Nilsson orbital | Nuclear Physics 101

Wolfram Notebook Archive

November 24, 2020

GoLuckyRyan Basic Nilsson orbital 2 Comments

The calculation of the Nilsson orbital using Mathematica is posted in the Notebook Archive.

Here is the link:

https://notebookarchive.org/nilsson-orbital-using-perturbation-method-using-spherical-harmonic-basis–2020-11-an9bzmn/

Visualize Nilsson Orbital

July 2, 2020

GoLuckyRyan Basic Nilsson orbital Leave a comment

The basis function is the hyper-spherical harmonic.

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

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

After solving the eigen system with

$\displaystyle H = H_0 - \beta H_p$

we have the eigen values $E_i$ and eigen vectors $v_i$ .

Thus, the Nilsson orbital is

$\displaystyle K[Nn_z \Lambda] = \sum_{jl} C_{jl} |Nljk\rangle$

To plot the orbital,

$\displaystyle f \left(K[Nn_z \Lambda] \right) = \left(K[Nn_z \Lambda] \right)\left(K[Nn_z \Lambda] \right)^*$

For N=1,

Annotation 2020-07-02 165527.png

At $\beta = 0.0$

Annotation 2020-07-02 172301

I am not sure why the orbital $|1p \frac{1}{2}\frac{1}{2} \rangle$ is spherical. The wave function is

$\displaystyle |1p \frac{1}{2}\frac{1}{2} \rangle = A R_2(r) \sqrt{\frac{1}{3}}\left( \sqrt{2} Y_{11}\chi_\downarrow - Y_{10} \chi_\uparrow \right)$

The amplitude is

$\displaystyle \left(|1p \frac{1}{2}\frac{1}{2} \rangle \right)^2 = A^2 R_2^2(r) \frac{1}{3}\left( 2 Y_{11}^2 + Y_{10}^2 \right)$

$\displaystyle Y_{11} = - \frac{1}{2} \sqrt{\frac{3}{2\pi}} \sin(\theta) e^{i \phi}$

$\displaystyle Y_{10} = - \frac{1}{2} \sqrt{\frac{3}{\pi}} \cos(\theta)$

$\displaystyle 2 Y_{11}^2 + Y_{10}^2 = \frac{3}{4\pi}$

So, mathematically, $|1p \frac{1}{2}\frac{1}{2} \rangle$ is spherical, But I am not sure why physically is like this.

At $\beta = 0.3$

Annotation 2020-07-02 165828.png

The orbital with the lowest energy is concentrate on the z-axis, as it should be. And the highest energy orbital is concentrated on the x-y plane.

The wave function, from highest energy to lowest:

$\displaystyle f \left(1/2[101] \right) = e^{-r^2}( 0.172 x^2 + 0.172 y^2 + 0.014 z^2)$

$\displaystyle f \left(3/2[101] \right) = 0.180 e^{-r^2}( x^2 + y^2)$

$\displaystyle f \left(1/2[110] \right) = e^{-r^2}( 0.007 x^2 + 0.007 y^2 + 0.345 z^2)$

N=2

Annotation 2020-07-02 170641.png

At $\beta = 0$

Annotation 2020-07-02 171545

For $k = 1/2$ , the wave function concentrates on the z-axis. At max $k$ , the wave function concentrates on the x-y plane.

At $\beta = 0.3$

Annotation 2020-07-02 171917.png

For the lowest energy state, the 1/2[220], it is very deformed, like a thick hamburger.

Notes on the original Nilsson paper

June 21, 2020

GoLuckyRyan Basic Laguerre, Nilsson orbital, paper reading, Wigner-D Leave a comment

The paper title is “Binding states of individual nucleons in strongly deformed nuclei” by Sven Gösta Nilsson on 1955.

In the introduction, the total nuclear wave function $\Psi$

$\Psi = \phi_\Omega D_{MK}^I$

where $D_{MK}^I$ is Wigner D-matrix for rotation, $\phi_\Omega$ is the intrinsic motion of all nucleons. The vibration component is skipped. Because of imcompressible nuclear matter, a vibration needs a lot of energy.

