A.42 Nu­clear forces

The pur­pose of this ad­den­dum is to ex­am­ine the na­ture of nu­clear forces some­what closer. The forces will be mod­eled us­ing the me­son ex­change idea. This idea il­lus­trates one pri­mary way that physi­cists cope with the fact that nu­clei are too com­plex to de­scribe ex­actly.

A.42.1 Ba­sic Yukawa po­ten­tial

As pointed out in chap­ter 7.5.2, the fun­da­men­tal forces of na­ture be­tween el­e­men­tary par­ti­cles are due to the ex­change of bosons be­tween these par­ti­cles. In those terms, nu­clei con­sist of quarks. The ex­change of glu­ons be­tween these quarks pro­duces the so-called color force. It is that force that holds nu­clei to­gether. Un­for­tu­nately, de­scrib­ing that math­e­mat­i­cally is not a prac­ti­cal propo­si­tion. Quan­tum chrom­e­dy­nam­ics is pro­hib­i­tively dif­fi­cult.

But you will never find quarks or glu­ons in iso­la­tion. Quarks and their glu­ons are al­ways con­fined in­side col­or­less com­bi­na­tions of two or three quarks. (To be painstak­ingly hon­est, there might be more ex­otic col­or­less com­bi­na­tions of quarks and glu­ons than that. But their en­ergy should be too high to worry about here.) What is ob­served phys­i­cally at the time of writ­ing, 2012, are groups of three quarks, (baryons), three an­ti­quarks, (an­tibaryons), and a quark and an an­ti­quark (mesons). An eas­ier de­scrip­tion of nu­clear forces can be based on these groups of quarks.

In this pic­ture, nu­clei can be taken to con­sist of nu­cle­ons. A nu­cleon con­sists of a group of three quarks, so it is a baryon. There are two types of nu­cle­ons: pro­tons and neu­trons. A pro­ton con­tains two up quarks, at elec­tric charge $\frac23e$ each, and one down quark, at $-\frac13e$. That makes the net charge of a pro­ton $\frac23e+\frac23e-\frac13e$ equal to $e$. A neu­tron has one up quark and two down ones, mak­ing its net charge $\frac23e-\frac13e-\frac13e$ equal to zero.

For both pro­tons and neu­trons, the group of three quarks is in its ground state, much like a he­lium atom is nor­mally in its ground state. Like sin­gle quarks, nu­cle­ons are fermi­ons with spin equal to $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$. (Roughly speak­ing, two of the three quarks in nu­cle­ons align their spins in op­po­site di­rec­tions, caus­ing them to can­cel each other.) Nu­cle­ons have pos­i­tive in­trin­sic par­ity. That means that their mere pres­ence does not pro­duce a change in sign in the wave func­tion when the co­or­di­nate sys­tem is in­verted, chap­ter 7.3. (Ac­tu­ally, there is some am­bi­gu­ity in the as­sign­ment of in­trin­sic par­ity to par­ti­cles. But a fermion and the cor­re­spond­ing an­tifermion must have op­po­site par­ity. Tak­ing the par­ity of fermi­ons pos­i­tive makes that of the cor­re­spond­ing an­tifermi­ons neg­a­tive.)

Pro­tons and neu­trons com­bine to­gether into nu­clei. How­ever, the pro­tons in nu­clei re­pel each other be­cause of their elec­tric charges. So there must be some com­pen­sat­ing force that keeps the nu­cle­ons to­gether any­way. This force is what is called the “nu­clear force.” The ques­tion in this ad­den­dum is how this nu­clear force can be de­scribed. Its phys­i­cal cause is still the force due to the ex­change of glu­ons be­tween quarks. But its math­e­mat­i­cal de­scrip­tion is go­ing to be dif­fer­ent. The rea­son is that by de­f­i­n­i­tion the nu­clear force is a net force on nu­cle­ons, i.e. on groups of quarks. And it is as­sumed to de­pend on the av­er­age po­si­tions, and pos­si­bly mo­menta, of these groups of quarks.

Note that there is some ap­prox­i­ma­tion in­volved here. Ex­actly speak­ing, the nu­clear forces should de­pend on the po­si­tions of the in­di­vid­ual quarks in the nu­cle­ons, not just on their av­er­age po­si­tion. That is a con­cern when two nu­cle­ons get very close to­gether. For one, then the dis­tinc­tion be­tween the two sep­a­rate groups of quarks must blur. Nu­cle­ons do re­pel one an­other strongly at very close dis­tances, much like atoms do due to Pauli re­pul­sion, chap­ter 5.10. But still their quan­tum un­cer­tainty in po­si­tion cre­ates a prob­a­bil­ity for them to be very close to­gether. For­tu­nately, typ­i­cal en­ergy lev­els in nor­mal nu­clear physics are low enough that this is not be­lieved to be a dom­i­nat­ing ef­fect. In­deed, the mod­els dis­cussed here are known to work very well at larger nu­cleon spac­ings. For smaller nu­cleon spac­ing how­ever, they be­come much more com­plex, and their ac­cu­racy much more un­cer­tain. And that hap­pens well be­fore the nu­cle­ons start in­trud­ing sig­nif­i­cantly on each oth­ers space. Lit­tle in life is ideal, isn’t it?

In a par­ti­cle ex­change ex­pla­na­tion of the nu­clear force, roughly speak­ing nu­cle­ons have to pop up par­ti­cles that other nu­cle­ons then ab­sorb and vice-versa. The first ques­tion is what these par­ti­cles would be. As al­ready men­tioned, only col­or­less com­bi­na­tions of quarks and their glu­ons are ob­served in iso­la­tion. There­fore only such col­or­less com­bi­na­tions can be ex­pected to be able to read­ily bridge the gap be­tween nu­cle­ons that are rel­a­tively far apart. The low­est en­ergy of these col­or­less com­bi­na­tions are the eas­i­est to pop up. And that are the pi­ons; a pion is a me­son con­sist­ing of a quark and an­ti­quark pair in its ground state.

There are three types of pi­ons. The $\pi^+$ pion con­sists of an up quark plus an an­ti­d­own quark. An­tipar­ti­cles have the op­po­site charge from the cor­re­spond­ing par­ti­cles, so the an­ti­d­own quark has charge $\frac13e$. That makes the net charge of the $\pi^+$ pion $\frac23e+\frac13e$ equal to $e$, the same as that of the pro­ton. The $\pi^-$ pion con­sists of an an­tiup quark plus a down quark, pro­duc­ing a net charge $-\frac23e-\frac13e$ equal to $\vphantom{0}\raisebox{1.5pt}{$-$}$$e$. That is as it should be since self-ev­i­dently the $\pi^-$ is the an­tipar­ti­cle of the $\pi^+$. The $\pi^0$ pion is a quan­tum su­per­po­si­tion of an up-an­tiup pair and a down-an­ti­d­own pair and is elec­tri­cally neu­tral.

Pi­ons are bosons of zero spin and neg­a­tive in­trin­sic par­ity. The neg­a­tive par­ity is due to the an­ti­quark, and zero spin is due to the fact that in pi­ons the quark and an­ti­quark align their spins in op­po­site di­rec­tions in a sin­glet state, chap­ter 5.5.6.

