A.41 Deuteron model

A very sim­ple model can be used to give some con­text to the data of the deuteron. This ad­den­dum de­scribes that model. Then it dis­cusses the var­i­ous ma­jor prob­lems of the model. Some pos­si­ble fixes for these prob­lems are in­di­cated.

In all cases it is as­sumed that the deuteron is mod­elled as a two-par­ti­cle sys­tem, a pro­ton and a neu­tron. Fur­ther­more, the pro­ton and neu­tron are as­sumed to have the same prop­er­ties in the deuteron as they have in free space. These as­sump­tions are not re­ally true. For one, the pro­ton and neu­tron are not el­e­men­tary par­ti­cles but com­bi­na­tions of quarks. How­ever, ig­nor­ing that is a rea­son­able start­ing point in try­ing to un­der­stand the deuteron.

A.41.1 The model

The deuteron con­tains two nu­cle­ons, a pro­ton and a neu­tron. The sim­ple model as­sumes that the po­ten­tial en­ergy of the deuteron only de­pends on the dis­tance $r$ be­tween the nu­cle­ons. More specif­i­cally, it as­sumes that the po­ten­tial en­ergy has some con­stant value $-V_0$ up to some spac­ing $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $d_0$. And it as­sumes that the po­ten­tial is zero for spac­ings larger than $d_0$. Fig­ure A.24 shows the idea.

This model is an­a­lyt­i­cally solv­able. First, the deuteron in­volves the mo­tion of two par­ti­cles, the pro­ton and the neu­tron. How­ever, the prob­lem may be sim­pli­fied to that of an imag­i­nary sin­gle re­duced mass en­cir­cling the cen­ter of grav­ity of the deuteron, ad­den­dum {A.5}.

The re­duced mass in the sim­pli­fied prob­lem is half the mass of the pro­ton or neu­tron. (That ig­nores the tiny dif­fer­ence in mass be­tween the pro­ton and neu­tron.) The po­ten­tial for the re­duced mass is $-V_0$ if the re­duced mass is within a dis­tance $d_0$ of the cen­ter of grav­ity and zero be­yond that. A po­ten­tial of this type is com­monly called a [spher­i­cal] [square] well po­ten­tial. Fig­ure A.24 shows the po­ten­tial in green.

Fig­ure A.24: Crude deuteron model. The po­ten­tial is in green. The rel­a­tive prob­a­bil­ity of find­ing the nu­cle­ons at a given spac­ing is in black.

The so­lu­tion for the re­duced mass prob­lem may be worked out fol­low­ing ad­den­dum {A.6}. Note that the model in­volves two un­known pa­ra­me­ters, the po­ten­tial $V_0$ and the dis­tance $d_0$. Two pieces of ex­per­i­men­tal in­for­ma­tion need to be used to fix val­ues for these pa­ra­me­ters.

First of all, the bind­ing en­ergy should match the ex­per­i­men­tal 2.2247 MeV. Also, the root-mean square ra­dial po­si­tion of the nu­cle­ons away from the cen­ter of the nu­cleus should be about 1.955 fm, [J.P. Mc­Tavish 1982 J. Phys. G 8 911; J.L. Friar et al 1984 Phys. Rev. C 30 1084]. (Based on elec­tron scat­ter­ing ex­per­i­ments, physi­cists are con­fi­dent that the root-mean-square ra­dial po­si­tion of the charge from the cen­ter of the deuteron is 2.14 fm. How­ever, this charge ra­dius is larger than the root mean square ra­dial po­si­tion of the nu­cle­ons. The main rea­son is that the pro­ton has a fi­nite size. For ex­am­ple, even in the hy­po­thet­i­cal case that the dis­tance of the two nu­cle­ons from the cen­ter of grav­ity would be zero, there would still be a pos­i­tive charge ra­dius; the one of the pro­ton. The pro­ton by it­self has a sig­nif­i­cant charge ra­dius, 0.88 fm.) The dis­tance $r$ of the re­duced mass from the ori­gin should match the dis­tance be­tween the nu­cle­ons; in other words it should be twice the 1.955 fm ra­dius.

Ta­ble A.4: Deuteron model data. The top half of the ta­ble al­lows some de­vi­a­tion from the ex­per­i­men­tal nu­cleon root-mean-square ra­dial po­si­tion. The bot­tom half al­lows some de­vi­a­tion from the ex­per­i­men­tal en­ergy.

