On the microscopic origin of deformation + paper reading

February 4, 2023

GoLuckyRyan Basic , , , , , , , Leave a comment

Nuclear deformation is usually mimicked by macroscopic deformation potential, for example, a quadrupole deformation potential as in the Nilsson model. This provides a description without understanding the origin of the deformation potential. This macroscopic description is particularly unsatisfactory for the deformed light nuclei. Also, it does not answer the question that, are the whole nuclei deformed? or just the surface?

The paper Towards a unified microscopic description of nuclear deformation by P. Federman and S. Pittel [Physic Letters B 69, 385 (1977)] provides an interesting and reasonable understanding of the microscopic description.

The paper starts by comparing 20Ne and 20O. 20Ne is well deformed with \beta_2 = 0.7 and 20O is less deformed with \beta_2 = 0.27. Let’s assume the 16O core is inert in both nuclei. The difference between 20Ne and 20O is two protons in the sd-shell are replaced with two neutrons from 20Ne to 20O. The paper said, “The conclusion seems clear. Deformation in light nuclei is due to the T=0 neutron-proton interaction.“.

As we know from the two-nucleon system, there are 4 possible combinations with 2 nucleons. In those combinations, the T=0, J=1 pn pair is the only bound pair due to the tensor force. And the pair is not spatial isotropic. In a many-nucleons nucleus, a similar thing happens for NN pair. In 20O, and 20Ne, all valence nucleons are in d5/2-orbitals in the simplest picture. A T=0 pn pair, the spin of the proton and neutron must be aligned, which means the proton and neutron are orbiting in the same direction, and the total spin of the pn pair is J = 5, which is very spatially deformed. And in 20O, all valence nucleons are neutrons, only able to form T=1, J=0 nn pairs, in which the neutrons are orbiting oppositely and spatially spherical. The result is 20Ne is deformed and 20O is spherical, and the ultimate reason is the tensor force tends to make T=0 pn pair.

Wait… the total spin of 20Ne and 20O are both Zero. If it is the T=0 pn pair and the spin is not Zero, would the spin of 20Ne be non-zero? For even-even nuclei, protons and neutrons are paired up and formed J=0 pp and nn pair. Where are the T=0 pn pairs? The key is that, although the pp and nn are paired up, it does not exclude the T=0 pn pair.

For simplicity, let’s check the Slater determinate for 3 fermions. Suppose the 3 nucleon wave functions are p_a, p_b, n_a, where p, n stands for proton and neutron, \alpha, \beta are spin-up and spin-down. And all wavefunctions are in the same orbital.

\displaystyle \Psi = \frac{1}{3!} \begin{vmatrix} p_a(1) & p_b(1) & n_a(1) \\ p_a(2) & p_b(2) & n_a(2) \\ p_a(3) & p_b(3) & n_a(3) \end{vmatrix}

without loss of generality, we can collect terms of the 1-th particle.

\displaystyle 3 \Psi = p_a(1) \frac{p_b(2) n_a(3) - p_b(3) n_a(2) }{2} + \\ ~~~~~~~~p_b(1) \frac{p_a(2) n_a(3) - p_a(3) n_a(2) }{2} + \\~~~~~~~~ n_a(1) \frac{p_a(2) p_b(3) - p_a(3) p_b(2) }{2}

\displaystyle 3 \Psi = p_a(1) ( pn, T=1, J = 0 ) + \\~~~~~~~~~ p_b(1) (pn, T=0, J =1) +\\~~~~~~~~~ n_a(1) (pp, T=1, J= 0)

Now, imagine it is the triton wave function. We can see that the total wave function contains a pp J=0 pair, but there are also pn J=0 and J=1 pair. The detailed coupling of the total spin of the total wave function involves CG coefficient, and the pn T=0, J=1 pair should be coupled to a spin-down proton (j=1/2) and form J = 1/2.

We can imagine that in a wavefunction with 2 protons and 2 neutrons, there will be T=0 pn pair and T=1 pp/nn pair. While a wavefunction with 4 neutrons can never form pn pair.