These pi­ons are the most im­por­tant par­ti­cles that pro­tons and neu­trons ex­change. The first ques­tion is then of course where they come from. How is it pos­si­ble that pi­ons just ap­pear out of noth­ing? Well, it is pos­si­ble due to a mix­ture of spe­cial rel­a­tiv­ity and the un­cer­tainty in­her­ent in quan­tum me­chan­ics.

The cre­ation of par­ti­cles out of en­ergy is al­lowed by spe­cial rel­a­tiv­ity. As dis­cussed in chap­ter 1.1.2, spe­cial rel­a­tiv­ity gives the en­ergy $E$ of a par­ti­cle as:

$$E = \sqrt{{\skew0\vec p}^{ 2}c^2 + \big(mc^2\big)^2}$$

Here $c$ is the speed of light, ${\skew0\vec p}$ the mo­men­tum of the par­ti­cle, and $m$ its mass (at rest). Ac­cord­ing to this ex­pres­sion, a par­ti­cle at rest rep­re­sents an amount of en­ergy equal to $mc^2$. This is the rest mass en­ergy. The charged $\pi^+$ and $\pi^-$ pi­ons have a rest mass en­ergy of about 140 MeV, and the neu­tral $\pi^0$ 135 MeV. So to cre­ate an ac­tual pion re­quires at least 135 MeV of en­ergy.

Quan­tum me­chan­ics re­places the mo­men­tum ${\skew0\vec p}$ in the en­ergy above by the op­er­a­tor $\hbar\nabla$$\raisebox{.5pt}{$/$}$​${\rm i}$ in or­der to find the Hamil­ton­ian. Then it ap­plies that Hamil­ton­ian to a pion wave func­tion $\varphi_\pi$. But the square root in the above ex­pres­sion is a prob­lem. For­tu­nately, for spin­less bosons like pi­ons an ac­cept­able so­lu­tion is easy: just square the en­ergy. Or rather, ap­ply the Hamil­ton­ian twice. That pro­duces the rel­a­tivis­tic so-called Klein-Gor­don eigen­value prob­lem

$$- \hbar^2 c^2 \nabla^2 \varphi_\pi + \Big(m_\pi c^2\Big)^2\varphi_\pi = E^2\varphi_\pi %$$ (A.260)

Now con­sider first a sin­gle nu­cleon lo­cated at the ori­gin. Sup­pos­edly this nu­cleon can pop up a pion. But where would the nu­cleon get the 135 MeV or more of en­ergy? Surely, if there was a prob­a­bil­ity of ac­tu­ally find­ing a 135 MeV pion well away from the nu­cleon, it would vi­o­late en­ergy con­ser­va­tion. But re­mark­ably, de­spite the pos­i­tive pion rest mass en­ergy, the Klein-Gor­don equa­tion has a sim­ple so­lu­tion where the to­tal pion en­ergy $E$ ap­pears to be zero:

$$\varphi_\pi = C \frac{e^{-r/R}}{r} \qquad R \equiv \frac{\hbar c}{m_\pi c^2} \approx 1.4\mbox{ fm}$$

Here $r$ is the dis­tance from the nu­cleon and $C$ an ar­bi­trary con­stant. In ef­fect, this so­lu­tion has a big neg­a­tive ki­netic en­ergy. You might say that a zero-en­ergy pion tun­nels out of the nu­cleon, chap­ter 7.12.2.

To check the above so­lu­tion, just plug it in the Klein-Gor­don equa­tion (A.260) with $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, us­ing the ex­pres­sion (N.5) for the Lapla­cian $\nabla^2$ found in the no­ta­tions. But to be true, this sub­sti­tu­tion is some­what mis­lead­ing. A more care­ful analy­sis shows that the left hand side in the Klein-Gor­don equa­tion does have a nonzero spike at $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, {D.2.2}. But there the pion will ex­pe­ri­ence an in­ter­ac­tion en­ergy with the nu­cleon.

Now as­sume that the nu­cleon does in­deed man­age to cre­ate a pion field around it­self. A field that acts as a po­ten­tial $\varphi$ that can pro­duce forces on other nu­cle­ons. That would be much like a charged par­ti­cle cre­ates an elec­tro­sta­tic po­ten­tial that can pro­duce forces on other charged par­ti­cles. Then it seems a plau­si­ble guess that the pion po­ten­tial $\varphi$ will vary with po­si­tion just like the zero-en­ergy wave func­tion $\varphi_\pi$ above. Look at elec­tro­mag­net­ics. The pho­ton of elec­tro­mag­net­ics has zero rest mass. And for zero rest mass, the zero-en­ergy wave func­tion above be­comes the cor­rect $C$$\raisebox{.5pt}{$/$}$​$r$ Coulomb po­ten­tial of elec­tro­mag­net­ics.

Ac­tu­ally, de­spite the fact that it works for elec­tro­mag­net­ics, the zero-en­ergy wave func­tion above does not quite give the right form for a pion po­ten­tial. But it does give the gen­eral idea. The cor­rect po­ten­tial is dis­cussed in the next sub­sec­tions. This sub­sec­tion will stick with the form above as qual­i­ta­tively OK.

Now con­sider a sec­ond nu­cleon. This nu­cleon will of course also cre­ate a pion po­ten­tial. That is just like if it was all by it­self. But in ad­di­tion, it will in­ter­act with the pion po­ten­tial cre­ated by the first nu­cleon. So there will be an en­ergy of in­ter­ac­tion be­tween the nu­cle­ons. Tak­ing an­other cue from elec­tro­mag­net­ics, this en­ergy should pre­sum­ably be pro­por­tional to the po­ten­tial that the first nu­cleon cre­ates at the po­si­tion of the sec­ond nu­cleon.

That idea then gives the in­ter­ac­tion en­ergy be­tween two nu­cle­ons as

$$\fbox{$\displaystyle V_{\rm Yukawa} = - C_{\rm Y} \frac{e^... ...R \equiv \frac{\hbar c}{m_\pi c^2} \approx 1.4\mbox{ fm} $} %$$ (A.261)

Here $r$ is the dis­tance be­tween the two nu­cle­ons, and $C_{\rm {Y}}$ is some pos­i­tive con­stant that must be de­ter­mined ex­per­i­men­tally. The above in­ter­ac­tion en­ergy is called the “Yukawa po­ten­tial” af­ter the Japan­ese physi­cist who first de­rived it. It is re­ally a po­ten­tial en­ergy, rather than a po­ten­tial. (At least in the ter­mi­nol­ogy of this book for fields. In physics, pretty much every­thing is called a po­ten­tial.)

The Yukawa po­ten­tial is at­trac­tive. This is in con­trast to the Coulomb po­ten­tial, which is re­pul­sive be­tween like charges. The best phys­i­cal ex­pla­na­tion for the dif­fer­ence may be the analy­sis in {A.22}, in par­tic­u­lar {A.22.5}. (There are many other ex­pla­na­tions that de­rive the dif­fer­ence us­ing an elec­tro­mag­netic Hamil­ton­ian or La­grangian that al­ready has the dif­fer­ence build in. But a de­riva­tion is not an ex­pla­na­tion.)

Note the ex­po­nen­tial in the Yukawa po­ten­tial. It will make the po­ten­tial neg­li­gi­bly small as soon as the dis­tance $r$ be­tween the nu­cle­ons is sig­nif­i­cantly greater than $R$. With ${\hbar}c$ about 197 MeV fm and the av­er­age pion rest mass en­ergy about 138 MeV, $R$ is about 1.4 fm (fem­tome­ter). So un­less the nu­cle­ons are within a dis­tance not much greater than 1.4 fm from each other, they do not ex­pe­ri­ence a nu­clear force from each other. Yukawa had de­rived the typ­i­cal range of the nu­clear force.

