Subsections


14.18 Isospin

Isospin is another way of thinking about the two types of nucleons. It has proved quite useful in understanding nuclei, as well as elementary particles.


14.18.1 Basic ideas

Normally, you think of nuclei as consisting of protons and neutrons. But protons and neutrons are very similar in properties, if you ignore the Coulomb force. They have almost the same mass. Also, according to charge independence, the nuclear force is almost the same whether it is protons or neutrons.

So suppose you define only one particle, called nucleon. Then you can give that particle an additional property called “nucleon type.” If the nucleon type is $\frac12$, it is a proton, and if the nucleon type is $-{\textstyle\frac{1}{2}}$ it is a neutron. That makes nucleon type a property that is mathematically much like the spin $S_z$ of a nucleon in a chosen $z$-​direction. But of course, there is no physical nucleon-type axis. Therefore nucleon type is conventionally indicated by the symbol $T_3$, not $T_z$, with no physical meaning attached to the 3-axis. In short, nucleon type is defined as:

\begin{displaymath}
\fbox{$\displaystyle
\mbox{proton: } T_3 = {\textstyle\f...
...\qquad
\mbox{neutron: } T_3 = -{\textstyle\frac{1}{2}}
$}
\end{displaymath} (14.40)

So far, all this it may seem like a stupid mathematical trick. And normally it would be. The purpose of mathematical analysis is to understand systems, not to make them even more incomprehensible.

But to the approximation that the nuclear force is charge-independent, nucleon type is not so stupid after all. If the nuclear force is charge-independent, and the Coulomb force is ignored, you can write down nuclear wave functions without looking at the nucleon type. Now suppose that in doing so, you find some energy eigenfunction of the form

\begin{displaymath}
\psi_{\rm {A}} = \psi_{\rm s}({\skew0\vec r}_2-{\skew0\vec r}_1) \big\vert 1\:1\big\rangle
\end{displaymath}

This is a wave function for two nucleons labeled 1 and 2. Assume here that the spatial part $\psi_{\rm {s}}$ is unchanged under nucleon exchange (swapping the nucleons):

\begin{displaymath}
\psi_{\rm s}({\skew0\vec r}_2-{\skew0\vec r}_1) = \psi_{\rm s}({\skew0\vec r}_1-{\skew0\vec r}_2)
\end{displaymath}

(This is equivalent to assuming that the wave function has even parity.) Further recall from chapter 5.5.6 that the triplet spin state $\big\vert 1\:1\big\rangle $ is unchanged under nucleon exchange too. So the total wave function $\psi_{\rm {A}}$ above is unchanged under nucleon exchange.

That is fine if nucleon 1 is a proton and nucleon 2 a neutron. Or vice-versa. But it is not OK if both nucleons are protons, or if they are both neutrons. Wave functions must change sign if two identical fermions are exchanged. That is the antisymmetrization requirement. The wave function above stays the same. So it is only acceptable for the deuteron, with one proton and one neutron.

Next suppose you could find a different wave function of the form

\begin{displaymath}
\psi_{\rm {B}} = \psi_{\rm s}({\skew0\vec r}_2-{\skew0\vec r}_1) \big\vert\:0\big\rangle
\end{displaymath}

(Here $\psi_s$ is not necessarily the same as before, but assume it still has even parity.) The singlet spin state $\big\vert\:0\big\rangle $ changes sign under nucleon exchange. Then so does the entire wave function $\psi_{\rm {B}}$. And that then means that $\psi_{\rm {B}}$ is acceptable even if the nucleons are both protons or both neutrons. This wave function works not just for the deuteron, but also for the diproton” and the “dineutron. (The prefix di means two.)

That would give nontrivial insight in nuclear energy levels. It would mean physically that the diproton, the dineutron, and the deuteron can be in an identical energy state. Such identical energy states, occurring for different nuclei, are called “isobaric analog (or analogue) states.” Or “charge states.” Or “isobaric multiplets.” Or “$T$-​multiplets.” Hey, don’t blame the messenger.

