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.