The total wave function should preserve parity, so, a complete wave function should be

$\displaystyle \Psi = \sqrt{\frac{2I+1}{16\pi^2}} \left( \phi_\Omega D_{MK}^I + (-1)^{I+K} \phi_{-\Omega} D_{M-K}^I \right)$

Below is a picture of the quantum number.

Annotation 2020-06-21 134519

In large deformation, $J$ is not a good quantum number, but $\Omega$ is always a good one.
At ground state, $K = \Omega$ , so, the rotation angular momentum is perpendicular to the body frame axis.

The single nucleon potential is the usual.

$\displaystyle \frac{H}{\hbar \omega_0} = -\frac{1}{2} \nabla^2 + \frac{1}{2}r^2 - \beta r^2 Y_{20} + C l\cdot s + D l^2$

The basis in Nilsson’s paper is the eigen state of harmonic oscillator in L-S representation. The Hamiltonian is

$\displaystyle \frac{H_0}{\hbar \omega_0} = -\frac{1}{2} \nabla^2 + \frac{1}{2}r^2$

with

$\displaystyle H_0 \left| N l m_l m_s \right> = \hbar \omega_0 \left( N + \frac{3}{2} \right)\left| N l m_l m_s \right>$

The original paper use $\Sigma = m_s , \Lambda = m_l$ . And the function solution is

$\displaystyle \left<r | N l m_l m_s \right> = A r^l e^{-\frac{r^2}{2}} F\left(- \frac{N-l}{2}, l+ \frac{3}{2}, r^2 \right) Y_{lm_l} \chi_{s m_s}$

where $F(a,b,x)$ is the confluent hypergeometric function, and it can be expressed as Laguerre polynomial. So, the solution can be rewrite as

$\displaystyle \left<r | N l m_l m_s \right> = A r^l e^{-\frac{r^2}{2}} L_{\frac{N-l}{2}}^{l+\frac{1}{2}}(r^2) Y_{lm_l} \chi_{s m_s}$

In contrast, in my calculation, I use the $J-\Omega$ presentation, that, the connection is

$\displaystyle \left| N l j \Omega \right> = \sum C_{l m_l s m_s}^{j \Omega } \left| N l m_l m_s \right>$

The different is that, Nilsson needs additional transformation to calculate the $C_{jl}$ coefficient, and the calculation of the $\left< l\cdot s \right>$ and $\left< l^2 \right>$ is a bit complicated, due to L-S is not a good quantum number when $l\cdot s$ interaction was included. Thus, in Nilsson paper, he spent sometimes to talk about the calculation of $\left< l\cdot s \right>$ and $\left< l^2 \right>$ .

Next, Nilsson gives the calculation parameters of $\kappa, \mu$ . And since he is using L-S representation, the Nilsson orbital is expressed in that basis. Here is a comparison between my calculation and Nilsson calculation.

Annotation 2020-06-21 141037

Next, he explained that for $\beta > 0$ , energy increase with increase $\Omega$ due to “surface coupling”. It can be imagine like this:

Annotation 2020-06-21 141313

In above picture, when $\Omega$ is smallest, $l$ is perpendicular to the body axis, so the nucleon has large overlap with the whole nucleus, thus it is most bounded.

For small deformation, the deformation field $r^2 Y_{20}$ is treated as a perturbation, so that $j$ is a good quantum number.

For large deformation, the spin-orbital interaction $l\cdot s, l^2$ is treated as perturbation, and the good quantum number is $N, n_z, m_l , m_s$ , since $\Omega = m_s + m_l$ , so $\Omega$ is also a good quantum number.

In our previous notation, Nilsson orbital is notated as $K[N n_z \lambda]$ or $[N n_z \lambda] K$ , which is equivalent in Nilsson as $[N n_z m_l] \Omega$

The wave function of many nucleons.

After established the single nucleon wave function. The receipt of the construction of many nucleons wave function is

select a set of Nilsson orbitals and form the Slater determent.
Minimum the total energy by adjusting the deformation parameter.

