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. )