In this simple demonstration, it is simply forming the wave function without considering the nuclear force. There are two questions: 1) In the Slater determinate, what is the percentage for T=0 and T=1 NN pairs? 2) with the nuclear force, what is that percentage?

The paper Probing Cold Dense Nuclear Matter by R. Subedi et al., Science 320, 1476 (2008), could give us some hints. The study found that in 12C, there are 18% pn pairs and only 2% of pp or nn pairs. There are more recent developments on the topic, for example, PRL 121, 242501 (2018), PLB 820, 10 (2021).

P. Federman and S. Pittel apply the same idea (deformation caused by T=0 pn pair from tensor force) on 100Zr, a very different and complex nucleus than 20Ne. 98Zr has Z = 40 and N = 58 and is spherical. The proton shell is semi-closed at 1p1/2 orbital. and the neutron shell is semi-closed at 2s1/2 ( on top of N=40, 0g9/2, 1d5/2, 2s1/2 ). But 100Zr is highly deformed with \beta_2 \approx 0.35 . The nuclear shape changed so much by just 2 neutron differences has drawn a lot of attention since its discovery in the 70s. The Interacting Boson Model and Shell model calculation has been tried to compute this sudden transition with quite good results, but there are still many discrepancies on the microscopic origin of the deformation. For example, is the deformation driven by protons or neutrons? and also what is the configuration.

As we mentioned before, the T=0 pair could be causing the deformation. In the case of 100Zr, it is the g-orbital pn pairs. Other proposed mechanisms are core polarization of 98Zr and the presence of a valence neutron in 0h11/2 orbital. In a recent experimental proposal, I wrote the following:

The above three mechanisms are intertwined. The interplay between these mechanisms is illustrated in Figure. 1. The neutrons in the 0g7/2 orbital lower the proton 0g9/2 binding energy while increasing the binding energy of the proton 1p1/2 and 0f5/2 with the action of the tensor force (attractive for J< − J> pair and repulsive for J< − J< or J> − J> pair), which reduces or even breaks the pf-g Z = 40 shell-gap, favoring the promotion of the protons to the 0g9/2 orbital from the pf-shell, and creates a core polarization and deformation. The core polarization of Z = 40 core promotes protons into the π0g9/2 orbital, enabling the coupling with the 0g7/2 neutron and forming T = 0 p-n pairs under the influence of tensor interaction. The g-orbital T = 0 p-n pair have their spins aligned and create a strong quadrupole deformation. Also, the presence of 0g9/2 protons lowers the effective single-particle energies (ESPEs) of the neutron 0g7/2 and 0h11/2 orbital via the so-called Type II shell evolution, which increases the occupancy for the neutron 0h11/2 orbital. The presence of 0h11/2 neutrons in turn provides a strong quadrupole deformation force. The deformation then enhances the occupation of valence orbitals and fragmentation in single-particle energies in return.

This is a bit complicated and hard to prove. But the tensor force plays an important role here. Without such, the chain reaction of the nucleon reconfiguration would not happen.

\beta_2 values around 100Zr.

We can see the A=100 nuclei, 100Mo has 2 protons at 0g9/2 and N=58, it is deformed with \beta_2 = 0.23. A deformation could promote neutrons to 0g7/2. 102Mo has 2 protons at 0g9/2 and 2 neutron at 0g7/2, so it is deformed with \beta_2 = 0.31 (Would 102Mo be more deformed?) 102Ru has 4 protons at 0g9/2, and N = 58. It is slightly deformed with \beta_2 = 0.17. 102Pd has 6 proton at 0g9/2 and N = 54 with \beta_2 = 0.14. In the opposite direction, 100Sr (Z=38, N=62) is very deformed with \beta_2 = 0.41.

Also, Z = 38 or Z = 42, all isotopes are deformed. Clearly, Z = 40 and N < 60 are NOT deformed and are the ANOMRALY. The shell Z = 40 closure clearly forbids proton-shell configuration mixing.