Ac­tu­ally, at the time that Yukawa did his work, the pion was un­known. But the range of the nu­clear force was fairly well es­tab­lished. So Yukawa re­ally pre­dicted the ex­is­tence, as well as the mass of the pion, a then un­known par­ti­cle! Af­ter a long and frus­trat­ing search, this par­ti­cle was even­tu­ally dis­cov­ered in cos­mic rays.

The Yukawa po­ten­tial also ex­plained why heavy nu­clei are un­sta­ble. Sup­pose that you keep stuff­ing nu­cle­ons, and in par­tic­u­lar pro­tons, into a nu­cleus. Be­cause of the ex­po­nen­tial in the Yukawa po­ten­tial, the nu­clear force is very short range. It is largely gone be­yond dis­tances of a cou­ple of fm. So a pro­ton gets pulled into the nu­cleus only by the nu­cle­ons in its im­me­di­ate vicin­ity. But the Coulomb re­pul­sion be­tween pro­tons does not have the ex­po­nen­tial de­cay. So the same pro­ton gets pushed out of the nu­cleus by pro­tons from all over the nu­cleus. If the nu­cleus is big enough, the push­ers sim­ply have to win be­cause of their much larger num­bers.

Putting a lot of neu­trons in the nu­cleus can help, be­cause they pro­duce nu­cleon at­trac­tion and no Coulomb re­pul­sion. But neu­trons by them­selves are un­sta­ble. Put too many neu­trons in a nu­cleus, and they will turn into pro­tons by beta de­cay. Ob­vi­ously, that de­feats the pur­pose. As a re­sult, be­yond a cer­tain size, the nu­cleus is go­ing to fall apart what­ever you do.

You can see why Yukawa would end up with the No­bel prize in physics.

A.42.2 OPEP po­ten­tial

The Yukawa po­ten­tial en­ergy (A.261) de­scribed in the pre­vi­ous sec­tion is not quite right yet. It does not give the true nu­clear force be­tween two nu­cle­ons pro­duced by pion ex­change.

In a more care­ful analy­sis, the po­ten­tial en­ergy de­pends crit­i­cally on the prop­er­ties of the ex­changed par­ti­cle. See the next sub­sec­tion for an ex­pla­na­tion of that. For a pion, the rel­e­vant prop­er­ties are that it has zero spin and neg­a­tive par­ity. Tak­ing that into ac­count pro­duces the so-called “one-pion ex­change po­ten­tial” en­ergy or “OPEP” for short:

$$\fbox{$\displaystyle V_{\rm OPEP} \sim \frac{g_\pi^2}{12}... ...\sigma_2 + S_{12} V_{\rm{T}} \Big] \frac{e^{-r/R}}{r/R} $} %$$ (A.262)

$$S_{12} \equiv \frac{3}{r^2}(\vec\sigma_1\cdot{\skew0\vec r... ...\qquad V_{\rm {T}} \equiv 1 + 3\frac{R}{r} + 3\frac{R^2}{r^2}$$

Here $r$ is again the dis­tance be­tween nu­cle­ons 1 and 2, equal to the length of the vec­tor ${\skew0\vec r}$ con­nect­ing the two nu­cle­ons, $R$ is again the typ­i­cal range of the nu­clear force, $m_\pi$ again the pion mass, while $m_{\rm p}$ is the nu­cleon mass:

$$r \equiv \vert{\skew0\vec r}\vert \equiv \vert{\skew0\vec r... ...rox 138 \mbox{ MeV} \quad m_{\rm p}c^2 \approx 938 \mbox{ MeV}$$

Also $g_\pi^2$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ 15 is an em­pir­i­cal con­stant, [36, p. 135], [5, p. 85]. Fur­ther $\vec\sigma_1$ and $\vec\sigma_2$ are the nu­cleon spins ${\skew 6\widehat{\vec S}}_1$ and ${\skew 6\widehat{\vec S}}_2$, nondi­men­sion­al­ized by di­vid­ing by $\frac12\hbar$.

Fi­nally the dot prod­uct $\vec\tau_1\cdot\vec\tau_2$ in­volves the so-called isospin of the nu­cle­ons. Isospin will be dis­cussed in chap­ter 14.18. There it will be ex­plained that it has noth­ing to do with spin. In­stead isospin is re­lated to nu­cleon type. In par­tic­u­lar, if both nu­cle­ons in­volved are pro­tons, or if both are neu­trons, then $\vec\tau_1\cdot\vec\tau_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.

If one nu­cleon is a pro­ton and the other a neu­tron, like in the deuteron, the value of $\vec\tau_1\cdot\vec\tau_2$ can vary. But in any case, it is re­lated to the sym­me­try of the spa­tial and spin states. In par­tic­u­lar, com­pare also chap­ter 5.5.6 and {A.10},

$$\begin{eqnarray*} \mbox{symmetric spatially:} &&\displaystyle \vec\sigma_1\cdo... ...w& \vec\tau_1\cdot\vec\tau_2 = \phantom{-}1 \end{array} \right. \end{eqnarray*}$$

For the deuteron, as well as for the hy­po­thet­i­cal dipro­ton and dineu­tron, the spa­tial state is sym­met­ric un­der nu­cleon ex­change. That is as you would ex­pect for a ground state, {A.8} and {A.9}. It then fol­lows from the above val­ues that the first, $\vec\sigma_1\cdot\vec\sigma_2$, term in the square brack­ets in the OPEP (A.262) pro­duces a neg­a­tive, at­trac­tive, po­ten­tial for these nu­clei. That is true re­gard­less whether the spin state is sin­glet or triplet.

The sec­ond, $S_{12}V_{\rm {T}}$, term in the OPEP is called a ten­sor po­ten­tial, {A.41.4}. This po­ten­tial can cre­ate un­cer­tainty in or­bital an­gu­lar mo­men­tum. As dis­cussed in {A.41.4}, hav­ing a ten­sor po­ten­tial is an es­sen­tial part in get­ting the deuteron bound. But the ten­sor po­ten­tial is zero for the sin­glet spin state. And the spin state must be the sin­glet one for the dipro­ton and dineu­tron, to meet the an­ti­sym­metriza­tion re­quire­ment for the two iden­ti­cal nu­cle­ons. So the dipro­ton and dineu­tron are not bound.

The deuteron how­ever can be in the triplet spin state. In that case the ten­sor po­ten­tial is not zero. To be sure, the ten­sor po­ten­tial does av­er­age out to zero over all di­rec­tions of ${\skew0\vec r}$. But that does not pre­vent at­trac­tive niches to be found. And note how big the mul­ti­ply­ing fac­tor $V_{\rm {T}}$ is for $R$ about 1.4 fm and nu­cleon spac­ings $r$ down to say 2 fm. The ten­sor po­ten­tial is big.

Of course, that also de­pends on $S_{12}$. But $S_{12}$ is not small ei­ther. For ex­am­ple, if ${\skew0\vec r}$ is in the $x$-​di­rec­tion, then $S_{12}$ times a triplet state is three times the op­po­site triplet state mi­nus the orig­i­nal one.

