Some thoughts on the quenching of spectroscopic factor

March 31, 2020

GoLuckyRyan Basic , , , , , Leave a comment

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.,

\displaystyle H = \sum_i^N \frac{P_i^2}{2m_i} + \sum_{i \neq j}^N V_{ij} = \sum_i^N  \left( \frac{P_i^2}{2m_i} + U \right) + \sum_{i\neq j}^N \left( V_{ij} - U \right) \\ = \sum_{i}^N h_i + H_R = H_0 + H_R

The effective single-particle Hamiltonian has solution:

\displaystyle h_i \phi_{nlj}(i) = \epsilon_{nlj} \phi_{nlj}(i)

where \epsilon_{nlj} is the single-particle energy. The solution for H_0 is

\displaystyle H_0 \Phi_k(N) = W_k \Phi_k(N)

\displaystyle \Phi_k(N)= \frac{1}{\sqrt{N!}}\begin{vmatrix} \phi_{p_1(k)}(1) & \phi_{p_1(k)}(2) & ... & \phi_{p_1(k)}(n) \\ \phi_{p_2(k)}(1) & \phi_{p_2(k)}(2) & ... & ... \\ ... & ... & ... & ... \\ \phi_{p_n(k)}(1) & \phi_{p_n(k)}(2) & ... & \phi_{p_N(k)}(N) \end{vmatrix}

\displaystyle W_k = \sum_i^N \epsilon_{p_i(k)}

where p_i(k) is the set of basis for state k from \phi_{nlj} , and W_k is the eigenenergy.

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

H \Psi_k(N) = W_k \Psi_k(N),    \Psi_k(N) = \sum_i \theta_{i}(k) \Phi_i(N) .

The normalization requires \sum_i \theta_{i}^2(k) = 1 .

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

\displaystyle \Phi_k(N) = \phi_{\mu} \otimes \Phi_{k}(N-1)

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

\displaystyle \Psi_k(N) = \sum_{\mu i} \theta_{\mu i}(k) \phi_{\mu} \otimes \Phi_i(N-1)

The \theta_{\mu i}^2 (k) 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.

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:

\displaystyle \Psi_k(N) =  \phi_{0} \otimes \left(\theta_{00}(k) \Phi_0(N-1) + \theta_{01}(k)\Phi_1(N-1) +....  \right) \\ + \phi_{1} \otimes \left( \theta_{10} \Phi_0(N-1) +...   \right) +...   .

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

\displaystyle  \left<  \phi_0 \Phi_0(N-1) | \Phi_0(N-1) \right> = \theta_{00}(0)

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

\displaystyle  \left<  \phi_0 \Phi_1(N-1) | \Phi_0(N-1) \right> = \theta_{01}(0)

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.

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”.

Annotation 2020-03-31 010604.png

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.

Annotation 2020-03-31 011328.png

Thus, the wavefunction of 19F is

\left|^{19}\textrm{F}\right> \approx  \sqrt{0.4} \left|1s_{1/2}\right> \otimes \left|^{18}\textrm{O}_{g.s.} \right> + \sqrt{0.6} \left|0d_{5/2} \right> \otimes \left|^{18}\textrm{O}(1.98) \right>

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,

\left|^{19}\textrm{F}\right> \approx  \sqrt{0.2} \left|1s_{1/2}\right> \otimes \left|^{18}\textrm{O}_{g.s.} \right> + \sqrt{0.3} \left|0d_{5/2} \right> \otimes \left|^{18}\textrm{O}(1.98) \right> + \sqrt{0.5} \Psi_k

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

f(x,y) \neq  g(x) h(y)

Am I misunderstood correlation?

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

In my naive understanding, the Slater determinant \Phi_k 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?

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 \Phi_k.
  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. \Phi_k(N-1) 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.

Simple model for 4He and NN-interaction

July 25, 2018

GoLuckyRyan Basic , , , , , , , Leave a comment

Starting from deuteron, the binding energy, or the p-n interaction is 2.2 MeV.

From triton, 3H, the total binding energy is 8.5 MeV, in which, there are only 3 interactions, two p-n and one n-n. Assume the p-n interaction does not change, the n-n interaction is 4.1 MeV.

The total binding energy of 3He is 7.7 MeV. The p-p interaction is 3.3 MeV.

Notices that we neglected the 3-body force in 3H and 3He. And it is strange that the n-n and p-p interaction is stronger then p-n interaction.

In 4He, the total binding energy becomes 28.3 MeV. I try to decompose the energy in term of 2-body, 3-body, and 4-body interaction.

If we only assume 2-body interaction, the interaction strength from n-p, n-n, and p-p are insufficient. One way to look is the 1-particle separation energy.

The neutron separation energy is 20.6 MeV = 2(p-n) + (n-n).
The proton separation energy is 19.8 MeV = 2(p-n) + (p-p).
The total energy is 28.3 MeV = 4(p-n) + (n-n) + (p-p).

