Alpha cluster and alpha separation energy

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 , the alpha-separation energy is,

.

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 .

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.

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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 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.

, 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]

Decoupling factor of 19F

From the last post, we have the formula for the rotation band energy and the decoupling factor.

Lets apply to 19F rotational band.

The and bands are straight forward. The fitting model,

The decoupling factor for the band is . which is consistence with zero.

For the band, we fit for all states,

The fit is not so good. Let only fit up to 9/2.

The fit is much better. The decoupling factor for both fit are 2.09 to 2.26.

 

 

Shell model calculation on 18O

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

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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

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,

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

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

where is the single particle energy.

Suppose the residual interaction is an attractive delta interaction

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Be fore we evaluate the general matrix element,

We have to for the wave function ,

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

Since the residual interaction is a delta function, the integral is evaluated at , thus the radial function and spherical harmonic can be pulled out in the 2-particle wave function at is

Using the product of spherical Harmonic,

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

where

which is equal to

For

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

with

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Return to the matrix element

Since the matrix element should not depends on , thus, we sum on M and divide by ,

with

( i give up, just copy the result ), for ,

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The block matrix are

To solve the eigen systems, it is better to find the 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,

the ground state of 18O is

Solving the , the eigen value are

Thus,

The solution for all status are

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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

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

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

,

where 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

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

And this energy is also equal to

Thus,

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

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

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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 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  . In here the basis wavefunctions are the “realistic” or “natural” basis.

Some thoughts on the quenching of spectroscopic factor

Spectroscopic factor plays the central role in unfolding the nuclear structure. In the simplest manner, the total Hamiltonian of the nucleus is transformed into a 1-body effective potential and the many-body residual interaction, i.e.,

The effective single-particle Hamiltonian has solution:

where is the single-particle energy. The solution for is

where is the set of basis for state from , and is the eigenenergy.

The residual interaction is minimized by adjusted the mean-field . Thus, the residual interaction can be treated as a perturbation. This perturbs the nuclear wave function

The normalization requires .

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In the Slater determinant , a single-particle wave function for a particular orbital can be pull out.

where is anti-symmetric, angular coupling operator. Thus,

The is the spectroscopic factor. There are another sum-rule for adding and removing a nucleon. so that the sum is equal to the number of particle in a particle orbital.

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I always imagine the quenching is because we did not sum-up the SFs from zero energy to infinity energy (really???), thus, we are always only observing a small fraction of the total wave function. For example,  the total wavefunction would look like this:

In experiment, we observe the overlap between ground-state to ground-state transition

for ground-state to 1st excited state transition for the same orbital is

And since we can only observed limited number of excited states, bounded by either or boht :

• experimental conditions, say incident energy
• the excited states that are beyond single-particle threshold.
• finite sensitivity of momentum

Thus, we cannot recover the full spectroscopic factor. This is what I believe for the moment.

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Experimentally, the spectroscopic factor is quenched by 40% to 50%. The “theory” is that, the short-range interaction quench ~25%, the long-range interaction quench ~20%. The long- and short-range interaction correlate the single-particle orbital and reduce the degree of “single-particle”.

The short-range interaction is mainly from the “hard-core” of nucleon, i.e., the interaction at range smaller than 1 fm. The long-range interaction is coupling with nearby vibration states of the rest of the nucleus.

For example, from the 19F(d,3He) reaction, the spectroscopic factor for 19F 1s1/2 state is ~0.4, and 0d5/2 is ~0.6.

Thus, the wavefunction of 19F is

It is worth to note that the above SFs is not re-analysised and the “quenching” is not shown. Many old data had been re-analysised using global optical model and the SF is reduced and show that the sum of SFs is ~ 0.55.

If it is the case for 19F, the wavefunction would become,

Here I use for the “correlated wavefunction” that the single-particle orbital cannot simply pull out. Nevertheless, if and are correlated,

Am I misunderstood correlation?

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My problem is, What does a correlated wave function look like?

