## Short-range interaction on same nucleons pair

From “Nuclear structure from a simple perspective” by Richard F. Casten, Chapter 4. The first half discusses the coupling between pp or nn T=1 isovector pair under δ-interaction, which is a good approximation to short-range interaction.

I found that the book makes it a bit complicated. The only 3 cases matter

The logic is follow. As the spatial part of the total wavefunction must be symmetric for the δ-interaction to be effective. The spin-isospin part must be antisymmetric as the total wavefunction must be antisymmetric due to identical fermions system. Also, the isospin part must be isovector or symmetric. Thus, the spin part must be antisymmetric or S = 0.

1. In other word, the δ-interaction has no effect on the S =1 state.
2. Also, the book said that $l_1 + l_2 - J = even$.
3. Most parallel states are either $J = J_{min} , J_{max}$

Using these 3 reasons, and consider these 3 cases, which actually covered all possible combinations and all cases. Point 2 also means that only half of the coupled states are affected. Since the number of state are even, when $J_{max} = even$ then $J_{min} = odd$ or via versa. If $J_{max}$ is formed by pure S = 1 state, thus, the most affected state is $J_{min}$ as $J_{min}$ is the other most parallel states, in which the overlap of the wavafunctions is maximum.

In the book, the example is d5/2 f7/2, J = 1, 2, 3, 4, 5, 6. This is belong to case 1, thus, the lowest state is J = 1. another example is d5/2 g7/2, which is case 2, the lowest state is J = 6.

For equivalent orbit, either case 1 or case 3. The lowest state is J = 0.

There are one interesting case. when coupling the s1/2. Since the s-orbit is isotropic, thus, no matter how the other orbit, the overlap is always maximum. Since pure S = 1 state is unaffected, thus, we can somehow, imagine the nucleon pair is forming S = 0 pair. In fact, for example, when s1/2 couples with d5/2,

$|J = 2 \rangle = \frac{1}{10} |20\rangle + \frac{1}{15\sqrt{2}} |21\rangle$

$|J = 3 \rangle = \frac{1}{6\sqrt{5}} |21\rangle$

The J = 3 state is unaffected by δ-interaction

## Sum rule of 9-j symbol

The 9-j symbol is the coupling coefficient when combining 2 nucleons with state $|l_1 s_1 j_1 \rangle$ and $|l_2 s_2 j_2 \rangle$ to from the state $|j_1 j_2 J M \rangle$ and $|L S J M \rangle$

$|L S J M \rangle = \sum_{j_1 j_2 } |j_1 j_2 J M \rangle \begin{pmatrix} l_1 & s_1 & j_1 \\ l_2 & s_2 & j_2 \\ L & S & J \end{pmatrix}$

or in simpler form

$|L S J M \rangle = \sum_{j_1 j_2 } |j_1 j_2 J M \rangle \langle j_1 j_2 | L S \rangle$

or reverse.

$|j_1 j_2 J M \rangle = \sum_{L S } |L S J M \rangle \langle L S | j_1 j_2 \rangle$

or to say, this is the coefficient between LS coupling and jj coupling scheme.

we can also see that

$\langle L S | j_1 j_2 \rangle = \begin{pmatrix} l_1 & l_2 & L \\ s_1 & s_2 & S \\ j_1 & j_2 & J \end{pmatrix}$

For example, when $p_{1/2}$ and $p_{3/2}$ coupled to $J = 1$  . We have

$\displaystyle|p_{1/2} p_{3/2} (J = 1) \rangle = \frac{1}{9} |01\rangle - \frac{1}{6\sqrt{6}} |10\rangle + \frac{1}{12} |11\rangle + \frac{1}{36} |21\rangle$

Since for each $|LS\rangle$, it is $(2L+1) (2S+1)$ degenerated. Thus, the sum rule is

$\displaystyle \sum_{LS} \left(\langle L S | j_1 j_2 \rangle\right)^2 (2L+1) (2S+1) = \frac{1}{(2j_1+1)(2j_2+1)}$

The sum is equal the a fraction, because the left-hand side is $(2j_1+1) (2j_2+1)$ degenerated for the state $|p_{1/2} p_{3/2} (J = 1) \rangle$.