$\displaystyle H_t = \sum_i T_i + \frac{1}{2} \sum_{i \neq j} V_{ij}$

It is interesting that the total wave function is not a mixture of various Slater determents from different combination of Nilsson orbitals, but rather a single Slater determent.

Ground state spin

Since each Nilsson orbital is degenerated to $\pm \Omega$ , which are rotate oppositely. For even-even nucleus, the ground state spin must be zero. For even-odd nucleus, the ground state spin is equal to the $\Omega$ of the last single nucleon.

For odd-odd nucleus, the ground state spin can be $\Omega = |\Omega_p \pm \Omega_n|$ , the p-n interaction decide which $\Omega$ is the ground state.

Decoupling parameter

For odd-A nuclei, the rotational energy is modified by a decoupling factor

$\displaystyle E_r = \frac{\hbar^2}{2 I_M} \left( I(I+1) + a (-1)^{I+\frac{1}{2}} \left(I+\frac{1}{2}\right) \right)$

And

$\displaystyle a = \sum_{j} (-1)^{j - \frac{1}{2}} \left( j+ \frac{1}{2} \right) |C_{jl}^2|$

Magnetic moments, EM transition probability, and ft-value for beta-decay are skipped.

Besides of the skipped material, it turns out the original Nilsson paper did not surprise me. And I still don’t really understand the “decoupling” thing.

Diagonal elements for Nilsson orbital in spherical-spin function

March 30, 2020

GoLuckyRyan Basic Laguerre, Nilsson orbital Leave a comment

In this post, we explain how to calculate the Nilsson orbital using perturbation method by compute the matrix elements using spherical-spin function. In that post, I said I will give the calculation for the perturbation element. Here we go for the diagonal elements

The diagonal matrix element,

$\displaystyle \left<Nljk|r^2 Y_{20}|Nljk\right>$

where the spherical-spin wave-function is

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

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

First, the radial part is

$\displaystyle R_{Nl}(r) = r^l e^{-\frac{r^2}{2}} L_{\frac{N-l}{2}}^{l + \frac{1}{2}}(r^2)$

The radial integral is easy, we can use the integration formula. The radial integration is,

$\displaystyle \int_0^{\infty} r^2 R_{Nl}^2(r) r^2 dr = \int_0^{\infty} r^{2l+4} e^{-r^2} \left(L_{\frac{N-l}{2}}^{l + \frac{1}{2}}(r^2)\right)^2 dr$

When using the integration formula, one has to be careful when changing variable $r^2 \rightarrow r$ . Since $dr^2 = 2 r dr$ , we have to pull $2r$ out to properly do the change of variable.

$\displaystyle \int_0^{\infty} R_{Nl}^2(r) r^2 dr = \frac{1}{2}\int_0^{\infty} r^{2(l+1/2 + 1)} e^{-r^2} \left(L_{\frac{N-l}{2}}^{l + \frac{1}{2}}(r^2)\right)^2 (2 r dr)$

Set $r^2 \rightarrow x$ , $l + 1/2 \rightarrow \alpha$ , and $n = \frac{N-l}{2}$

$\displaystyle = \frac{1}{2}\int_0^{\infty} x^{\alpha+1} e^{-x} \left(L_{N}^{\alpha}(x)\right)^2 (dx) = \frac{1}{2}\frac{(\alpha+n)!}{n!}(2n+\alpha+1)$

Thus,

$\displaystyle \int_0^{\infty} R_{Nl}^2(r) r^2 dr = \frac{1}{2}\frac{( \frac{N+l+1}{2})!}{(\frac{N-l}{2})!}(N+\frac{3}{2})$

The angular-spin part is

$\displaystyle S_{ljk}(\theta,\phi, m_s) = \sum_{m m_s} Y_{lm}(\theta, \phi) \chi_{\frac{1}{2} m_s} C_{lm\frac{1}{2} m_s}^{jk}$

This contains spatial and spin part.