Ta­ble A.4 shows that these two ex­per­i­men­tal con­straints are met when the dis­tance $d_0$ is 2.1 fm and the po­ten­tial $V_0$ about 35 MeV. The fact that the dis­tance $d_0$ matches the charge ra­dius is just a co­in­ci­dence.

There is some jus­ti­fi­ca­tion for this model. For one, it is well es­tab­lished that the nu­clear force very quickly be­comes neg­li­gi­ble be­yond some typ­i­cal spac­ing be­tween the nu­cle­ons. The above po­ten­tial re­flects that. Based on bet­ter mod­els, (in par­tic­u­lar the so-called OPEP po­ten­tial), the typ­i­cal range of the nu­clear force is roughly 1.5 fm. The po­ten­tial cut-off $d_0$ in the model is at 2.1 fm. Ob­vi­ously that is in the ball­park, though it seems a bit big. (For a full-po­ten­tial/zero-po­ten­tial cut-off.)

The fact that both the model and ex­act po­ten­tials van­ish at large nu­cleon spac­ings also re­flects in the wave func­tion. It means that the rate of de­cay of the wave func­tion at large nu­cleon spac­ings is cor­rectly rep­re­sented. The rate of de­cay de­pends only on the bind­ing en­ergy $E$.

To be more pre­cise, the model wave func­tion is, {A.6},

$$\psi = \frac{A_{\rm S}}{\sqrt{4\pi}} \frac{e^{-\sqrt{2m_{\rm red}\vert E\vert} r/\hbar}}{r} \qquad\mbox{for}\qquad r > d_0$$

where $A_{\rm {S}}$ is some con­stant. Un­like the model, the ex­per­i­men­tal wave func­tion has some an­gu­lar vari­a­tion. How­ever, if this vari­a­tion is av­er­aged away, the ex­per­i­men­tal wave func­tion de­cays just like the model for dis­tances much larger than 1.5 fm. In ad­di­tion, the model matches the ex­per­i­men­tal value for the con­stant $A_{\rm {S}}$, 0.88, ta­ble A.4.

To be fair, this good agree­ment does not ac­tu­ally sup­port the de­tails of the po­ten­tial as much as it may seem. As the dif­fer­ence be­tween $-V_0$ and the ex­pec­ta­tion po­ten­tial $\langle{V}\rangle$ in ta­ble A.4 shows, the nu­cle­ons are more likely to be found be­yond the spac­ing $d_0$ than be­low it. And the root mean square sep­a­ra­tion of the nu­cle­ons de­pends mostly on the wave func­tion at large val­ues of $r$. As a con­se­quence, if the model gets $A_S$ right, then the root mean square sep­a­ra­tion of the nu­cle­ons can­not be much wrong ei­ther. That is true re­gard­less of what ex­actly the po­ten­tial for $r$ $\raisebox{.3pt}{$<$}$ $d_0$ is. Still, the model does get the right value.

An­other point in fa­vor of the model is that the ki­netic en­ergy can­not be all wrong. In par­tic­u­lar, the Heisen­berg un­cer­tainty re­la­tion­ship im­plies that the ki­netic en­ergy of the deuteron must be at least 6.2 MeV. The sec­ond-last col­umn in the ta­ble shows the min­i­mum ki­netic en­ergy that is pos­si­ble for the root-mean-square ra­dial nu­cleon po­si­tion in the pre­vi­ous col­umn. It fol­lows that un­avoid­ably the ki­netic en­ergy is sig­nif­i­cantly larger than the bind­ing en­ergy. That re­flects the fact that the deuteron is only weakly bound. (For com­par­i­son, for the pro­ton-elec­tron hy­dro­gen atom the ki­netic en­ergy and bind­ing en­ergy are equal.)

The model also sup­ports the fact that there is only a sin­gle bound state for the deuteron. The sec­ond col­umn in the ta­ble gives the small­est value of $V_0$ for which there is a bound state at all. Clearly, the es­ti­mated val­ues of $V_0$ are com­fort­ably above this min­i­mum. But for a sec­ond bound state to ex­ist, the value of $V_0$ needs to ex­ceed the value in the sec­ond col­umn by a fac­tor 4. Ob­vi­ously, the es­ti­mated val­ues get nowhere close to that.