In the above, we always use \beta_2 as a measure or indicator for deformation. But the \beta_2 of 16O is 0.35. Is 16O not spherical? That is exactly the reason why deformation is “hard” to understand for light nuclei. In my opinion, 16O is not spherical, as \beta_2 is pretty much a measure of the geometrical shape. However, 16O is shell-closure and has almost no configuration mixing (there are ~ 10% sd-shell components). And the shape deformation is caused by the T=0 pn pairs. However, where is the rotational band of 16O? Another thing is, in light nuclei, deformation, and configuration mixing can be separated. Configuration mixing will lead to deformation, but the reverse is not always true.

In fact, all light nuclei are more or less deformed!! Within the sd-shell, beta_2 > 0.1 and the most spherical nuclei is 40Ca. Also also, I think all even-odd nuclei are deformed as the unpaired nucleon has a deformed orbital.

The known smallest beta_2 nucleus is 206Pb with a value of 0.03.

In the case of 20Ne, the deformation could have different effects on different orbitals. i.e. the mean field for each nucleon could be different.

How to apply this idea? and how to predict the degree of deformation? Is it the only mechanism?

I should calculate the beta_2 for each orbital….

Spin-spin (Tensor) interaction

October 14, 2020

GoLuckyRyan Basic , , , , 9 Comments

The spin-spin interaction, also called as tensor interaction, as the strength depends on the relative direction of the spins. The mathematical form of the interaction is

\displaystyle V_T(r) = \frac{1}{r^3} \frac{3(\vec{\sigma_1}\cdot\vec{r_1})(\vec{\sigma_2}\cdot\vec{r_2})-r^2(\vec{\sigma_1}\cdot\vec{\sigma_2})}{r^2}

For spin-1/2, we have.

\displaystyle \vec{\sigma} = \left(\sigma_x, \sigma_y, \sigma_z \right) = \left( \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & i \\ -i & 0 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \right)

The \vec{\sigma} is also called Pauli matrices.

The term \vec{\sigma}\cdot \vec{r} , we can expand,

\displaystyle \vec{\sigma}\cdot \vec{r} = \sigma_x x + \sigma_y y + \sigma_z z = \begin{pmatrix} z & x + i y \\ x-iy & -z \end{pmatrix}

If we express in spherical coordinate,

\displaystyle \vec{\sigma}\cdot \vec{r} =r \begin{pmatrix} \cos(\theta) & \sin(\theta)e^{i\phi} \\ \sin(\theta)e^{-i\phi} & - \cos(\theta) \end{pmatrix}

The eigen values are 1 and -1, with eigen vectors are

\displaystyle v_{-1} = (\sin(\theta), -e^{-i\phi}\cos(\theta) )

\displaystyle v_{1} = (\cos(\theta), e^{-i\phi}\sin(\theta) )

with

\displaystyle v_i \cdot v_j^* = \delta_{ij}

The v_{1} is a “condensed” vector of the \vec{r'} = (\cos(\theta), \sin(\theta) \sin(\phi), \sin(\theta) \cos(\phi)) = (z, y, x) . Since only the the z-component is known, the x,y-components condensed as a phase \phi.

The term (\vec{\sigma_1}\cdot\vec{r_1})(\vec{\sigma_2}\cdot\vec{r_2}) is a tricky one. It involves 2 spins in two spaces. The product has 9 terms, \sigma_1(i) \sigma_2(j) r_i r_j . The \sigma_1 and \sigma_2 act on particle-1 and particle-2 separately. In fact, when we talk about a 2-spin system, we can form a direct-product space, \sigma_1(i) \sigma_2(j) = \sigma_1(i) \otimes \sigma_2(j) , in matrix from, for example

\sigma_1(x) \otimes \sigma_2(y) = \begin{pmatrix} 0 & \sigma_y \\ \sigma_y & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & i \\ 0 & 0 & -i & 0 \\ 0 & i & 0 & 0 \\ -i & 0 & 0 & 0 \end{pmatrix}

The states in the direct-product space are

\displaystyle \left|\uparrow \uparrow \right> = (1, 0, 0, 0)

\displaystyle \left|\uparrow \downarrow \right> = (0, 1, 0, 0)

\displaystyle \left|\downarrow \uparrow \right> = (0, 0, 1, 0)

\displaystyle \left|\downarrow \downarrow \right> = (0, 0, 0, 1)

