Kinematics of Knockout reaction

Leave a comment

It is super surprise for me that there is no any calculation for knockout kinematics in this blog!!! Part of my PhD is about (p,2p) knockout reaction!


Assume the reaction is A(a,cd)B, so that A = B+b, and b becomes d after the reaction. The separation energy of b is S_p,

S_p + m_A = m_d + m_B

The master equation for the kinematic calculation is

P_A+P_a = P_c+P_d +P_B

where P_i is the 4-momentum of particle i. This equation is true for any reference frame.


We can rearrange the master equation

(P_A - P_B) +P_a = P_b + P_a =  P_c+P_d

Here P_b is the 4-momentum of the to-be-knockout particle b.  Let expand the master equation in A’s rest frame (or normal kinematic).

\displaystyle \begin{pmatrix} m_A  \\  0  \end{pmatrix} + \begin{pmatrix} m_a \gamma \\  m_a\gamma \beta \hat{z} \end{pmatrix} = \begin{pmatrix} E_c \\ \vec{k}_c  \end{pmatrix} + \begin{pmatrix} E_d \\ \vec{k}_d \end{pmatrix} + \begin{pmatrix} \sqrt{m_B^2+k^2} \\ -\vec{k}  \end{pmatrix}

where \beta is the Lorentz boost.

\displaystyle P_b = P_A-P_B = \begin{pmatrix} m_a - \sqrt{m_B^2+k^2} \\ \vec{k}  \end{pmatrix}

Since the introduction of the vector \vec{k} for the to-be-knockout particle, 3 extra degree of freedoms : |\vec{k}|, \theta_k, \phi_k

knockout.png


To calculate the 4-momenta particle c and d, transform to the NN-frame so that the total momentum is ZERO, by setting

P_c = P_b + P_a

extract the Lorentz boost to the NN-frame \vec{\beta}_c = \vec{k}_c/E_c . In the NN-frame, the incident channel is particle a and b, the exist channel is particle c and d. In this frame, introduce 2 degree of freedom \theta_{NN}, \phi_{NN} for the scattering angle. The problem becomes a 2-body scattering problem.

The total energy, or the total mass of the NN-frame is

M_c = E_{NN} = \sqrt{E_c^2 - |\vec{k}|^2}

The momentum of particle c and d is

\displaystyle |\vec{p}| = p = \frac{1}{2M_c} \sqrt{(M_c^2 - (m_b +m_c)^2)(M_c^2 - (m_b - m_c)^2)}

In this frame, the normal vectors of a and b are not necessary to be on the z-axis (the beam axis), but at some angle. Thus, the calculation of normal vectors of c and d has to be careful. After the calculation, transform the 4-momenta in to the desire frame.


Experimentally, we measure the energy and momentum of the scattered particle c and d and usually ignored the momentum of the particle B. How to calculate the separation energy ? There are two methods. the first one is calculate the 4-momenta of particle A, a, c, and d. And using

S_p = m_B + m_d - m_A

m_B^2 = (E_A + E_a - E_c -E_d)^2 - |\vec{k}_A+\vec{k}_a-\vec{k}_c-\vec{k}_d|^2

And 2nd method, particularly for inverse kinematic, in Lab’s frame, the sum of 4-momenta of c and b is

\displaystyle P_c + P_d = \begin{pmatrix} mc + T_c + m_d + T_d \\ \vec{k}_c + \vec{k}_d \end{pmatrix}

