14.17 Draft: Elec­tro­mag­netic Mo­ments

The most im­por­tant elec­tro­mag­netic prop­erty of nu­clei is their net charge. It is what keeps the elec­trons in atoms and mol­e­cules to­gether. How­ever, nu­clei are not re­ally elec­tric point charges. They have a small size. In a spa­tially vary­ing elec­tric field most re­spond some­what dif­fer­ent than a point charge. It is said that they have an elec­tric quadru­pole mo­ment. Also, most nu­clei act like lit­tle elec­tro­mag­nets. It is said that they have a “mag­netic di­pole mo­ment.” These prop­er­ties are im­por­tant for ap­pli­ca­tions like NMR and MRI, and for ex­per­i­men­tally ex­am­in­ing nu­clei.

14.17.1 Draft: Clas­si­cal de­scrip­tion

This sub­sec­tion ex­plains the mag­netic di­pole and elec­tric quadru­pole mo­ments from a clas­si­cal point of view.

14.17.1.1 Draft: Mag­netic di­pole mo­ment

The most ba­sic de­scrip­tion of an elec­tro­mag­net is charges go­ing around in cir­cles. It can be seen from ei­ther clas­si­cal or quan­tum elec­tro­mag­net­ics that the strength of an elec­tro­mag­net is pro­por­tional to the an­gu­lar mo­men­tum $\vec{L}$ of the charges times the ra­tio of their charge $q$ to their mass $m$, chap­ter 13.2 or 13.4.

This leads to the de­f­i­n­i­tion of the mag­netic di­pole mo­ment as

$$\vec \mu \equiv \frac{q}{2m} \vec L$$

In par­tic­u­lar, a mag­net wants to align it­self with an ex­ter­nal mag­netic field $\skew2\vec{\cal B}_{\rm {ext}}$. The en­ergy in­volved in this align­ment is

$$- \vec\mu \cdot \skew2\vec{\cal B}_{\rm ext}$$

14.17.1.2 Draft: Elec­tric quadru­pole mo­ment

Con­sider a nu­clear charge dis­tri­b­u­tion with charge den­sity $\rho_c$ placed in an ex­ter­nal elec­tri­cal po­ten­tial, or volt­age $\varphi$. The po­ten­tial en­ergy due to the ex­ter­nal field is

$$V = \int \varphi \rho_c { \rm d}^3{\skew0\vec r}$$

It may be noted that since nu­clear en­er­gies are of the or­der of MeV, an ex­ter­nal field is not go­ing to change the nu­clear charge dis­tri­b­u­tion $\rho$. It would need to have a mil­lion volt drop over a cou­ple of fem­tome­ters to make a dent in it. Un­less you shoot very high en­ergy charged par­ti­cles at the nu­cleus, that is not go­ing to hap­pen. Also, the cur­rent dis­cus­sion as­sumes that the ex­ter­nal field is steady or at least quasi-steady. That should be rea­son­able in many cases, as nu­clear in­ter­nal time scales are very fast.

Since nu­clei are so small com­pared to nor­mal ex­ter­nal fields, the elec­tric po­ten­tial $\varphi$ can be well rep­re­sented by a Tay­lor se­ries. That gives the po­ten­tial en­ergy as

$$V = \varphi_0 \int \rho_c { \rm d}^3{\skew0\vec r} + \sum... ...ial r_j}\right)_0 \int r_i r_j \rho { \rm d}^3{\skew0\vec r}$$

where $(r_1,r_2,r_3)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(x,y,z)$ are the three com­po­nents of po­si­tion and 0 in­di­cates that the de­riv­a­tive is eval­u­ated at the ori­gin, the cen­ter of the nu­cleus.

The first in­te­gral in the ex­pres­sion above is just the net nu­clear charge $q$. This makes the first term ex­actly the same as the po­ten­tial en­ergy of a point charge. The sec­ond in­te­gral de­fines the elec­tric di­pole mo­ment in the $i$-​di­rec­tion. It is nonzero if on av­er­age the charge is shifted some­what to­wards one side of the nu­cleus. But nu­clei do not have nonzero elec­tric di­pole mo­ments. The rea­son is that nu­clei have def­i­nite par­ity; the wave func­tion is ei­ther the same or the same save for a mi­nus sign when you look at the op­po­site side of the nu­cleus. Since the prob­a­bil­ity of a pro­ton to be found at a given po­si­tion is pro­por­tional to the square mag­ni­tude of the wave func­tion, it is just as likely to be found at one side as the other one. (That should re­ally be put more pre­cisely for the picky. The di­pole con­tri­bu­tion of any set of po­si­tions of the pro­tons is can­celed by an op­po­site con­tri­bu­tion from the set of op­po­site nu­cleon po­si­tions.)

The last in­te­gral in the ex­pres­sion for the po­ten­tial en­ergy de­fines the quadru­pole ma­trix or ten­sor. You may note a math­e­mat­i­cal sim­i­lar­ity with the mo­ment of in­er­tia ma­trix of a solid body in clas­si­cal me­chan­ics. Just like there, the quadru­pole ma­trix can be sim­pli­fied by ro­tat­ing the co­or­di­nate sys­tem to prin­ci­pal axes. That ro­ta­tion gets rid of the in­te­grals $\int{r}_ir_j\rho{\rm d}^3{\skew0\vec r}$ for $i$ $\raisebox{.2pt}{$\ne$}$ $j$, so what is left is

$$V = V_{\rm pc} + \frac12 \left(\frac{\partial^2\varphi}{\p... ...partial z^2}\right)_0 \int z^2 \rho { \rm d}^3{\skew0\vec r}$$

where the first term is the po­ten­tial of the point charge.

