A.45 Fermi the­ory

This note needs more work, but as far as I know is ba­si­cally OK. Un­for­tu­nately, a de­riva­tion of elec­tron cap­ture for zero spin tran­si­tions is not in­cluded.

This note de­rives the Fermi the­ory of beta de­cay. In par­tic­u­lar, it gives the ball­parks that were used to cre­ate fig­ure 14.54. It also de­scribes the Fermi in­te­gral plot­ted in fig­ure 14.52, as well as Fermi’s (sec­ond) golden rule. There is also a fi­nal sub­sec­tion on elec­tron cap­ture, A.45.7.

When beta de­cay was first ob­served, it was be­lieved that the nu­cleus sim­ply ejected an elec­tron. How­ever, prob­lems quickly arose with en­ergy and mo­men­tum con­ser­va­tion. To solve them, Pauli pro­posed in 1931 that in ad­di­tion to the elec­tron, also a neu­tral par­ti­cle was emit­ted. Fermi called it the neu­trino, for small neu­tral one. Fol­low­ing ideas of Pauli in 1933, Fermi in 1934 de­vel­oped a com­pre­hen­sive the­ory of beta de­cay. The the­ory jus­ti­fies the var­i­ous claims made about al­lowed and for­bid­den beta de­cays. It also al­lows pre­dic­tions of the de­cay rate and the prob­a­bil­ity that the elec­tron and an­ti­neu­trino will come out with given ki­netic en­er­gies. This note gives a sum­mary. The ball­parks as de­scribed in this note are the ones used to pro­duce fig­ure 14.54.

A large amount of work has gone into im­prov­ing the ac­cu­racy of the Fermi the­ory, but it is out­side the scope of this note. To get an idea of what has been done, you might start with [23] and work back­wards. One point to keep in mind is that the de­riva­tions be­low are based on ex­pand­ing the elec­tron and neu­trino wave func­tions into plane waves, waves of def­i­nite lin­ear mo­men­tum. For a more thor­ough treat­ment, it may be a bet­ter idea to ex­pand into spher­i­cal waves, be­cause nu­clear states have def­i­nite an­gu­lar mo­men­tum. That idea is worked out in more de­tail in the note on gamma de­cay, {A.25}. That is news to the au­thor. But it was sup­posed to be there, I think.

A.45.1 Form of the wave func­tion

A clas­si­cal quan­tum treat­ment will not do for beta de­cay. To see why, note that in a clas­si­cal treat­ment the wave func­tion state be­fore the de­cay is taken to be of the form

$$\psi_1({\skew0\vec r}_1,S_{z,1},{\skew0\vec r}_2,S_{z,2},\ldots,{\skew0\vec r}_A,S_{z,A})$$

where 1 through $A$ num­ber the nu­cle­ons. How­ever, the de­cay cre­ates an elec­tron and a an­ti­neu­trino out of noth­ing. There­fore, af­ter the de­cay the clas­si­cal wave func­tion is of the form

$$\psi_2({\skew0\vec r}_1,S_{z,1},{\skew0\vec r}_2,S_{z,2},\l... ...skew0\vec r}_e,S_{z,e},{\skew0\vec r}_{\bar\nu},S_{z,\bar\nu})$$

There is no way to de­scribe how $\psi_1$ could evolve into $\psi_2$. You can­not just scrib­ble in two more ar­gu­ments into a func­tion some­where half way dur­ing the evo­lu­tion. That would be voodoo math­e­mat­ics. And there is also a prob­lem with one nu­cleon turn­ing from a neu­tron into a pro­ton. You should re­ally cross out the ar­gu­ment cor­re­spond­ing to the old neu­tron, and write in an ar­gu­ment for the new pro­ton.

You might think that maybe the elec­tron and an­ti­neu­trino were al­ways there to be­gin with. But that has some ma­jor prob­lems. A lone neu­tron falls apart into a pro­ton, an elec­tron and an an­ti­neu­trino. So sup­pos­edly the neu­tron would con­sist of a pro­ton, an elec­tron, and an an­ti­neu­trino. But to con­fine light par­ti­cles like elec­trons and neu­tri­nos to the size of a nu­cleon would pro­duce huge ki­netic en­er­gies. Ac­cord­ing to the Heisen­berg un­cer­tainty re­la­tion $p$ $\vphantom0\raisebox{1.5pt}{$\sim$}$ $\hbar$$\raisebox{.5pt}{$/$}$​$\Delta{x}$, where the en­ergy for rel­a­tivis­tic par­ti­cles is about $pc$, so the ki­netic en­ergy of a light par­ti­cle con­fined to a 1 fm range is about 200 MeV. What con­ceiv­able force could be strong enough to hold elec­trons and neu­tri­nos that hot? And how come the ef­fects of this mys­te­ri­ous force never show up in the atomic elec­trons that can be very ac­cu­rately ob­served? How come that elec­trons come out in beta de­cays with only a few MeV, rather than 200 MeV?

Fur­ther, a high-en­ergy an­ti­neu­trino can re­act with a pro­ton to cre­ate a neu­tron and a positron. That neu­tron is sup­posed to con­sist of a pro­ton, an elec­tron, and an an­ti­neu­trino. So, fol­low­ing the same rea­son­ing as be­fore, the orig­i­nal pro­ton be­fore the re­ac­tion would con­sist of a positron, an elec­tron, and a pro­ton. That pro­ton in turn would sup­pos­edly also con­sist of a positron, an elec­tron, and an pro­ton. So the orig­i­nal pro­ton con­sists of a positron, an elec­tron, a positron, an elec­tron, and a pro­ton. And so on un­til a pro­ton con­sists of a pro­ton and in­fi­nitely many elec­tron / positron pairs. Not just one elec­tron with very high ki­netic en­ergy would need to be con­fined in­side a nu­cleon, but an in­fi­nite num­ber of them, and positrons to boot. And all these elec­trons and positrons would some­how have to be pre­vented from an­ni­hi­lat­ing each other.

It just does not work. There is plenty of solid ev­i­dence that neu­trons and pro­tons each con­tain three quarks, not other nu­cle­ons along with elec­trons, positrons, and neu­tri­nos. The elec­tron and an­ti­neu­trino are cre­ated out of pure en­ergy dur­ing beta de­cay, as al­lowed by Ein­stein’s fa­mous rel­a­tivis­tic ex­pres­sion $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $mc^2$. A rel­a­tivis­tic quan­tum treat­ment is there­fore nec­es­sary.

In par­tic­u­lar, it is nec­es­sary to deal math­e­mat­i­cally with the ap­pear­ance of the elec­tron and an an­ti­neu­trino out of noth­ing. To do so, a more gen­eral, more ab­stract way must be used to de­scribe the states that na­ture can be in. Con­sider a de­cay that pro­duces an elec­tron and an an­ti­neu­trino of spe­cific mo­menta ${\skew0\vec p}_e$, re­spec­tively ${\skew0\vec p}_{\bar\nu}$. The fi­nal state is writ­ten as

$$\psi_2 = \psi_{2,\rm nuc} {\left\vert 1e,{\skew0\vec p}_e\r... ...e} {\left\vert 1\bar\nu,{\skew0\vec p}_{\bar\nu}\right\rangle}$$ (A.274)

where $\psi_{2,\rm {nuc}}$ is the nu­clear part of the fi­nal wave func­tion. The elec­tron ket ${\left\vert 1e,{\skew0\vec p}_e\right\rangle}$ is a “Fock-space ket,” and should be read as “one elec­tron in the state with an­gu­lar mo­men­tum ${\skew0\vec p}_e$.” The an­ti­neu­trino ket should be read as “one an­ti­neu­trino in the state with an­gu­lar mo­men­tum ${\skew0\vec p}_{\bar\nu}$.”

Sim­i­larly, the state be­fore the de­cay is writ­ten as

$$\psi_1 = \psi_{1,\rm nuc} {\left\verte,{\skew0\vec p}_e\rig... ...gle} {\left\vert\bar\nu,{\skew0\vec p}_{\bar\nu}\right\rangle}$$ (A.275)

where ${\left\verte,{\skew0\vec p}_e\right\rangle}$ means “zero elec­trons in the state with an­gu­lar mo­men­tum ${\skew0\vec p}_e$,” and sim­i­lar for the an­ti­neu­trino ket. Writ­ten in this way, the ini­tial and fi­nal wave func­tions are no longer in­con­sis­tent. What is dif­fer­ent is not the form of the wave func­tion, but merely how many elec­trons and an­ti­neu­tri­nos are in the states with mo­men­tum ${\skew0\vec p}_e$, re­spec­tively ${\skew0\vec p}_{\bar\nu}$. Be­fore the de­cay, the “oc­cu­pa­tion num­bers” of these states are zero elec­trons and zero an­ti­neu­tri­nos. Af­ter the de­cay, the oc­cu­pa­tion num­bers are one elec­tron and one neu­trino. It is not that the ini­tial state does not have oc­cu­pa­tion num­bers for these states, (which would make $\psi_1$ and $\psi_2$ in­con­sis­tent), but merely that these oc­cu­pa­tion num­bers have the value zero, (which does not).

(You could also add kets for dif­fer­ent mo­men­tum states that the fi­nal elec­tron and an­ti­neu­trino are not in af­ter the de­cay. But states that have zero elec­trons and neu­tri­nos both be­fore and af­ter the con­sid­ered de­cay are phys­i­cally ir­rel­e­vant and can be left away.)

