[Papers Reading] On the recent progress of the quenching of spectroscopic factor

May 20, 2023

GoLuckyRyan papers reading , Leave a comment

The journey started as early as 1993. J.M. Udías et. al used (e,e’p) to study the spectroscopic factors (SFs) of 40Ca and 208Pb [ PRC 48, 2731 (1993) ]. The experiment used the Coulomb force (which is well-understood) to extract the SF and found that it is quenched, i.e. about 30% smaller than the SF calculated from the Shell model calculation. Although there were many experiments that also found the quenching of SF using nuclear force, for example, (d,n), (d,p) experiments, people had generally concerned that the hadron probe would introduce some quenching, for example, the sensitivity of the optical potentials. So, using (e,e’p) experiment with fully relativistic analysis confirmed the quenching.

In 2001, G.J. Kramer, H. P. Blokb, and L. Lapikás published a study [ NPA 679, 267 (2001) ] and found that the SF quenching from (e,e’p) and (d,3He) experiments are consistent. Later, J. Lee, M. B. Tsang, W. G. Lynch published another study [ PRC 75, 064320 (2007) ] that confirmed the quenching is consistent with (d,p) and (p,d) transfer reactions from Z = 3 – 24. Thus the SF quenching was considered a fact for stable nuclei.

The figure is taken from NPA 679, 267 (2001). It shows that the SF quenching is consistent for (e,e’p) and (d,3He) reactions for various stable nuclei. This founding confirmed the hardon probe is also a good probe given that a consistent analysis framework is used.

Meanwhile, many theoretical studies were published to understand the cause of the quenching. For example, W. H. Dickhoff and C. Barbieri [ Prog. part. Nucl. Phys. 52, 377 (2004) ] pointed out that the quenching is due to Short-range correlation and Long-range correlation. The correlation will virtually excite the nucleon out of its orbital, creating high-momentum pairs, and occupation in “unoccupied” orbitals. Another example is I. Sick, [ Quasi-free knockout reaction, QFS workshop at ECT, Trento, 2008 ] studied 208Pb. There are many recent progresses on nuclear correlation, but I am not so familiar.

  • Taken from Prog. Part. Nucl. Phys. 52, 377 (2004).
  • Taken from I. Sick, Quasi-free knockout reaction, QFS workshop at ECT, Trento, 2008.

In 2008, A. Gade et. al published a systematic study on the SF quenching [ PRC 77, 044306 (2008), updated on PRC 90, 057602 (2014) ] with decades of experimental data on unstable nuclei. They found that when plotting the SF vs the “boundness” of the knockout nucleon, the quenching is small for weekly bound nucleons and the quenching can be as large as 60% for deeply bounded nucleons. They suggested that the weekly bounded nucleons are more “free” and like “independent-particle” than those that are deeply bound. As the SF is a concept from the independent particle model, (the 1st order interaction on a nucleon is from a mean-field generated by other nucleons, a 2nd order interaction is the fine detail of NN-interaction, and effective non-1st order interaction is called residual interaction. This model formed the basis of the Shell model calculation.) that, a weekly bound nucleon has weaker residual interaction so it is mainly influenced by the mean field, and the quenching is small. And the metric of the boundness is from the difference between the proton-neutron Fermi surface, or more precisely, the difference between the separation energies.

Taken from PRC 90, 057602 (2014)

This result drew much attention and leaded 2 studies using (p,2p) knockout experiment on the oxygen isotope chain from GSI, Germany [ PRL 120, 052501 (2018) ], and RIKEN, Japan [ Prog. Theor. Exp. Phys. 2018, 021D01 (2018) ]. Both of them found that the quenching does almost not depend on the boundness from 14O ( \Delta S = -19 MeV ) to 24O ( \Delta S = 23 MeV ). The discrepancy between the (p,2p) studies and Gade et al. results in many debates and discussions. One focus is on the reaction mechanism. In the Gade plot, the data were produced from nucleon removal reaction at 100 MeV/u on 9Be target. Many suspected that the reaction theory using 9Be target may be not complete, while others suspect that the reaction theory for (p,2p) is based on the impulse approximation, which is still unclear. On top of that, the direct comparison with different reactions is also in question, for example, the treatment of final state interaction.

