20Ne | Nuclear Physics 101

On the microscopic origin of deformation + paper reading

February 4, 2023

GoLuckyRyan Basic 100Zr, 20Ne, 20O, deformation, paper reading, quadrapole, quadrupole, Tensor Leave a comment

Nuclear deformation is usually mimicked by macroscopic deformation potential, for example, a quadrupole deformation potential as in the Nilsson model. This provides a description without understanding the origin of the deformation potential. This macroscopic description is particularly unsatisfactory for the deformed light nuclei. Also, it does not answer the question that, are the whole nuclei deformed? or just the surface?

The paper Towards a unified microscopic description of nuclear deformation by P. Federman and S. Pittel [Physic Letters B 69, 385 (1977)] provides an interesting and reasonable understanding of the microscopic description.

The paper starts by comparing 20Ne and 20O. 20Ne is well deformed with $\beta_2 = 0.7$ and 20O is less deformed with $\beta_2 = 0.27$ . Let’s assume the 16O core is inert in both nuclei. The difference between 20Ne and 20O is two protons in the sd-shell are replaced with two neutrons from 20Ne to 20O. The paper said, “The conclusion seems clear. Deformation in light nuclei is due to the T=0 neutron-proton interaction.“.

As we know from the two-nucleon system, there are 4 possible combinations with 2 nucleons. In those combinations, the T=0, J=1 pn pair is the only bound pair due to the tensor force. And the pair is not spatial isotropic. In a many-nucleons nucleus, a similar thing happens for NN pair. In 20O, and 20Ne, all valence nucleons are in d5/2-orbitals in the simplest picture. A T=0 pn pair, the spin of the proton and neutron must be aligned, which means the proton and neutron are orbiting in the same direction, and the total spin of the pn pair is J = 5, which is very spatially deformed. And in 20O, all valence nucleons are neutrons, only able to form T=1, J=0 nn pairs, in which the neutrons are orbiting oppositely and spatially spherical. The result is 20Ne is deformed and 20O is spherical, and the ultimate reason is the tensor force tends to make T=0 pn pair.

Wait… the total spin of 20Ne and 20O are both Zero. If it is the T=0 pn pair and the spin is not Zero, would the spin of 20Ne be non-zero? For even-even nuclei, protons and neutrons are paired up and formed J=0 pp and nn pair. Where are the T=0 pn pairs? The key is that, although the pp and nn are paired up, it does not exclude the T=0 pn pair.

For simplicity, let’s check the Slater determinate for 3 fermions. Suppose the 3 nucleon wave functions are $p_a, p_b, n_a$ , where $p, n$ stands for proton and neutron, $\alpha, \beta$ are spin-up and spin-down. And all wavefunctions are in the same orbital.

$\displaystyle \Psi = \frac{1}{3!} \begin{vmatrix} p_a(1) & p_b(1) & n_a(1) \\ p_a(2) & p_b(2) & n_a(2) \\ p_a(3) & p_b(3) & n_a(3) \end{vmatrix}$

without loss of generality, we can collect terms of the 1-th particle.

$\displaystyle 3 \Psi = p_a(1) \frac{p_b(2) n_a(3) - p_b(3) n_a(2) }{2} + \\ ~~~~~~~~p_b(1) \frac{p_a(2) n_a(3) - p_a(3) n_a(2) }{2} + \\~~~~~~~~ n_a(1) \frac{p_a(2) p_b(3) - p_a(3) p_b(2) }{2}$

$\displaystyle 3 \Psi = p_a(1) ( pn, T=1, J = 0 ) + \\~~~~~~~~~ p_b(1) (pn, T=0, J =1) +\\~~~~~~~~~ n_a(1) (pp, T=1, J= 0)$

Now, imagine it is the triton wave function. We can see that the total wave function contains a pp J=0 pair, but there are also pn J=0 and J=1 pair. The detailed coupling of the total spin of the total wave function involves CG coefficient, and the pn T=0, J=1 pair should be coupled to a spin-down proton (j=1/2) and form J = 1/2.

We can imagine that in a wavefunction with 2 protons and 2 neutrons, there will be T=0 pn pair and T=1 pp/nn pair. While a wavefunction with 4 neutrons can never form pn pair.

