e}c^2}
\qquad
\tilde p...
...lde Q
\qquad
\tilde R \equiv \frac{m_{\rm e}c R}{\hbar} %
\end{displaymath} (A.289)