Nuclear Physics PHY303

2 Nuclear Forces

In discussing nuclear forces we first point to the differences between nucleons in nuclei and electrons in atoms. In the latter case the electrons are bound by the coulomb potential due to the electric charge of the nucleus while the electron-electron interactions can be treated as a perturbation. In addition the simplest system, the hydrogen atom, has an infinite number of energy levels which form a compehensive basis for comparisons between theory and experiment.

In contrast, the nucleons in a nucleus are bound by a potential which is due to the nucleons themselves. The problem is a many body one which only becomes tractable through simplification. This has led to the devising of quite different nuclear models to deal with different nuclear properties. Also the simplest nucleus, the deuteron, has only one bound state and so is rather limited as a test ground for theories.

2.1 Summary of characteristics - From very simple observations we can already write down quite a few of the characteristics of the nuclear force.

  1. It is a short range force. At distances of 1 fm or so it is much stronger than the coulomb force because we see that nuclei are held together against the disruptive coulomb repulsion between protons. But at distances of the order of the size of an atom the nuclear force is essentially zero. In a molecule for example the interaction between the nuclei of the different atoms is determined by their electric charges - no nuclear force is felt between them. The short range nature of the nuclear force is also indicated by the fact that the total binding energy of the nucleus is just proportional to the number of nucleons.Each nucleon just interacts with its nearest neighbours and so sees the same environment independent of how many nucleons there are at larger distances.
  2. It is an attractive force with a repulsive core. Nuclei are held together but they do not collapse. Recalling the observation made in Section 1.3 that the density of all nuclei is about the same, we see that nucleons bound in the nucleus tend to maintain the same average separation.
  3. Not all particles experience the nuclear force. This reflects the division of 'matter' into two classes of fundamental particle, quarks and leptons. The quarks are the basic entities from which nuclear matter is created. They are bound together by the strong force into hadrons like the proton, pion etc. The nuclear force is a 'left over' effect of this strong interaction - just as the inter molecular van der Waals force is a 'left over' effect of the electromagnetic interaction.The leptons (electrons, muons, tau particle and their neutrinos) do not participate in the strong interaction and consequently do not experience the nuclear force.
  4. The nucleon-nucleon force is the same irrespective of whether the nucleons are protons or neutrons - once the coulomb repulsion has been taken into account. Thus the nuclear force is charge symmetric or charge independent. For example the determination of nuclear sizes from the study of mirror nuclei is based on this charge independence of the nuclear force.
  5. The nuclear force depends on whether the nucleon spins are parallel or antiparallel. This is supported by the preponderance among stable nuclei of those with even Z and even N.
In spite of the fact that a nucleus is a many body object there is no evidence for many body forces as such. Nuclear binding can be understood in terms of two body forces and these are studied through the deuteron system and neutron-proton, proton-proton scattering.

2.2 The deuteron


Constituents 1 proton 1 neutron

Mass 2.014732 u

Binding energy 2.224589 +/- 0.000002 MeV

Angular momentum 1 (h-bar)

Magnetic moment 0.85741 +/- 0.00002 nuclear magnetons

Electric quadrupole moment +2.88 x 10-3 barn

RMS separation 4.2 fm

These properties alone tell us quite a bit about the nucleon-nucleon force. The first thing we should note about the deuteron itself is that the proton and neutron spend most of their time separated by a distance which is somewhat larger than our estimate for the range of the nuclear force. This indicates that the deuteron is rather loosely bound and such a conclusion is supported by the fact that the binding energy is very much less than the normal nuclear average B/A ~ 8 MeV per nucleon.

These points are illustrated by the figure below in which the deuteron potential and wavefunction are sketched. Note that the maximum of the wavefunction is only just inside the well so that there is a considerable exponential tail which falls outside - leading to a relatively large value (4.2 fm) for the RMS separation between the proton and neutron.

In analogy with the ground state of the hydrogen atom it is reasonable to assume that the ground state of the deuteron also has zero orbital angular momentum. Since the total angular momentum is one unit of (h-bar) it follows that this comes from the spin angular momentum which means that the proton and neutron spins are parallel. The implication is that two nucleons are not bound together if their spins are antiparallel and this is consistent with there being no proton-proton or neutron-neutron bound states. In the case of identical particles the bound parallel spin state is forbidden by the Pauli exclusion principle.

The nuclear force is thus seen to be spin dependent and this is also evident from the large difference between the neutron scattering cross-sections of ortho- and para-hydrogen.

On the assumption of no orbital angular momentum the magnetic moment of the deuteron should just be the sum of the proton and neutron magnetic moments.

