Nilsson Orbital using diagonalization method

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Long time ago, I tried to tackle the Nilsson orbital by solving the Hamiltonian analytically. However, the Hamiltonian is without LS coupling. This times, I redo the calculation according to the reference B. E. Chi, Nuclear Phyiscs 83 (1966) 97-144.

The Hamiltonian is

\displaystyle H = \frac{P^2}{2m} + \frac{1}{2}m\left( \omega_\rho^2 (x^2+y^2) + \omega_z^2 z^2 \right) + C L\cdot S + D L\cdot L


\displaystyle \omega_\rho = \omega_0 \left(1+\frac{2}{3}\delta\right)^{\frac{1}{2}} = \omega \left(\frac{3+2\delta}{3-4\delta}\right)^{1/6}

\displaystyle \omega_z = \omega_0 \left(1-\frac{4}{3}\delta\right)^{\frac{1}{2}} = \omega \left(\frac{3-4\delta}{3+2\delta}\right)^{1/3}

\displaystyle \beta = \frac{4}{3}\sqrt{\frac{\pi}{5}}\delta

\displaystyle r^2 Y_{20}(\theta, \phi) = \frac{1}{4}\sqrt{\frac{5}{\pi}} (3z^2-r^2)

The Hamiltonian becomes

\displaystyle H = -\frac{\hbar^2}{2m}\nabla^2 +\frac{1}{2} m \omega_0^2 r^2 - \frac{1}{2} m\omega_0^2 r^2 \beta Y_{20} + C L\cdot S + D L\cdot L

Set x_i^2 \rightarrow  x_i^2 \frac{\hbar}{m \omega_0} , and r^2 \rightarrow \rho^2 \frac{\hbar}{ m \omega_0}

\displaystyle \frac{H}{\hbar\omega_0} = \frac{1}{2}(-\nabla^2 + \rho^2)  - \rho^2 \beta Y_{20} - 2 \kappa L\cdot S - \mu \kappa L\cdot L


\displaystyle \frac{H_0}{\hbar\omega_0} = \frac{1}{2}(-\nabla^2 + \rho^2) - 2 \kappa L\cdot S - \mu \kappa L\cdot L

and the perturbation is

\displaystyle \frac{H_p}{\hbar\omega_0} =  - \rho^2 Y_{20}

The wavefunction for the spherical harmonic is

\displaystyle |Nljk\rangle = A r^l e^{-\frac{r^2}{2}} L_{\frac{N-l}{2}}^{l + \frac{1}{2}}(r^2) \sum_{m m_s} Y_{lm}(\theta, \phi) \chi_{\frac{1}{2} m_s} C_{lm\frac{1}{2} m_s}^{jk}

\displaystyle A = \sqrt{\frac{(\frac{N-l}{2})!(\frac{N+l}{2})! 2^{N+l+2}}{\sqrt{\pi} (N+l+1)!}}

The diagonal elements are

\displaystyle \frac{1}{\hbar \omega_0 }\langle Nljk|H_0|Nljk\rangle = N + \frac{3}{2} - \kappa \langle L\cdot S \rangle - \mu \kappa l(l+1)

where \langle L \cdot S \rangle = \frac{1}{2} ( j(j+1) - l(l+1) - \frac{3}{4} )

The off diagonal elements are

\displaystyle \frac{1}{\hbar \omega_0 }\langle Nljk|H_p|Nljk\rangle = - \langle Nljk| r^2 Y_{20}|Nljk\rangle

( I will evaluate this integral in future )

The rest is diagonalization the Hamiltonian

\displaystyle H = H_0 + \beta H_p

Here is the calculation for the 2nd harmonic for \kappa = 0.05, \mu = 0

Screen Shot 2019-07-25 at 18.25.45.png

The component of each orbital can be directly taken from the eigenvalue. Here is the [521]1/2 state. \kappa = 0.05, \mu(N=3) = 0.35, \mu(N=4) = 0.625, \mu(N=5) = 0.63

Screen Shot 2019-07-25 at 18.28.27.png


Shell model calculation and the USD, USDA, and USDB interaction

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Form the mean field calculation, the single particle energies are obtained. However, the residual interaction is still there that the actual state could be affected. Because the residual interaction produces the off-diagonal terms in the total Hamiltonian, and that mixed the single particle state.