In this simple demonstration, it is simply forming the wave function without considering the nuclear force. There are two questions: 1) In the Slater determinate, what is the percentage for T=0 and T=1 NN pairs? 2) with the nuclear force, what is that percentage?

The paper Probing Cold Dense Nuclear Matter by R. Subedi et al., Science 320, 1476 (2008), could give us some hints. The study found that in 12C, there are 18% pn pairs and only 2% of pp or nn pairs. There are more recent developments on the topic, for example, PRL 121, 242501 (2018), PLB 820, 10 (2021).

P. Federman and S. Pittel apply the same idea (deformation caused by T=0 pn pair from tensor force) on 100Zr, a very different and complex nucleus than 20Ne. 98Zr has Z = 40 and N = 58 and is spherical. The proton shell is semi-closed at 1p1/2 orbital. and the neutron shell is semi-closed at 2s1/2 ( on top of N=40, 0g9/2, 1d5/2, 2s1/2 ). But 100Zr is highly deformed with $\beta_2 \approx 0.35$ . The nuclear shape changed so much by just 2 neutron differences has drawn a lot of attention since its discovery in the 70s. The Interacting Boson Model and Shell model calculation has been tried to compute this sudden transition with quite good results, but there are still many discrepancies on the microscopic origin of the deformation. For example, is the deformation driven by protons or neutrons? and also what is the configuration.

As we mentioned before, the T=0 pair could be causing the deformation. In the case of 100Zr, it is the g-orbital pn pairs. Other proposed mechanisms are core polarization of 98Zr and the presence of a valence neutron in 0h11/2 orbital. In a recent experimental proposal, I wrote the following:

The above three mechanisms are intertwined. The interplay between these mechanisms is illustrated in Figure. 1. The neutrons in the 0g7/2 orbital lower the proton 0g9/2 binding energy while increasing the binding energy of the proton 1p1/2 and 0f5/2 with the action of the tensor force (attractive for J< − J> pair and repulsive for J< − J< or J> − J> pair), which reduces or even breaks the pf-g Z = 40 shell-gap, favoring the promotion of the protons to the 0g9/2 orbital from the pf-shell, and creates a core polarization and deformation. The core polarization of Z = 40 core promotes protons into the π0g9/2 orbital, enabling the coupling with the 0g7/2 neutron and forming T = 0 p-n pairs under the influence of tensor interaction. The g-orbital T = 0 p-n pair have their spins aligned and create a strong quadrupole deformation. Also, the presence of 0g9/2 protons lowers the effective single-particle energies (ESPEs) of the neutron 0g7/2 and 0h11/2 orbital via the so-called Type II shell evolution, which increases the occupancy for the neutron 0h11/2 orbital. The presence of 0h11/2 neutrons in turn provides a strong quadrupole deformation force. The deformation then enhances the occupation of valence orbitals and fragmentation in single-particle energies in return.

This is a bit complicated and hard to prove. But the tensor force plays an important role here. Without such, the chain reaction of the nucleon reconfiguration would not happen.

We can see the A=100 nuclei, 100Mo has 2 protons at 0g9/2 and N=58, it is deformed with $\beta_2 = 0.23$ . A deformation could promote neutrons to 0g7/2. 102Mo has 2 protons at 0g9/2 and 2 neutron at 0g7/2, so it is deformed with $\beta_2 = 0.31$ (Would 102Mo be more deformed?) 102Ru has 4 protons at 0g9/2, and N = 58. It is slightly deformed with $\beta_2 = 0.17$ . 102Pd has 6 proton at 0g9/2 and N = 54 with $\beta_2 = 0.14$ . In the opposite direction, 100Sr (Z=38, N=62) is very deformed with $\beta_2 = 0.41$ .

Also, Z = 38 or Z = 42, all isotopes are deformed. Clearly, Z = 40 and N < 60 are NOT deformed and are the ANOMRALY. The shell Z = 40 closure clearly forbids proton-shell configuration mixing.

In the above, we always use $\beta_2$ as a measure or indicator for deformation. But the $\beta_2$ of 16O is 0.35. Is 16O not spherical? That is exactly the reason why deformation is “hard” to understand for light nuclei. In my opinion, 16O is not spherical, as $\beta_2$ is pretty much a measure of the geometrical shape. However, 16O is shell-closure and has almost no configuration mixing (there are ~ 10% sd-shell components). And the shape deformation is caused by the T=0 pn pairs. However, where is the rotational band of 16O? Another thing is, in light nuclei, deformation, and configuration mixing can be separated. Configuration mixing will lead to deformation, but the reverse is not always true.

