## Spectroscopic factor (again)

In the last posts on SF, my mind have been biased to the mean field model. I started everything from the mean field. Now, lets forget the mean filed, just starting from experimental view point and any complete basis.

Suppose an arbitrary complete single-particle basis, say, a Woods-Saxon numerical basis, 3D spherical harmonic oscillator, or spherical Gaussian basis, and donate it as $\phi_i$. Using this basis,, we can form the Slater determinate of N-nucleons as

$\displaystyle \Phi_i = \frac{1}{\sqrt{N!}}\begin{vmatrix} \phi_1(1) & ... & ... \\ ... & \phi_r(s) & ... \\ ... & ... & \phi_N(N) \end{vmatrix}$.

The solution of any N-body Hamiltonian can be calculated by solving the eigen-system with the matrix elements

$\displaystyle H_{ij} = \left< \Phi_i |H| \Phi_j \right>$

Say, the solution is

$\displaystyle H \Psi_i = E_i \Psi_i$

Experimentally, for the A(d,p)B reaction, we are observing the transition probability

$\displaystyle T = \left< \chi_p \Psi_i(B) |V|\chi_d \Psi_0(A) \right>,$

where $\chi_k$ is the distorted plane wave of  particle $k$ and $V$ is the interaction during the reaction. The integration summed all internal degree of freedom, particularly,

$\displaystyle \left<\Psi_i(B) |V_{BA} | \Psi_0(A) \right > \approx \left<\Psi_i(B) | \Psi_0(A) \right > = \sum_i \alpha_i \phi_i = \psi$

where $\psi$ is called the quasi-particle wave function and it is expressed in the complete basis of $\phi_i$.

And then, the quasi-particle wave function will also be integrated.

$\displaystyle T = \left< \chi_p \psi |V'|\chi_d \right>$

In fact, it is the frame work of many calculations, that require 4 pieces of input,

1. the optical potential for the incoming channel to calculate the distorted wave of the deuteron,
2. the optical potential for the outgoing channel to calculate the distorted wave of the proton,
3. the quasi-particle wave function, and
4. the interaction.

If the Woods-Saxon basis is being used, then we can have shell model picture that, how the quasi-particle “populates” each Woods-Saxon states, and gives the spectroscopic factor.

The Woods-Saxon basis is somewhat “artificial”, it is an approximation. the nature does not care what the basis is, the transition probability is simple an overlap between two nuclei (given that the interaction is not strong, i.e. in the case of Direct reaction). So, the spectroscopic factor is model dependence. If a weird basis is used, the shell model picture is lost. In this sense, SF is pure artificial based on shell model.

In experiment, we mainly observe the scattered proton, and from the angular distribution, the orbital angular momentum is determined. And from the conservation of angular momentum, we will know which orbital the neutron is being added to nucleus A.

But it is somehow weird. Who said the quasi-particle must be in an orbital angular momentum but not many momenta? If the spin of nucleus A is zero, because of conservation of angular momentum, the spin of nucleus B is equal to that of added neutron. But if the spin of nucleus A is not zero,  the neutron can be in many difference spins at once. In general, the conservation of angular momentum,

$\displaystyle \left < \Psi_i(B) | \phi \Psi_0(A) \right >$

could be difference than that of

$\displaystyle \left < \chi_d \Psi_i(B) | V | \chi_p \Psi_0(A) \right >$

We discussed that the spectroscopic factor is artificial. However, if we use a self-consistence basis, i.e. the basis that describes both nucleus A and B very well (is such basis exist?) , then the quasi-particle wave function is being described using the basis of nucleus B. To be explicit, lets say, we have a basis can describe nucleus B as

$\Psi_0(A) = 0.8 \Phi_0 + 0.6 \Phi_1$

$\Psi_i(B) = \sum_{jk} \beta_{jk} \left( \phi_j \otimes \Psi_k(A) \right)$

where the single particle basis capture most of the nucleus A. is the spectroscopic factor meaningful in this sense?

So, is such basis exist? i.e, can it describe a nuclear wave function (almost) perfectly? In ab initial calculation, it can expend the nuclear wave function using some basis and give the overlap. Can the quasi-particle wave function be projected into the “single-particle state” ? How to define the “single-particle state” in ab initial calculation?

