A.40 Deuteron wave func­tion

This ad­den­dum ex­am­ines the form of the wave func­tion of the deuteron. It as­sumes that the deuteron can be de­scribed as a two par­ti­cle sys­tem; a pro­ton and a neu­tron. In re­al­ity, both the pro­ton and the neu­tron con­sist of three quarks. So the deuteron is re­ally a six par­ti­cle sys­tem. That will be ig­nored here.

Then the deuteron wave func­tion is a func­tion of the po­si­tions and spin an­gu­lar mo­menta of the pro­ton and neu­tron. That how­ever can be sim­pli­fied con­sid­er­ably. First of all, it helps if the cen­ter of grav­ity of the deuteron is taken as ori­gin of the co­or­di­nate sys­tem. In that co­or­di­nate sys­tem, the in­di­vid­ual po­si­tions of pro­ton and neu­tron are no longer im­por­tant. The only quan­tity that is im­por­tant is the po­si­tion vec­tor go­ing from neu­tron to pro­ton, {A.5}:

{\skew0\vec r}\equiv {\skew0\vec r}_{\rm {p}}-{\skew0\vec r}_{\rm {n}}

That rep­re­sents the rel­a­tive po­si­tion of the pro­ton rel­a­tive to the neu­tron.

Con­sider now the spin an­gu­lar mo­menta of pro­ton and neu­tron. The two have spin an­gu­lar mo­menta of the same mag­ni­tude. The cor­re­spond­ing quan­tum num­ber, called the spin for short, equals $s_{\rm {p}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $s_{\rm {n}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$. How­ever, the pro­ton and neu­tron can still have dif­fer­ent spin an­gu­lar mo­men­tum along what­ever is cho­sen to be the $z$-​axis. In par­tic­u­lar, each can have a spin $S_z$ along the $z$-​axis that is ei­ther $\frac12\hbar$ or $-\frac12\hbar$.

All to­gether it means that the deuteron wave func­tion de­pends non­triv­ially on both the nu­cleon spac­ing and the spin com­po­nents in the $z$-​di­rec­tion:

\psi = \psi\left({\skew0\vec r},S_{z,\rm p},S_{z,\rm n}\right)

The square mag­ni­tude of this wave func­tion gives the prob­a­bil­ity den­sity to find the nu­cle­ons at a given spac­ing ${\skew0\vec r}$ and with given spin val­ues along the $z$-​axis.

It is solidly es­tab­lished by ex­per­i­ments that the wave func­tion of the deuteron has net nu­clear spin $j_{\rm {N}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and even par­ity. The ques­tion to be ex­am­ined now is what that means for the or­bital an­gu­lar mo­men­tum and the spins of the pro­ton and neu­tron. To an­swer that, the wave func­tion needs to be writ­ten in terms of states that have def­i­nite com­bined or­bital an­gu­lar mo­men­tum and def­i­nite com­bined spin.

The con­di­tions for a state to have def­i­nite or­bital an­gu­lar mo­men­tum were dis­cussed in chap­ter 4.2. The an­gu­lar de­pen­dence of the state must be given by a spher­i­cal har­monic $Y_l^m(\theta,\phi)$. Here $\theta$ and $\phi$ are the an­gles that the vec­tor ${\skew0\vec r}$ makes with the axes of the cho­sen spher­i­cal co­or­di­nate sys­tem. The az­imuthal quan­tum num­ber $l$ de­scribes the mag­ni­tude of the or­bital an­gu­lar mo­men­tum. In par­tic­u­lar, the mag­ni­tude of the or­bital mo­men­tum is $\sqrt{l(l+1)}\hbar$. The mag­netic quan­tum num­ber $m$ de­scribes the com­po­nent of the or­bital an­gu­lar mo­men­tum along the cho­sen $z$-​axis. In par­tic­u­lar, that com­po­nent equals $m\hbar$. Both $l$ $\raisebox{-.5pt}{$\geqslant$}$ 0 and $\vert m\vert$ $\raisebox{-.3pt}{$\leqslant$}$ $l$ must be in­te­gers.