  • Taken from Prog. Theor. Exp. Phys. 2018, 021D01 (2018).
  • Taken from Phys. Rev. Lett. 120, 052501 (2018)

Recently, there are 2 studies using (d,p) reaction at 10 MeV/u [ Kay et. al PRL 129, 152501 (2022) ] and (p,pN) at 100 MeV/u [ Pohl et. al PRL 130, 172501 (2023) ] report that the quenching almost not depends on the boundness.

Kay et. al used a 14N and 14C mixed beam to perform (d,p) reaction simultaneously. The reaction mechanism for (d,p) reaction is well understood and is considered to be a reliable way to extract the SF. And the simultaneous reactions eliminated most of the systematic uncertainty. Pohl et. al used a proton target to perform nucleon knockout on 14O. The study found that when including the inelastic that the 14O is excited above the proton threshold and emits a proton, the quenching almost does not depend on the boundness. However, if the inelastic channel was ignored, the quenching does depend on the boundness. In other words, the study suggests that the reaction theory used in the Gade plot is not complete.

  • Taken from PRL 129, 152501 (2022)
  • Taken from PRL 130, 172501 (2023)

On the microscopic origin of deformation + paper reading

February 4, 2023

GoLuckyRyan Basic , , , , , , , 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.

\beta_2 values around 100Zr.

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….

Measuring ANC using (d,p) transfer reaction

August 24, 2021

GoLuckyRyan Basic, papers reading , Leave a comment

The reference of this post is A. M. Mukhamedzhanov, F. M. Nunes, and P. Mohr, Phys. Rev. C 77, 051601(R) (2008).

The transfer amplitude is, in a simple sense,

\displaystyle T = \left< \phi_p \phi_{B} | V | \phi_d \phi_A \right> = \left< \phi_p \psi | V | \phi_d \right>, ~~~ \left< \psi \right| = \left<B|A\right>

where \psi is the quasi-particle wave function. The function can be separated into short-range and long-range part.

\displaystyle \psi(r) \xrightarrow{r \rightarrow \infty } C \frac{W(r)}{r}

The cross section of the reaction is

\displaystyle \sigma^2 = |T|^2 = | T_{int} + T_{ext} |^2 = \left( \int_0^R \phi_p^* \psi^* V \phi_d + C \int_R^\infty \phi_p^* \frac{W^*(r)}{r} V \phi_d \right)^2

The idea is, “If a reaction is completely peripheral, it is possible to extract the nuclear ANC from the normalization to the data without any single-particle ambiguity…… For sub-Coulomb transfer reactions, one is only sensitive to the asymptotic part of the neutron wave function, and the ANC can be extracted virtually without theoretical uncertainties.“, from the reference.

Experimentally, the cross section is measured for a give reaction. The quasi-particle state is approximated by a single-particle wave function from Woods-Saxon potential.

\displaystyle \psi(r) \xrightarrow{r \rightarrow \infty } C \frac{W(r)}{r}

\displaystyle \phi(r) \xrightarrow{r \rightarrow \infty } b \frac{W(r)}{r}

Thus,

\displaystyle \psi(r) \approx \sqrt{S} \phi(r), ~~~ C(b) = \sqrt{S(b)}b

In this sense, the interior is approximated by the single-particle wave function, then, the nuclear ANC will depend on the single-particle wave function, and also the Woods-Saxon potential.

However, if the reaction is completely peripheral, the internal contribution is zero, and the nuclear ANC is independent of the Woods-Saxon parameters, or the single-particle ANC. Or, the change of the spectroscopic factor and the change of single-particle ANC will cancel out.

I think zero internal contribution and cancelation of SF and b are difference concepts. And in the DWBA framework, how to compute the internal part that it is small ? The quasi-particle state in the internal part is approximated by the single-particle wave function, and the internal part is an integral of many things, how it will not change with difference Woods-Saxon parameters?

And there is a difference between insensitive and not-contribute. If the internal part is only insensitive to the Woods-Saxon parameters, it can still be non-zero. If it is non-zero, the nuclear ANC has an offset.