_µd = µp + µn
_= (2.7928 - 1.9130) µN
_= 0.8798 µN

- which does not agree with the measured value quoted above with other deuteron properties and so throws some doubt on the assumption of no orbital motion.

Neither the proton nor the neutron has an electric quadrupole moment so that the non-zero measured value for the deuteron quoted above also points to their being some orbital motion present.

The discrepencies in the magnetic dipole and electric quadrupole moments can be overcome if the ground state is considered to be dominantly of zero orbital angular momentum (L= 0) with a small admixture of a state with non-zero orbital motion. Such a state must have the same parity (even) as the dominant (L= 0) state and also be capable of producing a total angular momentum of one unit of (h-bar) when combined with the nucleon spins. This latter eliminates any state with L> 2 and the parity requirement eliminates the odd parity L= 1 state - the only possibility is the L= 2 state. The deuteron wave function can therefore be expressed as

 d = a0(L= 0) + a2(L= 2)

and the magnetic moment as
µd = a02µ(L= 0) + a22µ(L= 2)

where µ(L= 0) = 0.8798 µN as above and µ(L= 2) = 0.3101 µN.

Using the observed value µd = 0.8574 µN and a02 + a22 = 1 we obtain the result that the ground state of the deuteron has approximately a 4% admixture of the L= 2 state. That is, for 4% of the time the orbital angular momentum switches from zero to a value of 2(h-bar) with the total angular momentum remaining fixed in magnitude and direction at I = (h-bar). In order to effect such a switching of the orbital angular momentum the nuclear force must apply a torque. The nucleon-nucleon potential is a function of angle as well as distance. It has a non-central or tensor component which is normally written

VT(r)[3(S.r)2/r2 - S.S]

The second term is included to make the average over all angles zero.

As shown at the left of the figure above, r is the vector separation of the two nucleons.The figure also illustrates the two types of non-spherical shape that would be generated by the tensor potential depending on the sign of VT(r). The deuteron has the cigar shape shown in the middle which gives a positive electric quadrupole moment as observed.

2.3 Isospin (also known as Isotopic or Isobaric spin) - The charge independence of nuclear forces has led to the treatment of the proton and neutron as two states of the same particle, the nucleon. A convenient way of dealing with this formally is to introduce the concept of isospin an abstract, extra degree of freedom which is analogous to spin. The nucleon states in this new subspace are equivalent to the "up" and "down" states of a particle with spin 1/2.

Whereas for atomic wave functions we have a space part and a spin part

= (r) spin

for the total nuclear wave function we include an isospin part

= (r) spin isospin

and apply a generalised antisymmetry condition. The total wave function including isospin must be antisymmetric under interchange of the coordinates of two fermions.

The letter T is used for the isospin of nuclei and t for that of individual nucleons. By definition the proton is the +1/2 state and the neutron the -1/2 state. This is represented by saying that these are the values of the third component of the isospin in each case - the proton has t3 = +1/2 etc.

From this we can define the third component of the isospin of a nucleus as

T3 = (Z - N)/2.

This is equivalent to the z-component in the normal spin case. Thus T3 is the projection onto the third axis in this abstract space of the isospin T which has maximum possible magnitude A/2 (total number of nucleons times 1/2). From this it can be seen that isobars have the same T (hence the term isobaric spin) and belong to an isospin multiplet with (2T + 1) members. Each nuclear state in this multiplet has the same space [ (r) ] and spin [ spin ] parts to their wave functions and consequently they all have about the same energy or mass.

The two nucleon system can be an isospin singlet (T = 0) or triplet (T = 1).

isospin triplet isospin singlet
(combines with parallel normal spin only) (combines with antiparallel normal spin only)
T = 1 T = 0
T3 = 1 (pp)
T3 = 0 (pn + np) T3 = 0 (pn - np)
T3 = -1 (nn)

Note the allowed spin states in each case. These are determined by the overall requirement that the total wave function must be antisymmetric. Isospin and spin singlets (triplets) are antisymmetric (symmetric) while the S state (L = 0) space wavefunction is symmetric.

The isospin of the nucleon and the isospin symmetry (charge independence) of nuclear forces arises from a corresponding symmetry in the quarks from which nucleons are made. The main fermion (spin 1/2) constituents of the proton and the neutron are the so-called u and d quarks. These light quarks are equivalent to each other with respect to strong interactions in the same way that the proton and the neutron are. The strong force which binds the quarks together depends upon spin but not on isospin. The u and d have different properties with respect to the electroweak interaction - e.g. their electric charges are +2/3 and -1/3 respectively.

2.4 Exchange forces - Modern field theory is based on a model in which fundamental forces are transmitted by the exchange of bosons (integer spin). The most obvious example is the electromagnetic interaction which is transmitted by photons.