A.42.3 Ex­pla­na­tion of the OPEP po­ten­tial

The pur­pose of this sub­sec­tion is to ex­plain the OPEP po­ten­tial be­tween nu­cle­ons as given in the pre­vi­ous sub­sec­tion phys­i­cally.

Note that the ob­jec­tive is not to give a rig­or­ous de­riva­tion of the OPEP po­ten­tial us­ing ad­vanced quan­tum field the­ory. Physi­cists pre­sum­ably al­ready got the OPEP right. They bet­ter, be­cause it is a stan­dard part of cur­rent nu­clear po­ten­tials. The ex­pla­na­tions here will be based on sim­ple phys­i­cal as­sump­tions. They fol­low the de­riva­tion of the Koulomb po­ten­tial in {A.22.1}. That de­riva­tion was clas­si­cal, al­though a sim­ple quan­tum field ver­sion can be found in {A.22.3}. Note that the orig­i­nal Yukawa de­riva­tion was clas­si­cal too. It was still worth a No­bel prize.

The ar­gu­ments here are loosely based on [16, p. 282-288]. How­ever, of­ten the as­sump­tions made in that ref­er­ence seem quite ar­bi­trary. To avoid that, the ex­po­si­tion be­low makes much more lib­eral use of quan­tum ideas. Af­ter all, in fi­nal analy­sis the clas­si­cal field is just a re­flec­tion of un­der­ly­ing quan­tum me­chan­ics. Hope­fully the quan­tum ar­gu­ments will show much more com­pellingly that things just have to be the way they are.

First of all, like in the first sub­sec­tion it will be as­sumed that every nu­cleon can gen­er­ate a pion po­ten­tial. Other nu­cle­ons can ob­serve that po­ten­tial and in­ter­act with it, pro­duc­ing forces be­tween the nu­cle­ons in­volved.

The net pion po­ten­tial pro­duced by all the nu­cle­ons will be called $\varphi$. It will be as­sumed that the en­ergy in the ob­serv­able pion field is given in terms of $\varphi$ as

$$E_\varphi = \frac{\epsilon_1}{2}\int \left\vert\frac{1}{c}... ...i c^2}{\hbar c} \varphi\right\vert^2 { \rm d}^3{\skew0\vec r}$$

Here $\epsilon_1$ is some em­pir­i­cal con­stant, $m_\pi$ the pion mass, and the in­te­gral is over all space. There should be some ex­pres­sion for the en­ergy in the ob­serv­able field, and the in­te­gral above is what the Klein-Gor­don equa­tion for free pi­ons pre­serves, {D.32}. So it seems the likely ex­pres­sion. Also, the above in­te­gral gives the cor­rect en­ergy in an elec­tro­sta­tic field, chap­ter 13.2 (13.11), tak­ing into ac­count that the pho­ton has no mass.

(Do note that there are some qual­i­fi­ca­tions to the state­ment that the above in­te­gral gives the cor­rect en­ergy in an elec­tro­sta­tic field. The elec­tro­mag­netic field is quite tricky be­cause, un­like the pion, the pho­ton wave func­tion is a rel­a­tivis­tic four-vec­tor. See {A.22} for more. But at the very least, the in­te­gral above gives the cor­rect ex­pres­sion for the ef­fec­tive en­ergy in the elec­tro­sta­tic field.)

Fi­nally it will be as­sumed that there is an in­ter­ac­tion en­ergy be­tween the ob­serv­able pion field and the nu­cle­ons. But the pre­cise ex­pres­sion for that in­ter­ac­tion en­ergy is not yet ob­vi­ous. Only a generic ex­pres­sion can rea­son­ably be pos­tu­lated at this stage. In par­tic­u­lar, it will be pos­tu­lated that the in­ter­ac­tion en­ergy of the pion field with an ar­bi­trary nu­cleon num­bered $i$ takes the form:

$$E_{\varphi i} = - \int \varphi f_i { \rm d}^3{\skew0\vec r}$$

The mi­nus sign is in­serted since the in­ter­ac­tion will pre­sum­ably lower the en­ergy. If it did not, there should be no pion field at all in the ground state. The fac­tor $f_i$ will be called the in­ter­ac­tion fac­tor of nu­cleon $i$.

It still needs to be fig­ured out what is the ap­pro­pri­ate form of this in­ter­ac­tion fac­tor. But it will be as­sumed that it in­volves the wave func­tion $\Psi_i$ of nu­cleon $i$ in some way. In par­tic­u­lar, in re­gions where the wave func­tion is zero, $f_i$ will be zero too. That means that where there is no prob­a­bil­ity of find­ing the nu­cleon, there is no in­ter­ac­tion of the nu­cleon with the field ei­ther. In other words, the in­ter­ac­tion is lo­cal, rather than long range; it oc­curs at the lo­ca­tion of the nu­cleon. One mo­ti­va­tion for this as­sump­tion is that long-range in­ter­ac­tions are just bound to pro­duce prob­lems with spe­cial rel­a­tiv­ity.

It will fur­ther be as­sumed that the wave func­tion of each nu­cleon $i$ is slighly spread out around some nom­i­nal po­si­tion ${\skew0\vec r}_i$. Af­ter all, if you want a po­ten­tial in terms of nu­cleon po­si­tions, then nu­cle­ons should at least ap­prox­i­mately have po­si­tions. One im­me­di­ate con­se­quence is then that the in­ter­ac­tion fac­tor $f_i$ is zero ex­cept close to the nom­i­nal po­si­tion ${\skew0\vec r}_i$ of the nu­cleon.

The ground state is the state in which the com­bined pion field and in­ter­ac­tion en­ergy is min­i­mal. To find the prop­er­ties of that state re­quires vari­a­tional cal­cu­lus. This is worked out in con­sid­er­able de­tail in {A.22.1} and {A.2}. (While those de­riva­tions do not in­clude the $m_\pi$ term, its in­clu­sion is triv­ial.) The analy­sis shows that the ob­serv­able po­ten­tial must sat­isfy

$$- \nabla^2 \varphi + \left(\frac{m_\pi c^2}{\hbar c}\right)^2\varphi = \frac{1}{\epsilon_1} \sum_i f_i %$$ (A.263)

As noted above, the in­ter­ac­tion fac­tors $f_i$ in the right hand side are zero away from the nu­cle­ons. And that means that away from the nu­cle­ons the po­ten­tial sat­is­fies the Klein-Gor­don eigen­value prob­lem with zero en­ergy. That was a good guess, in the first sub­sec­tion! But now the com­plete po­ten­tial can be fig­ured out, given the in­ter­ac­tion fac­tors $f_i$.

The vari­a­tional analy­sis fur­ther shows that the en­ergy of in­ter­ac­tion be­tween a nu­cleon num­bered $i$ and one num­bered $j$ is:

$$V_{ij} = - \int \varphi_i({\skew0\vec r}) f_j({\skew0\vec r}) { \rm d}^3{\skew0\vec r} %$$ (A.264)

Here $\varphi_i$ is the po­ten­tial caused by nu­cleon $i$. In other words, $\varphi_i$ is the so­lu­tion of (A.263) if only a sin­gle in­ter­ac­tion fac­tor $f_i$ in the sum in the right hand side is in­cluded.