So, the eigen states for total spin 1 are (1, 0, 0, 0), \frac{1}{\sqrt{2}}(0, 1, 1, 0), (0,0,0,1) , and for total spin 0 are \frac{1}{\sqrt{2}} (0, 1, -1, 0) . We set

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

In similar fashion, the term \vec{\sigma_1}\cdot \vec{\sigma_2} = \sigma_1(x)\otimes\sigma_2(x) + \sigma_1(y)\otimes\sigma_2(y)+\sigma_1(z)\otimes\sigma_2(z)

\displaystyle \vec{\sigma_1}\cdot \vec{\sigma_2} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 2 & 0 \\ 0 & 2 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}

Combine everything, we have

\displaystyle \frac{3(\vec{\sigma_1}\cdot\vec{r_1})(\vec{\sigma_2}\cdot\vec{r_2})-r^2(\vec{\sigma_1}\cdot\vec{\sigma_2})}{r^2} =\\ \frac{1}{r^2}\begin{pmatrix} 3z^2-r^2 & 3z(x+iy) & 3z(x+iy) & 3(r^2 - 2y^2-z^2+2ixy) \\ ... & r^2 - 3z^2 & r^2-3z^2 & -3z(x+iy) \\ ... & ... & r^2-3z^2 & -3z(x+iy) \\ ... & ... & ... & 3z^2 - r^2 \end{pmatrix}

Next, it is easy to calculate the matrix elements of the spin-spin interaction with the states \chi.

\displaystyle \left<\chi| V_T | \chi \right> = \frac{1}{r^2}\begin{pmatrix} 3z^2-r^2 & 3\sqrt{2} z(x+iy) & 3(x+iy)^2 & 0 \\ ... & 2(r^2 - 3z^2) & -3\sqrt{2}z (x+iy) & 0 \\ ... & ... & 3z^2-r^2 & 0 \\ ... & ... & ... & 0 \end{pmatrix}

We can see that, all matrix elements with spin-0 are 0. It is what we expect, as the spin-0 result no tensor force, as no spin in the system. For the diagonal elements:

\displaystyle \left< \chi_{Sm_S}| V_T | \chi_{Sm_S} \right> = \begin{cases} 2\left(1- 3 \frac{z^2}{r^2} \right), & S=1, m_S= 0 \\ 3 \frac{z^2}{r^2} -1, & S=1, m_S=\pm 1\\ 0, & S=0 \end{cases}

We can also calculate the matrix element of the \vec{\sigma_1}\cdot \vec{\sigma_2},

\left< \chi | \vec{\sigma_1}\cdot \vec{\sigma_2} |\chi \right> = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -3 \end{pmatrix}

we can also calculate in other way. First, we convert to the spinor that we familiar,

s_i = \frac{1}{2} \sigma_i

where the eigen value are 1/2, which is the correct eigen value for spin-1/2 system. And set

S = s_1 + s_2

so that the eigen values are 1 and 0.

\vec{\sigma_1}\cdot \vec{\sigma_2} = 4 s_1 \cdot s_2 = 2\left( S^2 - s_1^2 - s_2^2 \right)

Thus, we get

\displaystyle \left< \chi| \vec{\sigma_1}\cdot \vec{\sigma_2} | \chi\right> = \begin{cases} 1, & S=1\\ -3, & S=0 \end{cases} = 4 S - 3

This agrees with the matrix method.

The diagonal result can be written in spherical coordinate,

\displaystyle \left< \chi_{Sm_S}| V_T | \chi_{Sm_S} \right> = \begin{cases} 2\left(1- 3 \cos^2(\theta) \right), & S=1, m_S= 0 \\ 3 \cos^2(\theta) -1, & S=1, m_S=\pm 1\\ 0, & S=0 \end{cases}

We can see the angular dependent of the interaction. Let plot the 3 \cos^2(\theta) -1 ) for m_S = \pm 1 in below

At \theta = 0, \pi the force is strongest. This can picture as two magnets, aligned to the z-axis (m_S=\pm1). When the are aligned tail to head, or \theta = 0, \pi, the attraction is highest. At \theta = \pi/2 , when their head-to-head, tail-to-tail, they are repulsive.