As far as the com­bined spin an­gu­lar mo­men­tum is con­cerned, the pos­si­bil­i­ties were dis­cussed in chap­ter 5.5.6 and in more de­tail in chap­ter 12. First, the pro­ton and neu­tron spins can can­cel each other per­fectly, pro­duc­ing a state of zero net spin. This state is called the sin­glet state. Zero net spin has a cor­re­spond­ing quan­tum num­ber $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. And since the com­po­nent of the an­gu­lar mo­men­tum along any cho­sen $z$-​axis can only be zero, so is the spin mag­netic quan­tum num­ber $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.

The sec­ond pos­si­bil­ity is that the pro­ton and neu­tron align their spins in par­al­lel, crudely speak­ing. More pre­cisely, the com­bined spin has a mag­ni­tude given by quan­tum num­ber $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12+\frac12$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. The com­bined spin an­gu­lar mo­men­tum along the cho­sen $z$ di­rec­tion is $m_s\hbar$ where $m_s$ can be $\vphantom{0}\raisebox{1.5pt}{$-$}$1, 0, or 1.

The wave func­tion of the deuteron can be writ­ten as a com­bi­na­tion of the above states of or­bital and spin an­gu­lar mo­men­tum. It then takes the generic form:

\psi = \sum_{nlmsm_s}
c_{nlmsm_s} R_n(\vert{\skew0\vec r}\vert) Y_l^m(\theta,\phi) {\left\vert s\,m_s\right\rangle} %
\end{displaymath} (A.257)

Here the $c_{nlmsm_s}$ are con­stants. The func­tions $R_n$ are not of par­tic­u­lar in­ter­est here; any com­plete set of or­tho­nor­mal ra­dial func­tions will do. Note that the in­di­vid­ual terms in the sum above are not sup­posed to be en­ergy eigen­func­tions. They are merely cho­sen states of def­i­nite or­bital and spin an­gu­lar mo­men­tum. The ket ${\left\vert s\,m_s\right\rangle}$ is a way of in­di­cat­ing the com­bined spin state of the two nu­cle­ons. It is de­fined in terms of the sep­a­rate spins of the pro­ton and neu­tron in chap­ter 5.5.6 (5.26).

The above ex­pres­sion for the wave func­tion is quite gen­er­ally valid for a sys­tem of two fermi­ons. But it can be made much more spe­cific based on the men­tioned known prop­er­ties of the deuteron.

The sim­plest is the fact that the par­ity of the deuteron is even. Spher­i­cal har­mon­ics have odd par­ity if $l$ is odd, and even if $l$ is even, {D.14}. So there can­not be any odd val­ues of $l$ in the sum above. In other words, the con­stants $c_{nlmsm_s}$ must be zero for odd $l$.

Phys­i­cally, that means that the spa­tial wave func­tion is sym­met­ric with re­spect to re­plac­ing ${\skew0\vec r}$ by $\vphantom{0}\raisebox{1.5pt}{$-$}$${\skew0\vec r}$. It may be noted that this spa­tial sym­me­try and the cor­re­spond­ing even par­ity are ex­actly what is ex­pected the­o­ret­i­cally. The rea­sons were ex­plored ear­lier in {A.8} and {A.9}. The wave func­tion for any given spin state should not change sign, and odd par­ity can­not meet that re­quire­ment. How­ever, it should be noted that the ar­gu­ments in {A.8} and {A.9} are not valid if the po­ten­tial in­cludes terms of sec­ond or higher or­der in the mo­men­tum. Some more ad­vanced po­ten­tials that have been writ­ten down in­clude such terms.

The spa­tial sym­me­try also means that the wave func­tion is sym­met­ric with re­spect to swap­ping the two nu­cle­ons. That is be­cause ${\skew0\vec r}$ is the vec­tor from neu­tron to pro­ton, so swap­ping the two in­verts the sign of ${\skew0\vec r}$. This does as­sume that the small dif­fer­ence in mass be­tween the neu­tron and pro­ton is ig­nored. Oth­er­wise the swap would change the cen­ter of grav­ity. Re­call that the (part of the) wave func­tion con­sid­ered here is rel­a­tive to the cen­ter of grav­ity. In any case, the hy­po­thet­i­cal wave func­tions for a bound state of two pro­tons or one of two neu­trons would be ex­actly sym­met­ric un­der ex­change of the po­si­tions of the two iden­ti­cal par­ti­cles.

