Chapter 8 Nuclear Force

8.1 The deuteron

The simplest nucleus in which we can study the nuclear force is the deuteron (1 proton and 1 neutron), since there are only two nucleons with no Coulomb repulsion.

8.1.1 Properties of the deuteron

• Nuclear spin: $I=1$

• Parity: even $(\pi =1)$

• Magnetic dipole moment, $\mu$= 0.857 ${\mu }_N$

• Charge distribution, half radius = 2.1 fm (i.e. half charge within radius)

• Binding energy, $\mathrm{\Delta }$E = 2.224 MeV

• No excited states of deuteron exist

8.1.2 Deuteron spin

The deuteron spin $I=1$ represents the total angular momentum of the deuteron (note it is conventional to use $I$ to represent the total angular momentum of nuclei, not $j$). It includes the spins of the neutron and proton and their orbital angular momentum:

$$𝐈={𝐬}_{\mathrm{n}}+{𝐬}_{\mathrm{p}}+𝐥,$$ (8.1)

where ${𝐬}_{\mathrm{n}}$ is the neutron spin (with a quantum number $s_{\mathrm{n}}=\frac{1}{2}$), ${𝐬}_{\mathrm{p}}$ is the proton spin (with a quantum number $s_{\mathrm{p}}=\frac{1}{2}$) and $𝐥$ is the orbital angular momentum (with a quantum number $\mathrm{\ell }$). Note that this is a vector equation.

For $I=1$, we can have ${𝐬}_{\mathrm{n}}$ and ${𝐬}_{\mathrm{p}}$ parallel, i.e. $s=1$ with $\mathrm{\ell }=0$, 1 or 2 (Fig. 8.1). Alternatively, if ${𝐬}_{\mathrm{n}}$ and ${𝐬}_{\mathrm{p}}$ are anti-parallel, $s=0$ and $\mathrm{\ell }$ must be 1 (Fig. 8.2). Please refer to the slides for more detailed diagrams of this.

Figure 8.1: Possible combinations of $𝐥$ and $𝐬$ for $I=1$.

Figure 8.2: For $s=0$, $\mathrm{\ell }=1$.

Difficult 8.1.1.

Fig. 8.1 and Fig. 8.2 are also common sources of confusion. Recall Eqn. (8.1) is a vector equation. The figures show the combinations of angular momentum and spin consistent with this equation. The diagrams are labelled with their respective quantum numbers, which are always positive.

We can distinguish between these four possibilities on the basis of parity. The parity of the deuteron is inferred from its reactions to be even. It is given by:

$${\pi }_{\mathrm{D}}={\pi }_{\mathrm{n}}\times {\pi }_{\mathrm{p}}\times {(-1)}^{\mathrm{\ell }}.$$ (8.2)

${\pi }_{\mathrm{D}}$, ${\pi }_{\mathrm{n}}$ and ${\pi }_{\mathrm{p}}$ are all even, this implies $\mathrm{\ell }$ must be even, so the $\mathrm{\ell }=1$ options are excluded and we conclude ${𝐬}_{\mathrm{n}}$ and ${𝐬}_{\mathrm{p}}$ must be parallel ($s=1$) with $\mathrm{\ell }=0$ or $\mathrm{\ell }=2$. How can we distinguish between them?

The observed magnetic dipole moment of the deuteron $\mu =0.857{\mu }_{\mathrm{N}}$. The expected value for $\mathrm{\ell }$=0 is

$$\mu =\frac{1}{2}{\mu }_{\mathrm{N}}\left(g_{\mathrm{s}\mathrm{p}}+g_{\mathrm{s}\mathrm{n}}\right)=0.880{\mu }_{\mathrm{N}}.$$ (8.3)

The experimental value is consistent with the deuteron spending 96% of its time in the $l$=0 state (and 4% in the $l$=2 state). Mixing of $l$ values in the deuteron is the best evidence we have for the non-central character of the nuclear force.