That leaves the nu­clear part of the wave func­tion. You could use Fock-space kets to deal with the dis­ap­pear­ance of a neu­tron and ap­pear­ance of a pro­ton dur­ing the de­cay. How­ever, there is a neater way. The to­tal num­ber of nu­cle­ons re­mains the same dur­ing the de­cay. The only thing that hap­pens is that a nu­cleon changes type from a neu­tron into a pro­ton. The math­e­mat­i­cal trick is there­fore to take the par­ti­cles to be nu­cle­ons, in­stead of pro­tons and neu­trons. If you give each nu­cleon a nu­cleon type prop­erty, then the only thing that hap­pens dur­ing the de­cay is that the nu­cleon type of one of the nu­cle­ons flips over from neu­tron to pro­ton. No nu­cle­ons are cre­ated or de­stroyed. Nu­cleon type is typ­i­cally in­di­cated by the sym­bol $T_3$ and is de­fined to be ${\textstyle\frac{1}{2}}$ if the nu­cleon is a pro­ton and $-{\textstyle\frac{1}{2}}$ if the nu­cleon is a neu­tron. (Some older ref­er­ences may de­fine it the other way around.) The gen­eral form of the nu­clear wave func­tion there­fore be­comes

$$\Psi_N({\skew0\vec r}_1,S_{z,1},T_{3,1},{\skew0\vec r}_2,S_{z,2},T_{3,2}, \ldots,{\skew0\vec r}_A,S_{z,A},T_{3,A};t)$$

Dur­ing the de­cay, the $T_3$ value of one nu­cleon will change from $-{\textstyle\frac{1}{2}}$ to ${\textstyle\frac{1}{2}}$.

Of course, the name nu­cleon type for $T_3$ is not re­ally ac­cept­able, be­cause it is un­der­stand­able. In the old days, the names iso­baric spin” or “iso­topic spin were used, be­cause nu­cleon type has ab­solutely noth­ing to do with spin. How­ever, it was felt that these non­sen­si­cal names could cause some smart out­siders to sus­pect that the quan­tity be­ing talked about was not re­ally spin. There­fore the mod­ern term isospin was in­tro­duced. This term con­tains noth­ing to give the se­cret away that it is not spin at all.

A.45.2 Source of the de­cay

Next the source of the de­cay must be iden­ti­fied. Ul­ti­mately that must be the Hamil­ton­ian, be­cause the Hamil­ton­ian de­scribes the time evo­lu­tion of sys­tems ac­cord­ing to the Schrö­din­ger equa­tion.

In a spe­cific beta de­cay process, two states are in­volved. A state $\psi_1$ de­scribes the nu­cleus be­fore the de­cay, and a state $\psi_2$ de­scribes the com­bi­na­tion of nu­cleus, elec­tron, and an­ti­neu­trino af­ter the de­cay. That makes the sys­tem into a so-called “two state sys­tem.” The un­steady evo­lu­tion of such sys­tems was dis­cussed in chap­ter 7.6 and {D.38}. The key to the so­lu­tion were the Hamil­ton­ian co­ef­fi­cients. The first one is:

$$E_1 \equiv H_{11} \equiv \langle\psi_1\vert H\psi_1\rangle$$

where $H$ is the (rel­a­tivis­tic) Hamil­ton­ian. The value of $H_{11}$ is the ex­pec­ta­tion value of the en­ergy $E_1$ when na­ture is in the state $\psi_1$. As­sum­ing that the nu­cleus is ini­tially at rest, the rel­a­tivis­tic en­ergy is just the rest mass en­ergy of the nu­cleus. It is given in terms of its mass by Ein­stein’s fa­mous re­la­tion $E_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m_{{\rm {N}}1}c^2$.

The Hamil­ton­ian co­ef­fi­cient for the fi­nal state is sim­i­larly

$$E_2 \equiv H_{22} \equiv \langle\psi_2\vert H\psi_2\rangle$$

Us­ing the form given in the pre­vi­ous sec­tion for the fi­nal wave func­tion, that be­comes

$$E_2 = \langle 1\bar\nu,{\skew0\vec p}_{\bar\nu}\vert\langl... ...\vec p}_e\rangle\vert 1\bar\nu,{\skew0\vec p}_{\bar\nu}\rangle$$

It is the ex­pec­ta­tion value of en­ergy af­ter the de­cay. It con­sists of the sum of the rest mass en­er­gies of fi­nal nu­cleus, elec­tron, and an­ti­neu­trino, as well as their ki­netic en­er­gies.

The Hamil­ton­ian co­ef­fi­cient that de­scribes the in­ter­ac­tion be­tween the two states is cru­cial, be­cause it is the one that causes the de­cay. It is

$$H_{21} \equiv \langle\psi_2\vert H\psi_1\rangle$$

Us­ing the form for the wave func­tions given in the pre­vi­ous sec­tion:

$$H_{21} = \langle 1\bar\nu,{\skew0\vec p}_{\bar\nu}\vert\la... ...w0\vec p}_e\rangle\vert\bar\nu,{\skew0\vec p}_{\bar\nu}\rangle$$

If $H_{21}$ is zero, no de­cay will oc­cur. And most of the Hamil­ton­ian does not pro­duce a con­tri­bu­tion to $H_{21}$. But there is a small part of the Hamil­ton­ian, call it $H'$, that does pro­duce a nonzero in­ter­ac­tion. That part is due to the weak force.

Un­for­tu­nately, Fermi had no clue what $H'$ was. He as­sumed that beta de­cay would not be that much dif­fer­ent from the bet­ter un­der­stood de­cay of ex­cited atomic states in atoms. Gamma de­cay is the di­rect equiv­a­lent of atomic de­cay for ex­cited nu­clei. Beta de­cay is def­i­nitely dif­fer­ent, but maybe not that dif­fer­ent. In atomic de­cay an elec­tro­mag­netic pho­ton is cre­ated, rather than an elec­tron and an­ti­neu­trino. Still the gen­eral idea seemed sim­i­lar.

In atomic de­cay $H'$ is es­sen­tially pro­por­tional to the prod­uct of the charge of the ex­cited elec­tron, times the spa­tial eigen­state of the pho­ton, times a pho­ton cre­ation op­er­a­tor $\widehat a^\dagger$:

$$H' \propto e \psi_{\rm photon}({\skew0\vec r}) \widehat a^\dagger$$

In words, it says that the in­ter­ac­tion of the elec­tron with the elec­tro­mag­netic field can cre­ate pho­tons. The mag­ni­tude of that ef­fect is pro­por­tional to the am­pli­tude of the pho­ton at the lo­ca­tion of the elec­tron, and also to the elec­tric charge of the elec­tron. The elec­tric charge acts as a “cou­pling con­stant” that links elec­trons and pho­tons to­gether. If the elec­tron was un­charged, it would not be able to cre­ate pho­tons. So it would not be able to cre­ate an elec­tric field. Fur­ther, the fact that the cou­pling be­tween the elec­tron and the pho­ton oc­curs at the lo­ca­tion of the elec­tron elim­i­nates some prob­lems that rel­a­tiv­ity has with ac­tion at a dis­tance.

There is an­other term in $H'$ that in­volves an an­ni­hi­la­tion op­er­a­tor $\widehat a$ in­stead of a cre­ation op­er­a­tor. An an­ni­hi­la­tion op­er­a­tor de­stroys pho­tons. How­ever, that does not pro­duce a con­tri­bu­tion to $H_{21}$; if you try to an­ni­hi­late the nonex­ist­ing pho­ton in the ini­tial wave func­tion, you get a zero wave func­tion. On the other hand for the ear­lier term, the cre­ation op­er­a­tor is es­sen­tial. It turns the ini­tial state with no pho­ton into a state with one pho­ton. States with dif­fer­ent num­bers of par­ti­cles are or­thog­o­nal, so the Hamil­ton­ian co­ef­fi­cient $H_{12}$ would be zero with­out the cre­ation op­er­a­tor. Looked at the other way around, the pres­ence of the cre­ation op­er­a­tor in the Hamil­ton­ian en­sures that the fi­nal state must have one more pho­ton for the de­cay to oc­cur. (See ad­den­dum {A.15} for more de­tails on elec­tro­mag­netic in­ter­ac­tions, in­clud­ing a more pre­cise de­scrip­tion of $H'$. See also {A.25}.)

Fermi as­sumed that the gen­eral ideas of atomic de­cay would also hold for beta de­cay of nu­clei. Elec­tron and an­ti­neu­trino cre­ation op­er­a­tors in the Hamil­ton­ian would turn the zero-elec­tron and zero-an­ti­neu­trino kets into one-elec­tron and one-an­ti­neu­trino ones. Then the in­ner prod­ucts of the kets are equal to one pair­wise. There­fore both the cre­ation op­er­a­tors and the kets drop out of the fi­nal ex­pres­sion. In that way the Hamil­ton­ian co­ef­fi­cient sim­pli­fies to

$$H_{21} = \langle \psi_{2,\rm nuc} \vert \sum_{i=1}^A g h_i ... ... p}_{\bar\nu}}({\skew0\vec r}_i) \vert\psi_{1,\rm nuc}\rangle$$

where the in­dex $i$ is the nu­cleon num­ber and $gh_i$ is the re­main­ing still un­known part of the Hamil­ton­ian. In in­di­cat­ing this un­known part by $gh_i$, the as­sump­tion is that it will be pos­si­ble to write it as some generic di­men­sional con­stant $g$ times some sim­ple nondi­men­sion­al op­er­a­tor $h_i$ act­ing on nu­cleon num­ber $i$.

