Sub­sec­tions


D.43 Mul­ti­pole de­riva­tions

This de­rives the mul­ti­pole ma­trix el­e­ments cor­re­spond­ing to a sin­gle par­ti­cle in an atom or nu­cleus. These will nor­mally still need to be summed over all par­ti­cles.

Both a ba­sis of lin­ear mo­men­tum pho­ton wave func­tions and of an­gu­lar mo­men­tum ones are cov­ered. For the an­gu­lar mo­men­tum wave func­tions, the long wave length ap­prox­i­ma­tion will be made that $kR$ is small. Here $k$ is the pho­ton wave num­ber and $R$ the typ­i­cal size of atom or nu­cleus.

The de­riva­tions in­clude a term due to an ef­fect that was men­tioned in the ini­tial 1952 de­riva­tion by B. Stech, [43]. This ef­fect is not men­tioned in any text­book that the au­thor is aware off. That seems to be un­jus­ti­fied. The term does not ap­pear to be small for nu­clei, but at the very least com­pa­ra­ble to the usual elec­tric mul­ti­pole el­e­ment given.

The rules of en­gage­ment are as fol­lows:

The con­vo­luted de­riva­tions in this note make use of a trick. Since trick sounds too tricky, it will be re­ferred to as:

Lemma 1: This lemma al­lows you to get rid of de­riv­a­tives on the wave func­tion. The lemma as­sumes non­rel­a­tivis­tic par­ti­cles. It is a gen­er­al­iza­tion of a de­riva­tion of [16].

The lemma says that if $i$ is the num­ber of a par­ti­cle in the atom or nu­cleus, and if $F_i$ is any func­tion of the po­si­tion of that par­ti­cle $i$, then

\begin{displaymath}
\Big\langle\psi_{\rm {L}}\Big\vert
(\nabla_i F_i)\cdot\nab...
...m {L}}\Big\vert\nabla_i^2F_i\Big\vert\psi_{\rm {H}}\Big\rangle
\end{displaymath} (D.26)

Here $\nabla_i$ rep­re­sents the vec­tor of de­riv­a­tives with re­spect to the co­or­di­nates of par­ti­cle $i$, $V$ is the po­ten­tial, and $\psi_{\rm {L}}$ and $\psi_{\rm {H}}$ are the fi­nal and ini­tial atomic or nu­clear wave func­tions.

The en­ergy dif­fer­ence can be ex­pressed in terms of the en­ergy $\hbar\omega_0$ of the nom­i­nal pho­ton emit­ted in the tran­si­tion,

\begin{displaymath}
\fbox{$\displaystyle
\Big\langle\psi_{\rm{L}}\Big\vert
(\...
...}\Big\vert\nabla_i^2F_i\Big\vert\psi_{\rm{H}}\Big\rangle
$} %
\end{displaymath} (D.27)

The $\pm$ al­lows for the pos­si­bil­ity (in ab­sorp­tion) that $\psi_{\rm {L}}$ is ac­tu­ally the high en­ergy state. The nom­i­nal pho­ton fre­quency $\omega_0$ is nor­mally taken equal to the ac­tual pho­ton fre­quency $\omega$.

Note that none of my sources in­cludes the com­mu­ta­tor in the first term, not even [16]. (The orig­i­nal 1952 de­riva­tion by [43] used a rel­a­tivis­tic Dirac for­mu­la­tion, in which the term ap­pears in a dif­fer­ent place than here. The part in which it ap­pears there is small with­out the term and is not worked out with it in­cluded.) The com­mu­ta­tor is zero if the po­ten­tial $V$ only de­pends on the po­si­tion co­or­di­nates of the par­ti­cles. How­ever, nu­clear po­ten­tials in­clude sub­stan­tial mo­men­tum terms.

To prove the lemma, start with the left hand side

\begin{displaymath}
\langle\psi_{\rm {L}}\vert(\nabla_i F_i)\cdot\nabla_i\vert\...
...angle
\equiv \int \psi_{\rm {L}}^* F_{i,j} \psi_{{\rm {H}},j}
\end{displaymath}

where sub­scripts $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, 2, and 3 in­di­cates the de­riv­a­tives with re­spect to the three co­or­di­nates of par­ti­cle $i$. Sum­ma­tion over $j$ is to be un­der­stood. Av­er­age the above ex­pres­sion with what you get from do­ing an in­te­gra­tion by parts:

\begin{displaymath}
\langle\psi_{\rm {L}}\vert(\nabla_i F_i)\cdot\nabla_i\vert\...
...e\frac{1}{2}} \int (\psi_{\rm {L}}^* F_{i,j})_j \psi_{\rm {H}}
\end{displaymath}

or dif­fer­en­ti­at­ing out

\begin{displaymath}
\langle\psi_{\rm {L}}\vert(\nabla_i F_i)\cdot\nabla_i\vert\...
...tyle\frac{1}{2}} \int \psi_{\rm {L}}^* F_{i,jj} \psi_{\rm {H}}
\end{displaymath}

Com­bine the first two in­te­grals

\begin{displaymath}
\langle\psi_{\rm {L}}\vert(\nabla_i F_i)\cdot\nabla_i\vert\...
...yle\frac{1}{2}} \int \psi_{\rm {L}}^* F_{i,jj} \psi_{\rm {H}}
\end{displaymath}

and do an­other in­te­gra­tion by parts (I got this from [16], thanks):

\begin{displaymath}
\langle\psi_{\rm {L}}\vert(\nabla_i F_i)\cdot\nabla_i\vert\...
...tyle\frac{1}{2}} \int \psi_{\rm {L}}^* F_{i,jj} \psi_{\rm {H}}
\end{displaymath}