Note that the av­er­age of $x^2$, $y^2$, and $z^2$ is $\frac13r^2$. It is con­ve­nient to sub­tract that av­er­age in each in­te­gral. The sub­trac­tion does not change the value of the po­ten­tial en­ergy. The rea­son is that the sum of the three sec­ond or­der de­riv­a­tives of the ex­ter­nal field $\varphi$ is zero due to Maxwell”s first equa­tion, chap­ter 13.2. All that then leads to a de­f­i­n­i­tion of an elec­tric quadru­pole mo­ment for a sin­gle axis, taken to be the $z$-​axis, as

$$Q \equiv \frac{1}{e} \int (3 z^2 - r^2) \rho { \rm d}^3{\skew0\vec r}$$

For sim­plic­ity, the nasty frac­tions have been ex­cluded from the de­f­i­n­i­tion of $Q$. Also, it has been scaled with the charge $e$ of a sin­gle pro­ton.

That gives $Q$ units of square length, which is easy to put in con­text. Re­call that nu­clear sizes are of the or­der of a few fem­tome­ter. So the SI unit square fem­tome­ter, fm$\POW9,{2}$ or 10$\POW9,{-30}$ m$\POW9,{2}$, works quite nicely for the quadru­pole mo­ment $Q$ as de­fined. It is there­fore need­less to say that most sources do not use it. They use the “barn,” a non-SI unit equal to 10$\POW9,{-28}$ m$\POW9,{2}$. The rea­son is his­tor­i­cal; dur­ing the sec­ond world war some physi­cists fig­ured that the word barn would hide the fact that work was be­ing done on nu­clear bombs from the Ger­mans. Of course, that did not work since so few memos and re­ports are one-word ones. How­ever, physi­cists dis­cov­ered that it did help con­fuse stu­dents, so the term has be­come very widely used in the half cen­tury since then. Also, un­like a square fem­tome­ter, the barn is much too large com­pared to a typ­i­cal nu­clear cross sec­tion, pro­duc­ing all these so­phis­ti­cated look­ing tiny dec­i­mal frac­tions.

To bet­ter un­der­stand the likely val­ues of the quadru­pole mo­ment, con­sider the ef­fect of the charge dis­tri­b­u­tion of a sin­gle pro­ton. If the charge dis­tri­b­u­tion is spher­i­cally sym­met­ric, the av­er­ages of $x^2$, $y^2$ and $z^2$ are equal, mak­ing $Q$ zero. How­ever, con­sider the pos­si­bil­ity that the charge dis­tri­b­u­tion is not spher­i­cal, but an el­lip­soid of rev­o­lu­tion, a “spher­oid.”. If the axis of sym­me­try is the $z$-​axis, and the charge dis­tri­b­u­tion hugs closely to that axis, the spher­oid will look like a cigar or zep­pelin. Such a spher­oid is called“pro­late.” The value of $Q$ is then about ${\textstyle\frac{2}{5}}$ of the square nu­clear ra­dius $R$. If the charge dis­tri­b­u­tion stays close to the $xy$-​plane, the spher­oid will look like a flat­tened sphere. Such a spher­oid is called “oblate.” In that case the value of $Q$ is about $-{\textstyle\frac{2}{5}}$ of the square nu­clear ra­dius. Ei­ther way, the val­ues of $Q$ are no­tice­ably less than the square nu­clear ra­dius.

It may be noted that the quadru­pole in­te­grals also pop up in the de­scrip­tion of the elec­tric field of the nu­cleus it­self. Far from the nu­cleus, the de­vi­a­tions in its elec­tric field from that of a point charge are pro­por­tional to the same in­te­grals, com­pare chap­ter 13.3.3.

14.17.2 Draft: Quan­tum de­scrip­tion

Quan­tum me­chan­ics makes for some changes to the clas­si­cal de­scrip­tion of the elec­tro­mag­netic mo­ments. An­gu­lar mo­men­tum is quan­tized, and spin must be in­cluded.

14.17.2.1 Draft: Mag­netic di­pole mo­ment

As the clas­si­cal de­scrip­tion showed, the strength of an elec­tro­mag­net is es­sen­tially the an­gu­lar mo­men­tum of the charges go­ing around, times the ra­tio of their charge to their mass. In quan­tum me­chan­ics an­gu­lar mo­men­tum comes in units of $\hbar$. Also, for nu­clei the charged par­ti­cles are pro­tons with charge $e$ and mass $m_{\rm p}$. There­fore, a good unit to de­scribe mag­netic strengths in terms of is the so-called “nu­clear mag­ne­ton”

$$\fbox{$\displaystyle \mu_{\rm N}\equiv \frac{e\hbar}{2m_{\rm p}} $} %$$ (14.26)

In those terms, the mag­netic mag­netic di­pole mo­ment op­er­a­tor of a sin­gle pro­ton is

$$\frac{1}{\hbar} {\skew 4\widehat{\vec L}}_{\rm {p}} \mu_{\rm N}$$

But quan­tum me­chan­ics brings in a com­pli­ca­tion, chap­ter 13.4. Pro­tons have in­trin­sic an­gu­lar mo­men­tum, called spin. That also acts as an elec­tro­mag­net. In ad­di­tion the mag­netic strength per unit of an­gu­lar mo­men­tum is dif­fer­ent for spin than for or­bital an­gu­lar mo­men­tum. The fac­tor that it is dif­fer­ent is called the pro­ton $g$-​fac­tor $g_p$. That then makes the to­tal mag­netic di­pole mo­ment op­er­a­tor of a sin­gle pro­ton equal to

$${\skew 4\widehat{\skew{-.5}\vec\mu}}_{\rm p} = \frac{1}{\hb... ...hat{\vec S}}\right) \mu_{\rm N} \qquad g_{\rm p} \approx 5.59$$ (14.27)

The above value of the pro­ton $g$-​fac­tor is ex­per­i­men­tal.

