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). Since the spectroscopic factor is always in reference for a basis, i.e., the quasi-particle wave function has to be projected into a basis, 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.