In A’s rest frame, the Lorentz boost is -\beta, the energy part is

(P_c'+P_d)_E = \gamma(m_c + T_c + m_d + T_d) - \gamma \beta (\vec{k}_c + \vec{k}_d)\cdot \hat{z}

also

(P_c'+P_d')_E = (P_A' + P_a' - P_B')_E = m_A + \gamma m_a - E_B

using

\displaystyle  S_p = m_B + m_d - m_A = \sqrt{E_B^2 - |\vec{k}_b|^2} + m_d - m_A

to replace m_A

(P_c'+P_d')_E = \sqrt{E_B^2 - |\vec{k}_b|^2} + m_d - S_p + \gamma m_a - E_B

approximate

\displaystyle \sqrt{E_B^2 - |\vec{k}_b|^2} - E_B \approx  - \frac{|\vec{k}_b|^2}{2E_B}

put everything together,

\displaystyle S_p = m_d + \gamma(m_a - m_c -m_d) - \gamma(T_c + T_d) + \gamma \beta (\vec{k}_c + \vec{k}_d)\cdot \hat{z} - \frac{|\vec{k}_b|^2}{2E_B}

The last term is safe to be neglected as the orbital momentum is usually ~30 MeV/c and the total energy of particle B is close to the mass, which is so much better.

For (p,2p) reaction. m_a = m_b = m_c = m_p

\displaystyle S_p \approx (1-\gamma)m_p - \gamma(T_1 + T_2) + \gamma \beta (\vec{k}_1 + \vec{k}_2)\cdot \hat{z}

 

 

My PhD thesis

Leave a comment

Here https://doi.org/10.15083/00075194

or Here

The title is : Quasi-free proton knockout reaction of 23,25F


Abstract

The change of the neutron dripline from oxygen to fluorine indicates the 1d5/2 proton affects the neutron shell structure. We aim to know how the neutron sd-shell structure is changed by the 1d5/2 proton in neutron-rich 23F and 25F nucleus using proton spectroscopy. The spectroscopy is free from the effects of the proton shell structure, because the 1d5/2 proton in 23F or 25F is a single-particle state due to the 𝑍=8 magicity and even neutron number. If the neutron shell structure is not changed by the proton in 23,25F, after the sudden removal of that proton, the spectroscopic factor of that proton should be unity and not fragmented. Therefore, the effect on the neutron-shell from the proton will be shown on the spectroscopy.

The quasi-free 23,25F(p,2p) direct knockout reactions in inverse kinematics were performed in RIBF, RIKEN Nishina Center. Secondary beams of 23F and 25F were produced at ~280A MeV. The missing four-momentum of the residual oxygen (22O or 24O) was reconstructed using coincidence measurement of the incident nucleus and the two scattered protons. The excitation energy of the residue was then deduced.

thesis1.png
From the experimental results, the occupation number of the 1d5/2 proton of 25F was 0.1 ± 0.3 and the proton is indeed in single-particle state. Meanwhile, the spectroscopic strength of the 1d5/2 proton of 23F or 25F were fragmented. These pointed that the change of the sd-shell neutron structure due to the 1d5/2 proton is the reason of the fragmentation. The change of neutron shell suggests the disappearance of 𝑁=16 magicity. The nuclear structures of the 25F and 23F demonstrated the Type-1 shell evolution. The comparison with the present shell model interactions (SFO, USDB, and SDPF-MU) indicated that the tensor force should be stronger. Also, the spectroscopic strength of the p-orbit was ~0.8 in 23,25F, this shows that the short-range correlation in neutron-rich nuclei is as same as stable nuclei.

HELIOS for (p,2p) experiment

Leave a comment

The HELIOS concept is kinematics compression using a magnetic field, so that all charged particles spiral back to the z-axis. It is particularly useful in a special 2-body reaction where the degree of freedom is 2, which are center of mass angle \theta_{cm} and the excitation energy of the heavier recoil E_x. These 2 degree of freedom are mapped into the experimental space of the light recoil energy T and axial position of the light recoil z . i.e.

