N.14 The elec­tric mul­ti­pole prob­lem

There is a big prob­lem with elec­tric mul­ti­pole tran­si­tions in nu­clei. Elec­tric mul­ti­pole tran­si­tions arise from a ma­trix el­e­ment of the form

\begin{displaymath}
H_{21} =
\sum_i {\left\langle\psi_{\rm {L}}\hspace{0.3pt}\...
...t{\skew{-.5}\vec p}}_i {\left\vert\psi_{\rm {H}}\right\rangle}
\end{displaymath}

Here $\psi_{\rm {L}}$ is the fi­nal atomic or nu­clear state and $\psi_{\rm {H}}$ the ini­tial one. The sum is over the atomic or nu­clear par­ti­cles, with $i$ the par­ti­cle in­dex. Also

\begin{displaymath}
{\skew 4\widehat{\skew{-.5}\vec p}}_i = \frac{\hbar}{{\rm i...
...vec r}_i\times\nabla_i
j_\ell(kr_i) Y_\ell^m(\theta_i,\phi_i)
\end{displaymath}

where $j_\ell$ is a spher­i­cal Bessel func­tion and $Y_\ell^m$ a spher­i­cal har­monic.

Un­der the ap­prox­i­ma­tion that the atomic or nu­clear size is small com­pared to the wave length of the emit­ted or ab­sorbed pho­ton, this may be ap­prox­i­mated. In that ap­prox­i­ma­tion. the sin­gle-par­ti­cle elec­tric ma­trix el­e­ment is com­monly de­scribed as pro­por­tional to

\begin{displaymath}
{\left\langle\psi_{\rm {L}}\hspace{0.3pt}\right\vert} r_i^\...
...\rangle}
\qquad Y_{\ell,i}^m \equiv Y_\ell^m(\theta_i,\phi_i)
\end{displaymath}

where $r_i^{\ell}Y_{\ell,i}^m$ is a har­monic poly­no­mial. How­ever, the cor­rect in­ner prod­uct is

\begin{displaymath}
{\left\langle\psi_{\rm {L}}\hspace{0.3pt}\right\vert}
r_i^...
..._i^\ell Y_{\ell.i}^m]
{\left\vert\psi_{\rm {H}}\right\rangle}
\end{displaymath}

Here $\hbar\omega$ is the en­ergy of the pho­ton. The ad­di­tional com­mu­ta­tor term is not men­tioned in any other ba­sic text­book on nu­clei that this au­thor knows of. A de­riva­tion is in {D.43}.

If the po­ten­tial de­pends only on po­si­tion, the com­mu­ta­tor is zero, and there is no prob­lem. That is a valid ap­prox­i­ma­tion for the outer elec­trons in atoms. But nu­clear po­ten­tials in­clude sig­nif­i­cant mo­men­tum terms. These do not com­mute. One ex­am­ple is the spin-or­bit term in the shell model. Now con­sider the size of the com­mu­ta­tor term above com­pared to the first term. For a ball­park, note that it does not make much dif­fer­ence whether the or­bital an­gu­lar mo­men­tum in the spin-or­bit term acts on the wave func­tion $\psi_{\rm {H}}$ or on the har­monic poly­no­mial. So the rel­a­tive size of the com­mu­ta­tor term ball­parks to the ra­tio of the spin-or­bit en­ergy to the pho­ton en­ergy. That ra­tio can be big. For ex­am­ple, in the so-called “is­lands of iso­merism” tran­si­tions, one state has enough spin-or­bit en­ergy to cross a ma­jor shell bound­ary. But the en­ergy re­lease $\hbar\omega$ stays within the same shell and could be quite small.

As a check on this ball­park, con­sider the sim­plest pos­si­ble elec­tric mul­ti­pole tran­si­tion:

\begin{displaymath}
\psi_{\rm {L}} = R_{\rm {L}}(r_i) Y_{0,i}^0{\uparrow}\qquad...
...2} {\skew 4\widehat{\vec L}}_i\cdot{\skew 6\widehat{\vec S}}_i
\end{displaymath}

Here the part $V_0$ of the po­ten­tial $V$ rep­re­sents terms that only de­pend on po­si­tion, or do not in­volve the po­si­tion co­or­di­nates of par­ti­cle $i$. This part com­mutes with the har­monic poly­no­mial. Also $f$ is a func­tion of ra­dial po­si­tion of or­der 1. Then the con­stant $V_{\rm {so}}$ gives the mag­ni­tude of the spin or­bit en­ergy. In the above case, only the $z$ com­po­nents of the an­gu­lar mo­men­tum op­er­a­tors give a con­tri­bu­tion. Then it is eas­ily seen that the com­mu­ta­tor term pro­duces a con­tri­bu­tion that is larger in mag­ni­tude than the other term by a fac­tor of or­der $V_{\rm {so}}\ell$$\raisebox{.5pt}{$/$}$$\hbar\omega$.