## Sum rules of Clebsch-Gordon Coefficient

Since the CG coefficient is already normalized.Thus

$\displaystyle \sum_{m_1, m_2} \left(C^{j_1 m_1 j_2 m_2}_{JM}\right)^2 = 1$

Since the number of $M$ is $2J+1$, as $M = -J, -J+1, ... J$. Thus,

$\displaystyle \sum_M \sum_{m_1, m_2} \left(C^{j_1 m_1 j_2 m_2}_{JM}\right)^2 = 2J+1$

At last, the number of dimension of the coupled space or (tensor product space) is equation to $(2j_1 +1) (2j_2+1)$, i.e.

$\displaystyle \sum_J (2J+1) = (2j_1+1)(2j_2+1)$

Thus,

$\displaystyle \sum_{JM, m_1 m_2} \left(C^{j_1 m_1 j_2 m_2}_{JM}\right)^2 = (2j_1+1)(2j_2+1)$

## Winger 6-j and 9-j symbol

The meaning of 3-j symbol is same as Clebsch-Gordan coefficient. So, we skip in here.

I am not going to construct the 6-j symbol from 3-j symbol. In here, I just state the meaning and usage in Mathematica.

The 6-j symbol is the coupling between 3 angular momenta, $j_1, j_2, j_3$.

There are 2 ways to couple these 3 angular momenta. First,

$j_1 + j_2 + j_3 \rightarrow j_{12} + j_3 \rightarrow J$

the other way is

$j_ 1 + j_2 + j_3 \rightarrow j_1 + j_{23} \rightarrow J$

The 6-j symbol is

$\begin{pmatrix} j_1 & j_2 & j_{12} \\ j_3 & J & j_{23} \end{pmatrix}$

We can see that there are 4 vector-sum must satisfy.

$\Delta(j_1, j_2, j_{12})$

$\Delta(j_2, j_3, j_{23})$

$\Delta(j_1, j_{23}, J)$

$\Delta(j_{12}, j_3, J)$

If we draw a line to connect these 4 vector-sum, we have:

In Mathematica, there is a build in function

$\textrm{SixJSymbol}[ \left\{j_1, j_2, j_{23} \right\}, \left\{j_3, J , j_{23} \right\}]$

The 9-j symbol is the coupling between 4 angular momenta, $j_1, j_2, j_3, j_4$.

The 9-j symbol can be used in coupling 2 nucleons, $l_1, s_1, l_2, s_2$.

The 9-j symbol is

$\begin{pmatrix} l_1 & s_1 & j_1 \\ l_2 & s_2 & j_2 \\ L & S & J \end{pmatrix}$

We can see, each row and column must satisfy the vector-sum.

Unfortunately, there is no build in function in Mathematica. The formula for 9-j symbol is

$\displaystyle \begin{pmatrix} l_1 & s_1 & j_1 \\ l_2 & s_2 & j_2 \\ L & S & J \end{pmatrix} \\ = \sum_{g} (-1)^{2g} (2g+1) \begin{pmatrix} l_1 & s_1 & j_1 \\ j_2 & J & g \end{pmatrix} \begin{pmatrix} j_2 & s_2 & j_2 \\ s_1 & g & S \end{pmatrix} \begin{pmatrix} L & S & J \\ g & l_1 & l_2 \end{pmatrix}$

Where $g$ sum all possible value, which can be calculate using the 6 couplings inside the 3 6-j symbols.To check your result, the coupling between $d_{5/2}$ and $f_{7/2}$ to from a $L = 5, S = 0, J = 0$ state is

$\begin{pmatrix} 2 & 1/2 & 5/2 \\ 3 & 1/2 & 7/2 \\ 5 & 0 & 5 \end{pmatrix} = \frac{1}{2\sqrt{770}}$