To write ex­pres­sions for the wave func­tions of the elec­tron and an­ti­neu­trino, you face the com­pli­ca­tion that un­bound states in in­fi­nite space are not nor­mal­iz­able. That pro­duced math­e­mat­i­cal com­pli­ca­tions for mo­men­tum eigen­states in chap­ter 7.9.2, and sim­i­lar dif­fi­cul­ties resur­face here. To sim­plify things, the math­e­mat­i­cal trick is to as­sume that the de­cay­ing nu­cleus is not in in­fi­nite space, but in an ex­tremely large “pe­ri­odic box.” The as­sump­tion is that na­ture re­peats it­self spa­tially; a par­ti­cle that ex­its the box through one side reen­ters it through the op­po­site side. Space wraps around if you want, and op­po­site sides of the box are as­sumed to be phys­i­cally the same lo­ca­tion. It is like on the sur­face of the earth: if you move along the straight­est-pos­si­ble path on the sur­face of the earth, you travel around the earth along a big cir­cle and re­turn to the same point that you started out at. Still, on a lo­cal scale, the sur­face on the earth looks flat. The idea is that the empty space around the de­cay­ing nu­cleus has a sim­i­lar prop­erty, in each of the three Carte­sian di­men­sions. This trick is also com­monly used in solid me­chan­ics, chap­ter 10.

In a pe­ri­odic box, the wave func­tion of the an­ti­neu­trino is

$$\psi_{\bar\nu,{\skew0\vec p}_{\bar\nu}} = \frac{1}{\sqrt{{... ...rm i}(p_{x,\bar\nu}x + p_{y,\bar\nu}y + p_{z,\bar\nu}z)/\hbar}$$

where ${\cal V}$ is the vol­ume of the pe­ri­odic box. It is easy to check that this wave func­tion is in­deed nor­mal­ized. Also, it is seen that it is in­deed an eigen­func­tion of the $x$-​mo­men­tum op­er­a­tor $\hbar\partial$$\raisebox{.5pt}{$/$}$​${\rm i}\partial{x}$ with eigen­value $p_x$, and sim­i­lar for the $y$-​ and $z$-​mo­men­tum op­er­a­tors.

The wave func­tion of the elec­tron will be writ­ten in a sim­i­lar way:

$$\psi_{e,{\skew0\vec p}_e}({\skew0\vec r}) = \frac{1}{\sqrt... ...{{\cal V}}} e^{{\rm i}(p_{x,e}x + p_{y,e}y + p_{z,e}z)/\hbar}$$

This how­ever has an ad­di­tional prob­lem. It works fine far from the nu­cleus, where the mo­men­tum of the elec­tron is by de­f­i­n­i­tion the con­stant vec­tor ${\skew0\vec p}_e$. How­ever, near the nu­cleus the Coulomb field of the nu­cleus, and to some ex­tent that of the atomic elec­trons, af­fects the ki­netic en­ergy of the elec­tron, and hence its mo­men­tum. There­fore, the en­ergy eigen­func­tion that has mo­men­tum ${\skew0\vec p}$ far from the nu­cleus dif­fers sig­nif­i­cantly from the above ex­po­nen­tial closer to the nu­cleus. And this wave func­tion must be eval­u­ated at nu­cleon po­si­tions in­side the nu­cleus! The prob­lem is par­tic­u­larly large when the mo­men­tum ${\skew0\vec p}_e$ is low, be­cause then the elec­tron has lit­tle ki­netic en­ergy and the Coulomb po­ten­tial is rel­a­tively speak­ing more im­por­tant. The prob­lem gets even worse for low-en­ergy positron emis­sion, be­cause a pos­i­tively-charged positron is re­pelled by the pos­i­tive nu­cleus and must tun­nel through to reach it.

The usual way to deal with the prob­lem is to stick with the ex­po­nen­tial elec­tron wave func­tion for now, and fix up the prob­lem later in the fi­nal re­sults. The fix-up will be achieved by throw­ing in an ad­di­tional fudge fac­tor. While “Fermi fudge fac­tor” al­lit­er­ates nicely, it does not sound very re­spect­ful, so physi­cists call the fac­tor the Fermi func­tion.

The bot­tom line is that for now

$$H_{21} = \frac{g}{\sqrt{{\cal V}}} \langle \psi_{2,\rm nuc}... ...\nu}) \cdot {\skew0\vec r}_i/\hbar} \psi_{1,\rm nuc}\rangle %$$ (A.276)

That leaves the still un­known op­er­a­tor $gh_i$. The con­stant $g$ is sim­ply de­fined so that the op­er­a­tor $h_i$ has a mag­ni­tude that is of or­der one. That means that $\langle\psi_{2,\rm {nuc}}\vert h_i\psi_{1,\rm {nuc}}\rangle$ should never be greater than about one, though it could be much less if $\psi_{2,\rm {nuc}}$ and $h_i\psi_{1,\rm {nuc}}$ turn out to be al­most or­thog­o­nal. It is found that $g$ has a rough value of about 100 eV fm$\POW9,{3}$, de­pend­ing a bit on whether it is a Fermi or Gamow-Teller de­cay. Fig­ure 14.54 sim­ply used 100 MeV fm$\POW9,{3}$.

A.45.3 Al­lowed or for­bid­den

The ques­tion of al­lowed and for­bid­den de­cays is di­rectly re­lated to the Hamil­ton­ian co­ef­fi­cient $H_{21}$, (A.276), de­rived in the pre­vi­ous sub­sec­tion, that causes the de­cay.

First note that the emit­ted elec­tron and an­ti­neu­trino have quite small mo­men­tum val­ues, on a nu­clear scale. In par­tic­u­lar, in their com­bined wave func­tion

$$\frac{1}{{\cal V}} e^{{\rm i}({\skew0\vec p}_e + {\skew0\vec p}_{\bar\nu}) \cdot {\skew0\vec r}_i/\hbar}$$

the ar­gu­ment of the ex­po­nen­tial is small. Typ­i­cally, its mag­ni­tude is only a few per­cent, It is there­fore pos­si­ble to ap­prox­i­mate the ex­po­nen­tial by one, or more gen­er­ally by a Tay­lor se­ries:

$$e^{{\rm i}({\skew0\vec p}_e + {\skew0\vec p}_{\bar\nu}) \cd... ... p}_{\bar\nu})\cdot{\skew0\vec r}_i}{\hbar}\right)^2 + \ldots$$

Since the first term in the Tay­lor se­ries is by far the largest, you would ex­pect that the value of $H_{21}$, (A.276), can be well ap­prox­i­mated by re­plac­ing the ex­po­nen­tial by 1, giv­ing:

$$H_{21}^0 = \frac{g}{\sqrt{{\cal V}}} \langle \psi_{2,\rm nuc} \vert \sum_{i=1}^A h_i \psi_{1,\rm nuc}\rangle$$

How­ever, clearly this ap­prox­i­ma­tion does not work if the value of $H_{21}^0$ is zero for some rea­son:

If the sim­pli­fied co­ef­fi­cient $H_{21}^0$ is nonzero, the de­cay is al­lowed. If it is zero, the de­cay is for­bid­den.

If the de­cay is for­bid­den, higher or­der terms in the Tay­lor se­ries will have to be used to come up with a nonzero value for $H_{21}$. Since these higher or­der terms are much smaller, and $H_{21}$ dri­ves the de­cay, a for­bid­den de­cay will pro­ceed much slower than an al­lowed one.

Why would a de­cay not be al­lowed? In other words why would $\psi_{2,\rm {nuc}}$ and $h_i\psi_{1,\rm {nuc}}$ be ex­actly or­thog­o­nal? If you took two ran­dom wave func­tions for $\psi_{2,\rm {nuc}}$ and $\psi_{1,\rm {nuc}}$, they def­i­nitely would not be. But $\psi_{2,\rm {nuc}}$ and $\psi_{1,\rm {nuc}}$ are not ran­dom wave func­tions. They sat­isfy a sig­nif­i­cant amount of sym­me­try con­straints.

One very im­por­tant one is sym­me­try with re­spect to co­or­di­nate sys­tem ori­en­ta­tion. An in­ner prod­uct of two wave func­tions is in­de­pen­dent of the an­gu­lar ori­en­ta­tion of the co­or­di­nate sys­tem in which you eval­u­ate it. There­fore, you can av­er­age the in­ner prod­uct over all di­rec­tions of the co­or­di­nate sys­tem. How­ever, the an­gu­lar vari­a­tion of a wave func­tion is re­lated to its an­gu­lar mo­men­tum; see chap­ter 7.3 and its note. In par­tic­u­lar, if you av­er­age a wave func­tion of def­i­nite an­gu­lar mo­men­tum over all co­or­di­nate sys­tem ori­en­ta­tions, you get zero un­less the an­gu­lar mo­men­tum is zero. So, if it was just $\psi_{1,\rm {nuc}}$ in the in­ner prod­uct in $H_{21}^0$, the in­ner prod­uct would be zero un­less the ini­tial nu­cleus had zero spin. How­ever, the fi­nal state is also in the in­ner prod­uct, and be­ing at the other side of it, its an­gu­lar vari­a­tion acts to coun­ter­act that of the ini­tial nu­cleus. There­fore, $H_{21}^0$ will be zero un­less the ini­tial an­gu­lar mo­men­tum is ex­actly bal­anced by the net fi­nal an­gu­lar mo­men­tum. And that is an­gu­lar mo­men­tum con­ser­va­tion. The de­cay has to sat­isfy it.