By measuring the reaction at difference energy, the offset can be checked.

Using ANC to calculate the bound state radius?

August 23, 2021

GoLuckyRyan Basic, papers reading , , Leave a comment

The idea is from these papers :

  1. F. Carstoiu et al., Phys Rev. C 63, 054310 (2001)
  2. E. T. Li et al., Chinese Phys. C 40, 114104 (2016)

Here is the summary of the idea.

The mean-square radius of an orbital from nucleon transfer reaction A = B + N is

\displaystyle \left<r^2 \right> = \frac{\int_0^\infty r^4 I^2(r) dr}{\int_0^\infty r^2 I^2_{AB}(r) dr}

where I_{AB}(r) is the quasi-particle from the overlap of nuclei A and B, such that I_{AB}(r) = \left<A | B \right> . The quasi-particle state is normalized to the spectroscopic factor

\displaystyle S = \int_0^\infty r^2 I^2_{AB}(r) dr

The mean-square radius is then

\displaystyle \left<r^2 \right> = \frac{1}{S}\int_0^\infty r^4 I_{AB}^2(r) dr

We can split the integral into 2 parts, a short range and long range. The long range part should be close to the Whittaker function as the nuclear potential is essentially zero. And the quasi-particle is proportional to the Whittaker function by the ANC

\displaystyle I_{AB}(r) \xrightarrow{r\rightarrow \infty} C \frac{W(r)}{r}

The mean-square radius break down to

\displaystyle \left<r^2 \right> = \frac{1}{S} \left( \int_0^{R_n} r^4 I_{AB}^2(r) dr + C^2 \int_{R_n}^\infty r^2 W^2(r) dr \right)

Since the ANC is not sensitive to the nuclear potential and considered to be more reliable, thus, the rms radius can be extracted/calculated using the ANC and the bound state wave function, which is approximated and calculated by Woods-Saxon potential.

Approximate the normalized orbital or the bound state wave function to be

\displaystyle \phi (r) \approx \frac{1}{\sqrt{S}} I_{AB}(r)

The mean-square radius is

\displaystyle \left<r^2 \right> \approx \int_0^{R_n} r^4 \phi^2(r) dr + b^2 \int_{R_n}^\infty r^2 W^2(r) dr = \int_0^\infty r^4 \phi^2(r) dr

The last step used the fact that the bound state wave function must be approached to the Whittaker function with the single-particle ANC. The last step eliminated the need for the radius R_n . Since the bound state wave function is calculated from (usually) a Woods-Saxon potential, which reproduces the separation energy, in DWBA framework, the rms radius is directly calculated from the DWBA.

In the two references, the authors fitted the potential width and diffusiveness parameters, which reproduced the separation energy. They claimed that they found the best, model insensitive parameters, and using the ANC, to deduce the rms radius.

The formula of the rms radius still depends on the spectroscopic factor and bound state wave function. The references minimizing the difference between the bound state wave function and the Whittaker function for the long-range

\displaystyle \chi^2 = \int_{R_n}^\infty \left( I_{AB}(r) - C \frac{W(r)}{r} \right)^2 \approx \int_{R_n}^\infty \left( \sqrt{S} \phi(r) - C \frac{W(r)}{r} \right)

If one can have an independent value of the nuclear ANC C, then the minimization of the \chi^2 force the single-particle wave function, and the Woods-Saxon parameters reproduce the long-range behaviour. And give a consistent spectroscopic factor with the nuclear ANC.

For 16N,

Energy [MeV]Spin-paritysingle-particle ANCRMS radius [fm]RMS radius from reference 2 [fm]
-28.820s1/219.692.13
-15.270p3/28.552.80
-12.200p1/25.982.86
-2.340d5/20.433.763.85 +- 0.31
-2.161s1/2-2.184.814.82 +- 0.42
The Woods-Saxon parameters are V_0 = -49.57 MeV, r_0 = 1.25 fm, a_0 = a_{so} = 0.67 fm, V_{so} = 20.2 MeV, r_{so} = 1.20 fm.

