shell model calculation | Nuclear Physics 101

Shell model calculation on 16C

September 1, 2023

GoLuckyRyan Basic 16C, shell model calculation, TBME Leave a comment

Following the calculation of 18O in this post, it should be easy to have the calculation done for 16C, assuming 14C is an inert core.

From 15C, we know that the s-orbital is lower than the d-orbital,

$\displaystyle \epsilon_s = BE(15C) - BE(14C) = -1.2181~$ MeV

$\displaystyle \epsilon_d =\epsilon_s + 0.74 = -0.4781~$ MeV

The ground state of 16C with respect to 14C is

$\displaystyle E_0 = BE(16C) - BE(14C) = -5.4684~$ MeV.

with these numbers, we have $V_0\bar{R} = 1.00537~$ MeV, which is similar to the value for 18O, as it should be. The matrixes for 18O and 16C should be the same with the numerical values $\epsilon_d, \epsilon_s, V_0 \bar{R}$ different.

The states are

$\displaystyle E(J=0) = -5.468 = 0.000, 0.759 (d_{5/2})^2 + 0.652 (s_{1/2})^2$

$\displaystyle E(J=2) = -3.382 = 2.087, 0.465 (d_{5/2})^2 + 0.885 (d_{5/2})(s_{1/2})$

$\displaystyle E(J=0) = -1.945 = 3.523, -0.652 (d_{5/2})^2 + 0.759(s_{1/2})^2$

$\displaystyle E(J=3) = -1.696 = 3.772, (d_{5/2})(s_{1/2})$

$\displaystyle E(J=4) = -1.243 = 4.225, (d_{5/2})^2$

$\displaystyle E(J=2) = -1.167 = 4.302, -0.885 (d_{5/2})^2 + 0.465 (d_{5/2})(s_{1/2})$

Here are the energy levels

The calculation is far off from the experimental data.

There is a 15C(d,p) experiment A.H. Wuosmaa et al., PRL 105, 132501 (2010), extracted the SF for the low-lying states:

State of 16C	J-pi	SF
0.000	0+	0.6 +- 0.1
1.766	2+	0.5 +- 0.1
3.027	0+	1.4 +- 0.3
3.986 *	2+	< 0.34
4.088 *	3+	0.82 – 1.06

* cannot be resolved in the experiment.

The experimental data tell us the configuration of 16C, if 14C is an inert core, should be, (using the sum rule to normalize the SF)

$\displaystyle Ex(J=0) = 0.000, 0.83 (d_{5/2})^2 + 0.55 (s_{1/2})^2$

$\displaystyle Ex(J=2) = 1.766, 0.77 (d_{5/2})^2 + 0.64 (d_{5/2})(s_{1/2})$

$\displaystyle Ex(J=0) = 3.027, -0.54 (d_{5/2})^2 + 0.83 (s_{1/2})^2$

$\displaystyle Ex(J=3) = 4.088, (d_{5/2})(s_{1/2})$

$\displaystyle Ex(J=2) =3.986, -0.64 (d_{5/2})^2 + 0.77 (d_{5/2})(s_{1/2})$

The SFs of the $J = 0^+$ states are not too far off. while that of $J = 2^+$ states are very different. The $J= 3^+$ state is lower than the experimental value as expected. And the $J = 4^+$ state is more or less the same energy. Notice that, the experiment cannot resolve the 3+ state from the 2nd 2+ state and also the 4+ state due to energy resolution, so the SF may be not reliable.

The residual interaction for the $J = 0+$ states was also deduced from the experimental data.

$\displaystyle \left<s s| V| s s\right> = 0.92 \pm 0.28 \approx ? V_0 \bar{R} = 1.0054$

$\displaystyle \left<dd| V| s s\right> = 1.39 \pm 0.12 \approx ? \sqrt{3} V_0 \bar{R} = 1.741$

$\displaystyle \left<dd| V| d d\right> = 3.60 \pm 0.28 \approx ? 3 V_0 \bar{R} = 3.016$

If we take the $\left<s s| V| s s\right> = V_0 \bar{R} = 0.92$ , the off-diagonal term should be 1.59, and the d-orbital TBME should be 2.76. Using the experimental value, the energy and SF can be reproduced, as they should be.

The interesting thing is, if the SF for the $J=0^+$ states are similar to that of the experiment, why the energies are different? In 18O, the $J = 0^+$ states are well reproduced. Could it be because the 14C is not a good core?

Allowed Coupling for identical particles in the same orbital

August 18, 2023

GoLuckyRyan Basic shell model calculation Leave a comment

Let’s suppose the orbital has an angular momentum of $j$ . What is the allowed coupling for $n$ identical particles?

for $n=2$ , the problem is easy, using a simple symmetry argument, we know that only even $J = 0, 2, 4, ..., 2j-1$ are allowed.

for the general case, we need two theorems.

Theorem 1

For a multi-particle wave function $\Psi_M$ satisfy

$\displaystyle J_z \Psi_M = M\Psi_M$

and

$\displaystyle J_+ \Psi_M = 0$

where $J_+ = J_{1+}+ J_{2+} +... J_{n+}$ is the rising operator. The wave function $\Psi_M$ is an eigen function of $J^2$ , so that

$\displaystyle J^2 \Psi_M = M(M+1)\Psi_M$

i.e, the angular momentum of the wave function is $J = M$ .

Proof.

Using

$\displaystyle J^2 = \frac{1}{2}\left(J_+ J_- + J_-J_+ \right) + J_z^2$

$\displaystyle J_+J_- - J_- J_+ = 2J_z$

we have

$\displaystyle J^2 =2J_-J_+ + J_z + J_z^2$

Thus

$\displaystyle J^2 \Psi_M =2J_-J_+\Psi_M + (J_z + J_z^2)\Psi_M = M(M+1) \Psi_M$

Theorem 2

Suppose a given configuration $\Phi_M$ for $n$ identical particles, and there are $p$ distinguish Slater determinants $\Psi_M(i), i = 1, 2, ..., p$ so that $J_z\Psi_M(i) =M\Psi_M(i)$ . If there are $p+q$ linearly independent Slater determinants with $J_z \Psi_{M-1}(j) = (M-1) \Psi_{M-1}(j)$ , then there are $q$ linearly independent states with total angular momentum $J = M-1$ .