8.2 Properties of the nuclear force

The nuclear force has the following properties which should be reproduced by any theory describing it:

• Strong (100 $\times$ electromagnetic force)

• Short-range (cuts off at $\sim$2 fm)

• Attractive, with a repulsive core

• Charge independent (no distinction between proton and neutron)

• Saturation force (BE/nucleon is constant $\sim$8 MeV/nucleon)

8.3 The depth of the nuclear potential

8.3.1 Schrodinger’s equation for a spherically symmetric potential $V(r)$

The time-independent Schrodinger equation for a particle of mass $m$ in a potential $V(𝒓)$ is,

$$\left[-\frac{{\mathrm{\hslash }}^2}{2m}{\mathbf{\nabla }}^2+V(𝒓)\right]\mathrm{\Psi }(𝒓)=E\mathrm{\Psi }(𝒓),$$ (8.4)

where ${\mathbf{\nabla }}^2$ is the Laplacian. In spherical polar coordinates, this is

$-{\displaystyle \frac{{\mathrm{\hslash }}^2}{2m}}({\displaystyle \frac{{\partial }^2\mathrm{\Psi }}{\partial r^2}}+{\displaystyle \frac{2}{r}}{\displaystyle \frac{\partial \mathrm{\Psi }}{\partial r}}+{\displaystyle \frac{1}{r^2\sin \theta }}{\displaystyle \frac{\partial }{\partial \theta }}\left(\sin \theta {\displaystyle \frac{\partial \mathrm{\Psi }}{\partial \theta }}\right)$ (8.5)

$+{\displaystyle \frac{1}{r^2{\sin }^2\theta }}{\displaystyle \frac{{\partial }^2\mathrm{\Psi }}{\partial {\varphi }^2}})+V(r,\theta ,\varphi )\mathrm{\Psi }=E\mathrm{\Psi }$

For simplicity, we will assume that the deuteron is in the $\mathrm{\ell }=0$ state (all the time, rather than just 96%). In this case, the interaction potential $V$ is spherically symmetric and thus only a function of $r$.

We can then search for separable solutions of the form

$$\mathrm{\Psi }(r,\theta ,\psi )=\mathrm{\Psi }(r)\mathrm{\Theta }(\theta )\mathrm{\Phi }(\varphi ).$$ (8.6)

The angular terms are the spherical harmonics (see Krane page 27 for a discussion)

$$\mathrm{\Theta }(\theta )\mathrm{\Phi }(\varphi )=Y_{\mathrm{\ell }m}(\theta ,\varphi ).$$ (8.7)

The radial part $\mathrm{\Psi }(r)$ is

$$-\frac{{\mathrm{\hslash }}^2}{2m}\left(\frac{{\partial }^2\mathrm{\Psi }}{\partial r^2}+\frac{2}{r}\frac{\partial \mathrm{\Psi }}{\partial r}\right)+V(r)\mathrm{\Psi }=E\mathrm{\Psi }.$$ (8.8)

Recall this equation is for a single particle, but the deuteron has two nucleons. The Hamiltonian of a two-body problem can be reduced to a one-body problem if one uses the reduced mass, given for the deuteron by

$$m=\frac{m_n{m}_p}{m_n+m_p}\simeq \frac{m_{\mathrm{nucleon}}}{2}.$$ (8.9)

and $r$ is the distance between the two particles.

To solve Eqn. 8.8 we note

$$\frac{{\partial }^2\mathrm{\Psi }}{\partial r^2}+\frac{2}{r}\frac{\partial \mathrm{\Psi }}{\partial r}=\frac{1}{r}\frac{\partial }{\partial r}\left(\mathrm{\Psi }+r\frac{\partial \mathrm{\Psi }}{\partial r}\right)=\frac{1}{r}\frac{{\partial }^2}{\partial r^2}\left(r\mathrm{\Psi }\right).$$ (8.10)

If we make the change of variables $\mathrm{\Psi }(r)=u(r)/r$, Eqn. 8.8 then simplifies to

$$-\frac{{\mathrm{\hslash }}^2}{2m}\frac{{\partial }^{2}u}{\partial r^2}+V(r)u(r)=Eu(r).$$ (8.11)

8.3.2 Square well model of nuclear potential

Figure 8.3: 3D square well model of nuclear potential for the deuteron. Note there is no Coulomb barrier as the neutron is uncharged.

We assume the nuclear potential can be represented by a 3D square well potential of radius $R$ and depth $-V_0$, as illustrated in Fig. 8.3. Here $r$ represents the separation between the proton and the neutron, so $R$ is a measure of the diameter of the deuteron.

