Started from independent particle model, the Hamiltonian of a nucleus with mass number A is

H_A = \sum\limits_{i}^{A} h_i + \sum\limits_{i, j>i} V_{ij}

we can rewrite the Hamiltonian by isolating a nucleon

H_A = H_B + h_1 + V_{B1}

than, we can use the basis of H_B and h_1 to construct the wavefunction of the nucleus A as

\left|\Phi_A\right>_{J} = \sum\limits_{i,j} \beta_{ij} \left[ \left|\phi_i\right>\left|\Phi_{B}\right>_{j} \right]_{J}

where the square bracket is anti-symmetric angular coupling between single particle wavefunction \left|\phi_i\right> and wavefunction \left|\Phi_B\right>_{j}. The \beta_{ij} is the spectroscopic amplitude.

The square of the spectroscopic amplitude times number of particle n_i at state \left|\phi_i\right> is the spectroscopic factor of the nucleon at state \left|\phi_i\right> and nucleus B at state \left| \Phi_B\right>_j

S_{ij} = n_i \beta_{ij}^2

The occupation number is the sum of the spectroscopic factors of the nucleus B

\sum\limits_{j} S_{ij} = n_i

After the definition, we can see, when the nucleon-core interaction V_{B1} is neglected, the spectroscopic factor is 1 and the occupation number is also 1.

Since the \beta is coefficient for changing basis from \left|\Phi_A\right>_J into basis of  \left[ \left|\phi_i\right>\left|\Phi_{B}\right>_{j} \right]_{J} Thus, the matrix \beta is unitary

\beta\cdot \beta^\dagger = 1

thus, each column of row vector of \beta is normalized.

The properties of \beta can be found by solving the eigen system of the H_A from the core H_C = H_B + h_1. The core hamiltonian is diagonal. The nucleon-core interaction introduce diagonal terms and off-diagonal terms. When only diagonal terms  or monopole term exist, only the eigen energy changes but the eignestate unchanges. Therefore, the configuration mixing is due to the off-diagonal terms.

However, when there are off-diagonal terms, the change of diagonal terms will changes the mixing.

In degenerate 2 states system,  the Hamiltonian be

H= \begin{pmatrix} E_1 & V \\ V & E_2 \end{pmatrix}

The eigen energy are \bar{E} \pm \sqrt{ dE^2 + V^2} , where \bar{E} = (E_1+E_2)/2 and dE = (E_1 - E_2)/2 , The eigen vector are

\frac{1}{\sqrt{(dE \pm \sqrt{dE^2+V^2})^2 +V^2}} ( dE \pm \sqrt{dE^2 +V^2}, V)

When the states are degenerated, E_1 = E_2 = E, the eigen energy is E \pm V, and the eigen state is \frac{1}{\sqrt{2}} (1,1)

As we can see, the eigen state only depends on the difference of the energy level, thus, we can always subtract the core energy and only focus on a single shell. For example, when we consider 18O, we can subtract the 16O binding energy.


In above figure, we fixed the E_1 and E_2 . We can see the spectroscopic factor decreases for the lower energy state (red line), and the state mixing increase.


We now fixed the V and E2. When E_1 < E_2, the particle stays more on the E_1 state, which is lower energy. The below plot is the eigen energy. Even though the mixing is mixed more on the excited state, but the eigen energy did not cross.