The experimental spectroscopic factor is calculated by compare the experimental cross section with that from theory. In the theory, the cross section is calculated by assuming the bound state is from a Woods-Saxon potential without correlation. Thus. the spectroscopic factor is with respect to the Woods-Saxon potential, an approximation.

## 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.,

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

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”.

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

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

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.

## Diagonal elements for Nilsson orbital in spherical-spin function

In this post, we explain how to calculate the Nilsson orbital using perturbation method by compute the matrix elements using spherical-spin function. In that post, I said I will give the calculation for the perturbation element. Here we go for the diagonal elements

The diagonal matrix element,

$\displaystyle \left$

where the spherical-spin wave-function is

$\displaystyle |Nljk\rangle = A r^l e^{-\frac{r^2}{2}} L_{\frac{N-l}{2}}^{l + \frac{1}{2}}(r^2) \sum_{m m_s} Y_{lm}(\theta, \phi) \chi_{\frac{1}{2} m_s} C_{lm\frac{1}{2} m_s}^{jk}$

$\displaystyle A = \sqrt{\frac{(\frac{N-l}{2})!(\frac{N+l}{2})! 2^{N+l+2}}{\sqrt{\pi} (N+l+1)!}}$

$\displaystyle R_{Nl}(r) = r^l e^{-\frac{r^2}{2}} L_{\frac{N-l}{2}}^{l + \frac{1}{2}}(r^2)$

The radial integral is easy, we can use the integration formula. The radial integration is,

$\displaystyle \int_0^{\infty} r^2 R_{Nl}^2(r) r^2 dr = \int_0^{\infty} r^{2l+4} e^{-r^2} \left(L_{\frac{N-l}{2}}^{l + \frac{1}{2}}(r^2)\right)^2 dr$

When using the integration formula, one has to be careful when changing variable $r^2 \rightarrow r$. Since $dr^2 = 2 r dr$, we have to pull $2r$ out to properly do the change of variable.

$\displaystyle \int_0^{\infty} R_{Nl}^2(r) r^2 dr = \frac{1}{2}\int_0^{\infty} r^{2(l+1/2 + 1)} e^{-r^2} \left(L_{\frac{N-l}{2}}^{l + \frac{1}{2}}(r^2)\right)^2 (2 r dr)$

Set $r^2 \rightarrow x$, $l + 1/2 \rightarrow \alpha$, and $n = \frac{N-l}{2}$

$\displaystyle = \frac{1}{2}\int_0^{\infty} x^{\alpha+1} e^{-x} \left(L_{N}^{\alpha}(x)\right)^2 (dx) = \frac{1}{2}\frac{(\alpha+n)!}{n!}(2n+\alpha+1)$

Thus,

$\displaystyle \int_0^{\infty} R_{Nl}^2(r) r^2 dr = \frac{1}{2}\frac{( \frac{N+l+1}{2})!}{(\frac{N-l}{2})!}(N+\frac{3}{2})$

The angular-spin part is

$\displaystyle S_{ljk}(\theta,\phi, m_s) = \sum_{m m_s} Y_{lm}(\theta, \phi) \chi_{\frac{1}{2} m_s} C_{lm\frac{1}{2} m_s}^{jk}$

This contains spatial and spin part.

$\displaystyle S_{ljk}^*(\theta,\phi, m_s) Y_{20} S_{ljk}(\theta,\phi, m_s) \\=\sum_{m' m'_s} \sum_{m m_s} Y_{lm'}^*(\theta,\phi) Y_{20} Y_{lm}(\theta, \phi) \chi_{\frac{1}{2} m'_s} \cdot \chi_{\frac{1}{2} m_s} C_{lm'\frac{1}{2} m'_s}^{jk} C_{lm\frac{1}{2} m_s}^{jk}$

The dot-product of the spin part restricted the $m'_s$

$\displaystyle = \sum_{m' m m_s} Y_{lm'}^*(\theta,\phi) Y_{20} Y_{lm}(\theta, \phi) C_{lm'\frac{1}{2} m_s}^{jk} C_{lm\frac{1}{2} m_s}^{jk}$