Neu­trons do not have charge and there­fore their or­bital mo­tion cre­ates no mag­netic mo­ment. How­ever, neu­trons do cre­ate a mag­netic mo­ment through their spin:

$${\skew 4\widehat{\skew{-.5}\vec\mu}}_{\rm n} = \frac{1}{\hb... ...6\widehat{\vec S}}\mu_{\rm N} \qquad g_{\rm n} \approx - 3.83$$ (14.28)

The rea­son is that the neu­tron con­sists of three charged quarks; they pro­duce a net mag­netic mo­ment even if they do not pro­duce a net charge.

The net mag­netic di­pole mo­ment op­er­a­tor of the com­plete nu­cleus is

$${\skew 4\widehat{\skew{-.5}\vec\mu}}= \frac{1}{\hbar} \lef... ...}^A g_{\rm n} {\skew 6\widehat{\vec S}}_i \right] \mu_{\rm N}$$ (14.29)

where $i$ is the nu­cleon num­ber, the first $Z$ be­ing pro­tons and the rest neu­trons.

Now as­sume that the nu­cleus is placed in an ex­ter­nal mag­netic field ${\cal B}$ and take the $z$-​axis in the di­rec­tion of the field. Be­cause nu­clear en­er­gies are so large, ex­ter­nal elec­tro­mag­netic fields are far too weak to change the quan­tum struc­ture of the nu­cleus; its wave func­tion re­mains un­changed to a very good ap­prox­i­ma­tion. How­ever, the field does pro­duce a tiny change in the en­ergy lev­els of the quan­tum states. These may be found us­ing ex­pec­ta­tion val­ues:

$$\Delta E = \langle \Psi\vert{-}{\widehat\mu}_z {\cal B}\vert \Psi\rangle$$

The fact that that is pos­si­ble is a con­se­quence of small per­tur­ba­tion the­ory, as cov­ered in ad­den­dum {A.38}.

How­ever, it is not im­me­di­ately clear what nu­clear wave func­tion $\Psi$ to use in the ex­pec­ta­tion value above. Be­cause of the large val­ues of nu­clear en­er­gies, a nu­cleus is af­fected very lit­tle by its sur­round­ings. It be­haves es­sen­tially as if it is iso­lated in empty space. That means that while the nu­clear en­ergy may de­pend on the mag­ni­tude of the nu­clear spin ${\skew 6\widehat{\vec J}}$, (i.e. the net nu­clear an­gu­lar mo­men­tum), it does not de­pend on its di­rec­tion. In quan­tum terms, the en­ergy does not de­pend on the com­po­nent ${\widehat J}_z$ in the cho­sen $z$-​di­rec­tion. So, what should be used in the above ex­pec­ta­tion value to find the change in the en­ergy of a nu­cleus in a state of spin $j$? States with def­i­nite val­ues of $J_z$? Lin­ear com­bi­na­tions of such states? You get a dif­fer­ence an­swer de­pend­ing on what you choose.

Now a nu­cleus is a com­pos­ite struc­ture, con­sist­ing of pro­tons or neu­trons, each con­tribut­ing to the net mag­netic mo­ment. How­ever, the pro­tons and neu­trons them­selves are com­pos­ite struc­tures too, each con­sist­ing of three quarks. Yet at nor­mal en­ergy lev­els pro­tons and neu­trons act as el­e­men­tary par­ti­cles, whose mag­netic di­pole mo­ment is a scalar mul­ti­ple $g\mu_{\rm N}$ of their spin. Their en­er­gies in a mag­netic field split into two val­ues, one for the state with $J_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12\hbar$ and the other with $J_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-\frac12\hbar$. One state cor­re­sponds to mag­netic quan­tum num­ber $m_j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$, the other to $m_j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-\frac12$.

The same turns out to be true for nu­clei; they too be­have as el­e­men­tary par­ti­cles as long as their wave func­tions stay in­tact. In a mag­netic field, the orig­i­nal en­ergy level of a nu­cleus with spin $j$ splits into equally spaced lev­els cor­re­spond­ing to nu­clear mag­netic quan­tum num­bers $m_j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j,j{-}1,\ldots,-j$. The nu­mer­i­cal value of the mag­netic di­pole mo­ment $\mu$ is there­fore de­fined to be the ex­pec­ta­tion value of ${\widehat\mu}_z$ in the nu­clear state in which $m_j$ has its largest value $j$, call it the $\vert jj\rangle$ state:

$$\fbox{$\displaystyle \mu \equiv \langle jj \vert{\widehat\mu}_z \vert jj \rangle $} %$$ (14.30)

The fact that nu­clei would be­have so sim­ple is re­lated to the fact that nu­clei are es­sen­tially in empty space. That im­plies that the com­plete wave func­tion of a nu­cleus in the ground state, or an­other en­ergy eigen­state, will vary in a very sim­ple way with an­gu­lar di­rec­tion. Fur­ther­more, that vari­a­tion is di­rectly given by the an­gu­lar mo­men­tum of the nu­cleus. A brief dis­cus­sion can be found in chap­ter 7.3 and its note. See also the dis­cus­sion of the Zee­man ef­fect, and in par­tic­u­lar the weak Zee­man ef­fect, in ad­den­dum {A.38}.