Consider this distortion effect in deuteron. The proton and neutron couples to spin-1 and both are in the s-orbitals. Due to the tensor force, this s-orbital is no longer an eigenstate and it squeezes the orbital along the z-axis due to attraction and expands the x-y plan due to the repulsion. The total wave function would be oblate. Thus, the total wave function has to be mixed with the next orbital with the same parity, which is the d-orbital. And this is the case for m_s = \pm1.

For the case of m_s = 0 , the situation reversed, the tensor force expands the wavefunction along the z-axis and squeezes it on the x-y plane, making the total wavefunction be prolate. And since the magnitude is 2 times stronger for the m_s = 0 state than that of the m_s = \pm1 state. The deformation would be stronger.

[update 20230718]

In the case the total spin S= 1, m_S=0, the situation is like this:

This geometrical illustration explained why the tensor force change sign.

However, the x-y component of the total spin vector is unknown, so, the total spin can be rotated 90 degrees and look like this:

In this case, all tensor force is repulsive. Thus, the 2-spin system is not an axial deformed but also deformed on the x-y plane. However, after taking an average of all possible rotations, the strength on the x-y plane should be zero, and the deformation is axial.

We can see the tensor force, no matter how the spin is orientated, whenever tip to toe, it is attractive. whenever in parallel, repulsive. The m_S or the z-axis is artificially imposed so that we can do the calculation and make the tensor force operational. Or to say, the tensor force is rotationally symmetric.

The reason why the strength for the m_s = 0 case is twice that of the m_s = 1 is due to the internal degree of freedom on the x-y plane. When we sum up the strength, for m_s = 1, \theta = 0, \phi = 0, only 1 configuration is possible and has a strength of 2, but for m_s = 0, there are 4 configurations that have a strength of -1 each and sum up to -4. Similar for the \theta = \pi/2 as illustrated below.

But the problem is, the spin can rotate in any direction on the x-y plane, not just 4. Should it be 2\pi more instead of 2? Any better explanation?

In this post, the radial component of the tensor force AV18 p-n potential is negative, which explained why most nuclei tend to be prolated (longer in the z-axis) for axial deform.

Since different m_s states should degenerate without any external magnetic field. So that the different m_s states should be equally occupied. And the strength of the m_s = 0 state is twice that of the m_s = \pm 1 states, thus, the deformation effect along the z-axis of the tensor force should be canceled, but not on the x-y plane. As we discussed, for the m_s = 0 state, the x-y plane has no force on average due to rotation, but that is not true for the m_s = \pm1 states, the x-y plane is always attractive ( given the radial tensor force is negative), thus, the x-y plane is contracted, and make the whole nuclei prolate.

Alpha cluster and alpha separation energy

September 13, 2020

GoLuckyRyan Basic , , , , , Leave a comment

The alpha separation energy is the energy to add int a nucleus, so that it will break up into an alpha-particle and the rest of the nucleus. If we define the mass of a nucleus with mass number A and charge number Z as M(Z,A), the alpha-separation energy is,

\displaystyle S_\alpha(Z,A) = M(Z-2,A-4) + M(\alpha) - M(Z,A) .

The following plot the nuclides chart for alpha-separation energy,

The chart can be divided in 2 regions. In the upper region, nuclei have negative alpha-separation energies, i.e. nucleus gives out energy when emitting an alpha particle, thus, they are alpha-emitter. In the lower region, the alpha-separation energies are positive. And we can see that there are some local minimum for the S_\alpha.

According the alpha-decay theory, alpha-particle should be formed inside a nucleus before it tunnels through the nuclear potential and get out. the formation of the alpha-particle is described as the preformation factor.

It seems that, the alpha-separation energy, somehow, relates to the preformation factor.

The alpha cluster is studied for the many light nuclei, particularly on the 8Be, 12C, 16O, 20Ne, 24Mg, 28Si, 32S, 36Ar, 40Ca, and 44Ti. [Ref??]

From the above plot, it seems that the alpha-separation energies have no correlation with the magic number, and also no correlation with the Z=even. In a naive imagination, alpha-cluster could appear at all Z=even, A = 2Z nuclei. A trivial example is the the 8Be, its alpha-separation energy is -0.09 MeV and it will decay or split into 2 alpha particles. In 12C, the Hoyle state at 7.7 MeV is a triple-alpha state, note that S_\alpha(12C) = 7.37 MeV.