\displaystyle \begin{pmatrix} \theta_{cm} \\ E_x \end{pmatrix} \rightarrow \begin{pmatrix} T \\ z \end{pmatrix}

In (p,2p) reaction, the degree of freedom is 6, which are the momentum of the to-be-knockout proton (k_b, \theta_b, \phi_b) , the NN-scattering angle \theta_{NN}, \phi_{NN} , and the separation energy of the to-be-knockout proton S_p. Because of this, the scattered proton energy and axial position will be washed out. Below is an example of (p,2p) reaction. We can see that because of the orbital motion of the to-be-knockout proton, the scattered energy and angle of the scattered protons can have many value.

example of p2p reaction_2.png

Below is the energy vs z plot for a single proton. We can see that the usual straight line becomes diffused.

example of p2p reaction.png

However, if we measure the two protons at the same time, and sum up the energy and position, a magic appear.

example of p2p reaction_3.png

This is because the sum of energy and sum of angle is correlated even without magnetic field. When the energies summed up, the NN scattering angles are gone. This left 4 degree of freedom. (the following has to be checked.) And the axial position is related to the scattering angles, and they related to the angles of the to-be-knockout proton. Thus, only k_b and $ S_p$ is left.

In fact, the separation energy can be calculated by

\displaystyle S_p = (1-\gamma) m_p - \gamma(T_1 + T_2) + \beta \gamma (\vec{k_1} + \vec{k_2})\cdot \hat{z} - \frac{k_b^2}{2 E_B}

the term \vec{k}\cdot \hat{z} = \frac{cB}{2\pi} z_{cyc} . Therefore, the sum of proton energies is linearly related to the sum of axial positions.

The nuclear structure of 19F

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,

  1. Why deform? due to the single 1d5/2 proton? Suppose adding a proton on 18O, the proton fill on 1d5/2 shell, and the d-shell creates a deformation on the sd shell, that shift the energy lower by mixing with s-shell?
  2. in 19F(d, 3He) reaction, the sum of spectroscopic factor in sd-shell is just 1.54. This suggest large uncertainty. And the s-state SF is 0.38, almost a double for 18O(3He,d) reaction, How come?
  3. in 19F(d,3He) reaction, the s:d ratio is 0.4:0.6, this is similar to prediction of Nilsson model, but difference from USDB calculation.
  4. If the ground state has 1d5/2 proton, why the magnetic moment are so close to free proton? the l=2 should also contribute.
  5. Is neutron shell also 1s1/2 ?
  6. What is the \beta_2 ?
  7. in 20Ne, will the proton also in 1s1/2 shell? 20Ne has \beta_2 = 0.7 very deformed.
  8. Deformed DWBA?

The following is not organised thought.

According to the standard shell ordering, on top of 18O, an extra proton should fill up the 1d5/2 shell, and then the ground state spin of 19F should be 5/2. However, the ground state spin in 1/2. This is postulated to be due to deformation [mean field calculation, β2 = 0.275], 18O core excitation, or configuration mixing state [J.P. Elhot and A. M. Lane(1957)].

Under deformation, the conventional shell ordering is not suitable and may be an invalid picture to view the nucleus. So, talk about shell ordering is non-sense.

Since the 19F is 18O + 1s1/2 proton superposed with 1d5/2, there could be deformation. The spherical shape of 19F can be seen indirectly from the magnetic dipole momentum, the value is very close to that of a free proton of 2.78284734 μ0, only difference by 0.154 μ0, or 5.5%. How to solve this contradiction?

From the study of G. Th. Kaschl et al., the spectroscopic factor of the 19F(d,3He)18Og.s. channel is 0.38. The missing 1s1/2 strength most probably can be found in the higher excitation states. This indicates the ground state of 19F is a configuration mixing state. However, they also pointed out that caution is advisable with the absolute spectroscopic factor, this could be due to imperfect DWBA calculation.

The relative spectroscopic factors for the positive parity states, which normalised to the ground state, are agree with shell model prediction in sd-shell model space suggests that the core excitation should not play an important role.

 

From the USDB interaction, the shell ordering is normal, but the interaction result in a 1/2+ ground state. How?

What is the nature of the low lying 1/2- excited state in 19F?

20Ne(d,3He)19F reaction can populate this low lying state, suggests the p-shell proton pickup come from the nuclear surface.

( if (12C,13N) proton pickup reaction can populate this state, then, it can be confirmed that this is a surface p-shell proton, that it could be from 2p3/2. )