Note that the lin­ear mo­menta of the elec­tron and an­ti­neu­trino have be­come ig­nored in $H_{21}^0$. There­fore, their or­bital an­gu­lar mo­men­tum is ap­prox­i­mated to be zero too. Un­der these con­di­tion $H_{21}^0$ is zero un­less the an­gu­lar mo­men­tum of the fi­nal nu­cleus plus the spin an­gu­lar mo­men­tum of elec­tron and an­ti­neu­trino equals the an­gu­lar mo­men­tum of the orig­i­nal nu­cleus. Since the elec­tron and an­ti­neu­trino can have up to one unit of com­bined spin, the nu­clear spin can­not change more than one unit. That is the first se­lec­tion rule for al­lowed de­cays given in chap­ter 14.19.6.

An­other im­por­tant con­straint is sym­me­try un­der the par­ity trans­for­ma­tion ${\skew0\vec r}\to-{\skew0\vec r}$. This trans­for­ma­tion too does not af­fect in­ner prod­ucts, so you can av­er­age the val­ues be­fore and af­ter the trans­form. How­ever, a wave func­tion that has odd par­ity changes sign un­der the trans­form and av­er­ages to zero. So the in­ner prod­uct in $H_{21}^0$ is zero if the to­tal in­te­grand has odd par­ity. For a nonzero value, the in­te­grand must have even par­ity, and that means that the par­ity of the ini­tial nu­cleus must equal the com­bined par­ity of the fi­nal nu­cleus elec­tron, and an­ti­neu­trino.

Since the elec­tron and an­ti­neu­trino come out with­out or­bital an­gu­lar mo­men­tum, they have even par­ity. So the nu­clear par­ity must re­main un­changed un­der the tran­si­tion. (To be sure, this is not ab­solutely jus­ti­fied. Nu­clear wave func­tions ac­tu­ally have a tiny un­cer­tainty in par­ity be­cause the weak force does not con­serve par­ity, chap­ter 14.19.8. This ef­fect is usu­ally too small to be ob­served and will be ig­nored here.)

So what if ei­ther one of these se­lec­tion rules is vi­o­lated? In that case, maybe the sec­ond term in the Tay­lor se­ries for the elec­tron and an­ti­neu­trino wave func­tions pro­duces some­thing nonzero that can drive the de­cay. For that to be true,

$$H_{21}^1 = \frac{g}{\sqrt{{\cal V}}} \langle \psi_{2,\rm nu... ...bar\nu})\cdot{\skew0\vec r}_i}{\hbar} \psi_{1,\rm nuc}\rangle$$

has to be nonzero. If it is, the de­cay is a first-for­bid­den one. Now the spher­i­cal har­mon­ics $Y_1^m$ of or­bital an­gu­lar mo­men­tum are of the generic form, {D.14}

$$r Y_1^m = \sum_j c_j r_j$$

with the $c_j$ some con­stants. There­fore, the fac­tor ${\skew0\vec r}_i$ in $H_{21}^1$ brings in an­gu­lar vari­a­tion cor­re­spond­ing to one unit of an­gu­lar mo­men­tum. That means that the to­tal spin can now change by up to one unit, and there­fore the nu­clear spin by up to two units. That is in­deed the se­lec­tion rule for first for­bid­den de­cays.

And note that be­cause ${\skew0\vec r}_i$ changes sign when every ${\skew0\vec r}$ is re­placed by $\vphantom{0}\raisebox{1.5pt}{$-$}$${\skew0\vec r}$, the ini­tial and fi­nal nu­clear par­i­ties must now be op­po­site for $H_{21}^1$ not to be zero. That is in­deed the par­ity se­lec­tion rule for first-for­bid­den de­cays.

The higher or­der for­bid­den de­cays go the same way. For an $\ell$th-for­bid­den de­cay,

$$H_{21}^\ell = \frac{g}{\ell!\sqrt{{\cal V}}} \langle \psi_... ...\skew0\vec r}_i}{\hbar}\right)^\ell \psi_{1,\rm nuc}\rangle %$$ (A.277)

must be the first nonzero in­ner prod­uct. Note that an $\ell$th-for­bid­den de­cay has a co­ef­fi­cient $H_{21}$ pro­por­tional to a fac­tor of or­der $(pR/\hbar)^\ell$, with $R$ the nu­clear ra­dius. Since the de­cay rate turns out to be pro­por­tional to $\vert H_{21}\vert^2$, an $\ell$th-for­bid­den de­cay is slowed down by a fac­tor of or­der $(pR/\hbar)^{2\ell}$, mak­ing highly for­bid­den de­cays ex­tremely slow.

A.45.4 The nu­clear op­er­a­tor

This sub­sec­tion will have a closer look at the nu­clear op­er­a­tor $h_i$. While the dis­cus­sion will be kept sim­ple, hav­ing some idea about the na­ture of this op­er­a­tor can be use­ful. It can help to un­der­stand why some de­cays have rel­a­tively low de­cay rates that are not ex­plained by just look­ing at the elec­tron and an­ti­neu­trino wave func­tions, and the nu­clear spins and par­i­ties. The dis­cus­sion will mainly fo­cus on al­lowed de­cays.

Al­though Fermi did not know what $h_i$ was, Pauli had al­ready es­tab­lished the pos­si­ble generic forms for it al­lowed by rel­a­tiv­ity. It could take the form of a scalar (S), a vec­tor (V), an ax­ial vec­tor (A, a vec­tor like an­gu­lar mo­men­tum, one that in­verts when the physics is seen in a mir­ror), a pseu­doscalar (P, a scalar like the scalar triple prod­uct of vec­tors that changes sign when the physics is seen in the mir­ror), or a ten­sor (T, a mul­ti­ple-in­dex ob­ject like a ma­trix that trans­forms in spe­cific ways.) Fermi sim­ply as­sumed the in­ter­ac­tion was of the vec­tor, V, type in anal­ogy with the de­cay of ex­cited atoms.

Fermi ig­nored the spin of the elec­tron and an­ti­neu­trino. How­ever, Gamow & Teller soon es­tab­lished that to al­low for de­cays where the two come out with spin, (Gamow-Teller de­cays), $gh_i$ also should have terms with ax­ial, A, and/or ten­sor, T, char­ac­ter. Work of Fierz com­bined with ex­per­i­men­tal ev­i­dence showed that the Hamil­ton­ian could not have both S and V, nor both A and T terms. Ad­di­tional ev­i­dence nar­rowed $h_i$ down to STP com­bi­na­tions or VA ones.

Fi­nally, it was es­tab­lished in 1953 that the cor­rect one was the STP com­bi­na­tion, be­cause ex­per­i­men­tal ev­i­dence on RaE, (some physi­cists can­not spell bis­muth-210), showed that P was present. Un­for­tu­nately, it did not. For one, the con­clu­sion de­pended to an in­sane de­gree on the ac­cu­racy of a cor­rec­tion term.

How­ever, in 1955 it was es­tab­lished that it was STP any­way, be­cause ex­per­i­men­tal ev­i­dence on he­lium-6 clearly showed that the Gamow-Teller part of the de­cay was ten­sor. The ques­tion was there­fore solved sat­is­fac­to­rily. It was STP, or maybe just ST. Ex­per­i­men­tal ev­i­dence had re­deemed it­self.

How­ever, in 1958, a quar­ter cen­tury af­ter Fermi, it was found that beta de­cay vi­o­lated par­ity con­ser­va­tion, chap­ter 14.19.8, and the­o­ret­i­cally that was not re­ally con­sis­tent with STP. So ex­per­i­men­tal­ists had an­other look at their ev­i­dence and quickly came back with good news: “The he­lium-6 ev­i­dence does not show Gamow-Teller is ten­sor af­ter all.”

The fi­nal an­swer is that $gh_i$ is VA. Since so much of our knowl­edge about nu­clei de­pends on ex­per­i­men­tal data, it may be worth­while to keep this cau­tion­ary tale, taken from the Stan­ford En­cy­clo­pe­dia of Phi­los­o­phy, in mind.

It may next be noted that $gh_i$ will need to in­clude a isospin cre­ation op­er­a­tor to be able to turn a neu­tron into a pro­ton. In Fermi de­cays, $h_i$ is as­sumed to be just that op­er­a­tor. The con­stant of pro­por­tion­al­ity $g$, usu­ally called the cou­pling con­stant $g_{\rm {F}}$, de­scribes the strength of the weak in­ter­ac­tion. That is much like the unit elec­tric charge $e$ de­scribes the strength of the elec­tro­mag­netic in­ter­ac­tion be­tween charged par­ti­cles and pho­tons. In Fermi de­cays it is found that $g_{\rm {F}}$ is about 88 eV fm$\POW9,{3}$. Note that this is quite small com­pared to the MeV scale of nu­clear forces. If you ball­park rel­a­tive strengths of forces, [31, p. 285] the nu­clear force is strongest, the elec­tro­mag­netic force about hun­dred times smaller, the weak force an­other thou­sand times smaller than that, and fi­nally grav­ity is an­other 10$\POW9,{34}$ times smaller than that. The de­cay rates turn out to be pro­por­tional to the square of the in­ter­ac­tion, mag­ni­fy­ing the rel­a­tive dif­fer­ences.

