## Diagonal elements for Nilsson orbital in spherical-spin function

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, $\displaystyle \left$

where the spherical-spin wave-function is $\displaystyle |Nljk\rangle = A r^l e^{-\frac{r^2}{2}} L_{\frac{N-l}{2}}^{l + \frac{1}{2}}(r^2) \sum_{m m_s} Y_{lm}(\theta, \phi) \chi_{\frac{1}{2} m_s} C_{lm\frac{1}{2} m_s}^{jk}$ $\displaystyle A = \sqrt{\frac{(\frac{N-l}{2})!(\frac{N+l}{2})! 2^{N+l+2}}{\sqrt{\pi} (N+l+1)!}}$ $\displaystyle R_{Nl}(r) = r^l e^{-\frac{r^2}{2}} L_{\frac{N-l}{2}}^{l + \frac{1}{2}}(r^2)$

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

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

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 state and 1/2, In above gif, the parameter $\mu = 0.35 to 0.7$, the step is not even. At small $\mu < 0.50$, the 5/2 and 1/2 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.

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

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.     ## Sum rules of Nilsson orbital

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

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.

1. $\Lambda$ is the projection of orbital angular momentum, thus, $l \geq \Lambda$
2. $N = n_z$ gives the lowest energy state
3. Larger $n_z$, lower energy
4. $K = 1/2$ also gives the lowest energy state
5. From observation, it is fair to assume that $n_z + \Lambda = l$
6. Also, for $j = l \pm 1/2$, $K = \Lambda \pm 1/2$ Notice that, in some paper, for example the $2s_{1/2}$ state was assigned to be $1/2$. I think it is no correct, because for s-orbital, it is impossible to have $\Lambda = m_l = 1$.

## Nilsson Orbital using diagonalization method

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

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 $\kappa = 0.05, \mu = 0$ The component of each orbital can be directly taken from the eigenvalue. Here is the 1/2 state. $\kappa = 0.05, \mu(N=3) = 0.35, \mu(N=4) = 0.625, \mu(N=5) = 0.63$ 