The big ques­tion re­mains, what ex­actly is the in­ter­ac­tion fac­tor $f_i$ be­tween the pion field and a nu­cleon $i$? The first guess would be that the in­ter­ac­tion en­ergy at a given po­si­tion is pro­por­tional to the prob­a­bil­ity of find­ing the nu­cleon at that po­si­tion. In short,

$$f_{i,\rm fg} = g \vert\Psi_i\vert^2\quad\mbox{?}$$

where fg” stands for “first-guess and $g$ is some con­stant. This re­flects that the prob­a­bil­ity of find­ing the nu­cleon is given by its square wave func­tion $\vert\Psi_i\vert^2$. The above in­ter­ac­tion fac­tor is es­sen­tially what you would have in elec­tro­sta­t­ics. There $g$ would be the elec­tric charge, so for pi­ons you could call it the “mesic charge.” (Note that the square wave func­tion in­te­grates to 1 so the in­te­grated in­ter­ac­tion fac­tor above is $g$.)

Given the above first-guess in­ter­ac­tion fac­tor, ac­cord­ing to (A.263) a nu­cleon $i$ would cre­ate a first-guess po­ten­tial, {D.2.2},

$$\varphi_{i,\rm fg} = \frac{g}{4\pi\epsilon_1} \frac{e^{-r/R... ... \qquad\mbox{(except vanishingly close to the nucleon $i$)} %$$ (A.265)

Here $r$ is the dis­tance from the nu­cleon. If you as­sume for sim­plic­ity that the nu­cleon is at the ori­gin, $r$ is the dis­tance from the ori­gin.

The above po­ten­tial is spher­i­cally sym­met­ric; it is the same in all di­rec­tions. (That is true even if the nu­cleon wave func­tion is not spher­i­cally sym­met­ric. The wave func­tion is only nonzero very close to ${\skew0\vec r}_i$, so it looks like a sin­gle point away from the im­me­di­ate vicin­ity of the nu­cleon.)

The in­ter­ac­tion en­ergy with a sec­ond nu­cleon $j$ may now be found us­ing (A.264). In par­tic­u­lar, be­cause the wave func­tion of nu­cleon $j$ is only nonzero very close to its nom­i­nal po­si­tion ${\skew0\vec r}_j$, you can ap­prox­i­mate $\varphi_i({\skew0\vec r})$ in (A.264) as $\varphi_i({\skew0\vec r}_j)$. Then you can take it out of the in­te­gral. So the in­ter­ac­tion en­ergy is pro­por­tional to $\varphi_i({\skew0\vec r}_j)$. That is the po­ten­tial caused by nu­cleon $i$ eval­u­ated at the po­si­tion of nu­cleon $j$. That was an­other good guess, in the first sub­sec­tion! More pre­cisely, you get

$$V_{ij,\rm fg} = - \frac{g^2}{4\pi\epsilon_1} \frac{e^{-r_{ij}/R}}{r_{ij}}$$

where $r_{ij}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vert{\skew0\vec r}_j-{\skew0\vec r}_i\vert$ is the dis­tance be­tween the nu­cle­ons. This first guess po­ten­tial en­ergy is the Yukawa po­ten­tial of the first sub­sec­tion.

The Yukawa po­ten­tial would be ap­pro­pri­ate for a field of spin­less pi­ons with pos­i­tive in­trin­sic par­ity. And ex­cept for the sign prob­lem men­tioned in the first sub­sec­tion, it also gives the cor­rect Coulomb po­ten­tial en­ergy in elec­tro­sta­t­ics.

Un­for­tu­nately, as noted in the first sub­sec­tion, the pion has neg­a­tive in­trin­sic par­ity, not pos­i­tive. And that is a prob­lem. Imag­ine for a sec­ond that a nu­cleon pops up a pion. The nu­cleon has pos­i­tive par­ity. How­ever, the pion that pops up has neg­a­tive in­trin­sic par­ity. And par­ity is pre­served, chap­ter 7.3. If the in­trin­sic par­ity of the pion is neg­a­tive, its or­bital par­ity must be neg­a­tive too to main­tain a pos­i­tive com­bined sys­tem par­ity, chap­ter 7.4.2. Neg­a­tive or­bital par­ity means that the pion wave func­tion $\varphi_\pi$ must have op­po­site val­ues at ${\skew0\vec r}$ and $\vphantom{0}\raisebox{1.5pt}{$-$}$${\skew0\vec r}$. But as men­tioned, the first-guess po­ten­tial is spher­i­cally sym­met­ric; the val­ues at ${\skew0\vec r}$ and $\vphantom{0}\raisebox{1.5pt}{$-$}$${\skew0\vec r}$ are the same.

(Note that this ar­gu­ment blurs the dis­tinc­tion be­tween a pion wave func­tion $\varphi_\pi$ and an ob­serv­able pion po­ten­tial $\varphi$. But you would ex­pect them to be closely re­lated, {A.22.3}. In par­tic­u­larly, rea­son­ably speak­ing you would ex­pect that spher­i­cally sym­met­ric wave func­tions cor­re­spond to spher­i­cally sym­met­ric ob­serv­able po­ten­tials, as well as vice-versa.)

(You might also, cor­rectly, ob­ject to the in­ac­cu­rate pic­ture that the nu­cleon pops up a pion. The ground state of the nu­cleon-pi­ons sys­tem is a state of def­i­nite en­ergy. En­ergy states are sta­tion­ary, chap­ter 7.1.4. How­ever, in en­ergy states the com­plete nu­cleon-pi­ons sys­tem should have def­i­nite an­gu­lar mo­men­tum and par­ity, chap­ter 7.3. That is just like nu­clei in en­ergy states have def­i­nite an­gu­lar mo­men­tum and par­ity, chap­ter 14.1. The term in the nu­cleon-pi­ons sys­tem wave func­tion in which there is just the nu­cleon, with no pi­ons, al­ready sets the an­gu­lar mo­men­tum and par­ity. A dif­fer­ent term in the sys­tem wave func­tion, in par­tic­u­lar one in which there is a pion in a state of def­i­nite an­gu­lar mo­men­tum and par­ity, can­not have dif­fer­ent an­gu­lar mo­men­tum or par­ity. Oth­er­wise an­gu­lar mo­men­tum and par­ity would have un­cer­tainty.)

So how to fix this? Sup­pose that you dif­fer­en­ti­ate the first-guess po­ten­tial (A.265) with re­spect to, say, $x$. The dif­fer­en­ti­a­tion will bring in a fac­tor $x$ in the po­ten­tial,

$$\frac{\partial\varphi_{i,\rm fg}}{\partial x} = \frac{\partial\varphi_{i,\rm fg}}{\partial r} \frac{x}{r}$$

And that fac­tor $x$ will pro­duce an op­po­site sign at $\vphantom{0}\raisebox{1.5pt}{$-$}$${\skew0\vec r}$ com­pared to ${\skew0\vec r}$. That means that the par­ity is now neg­a­tive as it should be.

