## The nuclear structure of 19F

Some facts about 19F:

• Ground state spin-parity is 1/2+.
• Has low-lying 1/2- state at 197 keV.
• Magnetic dipole moment is 2.62885 μ0.
• 19F(p,2p) experiment reported only 2s-wave can fit the result. [M.D. High et al., PLB 41 (1972) 588]
• 19F(d, 3He) experiment reported the ground state is from 1s1/2 proton with spectroscopic factor of 0.38. [G. Th. Kaschl et al., NPA 155 (1970) 417]
• 18O(3He, d) experiment report the ground state is 1s1/2 proton with spectroscopic factor of 0.21. [C. Schmidt et al., NPA 155 (1970) 644]
• There is a rotational band of 19F [C. F. Williamson et al., PRL 40 (1978) 1702]

My understanding [2018-01-30]

The 19F is deformed. The deformation is confirmed from rotation band.

The deformation distorted the spherical basis into deformed basis. In the simplest deformed basis, the cylindrical basis, the $|Nn_z m_l K\rangle =|220(1/2)\rangle$ is mixed with 1s1/2 (~33%) and 0d5/2 (~66%) orbits.

The 19F wave function can be written as

$|^{19}F\rangle = \sqrt{0.2 \sim 0.4}|\pi 1s_{1/2} \times ^{18}O_{g.s}\rangle + \sqrt{0.8\sim0.6}|\pi 0d_{5/2} \times ^{18}O^*\rangle + ...$

Under proton transfer/pickup reactions, the selection of oxygen ground state force the transfer proton to be in 1s1/2 state. The founding of s-wave ground state make the association of the 19F ground state spin to be 1/2.

Using USDB interaction with pn formalism. The 18O, 19F ground state are

$|^{18}O\rangle = \sqrt{0.78} |(\nu0d_{5/2})^2 \times ^{16}O\rangle + \sqrt{0.17}|(\nu1s_{1/2})^2 \times ^{16}O\rangle + ...$

$|^{19}F\rangle = \\ \sqrt{0.22} |(\pi1s_{1/2})(\nu0d_{5/2})^2 \times ^{16}O\rangle + \\ \sqrt{0.17}|(\pi1s_{1/2})(\nu1s_{1/2})^2 \times ^{16}O\rangle + \\ \sqrt{0.27}|(\pi0d_{5/2})(\nu0d_{5/2})^2 \times ^{16}O\rangle + ...$

This (somehow) shows that the interaction accurately reproduce the shell configuration. The USDB interaction also suggest that, at the 19F ground state, the 1s1/2 orbit are only 41% filled and the 0d5/2 orbit are 47% filled.

The fact that the spectroscopic factor is much less then unity suggests the ground state configuration of 19F is not fit for single particle picture.

There are fill questions,

1. Why deform? due to the single 1d5/2 proton? Suppose adding a proton on 18O, the proton fill on 1d5/2 shell, and the d-shell creates a deformation on the sd shell, that shift the energy lower by mixing with s-shell?
2. in 19F(d, 3He) reaction, the sum of spectroscopic factor in sd-shell is just 1.54. This suggest large uncertainty. And the s-state SF is 0.38, almost a double for 18O(3He,d) reaction, How come?
3. in 19F(d,3He) reaction, the s:d ratio is 0.4:0.6, this is similar to prediction of Nilsson model, but difference from USDB calculation.
4. If the ground state has 1d5/2 proton, why the magnetic moment are so close to free proton? the l=2 should also contribute.
5. Is neutron shell also 1s1/2 ?
6. What is the $\beta_2$ ?
7. in 20Ne, will the proton also in 1s1/2 shell? 20Ne has $\beta_2 = 0.7$ very deformed.
8. Deformed DWBA?

The following is not organised thought.

According to the standard shell ordering, on top of 18O, an extra proton should fill up the 1d5/2 shell, and then the ground state spin of 19F should be 5/2. However, the ground state spin in 1/2. This is postulated to be due to deformation [mean field calculation, β2 = 0.275], 18O core excitation, or configuration mixing state [J.P. Elhot and A. M. Lane(1957)].