In fact, all light nuclei are more or less deformed!! Within the sd-shell, $beta_2 > 0.1$ and the most spherical nuclei is 40Ca. Also also, I think all even-odd nuclei are deformed as the unpaired nucleon has a deformed orbital.

The known smallest $beta_2$ nucleus is 206Pb with a value of 0.03.

In the case of 20Ne, the deformation could have different effects on different orbitals. i.e. the mean field for each nucleon could be different.

How to apply this idea? and how to predict the degree of deformation? Is it the only mechanism?

I should calculate the $beta_2$ for each orbital….

Alpha cluster and alpha separation energy

September 13, 2020

GoLuckyRyan Basic 19F, 20Ne, alpha, cluster, deformation, preformation Leave a comment

The alpha separation energy is the energy to add int a nucleus, so that it will break up into an alpha-particle and the rest of the nucleus. If we define the mass of a nucleus with mass number A and charge number Z as $M(Z,A)$ , the alpha-separation energy is,

$\displaystyle S_\alpha(Z,A) = M(Z-2,A-4) + M(\alpha) - M(Z,A)$ .

The following plot the nuclides chart for alpha-separation energy,

The chart can be divided in 2 regions. In the upper region, nuclei have negative alpha-separation energies, i.e. nucleus gives out energy when emitting an alpha particle, thus, they are alpha-emitter. In the lower region, the alpha-separation energies are positive. And we can see that there are some local minimum for the $S_\alpha$ .

According the alpha-decay theory, alpha-particle should be formed inside a nucleus before it tunnels through the nuclear potential and get out. the formation of the alpha-particle is described as the preformation factor.

It seems that, the alpha-separation energy, somehow, relates to the preformation factor.

The alpha cluster is studied for the many light nuclei, particularly on the 8Be, 12C, 16O, 20Ne, 24Mg, 28Si, 32S, 36Ar, 40Ca, and 44Ti. [Ref??]

From the above plot, it seems that the alpha-separation energies have no correlation with the magic number, and also no correlation with the Z=even. In a naive imagination, alpha-cluster could appear at all Z=even, A = 2Z nuclei. A trivial example is the the 8Be, its alpha-separation energy is -0.09 MeV and it will decay or split into 2 alpha particles. In 12C, the Hoyle state at 7.7 MeV is a triple-alpha state, note that $S_\alpha(12C) = 7.37$ MeV.

When we use the shell model to look at 8Be and 12C, we will found that the alpha-cluster is “not” making any sense. The protons and neutrons occupy the 0s1/2 and 0p3/2 orbitals. Since 9Be is stable and ground state spin is 3/2. A 8Be nucleus can be obtained by removing a 0p3/2 neutron. We may guess that, the 4 nucleons at the 0s1/2 orbital, which is an alpha-particle, somehow, escape from the 8Be, leaving the 4 0p3/2 nucleons behind. And the 4 0p3/2 nucleons “de-excite” back to the 0s1/2 shell and becomes another alpha-particle. But it seems that it does not make sense. Also, How to use shell model to describe the triplet-alpha cluster?

Above is the the alpha-separation energy for light nuclei. We can see that, there are few local minimum around 8Be, 20Ne, 40Ca, and 72Kr. Near 20Ne, the 18F, 19F, and 19Ne are having small alpha-separation energies, relative to the near by nuclei. For 20Ne, it could be understand why the alpha-separation energy is relatively smaller, as the 20Ne can be considered as 16O core with 4 nucleons in sd-shell. And 20Ne is well deformed that, there is a chance all 4 nucleons are in the 1s1/2 and form a quai-alpha particle. But the situation is a bit strange for 19Ne, 19F, and 18F. In order to form an alpha-cluster, or a quasi-alpha-particle, one or two nucleon from the p-shell has to excited to sd-shell. But the situation may be even more complicated. For some heavy alpha emitter, the s-orbital nucleons are tiny fraction compare to the rest of the nuclei, so, how the alpha-particle is formed before the decay is still unknown. This suggests that the involvement of s-orbital is not needed in alpha formation.