The Shell Model calculation can calculate the nuclear structure from another approach. It started from a assumed nuclear Hamiltonian, with a basis of wavefunctions. The Hamiltonian is diagonalized with the basis, then the eigenstates are the solution of the wavefunctions and the nuclear structure, both ground state and excited states. The basis is usually the spherical harmonic with some radial function. Or it could be, in principle, can take from the result of mean field calculation. Thus, the Shell Model calculation attacks the problem directly with only assumption of the nuclear interaction.

However, the dimension of the basis of the shell model calculation could be very huge. In principle, it should be infinitely because of the completeness of vector space. Fro practical purpose, the dimension or the number of the basis has to be reduced, usually take a major shell. for example the p-shell, s-d shell, p-f shell. However, even thought the model space is limited, the number of basis is also huge. “for ^{28}Si the 12-particle state with M=0 for the sum of the j_z quantum numbers and T_z=0 for the sum of the %Latex t_z$ quantum numbers has dimension 93,710 in the m-scheme” [B. A. Brown and B. H. Wildenthal, Ann. Rev. Nucl. Part. Sci. 38 (1998) 29-66]. Beside the huge dimensions and the difficult for diagonalizing the Hamiltonian, the truncation of the model space also affect the interaction.

We can imagine that the effective interaction is different from the actual nuclear interaction, because some energy levels cannot be reached, for example, the short range hard core could produce very high energy excitation. Therefore, the results of the calculation in the truncated model space must be “re-normalized”.

There are 4 problems in the shell model calculation:

  • the model space
  • the effective interaction
  • the diagonalization
  • the renormalization of the result

The shell model can also calculate the excited state with 1\hbar \omega (1 major shell). This requires combination of the interactions between 2 major shell.

For usage, say in the code OXBASH, user major concern is the choice of the interaction and model space. The shell model are able to calculate

  • The binding energy
  • The excitation energies
    • The nucleons separation energies
  • The configuration of each state
  • The magnetic dipole matrix elements
  • The Gamow-Teller (GT) transition
  • The spectroscopic factor
  • …… and more.


The W interaction (or the USD) for the s-d shell was introduced by B.H. Wildenthal around 1990s. It is an parametric effective interaction deduced from fitting experimental energy levels for some s-d shell nuclei. Before it, there are some theoretical interactions that require “no parameter”, for example the G-matrix interaction is the in-medium nucleon-nucleon interaction.

The problem for the USD interaction is the interpretation, because it is a black-box that it can reproduce most of the experimental result better than theoretical interactions, but no one know why and how. One possible way is translate the two-body matrix elements (TBME) back to the central, spin-orbit, tensor force. It found that the central and spin-orbit force are similar with the theoretical interactions, but the tensor force could be different. Also, there could be three-body force that implicitly included in the USD interaction.

In 2006, B.A. Brown and W.A. Richter improved the USD interaction with the new data from the past 20 years [B.A. Brown, PRC 74, 034315(2006)]. The new USD interaction is called USDA and USDB. The difference between USDA and USDB is the fitting (something like that, I am not so sure), but basically, USDA and USDB only different by very little. Since the USDB has better fitting, we will focus on the USDB interaction.

The single particle energy for the USDB is

  • 1d_{3/2} = 2.117
  • 2s_{1/2} = -3.2079
  • 1d_{5/2} = -3.9257

in comparison, the single particle energies of the neutron of 17O of 2s_{1/2} = -3.27 and 1d_{5/2} = -4.14.

Can to USD interaction predicts the new magic number N=16?