In above table, I calculate the RMS radius of 16N and compare to reference 2. They are consistent, which is no surprise. In reference 2, the nuclear ANC is deduced from spectroscopic factor, which is a fatal flaw in the logic. If the nuclear ANC is not independently deduced or based on spectroscopic factor, the \chi^2 is

\displaystyle \chi^2 \approx \int_{R_n}^\infty \left( \sqrt{S} \phi(r) - C(S_0) \frac{W(r)}{r} \right)

The S_0 indicate the nuclear ANC C has implicit dependence to the spectroscopic factor S_0, which depends on some Woods-Saxon parameters. The minimization is then nothing but forcing the fitting Woods-Saxon parameters to the Woods-Saxon parameters that gave S_0 .

And I bet that the Woods-Saxon parameters that gave S_0 are based on fitting the single-particle energies.

Paper reading : One-neutron halo structure in 15C

April 7, 2021

GoLuckyRyan papers reading , , Leave a comment

The paper DOI is https://doi.org/10.1103/PhysRevC.69.034613

This is my personal understanding of the paper.

Reaction : 14,15C one- or two- neutron removal on a carbon target (~ 375 mg/cm2)

Energy : 51 or 83 MeV (at the middle of the target), the 14,15C was produced using 110 A MeV 22Ne on 2 mm thick Be target.

Measurement : Reaction cross section and longitudinal momentum of carbon fragments (15C,14,13C) and (14C,13C) in the beam rest frame.

Result: the widths of the longitudinal momentum from (15C,14C) and (15C,13C) are 71 ± 9 MeV/c and 223 ± 28 MeV/c respectively, the width for that of (14C,13C) is 195 ± 21 MeV/c.

  • Due to the small neutron separation energy (Sn = -1.218 MeV) and the ground state spin-parity of 1/2+, 15C is thought to be a one-neutron halo.
  • The ground state spin parity of 1/2+ suggests that the valence neutron is in the 1s1/2 orbital.
  • 14C(d,p) reactions measured the spectroscopic factor for the s-wave around around 0.76 to 1.03, depending on the DWBA setting.

The authors motive is follow:

  • momentum distributions of (15C,14C) at beam energy ~ 100 MeV/u are measured at NSCL, and the result widths are smaller than the Glodhaber’s model. That suggests a larger spatial distribution, which is a characteristic of halo.
  • Another characteristic of halo is enhanced reaction cross section
  • The reaction cross section on carbon target (945 mb) at ~740 A MeV for 15C showed no enhancement, but the cross section (853, 862, 880, 1036, 1056, 1104, 1231, 1187 mb) for 12,13,14,16,17,18,19,20C are measured at ~950 A MeV [A. Ozawa, NPA, 693 (2001) 32]
  • The reaction cross section measured at ~ 60 MeV/u show some enhancement. [D.Q.Fang, PRC, 61 (2000) 064311]
  • The charge exchange cross section at 930 MeV/u show enhancement.

The authors stated that, low energy reaction is more sensitive to the tail of the nuclear density distribution than the high energy. They decided to measure the reaction cross section and the momentum distribution simultaneously at 51 and 83 MeV/u to find a consistence picture of the halo structure in 15C.

My comment : the high energy probe cannot see the enhancement is probably because of the probe is sensitive to all nucleus, or as the author suggested not sensitive to the tail. The high energy probe has de Broglie wave length shorter than nucleon can also able to removal the 0s-state. The 1s neutron only 1/9 of the all neutrons, so the effect of it is overwhelm by the other neutrons.

Also, may be I am not fully understood, I feel that the cross section is somewhat beyond intuition, for example, in every text book, the elastic cross section is 4 times bigger than the geometry cross section of the actual hard sphere, how and why ? I mean, I understand the math, but how to understand it in physical sense? And for the thermal neutron energy, the absorption cross sections for various elements do not make any sense. (do they?) I think the measured reaction cross section, not only depended on the nuclei size and the projectile de Broglie wave length, but also the nuclear interaction, and that makes the interpretation difficult and ambiguous.

The 1st part of the experimental result is the longitudinal momentum distribution. The widths of the longitudinal momentum from (15C,14C) and (15C,13C) are 71 ± 9 MeV/c and 223 ± 28 MeV/c respectively, the width for that of (14C,13C) is 195 ± 21 MeV/c.