And since $m+m_s = k$ and $m'+m_s =k$, therefore $m = m'$,

$\displaystyle = \sum_{m m_s} Y_{lm}^*(\theta,\phi) Y_{20} Y_{lm}(\theta, \phi) \left(C_{lm\frac{1}{2} m_s}^{jk}\right)^2$

And the integration of spherical harmonic gives,

$\displaystyle \int Y_{lm}^*(\theta,\phi) Y_{20} Y_{lm}(\theta, \phi) = \sqrt{\frac{5}{4\pi}} C_{20l0}^{l0} C_{20lm}^{lm}$

$\displaystyle C_{20l0}^{l0} = -\frac{l+1}{\sqrt{4l^3+8l^2+l-3}} , l > 0$

$\displaystyle C_{20l0}^{l0} = -\frac{l(l+1)-3m^2}{\sqrt{l(4l^3+8l^2+l-3)}} , l > 0$

Sum up everything,

$\displaystyle \left \\ = \frac{(\frac{N-l}{2})!(\frac{N+l}{2})! 2^{N+l+2}}{\sqrt{\pi} (N+l+1)!} \frac{1}{2}\frac{( \frac{N+l+1}{2})!}{(\frac{N-l}{2})!}(N+\frac{3}{2}) \sum_{m m_s} \sqrt{\frac{5}{4\pi}} C_{20l0}^{l0} C_{20lm}^{lm} \left(C_{lm\frac{1}{2} m_s}^{jk}\right)^2 \\ = \sqrt{\frac{5}{4}} \frac{(\frac{N+l}{2})! ( \frac{N+l+1}{2})! 2^{N+l+1}}{\pi (N+l+1)!}(N+\frac{3}{2}) \sum_{m m_s} C_{20l0}^{l0} C_{20lm}^{lm} \left(C_{lm\frac{1}{2} m_s}^{jk}\right)^2$

Note: I haven’t numerically check the formula. ( may be later )

For the off-diagonal element. The angular-spin part should be similar. The difficulty is the radial part, we have to evaluate the most general orthogonal relation of the Laguerre polynomial with weighting $x^{\frac{1}{2}(\alpha+\alpha'+2)} e^{-x}$.

We only knew the Laguerre polynomil is orthogonal with respect to $r^\alpha e^{-r}$, i.e.

$\displaystyle \int_0^{\infty} L_n^\alpha L_m^\alpha r^\alpha e^{-r} dr = \frac{(\alpha+n)!}{n!} \delta_{nm}$

But not this.

$\displaystyle \int_0^{\infty} L_n^\alpha L_m^\beta r^\frac{\alpha+\beta}{2} e^{-r} dr$

or this

$\displaystyle \int_0^{\infty} L_n^\alpha L_m^\beta r^{(\frac{\alpha+\beta}{2}+k)} e^{-r} dr$

But we are quite sure the last one, with $\alpha \neq \beta$ will not give zero, otherwise, The Nilsson orbital will be very simple and boring.

## Calculation of Knockout reaction

In the last post, we outline the calculation of knockout reaction. Here, we will give a detail calculation of the knockout reaction and compare with transfer reaction.

Suppose the reaction is A(a,12)B, where a = 1, and A = B + 2. Notice that the number 1, 2 are symbols and not represent any value.

The separation energy $S_p$ is

$\displaystyle S_p = m_B + m_2 - m_A$

In A’s rest frame, the 4-momenta of the 5 particles are

$\displaystyle P_A + P_a = P_1 + P_2 + P_B$

$\displaystyle \begin{pmatrix} m_A \\ 0 \end{pmatrix} + \begin{pmatrix} m_a + T \\ k_a \hat{z} \end{pmatrix} = \begin{pmatrix} \sqrt{m_1^2 + k_1^2} \\ \vec{k}_1 \end{pmatrix} + \begin{pmatrix} \sqrt{m_2^2 + k_2^2} \\ \vec{k}_2 \end{pmatrix} + \begin{pmatrix} \sqrt{m_B^2 + k^2} \\ -\vec{k} \end{pmatrix}$