Disappointingly, in real life there is no bound state of the form $\psi_{\rm {B}}$. Still the bottom line stays:

Within the approximations of charge independence and negligible Coulomb effects, whether a given state applies to a given set of nucleon types depends only on the antisymmetrization requirements.
Now for bigger systems of nucleons, the antisymmetrization requirements get much more complex. A suitable formalism for dealing with that has already been developed in the context of the spin of systems of identical fermions. It is convenient to adopt that formalism also for nucleon type.

As an example, consider the above three hypothetical isobaric analog states for the diproton, dineutron, and deuteron. They can be written out separately as, respectively,

\begin{displaymath}
\psi_{\rm s}({\skew0\vec r}_2-{\skew0\vec r}_1) \big\vert\...
...frac{\Uparrow_1\Downarrow_2+\Downarrow_1\Uparrow_2}{\sqrt{2}}
\end{displaymath}

Here $\Uparrow$ means the nucleon is a proton and $\Downarrow$ means it is a neutron. If you want, you can write out the above wave functions explicitly in terms of the nucleon type $T_3$ as

\begin{eqnarray*}
& \displaystyle
\psi_{\rm s}({\skew0\vec r}_2-{\skew0\vec ...
...rac{1}{2}}-{T_3}_1)({\textstyle\frac{1}{2}}+{T_3}_2)}{\sqrt{2}}
\end{eqnarray*}

Note, for example, that the first wave function is zero if either ${T_3}_1$ or ${T_3}_2$ is equal to $-\frac12$. So it is zero if either nucleon is a neutron. The only way to get something nonzero is if both nucleons are protons, with ${T_3}_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${T_3}_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$. Similarly, the second wave function is only nonzero if both nucleons are neutrons, with ${T_3}_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${T_3}_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-\frac12$.

The third wave function represents something of a change in thinking. It requires that one nucleon is a proton and the other a neutron. So it is a wave function for the deuteron. But the actual wave function above is a superposition of two states. In the first state, nucleon 1 is the proton and nucleon 2 the neutron. In the second state, nucleon 1 is the neutron and nucleon 2 the proton. In the combined state the nucleons have lost their identity. It is uncertain whether nucleon 1 is the proton and nucleon 2 the neutron, or vice-versa.

Within the formalism of identical nucleons that have an additional nucleon-type property, this uncertainty in nucleon types is unavoidable. The wave function would not be antisymmetric under nucleon exchange without it. But if you think about it, this may actually be an improvement in the description of the physics. Protons and neutrons do swap identities. That happens if they exchange a charged pion. Proton-neutron scattering experiments show that they can do that. For nucleons that have a probability of swapping type, assigning a fixed type in energy eigenstates is not right. Energy eigenstates must be stationary. And having a better description of the physics can affect what sort of potentials you would want to write down for the nucleons.

(You might think that without charge independence, the additional antisymmetrization requirement for identical nucleons would change the physics. But actually, it does not. The antisymmetrization requirement can be accomodated by uncertainty in which nucleon you label 1 and which 2. Consider some completely general proton-neutron wave function $\Psi({\skew0\vec r}_{\rm {p}},S_{z\rm {p}},{\skew0\vec r}_{\rm {n}},S_{z\rm {n}})$, one that would not be the same if you swap the proton and neutron. It might be a subsystem in some larger nucleus for which charge-independence is not a good approximation. The antisymmetrized identical-nucleon wave function is

\begin{displaymath}
\frac{1}{\sqrt{2}} \Psi({\skew0\vec r}_1,S_{z1},{\skew0\ve...
..._2,S_{z2},{\skew0\vec r}_1,S_{z1})
\Downarrow_1\Uparrow_2 %
\end{displaymath} (14.41)

This is a superposition of two parts. In the first part the proton is labeled 1 and the neutron 2. In the second part, the proton is labeled 2 and the neutron 1. The physics has stayed exactly the same. What has changed is that there is now confusion about whether the particle labeled 1 is the proton or the neutron. This illustrates that without charge independence, the mathematical trickery employed here is not really wrong. But it is entirely useless, as Wigner was the first to point out, [49, p. 4].)