Under deformation, the conventional shell ordering is not suitable and may be an invalid picture to view the nucleus. So, talk about shell ordering is non-sense.

Since the 19F is 18O + 1s1/2 proton superposed with 1d5/2, there could be deformation. The spherical shape of 19F can be seen indirectly from the magnetic dipole momentum, the value is very close to that of a free proton of 2.78284734 μ0, only difference by 0.154 μ0, or 5.5%. How to solve this contradiction?

From the study of G. Th. Kaschl et al., the spectroscopic factor of the 19F(d,3He)18Og.s. channel is 0.38. The missing 1s1/2 strength most probably can be found in the higher excitation states. This indicates the ground state of 19F is a configuration mixing state. However, they also pointed out that caution is advisable with the absolute spectroscopic factor, this could be due to imperfect DWBA calculation.

The relative spectroscopic factors for the positive parity states, which normalised to the ground state, are agree with shell model prediction in sd-shell model space suggests that the core excitation should not play an important role.

From the USDB interaction, the shell ordering is normal, but the interaction result in a 1/2+ ground state. How?

What is the nature of the low lying 1/2- excited state in 19F?

20Ne(d,3He)19F reaction can populate this low lying state, suggests the p-shell proton pickup come from the nuclear surface.

( if (12C,13N) proton pickup reaction can populate this state, then, it can be confirmed that this is a surface p-shell proton, that it could be from 2p3/2. )

## Momentum Matching in Transfer Reaction

In transfer reaction, the kinematics determines the scattering angles and momenta, but it does not tell the probability of the reaction. Not all angle have same cross section, because when a particle is being transfer to or pickup from a nucleus, the momentum has to be matched. It is like a spaceship need a proper angle and speed in order to orbit around the earth, it the incident angle or speed are not good, the spaceship will just go out or crash into the earth.

Suppose our reaction is $A(B,1)2$, $A$ is incident particle, $B$ is the target, $1, 2$ are the outgoing particles. The momentum transfer from $A$ via $2$ to $B$ or $1$ is

$\vec{q} = \vec{p}_2 - \vec{p}_A = \vec{p}_f - \vec{p}_i$

where $\vec{p}_i$ is initial momentum and $\vec{p}_f$ is final momentum.

Assume the reaction take place at the surface of nucleus $B$. The maximum angular momentum it can create is

$\vec{L} = \vec{r} \times \vec{q} \rightarrow L = r q$

In QM, the angular momentum must be $L = \sqrt{l(l+1)} \hbar$. Here we pause for unit conversion. In this blog, we usually using nuclear unit, that momentum in [MeV/c] and $\hbar = 197.327 \textrm{MeV fm/c}$, thus, to calculate the angular momentum, simple multiple the momentum in MeV/c and nuclear radius in fm. The radius of a nucleus is roughly $r = 1.25 A^{1/3} \textrm{fm}$.

So, if the momentum $\sqrt{(l+1)(l+2)} \hbar > r|q| > \sqrt{l(l+1)} \hbar$, the possible angular momentum is $\sqrt{l(l+1)} \hbar$ or the l-orbit.

In the transfer reaction, the equation for $\vec{p}_k$ and $\vec{p}_i$ are known. It is straight forward to calculate $r q / \hbar$, plot against scattering angle, and we look for the value for $\sqrt{l(l+1)}$.

Or, we can scale the y-axis by

$\displaystyle y' = \sqrt{\left(q\frac{r}{\hbar}\right)^2 + \frac{1}{4}} - \frac{1}{2}$

The new y-axis is in unit of $\sqrt{l(l+1)}$.

We can also plot the contour of momentum matching on the $p_k$ versus scatering angle. The momentum matching is

$\displaystyle q^2 = \frac{l(l+1)\hbar^2}{r^2} = p_f^2 + p_i^2 - 2 p_f p_i \cos(\theta)$

where $\theta$ is the scattering angle of particle $1$. Solve for $p_f$

$\displaystyle p_f = p_i \cos(\theta) \pm \sqrt{ \left( \frac{l(l+1) \hbar^2}{r^2} \right) - k_i^2 \sin^2(\theta) }$

We can plot the contour plot for difference $l$. and then overlap with the momentum and the scattering angle of the outgoing particle $2$.

a correction, the y-axis should be $p_2$, momentum of particle 2.