A fi­nal re­deem­ing fea­ture of the model is that the de­duced po­ten­tial $V_0$ is rea­son­able. In par­tic­u­lar, 35 MeV is a typ­i­cal po­ten­tial for a nu­cleon in­side a heavy nu­cleus. It is used as a ball­park in the com­pu­ta­tion of so-called al­pha de­cay of nu­clei, [31, p. 83, 252].

A.41.2 The re­pul­sive core

While the model of the deuteron de­scribed in the pre­vi­ous sub­sec­tion has sev­eral re­deem­ing fea­tures, it also has some ma­jor prob­lems. The prob­lem to be ad­dressed in this sub­sec­tion is that the nu­clear force be­comes re­pul­sive when the nu­cle­ons try to get too close to­gether. The model does not re­flect such a re­pul­sive core at all.

A sim­ple fix is to de­clare nu­cleon spac­ings be­low a cer­tain value $d_{\rm {min}}$ to be off-lim­its. Typ­i­cally, half a fem­tome­ter is used for $d_{\rm {min}}$. The po­ten­tial is taken to be in­fi­nite, rather than $-V_0$, for spac­ings be­low $d_{\rm {min}}$. That pre­vents the nu­cle­ons from get­ting closer than half a fem­tome­ter to­gether.

Ta­ble A.5: Deuteron model data with a re­pul­sive core of 0.5 fm.

The mod­i­fi­ca­tions needed to the math­e­mat­ics to in­clude this re­pul­sive core are mi­nor. Ta­ble A.5 sum­ma­rizes the re­sults. The value of $V_0$ for a sec­ond bound state would need to be about 160 MeV.

Note that the value of the po­ten­tial cut-off dis­tance $d_0$ has been re­duced from 2.1 fm to 1.7 fm. As dis­cussed in the pre­vi­ous sub­sec­tion, that can be taken to be an im­prove­ment. Also, the ex­pec­ta­tion po­ten­tial and ki­netic en­er­gies seem much bet­ter. A much more ad­vanced po­ten­tial, the so-called Ar­gonne $v_{18}$, gives 22 and 19.8 MeV for these ex­pec­ta­tion val­ues.

Fig­ure A.25: Crude deuteron model with a 0.5 fm re­pul­sive core. Thin grey lines are the model with­out the re­pul­sive core. Thin red lines are more or less com­pa­ra­ble re­sults from the Ar­gonne $v_{18}$ po­ten­tial.

Fig­ure A.25 shows the po­ten­tial and prob­a­bil­ity den­sity. The pre­vi­ous re­sults with­out re­pul­sive core are shown as thin grey lines for eas­ier com­par­i­son. Note that there are very dis­tinc­tive dif­fer­ences be­tween the wave func­tions with and with­out re­pul­sive core. But as­ton­ish­ingly, the val­ues for the root mean square nu­cleon sep­a­ra­tion $r_{\rm {rms}}$ are vir­tu­ally iden­ti­cal. The value of $r_{\rm {rms}}$ is not at all a good quan­tity to gauge the ac­cu­racy of the model.

Fig­ure A.25 also shows cor­re­spond­ing re­sults ac­cord­ing to the much more so­phis­ti­cated Ar­gonne $v_{18}$ model, [51]. The top red line shows the prob­a­bil­ity den­sity for find­ing the nu­cle­ons at that spac­ing. The lower curve shows an ef­fec­tive spher­i­cal po­ten­tial. A note of cau­tion is needed here; the true deuteron po­ten­tial has very large de­vi­a­tions from spher­i­cal sym­me­try. So the com­par­i­son of po­ten­tials is fishy. What is re­ally plot­ted in fig­ure A.25 is the ef­fec­tive po­ten­tial that in­te­grated against the prob­a­bil­ity den­sity pro­duces the cor­rect ex­pec­ta­tion po­ten­tial en­ergy.

It is in­ter­est­ing to see from fig­ure A.25 how small the 2.2 MeV bind­ing en­ergy of the deuteron re­ally is, as com­pared to the min­i­mum value of the po­ten­tial en­ergy.

A.41.3 Spin de­pen­dence

An big prob­lem with the model so far is that nu­cleon-nu­cleon in­ter­ac­tions de­pend strongly on the nu­cleon spins. Such an ef­fect also ex­ists for the pro­ton-elec­tron hy­dro­gen atom, {A.39.5}. How­ever, there the ef­fect is ex­tremely small. For the deuteron, the ef­fect of spin is dra­matic.