Now compare the trailing nucleon-type factors in the three hypothetical isobaric analog states above with the possible combined spin states of two spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ fermions. It is seen that the nucleon-type factors take the exact same form as the so-called triplet spin states, (5.26). So define similarly

\begin{displaymath}
\Uparrow_1\Uparrow_2 \equiv {\big\vert 1\:1\big\rangle }_T...
...\Uparrow_2}{\sqrt{2}}
\equiv {\big\vert 1\:0\big\rangle }_T
\end{displaymath} (14.42)

In those terms, the isobaric analog states are all three of the form

\begin{displaymath}
\psi_{\rm s}({\skew0\vec r}_2-{\skew0\vec r}_1) \big\vert\:0\big\rangle {\big\vert 1\:m_T\big\rangle }_T
\end{displaymath}

where $m_T$ is 1, $\vphantom0\raisebox{1.5pt}{$-$}$1, or 0 for the diproton, dineutron, and deuteron respectively. Note that that is just the sum of the $T_3$ values of the two nucleons,

\begin{displaymath}
m_T = {T_3}^{}_1 + {T_3}^{}_2
\end{displaymath}

Now reconsider the wave function for two nucleons that was written down first. The one that was only acceptable for the deuteron. In the same terminology, it can be written as

\begin{displaymath}
\psi_{\rm s}({\skew0\vec r}_2-{\skew0\vec r}_1) \big\vert ...
...frac{\Uparrow_1\Downarrow_2-\Downarrow_1\Uparrow_2}{\sqrt{2}}
\end{displaymath}

The formalism of identical nucleons with nucleon type forces again uncertainty in the nucleon types. But now there is a minus sign. That makes the nucleon-type state a singlet state in the terminology of spin.

Of course, all this raises the question what to make of the leading 0 in the singlet state ${\big\vert\:0\big\rangle }_T$, and the leading 1 in the triplet states $\big\vert 1\:m_T\big\rangle $? If this was spin angular momentum, the 0 or 1 would indicate the quantum number $s$ of the square spin angular momentum. Square spin angular momentum is the sum of the square spin components in the $x$, $y$, and $z$ directions. But at first that seems to make no sense for nucleon type. Nucleon type is just a simple number, not a vector. While it has been formally associated with some abstract 3-axis, there are no $T_1$” and “$T_2$ components.

However, it is possible to define such components in complete analogy with the $x$ and $y$ components of spin. In quantum mechanics the components of spin are the eigenvalues of operators. And using advanced concepts of angular momentum, chapter 12, the operators of $x$ and $y$ angular momenta can be found without referring explicitly to their axes. The same procedure can be followed for nucleon type.

To do so, first an operator ${\widehat T}_3$ for the nucleon type $T_3$ is defined as

\begin{displaymath}
\fbox{$\displaystyle
{\widehat T}_3 \Uparrow = {\textsty...
...at T}_3 \Downarrow = -{\textstyle\frac{1}{2}} \Downarrow
$}
\end{displaymath} (14.43)

In words, the proton state is an eigenstate of this operator with eigenvalue $\frac12$. The neutron state is an eigenstate with eigenvalue $-\frac12$. That follows the usual rules of quantum mechanics; observable quantities, (here nucleon type), are the eigenvalues of Hermitian operators.

Next a “charge creation operator” is defined by

\begin{displaymath}
\fbox{$\displaystyle
{\widehat T}^{+} \Downarrow = \Uparrow \qquad {\widehat T}^{+} \Uparrow = 0
$}
\end{displaymath} (14.44)

In words, it turns a neutron into a proton. In effect it adds a unit of charge to it. Since a nucleon with two units of charge does not exist, the operator needs to turn a proton state into zero. Similarly a “charge annihilation operator” is defined by
\begin{displaymath}
\fbox{$\displaystyle
{\widehat T}^{-} \Uparrow = \Downarrow \qquad {\widehat T}^{-} \Downarrow = 0
$}
\end{displaymath} (14.45)

Operators for nucleon type in the 1 and 2 directions can now be defined as
\begin{displaymath}
\fbox{$\displaystyle
{\widehat T}_1 = {\textstyle\frac{1...
... T}^{+} + {\textstyle\frac{1}{2}}{\rm i}{\widehat T}^{-}
$}
\end{displaymath} (14.46)

The eigenvalues of these operators are by definition the values of $T_1$ respectively $T_2$.