Now note the non­rel­a­tivis­tic eigen­value prob­lems for the two states

\begin{eqnarray*}
& \displaystyle
-\frac{\hbar^2}{2m_i} \nabla_i^2 \psi_{\rm {...
... \psi_{\rm {H}}
+ V \psi_{\rm {H}} = E_{\rm {H}} \psi_{\rm {H}}
\end{eqnarray*}

Here the sum is over the other par­ti­cles in the nu­cleus. These two eigen­value prob­lems are used to elim­i­nate the sec­ond or­der de­riv­a­tives in the in­te­gral above. The terms in­volv­ing the Lapla­cians with re­spect to the co­or­di­nates of the other par­ti­cles then drop out. The rea­son is that $F_i$ is just a con­stant with re­spect to those co­or­di­nates, and that Lapla­cians are Her­mit­ian. As­sum­ing that $V$ is at least Her­mit­ian, as it should, the $V$ terms pro­duce the com­mu­ta­tor in the lemma. And the right hand sides give the en­ergy-dif­fer­ence term. The re­sult is the lemma as stated.


D.43.1 Ma­trix el­e­ment for lin­ear mo­men­tum modes

This re­quires in ad­di­tion:

Lemma 2: This lemma al­lows you to ex­press a cer­tain com­bi­na­tion of de­riv­a­tives in terms of the an­gu­lar mo­men­tum op­er­a­tor. It will be as­sumed that vec­tor $\skew3\vec A_0$ is nor­mal to vec­tor ${\vec k}$.

In that case:

\begin{displaymath}
({\vec k}\cdot{\skew0\vec r}_i)(\skew3\vec A_0\cdot\nabla_i...
... \frac{{\rm i}}{\hbar}({\vec k}\times\skew3\vec A_0)\cdot\Lv_i
\end{displaymath}

The quick­est way to prove this is to take the $x$-​axis in the di­rec­tion of ${\vec k}$, and the $y$-​axis in the di­rec­tion of $\skew3\vec A_0$. (The ex­pres­sion above is also true if the two vec­tors are nor or­thog­o­nal. You can see that us­ing in­dex no­ta­tion. How­ever, that will not be needed.) The fi­nal equal­ity is just the de­f­i­n­i­tion of the an­gu­lar mo­men­tum op­er­a­tor.

The ob­jec­tive is now to use these lem­mas to work out the ma­trix el­e­ment

\begin{displaymath}
H_{21,i} = - \frac{q_i}{m_i}
\langle\psi_{\rm {L}}\vert\sk...
...\skew 4\widehat{\skew{-.5}\vec p}}_i\vert\psi_{\rm {H}}\rangle
\end{displaymath}

where ${\vec k}$ is the con­stant wave num­ber vec­tor and $\skew3\vec A_0$ is some other con­stant vec­tor nor­mal to ${\vec k}$. Also ${\skew0\vec r}_i$ is the po­si­tion of the con­sid­ered par­ti­cle, and ${\skew 4\widehat{\skew{-.5}\vec p}}_i$ is the mo­men­tum op­er­a­tor $\hbar\nabla_i$$\raisebox{.5pt}{$/$}$${\rm i}$ based on these co­or­di­nates.

To re­duce this, take the fac­tor $\hbar$$\raisebox{.5pt}{$/$}$${\rm i}$ out of ${\skew 4\widehat{\skew{-.5}\vec p}}_i$ and write the ex­po­nen­tial in a Tay­lor se­ries:

\begin{displaymath}
H_{21,i} = \sum_{n=0}^\infty \frac{{\rm i}q_i \hbar}{m_i}
...
...i)^n}{n!}\skew3\vec A_0\cdot\nabla_i\vert\psi_{\rm {H}}\rangle
\end{displaymath}

Take an­other messy fac­tor out of the in­ner prod­uct:

\begin{displaymath}
H_{21,i} = \sum_{n=0}^\infty \frac{{\rm i}(-{\rm i})^nq_i\h...
...c r}_i)^n\skew3\vec A_0\cdot\nabla_i\vert\psi_{\rm {H}}\rangle
\end{displaymath}

For brevity, just con­sider the in­ner prod­uct by it­self for now. It can triv­ially be rewrit­ten as a sum of two terms, ([16], not me):
$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\langle\psi_{\rm{L}}\vert({\v...
...{\skew0\vec r}_i){\vec k}\cdot\nabla_i]
\vert\psi_{\rm{H}}\rangle
$\hfill(1)}$

$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\qquad {} +
\langle\psi_{\rm...
...{\skew0\vec r}_i){\vec k}\cdot\nabla_i]
\vert\psi_{\rm{H}}\rangle
$\hfill(2)}$
Now on the first in­ner prod­uct (1), lemma 1 can be ap­plied with

\begin{displaymath}
F_i = ({\vec k}\cdot{\skew0\vec r}_i)^n(\skew3\vec A_0\cdot...
...t{\skew0\vec r}_i)^{n-2} (\skew3\vec A_0\cdot{\skew0\vec r}_i)
\end{displaymath}

(Re­call that $\skew3\vec A_0$ and ${\vec k}$ are or­thog­o­nal. Also note that the Lapla­cian of $F_i$ is of es­sen­tially the same form as $F_i$, just for a dif­fer­ent value of $n$.) On the sec­ond in­ner prod­uct (2), lemma 2 can be ap­plied.

