## Single-particle structure of 19F

About one year ago, I studied the nuclear structure of 19F. At that time, a lot of thing don’t understand and confused.  But now, I have a better understanding.

19F is always fascinating because the so complicated energy levels for such a relatively simple system with only 1 proton, 2 neutrons on top of a 16O core, which usually treated as double magic rigid core.

• It ground state is 1/2+, which is unusual.
• It is also well-deformed with $\beta_2 \sim 0.4$, deduced from rotational band.
• It also has very low lying negative parity state of 0.110 MeV. This state is also the band head of K=1/2- rotational band.
• The rotational band of K=1/2± does not following J(J+1) curve (see the last picture from this post), this indicate the rigid rotor assumption is not so good. 3 of the rotational bands of 19F, from M. Oyamada et al., PRC11 (1975) 1578

When looking the region around 19F. From 16O a double magic core, to 20Ne very deformed nucleus ( $\beta_2 \sim = 0.7$), and then go to 32Mg, the center of island of inversion. 16O, 18O, 18F, 20F are normal nuclei with normal shell order. 19F is a doorway nuclei for the complicated nuclear structure. That make the understanding the nuclear structure of 19F important.

1. it helps to understand how deformation happen in small system,
2. and how deformation can be described in particle-configuration.

The single-particle structure of the ground state of 19F has been studied using (p,2p) reaction. And found that the ground state to ground state transition only contain strength from 1s1/2 orbital. Thus, the wave function of 19F must bein the form $\displaystyle |^{19}F\rangle = \alpha |\pi 1s_{1/2} \rangle |^{18}O\rangle + \beta |\pi 0d_{5/2}\rangle |^{18}O^* \rangle + ...$

The single-particle structures of the ground state and excited states are best studied using single-nucleon transfer. There are at least 4 directions, neutron-removal from 20F, neutron-adding from 18O, proton-removal from 20Ne, and proton-adding from 18O.  Also, a proton/neutron-removal from 19F itself to study the ground state properties. In above figure, we also plotted the spectroscopic factors from 2 reactions, which is taken from G. Th. Kaschl et al., NPA155(1970)417, and M. Yasue et al., PRC 46 (1992) 1242. These are the most significant studies. It was suprised that the negative parity state of 0.110 meV can be populated from the 18O(3He,d) reaction. These suggest the 18O is not a good core that it has about 10% strength of two-nucleon hole in its ground state. From the 20Ne(d, 3He) reaction, the 0.110 MeV state is highly populated. Given that the ground state of 20Ne is mainly proton 1s1/2 strength due to deformation. These result indicated that the negative parity state of 19F is due to a proton-hole. Thus we have a picture for the 0.110 MeV state of 19F: The 18O(3He,d) reaction put a 0p1/2 proton in 18O 2-proton hole state to form the negative parity state. And 20Ne(d, 3He) reaction remove a 0p1/2 proton from 20Ne.

But this picture has a problem that, how come it is so easy to remove the 0p1/2 proton from 20Ne? the p-sd shell gap is known to be around 6 MeV! or, why the proton hole in 19F has such a small energy?

In current understanding, the 2-neutron are coupled to J=0 pair, and no contribution to low-lying states. But is it true?

I suspect, the underneath reason for the proton 1s1/2 orbital is lower is because the proton 0d5/2orbital was repealed by the 2-neutron in 0d5/2 orbital due to the tensor force. And somehow, the tensor force becomes smaller in 20F when the 0d5/2 orbital is half filled.

And because the proton is in 1s1/2 and the 0d5/2 is just 0.2 MeV away, a huge configuration mixing occurred. and then, a Nilsson orbit is formed with beta = 0.4. This is an example of NN-interaction driving deformation.