In my naive understanding, the Slater determinant is a complete basis for N-nucleon system. A particular single-particle orbital can ALWAYS be pull out from it. If it can not, therefore, the Slater determinant is NOT complete. The consequence is that all theoretical calculation is intrinsically missed the entire CORRELATED SPACE, an opposite of Slater determinant space (of course, due to truncation of vector space, it already missed somethings).

If the theory for correlation is correct, the short-range interaction is always there. Thus, the spectroscopic factor for deuteron 0s1/2 orbital is ~0.8, assuming no long-range correlation. However, we already knew that 96% of deuteron wavefunction is from 0s1/2 and  only 4% is from 1d5/2 due to tensor force. Is it not mean the spectroscopic factor of deuteron 0s1/2 state is 0.96?  Is deuteron is a special case that no media-modification of nuclear force? But, if the short-range correlation is due to the hard core of the nucleon, the media-modification is irrelevant. Sadly, there is no good data such as d(e,e’p) experiment. Another example is 4He(d,p)5He experiment. What is the spectroscopic factor for g.s. to g.s. transition, i.e. the 0p3/2 orbital? is it ~0.6 or ~ 1.0?

Since the experimental spectroscopic factor has model dependency (i.e. the optical potential). Could the quenching is due to incomplete treatment of the short- and long-range correlation during the interaction, that the theoretical cross section is always bigger?

In the very early days, people calibrate their optical potential using elastic scattering for both incoming and out-going channel, and using this to produce the inelastic one. At that time, the spectroscopic factors are close to ~1. But since each optical potential is specialized for each experiment. It is almost impossible to compare the SF from different experiments. Thus, people switch to a global optical potential. Is something wrong with the global optical potential? How is the deviation?

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Let me summarize in here.

1. The unperturbed wave function should be complete, i.e. all function can be expressed as a linear combination of them.
2. A particular single-particle orbital can be pull out from the Slater determinate .
3. The residual interaction perturbs the wave function. The short-/long-range correlation should be in the residual interaction by definition or by construction of the mean field.
4. The normalization of wave function required the sum of all SF to be 1.
5. Another sum rule of SFs equals to the number of particle.
6. By mean of the correlation, is that many excited states have to be included due to the residual interaction. No CORRELATED space, as the Slater determinant is complete. (pt. 1)
7. Above points (1) to (7) are solid mathematical statements, which are very hard to deny.
8. The logical result for the quenching of the observed SF is mainly due to not possible to sum up all SFs from all energy states for all momentum space.
9. The 2-body residual interaction can create virtual states. Are they the so called collective states?
10. But still, collective states must be able to express as the Slater determinant (pt. 1), in which a particular single-particle orbital can be pull out (pt. 2).
11. May be, even the particular single-particle orbital can be pull out, the rest cannot experimentally observed ? i.e. is not experimentally reachable. That go back to previous argument for limitation of experiments (pt. 8).
12. For some simple systems, say doubly magic +1, deuteron, halo-nucleon, very weakly bounded exited state, resonance state, the sum of SF could be close to 1. Isn’t it?
13. The theoretical cross section calculation that, the bound state wave function is obtained by pure single-particle orbital. I think it is a right thing to do.
14. The use of global optical potential may be, could be not a good thing to do. It may be the METHOD to deduce the OP has to be consistence, instead of the OP itself has to be universal. Need more reading from the past.

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 , 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 (), 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

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?

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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

In QM, that becomes

For axial symmetry,

Remember, in deformed nuclei, the projection of along the symmetry axis in the body-frame is . The expected value of the Hamiltonian with state in the body-frame is proportional to for and for . i.e.

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 . The problem of the following equation is that the J is not a good Q-number in Nilsson wavefunction )

However, the Wigner D-Matrix does not conserve parity transform:

In order to restored the parity, we need to include in the Lab-frame wave function.

where + for positive parity, – for negative parity.

From the above equation, for (), must be even (odd). For , .

We can see for , the are lower to the main sequence. This was explained by adding an extra term in the rotation Hamiltonian that connect .

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

A possible form of the . 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]

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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, 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)

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

The sum of 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 ?
7. in 20Ne, will the proton also in 1s1/2 shell? 20Ne has very deformed.
8. Deformed DWBA?

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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. )