Yes, in a report by O. Sorlin and M.-G. Porquet (Nuclear magic numbers: new features far from stability) They shows the effective single particle energy of oxygen and carbon using the monopole matrix elements of the USDB interaction. The new magic number N=16 can be observed.


Clebsch – Gordan Coefficient II

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As last post discussed, finding to CG coefficient is not as straight forward as text book said by recursion.
However, there are another way around, which is by diagonalization of J^2

first we use the identity:

J^2 = J_1^2+J_2^2 + 2 J_{1z} J_{2_z} + J_{1+} J_{2-} + J_{1-} J_{2+}

when we “matrix-lize” the operator. we have 2 choice of basis. one is \left| j_1,m_1;j_2;m_2 \right> , which give you non-diagonal matrix by the J_{\pm} terms. another one is \left|j,m\right>, which give you a diagonal matrix.

Thus, we have 2 matrixs, and we can diagonalized the non-diagonal. and we have the Unitary transform P, from the 2-j basis to j basis, and that is our CG coefficient.

oh, don’t forget the normalized the Unitary matrix.

i found this one is much easy to compute.

on Diagonalization (reminder)

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since i don’t have algebra book on my hand, so, it is just a reminder, very basic thing.

for any matrix M , it can be diagonalized by it eigenvalue \lambda_i  and eigen vector v_i , given that it eigenvectors span all the space. thus, the transform represented by the matrix not contractive, which is to say, the dimension of the transform space is equal to the dimension of the origin space.

Let denote, D before Diagonal matrix, with it elements are eigenvalues.

D_{ij} = \lambda_i \delta_{ij}

P be the matrix that collect the eigenvectors:

P_{i j} = \left( v_i \right)_j = \begin {pmatrix} v_1 & v_2 & ... & v_i \end {pmatrix}

Thus, the matrix M is :

M = P \cdot D \cdot P^{-1}

there are some special case. since any matrix can be rewritten by symmetric matrix S and anti-symmetric matrix A . so we turn our focus on these 2 matrices.

For symmetric matrix S , the transpose of P also work

S =P \cdot D \cdot P^{-1} = (P^T)^{-1} \cdot D \cdot P^T

which indicated that P^T = P^{-1} . it is because, for a symmetric matrix, M = M^T ,  the eigenvalues are all different, then all eigenvector are all orthogonal, thus P^T \cdot P = 1 .

For anti-symmetric matrix A

A = P \cdot D \cdot P^{-1}

since the interchange of row or column with corresponding exchange of eigenvalues in D still keep the formula working. Thus, the case P = P^T never consider.


For example, the Lorentz Transform

L = \gamma \begin {pmatrix} \beta & 1 \\ 1 & \beta \end {pmatrix}

which has eigenvalues:

D = \gamma \begin {pmatrix} \beta-1 & 0 \\ 0 & \beta+1 \end {pmatrix}

P = \begin {pmatrix} -1 & 1 \\ 1 & 1 \end {pmatrix}

the eigenvector are the light cone. because only light is preserved in the Lorentz Transform.

and it is interesting that

L = P \cdot D \cdot P^{-1} = P^{-1} \cdot D \cdot P = P^T \cdot D \cdot (P^T)^{-1} = (P^T)^{-1} \cdot D \cdot P^T

another example is the Rotation Matrix

R = \begin {pmatrix} cos(\theta) & - sin(\theta) \\ sin(\theta) & cos(\theta) \end{pmatrix}

D = \begin {pmatrix} Exp( - i \theta) & 0 \\ 0 & Exp(i \theta) \end {pmatrix}

P = \begin {pmatrix} -i & i \\ 1 & 1 \end{pmatrix}

the last example to give is the J_x of the spin-½ angular momentum

J_x = \begin {pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}

D = \begin {pmatrix} -1 & 0 \\ 0 & 1 \end {pmatrix}

P = \begin {pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix}