The nuclear forces are almost charge independent. If we assume they are, we can introduce a new quantum number which is conserved. For nucleons only, that is a proton and neutron, we can limit ourselves to two possible values which allow us to distinguish between the two particles. If we assign an isospin value of \( \tau=1/2 \) for protons and neutrons (they belong to an isospin doublet, in the same way as we discussed the spin \( 1/2 \) multiplet), we can define the neutron to have isospin projection \( \tau_z=+1/2 \) and a proton to have \( \tau_z=-1/2 \). These assignements are the standard choices in low-energy nuclear physics.
The aim is to give you an overview over central features of the nucleon-nucleon interaction and how it is constructed, with both technical and theoretical approaches.
Comparison of the binding energies of \( {}^2\mbox{H} \) (deuteron), \( {}^3\mbox{H} \) (triton), \( {}^4\mbox{He} \) (alpha - particle) show that the nuclear force is of finite range (\( 1-2 \) fm) and very strong within that range.
For nuclei with \( A>4 \), the energy saturates: Volume and binding energies of nuclei are proportional to the mass number \( A \) (as we saw from exercise 1).
Nuclei are also bound. The average distance between nucleons in nuclei is about \( 2 \) fm which must roughly correspond to the range of the attractive part.
Charge-symmetry breaking (CSB), after electromagnetic effects have been removed:
Here we display a typical way to parametrize (non-relativistic expression) the nuclear two-body force in terms of some operators, the central part, the spin-spin part and the central force. $$ V(\mathbf{r})= \left\{ C_c + C_\mathbf{\sigma} \mathbf{\sigma}_1\cdot\mathbf{\sigma}_2 + C_T \left( 1 + {3\over m_\alpha r} + {3\over\left(m_\alpha r\right)^2}\right) S_{12} (\hat r)\right. $$ $$ \left. + C_{SL} \left( {1\over m_\alpha r} + {1\over \left( m_\alpha r\right)^2} \right) \mathbf{L}\cdot \mathbf{S} \right\} \frac{e^{-m_\alpha r}}{m_\alpha r} $$ How do we derive such terms? (Note: no isospin dependence and that the above is an approximation)
To derive the above famous form of the nuclear force using field theoretical concepts, we will need some elements from relativistic quantum mechanics. These derivations will be given below. The material here gives some background to this. I know that many of you have not taken a course in quantum field theory. I hope however that you can see the basic ideas leading to the famous non-relativistic expressions for the nuclear force.
Furthermore, when we analyze nuclear data, we will actually try to explain properties like spectra, single-particle energies etc in terms of the various terms of the nuclear force. Moreover, many of you will hear about these terms at various talks, workshops, seminars etc. Then, it is good to have an idea of what people actually mean!!
| Baryons | Mass (MeV) | Mesons | Mass (MeV) |
|---|---|---|---|
| \( p,n \) | 938.926 | \( \pi \) | 138.03 |
| \( \Lambda \) | 1116.0 | \( \eta \) | 548.8 |
| \( \Sigma \) | 1197.3 | \( \sigma \) | \( \approx 550.0 \) |
| \( \Delta \) | 1232.0 | \( \rho \) | 770 |
| \( \omega \) | 782.6 | ||
| \( \delta \) | 983.0 | ||
| \( K \) | 495.8 | ||
| \( K^{\star} \) | 895.0 |
But before we proceed, we will look into specific quantum numbers of the relative system and study expectation vaues of the various terms of $$ V(\mathbf{r})= \left\{ C_c + C_\mathbf{\sigma} \mathbf{\sigma}_1\cdot\mathbf{\sigma}_2 + C_T \left( 1 + {3\over m_\alpha r} + {3\over \left(m_\alpha r\right)^2}\right) S_{12} (\hat r)\right. $$ $$ \left. + C_{SL} \left( {1\over m_\alpha r} + {1\over \left( m_\alpha r\right)^2} \right) \mathbf{L}\cdot \mathbf{S} \right\} \frac{e^{-m_\alpha r}}{m_\alpha r} $$
When solving the scattering equation or solving the two-nucleon problem, it is convenient to rewrite the Schroedinger equation, due to the spherical symmetry of the Hamiltonian, in relative and center-of-mass coordinates. This will also define the quantum numbers of the relative and center-of-mass system and will aid us later in solving the so-called Lippman-Schwinger equation for the scattering problem.