## Axial Harmonic Oscillator – Nilsson Orbit (III)

One of the problem (or difficulty) is the conversion between cylindrical quantum number $|Nn_zm\rangle$ and spherical quantum number $|Nlm\rangle$. In the limit of $\delta = 0$, all state with same $N$ has same energy and the conversion is not necessary.

With out LS coupling, the energy level can be found by the approximation

$\displaystyle E(N, n_z) = \hbar\omega \left( \frac{3}{2} + \left(1 + \frac{\delta}{3} \right) N - \delta n_z \right)$

We can see from the formula, for $\delta>0$, the largest $n_z$ has lowest energy. And the conversion to spherical harmonics can be done using projection method.

With LS coupling, the conversion becomes complicated. One issue is the coupling with the spin. Since the total angular momentum $J = L + S$ is not a conservative quantity due to deformation. In the body-fixed frame, the additional good quantum number is the z-projection of $J$, which denote as $K = |m_l \pm 1/2 |$, the quantum state becomes

$|Nn_z m_l K\rangle$

Note that each state will have only 2 degeneracy with negative $K$. On the spherical basis, the coupling with $J$ is a standard textbook content. the eigen state is

$|N l j mj \rangle = |N l j K\rangle$

The conversion get complicated, because there is no clear rule on how $| Nn_z m_l K\rangle \rightarrow |Nl j m_j \rangle$. When we want to calculate the energy level using cylindrical basis with the LS coupling, since the $J$ is not clear, even through we can write down the formula

$\displaystyle E( N, n_z, l, j) = \hbar \omega_z \left(\frac{1}{2} + n_z \right) + \hbar \omega_\rho \left(1 + N - n_z \right) \\ + a \frac{1}{2} ( j(j+1) - l(l+1) - s(s+1)) + b l^2$

It is difficult to know the $j , l$.

The go around this conversion, we can diagonal the Hamiltonian using spherical basis. The Hamiltonian is can be written as

$\displaystyle H = - \frac{\hbar^2}{2m} \nabla^2 + \frac{m\omega}{2}r^2 \left( 1 - \frac{2}{3}\sqrt{\frac{16\pi}{5}} Y_{20}(\theta, \phi) \delta \right) + a L\cdot S + b L^2$

The first 2 terms is the spherical harmonic oscillator. The energy is $\hbar\omega( 3/2 + N)$. The $Y_20$ restricts the coupling between $l, l \pm 2$ (see here.)

For example, when using the basis on $N = 1$, which are

$\displaystyle |N l j m_j \rangle = \left( |1 1 \frac{3}{2} \frac{3}{2}\rangle , |1 1 \frac{1}{2} \frac{1}{2}\rangle, |1 1 \frac{3}{2} \frac{1}{2}\rangle \right)$

at $\delta = 0.5$, the matrix is

$\begin{pmatrix} 2.633 & 0 & 0 \\ 0 & 2.233 & 0 \\ 0 & 0 & 2.233\end{pmatrix}$

The eigen values and eigen state are

$\displaystyle 2.633 , |1 1 \frac{3}{2} \frac{3}{2}\rangle$

$\displaystyle 2.233 , |1 1 \frac{1}{2} \frac{1}{2}\rangle , |1 1 \frac{3}{2} \frac{1}{2}\rangle$

The lower level has mixed $j$ !

Of course, a more complete diagonalization should be admixture from other major shell as well. When we disable the LS coupling, using 7 major shells, the Nilsson diagram look like this

This is consistence with $omega_0 = const.$, i.e. no volume conservation. With LS coupling switched on.

Although it is not so clear, we can see the state avoid “crossing”.  The s-state are unaffected by the LS coupling.

One deflect in the plot is that, the slope seem to be incorrect. And also, I tried to add volume conservation, but the result does not consistence….. (sad).