Most nuclear processes however can be understood in terms of what we have referred to as the 'left over' nuclear force.There is no need to deal directly with the fundamental strong force beween quarks. Nevertheless in working with the nuclear force it is still possible to use the concept of exchange and in fact one of the earliest applications of such a concept was made in this area.

At low energies, only the outer part of the force field which surrounds the nucleon is involved and this is dominated by single pion exchange. The pion is the lightest meson, it has zero intrinsic spin and belongs to an isospin triplet t = 1 with t3 = +1 (pi-plus), 0 (pi-zero), -1 (pi-minus).

As stated above the electromagnetic interaction is transmitted by the photon and this has zero rest mass. In the case where the exchanged boson has non-zero rest mass we can examine what effect this has on the shape of the field around a source, by drawing a parallel with electrostatics. In this latter (massless boson) case we have a potential which obeys Laplace's equation where for a point source or charge.

For a field transmitted by a particle with mass we start with the relativistic expression relating energy and momentum in this case

E2 = p2c2 + m2c4

Using the standard conversions and etc. to produce a wave equation we obtain

which for the static case () reduces to


Note that this becomes as m goes to zero which brings us back to the electrostatic case and gives some assurance that the argument we have used is reasonable.

For a point source we have spherical symmetry and the equation can reformulated as

For which the solution is with . If the particle exchanged is the pion then R = 1.4 fm which fits very well with the range of the nuclear force. A sketch of the potential is given below.

This is called the Yukawa potential since the original proposal that nuclear binding forces were due to particle exchange was made by the Hideki Yukawa in 1935, and from knowledge of the range of the force he was able to predict the mass of the particle using the sort of argument we have described above. Cecil Powell discovered the pion in 1947 in cosmic rays. Yukawa was awarded the Nobel prize in 1949 and Powell was awarded it in 1950.

We now examine what might be the result of exchange between two nucleons in terms of their spin and isospin. This is illustrated simply below.

The red disks represent protons and the blue disks are neutrons with spin states indicated by the attached arrows.

The effect of the exchange of a quantity between two particles can be represented by an operator P which operates on the wave function describing the two particle state. In this case the effective potential is written V(r)P where V(r) is a normal attractive central potential. Although such a potential has little meaning in classical mechanics it causes no difficulty in quantum mechanics since it is just incorporated into the Schrödinger equation.

We will look at the effect of P in each case illustrated above to get some idea of the different components of the nuclear potential. This is clearly much more complicated than the Coulomb potential of an atom.

In example a) of the above figure the system is left unchanged and so V(r)Pa = V(r). This is an ordinary attractive central force.

In example b) the spin orientations stay the same but the positions of the two particles are interchanged. This is equivalent to the parity operator which just Affects the space part of the wave function Pb(r) = +(-r). Thus such an exchange force would be attractive for even parity states ((-r) = +(r)) and repulsive for odd parity states ((-r) = -(r)). Since the parity is determined by the orbital angular momentum quantum number, the force is attractive for even L (S,D, etc) and repulsive for odd L (P,F, etc).

In example c) the spin orientations are interchanged and so Pc operates on the spin part of the wave function Pc12 = 21. The spin triplet state (parallel spins) is symmetric and consequently 21 = 12 while the spin singlet state (anti-parallel spins) is antisymmetric which means 21 = -12. From this we conclude that the potential is of opposite sign in these two cases which means that the force is in opposite directions. This force is attractive for spin triplets (S = 1) and repulsive for the spin singlet (S = 0).

In example d) both the positions and the spins are interchanged and this is equivalent to successive applications of Pb and Pc. Following the discussion above we have PbPc = + for (S = 1, L even - ie both symmetric) or (S = 0, L odd - ie both antisymmetric) while PbPc = - for (S = 1, L odd - ie symmetric & antisymmetric) or (S = 0, L even - ie antisymmetric & symmetric). Bearing in mind the overall antisymmetry requirement placed on the wavefunction  = (r) spin isospin it can be seen that the first set of conditions above imply isospin is antisymmetric (T = 0) and the second that isospin is symmetric (T = 1). Hence the force is attractive for isospin singlets (T = 0) and repulsive for isospin triplets (T = 1).

If we include the tensor force then we have a simple physical argument for considering a two nucleon interaction made up of at least six component potentials. This is not all since we should also include velocity dependent forces which are important in scattering processes but which also show effects in nuclei - for example the spin-orbit force.

Finally, as we have mentioned above, the nuclear force has a repulsive core which prevents the nuclei from collapsing in on themselves.

We conclude this section with a crude sketch of the nucleon-nucleon potential illustrating some of the types of particle exchange which are thought to contribute.

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