With these operators, square nucleon type can be defined just like square spin. All the mathematics has been forced to be the same.

The quantum number of square nucleon type will be indicated by $t_T$ in this book. Different sources use different notations. Many sources swap case, using lower case for the operators and upper case for the quantum numbers. Or they use lower case if it is for a single nucleon and upper case for the entire nucleus. They often do the same for angular momentum. Some sources come up with $I$ for the square nucleon type quantum number, using $J$ for the angular momentum one. However, this book cannot adopt completely inconsistent notations just for nuclear physics. Especially if there is no generally agreed-upon notation in the first place.

In any case there are three scaled operators whose definition and symbols are fairly standard in most sources. These are defined as

\begin{displaymath}
\fbox{$\displaystyle
\tau_1 = 2 {\widehat T}_1 \quad
\...
...at T}^{+} \qquad
\tau^{-} = 2 {\widehat T}^{-1} \qquad
$}
\end{displaymath} (14.47)

The first three are the direct equivalents of the famous Pauli spin matrices, chapter 12.10. Note that they simply scale away the factors $\frac12$ in the nucleon type. The Pauli spin matrices also scale away the factor $\hbar$ that appears in the values of spin.


14.18.2 Heavier nuclei

Now consider an example of isobaric analog states that actually exist. In this case the nucleons involved are carbon-14, nitrogen-14, and oxygen-14. All three have 14 nucleons, so they are isobars. However, carbon-14 has 6 protons and 8 neutrons, while oxygen-14 has 8 protons and 6 neutrons. Such pairs of nuclei, that have their numbers of protons and neutrons swapped, are called “conjugate” nuclei. Or “mirror” nuclei. Nitrogen-14 has 7 protons and 7 neutrons and is called “self-conjugate.”

Since $T_3$ values add up, carbon-14 with 6 protons at $\frac12$ each and 8 neutrons at $-\frac12$ each has net $T_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$1. Similarly, nitrogen-14 has $T_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, as any self-conjugate nucleus, while oxygen-14 has $T_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. In general,

\begin{displaymath}
\fbox{$\displaystyle
T_3 = {\textstyle\frac{1}{2}}(Z-N)
$} %
\end{displaymath} (14.48)

where $Z$ is the number of protons and $N$ the number of neutrons. Note that the value of $T_3$ is fixed for a given nucleus. It is minus half the neutron excess of the nucleus.

In general, 14 nucleons can have a maximum $T_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ 7, if all 14 are protons. The minimum is $\vphantom0\raisebox{1.5pt}{$-$}$7, if all 14 are neutrons.

Here is where the analogy with spin angular momentum gets interesting. Angular momentum is a vector. A given angular momentum vector can still have different directions. And different directions means different values of the $z$-​component of its spin $S_z$. In particular, the quantum numbers of the possible $z$ components are

\begin{displaymath}
m_s \equiv S_z/\hbar = -s,\ -s{+}1,\ \ldots,\ s{-}1,\ s
\end{displaymath}

Here $s$ is the quantum number of the square spin of the vector.

Since nucleon type has been defined to be completely equivalent to spin, essentially the same holds. A given nucleon energy state can still have different values of $T_3$:

\begin{displaymath}
\fbox{$\displaystyle
m_T \equiv T_3 = -t_T,\ -t_T{+}1,\ \ldots,\ t_T{-}1,\ t_T
$}
\end{displaymath} (14.49)

Here $t_T$ is the square nucleon-type quantum number of the state. The different values for $T_3$ above correspond to isobaric analog states for different nuclei. They are the same energy state, but for different nuclei.

You could say that isobaric analog states arise because rotating an energy state in the abstract 1,2,3-space defined above does not make a difference. And the reason it does not make a difference is charge independence.

Based on the values of $T_3$ above, consider the possible values of $t_T$ for 14 nucleons. The value of $t_T$ cannot be greater than 7. Otherwise there would be isobaric analog states with $T_3$ greater than 7, and that is not possible for 14 nucleons. As far as the lowest possible value of $t_T$ is concerned, it varies with nucleus. As the expression above shows, the value of $\vert T_3\vert$ cannot be greater than $t_T$. So $t_T$ cannot be less than $\vert T_3\vert$. Since carbon-14 and oxygen-14 have $\vert T_3\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, for these nuclei, $t_T$ cannot be less than 1. However, nitrogen-14, with $T_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, also allows states with $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.