## The nuclear structure of 19F

• Ground state spin-parity is 1/2+.
• Has low-lying 1/2- state at 197 keV.
• Magnetic dipole moment is 2.62885 μ0.
• 19F(p,2p) experiment reported only 2s-wave can fit the result. [M.D. High et al., PLB 41 (1972) 588]
• 19F(d, 3He) experiment reported the ground state is from 1s1/2 proton with spectroscopic factor of 0.38. [G. Th. Kaschl et al., NPA 155 (1970) 417]
• 18O(3He, d) experiment report the ground state is 1s1/2 proton with spectroscopic factor of 0.21. [C. Schmidt et al., NPA 155 (1970) 644]
• There is a rotational band of 19F [C. F. Williamson et al., PRL 40 (1978) 1702]

My understanding [2018-01-30]

The 19F is deformed. The deformation is confirmed from rotation band.

The deformation distorted the spherical basis into deformed basis. In the simplest deformed basis, the cylindrical basis, the $|Nn_z m_l K\rangle =|220(1/2)\rangle$ is mixed with 1s1/2 (~33%) and 0d5/2 (~66%) orbits.

The 19F wave function can be written as

$|^{19}F\rangle = \sqrt{0.2 \sim 0.4}|\pi 1s_{1/2} \times ^{18}O_{g.s}\rangle + \sqrt{0.8\sim0.6}|\pi 0d_{5/2} \times ^{18}O^*\rangle + ...$

Under proton transfer/pickup reactions, the selection of oxygen ground state force the transfer proton to be in 1s1/2 state. The founding of s-wave ground state make the association of the 19F ground state spin to be 1/2.

Using USDB interaction with pn formalism. The 18O, 19F ground state are

$|^{18}O\rangle = \sqrt{0.78} |(\nu0d_{5/2})^2 \times ^{16}O\rangle + \sqrt{0.17}|(\nu1s_{1/2})^2 \times ^{16}O\rangle + ...$

$|^{19}F\rangle = \\ \sqrt{0.22} |(\pi1s_{1/2})(\nu0d_{5/2})^2 \times ^{16}O\rangle + \\ \sqrt{0.17}|(\pi1s_{1/2})(\nu1s_{1/2})^2 \times ^{16}O\rangle + \\ \sqrt{0.27}|(\pi0d_{5/2})(\nu0d_{5/2})^2 \times ^{16}O\rangle + ...$

This (somehow) shows that the interaction accurately reproduce the shell configuration. The USDB interaction also suggest that, at the 19F ground state, the 1s1/2 orbit are only 41% filled and the 0d5/2 orbit are 47% filled.

The fact that the spectroscopic factor is much less then unity suggests the ground state configuration of 19F is not fit for single particle picture.

There are fill questions,

1. Why deform? due to the single 1d5/2 proton? Suppose adding a proton on 18O, the proton fill on 1d5/2 shell, and the d-shell creates a deformation on the sd shell, that shift the energy lower by mixing with s-shell?
2. in 19F(d, 3He) reaction, the sum of spectroscopic factor in sd-shell is just 1.54. This suggest large uncertainty. And the s-state SF is 0.38, almost a double for 18O(3He,d) reaction, How come?
3. in 19F(d,3He) reaction, the s:d ratio is 0.4:0.6, this is similar to prediction of Nilsson model, but difference from USDB calculation.
4. If the ground state has 1d5/2 proton, why the magnetic moment are so close to free proton? the l=2 should also contribute.
5. Is neutron shell also 1s1/2 ?
6. What is the $\beta_2$ ?
7. in 20Ne, will the proton also in 1s1/2 shell? 20Ne has $\beta_2 = 0.7$ very deformed.
8. Deformed DWBA?

The following is not organised thought.

According to the standard shell ordering, on top of 18O, an extra proton should fill up the 1d5/2 shell, and then the ground state spin of 19F should be 5/2. However, the ground state spin in 1/2. This is postulated to be due to deformation [mean field calculation, β2 = 0.275], 18O core excitation, or configuration mixing state [J.P. Elhot and A. M. Lane(1957)].