The solution to Eqn. 8.11 for is oscillatory

$u(r)=A\sin (k_{1}r)+\tilde{B}\cos (k_{1}r),$ (8.12)

where

$$k_1=\sqrt{\frac{2m(E-(-V_0))}{{\mathrm{\hslash }}^2}}.$$ (8.13)

For $r>R$ the solution is exponential

$u(r)=C{e}^{-k_{2}r}+D{e}^{k_{2}r},$ (8.14)

where

$$k_2=\sqrt{\frac{-2mE}{{\mathrm{\hslash }}^2}}.$$ (8.15)

Remember for a bound state, . We can argue that $\mathrm{\Psi }$(r) must remain finite both as $r\to \mathrm{\infty }$ and as $r\to 0$, so $D=0$ and $\tilde{B}=0$. We can also apply continuity conditions on both $u$ and $\frac{du}{dr}$ at $r=R$

$$u\to A\sin \left(k_{1}R\right)=C{e}^{-k_{2}R},$$ (8.16)

$$\frac{du}{dr}\to A{k}_1\cos \left(k_{1}R\right)=-k_{2}C{e}^{-k_{2}R}.$$ (8.17)

Dividing Eqn. (8.17) by Eqn. (8.16), we obtain

$$k_1\cot \left(k_{1}R\right)=-k_2.$$ (8.18)

This transcendental equation gives a relation between $V_0$ and $R$. From electron scattering experiments, the RMS charge radius of the deuteron is found to be $\sim 2.1$ fm. Solving Eqn. (8.18) numerically with $R=2.1$ fm and for $E=-2.224$ MeV, we can get $k_2$ and then substitute in to find $k_1$. We then find $V_0$ $\sim$35 MeV. This is the depth of the potential well holding the 2 nucleons together.

A 3D square potential can only have a bound state if it is sufficiently deep. As $u(r=0)=0$, the wave function must have enough curvature within the binding region to allow it to match a decreasing exponential for $r>R$ (see Figure 8.4.

Figure 8.4: Matching of $u(r)$ at edge of well.

8.3.3 Bound states of the deuteron

The bound deuteron is very close to the top of the potential well ($B=2.224\mathrm{MeV}$, $V_0$ $\sim 35$ MeV). If the nucleon-nucleon force were just a bit weaker, the bound state would not exist. (This would have profound consequences for nuclear synthesis, big bang/star burning etc.) No bound deuteron state exists for anti-parallel spins ($s=0$). Therefore the nuclear potential must be spin-dependent and appreciably weaker when the two nucleons interact with anti-parallel spins. This explains why there are no ‘di proton’ and ‘di neutron’ states. In both systems, the Pauli Exclusion Principle would require the spins to be anti-parallel and the potential is too weak to allow bound states, even for the uncharged ‘di neutron’. We can account for this by a spin-dependent term in the potential of the form ${𝐬}_1\cdot {𝐬}_2{V}_{\mathrm{s}}(r)$.

Example 8.3.1 (Probability Example).

In this example, let’s suppose the binding energy, $B$, of the deuteron is not given by its usual value, but instead by the condition $k_{1}R\approx \pi /2$. For $R=2.1\mathrm{fm}$ and $V_0=35\mathrm{MeV}$, (i) determine $B$; (ii) derive normalised wavefunctions for both and $r>R$; (iii) determine the probability that , i.e. the proton-neutron separation is less than the range of the strong-nuclear force.

Solution.

(i) $k_1=\sqrt{\frac{2m_{\star }}{{\mathrm{\hslash }}^2}(E+V_0)}=\sqrt{\frac{2m_{\star }}{{\mathrm{\hslash }}^2}(V_0-B)}.$

Assuming $k_{1}R=\pi /2$, we can rearrange to find

$B=V_0-\frac{1}{2m_{\star }{c}^2}{\left(\frac{\pi \mathrm{\hslash }c}{2R}\right)}^2.$

The reduced mass is

$m_{\star }=\frac{m_p{m}_n}{m_p+m_n}=\frac{938.3\times 939.6}{938.3+939.6}=469.5\mathrm{MeV}/c^2$

Plugging this in

$B=35-\frac{1}{2\times 469.5}{\left(\frac{\pi \times 197.3}{2\times 2.1}\right)}^2\mathrm{MeV}=11.8\mathrm{MeV}.$

(ii) To normalise we require ${\int }_0^{\mathrm{\infty }}{|\mathrm{\Psi }(r)|}^2{d}^3𝒓=1$. Since the wavefunction is only a function of radial co-ordinate then $4\pi {\int }_0^{\mathrm{\infty }}{|\mathrm{\Psi }(r)|}^2{r}^2𝑑r=1$.