My comment : The result could be well understand as the 1s1/2 spatial distribution, which is probed by the (15C,14C) alone, is larger than that of the p-wave, which is probed by (15C,13C) and (14C,13C). The (15C,13C) reaction should probe the mixed s- and p-wave distribution.

From the (15C,14C) distribution, they gave the s- and d- wave component of 66% and 4%, respectively, by assuming a fixed ratio between the s- and p- wave components. And I guess the remaining 30% is from the p-wave. This indicate the d-wave component is small, which is agreed with gamma-ray spectroscopy, that gave the d-wave is 2%.

The 2nd part of the experimental result is the reaction cross section. The result show a clearly enhancement on 15C at 81 MeV/u.

The Interesting part of the paper is the study of the relation between beam energy and reaction cross section using the finite-range Glauber model. The authors doing in this manner:

  • A harmonic-oscillator (HO) type density distribution were used to reproduced the cross section at relativistic energy (~ 950 meV/u)
  • The reaction cross section for lower energy was calculated using the high energy fitted parameter.
  • adding a p1/2 neutron tail for 14C, then fit the model with all energies.
  • Also tried HO-type + Yukawa tail, where the Yukawa tail gives a longer tail.

The study show that the Yukawa tail gives a better agreement to the 14C reaction cross section, although the other two methods also give similar result.

Then, they applied similar study on 15C. They found that a much longer tail will give much closer to the experimental reaction cross section.

In both 14C and 15C, the HO-type density ended at around 4 to 5 fm at density of 10-3 fm-3. For 14C, the tail extended to 12 fm at density of 10-7 fm-3, while for 15C, the tail extended to 20+ fm at density of 10-7 fm-3.

My comment : I am curious that can they reproduce the single-particle energy using the HO-type + Yukawa tail density. As in this post, I fitted the single-particle energy of 15C using ordinary Woods-Saxon potential, and also give the spatial and momentum distribution of the wave-function. The width (FWHM) of the momentum distribution for the 1s1/2 orbital is only ~ 60 MeV/c, smaller than 70 MeV/c. The p-wave width is about 180 MeV/c. Using 70% 1s-wave and 30% p-wave from the experiment, the simple mean 60 \times 0.7 + 180 \times 0.3 = 96 MeV/c, then the width will be larger than the experiment.

I guess one fundamental issue is, how the HO-type + Yukawa tail (or a Woods-Saxon) correctly describe the nuclear density for halo nuclei?

Paper reading: Statistical Models of Fragmentation Processes

April 5, 2021

GoLuckyRyan Basic, papers reading Leave a comment

The paper DOI is https://doi.org/10.1016/0370-2693(74)90388-8

This is a short paper, the author stated that relativistic ( GeV/u ) heavy ion 12C, 16O hit on difference target result similar momentum distribution in the frame of the ion, disregard of the mass of the fragment. Later, a detail study found out the momentum distribution depends on the mass of the fragment as

\displaystyle \sigma^2 = \sigma_0^2 \frac{K(A-K)}{A-1}

where K is the mass number of the fragment, and A is the mass of the projectile.

In order to understand this. the author asked and answered one question:

For a nuclei with A nucleons with the total momentum is \vec{p}_A = 0 , what is the mean square total momentum \vec{p}_K for arbitrary K nucleons?

Suppose each nucleon has momentum \vec{p}_i , since \vec{p}_A = 0 ,

\displaystyle \vec{p}_A \cdot \vec{p}_A = \left( \sum_{i}^A \vec{p}_i \right) \cdot \left( \sum_{i}^A \vec{p}_i \right) = \sum_{i}^{A} |\vec{p}_i|^2 + \sum_{i\neq j} \vec{p}_i \cdot \vec{p}_j = 0

The author assumes that the A nucleons has a mean square momentum \left< \vec{p}^2 \right> , i.e.

