Subsections


A.41 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.41.1 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 $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. (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 $\vphantom0\raisebox{1.5pt}{$-$}$$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:

\begin{displaymath}
E = \sqrt{{\skew0\vec p}^{\,2}c^2 + \big(mc^2\big)^2}
\end{displaymath}

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

\begin{displaymath}
- \hbar^2 c^2 \nabla^2 \varphi_\pi + \Big(m_\pi c^2\Big)^2\varphi_\pi
= E^2\varphi_\pi %
\end{displaymath} (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:

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

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

\begin{displaymath}
\fbox{$\displaystyle
V_{\rm Yukawa} = - C_{\rm Y} \frac{...
...\equiv \frac{\hbar c}{m_\pi c^2} \approx 1.4\mbox{ fm}
$} %
\end{displaymath} (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.41.2 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:

\begin{displaymath}
\fbox{$\displaystyle
V_{\rm OPEP} \sim
\frac{g_\pi^2}{...
...gma_2 + S_{12} V_{\rm{T}} \Big]
\frac{e^{-r/R}}{r/R}
$} %
\end{displaymath} (A.262)


\begin{displaymath}
S_{12} \equiv
\frac{3}{r^2}(\vec\sigma_1\cdot{\skew0\vec...
...quad
V_{\rm {T}} \equiv 1 + 3\frac{R}{r} + 3\frac{R^2}{r^2}
\end{displaymath}

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:

\begin{displaymath}
r \equiv \vert{\skew0\vec r}\vert \equiv \vert{\skew0\vec ...
...ox 138 \mbox{ MeV} \quad m_{\rm p}c^2 \approx 938 \mbox{ MeV}
\end{displaymath}

Also $g_\pi^2$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ 15 is an empirical constant, [35, 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\c...
...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.40.4}. This potential can create uncertainty in orbital angular momentum. As discussed in {A.40.4}, 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.41.3 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

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

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:

\begin{displaymath}
E_{\varphi i} = - \int \varphi f_i {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

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

\begin{displaymath}
- \nabla^2 \varphi + \left(\frac{m_\pi c^2}{\hbar c}\right)^2\varphi
= \frac{1}{\epsilon_1} \sum_i f_i %
\end{displaymath} (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:

\begin{displaymath}
V_{ij} = - \int \varphi_i({\skew0\vec r}) f_j({\skew0\vec r}) {\,\rm d}^3{\skew0\vec r} %
\end{displaymath} (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,

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

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},

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

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

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 $\vphantom0\raisebox{1.5pt}{$-$}$${\skew0\vec r}$. But as mentioned, the first-guess potential is spherically symmetric; the values at ${\skew0\vec r}$ and $\vphantom0\raisebox{1.5pt}{$-$}$${\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,

\begin{displaymath}
\frac{\partial\varphi_{i,\rm fg}}{\partial x} =
\frac{\partial\varphi_{i,\rm fg}}{\partial r} \frac{x}{r}
\end{displaymath}

And that factor $x$ will produce an opposite sign at $\vphantom0\raisebox{1.5pt}{$-$}$${\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

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

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

\begin{displaymath}
- \nabla^2 \varphi_{i,\rm sg} +
\left(\frac{m_\pi c^2}{\...
...i_{i,\rm sg} = \frac{\partial \varphi_{i,\rm fg}}{\partial x}
\end{displaymath}

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

\begin{displaymath}
c(r) Y_1^{1} - c(r) Y_1^{-1}
\end{displaymath}

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 $\vphantom0\raisebox{1.5pt}{$-$}$$\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 $\vphantom0\raisebox{1.5pt}{$-$}$$\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

\begin{displaymath}
f_{i,\rm tg} = g R \vec\sigma_i\cdot\nabla_i \vert\Psi_i\v...
..._z \frac{\partial}{\partial z_i}
\right] \vert\Psi_i\vert^2
\end{displaymath}

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

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

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 $f_i$:

\begin{displaymath}
\vec f_i = g R \vec\tau_i\;\;\vec\sigma_i\cdot\nabla_i \ve...
..._z \frac{\partial}{\partial z_i}
\right] \vert\Psi_i\vert^2
\end{displaymath}

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

\begin{displaymath}
\frac{gR}{4\pi\epsilon_1} \frac{{\rm d}e^{-r/R}/r}{{\rm d}...
...lon_1}
\left[\frac{1}{Rr^2}+\frac{1}{r^3}\right] e^{-r/R} x
\end{displaymath}

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

\begin{displaymath}
\varphi_i({\skew0\vec r}) = - \frac{gR}{4\pi\epsilon_1} \v...
...\frac{1}{r^3}\right] e^{-r/R} {\skew0\vec r}\cdot\vec\sigma_i
\end{displaymath}

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

\begin{displaymath}
\mbox{[some constant] }
\int \varphi_i({\skew0\vec r}) \...
...ial \vert\Psi_j\vert^2}{\partial x} {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

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

\begin{displaymath}
V_{ij} \sim \frac{g^2}{12\pi\epsilon_1R} \;\, \vec\tau_i\c...
...vec\sigma_j + S_{ij} V_{\rm {T}} \Big]
\frac{e^{-r/R}}{r/R}
\end{displaymath}

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


A.41.4 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, [30, p. 111], [35, 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 $\vphantom0\raisebox{1.5pt}{$-$}$$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 [35] picture the resonances as correlated multi-pion states. However, quantum chromedynamics has been associating actual particles with them. Take the rho, for example. In [35] 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, [35, 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 [35, 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, [31].