Since $\mathrm{\Psi }(r)=u(r)/r$ this means we require $4\pi {\int }_0^{\mathrm{\infty }}{u}^2(r)𝑑r=1$. This integral can be split into two parts,

$$4\pi \left[A^2{\int }_0^R{\sin }^2(k_{1}r)𝑑r+C^2{\int }_R^{\mathrm{\infty }}{e}^{-2k_{2}r}𝑑r\right]=1.$$

At the boundary, we have:

$$A\sin (k_{1}R)=C{e}^{-k_{2}R}.$$

Since we are assuming $k_{1}R=\pi /2$,

$$A=C{e}^{-k_{2}R},$$

or alternatively

$$C=A{e}^{k_{2}R}.$$

Substituting back into Eqn. (Solution), we get:

$$4\pi A^2\left[{\int }_0^R{\sin }^2(k_{1}r)𝑑r+{\int }_R^{\mathrm{\infty }}{e}^{-2k_2(r-R)}𝑑r\right]=1$$

The first integral is

$${\int }_0^R{\sin }^2(k_{1}r)𝑑r=\frac{R}{2}-\frac{\sin (2k_{1}R)}{4k_1},$$

and the second integral is

$${\int }_R^{\mathrm{\infty }}{e}^{-2k_2(r-R)}𝑑r=\frac{1}{2k_2}.$$

This fixes the normalisation $A$ by the condition

$$4\pi A^2\left[\frac{R}{2}+\frac{1}{2k_2}\right]=1.$$

(iii) The probability of finding the proton-neutron separation less than $R$ is

$$P=4\pi A^2{\int }_0^R{r}^2{\sin }^2(k_{1}r)𝑑r.$$

Using the result from part (ii)

$$P={\left(1+\frac{1}{k_{2}R}\right)}^{-1}.$$

$$k_2=\sqrt{\frac{2m_{\star }{c}^{2}B}{{\mathrm{\hslash }}^2{c}^2}}=\sqrt{\frac{2\times 469.5\times 11.8}{{197.3}^2}}{\text{ fm}}^{-1}=0.53{\text{ fm}}^{-1}.$$

This gives

$$P={\left(1+\frac{1}{0.53\times 2.1}\right)}^{-1}=0.53.$$

Example 8.3.2 (Bound States).

Consider the sum of two nucleon spins

$$𝐬={𝐬}_1+{𝐬}_2.$$ (8.19)

Calculate the change in the nuclear potential between singlet and triplet states.

Solution.

In analogy with the spin-orbit coupling of the shell model write

$${𝐬}_1\cdot {𝐬}_2=\frac{1}{2}\left({𝐬}^2-𝐬_1^{}{}_{}{}^2-𝐬_2^{}{}_{}{}^2\right).$$ (8.20)

Calculate the expectation value using

$$⟨{𝐬}^2⟩={\mathrm{\hslash }}^{2}s(s+1),$$ (8.21)

giving

$$⟨{𝐬}_1\cdot {𝐬}_2⟩=\frac{1}{2}\left[s(s+1)-s_1(s_1+1)-s_2(s_2+1)\right]{\mathrm{\hslash }}^2.$$ (8.22)

We have $s_1=s_2=1/2$. Allowed value of $s$ are $|s_1-s_2|=0$ (called a singlet state) and $|s_1+s_2|=1$ (called a triplet state). For the triplet state

$$⟨{𝐬}_1\cdot {𝐬}_2⟩=\frac{1}{2}\left[1(1+1)-1/2(1/2+1)-1/2(1/2+1)\right]{\mathrm{\hslash }}^2=\frac{1}{4}{\mathrm{\hslash }}^2.$$ (8.23)

For the single-state

$$⟨{𝐬}_1\cdot {𝐬}_2⟩=\frac{1}{2}\left[0(0+1)-1/2(1/2+1)-1/2(1/2+1)\right]{\mathrm{\hslash }}^2=-\frac{3}{4}{\mathrm{\hslash }}^2.$$ (8.24)

8.4 Scattering of neutrons by protons

Let us now turn our attention away from the bound deuteron state and consider the scattering of neutrons by protons. Experimental measurements of this scattering cross-section are shown in Figure 8.5.