Ac­cord­ing to (A.263), the first guess po­ten­tial sat­is­fies

$$- \nabla^2 \varphi_{i,\rm fg} + \left(\frac{m_\pi c^2}{\hb... ...varphi_{i,\rm fg} = \frac{1}{\epsilon_1} g \vert\Psi_i\vert^2$$

Dif­fer­en­ti­at­ing both sides with re­spect to $x$, you get for its $x$-​de­riv­a­tive, the sec­ond-guess po­ten­tial $\varphi_{i,\rm {sg}}$:

$$- \nabla^2 \varphi_{i,\rm sg} + \left(\frac{m_\pi c^2}{\hb... ...hi_{i,\rm sg} = \frac{\partial \varphi_{i,\rm fg}}{\partial x}$$

So ap­par­ently, if you put a $x$-​de­riv­a­tive on the square nu­cleon wave func­tion in the first-guess in­ter­ac­tion fac­tor $f_{i,\rm {fg}}$ you get a pion po­ten­tial con­sis­tent with par­ity con­ser­va­tion.

There are a cou­ple of new prob­lems. First of all, this po­ten­tial now has or­bital an­gu­lar mo­men­tum. If you check out the spher­i­cal har­mon­ics in ta­ble 4.3, you see that a spher­i­cally sym­met­ric wave func­tion has no or­bital an­gu­lar mo­men­tum. But the fac­tor $x$ pro­duces a wave func­tion of the form

$$c(r) Y_1^{1} - c(r) Y_1^{-1}$$

where $c(r)$ is some spher­i­cally sym­met­ric func­tion. The first term above has an­gu­lar mo­men­tum $\hbar$ in the $z$-​di­rec­tion. The sec­ond term has an­gu­lar mo­men­tum $\vphantom{0}\raisebox{1.5pt}{$-$}$$\hbar$ in the $z$-​di­rec­tion. So there is un­cer­tainty in an­gu­lar mo­men­tum, but it is not zero. The az­imuthal quan­tum num­ber of square or­bital an­gu­lar mo­men­tum, call it $l_\pi$, is 1 with no un­cer­tainty.

So where does this an­gu­lar mo­men­tum come from? An­gu­lar mo­men­tum should be pre­served. The pion it­self has no spin. So its or­bital an­gu­lar mo­men­tum will have to come from the half unit of nu­cleon spin. In­deed it is pos­si­ble for half a unit of nu­cleon spin, $s_i$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$, and one unit of pion or­bital an­gu­lar mo­men­tum, $l_\pi$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, to com­bine into still only half a unit of net an­gu­lar mo­men­tum $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$, 7.4.2.

But con­sider also the an­gu­lar mo­men­tum in the $z$-​di­rec­tion. If the pion is given $\hbar$ in the $z$-​di­rec­tion, then that must come from the fact that the nu­cleon spin changes from $\frac12\hbar$ in the $z$-​di­rec­tion to $-\frac12\hbar$. Con­versely, if the pion has $\vphantom{0}\raisebox{1.5pt}{$-$}$$\hbar$, then the nu­cleon must change from $-\frac12\hbar$ to $\frac12\hbar$. Ei­ther way, the nu­cleon spin in the $z$-​di­rec­tion must flip over.

In quan­tum terms, how does that hap­pen? Con­sider the scaled nu­cleon $z$ spin op­er­a­tor $\sigma_z$ for a sec­ond. If you ap­ply this op­er­a­tor on the spin-up state with $z$ spin $\frac12\hbar$, you get a mul­ti­ple of the same state back. (Ac­tu­ally, be­cause of the scal­ing, you get the ex­act same state back.) The spin-up state is an eigen­state of the op­er­a­tor $\sigma_z$ as it should. But the spin-up state is not an eigen­state of the op­er­a­tors $\sigma_x$ and $\sigma_y$. These op­er­a­tors do not com­mute with $\sigma_z$. So if you ap­ply $\sigma_x$ or $\sigma_y$ on the spin-up state, you will also get some of the $-\frac12\hbar$ spin-down state. In fact, if you look a bit closer at an­gu­lar mo­men­tum, chap­ter 12.10, you see that you get only a spin-down state. So both $\sigma_x$ and $\sigma_y$ do ex­actly what is needed; they flip spin-up over to spin-down. Sim­i­larly, they flip spin-down over to spin-up.

The sec­ond prob­lem has to do with the orig­i­nal no­tion of dif­fer­en­ti­at­ing the spher­i­cally sym­met­ric po­ten­tial with re­spect to $x$. Why not $y$ or $z$ or some oblique di­rec­tion? The pion field should not de­pend on how you have ori­ented your math­e­mat­i­cal axes sys­tem. But the $x$-​de­riv­a­tive does de­pend on it. A sim­i­lar prob­lem ex­ists of course with ar­bi­trar­ily choos­ing one of the op­er­a­tors $\sigma_x$ or $\sigma_y$ above.

Now dot prod­ucts are the same re­gard­less of how the co­or­di­nate sys­tem is ori­ented. That then sug­gests how both prob­lems above can be solved at the same time. In the first-guess in­ter­ac­tion fac­tor, add the dot prod­uct be­tween the scaled nu­cleon spin $\vec\sigma_i$ and the spa­tial dif­fer­en­ti­a­tion op­er­a­tor $\nabla_i$. That gives the third-guess in­ter­ac­tion fac­tor as

$$f_{i,\rm tg} = g R \vec\sigma_i\cdot\nabla_i \vert\Psi_i\ve... ...ma_z \frac{\partial}{\partial z_i} \right] \vert\Psi_i\vert^2$$

The fac­tor $R$ has been added to keep the units of the first guess in­tact.

Time for a re­al­ity check. Con­sider a nu­cleon in the spin-up state. If the mesic charge $g$ would be zero, there would be no pion field. There would just be this bare nu­cleon with half a unit of spin-up and pos­i­tive par­ity. Next as­sume that $g$ is not zero, but still small. Then the bare nu­cleon term should still dic­tate the spin and in­trin­sic par­ity. There will now also be terms with pi­ons in the com­plete sys­tem wave func­tion, but they must obey the same spin and par­ity. You can work out the de­tailed ef­fect of the third guess in­ter­ac­tion fac­tor above us­ing ta­ble 4.3 and chap­ter 12.10. If you do, you see that it as­so­ciates the spin-up nu­cleon with a state

$$\sqrt{4\pi}\frac{\partial\varphi_{i,\rm fg}}{\partial r} \... ...parrow}-\sqrt{{\textstyle\frac{2}{3}}}Y_1^1{\downarrow}\right)$$

where $Y_1^0$ and $Y_1^1$, ta­ble 4.3, de­scribe the spa­tial pion po­ten­tial and ${\uparrow}$ and ${\downarrow}$ nu­cleon spin-up, re­spec­tively spin-down. Loosely as­so­ci­at­ing the pion po­ten­tial with a pion wave func­tion, you can check from the Cleb­sch-Gor­don ta­bles 12.5 that the state in paren­the­ses obeys the spin and par­ity of the orig­i­nal bare nu­cleon.

So the third guess seems pretty good. But there is one more thing. Re­call that there are three dif­fer­ent pi­ons, with dif­fer­ent charges, So you would ex­pect that there are re­ally three dif­fer­ent func­tions $f_i$, one for each pion field. Al­ter­na­tively, the func­tion $f_i$ should be three-di­men­sion­al vec­tor. But what sort of vec­tor?