Before the proof, let’s demonstrate the theorems. Let’s say, in $j = 5/2$ orbital, there are 10 m-states or configurations regards the z-component. For 3 identical particles, we can construct 10 distinguish Slater determinants:

M-value	Slater determinant
M=9/2	1/2, 3/2, 5/2
M=7/2	-1/2, 3/2, 5/2
M=5/2	-3/2, 3/2, 5/2
	-1/2, 1/2, 5/2
M=3/2	-5/2, 3/2, 5/2
	-3/2, 1/2, 5/2
	-1/2, 1/2, 3/2
M=1/2	-5/2, 1/2, 5/2
	-3/2, 1/2, 5/2
	-3/2, 1/2, 3/2

Take $M=9/2$ and $M = 7/2$ , the number of $M=7/2$ Slater determinant is equal to that of $M=9/2$ , thus, according to the theorem, no state can be constructed with the total angular momentum of $J = 7/2$ .

Proof.

Let’s write down the linear combination of all Slater determinants as

$\displaystyle \Phi_{M-1} = \sum_{1,2,..., p+q} \alpha_i \Psi_{M-1}(i)$

Apply the rising operator $J_+$ , and we have

$\displaystyle J_+\Phi_{M-1} = \sum_{1,2,..., p} f(\alpha_1, \alpha_2, ..., \alpha_{p+q}) \Psi_M(i)$

If we want to construct a state with $J= M-1$ , we need $J_+ \Phi_{M-1} = 0$ according to Theorem 1.

Thus, we have $p$ equations with $p +q$ unknown, so that we can have $q$ solutions for states with $J = M-1$ .

So, what is the possible state for $n$ identical particle in $j$ orbital?

The middle part of the Theorem 2 is already demonstrated. We need to construct all possible $M$ configurations, starting from the highest to the lowest.

Here is a python code:

from itertools import combinations
from collections import Counter

def generate_combinations(numbers, n):
    valid_combinations = []
    
    for combo in combinations(numbers, n):
        if sum(combo) >= 0:
            valid_combinations.append(combo)
    
    return valid_combinations

def AllowedJ(j, n):
  number_list = list(range(-int(2*j), int(2*j)+1,2))
  valid_combinations = generate_combinations(number_list, n)

  print("Possible Slater determinants:")
  for combo in valid_combinations: print("   ",combo)
      
  print("Number of valid combinations:", len(valid_combinations))
  sum_counts = Counter(sum(combo) for combo in valid_combinations)
  sorted_sum_counts = sorted(sum_counts.items(), key=lambda item: item[1], reverse=True)[::-1]

  print("Distinct J and frequencies:")
  for (sum_val, count) in sorted_sum_counts:
      print(f"    J {sum_val}/2: Count {count}")

  print("Allowed J :")
  for index, x in enumerate(sorted_sum_counts):
    if index == 0 :
      print(f"    J {x[0]}/2: Count {x[1]}")
    else:
      if x[1]-sorted_sum_counts[index-1][1] > 0 :
        print(f"    J {x[0]}/2: Count {x[1]-sorted_sum_counts[index-1][1]}")

Simply type

AllowedJ(5/2, 3)

Gives

Possible Slater determinants:
    (-5, 1, 5)
    (-5, 3, 5)
    (-3, -1, 5)
    (-3, 1, 3)
    (-3, 1, 5)
    (-3, 3, 5)
    (-1, 1, 3)
    (-1, 1, 5)
    (-1, 3, 5)
    (1, 3, 5)
Number of valid combinations: 10
Distinct J and frequencies:
    J 9/2: Count 1
    J 7/2: Count 1
    J 5/2: Count 2
    J 3/2: Count 3
    J 1/2: Count 3
Allowed J :
    J 9/2: Count 1
    J 5/2: Count 1
    J 3/2: Count 1

In Mathematica

j = 5/2;
n = 3;
combinations = 
  Select[Subsets[Table[i, {i, -j, j}], {n}], 
   OrderedQ[#] && Total[#] >= 0 &];
Length[combinations]
distinctSums = Union[Total /@ combinations];
sumCounts = SortBy[Tally[Total /@ combinations], -#[[1]] &];
Print["Distinct Sums : ", distinctSums];
Print["Number of Distinct Sums : ", Length[distinctSums]];
Print["Distinct Sums and Their Frequencies : ", sumCounts];
allowedJ = 
  Select[Table[
    If[i == 1, 
     sumCounts[[1]], {sumCounts[[i, 1]], 
      sumCounts[[i, 2]] - sumCounts[[i - 1, 2]]}], {i, 1, 
     Length[sumCounts]}], #[[2]] != 0 &];
Print["Allowed J are :", allowedJ]

Here is a list of allowed J for $j = 3/2, 5/2, 7/2, 9/2$ , All allowed $J$ for $n = even$ are $J = 0,2,4, ... 2j-1$

Generating the list of two-body pair

August 5, 2023

GoLuckyRyan Basic shell model calculation, TBME 1 Comment

Isospin formalism

Let index the orbitals as 1, 2, 3, 4 …. for example, in p-shell,

p1/2
p3/2

for in p-sd shell

p1/2
p3/2
d3/2
d5/2
s1/2

There are 3 step to generate the list.

First, we need to generate a list of all possible index pair. For example, in p-shell, the possible pair is (1,1), (1,2), and (2,2).