## Axial Harmonic Oscillator – Nilsson Orbit (II)

This time we will project the deformed orbit into spherical orbit. The wavefunctions are stated in here again.

$\displaystyle \Psi^D_{N n_z m}(z, \rho, \phi) = |Nn_z m\rangle_D \\ = \sqrt{\frac{1}{\alpha_z \alpha^2}}\sqrt{\frac{ n_\rho !}{2^n_z n_z! (m + n_\rho)!\sqrt{\pi^3}}} H_{n_z}\left(\frac{z}{\alpha_z}\right) \\ \exp\left(- \frac{1}{2}\left(\frac{z^2}{\alpha_z^2}+\frac{\rho^2}{\alpha^2}\right)\right) \left(\frac{\rho}{\alpha}\right)^{m} L_{n_\rho}^{m} \left(\frac{\rho^2}{\alpha^2}\right) \exp(i m \phi)$

$\displaystyle \Psi^S_{Nlm}(r, \theta, \phi) = |N l m_l\rangle_S\\ =\sqrt{ \frac{1}{\sqrt{\pi}\alpha^{2l+3}} \frac{(\frac{N-l}{2})! (\frac{N+l}{2})! 2^{N+l+2}}{(N+l+1)!}} r^l \exp\left(-\frac{r^2}{2\alpha^2}\right) L_{k}^{l+\frac{1}{2}}\left( \frac{r^2}{\alpha^2} \right) Y_{lm}(\theta, \phi)$

Since both functions span the entire space and are basis, thus, we can related them as

$\displaystyle|Nn_z m \rangle_D = \sum_{N' l' m'} C_{N n_z m}^{N' l' m'} |N' l' m' \rangle_S$

where

$latex C_{N n_z m}^{N’ l’ m’} = \langle (N’l’m’)_S|(N n_z m)_D \rangle$

First thing we notice is that $m' = m$ . Because the $phi$ components are the same in both wave function. i.e. $\int \exp(- i m' \phi) \exp(i m \phi) d\phi = 0$ if $m' \neq m$. We can omit the $m, m'$, so that $C_{N n_z}^{N' l'}$

Second thing is the parity must be the same, thus when $N$ is even (or odd), $N'$ must be even (or odd).

For non-deformed $\delta = 0$, $N' = N$,, here are some results

$\displaystyle|0 0 0 \rangle_D = |0 0 0\rangle_S$

$\displaystyle|1 1 0 \rangle_D = |1 1 0 \rangle_S$

$\displaystyle|1 0 1 \rangle_D = |1 1 1 \rangle_S$

$\displaystyle|2 2 0 \rangle_D = \sqrt{\frac{2}{3}} |2 2 0 \rangle_S - \sqrt{\frac{1}{3}} |2 0 0 \rangle_S$

$\displaystyle|2 1 1 \rangle_D = |2 2 1 \rangle_S$

$\displaystyle|2 0 2 \rangle_D = |2 2 2 \rangle_S$

$\displaystyle|2 0 0 \rangle_D = \sqrt{\frac{1}{3}} |2 2 0 \rangle_S + \sqrt{\frac{2}{3}} |2 0 0 \rangle_S$

$\displaystyle|3 3 0 \rangle_D = \sqrt{\frac{2}{5}} |3 3 0 \rangle_S - \sqrt{\frac{3}{5}} |3 1 0 \rangle_S$

$\displaystyle|3 2 1 \rangle_D = -\sqrt{\frac{4}{5}} |3 3 1 \rangle_S + \sqrt{\frac{1}{5}} |3 1 1 \rangle_S$

$\displaystyle|3 1 2 \rangle_D = |3 3 2 \rangle_S$

$\displaystyle|3 0 3 \rangle_D = |3 3 3 \rangle_S$

$\displaystyle|3 1 0 \rangle_D = \sqrt{\frac{3}{5}} |3 3 0 \rangle_S + \sqrt{\frac{2}{5}} |3 1 0 \rangle_S$

$\displaystyle|3 0 1 \rangle_D = -\sqrt{\frac{2}{5}} |3 3 1 \rangle_S - \sqrt{\frac{4}{5}} |3 1 1 \rangle_S$