\displaystyle \frac{1}{A} \sum_{i}^{A} |\vec{p}_i|^2 = \left< \vec{p}^2 \right>

Thus,

\displaystyle A \left<\vec{p}^2 \right> + \sum_{i\neq j} \vec{p}_i \cdot \vec{p}_j = 0

For A nucleons, there are A(A-1) off-diagonal product, so, the “average” of \vec{p}_i \cdot \vec{p}_j is

\displaystyle A \left<\vec{p}^2 \right> + A(A-1) \left< \vec{p}_i \cdot \vec{p}_j \right> = 0

\displaystyle \left< \vec{p}_i \cdot \vec{p}_j \right> = \frac{\left<\vec{p}^2 \right>}{A-1}

Thus, for arbitrary K nucleon, the mean square of them is

\displaystyle \left<\vec{p}_K^2 \right> = \left<\left( \sum_{i}^K \vec{p}_i \right)^2 \right> = K \left< \vec{p}^2 \right> + K(K-1) \left< \vec{p}_i \cdot \vec{p}_j \right> \\= K \left< \vec{p}^2 \right> - \frac{K(K-1)}{A-1} \left< \vec{p}^2 \right> = \frac{K(A-K)}{A-1} \left< \vec{p}^2 \right>

This result is applied on the momentum distribution of the removal reaction, for example, 12C or 16O single nucleon removal. For single nucleon removal, K = 1 and then,

\displaystyle \left<\vec{p}_K^2 \right> = \frac{K(A-K)}{A-1} \left< \vec{p}^2 \right> = \left< \vec{p}^2 \right>

which is equal to the mean square of the A-nucleons system, as expected.

The author stated that the width of the longitudinal momentum distribution \sigma^2 follows

\displaystyle \sigma^2 = \sigma_0^2 \frac{K(A-K)}{A-1}

And the width should be

\displaystyle \sigma_0^2 = \frac{1}{3} \left<\vec{p}^2 \right>

due to the longitudinal momentum is only one-third of the total momentum, and \sigma_0 \approx 90 MeV/c.

From previous post on the momentum distribution from Woods-Saxon potential, the distribution width strongly depends on the orbital. And the mean square momentum depends on the energy level too. As the KE of a nucleon also depends on the energy level.

I suspect that, the mathematics formula is rigid, as we denote all nucleon, in average, has the same mean square momentum. And the relativitics fragmentation reaction, is statistically, removing random nucleon(s), that it average out the effect from difference orbital. A detail experiment on orbital identification should able to see the different momentum distributions from different orbitals.

Notes on the original Nilsson paper

June 21, 2020

GoLuckyRyan Basic , , , Leave a comment

The paper title is “Binding states of individual nucleons in strongly deformed nuclei” by Sven Gösta Nilsson on 1955.

In the introduction, the total nuclear wave function \Psi

\Psi = \phi_\Omega D_{MK}^I

where D_{MK}^I is Wigner D-matrix for rotation, \phi_\Omega is the intrinsic motion of all nucleons. The vibration component is skipped. Because of imcompressible nuclear matter, a vibration needs a lot of energy.

The total wave function should preserve parity, so, a complete wave function should be

\displaystyle \Psi = \sqrt{\frac{2I+1}{16\pi^2}} \left( \phi_\Omega D_{MK}^I + (-1)^{I+K} \phi_{-\Omega} D_{M-K}^I  \right)

Below is a picture of the quantum number.

Annotation 2020-06-21 134519

  1. In large deformation, J is not a good quantum number, but \Omega is always a good one.
  2. At ground state, K = \Omega , so, the rotation angular momentum is perpendicular to the body frame axis.

The single nucleon potential is the usual.

\displaystyle \frac{H}{\hbar \omega_0} = -\frac{1}{2} \nabla^2 + \frac{1}{2}r^2 - \beta r^2 Y_{20} + C l\cdot s + D l^2

The basis in Nilsson’s paper is the eigen state of harmonic oscillator in L-S representation. The Hamiltonian is

\displaystyle \frac{H_0}{\hbar \omega_0} = -\frac{1}{2} \nabla^2 + \frac{1}{2}r^2

with

\displaystyle H_0 \left| N l m_l m_s \right> = \hbar \omega_0 \left( N + \frac{3}{2} \right)\left| N l m_l m_s \right>

The original paper use \Sigma = m_s , \Lambda = m_l . And the function solution is