The most im­por­tant con­se­quence of those ideas is that

Nu­clei with spin zero do not have mag­netic di­pole mo­ments.

That is not very easy to see from the gen­eral ex­pres­sion for the mag­netic mo­ment, clut­tered as it is with $g$-​fac­tors. How­ever, zero spin means on a very fun­da­men­tal level that the com­plete wave func­tion of a nu­cleus is in­de­pen­dent of di­rec­tion, chap­ter 4.2.3. A mag­netic di­pole strength re­quires di­rec­tion­al­ity, there must be a north pole and a south pole. That can­not oc­cur for nu­clei of spin zero.

14.17.2.2 Draft: Elec­tric quadru­pole mo­ment

The de­f­i­n­i­tion of the elec­tric quadru­pole mo­ment fol­lows the same ideas as that of the mag­netic di­pole mo­ment. The nu­mer­i­cal value of the quadru­pole mo­ment is de­fined as the ex­pec­ta­tion value of $3z^2-r^2$, summed over all pro­tons, in the state in which the net nu­clear mag­netic quan­tum num­ber $m_j$ equals the nu­clear spin $j$:

$$\fbox{$\displaystyle Q \equiv \langle jj \vert\sum_{i=1}^Z 3 z_i^2 - r_i^2\vert jj \rangle $} %$$ (14.31)

Note that there is a close re­la­tion with the spher­i­cal har­mon­ics;

$$3 z^2 - r^2 = \sqrt{\frac{16\pi}{5}} r^2 Y_2^0$$ (14.32)

That is im­por­tant be­cause it im­plies that

Nu­clei with spin zero or with spin one-half do not have elec­tric quadru­pole mo­ments.

To see why, note that the ex­pec­ta­tion value in­volves the ab­solute square of the wave func­tion. Now if you mul­ti­ply two wave func­tions to­gether that have an an­gu­lar de­pen­dence cor­re­spond­ing to a spin $j$, math­e­mat­i­cally speak­ing you get pretty much the an­gu­lar de­pen­dence of two par­ti­cles of spin $j$. That can­not be­come more than an an­gu­lar de­pen­dence of spin $2j$, in other words an an­gu­lar de­pen­dence with terms pro­por­tional to $Y_{2j}^m$. Since the spher­i­cal har­mon­ics are mu­tu­ally or­tho­nor­mal, $Y_{2j}^m$ in­te­grates away against $Y_2^0$ for $j$ $\raisebox{-.3pt}{$\leqslant$}$ $\frac12$.

It makes nu­clei with spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ pop­u­lar for nu­clear mag­netic res­o­nance stud­ies. With­out the per­turb­ing ef­fects due to quadru­pole in­ter­ac­tion with the elec­tric field, they give nice sharp sig­nals. Also of course, analy­sis is eas­ier with only two spin states and no quadru­pole mo­ment.

14.17.2.3 Draft: Shell model val­ues

Ac­cord­ing to the odd-par­ti­cle shell model, all even-even nu­clei have spin zero and there­fore no mag­netic or elec­tric mo­ments. That is per­fectly cor­rect.

For nu­clei with an odd mass num­ber, the model says that all nu­cle­ons ex­cept for the last odd one are paired up in spher­i­cally sym­met­ric states of zero spin that pro­duce no mag­netic mo­ment. There­fore, the mag­netic mo­ment comes from the last pro­ton or neu­tron. To get it, ac­cord­ing to the sec­ond last sub­sub­sec­tion, what is needed is the ex­pec­ta­tion value of the mag­netic mo­ment op­er­a­tor ${\widehat\mu}_z$ as given there. As­sume the shell that the odd nu­cleon is in has sin­gle-par­ti­cle net mo­men­tum $j$. Ac­cord­ing to the de­f­i­n­i­tion of mag­netic mo­ment, the mag­netic quan­tum num­ber must have its max­i­mum value $m_j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j$. Call the cor­re­spond­ing state the $\psi_{nljj}$ one be­cause the spec­tro­scopic no­ta­tion is use­less as al­ways. In par­tic­u­lar for an odd-even nu­cleus,

$$\mu = \frac{1}{\hbar} \langle\psi_{nljj}\vert L_z + g_p {\widehat S}_z\vert\psi_{nljj}\rangle \mu_{\rm N}$$

while for an even-odd nu­cleus

$$\mu = \frac{1}{\hbar} \langle\psi_{nljj}\vert g_n {\widehat S}_z\vert\psi_{nljj}\rangle \mu_{\rm N}$$

The unit $\mu_{\rm N}$ is the nu­clear mag­ne­ton. The ex­pec­ta­tion val­ues can be eval­u­ated by writ­ing the state $\psi_{nljj}$ in terms of the com­po­nent states $\psi_{nlmm_s}$ of def­i­nite an­gu­lar mo­men­tum $\L _z$ and spin ${\widehat S}_z$ fol­low­ing chap­ter 12.8, 2.

It is then found that for an odd pro­ton, the mag­netic mo­ment is

$$\fbox{$\displaystyle \begin{array}{ll} j = l - \leavevmod... ...1}{2}} \left(2j - 1 + g_p\right) \mu_{\rm N} \end{array} $} %$$ (14.33)

while for an odd neu­tron

$$\fbox{$\displaystyle \begin{array}{ll} j = l - \leavevmod... ...2} = {\textstyle\frac{1}{2}} g_n \mu_{\rm N} \end{array} $} %$$ (14.34)

These are called the “Schmidt val­ues.”

