## Correlation energy

The correlation is from the off-diagonal terms of the residual interaction, which is the TBME of the shell model. Consider following Hamiltonian of these nuclei $H(Z,N-1) = H(Z,N-2) + h_n + R_n$ $H(Z,N) = H(Z,N-2) + h_{n_1} + h_{n_2} + R_{n_1} + R_{n_2} + R_{nn}$

the correlation energy between the two neutrons is the term $R_{nn}$. Be aware that this is a residual interaction, not the nucleon-nucleon interaction $V_{ij}$. $V_{ij}= \begin{pmatrix} V_{11} & V_{12} & V_{1C} \\ V_{21} & V_{22} & V_{2C} \\ V_{C1} & V_{C2} & V_{CC} \end{pmatrix}$

Thus, $R_{n} = \begin{pmatrix} V_{11} - U_1 & V_{1C} \\ V_{C1} & V_{CC}-U_C \end{pmatrix}$,

where $U_i$ is mean field.

The separation energies is proportional to the terms $S_n(Z,N-1) \sim h_n + V_n$ $S_{2n}(Z,N)\sim h_{n_1} + h_{n_2} + V_{n_1} + V_{n_2} + V_{nn}$

Thus, the neutron-neutron correlation energy is $\Delta_{pn}(N,Z) = 2*S_n(Z,N-1) - S_{2n}(N,Z)$

For 18O, $S_n(^{17}O) = 4.1431$ MeV, and $S_{2n}(^{18}O) = 12.1885$ MeV, thus, $\Delta_{2n}(^{18}O) = 3.9023$ MeV.

In the shell model calculation, the single particle energy of the 1d5/2 and 2s1/2 neutron are -4.143 MeV and -3.27 MeV respectively. The residual interaction is $V = \begin{pmatrix} -1.79 & -0.83 \\ -0.83 & -2.53\end{pmatrix}$

The eigenenergy of the ground state of 18O is -10.539 on top of 16O.

The non-correlated binding energy of 18O is  2*-4.143 = -8.286 MeV.

Therefore, the theoretical correlated energy is 2.258 MeV.

## Nuclear correlation & Spectroscopic factor

In the fundamental, correlation between two objects is $P(x,y) \neq P(x) P(y)$

where $P$ is some kind of function. To apply this concept on nuclear physics, lets take a sample from 18O. 18O can be treated as 16O + n + n. In the independent particle model (IPM), the wave function can be expressed as $\left|^{18}O\right>= \left|^{16}O\right>\left|\phi_a\right>\left|\phi_b\right>= \left|^{16}O\right>\left|2n\right>$

where the wave function of the two neutrons is expressed as a direct product of two IPM eigen wave functions, that they are un-correlated. Note that the anti-symmetry should be taken in to account, but neglected for simplicity.

We knew that IPM is not complete, the residual interaction has to be accounted. According to B.A. Brown, Lecture Notes in Nuclear Structure Physics , Chapter 22, the 1s1/2 state have to be considered. Since the ground state spins of 18O and 16O are 0, thus, the wavefunction of the two neutrons has to be spin 0, so that only both are in 1d5/2 or 2s1/2 orbit. Thus, the two neutrons wave function is $\left| 2n \right> = \alpha \left|\phi_1\right>\left|\phi_1\right>+\beta \left|\phi_2\right>\left|\phi_2\right>$

when either $\alpha$ or $\beta$ not equal 0, thus, the two neutrons are correlation. In fact, the $\alpha = 0.87$ and $\beta = 0.49$.

The spectroscopic factor of the sd-shell neutron is the coefficient of $\alpha$ times a isospin-coupling factor.

From the above example, the correlation is caused by the off-diagonal part of the residual interaction. To be more specific, lets take 18O as an example. The total Hamiltonian is $H_{18} = H_{16} + h_1 + h_2 + V$

where $h_1 = h_2 = h$ is the mean field or single particle Hamiltonian $h\left|\phi_i\right>= \epsilon_i\left|\phi_i\right>$

since $H_{16}$ is diagonal and not excited (if it excited, then it is called core polarization in shell model calculation, because the model space did not included 16O.), i.e. $H_{16} = \epsilon_{16} I$, we can neglect it in the diagonalization of the $h_1+h_2 + V$ and add back at the end. In the 1d5/2 and 2s1/2 model space, in order to form spin 0, there is only 2 basis, $\left|\psi_1\right> = \left|\phi_1\right>\left|\phi_1\right>$ and $\left|\psi_2\right> = \left|\phi_2\right>\left|\phi_2\right>$

express the Hamiltonian in these basis, $V = \begin{pmatrix} -1.79 & -0.83 \\ -0.83 & -2.53\end{pmatrix}$

Because of the diagonalization, the two states $\left|\psi_1\right>$ and $\left|\psi_2\right>$ are mixed, than the two neutrons are correlated. This is called configuration mixing.

According to B.A. Brown,

the configuration mixing on the above is long-ranged correlation (LRC). It is near the Fermi surface and the energy is up to 10 MeV.

The short-ranged correlation (SRC) is caused by the nuclear hard core that scattered a nucleon to highly single particle orbit up to 100 MeV.

The LRC is included in the two-body-matrix element. The SRC is included implicitly through re-normalization of the model space.

There is a correlation due to tensor force. Since the tensor force is also short-ranged, sometimes it is not clear what SRC is referring from the context. And the tensor force is responsible for the isoscalar pairing.

The discrepancy of the experimental spectroscopic factor and the shall model calculation is mainly caused by the SRC.

## Weighted mean and error

We have n values of $x_i$ and error $\sigma_i$,

With a weighting $w_i$, the uncorrelated weighted mean and error is $X= \sum x_i w_i / \sum w_i$ $S^2 = \sum w_i^2 \sigma_i^2 / (\sum w_i)^2$

when combining data, the weighting is $w_i = 1/\sigma_i^2$

and the weighted error becomes $S^2 = \sum{\frac{1}{\sigma_i^2}} / (\sum{\frac{1}{\sigma_i^2}})^2 = 1 / \sum \frac{1}{\sigma_i^2}$

Example.

we measured a quantity n times, we can assume the intrinsic error of the data is fixed. Thus, $w_i = 1/n$ $X = \sum x_i / n$ $S^2 = \sum \sigma_0^2/n^2 = \sigma_0^2 /n$

Therefore, when we take more and more data, the error is proportional to $1/\sqrt{n}$.

In normal distribution, the sample of size n, the estimator of the sample mean and sample variance are $X =\sum x_i/n$ $S^2 = \sum (x_i-X)^2 / (n-1)$

Don’t mix up the sample variance and intrinsic error, although they are very similar.

To explain the formula of the weighted variance, we have to go to the foundation of the algebra of distribution.

For a random variable follow a distribution with mean $\mu$ and variance $\sigma^2$, $X \sim D(\mu, \sigma^2)$

Another random variable built on it, $Z=aX+b \sim D(a \mu + b, a^2\sigma^2)$ $Z=aX + bY \sim D(a \mu_X + b \mu_Y, a^2\sigma_X^2 + b^2 \sigma_Y^2)$
But there is a catch, when the $\mu_X = \mu_Y$ and $\sigma_X = \sigma_Y$, The rule does not apply. But lets look back, if the mean and variance are the same, the two distribution does not really independent.