In Gamow-Teller de­cays, $h_i$ is as­sumed to con­sist of prod­ucts of isospin cre­ation op­er­a­tors times spin cre­ation or an­ni­hi­la­tion op­er­a­tors. The lat­ter op­er­a­tors al­low the spin of the neu­tron that con­verts to the pro­ton to flip over. Suit­able spin cre­ation and an­ni­hi­la­tion op­er­a­tors are given by the so-called “Pauli spin ma­tri­ces,” chap­ter 12.10 When they act on a nu­cleon, they pro­duce states with the spin in an or­thog­o­nal di­rec­tion flipped over. That al­lows the net spin of the nu­cleus to change by one unit. The ap­pro­pri­ate con­stant of pro­por­tion­al­ity $g_{GT}$ is found to be a bit larger than the Fermi one.

The rel­e­vant op­er­a­tors then be­come, [5],

$$h_i = \tau_1\pm{\rm i}\tau_2 \qquad h_i = (\tau_1\pm{\rm i}\tau_2) \sum_{j=1}^3 \sigma_j$$

for Fermi and Gamow-Teller de­cays re­spec­tively. Here the three $\sigma_j$ are the Pauli spin ma­tri­ces of chap­ter 12.10. The $\tau_i$ are the equiv­a­lents of the Pauli spin ma­tri­ces for isospin; in the com­bi­na­tions shown above they turn neu­trons into pro­tons, or vice-versa. Please ex­cuse: us­ing the clar­ity now made pos­si­ble by mod­ern phys­i­cal ter­mi­nol­ogy, they cre­ate, re­spec­tively an­ni­hi­late, isospin. The up­per sign is rel­e­vant for beta-mi­nus de­cay and the lower for beta-plus de­cay. The Gamow-Teller op­er­a­tor ab­sorbs the spin part of the elec­tron and an­ti­neu­trino wave func­tions, in par­tic­u­lar the av­er­ag­ing over the di­rec­tions of their spin.

So how do these nu­clear op­er­a­tors af­fect the de­cay rate? That is best un­der­stood by go­ing back to the more phys­i­cal shell-model pic­ture. In beta mi­nus de­cay, a neu­tron is turned into a pro­ton. That pro­ton usu­ally oc­cu­pies a dif­fer­ent spa­tial state in the pro­ton shells than the orig­i­nal neu­tron in the neu­tron shells. And dif­fer­ent spa­tial states are sup­pos­edly or­thog­o­nal, so the in­ner prod­uct $\langle\psi_{2,\rm {nuc}}\vert h_i\psi_{1,\rm {nuc}}\rangle$ will usu­ally be pretty small, if the de­cay is al­lowed at all. There is one big ex­cep­tion, though: mir­ror nu­clei. In a de­cay be­tween mir­ror nu­clei, a nu­cleus with a neu­tron num­ber $N_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $Z_1\pm1$ de­cays into one with neu­tron num­ber $N_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $Z_2\mp1$. In that case, the nu­cleon that changes type re­mains in the same spa­tial or­bit. There­fore, the Fermi in­ner prod­uct equals one, and the Gamow Teller one is max­i­mal too. Al­lowed de­cays of this type are called “su­per­al­lowed.” The sim­plest ex­am­ple is the beta de­cay of a free neu­tron.

If you al­low for beta de­cay to ex­cited states, more su­per­al­lowed de­cays are pos­si­ble. States that dif­fer merely in nu­cleon type are called iso­baric ana­log states, or isospin mul­ti­plets, chap­ter 14.18. There are about twenty such su­per­al­lowed de­cays in which the ini­tial and fi­nal nu­clei both have spin zero and pos­i­tive par­ity. These twenty are par­tic­u­larly in­ter­est­ing the­o­ret­i­cally, be­cause only Fermi de­cays are pos­si­ble for them. And the Fermi in­ner prod­uct is $\sqrt{2}$. (The rea­son that it is $\sqrt{2}$ in­stead of 1 like for mir­ror nu­clei can be seen from think­ing of isospin as if it is just nor­mal spin. Mir­ror nu­clei have an odd num­ber of nu­cle­ons, so the net nu­clear isospin is half in­te­ger. In par­tic­u­lar the net isospin will be ${\textstyle\frac{1}{2}}$ in the ground state. How­ever, nu­clei with zero spin have an even num­ber of nu­cle­ons, hence in­te­ger net isospin. The isospin of the twenty de­cays is one; it can­not be zero be­cause at least one nu­cleus must have a nonzero net nu­cleon type $T_{3,\rm {net}}$. The net nu­cleon type is only zero if the num­ber of pro­tons is the same as the num­ber of neu­trons. It is then seen from (12.9) and (12.10) in chap­ter 12 that the isospin cre­ation or an­ni­hi­la­tion op­er­a­tors will pro­duce a fac­tor $\sqrt{2}$.)

These de­cays there­fore al­low the value of the Fermi cou­pling con­stant $g_{\rm {F}}$ to be de­ter­mined from the de­cay rates. It turns out to be about 88 eV fm$\POW9,{3}$, re­gard­less of the par­tic­u­lar de­cay used to com­pute it. That seems to sug­gest that the in­ter­ac­tion with neigh­bor­ing nu­cle­ons in a nu­cleus does not af­fect the Fermi de­cay process. In­deed, if the value of $g_{\rm {F}}$ is used to an­a­lyze the de­cay rates of the mir­ror nu­clei, in­clud­ing the free neu­tron that has no neigh­bors, the data show no such ef­fect. The hy­poth­e­sis that neigh­bor­ing nu­cle­ons do not af­fect the Fermi de­cay process is known as the “con­served vec­tor cur­rent hy­poth­e­sis.” What name could be clearer than that? Un­like Fermi de­cays, Gamow-Teller de­cays are some­what af­fected by the pres­ence of neigh­bor­ing nu­clei.

Be­sides the spin and par­ity rules al­ready men­tioned, Fermi de­cays must sat­isfy the ap­prox­i­mate se­lec­tion rule that the mag­ni­tude of isospin must be un­changed. They can be slowed down by sev­eral or­ders of mag­ni­tude if that rule is vi­o­lated.

Gamow-Teller de­cays are much less con­fined than Fermi ones be­cause of the pres­ence of the elec­tron spin op­er­a­tor. As the shell model shows, nu­cleon spins are un­cer­tain in en­ergy eigen­states. There­fore, the nu­clear sym­me­try con­straints are a lot less re­stric­tive.

A.45.5 Fermi’s golden rule

The pre­vi­ous four sub­sec­tions have fo­cussed on find­ing the Hamil­ton­ian co­ef­fi­cients of the de­cay from a state $\psi_1$ to a state $\psi_2$. Most of the at­ten­tion was on the co­ef­fi­cient $H_{21}^\ell$ that dri­ves the de­cay. The next step is so­lu­tion of the Schrö­din­ger equa­tion to find the evo­lu­tion of the de­cay process.

The quan­tum am­pli­tude of the pre-de­cay state $\psi_1$ will be in­di­cated by $\bar{a}$ and the quan­tum am­pli­tude of the fi­nal de­cayed state $\psi_2$ by $\bar{b}$. The Schrö­din­ger equa­tion im­plies that $\bar{b}$ in­creases from zero ac­cord­ing to

$${\rm i}\hbar \dot{\bar b} = H_{21}^\ell e^{{\rm i}(E_2-E_1)t/\hbar} \bar a$$

(To use this ex­pres­sion, the quan­tum am­pli­tudes must in­clude an ad­di­tional phase fac­tor, but it is of no con­se­quence for the prob­a­bil­ity of the states. See chap­ter 7.6 and {D.38} for de­tails.)

Now pic­ture the fol­low­ing. At the ini­tial time there are a large num­ber of pre-de­cay nu­clei, all with $\bar{a}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. All these nu­clei then evolve ac­cord­ing to the Schrö­din­ger equa­tion, above, over a time in­ter­val $t_c$ that is short enough that $\bar{a}$ stays close to one. (Be­cause the per­tur­ba­tion of the nu­cleus by the weak force is small, the mag­ni­tudes of the co­ef­fi­cients only change slowly on the rel­e­vant time scale.) In that case, $\bar{a}$ can be dropped from the equa­tion and its so­lu­tion is then seen to be

$$\bar b = - H_{21}^\ell \frac{e^{{\rm i}(E_2-E_1)t_c/\hbar}-1}{(E_2-E_1)}$$

Half of the ex­po­nen­tial can be fac­tored out to pro­duce a real ra­tio:

$$\bar b = - H_{21}^\ell e^{{\rm i}\frac12(E_2-E_1)t_c/\hbar}... ...frac12(E_2-E_1)t_c/\hbar\Big)}{\frac12(E_2-E_1)t_c/\hbar} t_c$$

Then at the fi­nal time $t_c$, as­sume that the state of all the nu­clei is mea­sured. The macro­scopic sur­round­ings of the nu­clei es­tab­lishes whether or not elec­tron and an­ti­neu­trino pairs have come out. The prob­a­bil­ity that a give nu­cleus has emit­ted such a pair is given by the square mag­ni­tude $\vert\bar{b}\vert^2$ of the am­pli­tude of the de­cayed state. There­fore, a frac­tion $\vert\bar{b}\vert^2$ of the nu­clei will be found to have de­cayed and $1-\vert\bar{b}\vert^2$ will be found to be still in the pre-de­cay state $\psi_1$. Af­ter this mea­sure­ment, the en­tire process then re­peats for the re­main­ing $1-\vert\bar{b}\vert^2$ nu­clei that did not de­cay.