Odd-odd nu­clei are too messy to be cov­ered here, even if the Nord­heim rules would be re­li­able.

For the quadru­pole mo­ments of nu­clei of odd mass num­ber, filled shells do not pro­duce a quadru­pole mo­ment, be­cause they are spher­i­cally sym­met­ric. Con­sider now first the case that there is a sin­gle pro­ton in an oth­er­wise empty shell with sin­gle-par­ti­cle mo­men­tum $j$. Then the mag­netic mo­ment of the nu­cleus can be found as the one of that pro­ton:

$$Q = Q_{\rm p} = \langle\psi_{nljj}\vert 3 z^2 - r^2\vert\psi_{nljj}\rangle$$

Eval­u­a­tion, {D.78}, gives

$$\fbox{$\displaystyle Q_{\rm p} = - \frac{2j-1}{2j+2} \langle r^2\rangle $} %$$ (14.35)

where $\langle{r}^2\rangle$ is the ex­pec­ta­tion value of $r^2$ for the pro­ton. Note that this is zero as it should if the spin $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$. Since the spin $j$ must be half-in­te­ger, zero spin is not a con­sid­er­a­tion. For all other val­ues of $j$, the one-pro­ton quadru­pole mo­ment is neg­a­tive.

The ex­pec­ta­tion value $\langle{r}^2\rangle$ can hardly be much more than the square nu­clear ra­dius, ex­cept­ing maybe halo nu­clei. A rea­son­able guess would be to as­sume that the pro­ton is ho­mo­ge­neously dis­trib­uted within the nu­clear ra­dius $R$. That gives a ball­park value

$$\langle r^2\rangle \approx {\textstyle\frac{3}{5}} R^2$$

Next con­sider the case that there are not one but $I$ $\raisebox{-.5pt}{$\geqslant$}$ 3 pro­tons in the un­filled shell. The pic­ture of the odd-par­ti­cle shell model as usu­ally painted is: the first $I-1$ pro­tons are pair­wise com­bined in spher­i­cally sym­met­ric states and the last odd pro­ton is in a sin­gle par­ti­cle state, bliss­fully un­aware of the other pro­tons in the shell. In that case, the quadru­pole mo­ment would self ev­i­dently be the same as for one pro­ton in the shell. But as al­ready pointed out in sec­tion 14.12.4, the painted pic­ture is not re­ally cor­rect. For one, it does not sat­isfy the an­ti­sym­metriza­tion re­quire­ment for all com­bi­na­tions on pro­tons. There re­ally are $I$ pro­tons in the shell shar­ing one wave func­tion that pro­duces a net spin equal to $j$.

In par­tic­u­lar con­sider the case that there are $2j$ pro­tons in the shell. Then the wave func­tion takes the form of a filled shell, hav­ing no quadru­pole mo­ment, plus a hole, a state of an­gu­lar mo­men­tum $j$ for the miss­ing pro­ton. Since a pro­ton hole has mi­nus the charge of a pro­ton, the quadru­pole mo­ment for a sin­gle hole is op­po­site to that of one pro­ton:

$$\fbox{$\displaystyle Q_{2j\rm p} = - Q_{\rm p} $} %$$ (14.36)

In other words, the quadru­pole mo­ment for a sin­gle hole is pre­dicted to be pos­i­tive. For $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$, a sin­gle pro­ton also means a sin­gle hole, so the quadru­pole mo­ment must, once more, be zero. It has been found that the quadru­pole mo­ment changes lin­early with the odd num­ber of pro­tons, [31, p, 129]. There­fore for shells with more than one pro­ton and more than one hole, the quadru­pole mo­ment is in be­tween the one-pro­ton and one-hole val­ues. It fol­lows that the one-pro­ton value pro­vides an up­per bound to the mag­ni­tude of the quadru­pole mo­ment for any num­ber of pro­tons in the shell.

Since neu­trons have no charge, even-odd nu­clei would in the sim­plest ap­prox­i­ma­tion have no quadru­pole mo­ment at all. How­ever, con­sider the odd neu­tron and the spher­i­cal re­main­der of the nu­cleus as a two-body sys­tem go­ing around their com­mon cen­ter of grav­ity. In that pic­ture, the charged re­main­der of the nu­cleus will cre­ate a quadru­pole mo­ment. The po­si­tion vec­tor of the re­main­der of the nu­cleus is about 1$\raisebox{.5pt}{$/$}$​$A$ times shorter than that of the odd neu­tron, so qua­dratic lengths are a fac­tor 1$\raisebox{.5pt}{$/$}$​$A^2$ shorter. How­ever, the nu­cleus has $Z$ times as much charge as a sin­gle pro­ton. There­fore you ex­pect nu­clei with an odd neu­tron to have about $Z$$\raisebox{.5pt}{$/$}$​$A^2$ times the quadru­pole mo­ment of the cor­re­spond­ing nu­cleus with an odd pro­ton in­stead of an odd neu­tron. For heavy nu­clei, that would still be very much smaller than the mag­netic mo­ment of a sim­i­lar odd-even nu­cleus.