We define the center-of-mass (CoM) momentum as $$ \mathbf{K}=\sum_{i=1}^A\mathbf{k}_i, $$ with \( \hbar=c=1 \) the wave number \( k_i=p_i \), with \( p_i \) the pertinent momentum of a single-particle state. We have also the relative momentum $$ \mathbf{k}_{ij}=\frac{1}{2}(\mathbf{k}_i-\mathbf{k}_j). $$ We will below skip the indices \( ij \) and simply write \( \mathbf{k} \)
In a similar fashion we can define the CoM coordinate $$ \mathbf{R}=\frac{1}{A}\sum_{i=1}^{A}\mathbf{r}_i, $$ and the relative distance $$ \mathbf{r}_{ij}=(\mathbf{r}_i-\mathbf{r}_j). $$
With the definitions $$ \mathbf{K}=\sum_{i=1}^A\mathbf{k}_i, $$ and $$ \mathbf{k}_{ij}=\frac{1}{2}(\mathbf{k}_i-\mathbf{k}_j). $$ we can rewrite the two-particle kinetic energy (note that we use \( \hbar=c=1 \) as $$ \frac{\mathbf{k}_1^2}{2m_n}+\frac{\mathbf{k}_2^2}{2m_n}=\frac{\mathbf{k}^2}{m_n}+\frac{\mathbf{K}^2}{4m_n}, $$ where \( m_n \) is the average of the proton and the neutron masses.
Since the two-nucleon interaction depends only on the relative distance, this means that we can separate Schroedinger's equation in an equation for the center-of-mass motion and one for the relative motion.
With an equation for the relative motion only and a separate one for the center-of-mass motion we need to redefine the two-body quantum numbers.
Previously we had a two-body state vector defined as \( |(j_1j_2)JM_J\rangle \) in a coupled basis. We will now define the quantum numbers for the relative motion. Here we need to define new orbital momenta (since these are the quantum numbers which change). We define $$ \hat{l}_1+\hat{l}_2=\hat{\lambda}=\hat{l}+\hat{L}, $$ where \( \hat{l} \) is the orbital momentum associated with the relative motion and \( \hat{L} \) the corresponding one linked with the CoM. The total spin \( S \) is unchanged since it acts in a different space. We have thus that $$ \hat{J}=\hat{l}+\hat{L}+\hat{S}, $$ which allows us to define the angular momentum of the relative motion $$ { \cal J} = \hat{l}+\hat{S}, $$ where \( { \cal J} \) is the total angular momentum of the relative motion.
The total two-nucleon state function has to be anti-symmetric. The total function contains a spatial part, a spin part and an isospin part. If isospin is conserved, this leads to in case we have an \( s \)-wave with spin \( S=0 \) to an isospin two-body state with \( T=1 \) since the spatial part is symmetric and the spin part is anti-symmetric.
Since the projections for \( T \) are \( T_z=-1,0,1 \), we can have a \( pp \), an \( nn \) and a \( pn \) state.