$S_\alpha(19F) = 4.014 MeV$ , and there are many states around 4 MeV in 19F, I am wondering, one of these state is alpha cluster? If so, 19F(d,d’) reaction could excite those states and we will observed 15N + alpha. [need to check the data]. Similar experiment could be done on 18F, 19Ne, and 20Ne. If it is the case, that could provides some information on the alpha formation.

[need to check the present theory of alpha formation]

Shell model calculation on 18O

May 15, 2020

GoLuckyRyan Basic 18O, 19F, 20Ne, Clebsch-Gordon, Shell Model, shell model calculation, TBMEs Leave a comment

This post is copy from the book Theory Of The Nuclear Shell Model by R. D. Lawson, chapter 1.2.1

The model space is only the 0d5/2 and 1s1/2, and the number of valence nucleon is 2. The angular coupling of the 2 neutrons in these 2 orbitals are

$\displaystyle (0d5/2)^2 = 0, 2, 4$
$\displaystyle (0d5/2)(1s1/2) = 2,3$
$\displaystyle (1s1/2)^2 = 0$

Note that for identical particle, the allowed J coupled in same orbital must be even due to anti-symmetry of Fermion system.

The spin 3, and 4 can only be formed by (0d5/2)(1s1/2) and (0d5/2)^2 respectively.

Since the Hamiltonian commute with total spin, i.e., the matrix is block diagonal in J that the cross J matrix element is zero,

$\displaystyle H = h_1 + h_2 + V$

$\displaystyle \left< J | V|J' \right> = V_{JJ'} \delta_{JJ'}$

or to say, there is no mixture between difference spin. The Hamiltonian in matrix form is like,

$\displaystyle H = \begin{pmatrix} M_{J=0} (2 \times 2) & 0 & 0 & 0 \\ 0 & M_{J=2} (2 \times 2) & 0 & 0 \\ 0 & 0 & M_{J=3} (1 \times 1) & 0 \\ 0 & 0 & 0 & M_{J=4} (1 \times 1) \end{pmatrix}$

The metrix element of J=3 and J=4 is a 1 × 1 matrix or a scalar.

$\displaystyle M_{J=3,4} = \left<J| ( h_1 + h_2 + V) | J \right> = \epsilon_1 + \epsilon_2 + \left< j_1 j_2 | V | j_3 J_4 \right>$

where $\epsilon_i$ is the single particle energy.

Suppose the residual interaction is an attractive delta interaction

$\displaystyle V = - 4\pi V_0 \delta(r_i - r_j )$

Be fore we evaluate the general matrix element,

$\displaystyle \left< j_1 j_2 | V | j_3 j_4 \right>$

We have to for the wave function $\left|j_1 j_2 \right>$ ,

$\displaystyle \left| j_1 j_2 \right> = \\ \frac{1}{\sqrt{2(1+\delta_{j_1 j_2})}} \\ \sum_{m_1 m_2} C_{j_1 m_1 j_2 m_2}^{JM} \left( \phi_{j_1m_1}(1) \phi_{j_2m_2}(2) + (-1)^T \phi_{j_1m_1}(2) \phi_{j_2m_2}(1) \right)$

where $T$ is the isospin, and the single particle wave function is

$\displaystyle \phi_{jm} = R_l(r) \sum_{\kappa \mu} C_{l \kappa s \mu}^{jm} Y_{l \kappa} ( \hat{r} ) \chi_{\mu}$

Since the residual interaction is a delta function, the integral is evaluated at $r_1 = r_2 = r$ , thus the radial function and spherical harmonic can be pulled out in the 2-particle wave function $\left|j_1 j_2 \right>$ at $r_1 = r_2 = r$ is

$\displaystyle \left| j_1 j_2 \right> = \\ \frac{1}{\sqrt{2(1+\delta_{j_1 j_2})}} R_{l_1}(r) R_{l_2}(r) \\ \sum_{m_1 m_2 \kappa_1 \mu_1 \kappa_2 \mu_2} C_{j_1 m_1 j_2 m_2}^{JM} C_{l_1 \kappa_1 s \mu_1}^{j_1 m_1} C_{l_2 \kappa_2 s \mu_2}^{j_2 m_2} Y_{l_1 \kappa_1} ( \hat{r} ) Y_{l_2 \kappa_2} ( \hat{r} ) \\ \left( \chi_{\mu_1}(1) \chi_{\mu_2}(2) + (-1)^T \chi_{\mu_1}(2) \chi_{\mu_2}(1) \right)$