The bot­tom line is how­ever that a frac­tion $\vert\bar{b}\vert^2$ did. There­fore, the ra­tio $\vert\bar{b}\vert^2$$\raisebox{.5pt}{$/$}$​$t_c$ gives the spe­cific de­cay rate, the rel­a­tive amount of nu­clei that de­cay per unit time. Plug­ging in the above ex­pres­sion for $\bar{b}$ gives:

$$\lambda_{\rm single\ final\ state} = \frac{\vert H_{21}^\el... ...t_c/\hbar\Big)} {\Big(\frac12(E_2-E_1)t_c/\hbar\Big)^2} t_c %$$ (A.278)

To get the to­tal de­cay rate, you must still sum over all pos­si­ble fi­nal states. Most im­por­tantly, you need to sum the spe­cific de­cay rates to­gether for all pos­si­ble elec­tron and an­ti­neu­trino mo­menta.

And there may be more. If the fi­nal nu­clear state has spin you also need to sum over all val­ues of the mag­netic quan­tum num­ber of the fi­nal state. (The amount of nu­clear de­cay should not de­pend on the an­gu­lar ori­en­ta­tion of the ini­tial nu­cleus in empty space. How­ever, if you ex­pand the elec­tron and neu­trino wave func­tions into spher­i­cal waves, you need to av­er­age over the pos­si­ble ini­tial mag­netic quan­tum num­bers. It may also be noted that the to­tal co­ef­fi­cient $\vert H_{21}^\ell\vert$ for the de­cay $1\to2$ will not be the same as the one for $2\to1$: you av­er­age over the ini­tial mag­netic quan­tum num­ber, but sum over the fi­nal one.) If there are dif­fer­ent ex­ci­ta­tion lev­els of the fi­nal nu­cleus that can be de­cayed to, you also need to sum over these. And if there is more than one type of de­cay process go­ing on at the same time, they too need to be added to­gether.

How­ever, all these de­tails are of lit­tle im­por­tance in find­ing a ball­park for the dom­i­nant de­cay process. The real re­main­ing prob­lem is sum­ming over the elec­tron and an­ti­neu­trino mo­men­tum states. The to­tal ball­parked de­cay rate must be found from

$$\lambda = \sum_{{\rm all\ }{\skew0\vec p}_e,{\skew0\vec p}_... ...1)t_c/\hbar\Big)} {\Big(\frac12(E_2-E_1)t_c/\hbar\Big)^2} t_c$$

Based on en­ergy con­ser­va­tion, you would ex­pect that de­cays should only oc­cur when the to­tal en­ergy $E_2$ of the nu­cleus, elec­tron and an­ti­neu­trino af­ter the de­cay is ex­actly equal to the en­ergy $E_1$ of the nu­cleus be­fore the de­cay. How­ever, the sum­ma­tion above shows that that is not quite true. For a fi­nal state that has $E_2$ ex­actly equal to $E_1$, the last frac­tion in the sum­ma­tion is seen to be unity, us­ing l’Hos­pi­tal. For a fi­nal state with an en­ergy $E_2$ of, for ex­am­ple, $E_1+\hbar/t_c$, the ra­tio is quite com­pa­ra­ble. There­fore de­cay to such a state pro­ceeds at a com­pa­ra­ble rate as to a state that con­serves en­ergy ex­actly. There is slop in en­ergy con­ser­va­tion.

How can en­ergy not be con­served? The rea­son is that nei­ther the ini­tial state nor the fi­nal state is an en­ergy eigen­state, strictly speak­ing. En­ergy eigen­states are sta­tion­ary states. The very fact that de­cay oc­curs as­sures that these states are not re­ally en­ergy eigen­states. They have a small amount of un­cer­tainty in en­ergy. The nonzero value of the Hamil­ton­ian co­ef­fi­cient $H_{21}^\ell$ as­sures that, chap­ter 5.3, and there may be more de­cay processes adding to the un­cer­tainty in en­ergy. If there is some un­cer­tainty in en­ergy, then $E_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E_1$ is not an ex­act re­la­tion­ship.

To nar­row this ef­fect down more pre­cisely, the frac­tion is plot­ted in fig­ure 7.7. The spikes in the fig­ure in­di­cate the en­er­gies $E_2$ of the pos­si­ble fi­nal states. Now the en­ergy states are al­most in­fi­nitely densely spaced, if the pe­ri­odic box in which the de­cay is as­sumed to oc­cur is big enough. And the box must be as­sumed very big any­way, to sim­u­late de­cay in in­fi­nite space. There­fore, the sum­ma­tion can be re­placed by in­te­gra­tion, as fol­lows:

$$\lambda = \int_{{\rm all\ }E_2} \frac{\vert H_{21}^\ell\ve... ..._c/\hbar\Big)^2} t_c \frac{{\rm d}N}{{\rm d}E_2} {\,\rm d}E_2$$

where ${\rm d}{N}$$\raisebox{.5pt}{$/$}$​${\rm d}{E_2}$ is the num­ber of fi­nal states per unit en­ergy range, of­ten called the den­sity of states $\rho(E_2)$.

Now as­sume that the com­plete prob­lem is cut into bite-size pieces for each of which $\vert H_{21}^\ell\vert$ is about con­stant. It can then be taken out of the in­te­gral. Also, the range of en­ergy in fig­ure 7.7 over which the frac­tion is ap­pre­cia­ble, the en­ergy slop, is very small on a nor­mal nu­clear en­ergy scale: beta de­cay is a slow process, so the ini­tial and fi­nal states do re­main en­ergy eigen­states to a very good ap­prox­i­ma­tion. En­ergy con­ser­va­tion is al­most ex­actly sat­is­fied. Be­cause of that, the den­sity of states ${\rm d}{N}$$\raisebox{.5pt}{$/$}$​${\rm d}{E_2}$ will be al­most con­stant over the range where the in­te­grand is nonzero. It can there­fore be taken out of the in­te­gral too. What is left can be in­te­grated an­a­lyt­i­cally, [41, 18.36]. That gives:

$$\lambda = \frac{\vert H_{21}^\ell\vert^2}{\hbar^2} t_c \frac{{\rm d}N}{{\rm d}E} \frac{2\pi\hbar}{t_c}$$

That is Fermi’s (sec­ond) golden rule. It de­scribes how en­ergy slop in­creases the to­tal de­cay rate. It is not spe­cific to beta de­cay but also ap­plies to other forms of de­cay to a con­tin­uum of states. Note that it no longer de­pends on the ar­ti­fi­cial length $t_c$ of the time in­ter­val over which the sys­tem was sup­posed to evolve with­out mea­sure­ment. That is good news, since that time in­ter­val was ob­vi­ously poorly de­fined.

Be­cause of the as­sump­tions in­volved, like di­vid­ing the prob­lem into bite-size pieces, the above ex­pres­sion is not very in­tu­itive to ap­ply. It can be rephrased into a more in­tu­itive form that does not de­pend on such an as­sump­tion. The ob­tained de­cay rate is ex­actly the same as if in an en­ergy slop range

$$\Delta E_{\rm slop} = \frac{2\pi\hbar}{t_c}$$

all states con­tribute just as much to the de­cay as one that sat­is­fies en­ergy con­ser­va­tion ex­actly, while no states con­tribute out­side of that range.

(Note that if you ball­park $t_c$ as the half-life, then even a half-life as short as 10$\POW9,{-15}$ s gives an en­ergy slop of a few eV, al­most im­pos­si­ble to mea­sure. And any nu­clear de­cay with a half life of 10$\POW9,{-15}$ s should surely pro­duce an amount of en­ergy equal to very many MeV. Out­side the math­e­mat­ics of the Fermi the­ory, the en­ergy slop is im­per­cep­ti­ble un­der nor­mal con­di­tions.)

The good news is that phrased this way, it in­di­cates the rel­e­vant physics much more clearly than the ear­lier purely math­e­mat­i­cal ex­pres­sion for Fermi’s golden rule. The bad news is that it suf­fers es­thet­i­cally from still in­volv­ing the poorly de­fined time $t$, in­stead of al­ready hav­ing shoved $t$ un­der the mat. There­fore, it is more ap­peal­ing to write things in terms of the en­ergy slop al­to­gether:

$$\fbox{$\displaystyle \lambda_{\rm single\ final\ state} = ... ...quiv \varepsilon \qquad \varepsilon t_c \sim 2\pi\hbar $} %$$ (A.279)

Here $\varepsilon$ is the amount that en­ergy con­ser­va­tion seems to be vi­o­lated, and is re­lated to a typ­i­cal time $t_c$ be­tween col­li­sions by the en­ergy-time un­cer­tainty re­la­tion­ship shown.

It may be noted that the golden rule does not ap­ply if the evo­lu­tion is not to a con­tin­uum of states. It also does not ap­ply if the slop range $\varepsilon$ is so large that ${\rm d}{N}$$\raisebox{.5pt}{$/$}$​${\rm d}{E}$ is not con­stant over it. And it does not ap­ply for sys­tems that evolve with­out be­ing per­turbed over times long enough that the de­cay prob­a­bil­ity be­comes sig­nif­i­cant be­fore the sys­tem is mea­sured. (If $\bar{b}$ be­comes ap­pre­cia­ble, $\bar{a}$ can no longer be close to one since the prob­a­bil­i­ties $\vert\bar{a}\vert^2$ and $\vert\bar{b}\vert^2$ must add to one.) Mea­sure­ments, or rather in­ter­ac­tions with the larger en­vi­ron­ment, are called col­li­sions. Fermi’s golden rule ap­plies to so-called col­li­sion-dom­i­nated con­di­tions. Typ­i­cally ex­am­ples where the con­di­tions are not col­li­sion dom­i­nated are in NMR and atomic de­cays un­der in­tense laser light.