14.17.2.4 Draft: Val­ues for de­formed nu­clei

For de­formed nu­clei, part of the an­gu­lar mo­men­tum is due to ro­ta­tion of the nu­cleus as a whole. In par­tic­u­lar, for the ground state ro­ta­tional band of de­formed even-even nu­clei, all an­gu­lar mo­men­tum is in ro­ta­tion of the nu­cleus as a whole. This is or­bital an­gu­lar mo­men­tum. Pro­tons with or­bital an­gu­lar mo­men­tum pro­duce a mag­netic di­pole mo­ment equal to their an­gu­lar mo­men­tum, pro­vided the di­pole mo­ment is ex­pressed in terms of the nu­clear mag­ne­ton $\mu_{\rm N}$. Un­charged neu­trons do not pro­duce a di­pole mo­ment from or­bital an­gu­lar mo­men­tum. There­fore. the mag­netic di­pole mo­ment of the nu­cleus is about

$$\fbox{$\displaystyle \mbox{even-even, ground state band:} ... ...rm{R}} j \mu_{\rm N}\quad g_{\rm{R}} \approx \frac{Z}{A} $} %$$ (14.37)

where the $g$-​fac­tor re­flects the rel­a­tive amount of the nu­clear an­gu­lar mo­men­tum that be­longs to the pro­tons. This also works for vi­bra­tional nu­clei, since their an­gu­lar mo­men­tum too is in global mo­tion of the nu­cleus.

If a ro­ta­tional band has a min­i­mum spin $j_{\rm {min}}$ that is not zero, the di­pole mo­ment is, [40, p. 392],

$$\fbox{$\displaystyle \mu = \left[ g_{\rm{R}} j + \frac{j... ...kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em $} %$$ (14.38)

where $g_{\rm {int}}j_{\rm {min}}\mu_{\rm N}$ re­flects an in­ter­nal mag­netic di­pole strength. If $j_{\rm {min}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$, the top of the first ra­tio has an ad­di­tional term that has a mag­ni­tude pro­por­tional to $2j+1$ and al­ter­nates in sign.

The quadru­pole mo­ment of de­formed nu­clei is typ­i­cally many times larger than that of a shell model one. Ac­cord­ing to the shell model, all pro­tons ex­cept at most one are in spher­i­cal or­bits pro­duc­ing no quadru­pole mo­ment. But if the nu­cleus is de­formed, typ­i­cally into about the shape of some spher­oid in­stead of a sphere, then all pro­tons con­tribute. Such a nu­cleus has a very large “in­trin­sic” quadru­pole mo­ment $Q_{\rm {int}}$.

How­ever, that in­trin­sic quadru­pole mo­ment is not the one mea­sured. For ex­am­ple, many heavy even-even nu­clei have very dis­torted in­trin­sic shapes but all even-even nu­clei have a mea­sured quadru­pole mo­ment that is zero in their ground state. That is a pure quan­tum ef­fect. Con­sider the state in which the axis of the nu­cleus is aligned with the $z$-​di­rec­tion. In that state a big quadru­pole mo­ment would be ob­served due to the di­rec­tional charge dis­tri­b­u­tion. But there are also states in which the nu­cleus is aligned with the $x$-​di­rec­tion, the $y$-​di­rec­tion, and any other di­rec­tion for that mat­ter. No big deal clas­si­cally: you just grab hold of the nu­cleus and mea­sure its quadru­pole mo­ment. But quan­tum me­chan­ics makes the com­plete wave func­tion a lin­ear com­bi­na­tion of all these dif­fer­ent pos­si­ble ori­en­ta­tions; in fact an equal com­bi­na­tion of them by sym­me­try. If all di­rec­tions are equal, there is no di­rec­tion­al­ity left; the mea­sured quadru­pole mo­ment is zero. Also, di­rec­tion­al­ity means an­gu­lar mo­men­tum in quan­tum me­chan­ics; if all di­rec­tions are equal the spin is zero. Grab­bing hold of the nu­cleus means adding di­rec­tion­al­ity, adding an­gu­lar mo­men­tum. That cre­ates an ex­cited state.

A sim­ple known sys­tem that shows such ef­fects is the hy­dro­gen atom. Clas­si­cally the atom is just an elec­tron and a pro­ton at op­po­site sides of their cen­ter of grav­ity. If they are both on the $z$-​axis, say, that sys­tem would have a nonzero quadru­pole mo­ment. But such a state is not an ex­act en­ergy eigen­state, far from it. It in­ter­acts with states in which the di­rec­tion of the con­nect­ing line is dif­fer­ent. By sym­me­try, the ground state is the one in which all di­rec­tions have the same prob­a­bil­ity. The atom has be­come spher­i­cally sym­met­ric. Still, the atom has not be­come in­trin­si­cally spher­i­cally sym­met­ric; the wave func­tion is not of a form like $\psi_1(r_{\rm {e}})\psi_2(r_{\rm {p}})$. The po­si­tions of elec­tron and pro­ton are still cor­re­lated, {A.5}.

A model of a spher­oidal nu­cleus pro­duces the fol­low­ing re­la­tion­ship be­tween the in­trin­sic quadru­pole mo­ment and the one that is mea­sured:

$$\fbox{$\displaystyle Q = \frac{3j_{\rm min}^2 - j(j+1)}{(j+1)(2j+3)} Q_{\rm int} $} %$$ (14.39)

where $j_{\rm {min}}$ is the an­gu­lar mo­men­tum of the nu­cleus when it is not ro­tat­ing. De­riva­tions may be found in [40] or [36]. It can be seen that when the nu­cleus is not ro­tat­ing, the mea­sured quadru­pole mo­ment is much smaller than the in­trin­sic one un­less the an­gu­lar mo­men­tum is re­ally large. When the nu­cleus gets ad­di­tional ro­ta­tional an­gu­lar mo­men­tum, the mea­sured quadru­pole mo­ment de­creases even more and even­tu­ally ends up with the op­po­site sign.

