Subsections


A.40 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.40.1 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.
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...0,70){\makebox(0,0)[lb]{$r^2\vert\psi\vert^2$}}
\end{picture}
\end{figure}

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.
\begin{table}\begin{displaymath}
\begin{array}{ccccccccc}
\hline\hline
d_0...
....1 & 0.94 \\
\hline\hline
\end{array}
\end{displaymath}
\end{table}


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},

\begin{displaymath}
\psi = \frac{A_{\rm S}}{\sqrt{4\pi}}
\frac{e^{-\sqrt{2m_...
...d}\vert E\vert} r/\hbar}}{r}
\qquad\mbox{for}\qquad r > d_0
\end{displaymath}

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, [30, p. 83, 252].


A.40.2 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.
\begin{table}\begin{displaymath}
\begin{array}{ccccccccc}
\hline\hline
d_0...
....0 & 0.99 \\
\hline\hline
\end{array}
\end{displaymath}
\end{table}


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.
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...0,70){\makebox(0,0)[lb]{$r^2\vert\psi\vert^2$}}
\end{picture}
\end{figure}

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, [50]. 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.40.3 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.38.5}. However, there the effect is extremely small. For the deuteron, the effect of spin is dramatic.

The proton and neutron each have spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. 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

\begin{displaymath}
V = V_{\rm {t}}(r)
\left[\frac34+\frac{1}{\hbar^2}{\skew...
... S}}_{\rm {p}}\cdot{\skew 6\widehat{\vec S}}_{\rm {n}}\right]
\end{displaymath} (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.40.4 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.38}, 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:

\begin{displaymath}
r
\qquad
{\skew 6\widehat{\vec S}}_{\rm {n}}\cdot{\ske...
...quad
{\skew 6\widehat{\vec S}}_{\rm {p}}\cdot{\skew0\vec r}
\end{displaymath}

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

\begin{displaymath}
S_{12} V_{\rm {T}}(r)
\quad\mbox{with}\quad
S_{12} \eq...
...}_{\rm {n}}\cdot{\skew 6\widehat{\vec S}}_{\rm {p}}
\right)
\end{displaymath} (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.
\begin{figure}
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...x(0,0){0}}
\put(-90,64){\makebox(0,0){2}}
}
\end{picture}
\end{figure}

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, [50].

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.40.5 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.38}. 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

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

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.