Under deformation, the conventional shell ordering is not suitable and may be an invalid picture to view the nucleus. So, talk about shell ordering is non-sense.

Since the 19F is 18O + 1s1/2 proton superposed with 1d5/2, there could be deformation. The spherical shape of 19F can be seen indirectly from the magnetic dipole momentum, the value is very close to that of a free proton of 2.78284734 μ0, only difference by 0.154 μ0, or 5.5%. How to solve this contradiction?

From the study of G. Th. Kaschl et al., the spectroscopic factor of the 19F(d,3He)18Og.s. channel is 0.38. The missing 1s1/2 strength most probably can be found in the higher excitation states. This indicates the ground state of 19F is a configuration mixing state. However, they also pointed out that caution is advisable with the absolute spectroscopic factor, this could be due to imperfect DWBA calculation.

The relative spectroscopic factors for the positive parity states, which normalised to the ground state, are agree with shell model prediction in sd-shell model space suggests that the core excitation should not play an important role.

From the USDB interaction, the shell ordering is normal, but the interaction result in a 1/2+ ground state. How?

What is the nature of the low lying 1/2- excited state in 19F?

20Ne(d,3He)19F reaction can populate this low lying state, suggests the p-shell proton pickup come from the nuclear surface.

( if (12C,13N) proton pickup reaction can populate this state, then, it can be confirmed that this is a surface p-shell proton, that it could be from 2p3/2. )

## Momentum Matching in Transfer Reaction

In transfer reaction, the kinematics determines the scattering angles and momenta, but it does not tell the probability of the reaction. Not all angle have same cross section, because when a particle is being transfer to or pickup from a nucleus, the momentum has to be matched. It is like a spaceship need a proper angle and speed in order to orbit around the earth, it the incident angle or speed are not good, the spaceship will just go out or crash into the earth.

Suppose our reaction is $A(B,1)2$, $A$ is incident particle, $B$ is the target, $1, 2$ are the outgoing particles. The momentum transfer from $A$ via $2$ to $B$ or $1$ is

$\vec{q} = \vec{p}_2 - \vec{p}_A = \vec{p}_f - \vec{p}_i$

where $\vec{p}_i$ is initial momentum and $\vec{p}_f$ is final momentum.

Assume the reaction take place at the surface of nucleus $B$. The maximum angular momentum it can create is

$\vec{L} = \vec{r} \times \vec{q} \rightarrow L = r q$

In QM, the angular momentum must be $L = \sqrt{l(l+1)} \hbar$. Here we pause for unit conversion. In this blog, we usually using nuclear unit, that momentum in [MeV/c] and $\hbar = 197.327 \textrm{MeV fm/c}$, thus, to calculate the angular momentum, simple multiple the momentum in MeV/c and nuclear radius in fm. The radius of a nucleus is roughly $r = 1.25 A^{1/3} \textrm{fm}$.

So, if the momentum $\sqrt{(l+1)(l+2)} \hbar > r|q| > \sqrt{l(l+1)} \hbar$, the possible angular momentum is $\sqrt{l(l+1)} \hbar$ or the l-orbit.

In the transfer reaction, the equation for $\vec{p}_k$ and $\vec{p}_i$ are known. It is straight forward to calculate $r q / \hbar$, plot against scattering angle, and we look for the value for $\sqrt{l(l+1)}$.

Or, we can scale the y-axis by

$\displaystyle y' = \sqrt{\left(q\frac{r}{\hbar}\right)^2 + \frac{1}{4}} - \frac{1}{2}$

The new y-axis is in unit of $\sqrt{l(l+1)}$.

We can also plot the contour of momentum matching on the $p_k$ versus scatering angle. The momentum matching is

$\displaystyle q^2 = \frac{l(l+1)\hbar^2}{r^2} = p_f^2 + p_i^2 - 2 p_f p_i \cos(\theta)$

where $\theta$ is the scattering angle of particle $1$. Solve for $p_f$

$\displaystyle p_f = p_i \cos(\theta) \pm \sqrt{ \left( \frac{l(l+1) \hbar^2}{r^2} \right) - k_i^2 \sin^2(\theta) }$

We can plot the contour plot for difference $l$. and then overlap with the momentum and the scattering angle of the outgoing particle $2$.

a correction, the y-axis should be $p_2$, momentum of particle 2.