Plug­ging these re­sults back into the ex­pres­sion for the ma­trix el­e­ment, reno­tat­ing $n$ into $\ell-1$ for the first part of (1), into $\ell+1$ for the sec­ond part, which can then be com­bined with the first part, and into $\ell$ for (2), and clean­ing up gives the fi­nal re­sult:

\begin{displaymath}
H_{21,i} = - \frac{q_i}{m_i} \langle\psi_{\rm {L}}\vert
\s...
..._{\ell=1}^\infty H_{21,i}^{\rm E\ell} + H_{21,i}^{\rm M\ell 1}
\end{displaymath}

where

\begin{displaymath}
H_{21,i}^{\rm E\ell}
= {\rm i}q_i kc A_0 \frac{(-{\rm i}k)...
...ga,r_{i,k}^{\ell-1}r_{i,{\cal E}}]
\vert\psi_{\rm {H}}\rangle
\end{displaymath}

and

\begin{displaymath}
H_{21,i}^{\rm M\ell 1}
= {\rm i}\frac{q_i}{m_ic} kc A_0 \f...
...ert r_{i,k}^{\ell-1}\L _{i,{\cal B}}\vert\psi_{\rm {H}}\rangle
\end{displaymath}

Here $A_0$ is the mag­ni­tude of $\skew3\vec A_0$. Also $r_{i,k}$ is the com­po­nent of the po­si­tion ${\skew0\vec r}_i$ of par­ti­cle $i$ in the di­rec­tion of mo­tion of the elec­tro­mag­netic wave. The di­rec­tion of mo­tion is the di­rec­tion of ${\vec k}$. Sim­i­larly $r_{i,{\cal E}}$ is the com­po­nent of ${\skew0\vec r}_i$ in the di­rec­tion of the elec­tric field. The elec­tric field has the same di­rec­tion as $\skew3\vec A_0$. Fur­ther, $L_{i,{\cal B}}$ is the com­po­nent of the or­bital an­gu­lar mo­men­tum op­er­a­tor of par­ti­cle $i$ in the di­rec­tion of the mag­netic field. The mag­netic field is in the same di­rec­tion as ${\vec k}$ $\times$ $\skew3\vec A_0$. Fi­nally, the fac­tor $f$ is

\begin{displaymath}
f = \pm 1 + \frac{\hbar\omega}{2m_ic^2}\frac{\ell}{\ell+2} \approx \pm 1
\end{displaymath}

The ap­prox­i­ma­tion ap­plies be­cause nor­mally the en­ergy re­lease in a tran­si­tion is small com­pared to the rest mass en­ergy of the par­ti­cle. (And if it was not, the non­rel­a­tivis­tic elec­tro­mag­netic in­ter­ac­tion used here would not be valid in the first place.) For the emis­sion process cov­ered in {A.25}, the plus sign ap­plies, $f$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.

The com­mu­ta­tor is zero if the po­ten­tial de­pends only on po­si­tion. That is a valid ap­prox­i­ma­tion for elec­trons in atoms, but surely not for nu­clei. For these it is a real prob­lem, {N.14}.

For ad­den­dum {A.25}, the con­stant $A_0$ should be taken equal to $-{\cal E}_0$$\raisebox{.5pt}{$/$}$$\sqrt{2}{{\rm i}}kc$. Note also that the in­ter­ac­tion of the par­ti­cle spin with the mag­netic field still needs to be added to $H_{21,i}^{\rm {M}\ell1}$. This in­ter­ac­tion is un­changed from the naive ap­prox­i­ma­tion.


D.43.2 Ma­trix el­e­ment for an­gu­lar mo­men­tum modes

This sub­sec­tion works out the de­tails of the ma­trix el­e­ment when an­gu­lar mo­men­tum modes are used for the pho­ton wave func­tion.

The first ma­trix el­e­ment to find is

\begin{displaymath}
H_{21,i}^{\rm E\ell 1} = - \frac{q_i}{m_i}
\langle\psi_{\r...
...\skew 4\widehat{\skew{-.5}\vec p}}_i\vert\psi_{\rm {H}}\rangle
\end{displaymath}

where, {A.21.7},

\begin{displaymath}
\skew3\vec A_{\gamma i}^{\rm {E}}
= \nabla_i\times{\skew0\vec r}_i\times\nabla_i j_{\ell i} Y_{\ell i}^{m}
\end{displaymath}

is the elec­tric mul­ti­pole vec­tor po­ten­tial at the lo­ca­tion of par­ti­cle $i$. This uses the short hand

\begin{displaymath}
j_{\ell i} \equiv j_\ell(kr_i)
\qquad
Y_{\ell i}^{m} \equiv Y_\ell^m(\theta_i,\phi_i)
\end{displaymath}

where $\ell$ is the mul­ti­pole or­der or pho­ton an­gu­lar mo­men­tum, $k$ the pho­ton wave num­ber, $j_\ell$ a spher­i­cal Bessel func­tion, and $Y_\ell^m$ a spher­i­cal har­monic.