It turns out that light nuclei in their ground state generally have the smallest value of $t_T$ consistent with their value of $T_3$. One way to get some idea of why that would be so is to look at the antisymmetrization requirements. A set of states with $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 7 for 14 nucleons allows the state $T_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ 7, in which all 14 nucleons are protons. In that state, the wave function must be antisymmetric when any nucleon is interchanged with any other. On the other hand, a $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 state only needs to satisfy the antisymmetrization requirements for 7 protons and 7 neutrons. It does not have to be antisymmetric if a proton is exchanged with any one of the 7 neutrons. So antisymmetrization is less confining. In general, a state with $t_T$ greater than $\vert T_3\vert$ must work for more nuclei than one with $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vert T_3\vert$.

(Another argument, offered in literature, is essentially the reverse of the one that gives rise to the so-called Hund rule for atoms. Simply put, the Hund rule says that a couple of electrons maximize their spin, given the option between single-particle states of the same energy. The reason is that this allows electrons to stay farther apart, reducing their Coulomb repulsion, {D.56}. This argument reverses for nucleons, since they normally attract rather than repel each other. However, surely this is a relatively minor effect? Consider 3 nucleons. For these, the highest value $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac32$ allows the possibility that all 3 are protons. Within a single-particle-state picture, only one can go into the lowest energy state; the second must go into the second lowest energy state, and the third in the third lowest. On the other hand, for say 2 protons and 1 neutron, the neutron can go into the lowest energy state with the first proton. So the lower value $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$ should normally have significantly less energy.)

For the deuteron $T_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, so the lowest possible value of $t_T$ is 0. Then according to the general rule above, the ground state of the deuteron should have $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. That was already established above; it was the bound state not shared with the diproton and dineutron. Nitrogen-14 also has $T_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, which means it too must have $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 in its ground state. This lowest energy state cannot occur for carbon-14 or oxygen-14, because they have $\vert T_3\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. So nitrogen-14 should have less energy in its ground state than carbon-14 and oxygen-14. That seems at first surprising since nitrogen-14 has odd numbers of protons and neutrons, while carbon-14 and oxygen-14 have even numbers. Normally odd-odd nuclei are less tightly bound than even-even ones.

Figure 14.44: Isobaric analog states. [pdf]
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
% vertical spaci...
...9}{2}}$}}
\put(400,4){\makebox(0,0)[bl]{?}}
\end{picture}
\end{figure}

But it is true. Figure 14.44 shows the energy levels of carbon-14, nitrogen-14, and oxygen-14. More precisely, it shows their binding energy, relative to the ground state value for nitrogen-14. In addition the von Weizsäcker value for the Coulomb energy has been subtracted to more clearly isolate the nuclear force effects. (The net effect is simply to shift the normal spectra of carbon-14 and oxygen-14 up by 2.83, respectively 1.89 MeV.) It is seen that nitrogen-14 is indeed more tightly bound in its ground state than carbon-14 and oxygen-14. Qualitatively, since nitrogen-14 does not have 8 nucleons of the same kind, it has an easier job with satisfying the antisymmetrization requirements.

Traces of the lower energy of light nuclei with $T_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 can also be detected in figures like 14.2, and 14.3 through 14.6. In these figures $T_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 straight above the $Z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $N$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 helium nucleus. Note in particular a distinct dark/light discontinuity in figures 14.3 through 14.6 along this vertical line. This discontinuity is quite distinct both from the magic numbers and from the average stability line that curves away from it.

Carbon-14 and oxygen-14 are mirror nuclei, so you would expect them to have pretty much the same sort of energy levels. Indeed, any oxygen-14 state, having $T_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, must be part of a multiplet with $t_T$ at least 1. Such a multiplet must have an equivalent state with $T_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$1, which means an equivalent carbon-14 state. To the extent that the nuclear force is truly charge-independent, and the Coulomb effect has been properly removed, these two states should have the same energy. And indeed, the lowest four energy states of carbon-14 and oxygen-14 have identical spin and parity and similar energies. Also, both even-even nuclei have a 0$\POW9,{+}$ ground state, as even-even nuclei should. These ground states have $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, the smallest possible.