For $\delta = 0.3$

$\displaystyle|0 0 0 \rangle_D = 0.995|0 0 0 \rangle_S + 0.099|2 2 0 \rangle_S - 0.015|2 0 0 \rangle_S + 0.009|4 4 0 \rangle_S + ...$

$\displaystyle|1 1 0 \rangle_D = 0.987|1 1 0 \rangle_S + 0.132|3 1 0 \rangle_S - 0.014|5 3 0 \rangle_S + 0.014|5 5 0 \rangle_S + ...$

$\displaystyle|1 0 1 \rangle_D = 0.994|1 1 1 \rangle_S + 0.108|3 3 1 \rangle_S + 0.017|3 1 1 \rangle_S + 0.011|5 5 1 \rangle_S + ...$

$\displaystyle|2 2 0 \rangle_D = 0.790|2 2 0 \rangle_S - 0.564|2 0 0 \rangle_S - 0.153|4 2 0 \rangle_S + 0.139|4 4 0 \rangle_S + ...$

$\displaystyle|2 1 1 \rangle_D = 0.986|2 2 1 \rangle_S + 0.158|4 4 1 \rangle_S - 0.052|4 2 1 \rangle_S - 0.012|6 4 1 \rangle_S + ...$

For $\delta = 0.6$

$\displaystyle|0 0 0 \rangle_D = 0.958|0 0 0 \rangle_S + 0.258|2 2 0 \rangle_S - 0.084|2 0 0 \rangle_S + 0.061|4 4 0 \rangle_S + ...$

$\displaystyle|1 1 0 \rangle_D = 0.885|1 1 0 \rangle_S + 0.319|3 3 0 \rangle_S - 0.274|3 1 0 \rangle_S - 0.119|5 3 0 \rangle_S + ...$

$\displaystyle|1 0 1 \rangle_D = 0.955|1 1 1 \rangle_S + 0.281|3 3 1 \rangle_S + 0.078|5 5 1 \rangle_S - 0.043|5 3 1 \rangle_S + ...$

$\displaystyle|2 2 0 \rangle_D = 0.598|2 2 0 \rangle_S - 0.45|2 0 0 \rangle_S - 0.379|4 2 0 \rangle_S + 0.299|4 4 0 \rangle_S \\ - 0.259|0 0 0 \rangle_S + 0.198|4 0 0 \rangle_S + 0.173|6 2 0 \rangle_S - 0.168|6 4 0 \rangle_S + ...$

$\displaystyle|2 1 1 \rangle_D = 0.882|2 2 1 \rangle_S + 0.381|4 4 1 \rangle_S - 0.191|4 2 1 \rangle_S - 0.129|6 6 1 \rangle_S \\ - 0.114|6 4 1 \rangle_S + ...$

We can see, more deform, more higher angular momentum states are involved.

Also, for a pure state when non-deform, the mixing is still small.

## Axial Harmonic Oscillator – Nilsson Orbit

The Hamiltonian is

$\displaystyle H = -\frac{\hbar}{2m}\nabla^2 +\frac{m}{2}(\omega_\rho^2(x^2+y^2)+\omega_z z^2 )$

Use cylindrical coordinate, the Schrodinger equation is

$\displaystyle \left(-\frac{\hbar}{2m}\left(\frac{d^2}{dz^2} + \frac{1}{\rho}\frac{d}{d\rho}\left(\rho\frac{d}{d\rho}\right) + \frac{1}{\rho^2}\frac{d^2}{d\phi^2}\right) +\frac{m}{2}(\omega_\rho^2(\rho^2)+\omega_z z^2 ) \right) \Psi = E \Psi$

Where the energy is

$\displaystyle E = \hbar\omega_\rho(1+ n_x + n_y) + \hbar\omega_z\left(\frac{1}{2}+ n_z\right)$

Note that $n_x+n_y \neq n_\rho$. One of the reason is there are 2 degree of freedom, it cannot be solely expressed into 1 parameter.