For \( l=0 \) and \( S=1 \), a so-called triplet state, \( ^3S_1 \), we must have \( T=0 \), meaning that we have only one state, a \( pn \) state. For other partial waves, the following table lists states up to \( f \) waves. We can systemize this in a table as follows, recalling that \( |\mathbf{l}-\mathbf{S}| \le |\mathbf{J}| \le |\mathbf{l}+\mathbf{S}| \),
| \( ^{2S+1}l_J \) | \( J \) | \( l \) | \( S \) | \( T \) | \( \vert pp\rangle \) | \( \vert pn\rangle \) | \( \vert nn\rangle \) |
| \( ^{1}S_0 \) | 0 | 0 | 0 | 1 | yes | yes | yes |
| \( ^{3}S_1 \) | 1 | 0 | 1 | 0 | no | yes | no |
| \( ^{3}P_0 \) | 0 | 1 | 1 | 1 | yes | yes | yes |
| \( ^{1}P_1 \) | 1 | 1 | 0 | 0 | no | yes | no |
| \( ^{3}P_1 \) | 1 | 1 | 1 | 1 | yes | yes | yes |
| \( ^{3}P_2 \) | 2 | 1 | 1 | 1 | yes | yes | yes |
| \( ^{3}D_1 \) | 1 | 2 | 1 | 0 | no | yes | no |
| \( ^{3}F_2 \) | 2 | 3 | 1 | 1 | yes | yes | yes |
The tensor force is given by $$ S_{12} (\hat r) = \frac{3}{r^2}\left(\mathbf{\sigma}_1\cdot \mathbf{r}\right) \left(\mathbf{\sigma}_2\cdot \mathbf{r}\right) -\mathbf{\sigma}_1\cdot\mathbf{\sigma}_2$$ where the Pauli matrices are defined as $$ \sigma_x =\begin{Bmatrix} 0 & 1 \\ 1 & 0 \end{Bmatrix}, $$ $$ \sigma_y =\begin{Bmatrix} 0 & -\imath \\ \imath & 0 \end{Bmatrix}, $$ and $$ \sigma_z =\begin{Bmatrix} 1 & 0 \\ 0 & -1 \end{Bmatrix}, $$ with the properties \( \sigma = 2\mathbf{S} \) (the spin of the system, being \( 1/2 \) for nucleons), \( \sigma^2_x=\sigma^2_y=\sigma_z=\mathbf{1} \) and obeying the commutation and anti-commutation relations \( \{\sigma_x,\sigma_y\} =0 \) \( [\sigma_x,\sigma_y] =\imath\sigma_z \) etc.
When we look at the expectation value of \( \langle \mathbf{\sigma}_1\cdot\mathbf{\sigma}_2\rangle \), we can rewrite this expression in terms of the spin \( \mathbf{S}=\mathbf{s}_1+\mathbf{s}_2 \), resulting in $$ \langle\mathbf{\sigma}_1\cdot\mathbf{\sigma}_2\rangle=2(S^2-s_1^2-s_2^2)=2S(S+1)-3, $$ where we \( s_1=s_2=1/2 \) leading to $$ \left\{ \begin{array}{cc} \langle\mathbf{\sigma}_1\cdot\mathbf{\sigma}_2\rangle=1 & \mathrm{if} \hspace{0.2cm} S=1\\ \langle\mathbf{\sigma}_1\cdot\mathbf{\sigma}_2\rangle=-3 & \mathrm{if} \hspace{0.2cm} S=0\\\end{array}\right. $$
Similarly, the expectation value of the spin-orbit term is $$ \langle \mathbf{l}\mathbf{S} \rangle = \frac{1}{2}\left( J(J+1)-l(l+1)-S(S+1)\right), $$ which means that for \( s \)-waves with either \( S=0 \) and thereby \( J=0 \) or \( S=1 \) and \( J=1 \), the expectation value for the spin-orbit force is zero. With the above phenomenological model, the only contributions to the expectation value of the potential energy for \( s \)-waves stem from the central and the spin-spin components since the expectation value of the tensor force is also zero.
For \( s=1/2 \) spin values only for two nucleons, the expectation value of the tensor force operator is
| \( l' \) | |||
| \( l \) | \( J+1 \) | \( J \) | \( J-1 \) |
| \( J+1 \) | \( -\frac{2J(J+2)}{2J+1} \) | 0 | \( \frac{6\sqrt{J(J+1)}}{2J+1} \) |
| \( J \) | 0 | 2 | 0 |
| \( J-1 \) | \( \frac{6\sqrt{J(J+1)}}{2J+1} \) | 0 | \( -\frac{2(2J+1)}{2J+1} \) |
We will derive these expressions after we have discussed the Wigner-Eckart theorem.
