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.