Rearrange and set $P_k = P_A - P_B$

$\displaystyle (P_A - P_B) + P_a = P_k + P_a = P_1 + P_2$

$\displaystyle \begin{pmatrix} m_A - \sqrt{m_B^2 + k^2} \\ \vec{k} \end{pmatrix} + \begin{pmatrix} m_a + T \\ k_a \hat{z} \end{pmatrix} = \begin{pmatrix} \sqrt{m_1^2 + k_1^2} \\ \vec{k}_1 \end{pmatrix} + \begin{pmatrix} \sqrt{m_2^2 + k_2^2} \\ \vec{k}_2 \end{pmatrix}$

set

$\displaystyle P_k = \begin{pmatrix} m_A - \sqrt{m_B^2 + k^2} \\ \vec{k} \end{pmatrix} = \begin{pmatrix} \sqrt{m_k^2 + k^2} \\ \vec{k} \end{pmatrix}$

where $m_A - \sqrt{m_B^2 + k^2} = \sqrt{m_k^2 + k^2}$, simplify and get the effective orbital nucleon mass, that depends on the separation energy and momentum.

$\displaystyle m_k(S, k) = \sqrt{m_A^2 + m_B^2 - 2m_A\sqrt{m_B^2 + k^2}}$

Thus, we transformed the knockout reaction into quasi 2-body reaction.

$\displaystyle \begin{pmatrix} \sqrt{m_k^2 + k^2} \\ \vec{k} \end{pmatrix} + \begin{pmatrix} m_a + T \\ k_a \hat{z} \end{pmatrix} = \begin{pmatrix} \sqrt{m_1^2 + k_1^2} \\ \vec{k}_1 \end{pmatrix} + \begin{pmatrix} \sqrt{m_2^2 + k_2^2} \\ \vec{k}_2 \end{pmatrix}$

For 2-body reaction, we have to transform into the center of momentum frame. The 4-momentum of the center of momentum frame is

$\displaystyle P_c = P_k + P_a = \begin{pmatrix} \sqrt{m_k^2 + k^2} + m_a + T \\ \vec{k} + k_a \hat{z} \end{pmatrix}$

The Lorentz boost to the CM frame is

$\displaystyle \vec{\beta} = \frac{\vec{k} + k_a \hat{z}}{ \sqrt{m_k^2 + k^2} + m_a + T}$

and

$\displaystyle \gamma = \frac{1}{\sqrt{1-\beta^2}}$

A general Lorentz transformation in vector form is

$\displaystyle \begin{pmatrix} E' \\ \vec{k}' \end{pmatrix} = \begin{pmatrix} \gamma E \pm \gamma \vec{\beta}\cdot \vec{k} \\ \pm \gamma E \vec{\beta} + \vec{k} + (\gamma -1) (\hat{\beta}\cdot \vec{k}) \hat{\beta} \end{pmatrix}$

In the CM frame, the 4-momenta are

$\displaystyle P_k' = \begin{pmatrix} \gamma \sqrt{m_k^2 + k^2} - \gamma \vec{\beta} \cdot \vec{k} \\ - \gamma \sqrt{m_k^2+k^2} \vec{\beta} + \vec{k} + (\gamma -1 ) (\hat{\beta}\cdot \vec{k}) \hat{\beta} \end{pmatrix}$

$\displaystyle P_a' = \begin{pmatrix} \gamma (m_a+T) - \gamma k_a \vec{\beta} \cdot \hat{z} \\ - \gamma (m_a + T)\vec{\beta} + k_a\hat{z} + (\gamma -1 ) k_a (\hat{\beta}\cdot \hat{z}) \hat{\beta} \end{pmatrix}$

The total momentum should be zero,

$\displaystyle \vec{k}_c = - \gamma (m_a + T + \sqrt{m_k^2+k^2} )\vec{\beta} + \vec{k} + k_a\hat{z} + (\gamma -1 ) (\hat{\beta}\cdot ( \vec{k} + k_a \hat{z})) \hat{\beta}$

divide both side by $m_a + T + \sqrt{m_k^2+k^2} = E$,

$\displaystyle \frac{1}{E}\vec{k}_c = - \gamma \vec{\beta} + \vec{\beta} + (\gamma -1 ) (\hat{\beta}\cdot \vec{\beta} ) \hat{\beta} = 0$