Figure 8.5: Cross section for scattering of neutrons by protons.

At low energy, the cross-section is measured to be 20.4 b, whereas simple scattering theory predicts 4.5 b. What is the reason for the discrepancy?

The total spin for the proton and neutron, $s$, can be 0 or 1 (remember we are not considering the bound state of the deuteron here for which we argued we must have $s=1$). $s=0$ corresponds to a singlet state and $s=1$ corresponds to a triplet state.

As the incident neutron approaches the target proton, the probability of a triplet state is $3/4$ and a singlet state $1/4$. If the scattering cross-section is different for singlet and triplet states, then

$${\sigma }_{\mathrm{T}}=\frac{3}{4}{\sigma }_{\mathrm{triplet}}+\frac{1}{4}{\sigma }_{\mathrm{singlet}}.$$ (8.25)

The theoretical prediction of 4.5 b was based on parameters obtained from the deuteron which is in a $s=1$ state. If we therefore assume ${\sigma }_{\mathrm{triplet}}=4.5$ b, we can use the measured value of ${\sigma }_T=20.4$ b to deduce that ${\sigma }_{\mathrm{singlet}}=68.1$ b. This large difference re-affirms our assertion that the nuclear force is very spin-dependent.

Difficult 8.4.1.

The total spin for the proton and neutron, $s$, can be 0 or 1. The $s=0$ state only has 1 possible z component of spin ($m_s=0$) and is called a singlet state. You can visualize the spins of the two particles as in Fig. 8.6 (notice the spins are equal and opposite and the z component is 0).

Figure 8.6: Visualization of singlet state.

The $s=1$ state has 3 possible z components of spin ($m_s=-1,0,+1$) and is called a triplet state. You can visualize the spins of the two particles as in Fig. 8.7 (notice the spins add up to give a total spin of $s=1$ with different z components in each case).

Figure 8.7: Visualization of triplet state.

8.5 The exchange model

Extensive programmes of nucleon-nucleon scattering up to the GeV range have led to very detailed representations of the nuclear potential. Each new term improves the fit to experimental data, but at the expense of simplicity and has not necessarily improved our basic understanding. What is needed is a physical mechanism that yields the empirical potentials.

This mechanism turns out to be the exchange model. During interaction, the proton and neutron can ‘exchange places’, something that is observed in many nuclear reactions. We can associate this with the exchange quanta of the nuclear force. For a spin $1/2$ neutron to turn into a spin $1/2$ proton the exchange particle must have integral spin (0 or 1 is possible). It must carry electric charge, but for n-n or p-p interactions, there must also be an uncharged variety of the exchange particle.

8.5.1 Virtual particles

Based on the observed range of the nuclear force, $R$, we can estimate the mass, m_ex, of the exchange particle. The emission and re-absorption of the particle must take place within a short time, $\mathrm{\Delta }t$, such that we are ‘unaware’ that energy conservation has been violated. According to Heisenberg’s Uncertainty Principle:

$\mathrm{\Delta }E\mathrm{\Delta }t\sim \mathrm{\hslash },$ (8.26)

$m_{\mathrm{ex}}{c}^2\mathrm{\Delta }t\sim \mathrm{\hslash }.$

If we assume that the exchange particle travels at the velocity of light, $c$:

$$\mathrm{\Delta }t=\frac{R}{c},$$ (8.27)

and the rest mass energy of the exchange particle is:

$$m_{\mathrm{ex}}{c}^2\sim \frac{\mathrm{\hslash }c}{R}.$$ (8.28)

Example 8.5.1 (Virtual Particles).