Next, for each index pair, generate the possible angular momentum and isospin, and give a new list. For example, for the pair (1,1), which is p1/2 and p1/2 coupling, this can generate J = 0, 1. For J = 0, T = 1, and for J = 1, T = 0 due to the wavefunction of a nucleon pair must be anti-symmetric. Thus, we have

(1,1) –> p1/2 x p1/2 –> (J, T) = (0, 1), (1, 0)
(1,2) –> p1/2 x p3/2 –> (J, T) = (1, 0), (1, 1), (2, 0), (2, 1)
(2,2) –> p3/2 x p3/2 –> (J, T) = (0, 1), (1, 0), (2, 1), (3, 0)

Thus, we have a new list of all possible pair states, denoted as (i, j, J, T)

(1, 1, 0, 1)
(1, 1, 1, 0)
(1, 2, 1, 0)
(1, 2, 1, 1)
(1, 2, 2, 0)
(1, 2, 2, 1)
(2, 2, 0, 1)
(2, 2, 1, 0)
(2, 2, 2, 1)
(2, 2, 3, 0)

Last, pick up any 2 of them as long as the (J, T) are equal. for example, (1, 1, 0, 1) can pair with itself as the (J, T) is the same, also it can be paired with (2, 2, 0, 1), and that is all. So, we have the list of all possible two-body matrix pair, denoted as [i, j, k, l, J, T]

[1, 1, 1, 1, 0, 1]
[1, 1, 1, 1, 1, 0]
[1, 1, 1, 2, 1, 0]
[1, 1, 2, 2, 0, 1]
[1, 1, 2, 2, 1, 0]
[1, 2, 1, 2, 1, 0]
[1, 2, 1, 2, 1, 1]
[1, 2, 1, 2, 2, 0]
[1, 2, 1, 2, 2, 1]
[1, 2, 2, 2, 1, 0]
[1, 2, 2, 2, 2, 1]
[2, 2, 2, 2, 0, 1]
[2, 2, 2, 2, 1, 0]
[2, 2, 2, 2, 2, 1]
[2, 2, 2, 2, 3, 0]

For the p-shell, a total of 15 TBMEs.

For the pn formulism, the method is basically the same.

Here is the Python code for generating the list of two-body pair

def Coupling(i, j, orbitalList): # index of the orbital
  jList = [int(orb[1]) for orb in orbitalList]
  minJ = abs(jList[i]-jList[j])//2
  maxJ = (jList[i] + jList[j])//2
  #print(minJ, maxJ)
  if i == j:
    return [ [J, 1 - (J % 2) ] for J in range(minJ, maxJ+1)]
  else :
    out = [ [J, 0 ] for J in range(minJ, maxJ+1)]
    out += [ [J, 1 ] for J in range(minJ, maxJ+1)]
    return sorted(out, key=lambda x: x[0])

def GenIndexPair(indexList):
  combinations = []
  for i in range(len(indexList)):
    for j in range(i, len(indexList)):
        combination = [indexList[i], indexList[j]]
        combinations.append(combination)
  return combinations # return all possible index pair

def TBcouping(indexPairList, orbitalList):
  TB = []
  for a in indexPairList:
    scheme = Coupling(a[0], a[1], orbitalList)
    for b in scheme:
      TB.append(a + b )
  return TB ## array of [i, j, J, T]

def GenTwoBodyMatrix(TB):
  TBME = []
  for i in range(len(TB)):
    for j in range(i, len(TB)):
      if TB[i][2] != TB[j][2] or TB[i][3] != TB[j][3] : continue 
      TBME.append([TB[i][0]+1, TB[i][1]+1, TB[j][0]+1, TB[j][1]+1, TB[i][2], TB[i][3]])
  return TBME

def GenTBM(orbitalList):
  haha = GenIndexPair([i for i in range(len(orbitalList))])
  TB = TBcouping(haha, orbitalList)
  TBME = sorted(GenTwoBodyMatrix(TB), key=lambda x : (x[-1], x[-2]))
  print("Number of element : %d " %  len(TBME))
  return  TBME


def GenTBM_PN(orbitalList):
  orbtialListPN = orbitalList*2
  nOrb = len(orbtialListPN)

  indexPairList = [x for x in GenIndexPair([i for i in range(len(orbtialListPN))]) if x[0] < nOrb//2 and x[1] < nOrb//2 ]
  TB = TBcouping(indexPairList, orbtialListPN)
  TzPP = sorted([x[:-1] for x in GenTwoBodyMatrix(TB) if x[-1] == 1], key = lambda x : (x[0], x[1], x[2], x[3], x[4]))

  indexPairList = [x for x in GenIndexPair([i for i in range(len(orbtialListPN))]) if x[0] >= nOrb//2 and x[1] >= nOrb//2 ]
  TB = TBcouping(indexPairList, orbtialListPN)
  TzNN = sorted([x[:-1] for x in GenTwoBodyMatrix(TB) if x[-1] == 1], key = lambda x : (x[0], x[1], x[2], x[3], x[4]))

  indexPairList = [x for x in GenIndexPair([i for i in range(len(orbtialListPN))]) if x[0] < nOrb//2 and x[1] >= nOrb//2 ]
  TB = TBcouping(indexPairList, orbtialListPN)
  TzPN = sorted([x[:-1] for x in GenTwoBodyMatrix(TB) if x[-1] == 0], key = lambda x : (x[0], x[1], x[2], x[3], x[4]))

  print("(T = 1 [%d])x2  + (T = 0 [%d]) = %d" % (len(TzPP), len(TzPN), len(TzPP)*2 + len(TzPN)) )

  return TzPP + TzNN + TzPN

Here is an example

orbitalList = ['p1/2', 'p3/2', 'd3/2', 'd5/2', 's1/2']

TBME = sorted(GenTBM(orbitalList), key=lambda x : (x[0], x[1], x[2], x[3], x[4]))
TBME_PN = GenTBM_PN(orbitalList)

Configuration space	isospin	PN
0s1/2	2	4
0s1/2, 0p3/2, 0p1/2	48	120
p-shell	15	34
p-sd shell	342	940
sd-shell	63	158
sd-fp shell	1408	4016
fp-shell	195	518

KSHELL Shell model calculation

July 31, 2023

GoLuckyRyan Basic shell model calculation Leave a comment

I used to use OxBash [B. A. Brown et al., MSU-NSCL Report No. 1289] on Windows for shell model calculation and spectroscopic overlap between nuclei. I searched for some shell model code for LINUX or Mac. It comes up with the KSHELL [https://sites.google.com/alumni.tsukuba.ac.jp/kshell-nuclear/] developed by Prof. Noritaka Shimizu 清水則孝.

The advantage of the code is that it uses all cores for a 1 CPU machine. Also, the kshell_ui.py is very user-friendly. For example, using USDB interaction on 12C will tell you 12C is out of the configuration space. In addition, the calculation of spectroscopic factors is more effortless, all included in the kshell_ui.py. In comparison, OxBash required you to manually input the wavefunction overlap. Moreover, KSHELL always treats proton and neutron shells separately.