\displaystyle \left<r | N l m_l m_s \right> = A r^l e^{-\frac{r^2}{2}} F\left(- \frac{N-l}{2}, l+ \frac{3}{2}, r^2 \right) Y_{lm_l} \chi_{s m_s}

where F(a,b,x) is the confluent hypergeometric function, and it can be expressed as Laguerre polynomial. So, the solution can be rewrite as

\displaystyle \left<r | N l m_l m_s \right> = A r^l e^{-\frac{r^2}{2}} L_{\frac{N-l}{2}}^{l+\frac{1}{2}}(r^2) Y_{lm_l} \chi_{s m_s}

In contrast, in my calculation, I use the J-\Omega presentation, that, the connection is

\displaystyle \left| N l j \Omega \right> = \sum C_{l m_l s m_s}^{j \Omega } \left| N l m_l m_s \right>

The different is that, Nilsson needs additional transformation to calculate the C_{jl} coefficient, and the calculation of the \left< l\cdot s \right> and \left< l^2 \right> is a bit complicated, due to L-S is not a good quantum number when l\cdot s interaction was included. Thus, in Nilsson paper, he spent sometimes to talk about the calculation of \left< l\cdot s \right> and \left< l^2 \right> .

Next, Nilsson gives the calculation parameters of \kappa, \mu . And since he is using L-S representation, the Nilsson orbital is expressed in that basis. Here is a comparison between my calculation and Nilsson calculation.

Annotation 2020-06-21 141037

Next, he explained that for \beta > 0 , energy increase with increase \Omega due to “surface coupling”. It can be imagine like this:

Annotation 2020-06-21 141313

In above picture, when \Omega is smallest, l is perpendicular to the body axis, so the nucleon has large overlap with the whole nucleus, thus it is most bounded.

For small deformation, the deformation field r^2 Y_{20} is treated as a perturbation, so that j is a good quantum number.

For large deformation, the spin-orbital interaction l\cdot s, l^2 is treated as perturbation, and the good quantum number is N, n_z, m_l , m_s , since \Omega = m_s + m_l , so \Omega is also a good quantum number.

In our previous notation, Nilsson orbital is notated as K[N n_z \lambda] or [N n_z \lambda] K , which is equivalent  in Nilsson as [N n_z m_l] \Omega

The wave function of many nucleons.

After established the single nucleon wave function. The receipt of the construction of many nucleons wave function is

  1. select a set of Nilsson orbitals and form the Slater determent.
  2. Minimum the total energy by adjusting the deformation parameter.

\displaystyle H_t = \sum_i T_i + \frac{1}{2} \sum_{i \neq j} V_{ij}

It is interesting that the total wave function is not a mixture of various Slater determents from different combination of Nilsson orbitals, but rather a single Slater determent.

Ground state spin

Since each Nilsson orbital is degenerated to \pm \Omega, which are rotate oppositely. For even-even nucleus, the ground state spin must be zero. For even-odd nucleus, the ground state spin is equal to the \Omega of the last single nucleon.

For odd-odd nucleus, the ground state spin can be \Omega = |\Omega_p \pm \Omega_n|, the p-n interaction decide which \Omega is the ground state.

Decoupling parameter

For odd-A nuclei, the rotational energy is modified by a decoupling factor

\displaystyle E_r = \frac{\hbar^2}{2 I_M} \left( I(I+1) + a (-1)^{I+\frac{1}{2}} \left(I+\frac{1}{2}\right) \right)

And

\displaystyle a = \sum_{j} (-1)^{j - \frac{1}{2}} \left( j+ \frac{1}{2} \right) |C_{jl}^2|

Magnetic moments, EM transition probability, and ft-value for beta-decay are skipped.

Besides of the skipped material, it turns out the original Nilsson paper did not surprise me. And I still don’t really understand the “decoupling” thing.