The pro­ton and neu­tron each have spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$. They must align these spins in par­al­lel into a so-called triplet state of com­bined spin 1, chap­ter 5.5.6. If in­stead they align their spins in op­po­site di­rec­tions in a sin­glet state of zero net spin, the deuteron will not bind. The model as given so far does not de­scribe this.

One sim­ple way to fix this up is to write two dif­fer­ent po­ten­tials. One po­ten­tial $V_{\rm {t}}(r)$ is taken to ap­ply if the nu­cle­ons are in the triplet state. It can be mod­eled by the piece­wise con­stant po­ten­tial as dis­cussed so far. A sec­ond po­ten­tial $V_{\rm {s}}(r)$ is taken to ap­ply if the nu­cle­ons are in the sin­glet state. A suit­able form can be de­duced from ex­per­i­ments in which nu­cle­ons are scat­tered off each other. This po­ten­tial should not al­low a bound state.

That leaves only the prob­lem of how to write the com­plete po­ten­tial. The com­plete po­ten­tial should sim­plify to $V_{\rm {t}}$ for the triplet state and to $V_{\rm {s}}$ for the sin­glet state. A form that does that is

$$V = V_{\rm {t}}(r) \left[\frac34+\frac{1}{\hbar^2}{\skew 6... ...c S}}_{\rm {p}}\cdot{\skew 6\widehat{\vec S}}_{\rm {n}}\right]$$ (A.258)

The rea­son that this works is be­cause the dot prod­uct ${\skew 6\widehat{\vec S}}_{\rm {p}}\cdot{\skew 6\widehat{\vec S}}_{\rm {n}}$ be­tween the pro­ton and neu­tron spins is ${\textstyle\frac{1}{4}}\hbar^2$ in the triplet state and $-{\textstyle\frac{3}{4}}\hbar^2$ in the sin­glet state, {A.10}.

A.41.4 Non­cen­tral force

So far it has been as­sumed that the po­ten­tial in a given spin state only de­pends on the dis­tance $r$ be­tween the nu­cle­ons. If true, that would im­ply that the or­bital an­gu­lar mo­men­tum of the mo­tion of the nu­cle­ons is con­served. In terms of clas­si­cal physics, the forces be­tween the par­ti­cles would be along the con­nect­ing line be­tween the par­ti­cles. That does not pro­duce a mo­ment that can change or­bital an­gu­lar mo­men­tum.

In terms of quan­tum me­chan­ics, it gets phrased a lit­tle dif­fer­ently. A po­ten­tial that only de­pends on the dis­tance be­tween the par­ti­cles com­mutes with the or­bital an­gu­lar mo­men­tum op­er­a­tors. Then so does the Hamil­ton­ian. And that means that the en­ergy states can also be taken to be states of def­i­nite or­bital an­gu­lar mo­men­tum.

In par­tic­u­lar, in the ground state, the pro­ton and neu­tron should then be in a state of zero or­bital an­gu­lar mo­men­tum. Such a state is spher­i­cally sym­met­ric. There­fore the pro­ton charge dis­tri­b­u­tion should be spher­i­cally sym­met­ric too. All that would be just like for the elec­tron in the hy­dro­gen atom. See chap­ters 4.2, 4.3, 4.5, and 7.3, ad­den­dum {A.39}, and de­riva­tions {A.8} and {A.9} for more de­tails on these is­sues.

How­ever, the fact is that the charge dis­tri­b­u­tion of the deuteron is not quite spher­i­cally sym­met­ric. There­fore, the po­ten­tial can­not just de­pend on the dis­tance $r$ be­tween pro­ton and neu­tron. It must also de­pend on the di­rec­tion of the vec­tor ${\skew0\vec r}$ from neu­tron to pro­ton. In par­tic­u­lar, it must de­pend on how this vec­tor aligns with the nu­cleon spins. There are no other di­rec­tions to com­pare to in the deuteron be­sides the spins.