Note that the elec­tric mul­ti­pole vec­tor po­ten­tial is closely re­lated to the mag­netic one:

\begin{displaymath}
\skew3\vec A_{\gamma i}^{\rm {E}} = \nabla_i\times \skew3\v...
... = - \nabla_i\times j_{\ell i} Y_{\ell i}^{m} {\skew0\vec r}_i
\end{displaymath}

The ex­pres­sion for the elec­tric po­ten­tial can be sim­pli­fied for long pho­ton wave lengths. Note first that

\begin{displaymath}
\nabla_i\times \skew3\vec A_{\gamma i}^{\rm {E}}
= \nabla_...
... k^2 \nabla_i\times j_{\ell i} Y_{\ell i}^{m} {\skew0\vec r}_i
\end{displaymath}

where the sec­ond equal­ity ap­plied be­cause the vec­tor po­ten­tials are so­le­noidal and the stan­dard vec­tor iden­tity (D.1), while the third equal­ity is the en­ergy eigen­value prob­lem, {A.21}. It fol­lows that the elec­tric vec­tor po­ten­tial is of the form

\begin{displaymath}
\skew3\vec A_{\gamma i}^{\rm {E}} = - k^2 j_{\ell i} Y_{\ell i}^{m} {\skew0\vec r}_i + \nabla_i F_i
\end{displaymath}

be­cause vec­tor cal­cu­lus says that if the curl of some­thing is zero, it is the gra­di­ent of some scalar func­tion $F_i$. Here

\begin{displaymath}
F_i = \int_{{\underline{\skew0\vec r}}_i=0}^{{\skew0\vec r}...
...{m}{\skew0\vec r}_i] \cdot {\rm d}{\underline{\skew0\vec r}}_i
\end{displaymath}

The di­rec­tion of in­te­gra­tion in the ex­pres­sion for $F_i$ does not make a dif­fer­ence, so the sim­plest is to in­te­grate ra­di­ally out­wards. The ex­pres­sion for $\skew3\vec A_{{\gamma}i}^{\rm {E}}$ was given in {D.36.2}. That gives

\begin{displaymath}
F_i = \int_{{\underline r}_i=0}^{r_i}[-l(l+1)+k^2r^2]j_{\ell i}\frac{{\rm d}r}r\;Y_{\ell i}^{m}
\end{displaymath}

Long pho­ton wave length cor­re­sponds to small pho­ton wave num­ber $k$. All $k^2$ terms above can then be ig­nored and in ad­di­tion the fol­low­ing ap­prox­i­ma­tion for the Bessel func­tion ap­plies, {A.6},

\begin{displaymath}
j_{\ell i} \approx \frac{(kr_i)^\ell}{(2\ell+1)!!}
\end{displaymath}

This is read­ily in­te­grated to find

\begin{displaymath}
F_i \approx
-(\ell+1) \frac{(kr_i)^\ell}{(2\ell+1)!!} Y_{\ell i}^{m}
\end{displaymath}

and $\skew3\vec A_{{\gamma}i}^{\rm {E}}$ is the gra­di­ent.

That al­lows lemma 1 to be used to find the elec­tric ma­trix el­e­ment.

\begin{eqnarray*}
H_{21,i}^{\rm E\ell 1} & = & - \frac{q_i}{m_i}
\langle\psi_{...
...hbar\omega,r_i^\ell Y_{\ell i}^{m*}]
\vert\psi_{\rm {H}}\rangle
\end{eqnarray*}

This as­sumes $\psi_{\rm {L}}$ is in­deed the lower-en­ergy state. The value of $A_0$ (as de­fined here) to use in ad­den­dum {A.25} is $-\varepsilon_k^{\rm {E}}$$\raisebox{.5pt}{$/$}$$\sqrt{2}{{\rm i}}kc$.

The com­mu­ta­tor is again neg­li­gi­ble for atoms, but a big prob­lem for nu­clei, {N.14}.

There is also a term due to the in­ter­ac­tion of the spin with the mag­netic field, given by the curl of $\skew3\vec A_{{\gamma}i}^{\rm {E}}$ as al­ready found above,

\begin{displaymath}
H_{21,i}^{\rm E\ell 2}
= - \frac{q_i}{m_i} \frac{g_i}{2}
...
...})\cdot{\skew 6\widehat{\vec S}}_i
\vert\psi_{\rm {H}}\rangle
\end{displaymath}

Us­ing the prop­erty of the scalar triple prod­uct that the fac­tors can be in­ter­changed if a mi­nus sign is added, the ma­trix el­e­ment be­comes

\begin{displaymath}
H_{21,i}^{\rm E\ell 2} = \frac{q_i}{m_i} \frac{g_i}{2} k^2 ...
..._i\times{\skew 6\widehat{\vec S}}_i)\vert\psi_{\rm {H}}\rangle
\end{displaymath}

(Note that $\nabla_i$ only acts on the $j_{{\ell}i}Y_{{\ell}i}^{m*}$; $\skew3\vec A_{{\gamma}i}^{\rm {M}*}$ is a func­tion, not a dif­fer­en­tial op­er­a­tor.) In the long wave length ap­prox­i­ma­tion of the Bessel func­tion, that be­comes

\begin{displaymath}
H_{21,i}^{\rm E\ell 2} \approx q_i kc A_0 \frac{(\ell+1)k^\...
...}{\hbar}{\skew 6\widehat{\vec S}}_i)\vert\psi_{\rm {H}}\rangle
\end{displaymath}

The in­ner prod­uct should nor­mally be of the same or­der as the one of $H_{21,i}^{\rm {E}\ell1}$. How­ever, the sec­ond frac­tion above is nor­mally small; usu­ally the pho­ton en­ergy is small com­pared to the rest mass en­ergy of the par­ti­cles. (And if it was not, the non­rel­a­tivis­tic elec­tro­mag­netic in­ter­ac­tion used here would not be valid in the first place.) So this sec­ond term will be ig­nored in ad­den­dum {A.25}.