In Ubuntu 22.04 on a 1 CPU with a multi-core machine, the only required packages are

sudo apt-get install libblas-dev liblapack-dev

and also add a symbolic link at /usr/bin to set python to python3

/usr/bin>sudo ln -s python3 python

Download the package and go to the src file, modify the Makefile. In my case, I comment out the default and uncomment the following:

# --- GNU Fortran compiler, single node 
FC = gfortran
FFLAGS = -O3 -fopenmp -fallow-argument-mismatch
LIBS = -llapack -lblas -lm

After make, the executable will be placed in the bin folder.

It is nice to add the bin folder to the PATH

>export PATH=$PATH:<kshell-folder>/bin/

An example can be found in the English manual.

The output of the KSHELL contains 2 files.

summary_*.txt
log_*.txt

The summary_*.txt give the energy levels and the log_*.txt provides detail of the calculation. At the end of log_*.txt, the particle numbers for each orbital for each energy level are shown.

Convert the occupancies to wave function

KSHELL will give the particle numbers for each orbital for each energy level, which is convenient. Suppose we calculated the 18O using KSHELL with USDB interaction.

The ground state particle numbers are

$\displaystyle n(d_{3/2}) = 0.098$
$\displaystyle n(d_{5/2}) = 1.552$
$\displaystyle n(s_{1/2}) = 0.350$

Since the wave function for the ground state, 0+ must be

$\displaystyle \Psi = C (\alpha (d_{5/2})^2 + \beta (s_{1/2})^2 + \gamma (d_{3/2})^2 )$

where $C$ is the 16O core, and $\alpha^2 + \beta^2 +\gamma^2 = 1$

The number of particles can be extracted by using the Number operator $N_x$ , where $x$ is the orbital, so that $n_x = \left< \Psi | N_x | \Psi \right>$ . The number of particles in each orbital is then

$\displaystyle n(d_{5/2}) = 2\alpha^2 \rightarrow \alpha = 0.88$
$\displaystyle n(s_{1/2}) = 2\beta^2 \rightarrow \alpha = 0.42$
$\displaystyle n(d_{3/2}) = 2\gamma^2 \rightarrow \alpha = 0.22$

so, that the total number of particles is 2. Comparing this post, which only uses 2 orbitals, the numbers more or less agree.

For the 2+ state, the wave function is

$\displaystyle \Psi = C (\alpha (d_{5/2})^2 + \beta (d_{5/2}s_{1/2}) + \gamma (d_{3/2}s_{1/2}) )$

Then

$\displaystyle n(d_{5/2}) = 2\alpha^2 + \beta^2$
$\displaystyle n(s_{1/2}) = \beta^2 + \gamma^2$
$\displaystyle n(d_{3/2}) = \gamma^2$

Installation for Mac OSX Ventura 13.3

Although the gfortran is available in Mac OSX, but the crt0.o is removed. It is easier to download the Intel Fortran compiler.

Then we have the ifort to compile the code. We also need the Intel MATH Kernal Library.

After installed both, go to /opt/intel/oneapi/

/opt/intel/oneapi>source setvars.sh

My Makefile for MacOSX is

FC = ifort
FFLAGS = -mdynamic-no-pic -O3 -no-prec-div -fp-model fast=2 -xHost -qopenmp -no-ipo -Wl,-stack_size,40000000
LIBS = -qmkl

The code should be compiled without any problem. In Ventura 13.3, there is no python but python3. You can create a symbolic link in /usr/local/bin/. Also for some reason, when running the KSHELL, although the mkl library is installed, it searches for the wrong path for the lib, the bullforce way to change that is add all mkl dynamic library to the search path, my my case

/opt/intel/oneapi/compiler/latest/mac/compiler/lib>sudo ln -s /opt/intel/oneapi/mkl/latest/lib/*.dylib .

Then the KSHELL will run without issue.

Shell model calculation on 18O

May 15, 2020

GoLuckyRyan Basic 18O, 19F, 20Ne, Clebsch-Gordon, Shell Model, shell model calculation, TBMEs Leave a comment

This post is copy from the book Theory Of The Nuclear Shell Model by R. D. Lawson, chapter 1.2.1

The model space is only the 0d5/2 and 1s1/2, and the number of valence nucleon is 2. The angular coupling of the 2 neutrons in these 2 orbitals are

$\displaystyle (0d5/2)^2 = 0, 2, 4$
$\displaystyle (0d5/2)(1s1/2) = 2,3$
$\displaystyle (1s1/2)^2 = 0$

Note that for identical particle, the allowed J coupled in same orbital must be even due to anti-symmetry of Fermion system.

The spin 3, and 4 can only be formed by (0d5/2)(1s1/2) and (0d5/2)^2 respectively.

Since the Hamiltonian commute with total spin, i.e., the matrix is block diagonal in J that the cross J matrix element is zero,

$\displaystyle H = h_1 + h_2 + V$

$\displaystyle \left< J | V|J' \right> = V_{JJ'} \delta_{JJ'}$

or to say, there is no mixture between difference spin. The Hamiltonian in matrix form is like,

$\displaystyle H = \begin{pmatrix} M_{J=0} (2 \times 2) & 0 & 0 & 0 \\ 0 & M_{J=2} (2 \times 2) & 0 & 0 \\ 0 & 0 & M_{J=3} (1 \times 1) & 0 \\ 0 & 0 & 0 & M_{J=4} (1 \times 1) \end{pmatrix}$

The metrix element of J=3 and J=4 is a 1 × 1 matrix or a scalar.

$\displaystyle M_{J=3,4} = \left<J| ( h_1 + h_2 + V) | J \right> = \epsilon_1 + \epsilon_2 + \left< j_1 j_2 | V | j_3 J_4 \right>$

where $\epsilon_i$ is the single particle energy.