Now each of these multiplets should also have a version with $T_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, which means a nitrogen-14 state. So any state that carbon-14 and oxygen-14 have should also exist for nitrogen-14. For example, the ground states of carbon-14 and oxygen-14 should also appear as a 0$\POW9,{+}$ state with $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 in the nitrogen-14 spectrum. Indeed, if you inspect the energy levels for nitrogen-14 in figure 14.44, exactly halfway in between the carbon-14 and oxygen-14 ground state energies, there it is!

Ideally speaking, these three states should have the same height in the figure. But it would be difficult to remove the Coulomb effect completely. And charge independence is not exact either, even though it is quite accurate.

A similar $T_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 state can readily be found for the first three excited levels of carbon-14 and oxygen-14. In each case there is a nitrogen-14 state with exactly the same spin and parity and $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 right in between the matching carbon-14 and oxygen-14 levels. (To be sure, ENSDF does not list the $t_T$ values for carbon-14 above the ground state. But common sense says they must be the same as the corresponding states in nitrogen-14 and carbon-14. For the first excited state of carbon-14, this is confirmed in [49, p. 11].)

Figure 14.44 also shows that nitrogen-14 has a lot more low energy states than carbon-14 or oxygen-14. Square nucleon type can explain that too: all the low-lying states of nitrogen-14 that are not shared with carbon-14 and oxygen-14 are $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 states. These states are not possible for the other two nuclei.

Nothing is perfect, of course. The first state with nonzero $t_T$ in the nitrogen spectrum besides the mentioned four isobaric analog states is the 0$\POW9,{-}$, $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 state at 8.8 MeV, just below the 3$\POW9,{-}$ analog state. Carbon-14 has a 0$\POW9,{-}$ state immediately above the 3$\POW9,{-}$ state, but oxygen-14 has no obvious candidate.

Despite such imperfections, consideration of nucleon type is quite helpful for understanding the energy levels of light nuclei. And a lot of it carries over to heavier nuclei, [49, p. 12] and [35, p. 57]. While heavier nuclei have significant Coulomb energy, this long-range force is apparently often not that important here.

Now all that is needed is a good name. Nucleon type or nucleon class are not acceptable; they would give those hated outsiders and pesky students a general idea of what physicists were talking about. However, physicists noted that there is a considerable potential for confusion between nucleon type and spin, since both are described by the same mathematics. To maximize that potential for confusion, physicists decided that nucleon type should be called spin.

Of course, physicists themselves still have to know whether they are talking about nucleon type or spin. Therefore some physicists called nucleon type isobaric spin, because what differentiates isobars is the value of the net $T_3$. Other physicists talked about isotopic spin, because physicists like to think of isotopes, and hey, isotopes have nucleon type too. Some physicists took the isowhatever spin to be $\frac12$ for the proton, others for the neutron. However, that confused physicists themselves, so eventually it was decided that the proton has $\frac12$. Also, a great fear arose that the names might cause some outsiders to suspect that the spin being talked about was not really spin. If you think about it, isobaric angular momentum or “isotopic angular momentum” does not make much sense. So physicists shortened the name to isospin. Isospin means “equal spin” plain and simple; there is no longer anything to give the secret away that it is something completely different from spin. However, the confusion of having two different names for the same quantity was missed. Therefore, the alternate term i-spin was coined besides isospin. It too has nothing to give the secret away, and it restores that additional touch of confusion.

Isospin is conserved when only the nuclear force is relevant. As an example, consider the reaction in which a deuteron kicks an alpha particle out of an oxygen-16 nucleus:

\begin{displaymath}
\fourIdx{16}{8}{}{}{\rm O} +
\fourIdx{2}{1}{}{}{\rm H}
...
...d
\fourIdx{14}{7}{}{}{\rm N} +
\fourIdx{4}{2}{}{}{\rm He}
\end{displaymath}

The oxygen is assumed to be in the ground state. That is a $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 state, in agreement with the fact that oxygen-16 is a light nucleus with $T_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. The deuteron can only be in a $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 state; that is the only bound state. The alpha particle will normally be in the ground state, since it takes over 20 MeV to excite it. That ground state is a $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 one, since it is a light $T_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 nucleus. Conservation of isospin then implies that the nitrogen-14 must have $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 too. The nitrogen can come out excited, but it should not come out in its lowest excited state, the 0$\POW9,{+}$ $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 state shared with carbon-14 and oxygen-14 in figure 14.44. Indeed, experiments show that this lowest excited state is only produced in negligible amounts compared to the surrounding states.