The ori­en­ta­tion of the cho­sen co­or­di­nate sys­tem should not make a dif­fer­ence for the po­ten­tial en­ergy. From a clas­si­cal point of view, there are three nu­clear an­gles that are non­triv­ial. The first two are the an­gles that the vec­tor ${\skew0\vec r}$ from neu­tron to pro­ton makes with the neu­tron and pro­ton spins. The third is the an­gle be­tween the two spins. These three an­gles, plus the dis­tance be­tween the neu­tron and pro­ton, fully de­ter­mine the geom­e­try of the nu­cleus.

To check that, imag­ine a co­or­di­nate sys­tem with ori­gin at the neu­tron. Take the $x$-​axis along the con­nect­ing line to the pro­ton. Ro­tate the co­or­di­nate sys­tem around the $x$-​axis un­til the neu­tron spin is in the $xy$-​plane. What de­ter­mines the geom­e­try in this co­or­di­nate sys­tem is the an­gle in the $xy$-​plane be­tween the con­nect­ing line and the neu­tron spin. And the two an­gles that fix the di­rec­tion of the pro­ton spin; the one with the con­nect­ing line and the one with the neu­tron spin.

In quan­tum me­chan­ics, an­gles in­volv­ing an­gu­lar mo­men­tum vec­tors are not well de­fined. That is due to an­gu­lar mo­men­tum un­cer­tainty, chap­ter 4.2. How­ever, dot prod­ucts be­tween vec­tors can be used to sub­sti­tute for an­gles be­tween vec­tors, for given lengths of the vec­tors. Be­cause the spin vec­tors have a given length, there are four pa­ra­me­ters that fix the geom­e­try:

$$r \qquad {\skew 6\widehat{\vec S}}_{\rm {n}}\cdot{\skew 6... ...\qquad {\skew 6\widehat{\vec S}}_{\rm {p}}\cdot{\skew0\vec r}$$

The po­ten­tial en­ergy should de­pend on these four pa­ra­me­ters. Note that the ef­fect of the first two pa­ra­me­ters was al­ready mod­elled in the pre­vi­ous sub­sec­tions.

In or­der that or­bital an­gu­lar mo­men­tum is not con­served, the last two pa­ra­me­ters should be in­volved. But not sep­a­rately, be­cause they change sign un­der a par­ity trans­for­ma­tion or time re­ver­sal. It is known that to very good ap­prox­i­ma­tion, nu­clei re­spect the par­ity and time-re­ver­sal sym­me­tries. Terms qua­dratic in the last two pa­ra­me­ters are needed. And in par­tic­u­lar, the prod­uct of the last two pa­ra­me­ters is needed. If you just square ei­ther pa­ra­me­ter, you get a triv­ial mul­ti­ple of $r^2$. That can be seen from writ­ing the spins out in terms of the so-called Pauli ma­tri­ces, as de­fined in chap­ter 12.

The bot­tom line is that the needed ad­di­tional con­tri­bu­tion to the po­ten­tial is due to the prod­uct of the fi­nal two terms. This con­tri­bu­tion is called the “ten­sor po­ten­tial” for rea­sons that are not im­por­tant. By con­ven­tion, the ten­sor po­ten­tial is writ­ten in the form

$$S_{12} V_{\rm {T}}(r) \quad\mbox{with}\quad S_{12} \equiv... ...S}}_{\rm {n}}\cdot{\skew 6\widehat{\vec S}}_{\rm {p}} \right)$$ (A.259)

The choice of the sym­bol $S_{12}$ is an­other one of those con­fu­sion-prone physics con­ven­tions. One source uses the same sym­bol only a few pages away for a ma­trix el­e­ment. The di­vi­sion by $r^2$ is to make $S_{12}$ in­de­pen­dent of the dis­tance be­tween the nu­cle­ons. This dis­tance is sep­a­rately ac­counted for in the fac­tor $V_{\rm {T}}$. Also the sub­trac­tion of the dot prod­uct be­tween the spins makes the av­er­age of $S_{12}$ over all di­rec­tions of ${\skew0\vec r}$ zero. That means that $\Delta{V}$ only de­scribes the an­gu­lar vari­a­tion of the po­ten­tial. The an­gu­lar av­er­age must be de­scribed by other po­ten­tials such as the ones writ­ten down ear­lier.