Set $\Psi = Z P \Phi$,

$\displaystyle -\frac{\hbar}{2m}\left(\frac{d^2Z}{dz^2} P \Phi + \frac{1}{\rho}\frac{d}{d\rho}\left(\rho\frac{dP}{d\rho} \right) Z \Phi + \frac{1}{\rho^2}\frac{d^2\Phi}{d\phi^2} Z P\right) \\ + \frac{m}{2}\omega_\rho^2\rho^2 ZP\phi+\frac{m}{2}\omega_z z^2 ZP\Phi= E ZP\Phi$

$\displaystyle -\frac{\hbar}{2m}\left(\frac{d^2Z}{dz^2} \frac{1}{Z} + \frac{1}{\rho}\frac{d}{d\rho}\left(\rho\frac{dP}{d\rho} \right) \frac{1}{P} + \frac{1}{\rho^2}\frac{d^2\Phi}{d\phi^2} \frac{1}{\Phi}\right) \\ + \frac{m}{2}\omega_\rho^2\rho^2 +\frac{m}{2}\omega_z z^2 = E$

The angular part, we can set

$\displaystyle \frac{d^2\Phi}{d\phi^2} \frac{1}{\Phi} = -m_\phi^2$

The solution is $\Phi(\phi) = \exp(i m_\phi \phi)$

The z-part is usual 1D Harmonic oscillator

Thus the rest is

$\displaystyle -\frac{\hbar}{2m}\left( \frac{1}{\rho}\frac{d}{d\rho}\left(\rho\frac{dP}{d\rho} \right) \frac{1}{P} - \frac{m_\phi^2}{\rho^2}\right) + \frac{m}{2}\omega_\rho^2\rho^2 = \hbar\omega_\rho(1+ n_x + n_y)$

rearrange

$\displaystyle -\frac{\hbar}{2m}\left( \frac{1}{\rho}\frac{d}{d\rho}\left(\rho\frac{dP}{d\rho} \right) - \frac{m_\phi^2}{\rho^2}P\right) + \frac{m}{2}\omega_\rho^2\rho^2 P - \hbar\omega_\rho(1+ n_x + n_y) P = 0$

Set a dimensionless constant $\alpha^2 = \frac{\hbar}{m \omega_\rho}$, and $x = \rho/\alpha$

$\displaystyle \frac{1}{x}\frac{d}{dx}\left(x\frac{dP}{dx}\right) + \left( 2(1+n_x+n_y) - \frac{m_\phi^2}{x^2} - x^2 \right) P = 0$

Using the normalisation formula, $\int P^2 x dx = 1$, set $u = P \sqrt{x}$

$\displaystyle x^2\frac{d^2u}{dx^2}+ \left( 2(1+n_x+n_y) - \frac{m_\phi^2-\frac{1}{4}}{x^2} - x^2 \right) u = 0$

Because of the long-range and short-range behavior, set

$\displaystyle u(x) = f(x) x^{m_\phi+\frac{1}{2}} \exp\left(-\frac{x^2}{2}\right)$

$\displaystyle x\frac{d^2f}{dx^2} + (1+2m_\phi-2x^2)\frac{df}{dx} + 2(n_x+n_y-m_\phi)x f= 0$

Set $y = x^2$

$\displaystyle y \frac{d^2f}{dy^2} + (m_\phi+1-y)\frac{df}{dy} + \frac{n_x+n_y-m_\phi}{2} f= 0$

Define $n_\rho = \frac{n_x+n_y-m_\phi}{2} \rightarrow n_x+n_y = 2 n_\rho + m_\phi = N - n_z$.