$\displaystyle S_{ljk}^*(\theta,\phi, m_s) Y_{20} S_{ljk}(\theta,\phi, m_s) \\=\sum_{m' m'_s} \sum_{m m_s} Y_{lm'}^*(\theta,\phi) Y_{20} Y_{lm}(\theta, \phi) \chi_{\frac{1}{2} m'_s} \cdot \chi_{\frac{1}{2} m_s} C_{lm'\frac{1}{2} m'_s}^{jk} C_{lm\frac{1}{2} m_s}^{jk}$

The dot-product of the spin part restricted the $m'_s$

$\displaystyle = \sum_{m' m m_s} Y_{lm'}^*(\theta,\phi) Y_{20} Y_{lm}(\theta, \phi) C_{lm'\frac{1}{2} m_s}^{jk} C_{lm\frac{1}{2} m_s}^{jk}$

And since $m+m_s = k$ and $m'+m_s =k$ , therefore $m = m'$ ,

$\displaystyle = \sum_{m m_s} Y_{lm}^*(\theta,\phi) Y_{20} Y_{lm}(\theta, \phi) \left(C_{lm\frac{1}{2} m_s}^{jk}\right)^2$

And the integration of spherical harmonic gives,

$\displaystyle \int Y_{lm}^*(\theta,\phi) Y_{20} Y_{lm}(\theta, \phi) = \sqrt{\frac{5}{4\pi}} C_{20l0}^{l0} C_{20lm}^{lm}$

$\displaystyle C_{20l0}^{l0} = -\frac{l+1}{\sqrt{4l^3+8l^2+l-3}} , l > 0$

$\displaystyle C_{20l0}^{l0} = -\frac{l(l+1)-3m^2}{\sqrt{l(4l^3+8l^2+l-3)}} , l > 0$

Sum up everything,

$\displaystyle \left<Nljk|r^2 Y_{20}|Nljk\right> \\ = \frac{(\frac{N-l}{2})!(\frac{N+l}{2})! 2^{N+l+2}}{\sqrt{\pi} (N+l+1)!} \frac{1}{2}\frac{( \frac{N+l+1}{2})!}{(\frac{N-l}{2})!}(N+\frac{3}{2}) \sum_{m m_s} \sqrt{\frac{5}{4\pi}} C_{20l0}^{l0} C_{20lm}^{lm} \left(C_{lm\frac{1}{2} m_s}^{jk}\right)^2 \\ = \sqrt{\frac{5}{4}} \frac{(\frac{N+l}{2})! ( \frac{N+l+1}{2})! 2^{N+l+1}}{\pi (N+l+1)!}(N+\frac{3}{2}) \sum_{m m_s} C_{20l0}^{l0} C_{20lm}^{lm} \left(C_{lm\frac{1}{2} m_s}^{jk}\right)^2$

Note: I haven’t numerically check the formula. ( may be later )

For the off-diagonal element. The angular-spin part should be similar. The difficulty is the radial part, we have to evaluate the most general orthogonal relation of the Laguerre polynomial with weighting $x^{\frac{1}{2}(\alpha+\alpha'+2)} e^{-x}$ .

We only knew the Laguerre polynomil is orthogonal with respect to $r^\alpha e^{-r}$ , i.e.

$\displaystyle \int_0^{\infty} L_n^\alpha L_m^\alpha r^\alpha e^{-r} dr = \frac{(\alpha+n)!}{n!} \delta_{nm}$

But not this.

$\displaystyle \int_0^{\infty} L_n^\alpha L_m^\beta r^\frac{\alpha+\beta}{2} e^{-r} dr$

or this

$\displaystyle \int_0^{\infty} L_n^\alpha L_m^\beta r^{(\frac{\alpha+\beta}{2}+k)} e^{-r} dr$

But we are quite sure the last one, with $\alpha \neq \beta$ will not give zero, otherwise, The Nilsson orbital will be very simple and boring.