When we use the shell model to look at 8Be and 12C, we will found that the alpha-cluster is “not” making any sense. The protons and neutrons occupy the 0s1/2 and 0p3/2 orbitals. Since 9Be is stable and ground state spin is 3/2. A 8Be nucleus can be obtained by removing a 0p3/2 neutron. We may guess that, the 4 nucleons at the 0s1/2 orbital, which is an alpha-particle, somehow, escape from the 8Be, leaving the 4 0p3/2 nucleons behind. And the 4 0p3/2 nucleons “de-excite” back to the 0s1/2 shell and becomes another alpha-particle. But it seems that it does not make sense. Also, How to use shell model to describe the triplet-alpha cluster?

Above is the the alpha-separation energy for light nuclei. We can see that, there are few local minimum around 8Be, 20Ne, 40Ca, and 72Kr. Near 20Ne, the 18F, 19F, and 19Ne are having small alpha-separation energies, relative to the near by nuclei. For 20Ne, it could be understand why the alpha-separation energy is relatively smaller, as the 20Ne can be considered as 16O core with 4 nucleons in sd-shell. And 20Ne is well deformed that, there is a chance all 4 nucleons are in the 1s1/2 and form a quai-alpha particle. But the situation is a bit strange for 19Ne, 19F, and 18F. In order to form an alpha-cluster, or a quasi-alpha-particle, one or two nucleon from the p-shell has to excited to sd-shell. But the situation may be even more complicated. For some heavy alpha emitter, the s-orbital nucleons are tiny fraction compare to the rest of the nuclei, so, how the alpha-particle is formed before the decay is still unknown. This suggests that the involvement of s-orbital is not needed in alpha formation.

S_\alpha(19F) = 4.014 MeV, and there are many states around 4 MeV in 19F, I am wondering, one of these state is alpha cluster? If so, 19F(d,d’) reaction could excite those states and we will observed 15N + alpha. [need to check the data]. Similar experiment could be done on 18F, 19Ne, and 20Ne. If it is the case, that could provides some information on the alpha formation.

[need to check the present theory of alpha formation]

Rotation of deformed nucleus

July 8, 2020

GoLuckyRyan Basic , Leave a comment

The total Hamiltonian of a nuclear system must be rotational invariant because the state should not have a preferred direction. Thus, the total spin J of the wave function is a good quantum number.

There is a tricky thing should point out here. The spherical harmonic clearly has a preferred direction, it rotates around the “z-axis”. Since this “z-axis” is arbitrary, the wave function (or the spherical harmonic) still has no special preference and can point to any direction. i.e., the energy is the same in all directions. The nuclear spin is also the same. Nuclear spin is a mini magnetic dipole that is direction. The nuclear spin wave function is non-rotation symmetry but its energy is the same for any direction under a Hamiltonian without a magnetic field.

For axial deformed J = j = 0 nucleus, the nucleus is not rotated at the ground state and can orient in any direction. When this nucleus rotates, it has to rotate perpendicular to the body axis, because it is the only principal axis that exhibits a rotation (a stable rotation). And since the rotation should preserve parity (?), the rotation increases the total spin by 2 every step, i.e. J = R + j = R

For an axial deform nucleus with j > 0 , the ground state has to rotate so that the total spin J = R + j is constant of motion.In the Nilsson model, the intrinsic angular momentum is not a good quantum number, in other words, a Nilsson orbital is a superposition of many different spin j. And the spin j is precessing around the body (symmetry) axis. Thus, the introduction of rotation R is needed to make the total spin to be constant of motion. ( I am not sure this explanation is correct, it seems very artificial, and I don’t know how to construct the Hamiltonian)

In the previous paragraph, there are some exceptional cases for the maximum \Omega orbital, for example, the 3/2[101] is a pure p3/2. For such an orbital, the Nilsson orbital only contains one single j. In that case, the state for J = j need not be rotated.