Selection rules for which nuclear decays occur can also be formulated based on isospin. If the electromagnetic force plays a significant part, $T_3$ but not $\vec{T}$ is conserved. The weak force does not conserve $T_3$ either, as beta decay shows. For example, the ground states of oxygen-14 and carbon-14 in figure 14.44 will beta-decay to the ground state of nitrogen 14, changing both $T_3$ and $t_T$. (Oxygen-16 will also beta-decay to the corresponding isobaric analog state of nitrogen-14, a decay that is called superallowed, because it is unusually fast. It is much faster than to the ground state, even though decay to the ground state releases more energy. Carbon-14 has too little energy to decay to the analog state.)

Despite the lack of isospin conservation, isospin turns out to be very useful for understanding beta and gamma decay. See for example the discussion of superallowed beta decays in chapter 14.19, and the isospin selection rules for gamma decay in section 14.20.2.


14.18.3 Additional points

There are other particles besides nucleons that are also pretty much the same except for electric charge, and that can also be described using isospin. For example, the positive, neutral, and negatively charged pions form an isospin triplet of states with $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. Isospin was quite helpful in recognizing the existence of the more basic particles called quarks that make up baryons like nucleons and mesons like pions. In final analysis, the usefulness of isospin is a consequence of the approximate properties of these quarks.

Some sources incorrectly credit the concept of isospin to Heisenberg. But he did not understand what he was doing. Heisenberg did correctly guess that protons and neutrons might be described as two variants of the same particle. He then applied the only quantum approach for a two-state particle to it that he knew, that of spin. However, the mathematical machinery of spin is designed to deal with two-state properties that are preserved under rotations of an axis system, compare {A.19}. That is an inappropriate mathematical approach to describe nucleon type in the absence of charge independence. And at the time Heisenberg himself believed that the nuclear force was far from charge-independent.

(Because the nuclear force is in fact approximately charge-independent, unlike Heisenberg assumed, isospin is preserved under rotations of the abstract 1,2,3 coordinate system as defined in the first subsection. Phrased more simply, without charge independence, energy eigenfunctions would not have definite values of square isospin $t_T$. That would make isospin self-evidently entirely useless, as Wigner pointed out. This point is not very clear from the example of two nucleons in empty space, as discussed above. That is because there the spatial wave function happens to be symmetric under particle exchange even without charge independence. But if you express the isospin states in the general wave function (14.41) in terms of the singlet and triplet states, you quickly see the problem.)

The recognition that isospin was meaningful only in the presence of charge independence, and the proposal that the nuclear force is indeed quite accurately charge-​independent, was mostly due to Wigner, in part with Feenberg. Some initial steps had already been taken by other authors. In particular, Cassen & Condon had already proposed to write wave functions in a form to include isospin,

\begin{displaymath}
\psi = \psi({\skew0\vec r}_1,{S_z}_1,{T_3}_1,{\skew0\vec r}_2,{S_z}_2,{T_3}_2,\ldots)
\end{displaymath}

and proposed symmetry under particle exchange in that form. This is the form of wave functions as written down earlier for the two-nucleon system. Still Wigner is considered the founding father of the study of isospin. His identification of isospin for complex nuclei as we know it today, as a preserved quantum number due to charge independence, is the foundation charter of nuclear isospin. Wigner is also the infernal idiot who decided that nucleon type should be called spin.

See Wilkinson, [49, p. vi, 1-13], for a more extensive discussion of these historical issues. A very different history is painted by Henley in the next chapter in the same book. In this history, Heisenberg receives all the credit. Wigner does not exist. However, the author of this history implicitly admits that Heisenberg did think that the nuclear force was far from charge-independent. Maybe the author understood isospin too poorly to recognize that that is a rather big problem. Certainly there is no discussion. Or the author had a personal issue with Wigner and was willing to sacrifice his scientific integrity for it. Either way, the credibility of the author of this particular history is zero.