Ordering of the Nilsson orbital

March 9, 2020

GoLuckyRyan Basic Nilsson orbital Leave a comment

In this post and this post, the level ordering can be shifted by adjusting the parameter $\mu$ . And because of the ordering, the component of the harmonic oscillators strongly depends on it. For example, the 5/2[512] state and 1/2[521],

In above gif, the parameter $\mu = 0.35 to 0.7$ , the step is not even. At small $\mu < 0.50$ , the 5/2[512] and 1/2[521] are not crossing, it becomes crossed when $\mu > 0.5$ . We can also see the decomposition to the spherical harmonic oscillator also change by a lot.

Nilsson Diagram from N=1 to N=6.

October 12, 2019

GoLuckyRyan Basic deformation, Nilsson orbital Leave a comment

The calculation use 84 Nilsson basis from 0s1/2, up to 6i13/2. Although some lines are broken, it is kind of nice. And we notices that, when the same j orbitals approach each other, they repulse. The straight line states are the (almost) pure state, which only consist with 1 spherical orbital.

Int the calculation, $\kappa = 0.05$ and

$\mu(N) = \begin{pmatrix} 0 & N=1 \\ 0 & N=2 \\ 0.35 & N=3 \\ 0.625 & N = 4 \\ 0.625 & N=5 \\ 0.63 & N=6 \\ 0.63 & N=7 \end{pmatrix}$

The parameter $\mu$ is for adjusting the energy to match with experimental data for spherical nuclei.

The spherical energy, which is the diagonal element of the Hamiltonian of spherical basis, is

$\displaystyle E_0(n,l,j,k) = n + \frac{3}{2} - \kappa(2 l\cdot s + \mu l(l+1))$

$\displaystyle l \cdot s = \frac{1}{2}(j(j+1) - l(l+1) - s(s+1))$

N=1-6.png

Nilsson energy-deformation plots

October 11, 2019

GoLuckyRyan Basic deformation, Nilsson orbital Leave a comment

From the previous post, we used Mathematica to calculate the Nilsson orbitals with diagonalization method. At that time, we had a problem that Mathematica will sort the eigen energies, that create a problem that it is very difficult to track the Nilsson orbital. But now, this problem was solved by using the orthogonal property of the eigenstate. At small deformation, $\beta$ and $\beta+\delta \beta$ , when the $\delta \beta$ is small enough, the eigenstates for these two deformation would be almost perpendicular. When the ordering of the eigenstates changed due to energy sorting, the dot product of the eigenstates matrix would have off-diagonal elements. e.g.

$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$

This, we can use this matrix to change the order array.

Suppose at the beginning, the order array is {1,2,3}. After this matrix, the order array becomes {1,3,2}.

The tricky part is that, the element of the order array is the record for the position of states. For example, {1,2,3,5,7,4,8,6} tells us the position of the 7th state is at the 5-th position. In order to rearrange the energy so that the n-th state is placed at the n-th position, we have to find the position of the n-th state in the order array. The position array is {1,2,3,6,4,8,5,7}, and the energy of the 7-th state can be obtained from 5th position, as we expected. What we did here, can be better illustrate in following,

Suppose the state is notated using letter, e.g. {a,b,c,d,e,f,g,h}. For an order array, such that, {a,b,c,e,g,d,h,f}, we want to find the position array to tell us the position for the x-state. Thus, we have the position array {1,2,3,6,4,8,5,7}. This transform is between the position and the state.

The letter symbols in the order array represent the states and the position of the order array is the “position” or “order” of the state energies. The number in the position array represent the “position” or “order”, and the position of the position array represents each state in order. If we use {1,a} ordered array to represent the a-state is in position 1. Thus, {a,b,c,e,g,d,h,f} can be written as {{1,a},{2,b},{3,c},{4,e},{5,g},{6,d},{7,h},{8,f}}. And {1,2,3,6,4,8,5,7} = {{1,a},{2,b},{3,c},{6,d},{4,e},{8,f},{5,g},{7,h}}. On the other hands, it is a sorting with position to sorting with state.

The matrix of the eigenvectors from $\beta$ and $\beta+\delta \beta$ interchange the order array, or the states. And for plotting with same color for the same state, we want the state is in order. Thus, we use the matrix to change the order of the states with respect to the position, after that, we have to sort it back according to states.