Math­e­mat­i­cally, the con­di­tions for Fermi’s golden rule can be writ­ten as

$$\vert H_{21}\vert \ll \varepsilon \ll E \qquad \varepsilon \equiv \frac{2\pi\hbar}{t_c}$$ (A.280)

The first in­equal­ity means that the per­tur­ba­tion caus­ing the de­cay must be weak enough that there is only a small chance of de­cay be­fore a col­li­sion oc­curs. The sec­ond in­equal­ity means that there must be enough time be­tween col­li­sions that an ap­par­ent en­ergy con­ser­va­tion from ini­tial to fi­nal state ap­plies. Roughly speak­ing, col­li­sions must be suf­fi­ciently fre­quent on the time scale of the de­cay process, but rare on the quan­tum time scale $\hbar$$\raisebox{.5pt}{$/$}$​$E$.

It should also be noted that the rule was de­rived by Dirac, not Fermi. The way Fermi got his name on it was that he was the one who named it a golden rule. Fermi had a flair for find­ing mem­o­rable names. God knows how he ended up be­ing a physi­cist.

A.45.6 Mop­ping up

The pre­vi­ous sub­sec­tions de­rived the ba­sics for the rate of beta de­cay. The pur­pose of this sec­tion is to pull it all to­gether and get some ac­tual ball­park es­ti­mates for beta de­cay.

First con­sider the pos­si­ble val­ues for the mo­menta ${\skew0\vec p}_e$ and ${\skew0\vec p}_{\bar\nu}$ of the elec­tron and an­ti­neu­trino. Their wave func­tions were ap­prox­i­mately of the form

$$\psi_{{\skew0\vec p}} = \frac{1}{\sqrt{{\cal V}}} e^{{\rm i}(p_x x + p_y y + p_z z)/\hbar}$$

where ${\cal V}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell^3$ is the vol­ume of the pe­ri­odic box in which the de­cay is as­sumed to oc­cur.

In a pe­ri­odic box the wave func­tion must be the same at op­po­site sides of the box. For ex­am­ple, the ex­po­nen­tial fac­tor $e^{{\rm i}{p}_xx/\hbar}$ is 1 at $x$=0, and it must be 1 again at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell$. That re­quires ${p}_x\ell$$\raisebox{.5pt}{$/$}$​$\hbar$ to be a whole mul­ti­ple of $2\pi$. There­fore $p_x$ must be a whole mul­ti­ple of $2\pi\hbar$$\raisebox{.5pt}{$/$}$​$\ell$. Suc­ces­sive pos­si­ble $p_x$ val­ues are there­fore spaced the fi­nite amount $2\pi\hbar$$\raisebox{.5pt}{$/$}$​$\ell$ apart. And so are suc­ces­sive $p_y$ and $p_z$ val­ues.

Fig­ure A.27: Pos­si­ble mo­men­tum states for a par­ti­cle con­fined to a pe­ri­odic box. The states are shown as points in mo­men­tum space. States that have mo­men­tum less than some ex­am­ple max­i­mum value are in red.

Graph­i­cally this can by vi­su­al­ized by plot­ting the pos­si­ble mo­men­tum val­ues as points in a three-di­men­sion­al $p_x,p_y,p_z$ axis sys­tem. That is done in fig­ure A.27. Each point cor­re­spond to one pos­si­ble mo­men­tum state. Each point is the cen­ter of its own lit­tle cube with sides $2\pi\hbar$$\raisebox{.5pt}{$/$}$​$\ell$. The vol­ume (in this three-di­men­sion­al mo­men­tum plot, not phys­i­cal vol­ume) of that lit­tle cube is $(2\pi\hbar/\ell)^3$. Since $\ell^3$ is the phys­i­cal vol­ume ${\cal V}$ of the pe­ri­odic box, the vol­ume in mo­men­tum space taken up by each mo­men­tum state is $(2\pi\hbar)^3$$\raisebox{.5pt}{$/$}$​${\cal V}$

That al­lows the num­ber of dif­fer­ent mo­men­tum states to be com­puted. In par­tic­u­lar, con­sider how many states have mag­ni­tude of mo­men­tum $\vert{\skew0\vec p}\,'\vert$ less than some max­i­mum value $p$. For some ex­am­ple value of $p$, these are the red states in fig­ure A.27. Note that they form a sphere of ra­dius $p$. That sphere has a vol­ume equal to $\frac43\pi{p}^3$. Since each state takes up a vol­ume $(2\pi\hbar)^2$$\raisebox{.5pt}{$/$}$​${\cal V}$, the num­ber of states $N$ is given by the num­ber of such vol­umes in the sphere:

$$N_{\vert{\skew0\vec p}\,'\vert\mathrel{\raisebox{-.5pt}{$\s... ...le\leqslant$}}p}=\frac{\frac43\pi p^3}{(2\pi\hbar)^3/{\cal V}}$$

The num­ber of elec­tron states that have mo­men­tum in a range from $p_e$ to $p_e+{\rm d}{p_e}$ can be found by tak­ing a dif­fer­en­tial of the ex­pres­sion above:

$${\rm d}N_e = \frac{{\cal V}p_e^2}{2 \pi^2\hbar^3} {\rm d}p_e$$

(Here the range ${\rm d}{p}_e$ is as­sumed small, but not so small that the fact that the num­ber of states is dis­crete would show up.) Each mo­men­tum state still needs to be mul­ti­plied by the num­ber of cor­re­spond­ing an­ti­neu­trino states to find the num­ber of states of the com­plete sys­tem.

Now the ki­netic en­ergy of the an­ti­neu­trino $T_{\bar\nu}$ is fixed in terms of that of the elec­tron and the en­ergy re­lease of the de­cay $Q$ by:

$$T_{\bar\nu} = Q - T_e$$

Here the ki­netic en­ergy of the fi­nal nu­cleus is ig­nored. The heavy fi­nal nu­cleus is un­selfish enough to as­sure that mo­men­tum con­ser­va­tion is sat­is­fied for what­ever the mo­menta of the elec­tron and an­ti­neu­trino are, with­out de­mand­ing a no­tice­able share of the en­ergy for it­self. That is much like Mother Earth does not take any of the ki­netic en­ergy away if you shoot rock­ets out to space from dif­fer­ent lo­ca­tions. You might write equa­tions down for it, but the only thing they are go­ing to tell you is that it is true as long as the speed of the nu­cleus does not get close to the speed of light. Beta de­cays do not re­lease that much en­ergy by far.

The elec­tron and an­ti­neu­trino ki­netic en­er­gies are re­lated to their mo­menta by Ein­stein’s rel­a­tivis­tic ex­pres­sion, chap­ter 1.1.2:

$$T_e = \sqrt{(m_{\rm e}c^2)^2 + p_e^2 c^2} - m_{\rm e}c^2 \qquad T_{\bar\nu} = p_{\bar\nu} c %$$ (A.281)

where $c$ is the speed of light and the ex­tremely small rest mass of the neu­trino was ig­nored. With the neu­trino en­ergy fixed, so is the mag­ni­tude of the neu­trino mo­men­tum:

$$p_{\bar\nu} = \frac{1}{c}(Q - T_e)$$

These re­sult shows that the neu­trino mo­men­tum is fixed for given elec­tron mo­men­tum $p_e$. There­fore there should not be a neu­trino mo­men­tum range ${\rm d}{p}_{\bar\nu}$ and so no neu­trino states. How­ever, Fermi’s golden rule says that the the­o­ret­i­cal en­ergy af­ter the de­cay does not need to be ex­actly the same as the one be­fore it, be­cause both en­er­gies have a bit of un­cer­tainty. This slop in the en­ergy con­ser­va­tion equa­tion al­lows a range of en­er­gies

$$\Delta E_{\rm slop} = \Delta T_{\bar\nu} = \Delta p_{\bar\nu} c \equiv \varepsilon$$

There­fore the to­tal amount of neu­trino states for a given elec­tron mo­men­tum is not zero, but

$$\Delta N_{\bar\nu} = \frac{{\cal V}p_{\bar\nu}^2}{2 \pi^2\h... ...ac{1}{c} \varepsilon \qquad p_{\bar\nu} = \frac{1}{c} (Q-T_e)$$

The num­ber of com­plete sys­tem states in an elec­tron mo­men­tum range ${\rm d}{p}_e$ is the prod­uct of the num­ber of elec­tron states times the num­ber of an­ti­neu­trino states:

$${\rm d}N = {\rm d}N_e \Delta N_{\bar\nu} = \frac{{\cal V}^... ...i^4\hbar^6 c}\, p_e^2 p_{\bar\nu}^2\, \varepsilon {\,\rm d}p_e$$

Each of these states adds a con­tri­bu­tion to the spe­cific de­cay rate given by

$$\lambda_{\rm single\ final\ state} = \frac{2\pi}{\hbar\varepsilon} \vert H_{21}^\ell\vert^2$$

There­fore the to­tal spe­cific de­cay rate is

$$\lambda = \int_{p_e=0}^{p_{e,\rm max}} \frac{{\cal V}^2\ve... ...ell\vert^2}{2 \pi^3\hbar^7 c} p_e^2 p_{\bar\nu}^2 {\,\rm d}p_e$$

where the max­i­mum elec­tron mo­men­tum $p_{e,max}$ can be com­puted from the $Q$-​value of the de­cay us­ing (A.281). (For sim­plic­ity it will be as­sumed that $\vert H_{21}^\ell\vert^2$ has al­ready been av­er­aged over all di­rec­tions of the elec­tron and an­ti­neu­trino mo­men­tum.)