Taking $R=2\mathrm{fm}$, we find

$m_{\mathrm{ex}}{c}^2\sim {\displaystyle \frac{1\times {10}^{-34}\times (3\times {10}^8)}{2\times {10}^{-15}}}\mathrm{J}\sim 100\mathrm{MeV}.$ (8.29)

Such exchange particles are known as ‘virtual’ particles because they only exist within the constraints of the Uncertainty Principle.

8.5.2 Pion exchange theory

The exchange particles are identified as pions ($\pi$-mesons). To satisfy all of the exchanges there must be three types, carrying -1, 0, +1 units of charge.

The pions have spin 0 and rest mass:

• 139.6 MeV/c² (${\pi }^{+},{\pi }^{-}$)

• 135.0 MeV/c² (${\pi }^0$)

The pion exchange theory of the nuclear force was developed by Yukawa in 1935. It is shown diagrammatically for the different types of nucleon-nucleon interaction in Figure 8.8.

Figure 8.8: Yukawa’s pion exchange model of the nuclear force. Incoming particles are shown on the left, outgoing particles on the right.

In charge exchange diagrams it is irrelevant whether we have a ${\pi }^{+}$ moving one way or ${\pi }^{-}$ moving the other way. Pion exchange can be represented by a potential of the form

$$V(r)\sim \frac{e^{-r/R}}{r}$$ (8.30)

Other properties of the nucleon-nucleon potential – spin dependence, tensor component etc. can be understood in various versions of this theory. However, quantitative calculations are not very successful in accounting for experimental results and the theory has been discarded in favour of quark theory. Nevertheless, investigating the properties of pions led to the discovery of many new particles and to the beginning of an understanding of the structure of nucleons.

8.6 Exercises

Example 8.6.1.

Assume the combined neutron and proton spin is $𝐬={𝐬}_{𝐧}+{𝐬}_{𝐩}$ and they have the same spin g-factor $g_S$. The magnetic moment of the deuteron is, in terms of ${\mu }_N$,

$$\mu =\frac{g_L𝐥\cdot 𝐈+g_S𝐬\cdot 𝐈}{I(I+1)}.$$ (8.31)

Evaluate the expectation value of this for when the deuteron spin is $I=1$, the neutron and proton spins are parallel ($s=1$) and (i) $\mathrm{\ell }=0$ and (ii) $\mathrm{\ell }=2$.

Solution.

We need to evaluate $𝐥\cdot 𝐈$ and $𝐬\cdot 𝐈$. This can be done by using the usual trick of writing $𝐈=𝐬+𝐥$, squaring and rearranging to obtain $𝐥\cdot 𝐬=\frac{1}{2}\left({𝐈}^2-{𝐬}^2-{𝐥}^2\right)$. Then

$$𝐥\cdot 𝐈={𝐥}^2+𝐥\cdot 𝐬=\frac{1}{2}\left({𝐈}^2-{𝐬}^2+{𝐥}^2\right),$$

and similarly

$$𝐬\cdot 𝐈=\frac{1}{2}\left({𝐈}^2+{𝐬}^2-{𝐥}^2\right).$$

The expectation values are

$$⟨𝐥\cdot 𝐈⟩=\frac{{\mathrm{\hslash }}^2}{2}\left(I(I+1)-s(s+1)+\mathrm{\ell }(\mathrm{\ell }+1)\right),$$

and

$$⟨𝐬\cdot 𝐈⟩=\frac{{\mathrm{\hslash }}^2}{2}\left(I(I+1)+s(s+1)-\mathrm{\ell }(\mathrm{\ell }+1)\right),$$

For the case $\mathrm{\ell }=0$ we have $⟨𝐥\cdot 𝐈⟩=0$ and $⟨𝐬\cdot 𝐈⟩=2{\mathrm{\hslash }}^2$. This gives $\mu =g_s$ (setting $\mathrm{\hslash }=1$). For the case $\mathrm{\ell }=2$ we have $⟨𝐥\cdot 𝐈⟩=3{\mathrm{\hslash }}^2$ and $⟨𝐬\cdot 𝐈⟩=-{\mathrm{\hslash }}^2$. This gives $\mu =3g_L/2-g_s/2$ (again setting $\mathrm{\hslash }=1$).