If we now add isospin to our simple \( V_4 \) interaction model, we end up with \( 8 \) operators, popularly dubbed \( V_8 \) interaction model. The explicit form reads $$ V(\mathbf{r})= \left\{ C_c + C_\mathbf{\sigma} \mathbf{\sigma}_1\cdot\mathbf{\sigma}_2 + C_T \left( 1 + {3\over m_\alpha r} + {3\over \left(m_\alpha r\right)^2}\right) S_{12} (\hat r)\right. $$ $$ \left. + C_{SL} \left( {1\over m_\alpha r} + {1\over \left( m_\alpha r\right)^2} \right) \mathbf{L}\cdot \mathbf{S} \right\} \frac{e^{-m_\alpha r}}{m_\alpha r} $$ $$ + \left\{ C_{c\tau} + C_{\sigma\tau}\mathbf{\sigma}_1\cdot\mathbf{\sigma}_2 + C_{T\tau} \left( 1 + {3\over m_\alpha r} + {3\over \left(m_\alpha r\right)^2}\right) S_{12} (\hat r)\right. $$ $$ \left. + C_{SL\tau} \left( {1\over m_\alpha r} + {1\over \left( m_\alpha r\right)^2} \right) \mathbf{L}\cdot \mathbf{S} \right\}\mathbf{\tau}_1\cdot\mathbf{\tau}_2 \frac{e^{-m_\alpha r}}{m_\alpha r} $$
From 1950 till approximately 2000: One-Boson-Exchange (OBE) models dominate. These are models which typically include several low-mass mesons, that is with masses below 1 GeV. Potentials which are based upon the standard non-relativistic operator structure are called "Phenomenological Potentials" Some historically important examples are
The total two-nucleon state function has to be anti-symmetric. The total function contains a spatial part, a spin part and an isospin part. If isospin is conserved, this leads to in case we have an \( s \)-wave with spin \( S=0 \) to an isospin two-body state with \( T=1 \) since the spatial part is symmetric and the spin part is anti-symmetric.
Since the projections for \( T \) are \( T_z=-1,0,1 \), we can have a \( pp \), an \( nn \) and a \( pn \) state.
For \( l=0 \) and \( S=1 \), a so-called triplet state, \( ^3S_1 \), we must have \( T=0 \), meaning that we have only one state, a \( pn \) state. For other partial waves, see exercises below.
The one-pion exchange contribution (see derivation below), can be written as $$ V_{\pi}(\mathbf{r})= -\frac{f_{\pi}^{2}}{4\pi m_{\pi}^{2}}\mathbf{ \tau}_1\cdot\mathbf{\tau}_2 \frac{1}{3}\left\{\mathbf{ \sigma}_1\cdot\mathbf{ \sigma}_2+\left( 1 + {3\over m_\pi r} + {3\over\left(m_\pi r\right)^2}\right) S_{12} (\hat r)\right\} \frac{e^{-m_\pi r}}{m_\pi r}. $$ Here the constant \( f_{\pi}^{2}/4\pi\approx 0.08 \) and the mass of the pion is \( m_\pi\approx 140 \) MeV/$\mbox{c}^2$.
Let us look closer at specific partial waves for which one-pion exchange is applicable. If we have \( S=0 \) and \( T=0 \), the orbital momentum has to be an odd number in order for the total anti-symmetry to be obeyed. For \( S=0 \), the tensor force component is zero, meaning that the only contribution is $$ V_{\pi}(\mathbf{r})=\frac{3f_{\pi}^{2}}{4\pi m_{\pi}^{2}}\frac{e^{-m_\pi r}}{m_\pi r}, $$ since \( \langle\mathbf{ \sigma}_1\cdot\mathbf{ \sigma}_2\rangle=-3 \), that is we obtain a repulsive contribution to partial waves like \( ^1P_0 \).