14.17.3 Draft: Mag­netic mo­ment data

Fig­ure 14.42 shows ground state mag­netic mo­ments in units of the nu­clear mag­ne­ton $\mu_{\rm N}$. Even-even nu­clei do not have mag­netic mo­ments in their ground state, so they are not shown. The red and blue hor­i­zon­tal lines are the Schmidt val­ues pre­dicted by the shell model. They dif­fer in whether spin sub­tracts from or adds to the net an­gu­lar mo­men­tum $j$ to pro­duce the or­bital mo­men­tum $l$. Red dots should be on the red lines, blue dots on the blue lines. For black dots, no con­fi­dent pre­dic­tion of the or­bital an­gu­lar mo­men­tum could be made. The val­ues have an er­ror of no more than about 0.1 $\mu_{\rm N}$, based on a sub­jec­tive eval­u­a­tion of both re­ported er­rors as well as dif­fer­ences be­tween re­sults ob­tained by dif­fer­ent stud­ies for the same num­ber. These dif­fer­ences are of­ten much larger than the re­ported er­rors for the in­di­vid­ual num­bers.

Fig­ure 14.42: Mag­netic di­pole mo­ments of the ground-state nu­clei. [pdf]

One good thing to say about it all is that the gen­eral mag­ni­tude is well pre­dicted. Few nu­clei end up out­side the Schmidt lines. (Rhodium-103, a sta­ble odd-even $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ nu­cleus, is a no­table ex­cep­tion.) Also, some nu­clei are ac­tu­ally on their line. And the oth­ers tend to at least be on the right side of the cloud. The bad news is, of course, that the agree­ment is only qual­i­ta­tively.

The main ex­cuses that are of­fered are:

1.

The $g$-​fac­tors $g_{\rm {p}}$ and $g_{\rm {n}}$ de­scribe the ef­fec­tive­ness of pro­ton and neu­tron spins in gen­er­at­ing mag­netic mo­ments in free space. They may be re­duced when these nu­cle­ons are in­side a nu­cleus. In­deed, it seems rea­son­able enough to as­sume that the mo­tion of the quarks that make up the pro­tons and neu­trons could be af­fected if there are other quarks nearby. Re­duc­tion of the $g$-​fac­tors dri­ves the Schmidt lines to­wards each other, and that can clearly re­duce the av­er­age er­rors. Un­for­tu­nately, dif­fer­ent nu­clei would need dif­fer­ent re­duc­tions to ob­tain quan­ti­ta­tive agree­ment.

2.

Col­lec­tive mo­tion. If some of the an­gu­lar mo­men­tum is into col­lec­tive mo­tion, it tends to drift the mag­netic mo­ment to­wards about $\frac12{j}\mu_{\rm N}$, com­pare (14.38). To com­pute the ef­fect re­quires the in­ter­nal mag­netic mo­ment of the nu­cleus to be known. For some nu­clei, fairly good mag­netic mo­ments can be ob­tained by us­ing the Schmidt val­ues for the in­ter­nal mag­netic mo­ment, [40, p. 393].

For odd-odd nu­clei, the data av­er­age out to about $0.5j$ nu­clear mag­ne­tons, with a stan­dard de­vi­a­tion of about one mag­ne­ton. These av­er­age val­ues are shown as yel­low lines in fig­ure 14.42. In­ter­est­ingly enough, the av­er­age is like a col­lec­tive ro­ta­tion, (14.37).

Ac­cord­ing to the shell model, two odd par­ti­cles con­tribute to the spin and mag­netic mo­ment of odd-odd nu­clei. So they could have sig­nif­i­cantly larger spins and mag­netic mo­ments than odd mass nu­clei. Note from the data in fig­ure 14.42 that that just does not hap­pen.

Fig­ure 14.43: 2$\POW9,{+}$ mag­netic mo­ment of even-even nu­clei. [pdf][con]

Even-even nu­clei do not have mag­netic mo­ments in their ground state. Fig­ure 14.43 shows the mag­netic mo­ments of the first ex­ited 2$\POW9,{+}$ state of these nu­clei. The val­ues are in fairly good agree­ment with the pre­dic­tion (14.37) of col­lec­tive mo­tion that the mag­netic mo­ment equals $Zj$$\raisebox{.5pt}{$/$}$​$A$ nu­clear mag­ne­tons. Bright green squares are cor­rect. Big de­vi­a­tions oc­cur only near magic num­bers. The max­i­mum er­ror in the shown data is about a quar­ter of a nu­clear mag­ne­ton, sub­jec­tively eval­u­ated.

14.17.4 Draft: Quadru­pole mo­ment data

If you love the shell model, you may want to skip this sub­sec­tion. It is go­ing to get a beat­ing.

Fig­ure 14.44: Elec­tric quadru­pole mo­ment. [pdf][con]

The pre­dic­tion of the shell model is rel­a­tively straight­for­ward. The elec­tric quadru­pole mo­ment of a sin­gle pro­ton in an un­filled shell of high an­gu­lar mo­men­tum can quite well be ball­parked as

$$Q_{\rm p ballpark} \sim {\textstyle\frac{3}{5}} R^2$$

where $R$ is the nu­clear ra­dius com­puted from (14.9). This value cor­re­sponds to the area of the square marked “a pro­ton’s” in the leg­end of fig­ure 14.44. As dis­cussed in sub­sec­tion 14.17.2.3, if there are more pro­tons in the shell, the mag­ni­tude is less, though the sign will even­tu­ally re­verse. If the an­gu­lar mo­men­tum is not very high, the mag­ni­tude is less. If there is no odd pro­ton, the mag­ni­tude will be al­most zero. So, es­sen­tially all squares in fig­ure 14.44 must be smaller, most a lot smaller, and those on lines of even $Z$ very much smaller, than the sin­gle pro­ton square in the leg­end...