Suppose the residual interaction is an attractive delta interaction

$\displaystyle V = - 4\pi V_0 \delta(r_i - r_j )$

Be fore we evaluate the general matrix element,

$\displaystyle \left< j_1 j_2 | V | j_3 j_4 \right>$

We have to for the wave function $\left|j_1 j_2 \right>$ ,

$\displaystyle \left| j_1 j_2 \right> = \\ \frac{1}{\sqrt{2(1+\delta_{j_1 j_2})}} \\ \sum_{m_1 m_2} C_{j_1 m_1 j_2 m_2}^{JM} \left( \phi_{j_1m_1}(1) \phi_{j_2m_2}(2) + (-1)^T \phi_{j_1m_1}(2) \phi_{j_2m_2}(1) \right)$

where $T$ is the isospin, and the single particle wave function is

$\displaystyle \phi_{jm} = R_l(r) \sum_{\kappa \mu} C_{l \kappa s \mu}^{jm} Y_{l \kappa} ( \hat{r} ) \chi_{\mu}$

Since the residual interaction is a delta function, the integral is evaluated at $r_1 = r_2 = r$ , thus the radial function and spherical harmonic can be pulled out in the 2-particle wave function $\left|j_1 j_2 \right>$ at $r_1 = r_2 = r$ is

$\displaystyle \left| j_1 j_2 \right> = \\ \frac{1}{\sqrt{2(1+\delta_{j_1 j_2})}} R_{l_1}(r) R_{l_2}(r) \\ \sum_{m_1 m_2 \kappa_1 \mu_1 \kappa_2 \mu_2} C_{j_1 m_1 j_2 m_2}^{JM} C_{l_1 \kappa_1 s \mu_1}^{j_1 m_1} C_{l_2 \kappa_2 s \mu_2}^{j_2 m_2} Y_{l_1 \kappa_1} ( \hat{r} ) Y_{l_2 \kappa_2} ( \hat{r} ) \\ \left( \chi_{\mu_1}(1) \chi_{\mu_2}(2) + (-1)^T \chi_{\mu_1}(2) \chi_{\mu_2}(1) \right)$

Using the product of spherical Harmonic,

$\displaystyle Y_{l_1 \kappa_1}(\hat{r})Y_{l_2 \kappa_2}(\hat{r}) = \sum_{LM} \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi (2L+1)}} C_{l_1 0 l_2 0 }^{L0} C_{l_1 \kappa_1 l_2 \kappa_2}^{LM} Y_{LM}(\hat{r})$

using the property of Clebsch-Gordon coefficient for spin half system

$\displaystyle \chi_{SM_S} = \sum_{\mu_1 \mu_2} \chi_{\mu_1}(1) \chi_{\mu_2}(2)$

where

$\displaystyle \chi_{0,0} = \frac{1}{\sqrt{2}} \left( \chi_{1/2}(1) \chi_{-1/2}(2) - \chi_{-1/2}(1) \chi_{1/2}(2) \right)$

which is equal to $T = 1$

For $T = 0$

$\displaystyle \chi_{1,1} = \chi_{1/2}(1) \chi_{1/2}(2)$

With some complicated calculation, the J-J coupling scheme go to L-S coupling scheme that

$\displaystyle \left| j_1 j_2 \right> =\sum_{L S M_L M_S} \alpha_{LS}(j_1 j_2 JT ) C_{LM_L S M_S}^{LS} Y_{LM_L}(\hat{r}) \chi_{S M_S} R_{l_1}(r) R_{l_2}(r)$

with

$\displaystyle \alpha_{LS}(j_1j_2 JT) = \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi (2L+1)}} \\ \frac{1-(-1)^{S+T}}{\sqrt{2(1+\delta_{j_1 j_2} \delta_{l_1 l_2})}} C_{l_1 0 l_2 0}^{L 0} \gamma_{LS}^{J}(j_1 l_1;j_2 l_2)$

$\displaystyle \gamma_{LS}^{J}(j_1 l_1;j_2 l_2) = \sqrt{(2j_1+1)(2j_2+1)(2L+1)(2S+1)} \\ \begin{Bmatrix} l_1 & s & j_1 \\ l_2 & s & j_2 \\ L & S & J \end{Bmatrix}$

Return to the matrix element

$\displaystyle \left< j_1 j_2 J M | V | j_3 j_4 J M \right>$

Since the matrix element should not depends on $M$ , thus, we sum on M and divide by $(2J+1)$ ,

$\displaystyle \left< j_1 j_2 J M | V | j_3 j_4 J M \right> = \frac{1}{2J+1} \sum_M \left< j_1 j_2 J M | V | j_3 j_4 J M \right>$

$\displaystyle \left< j_1 j_2 J M | V | j_3 j_4 J M \right> = \frac{-4\pi V_0 \bar{R}}{2J+1} \sum_{LS} \alpha_{LS}(j_1j_2JT) \alpha_{LS}(j_3j_4JT) \\ \sum_{M M_L M_S} (C_{LM_SSM_S}^{JM})^2 \int Y_{LM}^* Y_{LM} d\hat{r}$

with

$\displaystyle \bar{R} = \int R_{j_1} R_{j_2} R_{j_3} R_{j_4} dr$

( i give up, just copy the result ), for $T = 1$ ,

$\displaystyle \left< j_1 j_2 J M | V | j_3 j_4 J M \right> = (-1)^{j_1+j_3+l_2+l_4 + n_1+n_2+n_3+n_4}\\ (1+(-1)^{l_1+l_2+l_3+l_4}) (1 + (-1)^{l_3+l_4+J}) \\ \frac{V_0 \bar{R}}{4)2J+1)} \sqrt{\frac{(2j_1+1)(2j_2+1)(2j_3+1)(2j_4+1)}{(1+\delta_{j_3j_4})(1+\delta_{j_1j_2})}} \\ C_{j_1(1/2)j_2(-1/2)}^{J0} C_{j_3(1/2)j_4(-1/2)}^{J0}$

The block matrix are

$\displaystyle M_{J=0} = \begin{pmatrix} 2 \epsilon_d - 3 V_0 \bar{R} & -\sqrt{3} V_0 \bar{R} \\ -\sqrt{3} V_0 \bar{R} & 2 \epsilon_s - V_0 \bar{R} \end{pmatrix}$

$\displaystyle M_{J=2} = \begin{pmatrix} 2 \epsilon_d - \frac{24}{35} V_0 \bar{R} & -\frac{12\sqrt{7}}{35} V_0 \bar{R} \\ -\frac{12\sqrt{7}}{35} V_0 \bar{R} & \epsilon_d + \epsilon_s - \frac{6}{5}V_0 \bar{R} \end{pmatrix}$