The third ma­trix el­e­ment to find is the mag­netic mul­ti­pole one

\begin{displaymath}
H_{21,i}^{\rm M\ell 1} = - \frac{q_i}{m_i}
\langle\psi_{\r...
...\skew 4\widehat{\skew{-.5}\vec p}}_i\vert\psi_{\rm {H}}\rangle
\end{displaymath}

Note that in in­dex no­ta­tion

\begin{displaymath}
\skew3\vec A_{\gamma i}^{\rm {M}}\cdot{\skew 4\widehat{\ske...
...{\ell i} Y_{\ell i}^{m})_{\overline{\jmath}}{\widehat p}_{i,j}
\end{displaymath}

where ${\overline{\jmath}}$ fol­lows $j$ in the cyclic se­quence $\ldots123123\ldots$ and ${\overline{\overline{\jmath}}}$ pre­cedes $j$. By a triv­ial reno­ta­tion of the sum­ma­tion in­dices,

\begin{displaymath}
\skew3\vec A_{\gamma i}^{\rm {M}*}\cdot{\skew 4\widehat{\sk...
...ath}}}}
= - (\nabla_i j_{\ell i} Y_{\ell i}^{m*}) \cdot \Lv_i
\end{displaymath}

where ${\skew 4\widehat{\vec L}}$ is the or­bital an­gu­lar mo­men­tum op­er­a­tor. Note that the par­en­thet­i­cal term com­mutes with this op­er­a­tor, some­thing not men­tioned in [32, p. 874].

It fol­lows that

\begin{displaymath}
H_{21,i}^{\rm M\ell 1} = - \frac{q_i}{m_i}
\langle\psi_{\r...
...ll i} Y_{\ell i}^{m*}) \cdot \Lv_i
\vert\psi_{\rm {H}}\rangle
\end{displaymath}

or in the long wave length ap­prox­i­ma­tion

\begin{displaymath}
H_{21,i}^{\rm M\ell 1} = - \frac{q_i}{m_i}
\langle\psi_{\r...
...^\ell Y_{\ell i}^{m*}) \cdot \Lv_i
\vert\psi_{\rm {H}}\rangle
\end{displaymath}

There is also a term due to the in­ter­ac­tion of the spin with the mag­netic field, given by the curl of $\skew3\vec A_{{\gamma}i}^{\rm {B}}$, which equals $\skew3\vec A_{{\gamma}i}^{\rm {E}}$,

\begin{displaymath}
H_{21,i}^{\rm M\ell 2} = - \frac{q_i}{m_i} \frac{g_i}{2}
\...
...E}*}\cdot{\skew 6\widehat{\vec S}}_i\vert\psi_{\rm {H}}\rangle
\end{displaymath}

Us­ing the same long wave length ap­prox­i­ma­tion for $\skew3\vec A_{{\gamma}i}^{\rm {E}}$ as be­fore, that be­comes

\begin{displaymath}
H_{21,i}^{\rm M\ell 2} \approx \frac{q_i}{m_i} A_0
\frac{(...
...m*})\cdot{\skew 6\widehat{\vec S}}_i\vert\psi_{\rm {H}}\rangle
\end{displaymath}

The or­bital and spin ma­trix el­e­ments may be com­bined into one as

\begin{displaymath}
H_{21,i}^{\rm M\ell} \approx \frac{q_i}{2 m_i} A_0
\frac{(...
..._i{\skew 6\widehat{\vec S}}_i\Big)
\vert\psi_{\rm {H}}\rangle
\end{displaymath}

The value of $A_0$ to use in ad­den­dum {A.25} is $-\varepsilon_k^{\rm {E}}$$\raisebox{.5pt}{$/$}$$\sqrt{2}{{\rm i}}c$.


D.43.3 Weis­skopf and Moszkowski es­ti­mates

This sub­sec­tion ex­plains where the ra­dial, an­gu­lar, and mo­men­tum fac­tors in the Weis­skopf and Moszkowski es­ti­mates come from. These fac­tors rep­re­sent the nondi­men­sion­al­ized ma­trix el­e­ments.

The elec­tric ma­trix el­e­ment is sim­plest. It is, writ­ten out in spher­i­cal co­or­di­nates us­ing the as­sumed wave func­tions,

\begin{displaymath}
\vert h_{21}^{\rm E\ell}\vert \approx
\int R_{\rm {L}}(r_i...
...}i}^{m_{j\rm {H}}}
\sin^2\theta_i{\rm d}\theta_i{\rm d}\phi_i
\end{displaymath}

The Weis­skopf and Moszkowski es­ti­mates as­sume that the ra­dial parts of wave func­tions equal a con­stant $C$ un­til the nu­clear edge $R$ and are zero out­side the nu­cleus. To per­form the ra­dial in­te­gral is then straight­for­ward:

\begin{displaymath}
\int R_{\rm {L}}(r_i)^* (r_i/R)^\ell R_{\rm {H}}(r_i) r_i^2...
...rm d}r_i}{\int_0^R C^2 r_i^2 {\,\rm d}r_i}
= \frac{3}{\ell+3}
\end{displaymath}

The first equal­ity is true be­cause the in­te­gral in the de­nom­i­na­tor is 1 on ac­count of the nor­mal­iza­tion con­di­tion of wave func­tions. The sec­ond in­equal­ity fol­lows from in­te­grat­ing.