Well, you might be able to find a smaller square some­where. For ex­am­ple, the square for lithium-6, straight above dou­bly-magic ${}\fourIdx{4}{2}{}{}{\rm {He}}$, has about the right size and the right color, blue. The data shown have a sub­jec­tively es­ti­mated er­ror of up to 40%, [sic], and the area of the squares gives the scaled quadru­pole mo­ment. Ni­tro­gen-14, straight be­low dou­bly-magic ${}\fourIdx{16}{8}{}{}{\rm {O}}$, has a suit­ably small square of the right color, red. So does potas­sium-39 with one pro­ton less than dou­bly-magic ${}\fourIdx{40}{20}{}{}{\rm {Ca}}$. Bis­muth-209, with one more pro­ton than ${}\fourIdx{208}{82}{}{}{\rm {Pb}}$ has a rel­a­tively small square of the right color. Some nu­clei on magic pro­ton num­ber lines have quite small scaled quadru­pole mo­ments, though hardly al­most zero as they should. Nu­clei one pro­ton above magic pro­ton num­bers tend to be of the right color, blue, as long as their squares are small. Nu­clei one pro­ton be­low the magic pro­ton num­bers should be red; how­ever, pro­mo­tion can mess that up.

Back to re­al­ity. Note that many nu­clei in the $Z$ $\raisebox{.3pt}{$<$}$ 82, $N$ $\raisebox{.3pt}{$>$}$ 82 wedge, and above $Z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 82, as well as var­i­ous other nu­clei, es­pe­cially away from the sta­ble line, have quadru­pole mo­ments that are very many times larger than the ball­park for a sin­gle pro­ton. That is sim­ply not pos­si­ble un­less many or all pro­tons con­tribute to the quadru­pole mo­ment. The odd-par­ti­cle shell model pic­ture of a spher­i­cally sym­met­ric nu­clear core plus an odd pro­ton, and maybe a neu­tron, in non­spher­i­cal or­bits hang­ing on is com­pletely wrong for these nu­clei. These nu­clei have a global shape that sim­ply is not spher­i­cal. And be­cause the shell model was de­rived based on a spher­i­cal po­ten­tial, its re­sults are in­valid for these nu­clei. They are the de­formed nu­clei that also showed up in fig­ures 14.19 and 14.22. It is the quadru­pole mo­ment that shows that it was not just an empty ex­cuse to ex­clude these nu­clei in shell model com­par­isons. The mea­sured quadru­pole mo­ments show with­out a shadow of a doubt that the shell model can­not be valid.

You might how­ever won­der about the ap­par­ently large amount in ran­dom scat­ter in the quadru­pole mo­ments of these nu­clei. Does the amount of de­for­ma­tion vary that ran­domly? Be­fore that can be an­swered, a cor­rec­tion to the data must be ap­plied. Mea­sured quadru­pole mo­ments of a de­formed nu­cleus are of­ten much too small for the ac­tual nu­clear de­for­ma­tion. The rea­son is un­cer­tainty in the an­gu­lar ori­en­ta­tion of these nu­clei. In par­tic­u­lar, nu­clei with spin zero have com­plete un­cer­tainty in ori­en­ta­tion. Such nu­clei have zero mea­sured quadru­pole mo­ment re­gard­less how big the de­for­ma­tion of the nu­cleus is. Nu­clei with spin one-half still have enough un­cer­tainty in ori­en­ta­tion to mea­sure as zero.

Fig­ure 14.45: Elec­tric quadru­pole mo­ment cor­rected for spin. [pdf][con]

Fig­ure 14.45 shows what hap­pens if you try to es­ti­mate the in­trin­sic quadru­pole mo­ment of the nu­clei in ab­sence of un­cer­tainty in an­gu­lar ori­en­ta­tion. For nu­clei whose spin is at least one, the es­ti­mate was made based on the mea­sured value us­ing (14.39), with both $j_{\rm {min}}$ and $j$ equal to the spin. This as­sumes that the in­trin­sic shape is roughly spher­oidal. For shell-model nu­clei, this also roughly cor­rects for the spin ef­fect, though it over­cor­rects to some ex­tent for nu­clei of low spin.

To es­ti­mate the in­trin­sic quadru­pole mo­ment of nu­clei with zero ground state spin, in­clud­ing all even-even nu­clei, the quadru­pole mo­ment of the low­est ex­cited 2$\POW9,{+}$ state was used, if it had been mea­sured. For spin one-half the low­est $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ state was used. In ei­ther case, $j_{\rm {min}}$ was taken to be the spin of the ground state and $j$ that of the ex­cited state. Re­gret­tably, these es­ti­mates do not make much sense if the nu­cleus is not a ro­tat­ing one.

Note in fig­ure 14.45 how much more uni­form the squares in the re­gions of de­formed nu­clei have be­come. And that the squares of nu­clei of spin zero and one-half have sim­i­lar sizes. These nu­clei were not re­ally more spher­i­cal; it was just hid­den from ex­per­i­ments.

The ob­served in­trin­sic quadru­pole mo­ments in the re­gions of de­formed nu­clei cor­re­spond to roughly 20% ra­dial de­vi­a­tion from the spher­i­cal value. Clearly, that means quite a large change in shape.

It may be noted that fig­ure 14.44 leaves out mag­ne­sium-23, whose re­ported quadru­pole mo­ment of 1.25 barn is far larger that that of sim­i­lar nu­clei. If this value is cor­rect, clearly mag­ne­sium-23 must be a halo nu­cleus with two pro­tons out­side a neon-21 core.