$\displaystyle M_{J=3} = \epsilon_d + \epsilon_s$

$\displaystyle M_{J=4} = 2 \epsilon_d - \frac{2}{7} V_0 \bar{R}$

To solve the eigen systems, it is better to find the $\epsilon_d, \epsilon_s, V_0 \bar{R}$ from experimental data. The single particle energy of the d and s-orbtial can be found from 17O, We set the reference energy to the binding energy of 16O,

$\epsilon_d = BE(17O) - BE(16O) = -4.143 \textrm{MeV}$

$\epsilon_s = -4.143 + 0.871 = -3.272 \textrm{MeV}$

the ground state of 18O is

$E_0 = BE(18O) - BE(16O) = -12.189 \textrm{MeV}$

Solving the $M_{J=0}$ , the eigen value are

$\displaystyle \epsilon_d + \epsilon_s - 2 (V_0 \bar{R}) \pm \sqrt{ (\epsilon_s - \epsilon_d + V_0 \bar{R})^2 + 3 (V_0 \bar{R})^2 }$

Thus,

$V_0 \bar{R} = 1.057 \textrm{MeV}$

The solution for all status are

$E(j=0) = -12.189 ( 0 ) , 0.929 (0d_{5/2})^2 + 0.371 (1s_{1/2})^2$

$E(j=2) = -9.820 ( 2.368 ) , 0.764 (0d_{5/2})^2 + 0.645 (0d_{5/2})(1s_{1/2})$

$E(j=4) = -8.588 ( 3.600) , (0d_{5/2})^2$

$E(j=2) = -7.874 ( 4.313) , 0.645 (0d_{5/2})^2 -0.764 (0d_{5/2})(1s_{1/2})$

$E(j=3) = -7.415 (4.773) , (0d_{5/2})(1s_{1/2})$

$E(j=0) = -6.870 (5.317 ) , 0.371 (0d_{5/2})^2 - 0.929 (1s_{1/2})^2$

Annotation 2020-05-14 235102

The 2nd 0+ state is missing in above calculation. This is due to core-excitation that 2 p-shell proton promotes to d-shell.

In the sd- shell, there are 2 protons and 2 neutrons coupled to the lowest state. which is the same s-d shell configuration as 20Ne. The energy is

$E_{sd} = B(20Ne) - B(16O) = -33.027 \textrm{MeV}$

In the p-shell, the configuration is same as 14C, the energy is

$E_{p} = B(14C) - B(16O) = 22.335 \textrm{MeV}$

Thus, the energy for the 2-particle 2-hole of 18O is

$E(0^+_2) = -10.692 + 8 E' \textrm{MeV}$ ,

where $E'$ is the p-sd interaction, there are 4 particle in sd-shell and 2 hole in p-shell, thus, total of 8 particle-hole interaction.

The particle-hole can be estimate using 19F 1/2- state, This state is known to be a promotion of a p-shell proton into sd-shell.

In the sd-shell of 19F, the configuration is same as 20Ne. In the p-shell of 19F , the configuration is same as 15N, the energy is

$E_{p} = B(15N) - B(16O) = 12.128 \textrm{MeV}$

Thus, the energy for the 1/2- state of 19F relative to 16O is

$E_{1/2} = -20.899 + 4 E' \textrm{MeV}$

And this energy is also equal to

$E_{1/2} = -20.899 + 4 E' \textrm{MeV} = BE(19F) - BE(16O) + 0.110 = -20.072 \textrm{MeV}$

Thus,

$E' = 0.20675 \textrm{MeV}$

Therefore, the 2nd 0+ energy of 18O is

$E(0^+_2) = -9.038' \textrm{MeV} = 3.151 \textrm{MeV}$

Compare the experimental value of 3.63 MeV, this is a fair estimation.

It is interesting that, we did not really calculate the radial integral, and the angular part is calculated solely base on the algebra of J-coupling and the properties of delta interaction.

And since the single particle energies and residual interaction $V_0 \bar{R}$ are extracted from experiment. Thus, we can think that the basis is the “realistic” orbital of d and s -shell.

The spectroscopic strength and the wave function of the 18O ground state is $0.929 (0d_{5/2})^2 + 0.371 (1s_{1/2})^2$ . In here the basis wavefunctions $0d_{5/2}, 1s_{1/2}$ are the “realistic” or “natural” basis.

Construction of many-body wavefunction

May 8, 2020

GoLuckyRyan Basic Clebsch-Gordon, Shell Model, shell model calculation Leave a comment

This post is an extraction from “Theory of the nuclear shell model” by R.D. Lawson, Chapter 1.1.2

In a simple construction, the single-particle wave function is

$\displaystyle \phi_{jm}(\vec{r}) = R_l(r) \sum_{m_l m_s} Y_{l m_l}(\hat{r}) \chi_{m_s}$

And the many body Slater determinants is

$\displaystyle \Phi_{JM}(p_1, p_2, ...., p_n) = \frac{1}{\sqrt{N!}} \begin{vmatrix} \phi_{p_1}(1) & \phi_{p_1}(2) & ... & \phi_{p_1}(n) \\ \phi_{p_2}(1) & \phi_{p_2}(2) & ... & \phi_{p_2}(n) \\ ... & ... & \phi_{p_r}(k) & ... \\ \phi_{p_n}(1) & \phi_{p_n}(2) & ... & \phi_{p_n}(n) \end{vmatrix}$

where $p_r$ is a set of n states from the model space. Thus, for given $J, M$ , there are many possible choices of $p_r$ .

And the Hamiltonian is expanded into the space of $\Phi_{JM}(p_1, p_2, ..., p_n)$ .

I thought this is the end of the story, until I read the Chapter 1.1.2

To continuous further, we need 2 theorems. The 1st one is that.

If $J_+ \Psi_{JM} = 0$ , then, $M=J$ and $J^2 \Psi_{JM} = M(M+1) \Psi_{JM}$

This is kind of trivial, as the Ladder operator cannot promote the m-state any further, the m-state must be maximized and equal to J.

Before go to the 2nd theorem, let introduce the creation operator $a_{jm}^\dagger$ , and denote the Slater determinant as

$\displaystyle \frac{1}{\sqrt{N!}}\begin{vmatrix} \phi_{p_1}(1) & \phi_{p_1}(2) & ... & \phi_{p_1}(n) \\ \phi_{p_2}(1) & \phi_{p_2}(2) & ... & \phi_{p_2}(n) \\ ... & ... & \phi_{p_r}(k) & ... \\ \phi_{p_n}(1) & \phi_{p_n}(2) & ... & \phi_{p_n}(n) \end{vmatrix} = a_{p_1}^\dagger a_{p_2}^\dagger ... a_{p_n}^\dagger \left|0\right>$

The 2nd one is a bit complicated.