Note also that the ef­fect is es­pe­cially counter-in­tu­itive for elec­tric di­pole tran­si­tions. It would seem log­i­cal to think that such tran­si­tions could be ap­prox­i­mately de­scribed by a straight­for­ward in­ter­ac­tion of the par­ti­cles with the elec­tric field. How­ever, the com­mu­ta­tor need not be zero. So the elec­tric field could be dwarfed by a larger ad­di­tional field. That field is then a con­se­quence of the fact that quan­tum me­chan­ics uses the vec­tor po­ten­tial rather than the clas­si­cal elec­tric field.

Which brings up the next prob­lem. The com­mu­ta­tor term will not be there if you use the gauge prop­erty of the elec­tro­mag­netic field to ap­prox­i­mate the lead­ing or­der elec­tric field through an elec­tro­sta­tic po­ten­tial. So should the com­mu­ta­tor be there or not to get the best so­lu­tion? As far as this au­thor can see, there is no way to tell. And cer­tainly for higher mul­ti­pole or­ders you can­not even ig­nore the prob­lem by us­ing the elec­tro­sta­tic po­ten­tial.

(Note that the real prob­lem is the fact that you get dif­fer­ent an­swers de­pend­ing on how you se­lect the gauge. If the nu­clear Hamil­ton­ian (A.169) re­spected the quan­tum gauge prop­erty of {A.19.5}, the com­mu­ta­tor would be zero. That can be seen by sub­sti­tut­ing in the gauge prop­erty: it shows that for any par­ti­cle the po­ten­tial must com­mute with $e^{-{{\rm i}}q_i\chi_i/\hbar}$, and the ex­po­nen­tial is a com­pletely ar­bi­trary func­tion of the par­ti­cle co­or­di­nates. But the fact that the com­mu­ta­tor should be zero does not take away the fact that it is not zero. Pre­sum­ably, if you de­scribed the nu­cleus in terms of the in­di­vid­ual quarks, you could write down a po­ten­tial that re­spected the gauge prop­erty. But us­ing quarks is not an op­tion. The re­al­ity is that you must use pro­tons and neu­trons. And in those terms, a po­ten­tial that does a de­cent job of de­scrib­ing the nu­cleus will sim­ply not re­spect the gauge prop­erty. Yes, this does sug­gest that de­scrib­ing gamma de­cay us­ing pro­tons and neu­trons in­stead of quarks is an in­her­ently fishy pro­ce­dure.)

This au­thor knows not a sin­gle ref­er­ence that gives a de­cent de­scrip­tion of the above is­sue. Many sim­ply do not men­tion the prob­lem at all and just omit the com­mu­ta­tor. Some sim­ply state that the po­ten­tial is as­sumed to de­pend on the par­ti­cle po­si­tions only, like [32]. Surely it ought to be men­tioned ex­plic­itly that the lead­ing elec­tric mul­ti­pole op­er­a­tor as listed may well be no good at all? If it is listed, peo­ple will as­sume that is is mean­ing­ful un­less stated oth­er­wise. As [32] notes, “sig­nif­i­cant in­for­ma­tion re­gard­ing nu­clear wave func­tions can be ob­tained from a com­par­i­son of ex­per­i­men­tal $\gamma$-​de­cay tran­si­tion prob­a­bil­i­ties with the­o­ret­i­cal val­ues cal­cu­lated on ba­sis of spe­cific mod­els of the nu­cleus.” If one is not aware of the pos­si­bil­ity of the ad­di­tional com­mu­ta­tor as a lead­ing or­der ef­fect, one might in­cor­rectly con­clude that a nu­clear wave func­tion is poor where the real prob­lem is the ig­nored com­mu­ta­tor.

At least [11, p.9-172] and [5] can be read to say that there might be a non­triv­ial prob­lem. The orig­i­nal rel­a­tivis­tic de­riva­tion of Stech, [43], men­tions the is­sue, but no ball­park is given. (It is how­ever noted that the spin-or­bit term might be sig­nif­i­cant for mag­netic tran­si­tions. The purely non­rel­a­tivis­tic analy­sis used here does not show such an ef­fect. The present au­thor sus­pects that the dif­fer­ence is that the rel­a­tivis­tic de­riva­tion of Stech in­her­ently as­sumes that the gauge prop­erty is valid. Surely there must be ways to do the rel­a­tivis­tic analy­sis such that, in say the elec­tric di­pole case, both the re­sults with and with­out com­mu­ta­tor are re­pro­duced.)

To be sure, the au­thor no longer be­lieves that this is a po­ten­tial ex­pla­na­tion why ${\rm {E}}1$ tran­si­tions are so much slower than the Weis­skopf es­ti­mate. The de­vi­a­tions in chap­ter 14.20.5 fig­ures 14.63 and 14.64 seem much too big and sys­tem­atic to be ex­plained by this mech­a­nism. And it does not seem to ad­dress the con­cerns about mixed tran­si­tions that are men­tioned there.