Since \( S=0 \) yields always a zero tensor force contribution, for the combination of \( T=1 \) and then even \( l \) values, we get an attractive contribution $$ V_{\pi}(\mathbf{r})=-\frac{f_{\pi}^{2}}{4\pi m_{\pi}^{2}}\frac{e^{-m_\pi r}}{m_\pi r}. $$ With \( S=1 \) and \( T=0 \), \( l \) can only take even values in order to obey the anti-symmetry requirements and we get $$ V_{\pi}(\mathbf{r})= -\frac{f_{\pi}^{2}}{4\pi m_{\pi}^{2}} \left(1+( 1 + {3\over m_\pi r} + {3\over\left(m_\pi r\right))^2}) S_{12} (\hat r)\right) \frac{e^{-m_\pi r}}{m_\pi r}, $$ while for \( S=1 \) and \( T=1 \), \( l \) can only take odd values, resulting in a repulsive contribution $$ V_{\pi}(\mathbf{r})= \frac{1}{3}\frac{f_{\pi}^{2}}{4\pi m_{\pi}^{2}}\left(1+( 1 + {3\over m_\pi r} + {3\over\left(m_\pi r\right)^2}) S_{12} (\hat r)\right) \frac{e^{-m_\pi r}}{m_\pi r}. $$
The central part of one-pion exchange interaction, arising from the spin-spin term, is thus attractive for \( s \)-waves and all even \( l \) values. For \( p \)-waves and all other odd values it is repulsive. However, its overall strength is weak. This is discussed further in one of exercises below.
a) List all allowed according to the Pauli principle partial waves with isospin \( T \), their projection \( T_z \), spin \( S \), orbital angular momentum \( l \) and total spin \( J \) for \( J\le 3 \). Use the standard spectroscopic notation \( ^{2S+1}L_J \) to label different partial waves. A proton-proton state has \( T_Z=-1 \), a proton-neutron state has \( T_z=0 \) and a neutron-neutron state has \( T_z=1 \).
a) Find the closed form expression for the spin-orbit force. Show that the spin-orbit force {\bf LS} gives a zero contribution for \( S \)-waves (orbital angular momentum \( l=0 \)). What is the value of the spin-orbit force for spin-singlet states (\( S=0 \))?
b) Find thereafter the expectation value of \( \mathbf{\sigma}_1\cdot\mathbf{\sigma}_2 \), where \( \mathbf{\sigma}_i \) are so-called Pauli matrices.
c) Add thereafter isospin and find the expectation value of \( \mathbf{\sigma}_1\cdot\mathbf{\sigma}_2\mathbf{\tau}_1\cdot\mathbf{\tau}_2 \), where \( \mathbf{\tau}_i \) are also so-called Pauli matrices. List all the cases with \( S=0,1 \) and \( T=0,1 \).
A simple parametrization of the nucleon-nucleon force is given by what is called the \( V_8 \) potential model, where we have kept eight different operators. These operators contain a central force, a spin-orbit force, a spin-spin force and a tensor force. Several features of the nuclei can be explained in terms of these four components. Without the Pauli matrices for isospin the final form of such an interaction model results in the following form: $$ V(\mathbf{r})= \left\{ C_c + C_\mathbf{\sigma} \mathbf{\sigma}_1\cdot\mathbf{\sigma}_2 + C_T \left( 1 + {3\over m_\alpha r} + {3\over\left(m_\alpha r\right)^2}\right) S_{12} (\hat r)\right. $$ $$ \left. + C_{SL} \left( {1\over m_\alpha r} + {1\over \left( m_\alpha r\right)^2} \right) \mathbf{L}\cdot \mathbf{S} \right\} \frac{e^{-m_\alpha r}}{m_\alpha r} $$ where \( m_{\alpha} \) is the mass of the relevant meson and \( S_{12} \) is the familiar tensor term. The various coefficients \( C_i \) are normally fitted so that the potential reproduces experimental scattering cross sections. By adding terms which include the isospin Pauli matrices results in an interaction model with eight operators.