$\displaystyle y \frac{d^2f}{dy^2} + (m_\phi+1-y)\frac{df}{dy} + n_\rho f= 0$

This is our friend again! The complete solution is

$\displaystyle \Psi_{n_z n_\rho m_\phi}(z, \rho, \phi) \\ = \sqrt{\frac{1}{\alpha_z \alpha^2}}\sqrt{\frac{ n_\rho !}{2^n_z n_z! (m_\phi + n_\rho)!\sqrt{\pi^3}}} H_{n_z}\left(\frac{z}{\alpha_z}\right) \\ \exp\left(- \frac{1}{2}\left(\frac{z^2}{\alpha_z^2}+\frac{\rho^2}{\alpha^2}\right)\right) \left(\frac{\rho}{\alpha}\right)^{m_\phi} L_{n_\rho}^{m_\phi} \left(\frac{\rho^2}{\alpha^2}\right) \exp(i m_\phi \phi)$

where $\alpha_z^2 = \hbar/m/\omega_z$

The energy is

$\displaystyle E = \hbar\omega_z \left(\frac{1}{2} + n_z \right) + \hbar\omega_\rho \left( 2n_\rho + m_\phi + 1 \right)$

The quantum number $m_\phi$ has same meaning as $m_l$ in spherical case.

The notation for the state is

$|Nn_z m_l \rangle$

with the spin, the only good quantum number is the z-component of the total angular momentum $J = L+S$  long the body axis, i.e. $K = m_\phi \pm 1/2$, thus, the state is

$|Nn_z m_l K \rangle$

Note that the total angular momentum $J^2$ is not a good quantum number in deformation, as the rotational symmetry is lost. However, the quantum number $K$ is linked with the angular momentum of the Nilsson single particle orbit. It is because when a particle has $m_j = K$, the angular momentum must at least $j \geq K$.

The above is a general solution for the harmonic oscillator in cylindrical coordinate. When $\omega_z = \omega_\rho$, it reduce to spherical case.

According to P. Ring & P. Schuck (2004) (The Nuclear Many-Body Problem, P.68), the Hamiltonian can be expressed as a quadruple deform field by setting

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

$\displaystyle H = -\frac{\hbar^2}{2m}\nabla^2 + \frac{1}{2} m \omega^2 \left(r^2- \frac{2}{3}\sqrt{\frac{16\pi}{5}} \delta r^2 Y_{20}(\theta,\phi) \right)$

This Hamiltonian has similarity with the deformation

$\displaystyle R(\theta, \phi) = R_0 \left(1 + \alpha_00 + \sum_{\lambda=1}^{\infty}\sum_{\mu=-\lambda}^{\lambda} \alpha_{\lambda \mu} Y_{\lambda \mu}(\theta, \phi) \right)$

take the first quadruple term, and calculate $R^2(\theta, \phi)$

$R^2(\theta, \phi) = R_0^2 (1 + 2 \beta Y_{20}(\theta,\phi) )$

Compare, we have

$\displaystyle \beta = \frac{1}{3}\sqrt{\frac{16\pi}{5}}\delta \approx 1.05689 \delta$

The ratio

$\displaystyle \frac{\omega_z}{\omega_\rho} = \left(\frac{\alpha_\rho}{\alpha_z} \right) = \sqrt{\frac{1-\frac{4}{3}\delta}{1+\frac{2}{3}\delta}} = 1- \delta + \frac{1}{6}\delta^2 - \frac{5}{18}\delta^3...$

We need volume conservation

$\omega_x \omega_y \omega_z = \omega_0^3$

Thus,

$\displaystyle \omega = \omega_0 \left(\frac{1}{(1+2/3\delta)(1-4/3\delta)} \right)^{\frac{1}{6}} \approx \omega_0 \left(1+\frac{2}{9}\delta^2 + \frac{8}{81}\delta^3 \right)$

The energy is

$\displaystyle E = \hbar\omega \left( \sqrt{1-\frac{4}{3}\delta}\left(\frac{1}{2} + n_z \right) + \sqrt{1+\frac{2}{3}\delta} \left( N - n_z + 1 \right) \right) \\ \approx \hbar \omega_0 \left( \frac{3}{2} + \left(1 + \frac{1}{3}\delta\right) N - \delta n_z \right)$

Here is some plots with various $\delta = 0, 0.3, 0.5$

Next time, I will add the LS coupling and L-square term to recreate the Nilsson diagram. Also I will expand the solution from cylindrical coordinate into spherical coordinate. This unitary transform is the key to understand the single particle-ness.