N=6.png

Screen Shot 2019-11-22 at 09.21.32.png

N=4.png

N=3.png

N=2.png

Sum rules of Nilsson orbital

August 5, 2019

GoLuckyRyan Basic Nilsson orbital, Sum rules Leave a comment

The Nilsson orbital can be decomposed into series of orbitals of 3D-harmonic oscillator, such that

$\displaystyle [Nn_z\lambda]K = \sum_{N'lj} C_{N'lj}^{N n_z \lambda} |N'ljK\rangle$

with eigen energy $\epsilon_{Nn_z\lambda K} (\beta)$ and

$n_z + \lambda = l$

$n_z + K = j$

Since the Nilsson orbital is normalized

$\displaystyle \sum_{N'lj} \left(C_{N'lj}^{N n_z \lambda}\right)^2 = 1$

Since the number of orbital for fixed $l,j$ is $2j+1$ , thus using an inverse transformation from spherical orbital to Nilsson orbital, we have,

$\displaystyle \sum_{N n_z \lambda} \left(C_{N'lj}^{N n_z \lambda}\right)^2 = 2j+1$

I cannot prove it, but

$\displaystyle \sum_{N n_z \lambda} \epsilon_{N n_z \lambda K}(\beta) \left(C_{N'lj}^{N n_z \lambda}\right)^2 = \epsilon_{N n_z \lambda K} (0)$

Thus, the single-particle energy fro Nilsson orbital is as same as the spherical orbital !!!

Quantum number of Nilsson orbital

July 26, 2019

GoLuckyRyan Basic Nilsson orbital 4 Comments

Using the diagonalization method using 3D harmonic oscillator as a basis to expend or approximate the Nilsson orbital, it is not easy to know the good quantum number, i.e. $K[Nn_z\Lambda]$ , here $\Lambda = m_l$ and $K$ is the total spin projected on the body symmetric axis.

One way to find out is using the dominant component $|NljK\rangle$ for small deformation and track the energy curve to the large deformation.

When I listed all the Nilsson orbital in below, it shows a clear pattern.

$\Lambda$ is the projection of orbital angular momentum, thus, $l \geq \Lambda$
$N = n_z$ gives the lowest energy state
Larger $n_z$ , lower energy
$K = 1/2$ also gives the lowest energy state
From observation, it is fair to assume that $n_z + \Lambda = l$
Also, for $j = l \pm 1/2$ , $K = \Lambda \pm 1/2$

NilssonOrbit to spherical.png

Notice that, in some paper, for example the $2s_{1/2}$ state was assigned to be $1/2[211]$ . I think it is no correct, because for s-orbital, it is impossible to have $\Lambda = m_l = 1$ .

[update on 20240419] Comment from Randomizer

Here is the calculation using the N=4, N=6, and N=8 $\hbar \omega$ , and the states are labeled using the “rules” summarized in this post. Hope can help.

Nilsson Orbital using diagonalization method

July 25, 2019

GoLuckyRyan Basic Clebsch-Gordon, diagonalization, Harmonic Oscillator, harmonics, Laguerre polynomial, Nilsson orbital, Variational method Leave a comment

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

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

using

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

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

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

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

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 $r^2 \rightarrow \rho^2 \frac{\hbar}{ m \omega_0}$

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

Set

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

and the perturbation is

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

with

$\displaystyle \hbar \omega_0 \approx 45 A^{-1/3} - 25 A^{-2/3}$

For example, when A = 12, $\hbar \omega_0 \approx 15$ MeV, when A = 132, $\hbar \omega_0 \approx 8$ MeV

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

Nuclear Physics 101

Wolfram Notebook Archive

Visualize Nilsson Orbital

Notes on the original Nilsson paper

Diagonal elements for Nilsson orbital in spherical-spin function

Ordering of the Nilsson orbital

Nilsson Diagram from N=1 to N=6.

Nilsson energy-deformation plots

Sum rules of Nilsson orbital

Quantum number of Nilsson orbital

Nilsson Orbital using diagonalization method

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