The con­di­tion that the nu­clear spin $j_{\rm {N}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 is a bit more com­plex. First a brief re­view is needed into how an­gu­lar mo­menta com­bine in quan­tum me­chan­ics. (For a more com­plete de­scrip­tion, see chap­ter 12.) A state with def­i­nite quan­tum num­bers $l$ and $s$ has in gen­eral quan­tum un­cer­tainty in the net nu­clear spin $j_{\rm {N}}$. But the val­ues of $j_{\rm {N}}$ can­not be com­pletely ar­bi­trary. The only val­ues that can have nonzero prob­a­bil­ity are in the range

\vert l-s\vert \mathrel{\raisebox{-.7pt}{$\leqslant$}}j_{\rm {N}} \mathrel{\raisebox{-.7pt}{$\leqslant$}}l+s

The key is now that un­less a state $l,s$ has a nonzero prob­a­bil­ity for $j_{\rm {N}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, it can­not ap­pear in the deuteron wave func­tion at all. To ver­ify that, take an in­ner prod­uct of the state with the rep­re­sen­ta­tion (A.257) of the deuteron wave func­tion. In the left hand side, you get zero be­cause the deuteron wave func­tion has $j_{\rm {N}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and states of dif­fer­ent $j_{\rm {N}}$ are or­thog­o­nal. In the right hand side, all terms ex­cept one drop out be­cause the states in the sum are or­tho­nor­mal. The one re­main­ing term is the co­ef­fi­cient of the con­sid­ered state. Then this co­ef­fi­cient must be zero since the left hand side is.

Us­ing the above cri­te­rion, con­sider which states can­not ap­pear in the deuteron wave func­tion. First of all, states with $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 are ac­cord­ing to the in­equal­i­ties above states of nu­clear spin $j_{\rm {N}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l$. That can­not be 1, since $l$ had to be even be­cause of par­ity. So states with $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 can­not ap­pear in the deuteron wave func­tion. It fol­lows that the deuteron wave func­tion has a com­bined nu­cleon spin $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 with­out quan­tum un­cer­tainty.

Sec­ondly, states with $l$ $\raisebox{-.5pt}{$\geqslant$}$ 4 have $j_{\rm {N}}$ at least equal to 3 ac­cord­ing to the above in­equal­i­ties. So these states can­not ap­pear. That leaves only states with $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 or 2 and $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 as pos­si­bil­i­ties.

Now states with $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 and $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 are states with $j_{\rm {N}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. Any such state can ap­pear in the deuteron wave func­tion. To what amount re­mains un­known. That would only be an­swer­able if an ex­act so­lu­tion to the pro­ton-neu­tron deuteron would be avail­able. But surely, based on ar­gu­ments like those in {A.8} and {A.9}, it is to be ex­pected that there is a sig­nif­i­cant $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 com­po­nent.

States with $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 and $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 are also a pos­si­bil­ity. But they can­not ap­pear in ar­bi­trary com­bi­na­tions. Any such state has mul­ti­ple pos­si­ble val­ues for $j_{\rm {N}}$ in the range from 1 to 3. That un­cer­tainty must be elim­i­nated be­fore the states are ac­cept­able for the deuteron wave func­tion. It turns out that pure $j_{\rm {N}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 states can be ob­tained by tak­ing spe­cific com­bi­na­tions of states. In par­tic­u­lar, groups of states that vary only in the quan­tum num­bers $m$ and $m_s$ can be com­bined into states with $j_{\rm {N}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. (For the cu­ri­ous, the spe­cific com­bi­na­tions needed can be read off in fig­ure 12.6).

The bot­tom line is that the deuteron wave func­tion can have un­cer­tainty in the or­bital an­gu­lar mo­men­tum. In par­tic­u­lar, both or­bital an­gu­lar mo­men­tum num­bers $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 and $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 can and do have a nonzero prob­a­bil­ity.