## 3D Harmonic Oscillator – Cartesian to Spherical Coordinate

We have two solutions for the 3D harmonic oscillator, one is in Cartesian coordinate, and the other is in spherical coordinate.

$\Psi^C_{n_x n_y n_z}(x,y,z) = N_C H_{n_x}(x)H_{n_y}(y)H_{n_z}(z) \exp(-r^2/2)$

$\Psi^S_{n, l, m}(r,\theta,\phi) = N_S r^l \exp(-r^2/2) L_{\frac{n-l}{2}}^{l+\frac{1}{2}}(r^2)$

We can project the $\Psi^C$ onto $\Psi^S$ or,

$\displaystyle \Psi^C_{n_x n_y n_z} = \sum_{l,m} C_{lm} \Psi^S_{nlm}$

Replace $(x,y,z) \rightarrow (r, \theta, \phi)$ in the integration.Here are some results

$n = 0 \rightarrow C_{00} = 1$

$(n_x, n_y, n_z) = (0,0,1) \rightarrow C_{1,0} = 1$

$(n_x, n_y, n_z) = (1,0,0) \rightarrow C_{1,-1} = 1/\sqrt{2} , C_{1,1}= -1/\sqrt{2}$

$(n_x, n_y, n_z) = (0,1,0) \rightarrow C_{1,-1} = -i/\sqrt{2} , C_{1,1}= i/\sqrt{2}$

$(n_x, n_y, n_z) = (0,0,2) \rightarrow C_{0,0} = -1/\sqrt{3} , C_{2,0}= \sqrt{2/3}$

$(n_x, n_y, n_z) = (1,0,1) \rightarrow C_{2,-1} = 1/\sqrt{2} , C_{2,1}= 1/\sqrt{2}$

$(n_x, n_y, n_z) = (1,1,0) \rightarrow C_{2,-2} = -i/\sqrt{2} , C_{2,2}= i/\sqrt{2}$

$(n_x, n_y, n_z) = (0,0,3) \rightarrow C_{1,0} = -\sqrt{3/5} , C_{3,0}= \sqrt{2/5}$

$(n_x, n_y, n_z) = (0,0,4) \rightarrow C_{0,0} = \sqrt{1/5} , C_{2,0}= -\sqrt{4/7} , C_{4,0}= \sqrt{8/35}$

## Radial Density of 3D Harmonic Oscillator

From the previous post, we have the radial formula for the 3D harmonic oscillator.

$\displaystyle R_{nl}(r) \\ =\sqrt{ \frac{1}{\sqrt{\pi}\alpha^{2l+3}} \frac{(\frac{n-l}{2})! (\frac{n+l}{2})! 2^{n+l+2}}{(n+l+1)!}} r^l \exp\left(-\frac{r^2}{2\alpha^2}\right) L_{k}^{l+\frac{1}{2}}\left( \frac{r^2}{\alpha^2} \right)$

For $\alpha = 1 fm$, the $R^2$ are plotted below.

Assume all orbits are fully filled.

$\displaystyle \rho_n = \sum_{l}R^2_{nl} (2l+1)$

The red is $n=0$, the pink is $n = 1$, the blue is $n = 2$, the cyan is $n = 3$, and the green is $n = 4$.

Next, we are summing all the shells to get the nucleon density.

In here, the red is the 4He nucleon density function, the pink is the nucleus with all 1p-shell are filled, that is 16O. The blue is the sd-shell are filled, that is 40Ca. The cyan is fp-shell filled.

We can see some systematic trend there. The $n = odd$ nuclei has shape, which is somewhat similar to the Wood-Saxon shape. But for the $n=even$ nuclei, the shape is like a V.

Note that the $\alpha$ is fixed in this calculation. In fact, from Samuel Wong (2004) (Introductory Nuclear Physics),

$\displaystyle \hbar \omega \approx 41 A^{-1/3} \textrm{MeV}$

$\displaystyle \alpha = \sqrt{\frac{\hbar}{m \omega}} = \sqrt{\frac{\hbar^2 c^2}{mc^2 \hbar \omega}} \approx A^{1/6} \textrm{fm}$