Note that charge is pre­served. If a pro­ton pops up a pos­i­tively charged $\pi^+$ pion, it must it­self change into a un­charged neu­tron. And if a neigh­bor­ing neu­tron ab­sorbs that $\pi^+$, it ac­quires its pos­i­tive charge and turns into a pro­ton. The same thing hap­pens when a neu­tron emits a neg­a­tively charged $\pi^-$ that a pro­ton ab­sorbs. When­ever a charged par­ti­cle is ex­changed be­tween a pro­ton and a neu­tron, both change type. (Charged par­ti­cles can­not be ex­changed be­tween nu­cle­ons of the same type be­cause there are no nu­cle­ons with neg­a­tive charge or with two units of pos­i­tive charge.)

So, it is nec­es­sary to de­scribe change of nu­cleon type. Physi­cists do that in a very weird way; they pat­tern the math­e­mat­ics on that of spin, chap­ter 14.18. First a com­pletely ab­stract 123 co­or­di­nate sys­tem is in­tro­duced. If a nu­cleon is a pro­ton, then it is said that the nu­cleon has a com­po­nent $\frac12$ along the ab­stract 3-axis. If a nu­cleon is a neu­tron, it is said that it has a com­po­nent $-\frac12$ along the 3-axis.

Com­pare that with spin. If a nu­cleon is spin-up, it has a spin com­po­nent $\frac12\hbar$ along the phys­i­cal $z$-​axis. If it is spin-down, it has a spin com­po­nent $-\frac12\hbar$ along the $z$-​axis. The idea is very sim­i­lar.

Now re­call from above that the op­er­a­tors $\sigma_x$ and $\sigma_y$ flip over the spin in the $z$-​di­rec­tion. In 123-space, physi­cist de­fine ab­stract op­er­a­tors $\tau_1$ and $\tau_2$ that do a sim­i­lar thing: they flip over the value along the 3-axis. And that means that these op­er­a­tors change pro­tons into neu­trons or vice-versa. So they do ex­actly what is needed in ex­changes of charged pi­ons. Physi­cist also de­fine an op­er­a­tor $\tau_3$, anal­o­gous to $\sigma_z$, which does not change the 3-com­po­nent.

Of course, all this may seem an ex­tremely round-about way of do­ing some­thing sim­ple: de­fine op­er­a­tors that flip over nu­cleon type. And nor­mally it re­ally would be. But if it is as­sumed that nu­clear forces are charge-in­de­pen­dent, (which is a rea­son­able ap­prox­i­ma­tion), things change. In that case it turns out that the physics must re­main the same un­der ro­ta­tions of this ab­stract 123-co­or­di­nate sys­tem. And that re­quire­ment can again be met by form­ing a dot prod­uct, this time be­tween $\vec\tau$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(\tau_1,\tau_2,\tau_3)$ vec­tors.

That idea then gives the fi­nal ex­pres­sion for the func­tions $f_i$:

$$\vec f_i = g R \vec\tau_i\;\;\vec\sigma_i\cdot\nabla_i \ver... ...ma_z \frac{\partial}{\partial z_i} \right] \vert\Psi_i\vert^2$$

Note that $\vec{f}_i$ is now a three-di­men­sion­al vec­tor be­cause $\vec\tau_i$ is. In the fi­nal po­ten­tial, $\vec\tau_i$ gets into a dot prod­uct with $\vec\tau_j$ of the other nu­cleon. That makes the com­plete po­ten­tial the same re­gard­less of ro­ta­tion of the ab­stract 123-co­or­di­nate sys­tem as it should.

Now it is just a mat­ter of work­ing out the fi­nal po­ten­tial. Do one thing at a time. Re­call first the ef­fect of the $x$-​de­riv­a­tive on the nu­cleon wave func­tion. It pro­duces a po­ten­tial that is the $x$-​de­riv­a­tive of the spher­i­cally sym­met­ric first-guess po­ten­tial (A.265). That works out to

$$\frac{gR}{4\pi\epsilon_1} \frac{{\rm d}e^{-r/R}/r}{{\rm d}r... ...silon_1} \left[\frac{1}{Rr^2}+\frac{1}{r^3}\right] e^{-r/R} x$$

Of course, there ares sim­i­lar ex­pres­sions for the de­riv­a­tives in the other two di­rec­tions. So the po­ten­tial pro­duced by nu­cleon $i$ at the ori­gin is

$$\varphi_i({\skew0\vec r}) = - \frac{gR}{4\pi\epsilon_1} \ve... ...+\frac{1}{r^3}\right] e^{-r/R} {\skew0\vec r}\cdot\vec\sigma_i$$

Now the in­ter­ac­tion po­ten­tial with an­other nu­cleon fol­lows from (A.264). But here you need to be care­ful. The in­te­gral will in­volve terms like

$$\mbox{[some constant] } \int \varphi_i({\skew0\vec r}) \fr... ...tial \vert\Psi_j\vert^2}{\partial x} { \rm d}^3{\skew0\vec r}$$

In this case, you can­not just ap­prox­i­mate ${\skew0\vec r}$ in $\varphi_i({\skew0\vec r})$ as the nom­i­nal po­si­tion ${\skew0\vec r}_j$ of nu­cleon $j$. That is not ac­cu­rate. Since the $x$-​de­riv­a­tive works on a very con­cen­trated wave func­tion, it will pro­duce large neg­a­tive and pos­i­tive val­ues, and er­rors will ac­cu­mu­late. The so­lu­tion is to per­form an in­te­gra­tion by parts in the $x$-​di­rec­tion. That puts the de­riv­a­tive on the po­ten­tial in­stead of the wave func­tion and adds a mi­nus sign. Then you can eval­u­ate this neg­a­tive de­riv­a­tive of the po­ten­tial at the nom­i­nal po­si­tion of nu­cleon ${\skew0\vec r}_j$.

Dif­ferentat­ing the po­ten­tial is a bit of a mess, but straight­for­ward. Then the po­ten­tial be­comes

$$V_{ij} \sim \frac{g^2}{12\pi\epsilon_1R} \; \vec\tau_i\cd... ...t\vec\sigma_j + S_{ij} V_{\rm {T}} \Big] \frac{e^{-r/R}}{r/R}$$

If you de­fine the con­stant $g_\pi$ ap­pro­pri­ately, this gives the OPEP po­ten­tial (A.262).

A.42.4 Mul­ti­ple pion ex­change and such

Un­for­tu­nately, the nu­clear force is not just a mat­ter of the ex­change of sin­gle pi­ons. While the OPEP works very well at nu­cleon dis­tances above 3 fm, at shorter ranges other processes be­come im­por­tant.

The most im­por­tant range is the one of the pri­mary nu­cleon at­trac­tions. Con­ven­tion­ally, this range is ball­parked as nu­cleon dis­tances in the range 1 $\raisebox{.3pt}{$<$}$ r $\raisebox{.3pt}{$<$}$ 2 fm, [5, p. 91], [[3]]. (Ref­er­ences vary about the ac­tual range how­ever, [31, p. 111], [36, pp. 177].) In this range, two-pion ex­changes dom­i­nate. In such ex­changes two pi­ons ap­pear dur­ing the course of the in­ter­ac­tion. Since this re­quires dou­ble the un­cer­tainty in en­ergy, the typ­i­cal range is cor­re­spond­ingly smaller than for one-pion ex­changes.

Two-pion ex­changes are much more dif­fi­cult to crunch out than one-pion ones. In ad­di­tion, it turns out that straight­for­ward two-pion ex­changes are not enough, [[3]]. The in­ter­ac­tions also have to in­clude var­i­ous so-called res­o­nances.

