D.44 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, [42]. 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 considered particle will be indicated by a subscript $i$.
• The Cartesian axes are numbered using an index $j$, with $j$ = 1, 2, and 3 for $x_i$, $y_i$, and $z_i$ respectively.
• Also, $r_{i,j}$ indicates the coordinate in the $j$ direction, $x_i$, $y_i$, or $z_i$.
• Derivatives with respect to a coordinate $r_{i,j}$ are indicated by a simple subscript $j$.
• If the quantity being differentiated is a vector, a comma is used to separate the vector index from differentiation ones.
• A bare $\int$ integral sign is assumed to be an integration over all nuclear coordinates.
• A superscript $^*$ indicates a complex conjugate.

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 [15].

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

$$\Big\langle\psi_{\rm {L}}\Big\vert (\nabla_i F_i)\cdot\n... ...\Big\vert(\nabla_i^2F_i) \Big\vert\psi_{\rm {H}}\Big\rangle$$ (D.24)

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,

$$\fbox{$\displaystyle \Big\langle\psi_{\rm{L}}\Big\vert ... ...e(\nabla_i^2F_i)} \Big\vert\psi_{\rm{H}}\Big\rangle $} %$$ (D.25)

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 [15]. (The original 1952 derivation by [42] 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

$$\langle\psi_{\rm {L}}\vert(\nabla_i F_i)\cdot\nabla_i\vert... ...gle \equiv \int \psi_{\rm {L}}^* F_{i,j} \psi_{{\rm {H}},j}$$

where subscripts $j$ = 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:

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

or differentiating out

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

Combine the first two integrals

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

and do another integration by parts (I got this from [15], thanks):

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

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.44.1 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:

$$({\vec k}\cdot{\skew0\vec r}_i)(\skew3\vec A_0\cdot\nabla_... ...{\vec k}\times\skew3\vec A_0)\cdot{\skew 4\widehat{\vec L}}_i$$

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

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

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 $\hbar\nabla_i/{\rm i}$ based on these coordinates.

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

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

Take another messy factor out of the inner product:

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

For brevity, just consider the inner product by itself for now. It can trivially be rewritten as a sum of two terms, ([15], not me):

$\parbox{404pt}{\hspace{15pt}\hfill$\displaystyle \langle\psi_{\rm{L}}\vert({\... ...ew0\vec r}_i){\vec k}\cdot\nabla_i] \vert\psi_{\rm{H}}\rangle $\hfill(1)}$

$\parbox{404pt}{\hspace{15pt}\hfill$\displaystyle \qquad {} + \langle\psi_{\... ...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

$$F_i = ({\vec k}\cdot{\skew0\vec r}_i)^n(\skew3\vec A_0\cdo... ...{\skew0\vec r}_i)^{n-2} (\skew3\vec A_0\cdot{\skew0\vec r}_i)$$

(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:

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

where

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

and

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

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

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

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.23}, the plus sign applies, $f$ = 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.23}, the constant $A_0$ should be taken equal to $-{\cal E}_0/\sqrt{2}{{\rm i}}kc$. 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.44.2 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

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

where, {A.19.7},

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

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

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

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:

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

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

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

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.19}. It follows that the electric vector potential is of the form

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

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

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

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.37.2}. That gives

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

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},

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

This is readily integrated to find

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

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

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

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

This assumes $\psi_{\rm {L}}$ is indeed the lower-energy state. The value of $A_0$ (as defined here) to use in addendum {A.23} is $-\varepsilon_k^{\rm {E}}/\sqrt{2}{{\rm i}}kc$.

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,

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

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

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

(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

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

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.23}.

The third matrix element to find is the magnetic multipole one

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

Note that in index notation

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

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,

$$\skew3\vec A_{\gamma i}^{\rm {M}*}\cdot{\skew 4\widehat{\s... ...j_{\ell i} Y_{\ell i}^{m*}) \cdot {\skew 4\widehat{\vec L}}_i$$

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 [31, p. 874].

It follows that

$$H_{21,i}^{\rm M\ell 1} = - \frac{q_i}{m_i} \langle\psi_{... ...cdot {\skew 4\widehat{\vec L}}_i \vert\psi_{\rm {H}}\rangle$$

or in the long wave length approximation

$$H_{21,i}^{\rm M\ell 1} = - \frac{q_i}{m_i} \langle\psi_{... ...cdot {\skew 4\widehat{\vec L}}_i \vert\psi_{\rm {H}}\rangle$$

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 $\skew3\vec A_{\gamma i}^{\rm {E}}$,

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

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

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

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

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

The value of $A_0$ to use in addendum {A.23} is $-\varepsilon_k^{\rm {E}}/\sqrt{2}{{\rm i}}c$.

D.44.3 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,

$$\vert h_{21}^{\rm E\ell}\vert \approx \int R_{\rm {L}}(r... ...}^{m_{j\rm {H}}} \sin^2\theta_i{\rm d}\theta_i{\rm d}\phi_i$$

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:

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

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 $\ell=\vert j_{\rm {H}}-j_{\rm {L}}\vert$. (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 $\vert l_{\rm {H}}-l_{\rm {L}}\vert=\ell$.

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

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

which have

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

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

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

Now $Y_0^0$ = $1/\sqrt{4\pi}$ 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 $m$ = $\ell$. 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

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

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 $l_{\rm {H}}$ = $\ell$ and total angular momentum $j_{\rm {H}}$ = $\ell+\frac12$. 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 $m_{j\rm {H}}$ = $\ell+\frac12$.

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:

$$\Theta_{l_{\rm {H}}j_{\rm {H}}i}^{m_{j\rm {H}}} = \Theta... ...c12}{\uparrow}+ c_2 Y_\ell^{m_{j\rm {H}}+\frac12}{\downarrow}$$

Now the first term produces decays to $Y_0^0{\uparrow}$ by the emission of a photon with $m_\ell$ = $\ell-\frac12$. 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 $m_\ell$ = $\ell+\frac12$. This decay rate is a factor $c_2^2$ less than the Weisskopf unit. Since $c_1^2+c_2^2$ = 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 $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, 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.174) on the orbital angular momentum change in magnetic transitions. Therefore the simplest possible initial and final states have

$$l_{\rm {L}} = 1 \quad j_{\rm {L}} = {\textstyle\frac{1}{2}... ...{1}{2}} \quad m_{j{\rm {H}}} = \ell+{\textstyle\frac{1}{2}}$$

For these quantum numbers, the initial and final states are

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

where the square roots come from figure 12.5 in the $j_a,j_b=1,\frac12$ tabulation.

Now consider the form of the magnetic matrix element (A.180). First note, {D.44.2}, 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 $m_{\rm {L}}$ = 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

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

as seen in table 4.3 or {D.66}. 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 $\ell/(x+{{\rm i}}y)$ to it. Similarly, the $y$-derivative simply adds a factor ${{\rm i}}\ell/(x+{{\rm i}}y)$. 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

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

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 $m_{j\rm {L}}=-\frac12$.

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 $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, 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 $\vert j_{\rm {H}}-j_{\rm {L}}\vert=\ell$ and that $\vert l_{\rm {H}}-l_{\rm {L}}\vert=\ell-1$

The momentum factors (A.188) 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.