The an­gu­lar in­te­gral above is more tricky to ball­park. First of all, it will be as­sumed that the ma­trix el­e­ment of in­ter­est is the low­est mul­ti­pole or­der al­lowed by an­gu­lar mo­men­tum con­ser­va­tion. That seems rea­son­able, given that nor­mally higher mul­ti­pole tran­si­tions will be very much slower. It fol­lows that $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vert j_{\rm {H}}-j_{\rm {L}}\vert$. (The pos­si­bil­ity that the ini­tial and fi­nal an­gu­lar mo­menta are equal will be ig­nored.)

The change in or­bital an­gu­lar mo­menta could in prin­ci­ple be up to one unit dif­fer­ent from the change in net an­gu­lar mo­menta be­cause of the spins. But par­ity con­ser­va­tion al­lows only $\vert l_{\rm {H}}-l_{\rm {L}}\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell$.

To sim­plify even fur­ther, as­sume the fol­low­ing spe­cific an­gu­lar states:

\begin{displaymath}
\Theta_{l_{\rm {L}}j_{\rm {L}}i}^{m_{j\rm {L}}} = Y_{0 i}^0...
... {H}}j_{\rm {H}}i}^{m_{j\rm {H}}} = Y_{\ell i}^\ell {\uparrow}
\end{displaymath}

which have

\begin{displaymath}
l_{\rm {L}} = 0 \quad j_{\rm {L}} = {\textstyle\frac{1}{2}}...
...ac{1}{2}}
\quad m_{j{\rm {H}}} = \ell+{\textstyle\frac{1}{2}}
\end{displaymath}

If these states are sub­sti­tuted into the an­gu­lar in­te­gral, the prod­uct of the spin states is 1 be­cause spin states are or­tho­nor­mal. What is left is

\begin{displaymath}
\sqrt{4\pi} \int Y_{0 i}^{0*} Y_{\ell i}^{m*} Y_{\ell i}^{\ell}
\sin^2\theta_i{\rm d}\theta_i{\rm d}\phi_i
\end{displaymath}

Now $Y_0^0$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1$\raisebox{.5pt}{$/$}$$\sqrt{4\pi}$ which is just a con­stant that can be taken out of the in­te­gral. There it can­cels the cor­re­spond­ing square root in the de­f­i­n­i­tion of the ma­trix el­e­ment. Then it is seen that the tran­si­tion can only cre­ate a pho­ton for which $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell$. The rea­son is that spher­i­cal har­mon­ics are or­tho­nor­mal; the in­ner prod­uct is only nonzero if the two spher­i­cal har­mon­ics are equal, and then it is 1. So the con­clu­sion is that for the given states

\begin{displaymath}
\sqrt{4\pi} \int \Theta_{l_{\rm {L}}j_{\rm {L}}i}^{m_{j\rm ...
...m_{j\rm {H}}}
\sin^2\theta_i{\rm d}\theta_i{\rm d}\phi_i
= 1
\end{displaymath}

The an­gu­lar in­te­gral is 1. That makes the de­cay rate ex­actly 1 Weis­skopf unit.

One glar­ing de­fi­ciency in the above analy­sis was the as­sump­tion that the ini­tial pro­ton state was a $Y_\ell^\ell{\uparrow}$ one. It would cer­tainly be rea­son­able to have an ini­tial nu­clear state that has or­bital an­gu­lar mo­men­tum $l_{\rm {H}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell$ and to­tal an­gu­lar mo­men­tum $j_{\rm {H}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell+\frac12$. But a bunch of these nu­clei would surely each be ori­ented in its own ran­dom di­rec­tion. So they would have dif­fer­ent mag­netic quan­tum num­bers $m_{j\rm {H}}$. They would not all have $m_{j\rm {H}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell+\frac12$.

For­tu­nately, it turns out that this makes no dif­fer­ence. For ex­am­ple, by sym­me­try the state $Y_\ell^{-\ell}{\downarrow}$ de­cays just as hap­pily to $Y_0^0{\downarrow}$ as $Y_\ell^\ell{\uparrow}$ does to $Y_0^0{\uparrow}$. For other val­ues of $m_{j\rm {H}}$ it is a bit more nu­anced. They pro­duce an ini­tial state of the form:

\begin{displaymath}
\Theta_{l_{\rm {H}}j_{\rm {H}}i}^{m_{j\rm {H}}}
= \Theta_{...
...ac12}{\uparrow}+ c_2 Y_\ell^{m_{j\rm {H}}+\frac12}{\downarrow}
\end{displaymath}

Now the first term pro­duces de­cays to $Y_0^0{\uparrow}$ by the emis­sion of a pho­ton with $m_\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell-\frac12$. How­ever, be­cause of the fac­tor $c_1$ the num­ber of such de­cays that oc­cur per sec­ond is a fac­tor $c_1^2$ less than the Weis­skopf unit. But the sec­ond term pro­duces de­cays to $Y_0^0{\downarrow}$ by the emis­sion of a pho­ton with $m_\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell+\frac12$. This de­cay rate is a fac­tor $c_2^2$ less than the Weis­skopf unit. Since $c_1^2+c_2^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, (the nor­mal­iza­tion con­di­tion of the state), the to­tal de­cay rate is still 1 Weis­skopf unit.

So as long as the fi­nal state $\psi_{\rm {L}}$ has zero or­bital an­gu­lar mo­men­tum, the de­cay is at 1 Weis­skopf unit. The ori­en­ta­tion of the ini­tial state makes no dif­fer­ence. That is re­flected in ta­ble A.3. This ta­ble lists the an­gu­lar fac­tors to be ap­plied to the Weis­skopf unit to get the ac­tual de­cay rate. The first row shows that, in­deed, when the fi­nal an­gu­lar mo­men­tum is $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, as oc­curs for zero an­gu­lar mo­men­tum, and the ini­tial an­gu­lar mo­men­tum is $\ell+\frac12$, then no cor­rec­tion is needed. The cor­rec­tion fac­tor is 1.