The center of mass energy or total rest mass is

$\displaystyle E_c = M_c = \gamma (m_a + T + \sqrt{m_k^2 + k^2} ) - \gamma \vec{\beta} \cdot (\vec{k} + k_a \hat{z} ) \\ = \gamma E ( 1 - \beta^2) = \frac{E}{\gamma}$

Starting from here, thing will be the same as 2-body transfer reaction. The momentum after scattering is

$\displaystyle p = \frac{1}{2E_c} \sqrt{(E_c^2 - (m_1+m_2)^2)(E_c^2-(m_1-m_2)^2)}$

The scattered 4-momenta are

$\displaystyle P_1' = (\sqrt{m_1^2+p^2}, \vec{p})$
$\displaystyle P_2' = (\sqrt{m_2^2+p^2}, -\vec{p})$

In here, the direction of the vector $\vec{p}$ could be tricky, as the momenta are not parallel to z-axis.

Transform it back to A’s rest frame.

$\displaystyle P_1 = \begin{pmatrix} \gamma \sqrt{m_1^2 +p^2} + \gamma \vec{\beta}\cdot \vec{p} \\ \gamma \sqrt{m_1^2 +p^2} \vec{\beta} + \vec{p} + (\gamma -1) (\hat{\beta}\cdot \vec{p}) \hat{\beta} \end{pmatrix}$

$\displaystyle P_2 = \begin{pmatrix} \gamma \sqrt{m_2^2 +p^2} - \gamma \vec{\beta}\cdot \vec{p} \\ \gamma \sqrt{m_2^2 +p^2} \vec{\beta} - \vec{p} - (\gamma -1) (\hat{\beta}\cdot \vec{p}) \hat{\beta} \end{pmatrix}$

Or, we can use another Lorentz boost to the a’s rest frame by

$\displaystyle \vec{\beta}_{ca} = \frac{ - \gamma (m_a + T)\vec{\beta} + k_a\hat{z} + (\gamma -1 ) k_a (\hat{\beta}\cdot \hat{z}) \hat{\beta}}{\gamma (m_a+T) - \gamma k_a \vec{\beta} \cdot \hat{z}}$

but the form for $P_1$ and $P_2$ will keep in the same.

The difference from transfer reaction is that, the CM frame energy $E_c$ is a function of $S_p, \vec{k}$. So, it is not a reaction constant anymore. In summary,

$\displaystyle P_1 = \begin{pmatrix} \gamma \sqrt{m_1^2 +p^2} + \gamma \vec{\beta}\cdot \vec{p} \\ \gamma \sqrt{m_1^2 +p^2} \vec{\beta} + \vec{p} + (\gamma -1) (\hat{\beta}\cdot \vec{p}) \hat{\beta} \end{pmatrix}$

$\displaystyle P_2 = \begin{pmatrix} \gamma \sqrt{m_2^2 +p^2} - \gamma \vec{\beta}\cdot \vec{p} \\ \gamma \sqrt{m_2^2 +p^2} \vec{\beta} - \vec{p} - (\gamma -1) (\hat{\beta}\cdot \vec{p}) \hat{\beta} \end{pmatrix}$

Unwrapping,

$\displaystyle p = \frac{1}{2E_c} \sqrt{(E_c^2 - (m_1+m_2)^2)(E_c^2-(m_1-m_2)^2)}$

$\displaystyle E_c = M_c = \frac{E}{\gamma}$

$\displaystyle E = m_a + T + \sqrt{m_k^2+k^2} = m_A + m_a + T - \sqrt{m_B^2 + k^2}$

$\displaystyle S_p = m_B + m_2 - m_A$

The Lorentz boost can be either to A’s rest frame (normal kinematics)

$\displaystyle \vec{\beta} = \frac{\vec{k} + k_a \hat{z}}{ \sqrt{m_k^2 + k^2} + m_a + T} = \frac{\vec{k} + k_a \hat{z}}{E}$

or to a’s rest frame (inverse kinematics)

$\displaystyle \vec{\beta}_{ca} = \frac{ - \gamma (m_a + T)\vec{\beta} + k_a\hat{z} + (\gamma -1 ) k_a (\hat{\beta}\cdot \hat{z}) \hat{\beta}}{\gamma (m_a+T) - \gamma k_a \vec{\beta} \cdot \hat{z}}$

$\displaystyle \gamma = \frac{1}{\sqrt{1-\beta^2}}$

## Trapezoid filter

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.