It turns out that $S_{12}$ com­mutes with the op­er­a­tor of the square net nu­cleon spin but not with the op­er­a­tor of or­bital an­gu­lar mo­men­tum. That is con­sis­tent with the fact that the deuteron has def­i­nite net nu­cleon spin $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 but un­cer­tain or­bital an­gu­lar mo­men­tum. Its quan­tum num­ber of or­bital an­gu­lar mo­men­tum can be $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 or 2.

It should also be noted that $S_{12}$ pro­duces zero when ap­plied on the sin­glet nu­cleon spin state. So the term has no ef­fect on sin­glet states. These prop­er­ties of $S_{12}$ may be ver­i­fied by crunch­ing it out us­ing the prop­er­ties of the Pauli spin ma­tri­ces, chap­ter 12.10, in­clud­ing that $\sigma_i^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $I$ and $\sigma_i\sigma_{\overline{\imath}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-\sigma_{\overline{\imath}}\sigma_i$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\rm i}\sigma_{\overline{\overline{\imath}}}$ with $i,{\overline{\imath}},{\overline{\overline{\imath}}}$ suc­ces­sive num­bers in the cyclic se­quence $\ldots1231231\ldots$.

Fig­ure A.26: Ef­fects of un­cer­tainty in or­bital an­gu­lar mo­men­tum.

Fig­ure A.26 il­lus­trates the ef­fects of the un­cer­tainty in or­bital an­gu­lar mo­men­tum on the deuteron. The data are taken from the Ar­gonne $v_{18}$ po­ten­tial, [51].

The black curve is the prob­a­bil­ity den­sity for find­ing the nu­cle­ons at that spac­ing $r$. Most of this prob­a­bil­ity den­sity is due to the spher­i­cally sym­met­ric, $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, part of the wave func­tion. This con­tri­bu­tion is shown as the grey curve la­belled 0. The con­tri­bu­tion due to the $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 state is the grey curve la­belled 2. The to­tal prob­a­bil­ity of the $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 state is only 5.8% ac­cord­ing to the Ar­gonne $v_{18}$ po­ten­tial.

That might sug­gest that the ef­fect of the $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 state on the deuteron bind­ing could be ig­nored. But that is un­true. If the deuteron wave func­tion was com­pletely spher­i­cally sym­met­ric, the po­ten­tial would given by the thin green curve la­beled 0. The bind­ing en­ergy from this po­ten­tial is sig­nif­i­cantly less than that of the hy­po­thet­i­cal dineu­tron, shown in blue. And the dineu­tron is not even bound. If the deuteron was in a pure $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 state, the bind­ing would be less still ac­cord­ing to the thin green line la­belled 2. How­ever, the deuteron is in a com­bi­na­tion of the $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 and $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 states. The en­ergy of the in­ter­ac­tion of these two states low­ers the po­ten­tial en­ergy greatly. It pro­duces the com­bined po­ten­tial en­ergy curve shown as the thick green line.

In terms of chap­ter 5.3, the low­er­ing of the po­ten­tial en­ergy is a twi­light ef­fect. Even for an $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 state prob­a­bil­ity of only 5.8%, a twi­light ef­fect can be quite large. That is be­cause it is pro­por­tional to the square root of the 5.8% prob­a­bil­ity, which is a quar­ter. In ad­di­tion, the fac­tor $V_{\rm {T}}$ in the ten­sor po­ten­tial turns out to be quite large.

A.41.5 Spin-or­bit in­ter­ac­tion

So far, the as­sump­tion has been that the po­ten­tial of the deuteron only de­pends on its geom­e­try. But scat­ter­ing data sug­gests that the po­ten­tial also de­pends on the mo­tion of the nu­cle­ons. A sim­i­lar ef­fect, the spin-or­bit cou­pling oc­curs for the pro­ton-elec­tron hy­dro­gen atom, ad­den­dum {A.39}. How­ever, there the ef­fect is very small. Spin-or­bit in­ter­ac­tion is pro­por­tional to the dot prod­uct of net or­bital an­gu­lar mo­men­tum and net nu­cleon spin. In par­tic­u­lar, it pro­duces a term in the po­ten­tial of the form

$$V_{\rm {so}}(r) {\skew 4\widehat{\vec L}}\cdot {\skew 6\widehat{\vec S}}$$

This term does not con­tribute to the un­cer­tainty in or­bital an­gu­lar mo­men­tum. But it can ex­plain how nu­cleon beams can get po­lar­ized with re­spect to spin in scat­ter­ing ex­per­i­ments.