More in­ter­est­ing is the pos­si­bil­ity that the two states are swapped. Then the ini­tial state is the one with zero or­bital an­gu­lar mo­men­tum. It might at first seem that that will not make a dif­fer­ence ei­ther. Af­ter all, de­cay rates be­tween spe­cific states are ex­actly the same.

But there is in fact a dif­fer­ence. Pre­vi­ously, each ini­tial nu­cleus had only two states to de­cay to: the spin-up and the spin-down ver­sion of the fi­nal state. Now how­ever, each ini­tial nu­cleus has $2j_{\rm {L}}+1$, i.e. $2\ell+2$ fi­nal states it can de­cay to, cor­re­spond­ing to the pos­si­ble val­ues of the fi­nal mag­netic quan­tum num­ber $m_{\rm {L}}$. That will in­crease the to­tal de­cay rate cor­re­spond­ingly. In fact, sup­pose that the ini­tial nu­clei come in spin-up and spin-down pairs. Then each pair will de­cay at a rate of one Weis­skopf unit to each pos­si­ble fi­nal state. That is be­cause this pic­ture is the ex­act re­verse of the de­cay of the fi­nal state. So the pairs would de­cay at a rate $2\ell+2$ faster than the Weis­skopf unit. So by sym­me­try each nu­cleus of the pair de­cays $\ell+1$ times faster than the Weis­skopf unit. That is re­flected in the first col­umn of ta­ble A.3. (Re­call that $\ell$ is the dif­fer­ence in the $j$ val­ues.)

If nei­ther the ini­tial nor fi­nal state has zero or­bital an­gu­lar mo­men­tum, it gets more messy. Fig­ur­ing out the cor­rec­tion fac­tor in that case is some­thing for those who love ab­stract math­e­mat­ics.

Next con­sider mag­netic mul­ti­pole tran­si­tions. They are much messier to ball­park. It will again be as­sumed that the mul­ti­pole or­der is the small­est pos­si­ble. Un­for­tu­nately, now the fi­nal or­bital an­gu­lar mo­men­tum can­not be zero. Be­cause of par­ity, that would re­quire that the ini­tial or­bital an­gu­lar mo­men­tum would be $\ell+1$. But that is too large be­cause of the lim­i­ta­tion (A.175) on the or­bital an­gu­lar mo­men­tum change in mag­netic tran­si­tions. There­fore the sim­plest pos­si­ble ini­tial and fi­nal states have

\begin{displaymath}
l_{\rm {L}} = 1 \quad j_{\rm {L}} = {\textstyle\frac{1}{2}}...
...ac{1}{2}}
\quad m_{j{\rm {H}}} = \ell+{\textstyle\frac{1}{2}}
\end{displaymath}

For these quan­tum num­bers, the ini­tial and fi­nal states are

\begin{displaymath}
\psi_{\rm {L}} = R_{{\rm {L}},i} \Theta_{l_{\rm {L}}j_{\rm ...
...}^{m_{j\rm {H}}}
= R_{{\rm {H}},i} Y_{\ell i}^\ell {\uparrow}
\end{displaymath}

where the square roots come from fig­ure 12.5 in the $j_a,j_b$ $\vphantom0\raisebox{1.5pt}{$=$}$ $1,\frac12$ tab­u­la­tion.

Now con­sider the form of the mag­netic ma­trix el­e­ment (A.181). First note, {D.43.2}, that the an­gu­lar mo­men­tum and gra­di­ent fac­tors com­mute. That helps be­cause then the an­gu­lar mo­men­tum op­er­a­tors, be­ing Her­mit­ian, can be ap­plied on the eas­ier state $\psi_{\rm {L}}$.

The $z$-​com­po­nent part of the dot prod­uct in the ma­trix el­e­ment is then the eas­i­est. The $z$ com­po­nents of the an­gu­lar mo­men­tum op­er­a­tors leave the state $\psi_{\rm {L}}$ es­sen­tially un­changed. They merely mul­ti­ply the two terms by the eigen­value $m_l\hbar$ re­spec­tively $m_s\hbar$.

Next, this gets mul­ti­plied by the $z$-​com­po­nent of the gra­di­ent. But mul­ti­ply­ing by the gra­di­ent can­not change the spin. So the spin-down first term in $\psi_{\rm {L}}$ stays spin-down. That can­not match the spin-up of $\psi_{\rm {H}}$. So the first term does not pro­duce a con­tri­bu­tion.

The sec­ond term in $\psi_{\rm {L}}$ has the right spin. Since spin states are or­tho­nor­mal, their in­ner prod­uct pro­duces 1. But now there is a prob­lem of match­ing the mag­netic quan­tum num­ber of $\psi_{\rm {H}}$. In par­tic­u­lar, con­sider the har­monic poly­no­mial $r^{\ell}Y_\ell^{m_l}$ in the gra­di­ent. The gra­di­ent re­duces it to a com­bi­na­tion of har­monic poly­no­mi­als of one de­gree less, in other words, to $r^{\ell-1}Y_{\ell-1}^{m_l}$ poly­no­mi­als. That lim­its $m_l$ to a value no larger than $\ell-1$, and since the sec­ond term in $\psi_{\rm {L}}$ has mag­netic quan­tum num­ber 0, the value $\ell$ in $\psi_{\rm {H}}$ can­not be matched. The bot­tom line is that the $z$-​com­po­nent terms in the in­ner prod­uct of the ma­trix el­e­ment do not pro­duce a con­tri­bu­tion.