There is no solution for above 3 equations. Thus, only consider 2-body interaction is not enough.

The neutron separation energy is 20.6 MeV = 2(p-n) + (n-n) + 2(n-n-p) + (n-p-p)
The proton separation energy is 19.8 MeV = 2(p-n) +(p-p) + 2(n-p-p) + (n-n-p).
The total energy is 28.3 MeV = 4(p-n) + (n-n) + (p-p) + 2(n-n-p) + 2(n-p-p) + (n-n-p-p).

Assuming the 2-body terms are the same in 2H, 3H, and 3He, the (n-p-p) and (n-n-p) is 4.03 MeV, which is strange again, as the Coulomb repulsion should make the (n-p-p) interaction smaller then the (n-n-p) interaction. The (n-n-p-p) interaction is -4.03 MeV.

Lets also add 3-body force in 3H and 3He.

The neutron separation energy of 3H is 6.3 MeV = (p-n) + (n-n) + (n-n-p)
The toal energy of 3H is 8.5 MeV = 2(p-n) + (n-n) + (n-n-p)
The toal energy of 3He is 7.7 MeV = 2(p-n) + (p-p) + (n-p-p)
The neutron separation energy of 4He is 20.6 MeV = 2(p-n) + (n-n) + 2(n-n-p) + (n-p-p)
The proton separation energy is 4He 19.8 MeV = 2(p-n) +(p-p) + 2(n-p-p) + (n-n-p).
The total energy of 4He is 28.3 MeV = 4(p-n) + (n-n) + (p-p) + 2(n-n-p) + 2(n-p-p) + (n-n-p-p).

We have 6 equations, with 6 unknown [ (p-n) , (n-n) , (p-p) , (n-n-p), (n-p-p), and (n-n-p-p)]. Notice that the equation from proton separation energy of 3He is automatically satisfied. The solution is

(p-n) = 2.2 MeV
(p-p) = – 4.7 – (n-n) MeV
(n-n-p)  = 4.1 – (n-n) MeV
(n-p-p) = 8 + (n-n) MeV
(n-n-p-p) = 0 MeV

It is interesting that there is redundant equation. But still, the (p-n) interaction is 2.2 MeV, and 4-body (n-n-p-p) becomes 0 MeV. Also the (n-p-p) is more bound than (n-n-p) by 3.9 + 2(n-n) MeV. If (n-p-p) should be more unbound, than (n-n) must be negative and smaller than -1.95 MeV.

Since the interaction strength has to be on the s-orbit (mainly), by considering 2H, 3H, 3He, and 4He exhausted all possible equations (I think). We need other way to anchor either (n-n), (p-p), (n-p-p), and (n-n-p) interactions.

Use the Coulomb interaction, the Coulomb interaction should add -1.44 MeV on the NN pair (assuming the separation is a 1 fm). Lets assume the (n-n) – (p-p) = 1.44 MeV

(n-n) = -1.63 MeV
(p-p) = – 3.07 MeV
(n-n-p)  = 5.73 MeV
(n-p-p) = 6.37  MeV

The (p-p) is more unbound than (n-n) as expected, but the (n-p-p) is more bound than (n-n-p) by 0.64 MeV. This is surprising! We can also see that, the 3-body interaction play an important role in nuclear interaction.

According to this analysis, the main contribution of the binding energies of 3H and 3He are the 3-body force.

In 3H:  (n-n) + 2(n-p) + (n-n-p) = -1.6 + 4.4 + 5.7 = 8.5 MeV
In 3He: (p-p) + 2(n-p) + (n-p-p) = -3.1 + 4.4 + 6.4 = 7.7 MeV

Worked on the algebra, when ever the difference  (n-n) – (p-p)  > 0.8 MeV, the (n-p-p) will be more bound that (n-n-p). Thus, the average protons separation should be more than 1.8 fm. I plot the interactions energies with the change of Coulomb energy below.

nn-pp.PNG

The (n-n) and (p-p) are isoscalar pair, where tensor force is zero. While the (n-p) quasi-deuteron is isovector pair. Thus, the difference between (n-n) and (n-p) reflect the tensor force in s-orbit, which is 3.8 MeV. In s-orbit, there is no spin-orbital interaction, therefore, we can regard the tensor force is 3.8 MeV for (n-p) isovector pair.

Following this method, may be, we can explore the NN interaction in more complex system, say the p-shell nuclei. need an automatic method. I wonder the above analysis agreed present interaction theory or not. If not, why? 

Radial Density of 3D Harmonic Oscillator

January 19, 2018

GoLuckyRyan Basic , Leave a comment

From the previous post, we have the radial formula for the 3D harmonic oscillator.