Using the product of spherical Harmonic,

$\displaystyle Y_{l_1 \kappa_1}(\hat{r})Y_{l_2 \kappa_2}(\hat{r}) = \sum_{LM} \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi (2L+1)}} C_{l_1 0 l_2 0 }^{L0} C_{l_1 \kappa_1 l_2 \kappa_2}^{LM} Y_{LM}(\hat{r})$

using the property of Clebsch-Gordon coefficient for spin half system

$\displaystyle \chi_{SM_S} = \sum_{\mu_1 \mu_2} \chi_{\mu_1}(1) \chi_{\mu_2}(2)$

where

$\displaystyle \chi_{0,0} = \frac{1}{\sqrt{2}} \left( \chi_{1/2}(1) \chi_{-1/2}(2) - \chi_{-1/2}(1) \chi_{1/2}(2) \right)$

which is equal to $T = 1$

For $T = 0$

$\displaystyle \chi_{1,1} = \chi_{1/2}(1) \chi_{1/2}(2)$

With some complicated calculation, the J-J coupling scheme go to L-S coupling scheme that

$\displaystyle \left| j_1 j_2 \right> =\sum_{L S M_L M_S} \alpha_{LS}(j_1 j_2 JT ) C_{LM_L S M_S}^{LS} Y_{LM_L}(\hat{r}) \chi_{S M_S} R_{l_1}(r) R_{l_2}(r)$

with

$\displaystyle \alpha_{LS}(j_1j_2 JT) = \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi (2L+1)}} \\ \frac{1-(-1)^{S+T}}{\sqrt{2(1+\delta_{j_1 j_2} \delta_{l_1 l_2})}} C_{l_1 0 l_2 0}^{L 0} \gamma_{LS}^{J}(j_1 l_1;j_2 l_2)$

$\displaystyle \gamma_{LS}^{J}(j_1 l_1;j_2 l_2) = \sqrt{(2j_1+1)(2j_2+1)(2L+1)(2S+1)} \\ \begin{Bmatrix} l_1 & s & j_1 \\ l_2 & s & j_2 \\ L & S & J \end{Bmatrix}$

Return to the matrix element

$\displaystyle \left< j_1 j_2 J M | V | j_3 j_4 J M \right>$

Since the matrix element should not depends on $M$ , thus, we sum on M and divide by $(2J+1)$ ,

$\displaystyle \left< j_1 j_2 J M | V | j_3 j_4 J M \right> = \frac{1}{2J+1} \sum_M \left< j_1 j_2 J M | V | j_3 j_4 J M \right>$

$\displaystyle \left< j_1 j_2 J M | V | j_3 j_4 J M \right> = \frac{-4\pi V_0 \bar{R}}{2J+1} \sum_{LS} \alpha_{LS}(j_1j_2JT) \alpha_{LS}(j_3j_4JT) \\ \sum_{M M_L M_S} (C_{LM_SSM_S}^{JM})^2 \int Y_{LM}^* Y_{LM} d\hat{r}$

with

$\displaystyle \bar{R} = \int R_{j_1} R_{j_2} R_{j_3} R_{j_4} dr$

( i give up, just copy the result ), for $T = 1$ ,

$\displaystyle \left< j_1 j_2 J M | V | j_3 j_4 J M \right> = (-1)^{j_1+j_3+l_2+l_4 + n_1+n_2+n_3+n_4}\\ (1+(-1)^{l_1+l_2+l_3+l_4}) (1 + (-1)^{l_3+l_4+J}) \\ \frac{V_0 \bar{R}}{4)2J+1)} \sqrt{\frac{(2j_1+1)(2j_2+1)(2j_3+1)(2j_4+1)}{(1+\delta_{j_3j_4})(1+\delta_{j_1j_2})}} \\ C_{j_1(1/2)j_2(-1/2)}^{J0} C_{j_3(1/2)j_4(-1/2)}^{J0}$