Res­o­nances are ex­tremely short-lived ex­cited states of baryons and mesons. They de­cay through the strong force, which typ­i­cally takes on the or­der of 10$\POW9,{-23}$ s. A par­ti­cle mov­ing near the speed of light will only travel a dis­tance of the or­der of a fem­tome­ter dur­ing such a time. There­fore res­o­nances are not ob­served di­rectly. In­stead they are de­duced from ex­per­i­ments in which par­ti­cles are scat­tered off each other. Ex­cited states of nu­cle­ons can be de­duced from pre­ferred scat­ter­ing of par­tic­u­lar fre­quen­cies. More or less bound states of pi­ons can be de­duced from col­li­sion dy­nam­ics ef­fects. Col­li­sions in­volv­ing three par­ti­cles are quite dif­fer­ent if two of the three par­ti­cles stick to­gether, even briefly, than if all three go off in sep­a­rate di­rec­tions.

The low­est en­ergy ex­cited state for nu­cle­ons is a set of res­o­nances called the “delta par­ti­cles,” $\Delta^{++}$, $\Delta^+$, $\Delta^0$, and $\Delta^-$. In the deltas, the three con­stituent quarks of the nu­cle­ons align their spins in the same di­rec­tion for a net spin of $\frac32$. The state fur­ther has enough an­ti­sym­me­try to al­low three quarks to be equal. That ex­plains the nu­cleon charge 2$e$ of the $\Delta^{++}$, con­sist­ing of three up quarks at ${\textstyle\frac{2}{3}}e$ each, and the charge $\vphantom{0}\raisebox{1.5pt}{$-$}$$e$ of the $\Delta^-$, con­sist­ing of three down quarks at $-{\textstyle\frac{1}{3}}e$ each. The delta res­o­nances are of­ten in­di­cated by $\Delta(1232)$, where the quan­tity be­tween paren­the­ses is the nom­i­nal rest mass en­ergy in MeV. That al­lows ex­cited states of higher en­ergy to be ac­com­mo­dated. If the ex­cited states al­low no more than two quarks to be equal, like the nor­mal nu­cle­ons, they are in­di­cated by $N$ in­stead of $\Delta$. In those terms, the nor­mal pro­ton and neu­tron are $N(939)$ states. (The rest mass en­er­gies are nom­i­nal be­cause res­o­nances have a tremen­dous un­cer­tainty in en­ergy. That is to be ex­pected from their short life time on ac­count of the en­ergy-time un­cer­tainty re­la­tion­ship. The width of the delta en­ergy is over 100 MeV.)

Pion res­o­nances of in­ter­est in­volve the 775 MeV rho ($\rho$), and the 783 MeV omega ($\omega$) res­o­nances. Both of these states have spin 1 and odd par­ity. The 550 MeV eta ($\eta$) par­ti­cle is also of im­por­tance. This par­ti­cle has spin 0 and odd par­ity like the pi­ons. The eta is not re­ally a res­o­nance, based on its rel­a­tively long 0.5 10$\POW9,{-18}$ s life time.

Older ref­er­ences like [36] pic­ture the res­o­nances as cor­re­lated multi-pion states. How­ever, quan­tum chrom­e­dy­nam­ics has been as­so­ci­at­ing ac­tual par­ti­cles with them. Take the rho, for ex­am­ple. In [36] it is pic­tured as a two-pion cor­re­lated state. (A true bound state of two 140 MeV pi­ons should have an en­ergy less than 280 MeV, not 775 MeV.) How­ever, quan­tum chrom­e­dy­nam­ics iden­ti­fies a rho as a sin­gle ex­cited pion with a 775 MeV rest mass. It does de­cay al­most in­stantly into two pi­ons. The omega, pic­tured as a three-pion cor­re­lated state, is ac­cord­ing to quan­tum chrom­e­dy­nam­ics a quan­tum su­per­po­si­tion of half an up-an­tiup and half a down-an­ti­d­own quark pair, not un­like the neu­tral rho. It usu­ally de­cays into three pi­ons. Quan­tum chrom­e­dy­nam­ics de­scribes the $\eta$ as a me­son hav­ing a strange-an­ti­s­trange quark com­po­nent.

The rho and omega res­o­nances ap­pear to be im­por­tant for the nu­cleon re­pul­sions at short range. And 3 and 4 pion ex­changes have about the same range as the $\omega$. So if the omega is in­cluded, as it nor­mally is, it seems that multi-pion ex­changes should be in­cluded too. Crunch­ing out com­plete multi-pion pion ex­changes, with the ad­di­tional com­pli­ca­tions of the men­tioned res­o­nances, is a for­mi­da­ble task.

One-me­son ex­changes are much eas­ier to an­a­lyze than multi-me­son ones. There­fore physi­cists may model the multi-pion processes as the ex­change of one com­bined bo­son, rather than of mul­ti­ple pi­ons. That pro­duces so-called “one-bo­son ex­change po­ten­tials,” or “OBEP”s for short. They work sur­pris­ingly well.

The pre­cise Yukawa po­ten­tial that is pro­duced de­pends on the spin and par­ity of the ex­changed bo­son, [36, pp. 176ff], [[3]]. The pion has zero spin and neg­a­tive par­ity. Such a par­ti­cle is of­ten called “pseu­doscalar.” Scalar means that its wave func­tion at each point is a just a num­ber. How­ever, nor­mal num­bers, like say a mass, do not change sign if the di­rec­tions of the axes are in­verted. The eta is a 0$\POW9,{-}$ pseu­doscalar like the pion. It pro­duces a sim­i­lar po­ten­tial as the OPEP.

How­ever, the rho and omega are 1$\POW9,{-}$ bosons. Such bosons are of­ten called “”vec­tor par­ti­cles.” Their wave func­tion at each point is a three-di­men­sion­al vec­tor, {A.20}. And nor­mal vec­tors do change sign if the di­rec­tions of the axes are in­verted, so the rho and omega are not pseudovec­tors. Vec­tor bosons gen­er­ate a re­pul­sive po­ten­tial, among var­i­ous other ef­fects. That can take care of the needed re­pul­sive short range forces quite nicely.

Un­for­tu­nately, to de­scribe the at­trac­tive forces in the in­ter­me­di­ate range, OBEP mod­els need a roughly 600 MeV 0$\POW9,{+}$ scalar bo­son. In fact, many OBEP mod­els use both a 500 MeV and a 700 MeV scalar bo­son. The ex­is­tence of such scalar res­o­nances has never been ac­cepted. While an older ref­er­ence like [36, pp. 172] may point to a per­ceived very wide res­o­nance at 700 MeV, how con­vinc­ing can a 700 MeV res­o­nance with a width of at least 600 MeV be? This lack of phys­i­cal jus­ti­fi­ca­tion does de­tract from the OBEP po­ten­tials.

And of course, they are ap­prox­i­ma­tions in any case. There are im­por­tant is­sues like multi-nu­cleon in­ter­ac­tions and elec­tro­mag­netic prop­er­ties that prob­a­bly only a com­pre­hen­sive de­scrip­tion of the ac­tual ex­change processes can cor­rectly de­scribe, [[3]]. De­spite much work, nu­clear po­ten­tials re­main an ac­tive re­search area. One au­thor al­ready thinks in terms of mil­len­nia, [32].