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.

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

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.

## Ordering of the Nilsson orbital

In this post and this post, the level ordering can be shifted by adjusting the parameter $\mu$. And because of the ordering, the component of the harmonic oscillators strongly depends on it. For example, the 5/2[512] state and 1/2[521],

In above gif, the parameter $\mu = 0.35 to 0.7$, the step is not even. At small $\mu < 0.50$, the 5/2[512] and 1/2[521] are not crossing, it becomes crossed when $\mu > 0.5$. We can also see the decomposition to the spherical harmonic oscillator also change by a lot.

## Coulomb Energy for two protons system

From this post and this post,  we already have the framework to calculate the energy. The 1-body intersection is zero, as we do not interested on the kinematics energy. And also, suppose we already have the solution for the wave function for a fixed mean-field. Although it is unrealistic, it is a good starting point.

In fact, we have to solve the Schrödinger equation (using Hartree-Fock or numerical solution) with a realistic NN-interaction with and without Coulomb potential, and subtract the difference to get the Coulomb energy.

We have to evaluate the integral

$\displaystyle W = \left<12||12\right> - \left<12||21\right>$

When the 2 protons are in same orbital, the total wavefunction is

$\displaystyle \Psi(1,2) = \frac{1}{\sqrt{2}}\left(\uparrow \downarrow - \downarrow \uparrow \right) \phi(r_1) \phi(r_2)$

The exchange term is zero as the Coulomb operator does not act on the spin part.

$\displaystyle \left<12||12\right> = \sum_{L}^{\infty} \sum_{M=-L}^{L} \frac{4\pi}{2L+1}\int_{0}^{\infty} \int_{0}^{\infty} \phi^2(x) \phi^2(y) \frac{r_<^L}{r_>^{L+1}} x^2 y^2 dx dy \\ \int Y_{lm}^*(1) Y_{LM}^*(1) Y_{lm}(1) d\Omega_1 \int Y_{lm}^*(2) Y_{LM}(2) Y_{lm}(2) d\Omega_2$

For the angular part, from this post or this post, we have

$\displaystyle \int Y_{l_1m_1}^* Y_{LM}^* Y_{l_2m_2} d\Omega = \sqrt{\frac{(2L+1)(2l_1+1)}{4\pi(2l_2+1)}} C_{L0l_10}^{l_20} C_{LMl_1m_1}^{l_2m_2}$

For $l_1 = l_2$, the Clebsh-Gordon coefficient dictated that $2l \geq L$, $L = even$, $M = 0$.

We use the potential is 3D spherical infinite well with radius $a = 2.2$ fm. the wave function is spherical Bessel function ,

$\phi_n(r ) = \begin{matrix} A j_n(k r) & r < a \\ 0 & r > a \end{matrix}, j_n(ka) = 0$,

where $A$ is the normalization factor.

The first non-zero root of spherical Bessel Function take the form,

$r_0(n) = 2.96519 + 0.701921\sqrt{n} + 0.993301 n, n < 500$

The energy for the $n$-th s-state is

$\displaystyle E_n = \frac{\hbar k^2}{2 m_p^2}, k a = r_0(n)$

In Mathematica, the radial calculation can be done as

a=2.2;
R[r_]:=Piecewise[{{SphericalBesselJ[n, r0[n] r], 0 < r < a}, {0, r > 0}}];
A=(NIntegrate[(R[r])^2 r^2, {r, 0, ∞}]])^(-½);
w = NIntegrate[(R[x]R[y])^2 y^L/x^(L+1) x^2 y^2 , {x, 0, ∞}, {y,0, x}] + NIntegrate[(R[x]R[y])^2 x^L/y^(L+1) x^2 y^2 , {x, 0, ∞}, {y,x, ∞}]
W=1.44 A^4 w

The result for first n s-orbital is

I also extract the radius for uniformly charged sphere. The trend is reasonable as higher principle number, the wave function is close to the boundary $a = 2.2 \textrm{fm}$.

Next time, we will work on Woods-Saxon potential.