\displaystyle R_{nl}(r) \\ =\sqrt{ \frac{1}{\sqrt{\pi}\alpha^{2l+3}} \frac{(\frac{n-l}{2})! (\frac{n+l}{2})! 2^{n+l+2}}{(n+l+1)!}} r^l \exp\left(-\frac{r^2}{2\alpha^2}\right) L_{k}^{l+\frac{1}{2}}\left( \frac{r^2}{\alpha^2} \right)

For \alpha = 1 fm , the R^2 are plotted below.

Screen Shot 2018-01-19 at 17.10.27.png

Assume all orbits are fully filled.

\displaystyle \rho_n = \sum_{l}R^2_{nl} (2l+1)

Screen Shot 2018-01-19 at 17.13.16.png

The red is n=0, the pink is n = 1, the blue is n = 2, the cyan is n = 3, and the green is n = 4.

Next, we are summing all the shells to get the nucleon density.

Screen Shot 2018-01-19 at 17.15.37.png

In here, the red is the 4He nucleon density function, the pink is the nucleus with all 1p-shell are filled, that is 16O. The blue is the sd-shell are filled, that is 40Ca. The cyan is fp-shell filled.

We can see some systematic trend there. The n = odd nuclei has shape, which is somewhat similar to the Wood-Saxon shape. But for the n=even nuclei, the shape is like a V.

Note that the \alpha is fixed in this calculation. In fact, from Samuel Wong (2004) (Introductory Nuclear Physics),

\displaystyle \hbar \omega \approx 41 A^{-1/3} \textrm{MeV}

\displaystyle \alpha = \sqrt{\frac{\hbar}{m \omega}} = \sqrt{\frac{\hbar^2 c^2}{mc^2 \hbar \omega}} \approx A^{1/6} \textrm{fm}

 

Differential Cross Section II

February 6, 2011

GoLuckyRyan Basic, no math, operational, Scattering, theory , , , , , , , , , , , Leave a comment

Last time, the differential cross section discussion is based on quantum mechanics. This time, i try to explain it will out any math. so, that my mum ask me, i can tell her and make her understand. :)

in a scattering experiment, think about a target, say, a proton fixed in the center, it is positive charged. if another proton coming with some energy. it will get repelled, due to the repulsive nature of Coulomb force of same charge. it should be easy to understand, if the proton coming with high energy, it will get closer to the target, or even enter inside the target.

the repelling angle of the proton is not just depend on the energy it carry, but also on the impact parameter ( we usually call it b , but i like to call it r). the impact parameter is the shortest distance between the target and the line of the moving direction of the proton at long long away.

if the impact parameter is large, the proton miss the target. it almost cannot feel the target affection. thus, it go straight and unaffected. when the r is zero, it will hit the target head on head. and due to the repulsion. it will return back.  so, we can understand. the smaller the impact parameter, the deflection will be larger. since it can feel the force stronger.

For same impact parameter, the higher energy proton will have less deflection, since it travel faster, spend less time by the force, and the deflection get less.

Thus, we have an idea that the angle of deflection is high for small impact parameter and high energy. And most important, it only depends on these 2 factors and the effect from the target.

since our detector can only detect some small angle over some small area. So, we can place out detector on some angle, get the yield, and this is the name –  differential cross section come from.

Now, we have a uniform flow of particle with energy E. they will be deflected by the target and go to some angles. If we detect at the deflection angle, see how many particles ( the yield ) we can get in each angle. we can calculate back the effect of the target. For example, for a small angle, the particle get little defected, and this means the particle is from large impact parameter. for a large angle, the particles are from small impact parameter.

In some cases, the number of particle detected will be very high at some particular angle then others angles and this means, the cross section is large. and this means something interesting.

Moreover, don’t forget we can change the energy of the beam. for some suitable energy, the particle will being absorbed or resonance with the target. that given us low or high cross section on the energy spectrum.

( the graph is an unauthorized from the link: http://www.astm.org/Standards/E496.htm )

The above diagram is the differential cross section obtained from a Deuteron to a Tritium ( an isotope of Hydrogen with 2 neutrons and 1 proton) target, and the reaction change the Tritium into Helium and a neutron get out.

The reaction notation is

X(a,b)Y

where a is incident particle, X is target, b is out come particle, and Y is the residual particle.

the horizontal axis is detector angle at lab-frame. and the vertical axis is energy of Tritium. we always neglect the angle 0 degree, because it means no deflection and the particle does not “see” the target. at low energy, the d.c.s. is just cause by Coulomb force. but when the energy gets higher and higher, there is a peak around 60 degree. this peak is interesting, because it penetrated into the Helium and reveal the internal structure of it. it tells us, beside of the Coulomb force, there are another force inside. that force make the particle deflects to angle 60 degree. for more detail analysis, we need mathematic. i wish someday, i can explain those mathematics in a very simple way.

Therefore, we can think that, for higher energy beam, the size we can “see” will be smaller. if we think a particle accelerator is a microscope. higher energy will have larger magnification power. That’s why we keep building large and larger machines.