14.18 Draft: 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 Draft: 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 ${\textstyle\frac{1}{2}}$, 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:

$$\fbox{$\displaystyle \mbox{proton: } T_3 = {\textstyle\fra... ...}} \qquad \mbox{neutron: } T_3 = -{\textstyle\frac{1}{2}} $}$$ (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

$$\psi_{\rm {A}} = \psi_{\rm s}({\skew0\vec r}_2-{\skew0\vec r}_1) {\left\vert 1\:1\right\rangle}$$

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):

$$\psi_{\rm s}({\skew0\vec r}_2-{\skew0\vec r}_1) = \psi_{\rm s}({\skew0\vec r}_1-{\skew0\vec r}_2)$$

(This is equiv­a­lent to as­sum­ing that the wave func­tion has even par­ity.) Fur­ther re­call from chap­ter 5.5.6 that the triplet spin state ${\left\vert 1\:1\right\rangle}$ 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

$$\psi_{\rm {B}} = \psi_{\rm s}({\skew0\vec r}_2-{\skew0\vec r}_1) {\left\vert\:0\right\rangle}$$

(Here $\psi_s$ is not nec­es­sar­ily the same as be­fore, but as­sume it still has even par­ity.) The sin­glet spin state ${\left\vert\:0\right\rangle}$ 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,

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

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

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

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

$$\frac{1}{\sqrt{2}} \Psi({\skew0\vec r}_1,S_{z1},{\skew0\vec... ...r}_2,S_{z2},{\skew0\vec r}_1,S_{z1}) \Downarrow_1\Uparrow_2 %$$ (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, [50, 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 $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ 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

$$\Uparrow_1\Uparrow_2 \equiv {{\left\vert 1\:1\right\rangle}... ...parrow_2}{\sqrt{2}} \equiv {{\left\vert 1\:0\right\rangle}}_T$$ (14.42)

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

$$\psi_{\rm s}({\skew0\vec r}_2-{\skew0\vec r}_1) {\left\vert\:0\right\rangle} {{\left\vert 1\:m_T\right\rangle}}_T$$

where $m_T$ is 1, $\vphantom{0}\raisebox{1.5pt}{$-$}$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,

$$m_T = {T_3}^{}_1 + {T_3}^{}_2$$

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

$$\psi_{\rm s}({\skew0\vec r}_2-{\skew0\vec r}_1) {\left\vert... ...\frac{\Uparrow_1\Downarrow_2-\Downarrow_1\Uparrow_2}{\sqrt{2}}$$

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 ${{\left\vert\:0\right\rangle}}_T$, and the lead­ing 1 in the triplet states ${\left\vert 1\:m_T\right\rangle}$? 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

$$\fbox{$\displaystyle {\widehat T}_3 \Uparrow = {\textstyle... ...ehat T}_3 \Downarrow = -{\textstyle\frac{1}{2}} \Downarrow $}$$ (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

$$\fbox{$\displaystyle {\widehat T}^{+} \Downarrow = \Uparrow \qquad {\widehat T}^{+} \Uparrow = 0 $}$$ (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

$$\fbox{$\displaystyle {\widehat T}^{-} \Uparrow = \Downarrow \qquad {\widehat T}^{-} \Downarrow = 0 $}$$ (14.45)

Op­er­a­tors for nu­cleon type in the 1 and 2 di­rec­tions can now be de­fined as

$$\fbox{$\displaystyle {\widehat T}_1 = {\textstyle\frac{1}{... ...at T}^{+} + {\textstyle\frac{1}{2}}{\rm i}{\widehat T}^{-} $}$$ (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

$$\fbox{$\displaystyle \tau_1 = 2 {\widehat T}_1 \quad \tau... ...dehat T}^{+} \qquad \tau^{-} = 2 {\widehat T}^{-1} \qquad $}$$ (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 Draft: 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 $T_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$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,

$$\fbox{$\displaystyle T_3 = {\textstyle\frac{1}{2}}(Z-N) $} %$$ (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 $\vphantom{0}\raisebox{1.5pt}{$-$}$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

$$m_s \equiv S_z/\hbar = -s, -s{+}1, \ldots, s{-}1, s$$

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 $T_3$:

$$\fbox{$\displaystyle m_T \equiv T_3 = -t_T, -t_T{+}1, \ldots, t_T{-}1, t_T $}$$ (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, {A.34}. 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 14.46: Iso­baric ana­log states. [pdf]

But it is true. Fig­ure 14.46 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 com­bined 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 14.4, and 14.5 through 14.8. 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 14.5 through 14.8 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 $T_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$1, which means an equiv­a­lent car­bon-14 state. To the ex­tent that the nu­clear force is truly charge-in­de­pen­dent, and the Coulomb ef­fect has been prop­erly re­moved, these two states should have the same en­ergy. And in­deed, the low­est four en­ergy states of car­bon-14 and oxy­gen-14 have iden­ti­cal spin and par­ity and sim­i­lar en­er­gies. Also, both even-even nu­clei have a 0$\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 14.46, 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 [50, p. 11].)

Fig­ure 14.46 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, [50, p. 12] and [36, 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:

$$\fourIdx{16}{8}{}{}{\rm O} + \fourIdx{2}{1}{}{}{\rm H} \q... ...quad \fourIdx{14}{7}{}{}{\rm N} + \fourIdx{4}{2}{}{}{\rm He}$$

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 14.46. 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 14.46 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 Draft: 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,

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

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, [50, 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 Draft: 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 $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ fermi­ons. Com­pare the dot prod­uct of their spins to the op­er­a­tor $\widehat{P}^s_{12}$ that ex­changes their spins:

$$\begin{array}{c} \;\;\:{\skew 6\widehat{\vec S}}_1\cdot{\s... ...s\right\rangle} = {\left\vert 1\:m_s\right\rangle} \end{array}$$

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:

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

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

$${\skew 6\widehat{\vec S}}^{ 2} \equiv \left(\sum_{i=1}^I ... ...idehat{\vec S}}_i\cdot{\skew 6\widehat{\vec S}}_{\underline i}$$

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

$${\skew 6\widehat{\vec S}}^{ 2} = \sum_{i=1}^I {\skew 6\wid... ...idehat{\vec S}}_i\cdot{\skew 6\widehat{\vec S}}_{\underline i}$$

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:

$${\skew 6\widehat{\vec S}}^{ 2} = \hbar^2(I-{\textstyle\fra... ...}^I \sum_{{\underline i}=i+1}^I \widehat P^s_{i{\underline i}}$$

Sim­i­larly then for isospin as de­fined in the first sub­sec­tion,

$${\skew 4\widehat{\skew 0\vec T}}^{ 2} = (I-{\textstyle\fra... ...}^I \sum_{{\underline i}=i+1}^I \widehat P^T_{i{\underline i}}$$

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.