The block matrix are

$\displaystyle M_{J=0} = \begin{pmatrix} 2 \epsilon_d - 3 V_0 \bar{R} & -\sqrt{3} V_0 \bar{R} \\ -\sqrt{3} V_0 \bar{R} & 2 \epsilon_s - V_0 \bar{R} \end{pmatrix}$

$\displaystyle M_{J=2} = \begin{pmatrix} 2 \epsilon_d - \frac{24}{35} V_0 \bar{R} & -\frac{12\sqrt{7}}{35} V_0 \bar{R} \\ -\frac{12\sqrt{7}}{35} V_0 \bar{R} & \epsilon_d + \epsilon_s - \frac{6}{5}V_0 \bar{R} \end{pmatrix}$

$\displaystyle M_{J=3} = \epsilon_d + \epsilon_s$

$\displaystyle M_{J=4} = 2 \epsilon_d - \frac{2}{7} V_0 \bar{R}$

To solve the eigen systems, it is better to find the $\epsilon_d, \epsilon_s, V_0 \bar{R}$ from experimental data. The single particle energy of the d and s-orbtial can be found from 17O, We set the reference energy to the binding energy of 16O,

$\epsilon_d = BE(17O) - BE(16O) = -4.143 \textrm{MeV}$

$\epsilon_s = -4.143 + 0.871 = -3.272 \textrm{MeV}$

the ground state of 18O is

$E_0 = BE(18O) - BE(16O) = -12.189 \textrm{MeV}$

Solving the $M_{J=0}$ , the eigen value are

$\displaystyle \epsilon_d + \epsilon_s - 2 (V_0 \bar{R}) \pm \sqrt{ (\epsilon_s - \epsilon_d + V_0 \bar{R})^2 + 3 (V_0 \bar{R})^2 }$

Thus,

$V_0 \bar{R} = 1.057 \textrm{MeV}$

The solution for all status are

$E(j=0) = -12.189 ( 0 ) , 0.929 (0d_{5/2})^2 + 0.371 (1s_{1/2})^2$

$E(j=2) = -9.820 ( 2.368 ) , 0.764 (0d_{5/2})^2 + 0.645 (0d_{5/2})(1s_{1/2})$

$E(j=4) = -8.588 ( 3.600) , (0d_{5/2})^2$

$E(j=2) = -7.874 ( 4.313) , 0.645 (0d_{5/2})^2 -0.764 (0d_{5/2})(1s_{1/2})$

$E(j=3) = -7.415 (4.773) , (0d_{5/2})(1s_{1/2})$

$E(j=0) = -6.870 (5.317 ) , 0.371 (0d_{5/2})^2 - 0.929 (1s_{1/2})^2$

Annotation 2020-05-14 235102

The 2nd 0+ state is missing in above calculation. This is due to core-excitation that 2 p-shell proton promotes to d-shell.

In the sd- shell, there are 2 protons and 2 neutrons coupled to the lowest state. which is the same s-d shell configuration as 20Ne. The energy is

$E_{sd} = B(20Ne) - B(16O) = -33.027 \textrm{MeV}$

In the p-shell, the configuration is same as 14C, the energy is

$E_{p} = B(14C) - B(16O) = 22.335 \textrm{MeV}$

Thus, the energy for the 2-particle 2-hole of 18O is

$E(0^+_2) = -10.692 + 8 E' \textrm{MeV}$ ,

where $E'$ is the p-sd interaction, there are 4 particle in sd-shell and 2 hole in p-shell, thus, total of 8 particle-hole interaction.

The particle-hole can be estimate using 19F 1/2- state, This state is known to be a promotion of a p-shell proton into sd-shell.

In the sd-shell of 19F, the configuration is same as 20Ne. In the p-shell of 19F , the configuration is same as 15N, the energy is

$E_{p} = B(15N) - B(16O) = 12.128 \textrm{MeV}$

Thus, the energy for the 1/2- state of 19F relative to 16O is

$E_{1/2} = -20.899 + 4 E' \textrm{MeV}$

And this energy is also equal to

$E_{1/2} = -20.899 + 4 E' \textrm{MeV} = BE(19F) - BE(16O) + 0.110 = -20.072 \textrm{MeV}$

Thus,

$E' = 0.20675 \textrm{MeV}$

Therefore, the 2nd 0+ energy of 18O is

$E(0^+_2) = -9.038' \textrm{MeV} = 3.151 \textrm{MeV}$

Compare the experimental value of 3.63 MeV, this is a fair estimation.