PRL paper

May 26, 2020

GoLuckyRyan Basic , , , , Leave a comment

My PhD work is finally published today!

https://doi.org/10.1103/PhysRevLett.124.212502

https://phys.org/news/2020-05-proton-world-difference.html

There are few messages in this paper:

  1. Since the 24O is a doubly magic, by adding an extra proton, a conventional thinking would be no change of the magicity. i.e. when a proton is being knockout from 25F, the SF of 24O ground state would be large and not fragmented.
  2. But experiment on 25F(p,2p) tell the otherwise, the strength of 24O is small comparing to 24O excited state. That means the 25F composites a proton + (24O ground state and excited states ).
  3. Thus, there is large configuration mixing in the neutron sd-shell of 25F. This could means the end of N=16 magicity
  4. Using (p,2p) reaction to study neutron shell is a new method. This methodology is working because of the proton shell is a magic + 1 structure. Thus, the same method could be used to study 41Sc, 49Sc, etc.

Trapezoid filter

March 20, 2020

GoLuckyRyan Basic Leave a comment

The Trapezoid filter is one of the signal processing that used in CAEN digitizer.

The trapezoid filter is linear.

In the filter, there are few parameters, the “shaping” parameters are

  • trapezoid rise time – It has to be longer than the rise time of the signal.
  • trapezoid flat top
  • decay time constant – This is the decay time of the signal, it has to be matched.

In the following, we have a signal has rise time very short, only 1 time unit, and the decay time is 1000 time unit. The trapezoid setting is rise time = 100 unit, flat-top is 150 unit, and the decay time is 1000 unit. The Trapezoid is symmetry by design.

Screenshot 2020-03-20 at 19.31.16.png

The origin of the trapezoid filter is from this paper (https://doi.org/10.1016/0168-9002(94)91011-1). I do not fully understand the paper, for example, “the response of the system” I have no idea what does it mean. However, the equation (28), (30), and (31) gives the formula for the trapezoid filter.

Suppose the digital signal is v(n), when n < 0, v(n) = 0 , set

\displaystyle d^{m,l}(n) = v(n) - v(n-m-l) - v(n-l) + v(n - m - 2l)

\displaystyle p(n) = p(n-1)  + d^{m,l}(n), n\geq 0

\displaystyle s(n) = s(n-1) + p(n)  + d^{m,l}(n) M, n\geq 0

where l is the trapezoid rise time, m is the trapezoid flat top, M should be the signal decay time, and p(n) = s(n) = 0 for n <0 .

There are two notes should be taken

First, since the slope of the trapezoid is equal to the decay time. Thus, we need to re-scale the trapezoid by factor $latex  1/l/M$.

Second, for a constant non-zero signal, The trapezoid filter will treat the signal as a square pulse with infinite decay time. Thus, it is also useful to calculate the offset or baseline and subtract it.

Usually, the rise time is not that sharp, for a longer rise time, the edge of the Trapezoid filter will be rounded.

Screenshot 2020-03-20 at 19.45.19.png

Also, when the parameter M does not match the decay time, the trapezoid becomes asymmetric. This is also called pole-zero.

Screenshot 2020-03-20 at 19.50.17.png

One good thing for the Trapezoid filter is an easy discriminate a pile up signal and extract the pile up energy due to the filter is linear.

In the below plot, we have 2 pile up signals, the later one has size twice the first one.

Screenshot 2020-03-20 at 19.55.19.png

See this post for another alternative and a clear explanation.

PRL paper

February 13, 2020

GoLuckyRyan Basic , , Leave a comment

Haha!

https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.124.062502

Here are two news for this subject.

https://phys.org/news/2020-02-isolde-unexplored-region-nuclear-exotic.html

https://home.cern/news/news/physics/isolde-steps-unexplored-region-nuclear-chart

The experiment is simple and elegant. A neutron is put on top of 206Hg, which has N=126 and magic. This extra neutron will populate the next major shell ( N > 126). From the energy and angular distribution measured. We can identify the single-particle energy level, the orbital angular momentum. With the help of DWBA calculation, the occupation or the spectroscopic number can be extracted.

We deduced the single-particle energies and compared with Woods-Saxon calculation. Together with the 209Pb single particle energies. We extrapolating the single-particle energy to the neutron-dripline, where the r-process for heavy nuclear synthesis expected to take place. The result agree with most of the present theoretical calculations. In other word, the present theoretical calculations are seem to be fine.

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