In a given model space ( j_1, j_2, …., j_k) with n particles. For a p distinct Slater determinants $\Phi_{M}(i), i = 1, ..., p$ with $J_z \Phi_{M}(i) = M \Phi_{M}(i)$ . Suppose there are another p+q distinct Slater determinants $\Phi_{M-1}(i)$ , so that $J_z \Phi_{M-1}(i) = (M-1) \Phi_{M-1}(i)$ , thus, q states with angular momentum J= M-1 can be constructed$.

Lets use an example to illustrate, say, we have a model space of 0d5/2 with 2 particles, we can construct the $J = 4, M = 4$ state as,

$\Phi_{M=4} = a_{5/2}^\dagger a_{3/2}^\dagger \left|0\right>$

And for $M= 3$ , the only possible state is

$\Phi_{M=3} = a_{5/2}^\dagger a_{1/2}^\dagger \left|0\right>$

Since the number of state for $M=4$ is 1, and so $p = 1$ . The number of state for $M=3$ is also 1, thus $p+q = 1$ . Therefore, according to the theorem 2, there is no state for $J =3$ . And we knew that, from a 2 particles in d5/2 shell can only form J = 0, 2, 4, due to the anti-symmetric of Fermions system.

we can continuous, $M= 2$ , there are 2 possible states,

$\Phi_{M=2}(1) = a_{5/2}^\dagger a_{-1/2}^\dagger \left|0\right>$
$\Phi_{M=2}(2) = a_{3/2}^\dagger a_{1/2}^\dagger \left|0\right>$

Thus, this time, $p = 1$ (as M=3 has 1 state) and $p+q = 2$ , therefore, there is 1 state of $J=2$ can be constructed.

For $M=1$ ,

$\Phi_{M=1}(1) = a_{5/2}^\dagger a_{-3/2}^\dagger \left|0\right>$
$\Phi_{M=1}(2) = a_{3/2}^\dagger a_{-1/2}^\dagger \left|0\right>$

Thus, the number of $J = 1$ state is 0.

For $M=0$ ,

$\Phi_{M=0}(1) = a_{5/2}^\dagger a_{-5/2}^\dagger \left|0\right>$
$\Phi_{M=0}(2) = a_{3/2}^\dagger a_{-3/2}^\dagger \left|0\right>$
$\Phi_{M=0}(3) = a_{1/2}^\dagger a_{-1/2}^\dagger \left|0\right>$

Therefore, the number of $J=0$ state is 1. In summary,

The number of J = 4 state = 1.
The number of J = 3 state = 0.
The number of J = 2 state = 1.
The number of J = 1 state = 0.
The number of J = 0 state = 1.

Another way to do so is draw a table

Ignore the negative $\sum m$ , and we can deduce the number of state for $\Psi_{JJ}$ .

The prove of the 2nd theorem is skipped. And I like to add an auxiliary theorem that for n identical particle in the j-orbital, the maximum spin is

$\displaystyle J(j^n) \leq n\left(j - \frac{n-1}{2} \right)$

This theorem can be proved by using the 1st theorem. The reason that this theorem is included is that, the anti-symmetry property implicitly contains in this theorem, and I see the elegant and beauty in it. Also, this also an example that theorem is trivial but useful.

Back to the wave function construction, the wave function,

$\displaystyle \Phi_{JM}(p_1, p_2, ...., p_n) = a_{p_1}^\dagger a_{p_2}^\dagger ... a_{p_n}^\dagger \left|0\right>$

is only a component, or part of the total wave function of spin J. For example, the coupling of 0d5/2 and 1s1/2 states, we can construct

$\displaystyle \Phi_{JM}\left( \frac{5}{2} m_1 \frac{1}{2} m_2 \right) = a_{\frac{5}{2}m_1}^\dagger a_{\frac{1}{2}m_2}^\dagger \left|0\right>$

But hold-on, we missed the coupling coefficient, without the coupling coefficient, we actually did not know the total J!! The proper wave function should be

$\displaystyle \Psi_{JM}\left( \frac{5}{2} m_1 \frac{1}{2} m_2 \right) = \sum_{m_1 m_2} C_{ \frac{5}{2} m_1 \frac{1}{2} m_2}^{JM} a_{\frac{5}{2}m_1}^\dagger a_{\frac{1}{2}m_2}^\dagger \left|0\right>$

Up to now, we should separate two wave function, $\Phi_{M}$ and $\Psi_{JM}$ . And more important, we drop the J in $\Phi_{M}$ .

$\phi_{jm}$ — single particle orbital
$\Phi_{M} = a_{p_1}^\dagger a_{p_2}^\dagger ... a_{p_n}^\dagger \left|0\right>$ — m-state
$\Psi_{JM} = \sum_{i} \alpha_i \Phi_{M}(i)$ — many-body wave function with spin J

when only 2 particles, the $\sqrt{2}\alpha_i$ is the Clebsch-Gordon coefficient.

We now give an example on (0d5/2)^2 configuration. For J = 4, only 1 state is possible, this

$\displaystyle \Psi_{44} = \Phi_{4} = a_{5/2}^\dagger a_{3/2}^\dagger \left|0\right> = \frac{1}{\sqrt{2}}\left( \phi_{\frac{5}{2}}(1)\phi_{\frac{3}{2}}(2) - \phi_{\frac{5}{2}}(2)\phi_{\frac{3}{2}}(1) \right)$

Using the Ladder operator, we can find all $\Psi_{4M}$ .