It is interesting that, we did not really calculate the radial integral, and the angular part is calculated solely base on the algebra of J-coupling and the properties of delta interaction.

And since the single particle energies and residual interaction $V_0 \bar{R}$ are extracted from experiment. Thus, we can think that the basis is the “realistic” orbital of d and s -shell.

The spectroscopic strength and the wave function of the 18O ground state is $0.929 (0d_{5/2})^2 + 0.371 (1s_{1/2})^2$ . In here the basis wavefunctions $0d_{5/2}, 1s_{1/2}$ are the “realistic” or “natural” basis.

Single-particle structure of 19F

January 28, 2019

GoLuckyRyan Basic 18O, 19F, 20Ne, Nilsson orbital, rotational band 2 Comments

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.

it helps to understand how deformation happen in small system,
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.

The nuclear structure of 19F

January 30, 2018

GoLuckyRyan Basic 18O, 19F, 20Ne, 3Hed, d3He, Nilsson orbital, p2p Leave a comment

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,

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?
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?
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.
If the ground state has 1d5/2 proton, why the magnetic moment are so close to free proton? the l=2 should also contribute.
Is neutron shell also 1s1/2 ?
What is the $\beta_2$ ?
in 20Ne, will the proton also in 1s1/2 shell? 20Ne has $\beta_2 = 0.7$ very deformed.
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. )

a GEANT4 Simulation

January 30, 2016

GoLuckyRyan Basic, computer, simulation 20Ne, C++, Cross section, GEANT4, neutron, positron 5 Comments

The GEANT4 program structure was borrow from the example B5. I found that the most confusing part is the Action. Before that, let me start with the main().

GEANT4 is a set of library and toolkit, so, to use it, basically, you add a alot GEANT4 header files on your c++ programs. And, every c++ program started with main(). I write the tree diagram for my simplified exampleB5,

main()
 +- DetectorConstruction.hh
    +- Construct()
       +- HodoscopeSD.hh     // SD = sensitive detector
          +- HodoscopeHit.hh //information for a hit
          +- ProcessHits()   //save the hit information
       +- ConstructSDandField() //define which solid is SD
       +- ConstructMaterials()  //define material
+- PhysicsList.hh  // use FTFP_BERT, a default physics for high energy physics
+- ActionInitialization.hh
   +- PrimaryGeneratorAction.hh // define the particle gun
   +- RunAction.hh // define what to do when a Run start, like define root tree
   +- Analysis.h  // call for g4root.h
   +- EventAction.hh //fill the tree and show some information during an event

A GEANT4 program contains 3 parts: Detector Construction, Physics, and Action. The detector construction is very straight forward. GEANT4 manual is very good place to start. The physics is a kind of mystery for me. The Action part can be complicated, because there could be many things to do, like getting the 2ndary beam, the particle type, the reaction channel, energy deposit, etc…

Anyway, I managed to do a simple scattering simulation, 6Li(2mm) + 22Ne(60A MeV) scattering in vacuum. A 100 events were drawn. The detector is a 2 layers plastic hodoscope, 1 mm for dE detector, 5 mm for E detector. I generated 1 million events. The default color code is Blue for positive charge, Green for neutral, Red for negative charge. So, the green rays could be gamma or neutron. The red rays could be positron, because it passed through the dE detector.

Screenshot from 2016-01-30 00:34:34.png

The histogram for the dE-TOF is Screenshot from 2016-01-29 23:26:34.png

We can see a tiny spot on (3.15,140), this is the elastics scattered beam, which is 20Ne. We can see 11 loci, started from Na on the top, to H at the very bottom.

The histogram of dE-E

Screenshot from 2016-01-29 23:30:38.png

For Mass identification, I follow the paper T. Shimoda et al., Nucl. Instrum. Methods 165, 261 (1979).

Screenshot from 2016-01-30 00:06:02.png

I counted the 20Na from 0.1 billion beam, the cross section is 2.15 barn.

Nuclear Physics 101

On the microscopic origin of deformation + paper reading

Alpha cluster and alpha separation energy

Shell model calculation on 18O

Single-particle structure of 19F

The nuclear structure of 19F

a GEANT4 Simulation

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