## Axial Harmonic Oscillator – Nilsson Orbit (III)

One of the problem (or difficulty) is the conversion between cylindrical quantum number $|Nn_zm\rangle$ and spherical quantum number $|Nlm\rangle$. In the limit of $\delta = 0$, all state with same $N$ has same energy and the conversion is not necessary.

With out LS coupling, the energy level can be found by the approximation

$\displaystyle E(N, n_z) = \hbar\omega \left( \frac{3}{2} + \left(1 + \frac{\delta}{3} \right) N - \delta n_z \right)$

We can see from the formula, for $\delta>0$, the largest $n_z$ has lowest energy. And the conversion to spherical harmonics can be done using projection method.

With LS coupling, the conversion becomes complicated. One issue is the coupling with the spin. Since the total angular momentum $J = L + S$ is not a conservative quantity due to deformation. In the body-fixed frame, the additional good quantum number is the z-projection of $J$, which denote as $K = |m_l \pm 1/2 |$, the quantum state becomes

$|Nn_z m_l K\rangle$

Note that each state will have only 2 degeneracy with negative $K$. On the spherical basis, the coupling with $J$ is a standard textbook content. the eigen state is

$|N l j mj \rangle = |N l j K\rangle$

The conversion get complicated, because there is no clear rule on how $| Nn_z m_l K\rangle \rightarrow |Nl j m_j \rangle$. When we want to calculate the energy level using cylindrical basis with the LS coupling, since the $J$ is not clear, even through we can write down the formula

$\displaystyle E( N, n_z, l, j) = \hbar \omega_z \left(\frac{1}{2} + n_z \right) + \hbar \omega_\rho \left(1 + N - n_z \right) \\ + a \frac{1}{2} ( j(j+1) - l(l+1) - s(s+1)) + b l^2$

It is difficult to know the $j , l$.

The go around this conversion, we can diagonal the Hamiltonian using spherical basis. The Hamiltonian is can be written as

$\displaystyle H = - \frac{\hbar^2}{2m} \nabla^2 + \frac{m\omega}{2}r^2 \left( 1 - \frac{2}{3}\sqrt{\frac{16\pi}{5}} Y_{20}(\theta, \phi) \delta \right) + a L\cdot S + b L^2$

The first 2 terms is the spherical harmonic oscillator. The energy is $\hbar\omega( 3/2 + N)$. The $Y_20$ restricts the coupling between $l, l \pm 2$ (see here.)

For example, when using the basis on $N = 1$, which are

$\displaystyle |N l j m_j \rangle = \left( |1 1 \frac{3}{2} \frac{3}{2}\rangle , |1 1 \frac{1}{2} \frac{1}{2}\rangle, |1 1 \frac{3}{2} \frac{1}{2}\rangle \right)$

at $\delta = 0.5$, the matrix is

$\begin{pmatrix} 2.633 & 0 & 0 \\ 0 & 2.233 & 0 \\ 0 & 0 & 2.233\end{pmatrix}$

The eigen values and eigen state are

$\displaystyle 2.633 , |1 1 \frac{3}{2} \frac{3}{2}\rangle$

$\displaystyle 2.233 , |1 1 \frac{1}{2} \frac{1}{2}\rangle , |1 1 \frac{3}{2} \frac{1}{2}\rangle$

The lower level has mixed $j$ !

Of course, a more complete diagonalization should be admixture from other major shell as well. When we disable the LS coupling, using 7 major shells, the Nilsson diagram look like this

This is consistence with $omega_0 = const.$, i.e. no volume conservation. With LS coupling switched on.

Although it is not so clear, we can see the state avoid “crossing”.  The s-state are unaffected by the LS coupling.

One deflect in the plot is that, the slope seem to be incorrect. And also, I tried to add volume conservation, but the result does not consistence….. (sad).