The de­rived ex­pres­sion (A.277) for the Hamil­ton­ian co­ef­fi­cient $H_{21}^\ell$ can be writ­ten in the form

$$\vert H_{21}^\ell\vert^2 = \frac{g^2}{{\cal V}^2} \frac{1}{... ...c{\sqrt{p_e^2+p_{\bar\nu}^2} R}{\hbar}\right)^{2\ell} C_N^\ell$$

$$C_N^\ell \equiv \overline{ \bigg\vert\bigg\langle \psi_{2... ...} R} \bigg)^\ell \psi_{1,\rm nuc}\bigg\rangle\bigg\vert^2 }$$

where the over­line in­di­cates some suit­able av­er­age over the di­rec­tions of the elec­tron and an­ti­neu­trino mo­menta.

It is not easy to say much about $C_N^\ell$ in gen­eral, be­yond the fact that its mag­ni­tude should not be much more than one. This book will es­sen­tially ig­nore $C_N^\ell$ to ball­park the de­cay rate, as­sum­ing that its vari­a­tion will surely be much less than that of the beta de­cay life­times, which vary from mil­lisec­onds to 10$\POW9,{17}$ year.

The de­cay rate be­comes af­ter clean up

$$\lambda = \frac{1}{2\pi^3} \frac{g^2m_{\rm e}^4 c^2}{\hba... ... \tilde p_e^2 \tilde p_{\bar\nu}^2 F^\ell {\rm d}\tilde p_e %$$ (A.282)

where the $\tilde{p}$ in­di­cate the elec­tron and an­ti­neu­trino mo­menta nondi­men­sion­al­ized with ${m_{\rm e}}c$. Also,

$$\tilde Q \equiv \frac{Q}{m_{\rm e}c^2} \quad \tilde p_{e,... ..._e^2} + 1 \quad \tilde R \equiv \frac{m_{\rm e}c R}{\hbar} %$$ (A.283)

Here $\tilde{Q}$ is the $Q$-​value or ki­netic en­ergy re­lease of the de­cay in units of the elec­tron rest mass, and the next two re­la­tions fol­low from the ex­pres­sion for the rel­a­tivis­tic ki­netic en­ergy. The vari­able $\tilde{R}$ a suit­ably nondi­men­sion­al­ized nu­clear ra­dius, and is small.

The fac­tor $F^\ell$ that popped up out of noth­ing in the de­cay rate is thrown in to cor­rect for the fact that the wave func­tion of the elec­tron is not re­ally just an ex­po­nen­tial. The nu­cleus pulls on the elec­tron with its charge, and so changes its wave func­tion lo­cally sig­nif­i­cantly. The cor­rec­tion fac­tor $F^0$ for al­lowed de­cays is called the “Fermi func­tion” and is given by

$$F(\tilde p_e,Z_2,A) = \frac{2(1+\xi)}{\Gamma^2(1+2\xi)} \f... ...R)^{2-2\xi}} e^{\pi\eta}\vert\Gamma(\xi+{\rm i}\eta)\vert^2 %$$ (A.284)

$$\xi \equiv \sqrt{1-(\alpha Z_2)^2} \quad \eta \equiv \alp... ...\alpha = \frac{e^2}{4\pi\epsilon_0\hbar c} \approx \frac1{137}$$

where $\alpha$ is the fine struc­ture con­stant and $\Gamma$ the gamma func­tion. The non­rel­a­tivis­tic ver­sion fol­lows for let­ting the speed of light go to in­fin­ity, while keep­ing $p_e$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${m_{\rm e}}c\tilde{p}_e$ fi­nite. That gives $\xi$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and

$$F(p_e,Z_2) = \frac{2\pi\eta}{1-e^{-2\pi\eta}} \qquad \eta = \frac{e^2}{4\pi\epsilon_0\hbar} \frac{m_{\rm e}}{p_e}$$

For beta-plus de­cay, just re­place $Z_2$ by $-Z_2$, be­cause an elec­tron is just as much re­pelled by a neg­a­tively charged nu­cleus as a positron is by a pos­i­tively charged one.

To ball­park the ef­fect of the nu­clear charge on the elec­tron wave func­tion, this book will use the rel­a­tivis­tic Fermi func­tion above whether it is an al­lowed de­cay or not.

For al­lowed de­cays, the fac­tor in the de­cay rate that is gov­erned by the $Q$-​value and nu­clear charge is

$$f = \int_{\tilde p_e=0}^{\tilde p_{e,\rm max}} \tilde p_e^2 \tilde p_{\bar\nu}^2 F {\rm d}\tilde p_e %$$ (A.285)

This quan­tity is known as the “Fermi in­te­gral.” Typ­i­cal val­ues are shown in fig­ure 14.52.

Note that $f$ also de­pends a bit on the mass num­ber through the nu­clear ra­dius in $F$. The fig­ure used

$$A = 1.82+1.9\,Z_2 + 0.012\,71\,Z_2^2- 0.000\,06\,Z_2^3$$ (A.286)

for beta-mi­nus de­cay and

$$A = -1.9+1.96\,Z_2 + 0.007\,9\,Z_2^2- 0.000\,02\,Z_2^3$$ (A.287)

for beta-plus de­cay, [23].

A.45.7 Elec­tron cap­ture

Elec­tron cap­ture is much more sim­ply to an­a­lyze than beta de­cay, be­cause the cap­tured elec­tron is in a known ini­tial state.

It will be as­sumed that a 1s, or K-shell, elec­tron is cap­tured, though L-shell cap­ture may also con­tribute to the de­cay rate for heavy nu­clei. The Hamil­ton­ian co­ef­fi­cient that dri­ves the de­cay is

$$H_{21} = \langle 1\bar\nu,{\skew0\vec p}_{\bar\nu}\vert\la... ...1e,1{\rm s}\rangle\vert\bar\nu,{\skew0\vec p}_{\bar\nu}\rangle$$

In this case, it is an elec­tron an­ni­hi­la­tion term in the Hamil­ton­ian that will pro­duce a nonzero term. How­ever, the re­sult will be the pretty much same; the Hamil­ton­ian co­ef­fi­cient sim­pli­fies to

$$H_{21} = \frac{g}{\sqrt{{\cal V}}}\langle \psi_{2,\rm nuc} ... ...{\bar\nu}\cdot{\skew0\vec r}_i/\hbar} \psi_{1,\rm nuc}\rangle$$

Here $\psi_{100}$ is the hy­dro­gen ground state wave func­tion, but rescaled for a nu­cleus of charge $Ze$ in­stead of $e$. It does not con­tribute to mak­ing for­bid­den de­cays pos­si­ble, be­cause $\psi_{100}$ is spher­i­cally sym­met­ric. In other words, the 1s elec­tron has no or­bital an­gu­lar mo­men­tum and so can­not con­tribute to con­ser­va­tion of an­gu­lar mo­men­tum and par­ity. There­fore, $\psi_{100}$ can safely be ap­prox­i­mated by its value at the ori­gin, from chap­ter 4.3,

$$\psi_{100}({\skew0\vec r}_i) \approx \frac{1}{\sqrt{\pi a_0... ...\hbar^2}{m_{\rm e}e^2Z_1} = \frac{\hbar}{m_{\rm e}c\alpha Z_1}$$

where $\alpha$ is the fine-struc­ture con­stant.

The square Hamil­ton­ian co­ef­fi­cient for $\ell$th-for­bid­den de­cays then be­comes

$$\vert H_{21}^\ell\vert^2 = \frac{g^2}{{\cal V}} \frac{m_{... ...)^2} \left(\frac{p_{\bar\nu}R}{\hbar}\right)^{2\ell} C_N^\ell$$

$$C_N^\ell = \equiv \overline{ \bigg\vert\bigg\langle \psi_... ...u} R} \bigg)^\ell \psi_{1,\rm nuc}\bigg\rangle\bigg\vert^2 }$$

The de­cay rate for elec­tron cap­ture is

$$\lambda = \frac{2\pi}{\hbar\varepsilon}\vert H_{21}^\ell\v... ... p_{\bar\nu} \qquad \Delta p_{\bar\nu} = \frac{\varepsilon}{c}$$

where the first ra­tio is the de­cay rate of a sin­gle state, with $\varepsilon$ the en­ergy slop im­plied by Fermi’s golden rule.

Put it all to­gether, in­clud­ing the fact that there are two K elec­trons, and the elec­tron-cap­ture de­cay rate be­comes

$$\lambda = \frac{2}{\pi^2} \frac{g^2m_{\rm e}^4c^2}{\hbar^... ... C_N^\ell \tilde p_{\bar\nu}^{2\ell} \tilde p_{\bar\nu}^2 %$$ (A.288)

where the $\tilde{p}_{\bar\nu}$ in­di­cate the neu­trino mo­men­tum nondi­men­sion­al­ized with ${m_{\rm e}}c$. Also,

$$\tilde Q \equiv \frac{Q}{m_{\rm e}c^2} \qquad \tilde p_{\... ...\tilde Q \qquad \tilde R \equiv \frac{m_{\rm e}c R}{\hbar} %$$ (A.289)

for the co­ef­fi­cients.