The expectaction value of the tensor operator is non-zero only for \( S=1 \). We will show this in a forthcoming lecture, after that we have derived the Wigner-Eckart theorem. Here it suffices to know that the expectaction value of the tensor force for different partial values is (with \( l \) the orbital angular momentum and \( {\cal J} \) the total angular momentum in the relative and center-of-mass frame of motion) $$ \langle l {\cal J}S=1| S_{12} | l' {\cal J}S=1\rangle = -\frac{2{\cal J}({\cal J}+2)}{2{\cal J}+1} \hspace{0.5cm} l= {\cal J}+1 \hspace{0.1cm}\mathrm{and} \hspace{0.1cm} l'={\cal J}+1, $$ $$ \langle l {\cal J}S=1| S_{12} | l' {\cal J}S=1\rangle = \frac{6\sqrt{{\cal J}({\cal J}+1)}}{2{\cal J}+1} \hspace{0.5cm} l= {\cal J}+1 \hspace{0.1cm}\mathrm{and} \hspace{0.1cm} l'={\cal J}-1, $$ $$ \langle l {\cal J}S=1| S_{12} | l' {\cal J}S=1\rangle = \frac{6\sqrt{{\cal J}({\cal J}+1)}}{2{\cal J}+1} \hspace{0.5cm} l= {\cal J}-1 \hspace{0.1cm}\mathrm{and} \hspace{0.1cm} l'={\cal J}+1, $$ $$ \langle l {\cal J}S=1| S_{12} | l' {\cal J}S=1\rangle = -\frac{2({\cal J}-1)}{2{\cal J}+1} \hspace{0.5cm} l= {\cal J}-1 \hspace{0.1cm}\mathrm{and} \hspace{0.1cm} l'={\cal J}-1, $$ $$ \langle l {\cal J}S=1| S_{12} | l' {\cal J}S=1\rangle = 2 \hspace{0.5cm} l= {\cal J} \hspace{0.1cm}\mathrm{and} \hspace{0.1cm} l'={\cal J}, $$ and zero else.
In this exercise we will focus only on the one-pion exchange term of the nuclear force, namely $$ V_{\pi}(\mathbf{r})= -\frac{f_{\pi}^{2}}{4\pi m_{\pi}^{2}}\mathbf{ \tau}_1\cdot\mathbf{\tau}_2 \frac{1}{3}\left\{\mathbf{ \sigma}_1\cdot\mathbf{ \sigma}_2+\left( 1 + {3\over m_\pi r} + {3\over\left(m_\pi r\right)^2}\right) S_{12} (\hat r)\right\} \frac{e^{-m_\pi r}}{m_\pi r}. $$ Here the constant \( f_{\pi}^{2}/4\pi\approx 0.08 \) and the mass of the pion is \( m_\pi\approx 140 \) MeV/c${}^{2}$.
a) Compute the expectation value of the tensor force and the spin-spin and isospin operators for the one-pion exchange potential for all partial waves you found in exercise 9. Comment your results. How does the one-pion exchange part behave as function of different \( l \), \( {\cal J} \) and \( S \) values? Do you see some patterns?
b) For the binding energy of the deuteron, with the ground state defined by the quantum numbers \( l=0 \), \( S=1 \) and \( {\cal J}=1 \), the tensor force plays an important role due to the admixture from the \( l=2 \) state. Use the expectation values of the different operators of the one-pion exchange potential and plot the ratio of the tensor force component over the spin-spin component of the one-pion exchange part as function of \( x=m_\pi r \) for the \( l=2 \) state (that is the case \( l,l'={\cal J}+1 \)). Comment your results.