How­ever, the $x$-​ and $y$-​com­po­nent terms are an­other story. The an­gu­lar mo­men­tum op­er­a­tors in these di­rec­tions change the cor­re­spond­ing mag­netic quan­tum num­bers, chap­ter 12.11. In gen­eral, their ap­pli­ca­tion pro­duces a mix­ture of $m+1$ and $m-1$ states. In par­tic­u­lar, the $x$ and $y$ com­po­nents of spin will pro­duce a spin-up ver­sion of the first term in $\psi_{\rm {L}}$. That now matches the spin in $\psi_{\rm {H}}$ and a nonzero con­tri­bu­tion re­sults. Sim­i­larly, the or­bital an­gu­lar mo­men­tum op­er­a­tors will pro­duce an $m_{\rm {L}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 ver­sion of the sec­ond term in $\psi_{\rm {L}}$. Com­bined with the $\ell-1$ units from the gra­di­ent, that is enough to match the mag­netic quan­tum num­ber of $\psi_{\rm {H}}$. So there is a to­tal of four nonzero con­tri­bu­tions to the ma­trix el­e­ment.

Now it is just a mat­ter of work­ing out the de­tails to get the com­plete ma­trix el­e­ment. The in­for­ma­tion in chap­ter 12.11 can be used to find the ex­act states pro­duced from $\Theta_{l_{\rm {H}}j_{\rm {H}}i}^{m_{j\rm {H}}}$ by the $x$ and $y$ an­gu­lar mo­men­tum op­er­a­tors. Each state is a mul­ti­ple of the $Y_1^1{\uparrow}$ state. As far as the gra­di­ent term is con­cerned, the har­monic poly­no­mi­als are of the gen­eral form

\begin{displaymath}
r^\ell Y_\ell^\ell = C (x+{\rm i}y)^\ell \qquad
r^\ell Y_\ell^{\ell-1} = D z (x+{\rm i}y)^{\ell-1} \qquad \ldots
\end{displaymath}

as seen in ta­ble 4.3 or {D.64}. The con­stants $C,D,\ldots$ are of no im­por­tance here. The $x$ and $y$ de­riv­a­tives of the first har­monic poly­no­mial will give the needed $Y_{\ell-1}^{\ell-1}$ har­monic. (For val­ues of $\ell$ greater than 1, the third har­monic could also make a con­tri­bu­tion. How­ever, it turns out that here the $x$ and $y$ con­tri­bu­tions can­cel each other.) The ef­fect of the $x$-​de­riv­a­tive on the first har­monic is sim­ply to add a fac­tor $\ell$$\raisebox{.5pt}{$/$}$$(x+{{\rm i}}y)$ to it. Sim­i­larly, the $y$-​de­riv­a­tive sim­ply adds a fac­tor ${{\rm i}}\ell$$\raisebox{.5pt}{$/$}$$(x+{{\rm i}}y)$. Now if you look up $Y_1^1$ in ta­ble 4.3, you see it is a mul­ti­ple of $x+{{\rm i}}y$. So the prod­uct with the gra­di­ent term pro­duces a sim­ple mul­ti­ple of $Y_\ell^\ell{\uparrow}$. The in­ner prod­uct with $\psi_{\rm {H}}$ then pro­duces that mul­ti­ple (which still de­pends on $r_i$ of course.) Iden­ti­fy­ing and adding the four mul­ti­ples pro­duces

\begin{displaymath}
h_{21}^{\rm M\ell} = - \left(g_i\ell - \frac{2\ell}{\ell+1}...
...}}(r_i)^* (r_i/R)^{\ell-1} R_{\rm {H}}(r_i) r_i^2 {\,\rm d}r_i
\end{displaymath}

The re­main­ing ra­dial in­te­gral may be ball­parked ex­actly the same as for the elec­tric case. The only dif­fer­ence is that the power of $r_i$ is one unit smaller.

A sim­i­lar analy­sis shows that the given ini­tial state can­not de­cay to the ver­sion of the fi­nal state with neg­a­tive mag­netic quan­tum num­ber $m_{j\rm {L}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-\frac12$.

And of course, if the ini­tial and fi­nal states are swapped, there is again a fac­tor $\ell+1$ in­crease in de­cay rate.

More in­ter­est­ingly, the same ex­pres­sion turns out to hold if nei­ther the ini­tial nor the fi­nal an­gu­lar mo­men­tum equals $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, us­ing the cor­rec­tion fac­tor of ta­ble A.3. But the ob­tained mag­netic mul­ti­pole de­cay rate is more lim­ited than the elec­tric one. It does re­quire that $\vert j_{\rm {H}}-j_{\rm {L}}\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell$ and that $\vert l_{\rm {H}}-l_{\rm {L}}\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell-1$

The mo­men­tum fac­tors (A.189) were iden­ti­fied us­ing a com­puter pro­gram. This pro­gram crunched out the com­plete ma­trix el­e­ments us­ing pro­ce­dures ex­actly like the ones above. This pro­gram was also used to cre­ate ta­ble A.3 of an­gu­lar fac­tors. This guards against ty­pos and pro­vides an in­de­pen­dent check on the Cleb­sch-Gor­dan val­ues.