For J = 2, there are 2 m-states,

$\displaystyle \Psi_{22} = a \Phi_{2}(1) + b \Phi_{2}(2)$

In order to find the coefficient $a, b$ , we have the normalization condition $a^2 + b^2 = 1$ , and we have to use the theorem 1, and apply the $J_+ = j_+(1) + j_+(2)$ operator and require that

$\displaystyle J_+ \Psi_{22} = 0$

$\displaystyle j_{+} \phi_{jm} = \sqrt{(j-m)(j+m+1)} \phi_{j,m+1}$

Thus,

$\displaystyle J_+ \Psi_{22} = (3a + \sqrt{5} b) a_{5/2}^\dagger a_{1/2}^\dagger \left|0\right>$

solve

$3a + \sqrt{5} b = 0$
$a^2 + b^2 = 1$

$\displaystyle \Psi_{22} = \sqrt{\frac{5}{14}} \Phi_{2}(1) - \sqrt{\frac{9}{14}} \Phi_{2}(2)$

In fact,

$\displaystyle C_{\frac{5}{2}\frac{5}{2}\frac{5}{2}\frac{-1}{2}}^{22} = \sqrt{\frac{5}{28}} = \frac{a}{\sqrt{2}}$
$\displaystyle C_{\frac{5}{2}\frac{3}{2}\frac{5}{2}\frac{1}{2}}^{22} = -\sqrt{\frac{9}{28}} = \frac{b}{\sqrt{2}}$

In fact, for 2 particle wave function

$\displaystyle \Psi_{JM}(j_1 j_2 ) = \sum_{m_1 m_2} C_{j_1m_1 j_2m_2}^{JM} \phi_{j_1m_1}(1) \phi_{j_2m_2}(2)$

The power of this construction method is the many-body wave function. we now demonstrate 4 identical particle in 0d5/2 orbital,

The maximum J is

$\displaystyle J_{max} = 4 \left( \frac{5}{2} - \frac{4-1}{2} \right) = 4$ .

the M-state are

For $M = 4$

$\Phi_{M=4} = a_{5/2}^\dagger a_{3/2}^\dagger a_{1/2}^\dagger a_{-1/2}^\dagger \left|0\right>$

For $M = 3$

$\Phi_{M=3} = a_{5/2}^\dagger a_{3/2}^\dagger a_{1/2}^\dagger a_{-3/2}^\dagger \left|0\right>$

For $M = 2$

$\Phi_{M=2}(1)= a_{5/2}^\dagger a_{3/2}^\dagger a_{1/2}^\dagger a_{-5/2}^\dagger \left|0\right>$
$\Phi_{M=2}(2)= a_{5/2}^\dagger a_{3/2}^\dagger a_{-1/2}^\dagger a_{-3/2}^\dagger \left|0\right>$

For $M = 1$

$\Phi_{M=1}(1)= a_{5/2}^\dagger a_{3/2}^\dagger a_{-1/2}^\dagger a_{-5/2}^\dagger \left|0\right>$
$\Phi_{M=1}(2)= a_{5/2}^\dagger a_{1/2}^\dagger a_{-1/2}^\dagger a_{-3/2}^\dagger \left|0\right>$

For $M = 0$

$\Phi_{M=0}(1)= a_{5/2}^\dagger a_{3/2}^\dagger a_{-3/2}^\dagger a_{-5/2}^\dagger \left|0\right>$
$\Phi_{M=0}(2)= a_{5/2}^\dagger a_{1/2}^\dagger a_{-1/2}^\dagger a_{-5/2}^\dagger \left|0\right>$
$\Phi_{M=0}(3)= a_{3/2}^\dagger a_{1/2}^\dagger a_{-1/2}^\dagger a_{-3/2}^\dagger \left|0\right>$

or, in table

From the theorem 2, the number of m-state tell us the number of state for J are

$J = 4 , 1$
$J = 2 , 1$
$J = 0 , 1$

To construct $J = 4$ state,

$\displaystyle \Psi_{44} = a_{5/2}^\dagger a_{3/2}^\dagger a_{1/2}^\dagger a_{-1/2}^\dagger \left|0\right>$

$J = 2$ , using the same method

$\displaystyle \Psi_{22} = \sqrt{\frac{9}{14}} \Phi_{2}(1) - \sqrt{\frac{5}{14}} \Phi_{2}(2)$

$J = 0$

$\displaystyle \Psi_{00} = \sqrt{\frac{1}{3}} \left(\Phi_{0}(1) - \Phi_{0}(2) + \Phi_{0}(3) \right)$

I suspect the coefficient $\alpha_i$ is deeply related to n-j symbol. And I am sure the Group theory can provide an elegant solution and display for that.

At last, in the construction of m-state, we restricted the order, therefore, state like

$\displaystyle a_{3/2}^\dagger a_{5/2}^\dagger \left|0\right> = - a_{5/2}^\dagger a_{3/2}^\dagger \left|0\right>$

will be discarded.

But in general 2-body wave function,

$\displaystyle \Psi_{JM}\left(j_1 m_1 j_2 m_2 \right) = \sum_{m_1 m_2} C_{ j_1 m_1 j_2 m_2}^{JM} a_{\ j_1 m_1}^\dagger a_{j_2 m_2}^\dagger \left|0\right> \\ = \sum_{m_1 m_2} C_{j_1m_1 j_2m_2}^{JM} \phi_{j_1m_1}(1) \phi_{j_2m_2}(2)$

there is no $m_1 > m_2$ restriction.

Nuclear correlation & Spectroscopic factor

April 25, 2016

GoLuckyRyan Basic 18O, configuration, configuration mixing, core polarization, correlation, long-ranged correlation, neutron, residual interaction, shell model calculation, short-ranged correlation, spectroscopic factor, TBMEs Leave a comment

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 [2011], 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

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.

Nuclear Physics 101

Shell model calculation on 16C

Allowed Coupling for identical particles in the same orbital

Generating the list of two-body pair

KSHELL Shell model calculation

Shell model calculation on 18O

Construction of many-body wavefunction

Nuclear correlation & Spectroscopic factor

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j	n	Allowed $J$
3/2	3	3/2
5/2	3	9/2, 5/2, 3/2
	5	5/2
7/2	3	15/2, 11/2, 9/2, 7/2, 5/2, 3/2
	5	15/2, 11/2, 9/2, 7/2, 5/2, 3/2
	7	7/2
9/2	3	21/2, 17/2, 15/2, 13/2, 11/2, 9/2, 7/2, 5/2, 3/2
	5	25/2, 21/2, 19/2, 17/2, 15/2, 13/2, 11/2, 9/2, 7/2, 5/2, 3/2, 1/2
	7	21/2, 17/2, 15/2, 13/2, 11/2, 9/2, 7/2, 5/2, 3/2
	9	9/2