14.18.4 Why does this work?

It may seem astonishing that all this works. Why would nucleon type resemble spin? Spin is a vector in three-di­men­sion­al space, not a simple number. Why would energy eigenstates be unchanged under rotations in some weird abstract space?

The simplest and maybe best answer is that nature likes this sort of mathematics. Nature just loves creation and annihilation operators. But still, why would that lead to preserved lengths of vectors in an abstract spaces?

An answer can be obtained by looking a bit closer at square spin. Consider first two spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ fermions. Compare the dot product of their spins to the operator $\widehat{P}^s_{12}$ that exchanges their spins:

\begin{displaymath}
\begin{array}{c}
\;\;\:{\skew 6\widehat{\vec S}}_1\cdot{...
...\:m_s\big\rangle = \big\vert 1\:m_s\big\rangle
\end{array}
\end{displaymath}

The first set of relations is derived in {A.10}. The second set can be verified by looking at the expressions of the spin states (5.26).

Comparing the two sets of relations, it is seen that the dot product of two spins is closely related to the operator that exchanges the two spins:

\begin{displaymath}
{\skew 6\widehat{\vec S}}_1\cdot{\skew 6\widehat{\vec S}}_2 = {\textstyle\frac{1}{4}} \hbar^2 (2 \widehat P^s_{12}-1)
\end{displaymath}

Now consider the square spin of a system of $I$ fermions. By definition

\begin{displaymath}
{\skew 6\widehat{\vec S}}^{\,2} \equiv
\left(\sum_{i=1}^...
...dehat{\vec S}}_i\cdot{\skew 6\widehat{\vec S}}_{\underline i}
\end{displaymath}

Split up the sum into terms that have $i$ and ${\underline i}$ equal, respectively not equal:

\begin{displaymath}
{\skew 6\widehat{\vec S}}^{\,2} = \sum_{i=1}^I {\skew 6\wi...
...dehat{\vec S}}_i\cdot{\skew 6\widehat{\vec S}}_{\underline i}
\end{displaymath}

The first sum is just the square spin angular momentum of the individual fermions. The second sum can be written in terms of the exchange operators using the expression above. Doing so and cleaning up gives:

\begin{displaymath}
{\skew 6\widehat{\vec S}}^{\,2} = \hbar^2(I-{\textstyle\fr...
...^I \sum_{{\underline i}=i+1}^I \widehat P^s_{i{\underline i}}
\end{displaymath}

Similarly then for isospin as defined in the first subsection,

\begin{displaymath}
{\skew 4\widehat{\skew 0\vec T}}^{\,2} = (I-{\textstyle\fr...
...^I \sum_{{\underline i}=i+1}^I \widehat P^T_{i{\underline i}}
\end{displaymath}

Square isospin by itself does not have direct physical meaning. However, the exchange operators do. In particular, charge independence means that exchanging nucleon types does not make a difference for the energy. That then means that ${\skew 4\widehat{\skew 0\vec T}}^2$ commutes with the Hamiltonian. That makes it a conserved quantity according to the rules of quantum mechanics, chapter 4.5.1 and/or {A.19}.

It may be noted that the exchange operators do not commute among themselves. That makes the symmetry requirements so messy. However, it is possible to restrict consideration to exchange operators of the form $\widehat{P}_{i\,i+1}$. See [14] for more.

Infinitesimal rotations of a state in 1,2,3 isospin state correspond to applying small multiples of the operators ${\widehat T}_1$, ${\widehat T}_2$ and ${\widehat T}_3$, compare {A.19}. According to the definitions of ${\widehat T}_1$ and ${\widehat T}_2$, this corresponds to applying small multiples of the charge creation and annihilation operators. So it amounts to gradually changing protons into neutrons and vice-versa. As a simple example, a 180$\POW9,{\circ}$ rotation around the 1 or 2 axis inverts the 3-component of every nucleon. That turns every proton into a neutron and vice-versa.