## Rotational Band

For deformed nuclei, it can be rotated in various angular momentum in Laboratory frame. Assume rigid body rotation, the energy is $\displaystyle E_J = \frac{1}{2}I\omega^2 = \frac{1}{2I}J^2$

In QM, that becomes $\displaystyle H = \sum_{i=1}^{3} \frac{\hbar^2}{2I_i} J_i^2$

For axial symmetry, $I_1 = I_2 = I$ $\displaystyle H = \frac{\hbar^2}{2I} (J^2 - J_3^2) + \frac{\hbar^2}{2I_3}J_3^2$

Remember, in deformed nuclei, the projection of $J$ along the symmetry axis in the body-frame is $K$ . The expected value of the Hamiltonian with state $|Nn_z m_l K \rangle$ in the body-frame is proportional to $J(J+1)$ for $J^2$ and $K$ for $J_3$. i.e. $\displaystyle E_J = \frac{\hbar^2}{2I} J(J+1) + E_K$

From body-frame to Lab-frame, we should apply the Wigner D-Matrix to the intrinsic wave function. ( I am not sure the following equation is correct, but the idea is rotating the body-frame wavefunction with Wigner D-Matrix to get the Lab-frame wave function. In Lab frame the total angular momentum must be a good Quantum number as rotational symmetry restored, so as $J_z = M$. The problem of the following equation is that the J is not a good Q-number in Nilsson wavefunction ) $\displaystyle |JMK\rangle = \sum_{M} D_{MK}^{J} |Nn_zm_lK\rangle$

However, the Wigner D-Matrix does not conserve parity transform: $\displaystyle D_{MK}^J \rightarrow (-1)^{J+K} D_{M-K}^{J}$

In order to restored the parity, we need to include $\pm K$ in the Lab-frame wave function. $\displaystyle |JMK\rangle = \sum_{M} \left( D_{MK}^J \pm (-1)^{J+K} D_{M-K}^J \right) |Nn_zm_lK\rangle$

where + for positive parity, – for negative parity.

From the above equation, for $K^\pi = 0^+$ ( $0^-$), $J$ must be even (odd). For $K > 0$, $J = K, K+1, K+2, ...$.   We can see for $K = 1/2$, the $J = 5/2, 9/2, 11/2$ are lower to the main sequence. This was explained by adding an extra term in the rotation Hamiltonian that connect $\Delta K = 1$. $\displaystyle \langle JMK | H'(\Delta K = 1) |JMK \rangle$ $\displaystyle \rightarrow \langle D_{MK}^J | H' | D_{MK}^J \rangle+ \langle D_{M-K}^J |H'| D_{MK}^J \rangle + \langle D_{MK}^J | H' | D_{M-K}^J \rangle+ \langle D_{M-K}^J |H'| D_{M-K}^J \rangle$

The term with $\Delta K = 0$ vanished. And since $\latex K = 1$, the only non-zero case is $K = 1/2$.

A possible form of the $H' (\Delta K = 1) = \frac{1}{2} \omega (J_+ + J_-)$. These are the ladder operator to rise or lower the m-component by 1. In 19F case, we can think it is a single proton on top of 18O core.  A rotation core affect the proton with an additional force, similar to Coriolis force on earth.

## The nuclear structure of 19F

Some facts about 19F:

• Ground state spin-parity is 1/2+.
• Has low-lying 1/2- state at 110 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 lowest s-d shell state is $|Nn_z m_l K\rangle =|220(1/2)\rangle$, which is mixed with 1s1/2 (~33%) and 0d5/2 (~66%) orbits and the K, the intrinsic spin, is 1/2. Thus, the 19F ground state spin must be 1/2.

The 19F wave function can be written as (approximately) $|^{19}F\rangle = \sqrt{0.2 \sim 0.4}|\pi 1s_{1/2} \times ^{18}O_{g.s}\rangle + \sqrt{0.8\sim0.6}|\pi 0d_{5/2} \times ^{18}O^*\rangle + ...$

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 spin of 1/2 of 19F agrees with this picture.

Using USDB interaction with pn formalism. The 18O, 19F ground state are $|^{18}O\rangle = \sqrt{0.78} |(\nu0d_{5/2})^2 \times ^{16}O\rangle + \sqrt{0.17}|(\nu1s_{1/2})^2 \times ^{16}O\rangle + ...$ $|^{19}F\rangle = \\ \sqrt{0.22} |(\pi1s_{1/2})(\nu0d_{5/2})^2 \times ^{16}O\rangle + \\ \sqrt{0.17}|(\pi1s_{1/2})(\nu1s_{1/2})^2 \times ^{16}O\rangle + \\ \sqrt{0.27}|(\pi0d_{5/2})(\nu0d_{5/2})^2 \times ^{16}O\rangle + ...$

The sum of $\pi1s_{1/2}$ is 0.39, and the ground state is 0.17. This close to the 18O(3He, d) experiment result, so that the interaction accurately reproduce the shell configuration.

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. )