Annotation 2020-07-08 222954.png

In the above picture, the left illustrates the relation between different quantum numbers. on the right, it is more interesting. As we mansion before, the spin j is precessing around the body axis. And the rotation R must change in both direction and magnitude with the spin j to keep the total spin J to be constant. So, the rotation of a deformed nucleus can be very complicated for j > 0 . One interesting and simple case is that the J = 1/2, which is aligned on the body axis. In that case, the rotation direction R is also rotating with a constant magnitude.  Nevertheless, a Nilsson orbital usually contains many j, and each of them is coupled to different rotations to form a constant J. Therefore, when a deformed nucleus starts to rotate, many rotations are happening at the same time.

Because of this complicated motion ( although it is quite a classical point of view ), the nucleon is also moving in this rotating body, so the Coriolis force will be there. This will be the next topic.

Rotational energy is

\displaystyle E(J) = \frac{J^2}{2 I} = J(J+1)\frac{\hbar^2}{2I}

where I is the moment of inertia. For an ellipsoid with a long axis length is a and a short axis length is b is

\displaystyle I = \frac{m}{5}(a^2 + b^2 )

So,

\displaystyle E(J) = J(J+1)\frac{5\hbar^2}{2 m (a^2 +b^2)} = J(J+1) E_0

The term a^2 + b^2 \propto 1.25^2 A^{4/3}, substitute \hbar c = 197~\textrm{MeV~fm}, the mass of the nucleus is m c^2 \propto A .

\displaystyle E_0 \approx \frac{X}{A^{7/3}} ~\textrm{MeV}

For 180Hf, the estimated E_0 \approx 0.016 MeV, so the 2+ excited state energy is 0.098 MeV. and the 4+ excited state energy is 0.330 MeV. which is not far away from the experimental value.

If the nucleus is rigid enough during rotation, the ratio between E(4+) and E(2+) should be a simple ratio 4(5)/2/3 = 3.33.

Decoupling factor of 19F

June 22, 2020

GoLuckyRyan Basic , , , Leave a comment

From the last post, we have the formula for the rotation band energy and the decoupling factor.

\displaystyle E_R = \frac{\hbar^2}{2I_M}\left( J(J+1) + a (-1)^{J+\frac{1}{2}}\left( J + \frac{1}{2} \right)  \right)

Lets apply to 19F rotational band.

19f_rotational_bands

The K = 1/2^- and K = 3/2^+ bands are straight forward. The fitting model,

\displaystyle E_R(J) =B + A\left( J(J+1) + a (-1)^{J+\frac{1}{2}}\left( J + \frac{1}{2} \right)  \right)

Annotation 2020-06-22 212329

Annotation 2020-06-22 212347

The decoupling factor for the K = 3/2^+ band is a = -0.015 \pm 0.020. which is consistence with zero.

For the K = 1/2^+ band, we fit for all states,

Annotation 2020-06-22 212508

The fit is not so good. Let only fit up to 9/2.

Annotation 2020-06-22 212529

The fit is much better. The decoupling factor for both fit are 2.09 to 2.26.

 

 

Nilsson Diagram from N=1 to N=6.

October 12, 2019

GoLuckyRyan Basic , 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 , 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

Rotational Band

January 18, 2019

GoLuckyRyan Basic , , , , , , 2 Comments

For deformed nuclei, it can be rotated in various angular momentum in Laboratory frame. Assume rigid body rotation, the energy is

\displaystyle E_J = \frac{1}{2}I\omega^2 = \frac{1}{2I}J^2

In QM, that becomes

\displaystyle H = \sum_{i=1}^{3} \frac{\hbar^2}{2I_i} J_i^2

For axial symmetry, I_1 = I_2 = I

\displaystyle H = \frac{\hbar^2}{2I} (J^2 - J_3^2) + \frac{\hbar^2}{2I_3}J_3^2

Remember, in deformed nuclei, the projection of J along the symmetry axis in the body-frame is K . The expected value of the Hamiltonian with state |Nn_z m_l K \rangle in the body-frame is proportional to J(J+1) for J^2 and K for J_3. i.e.

\displaystyle E_J = \frac{\hbar^2}{2I} J(J+1) + E_K

From body-frame to Lab-frame, we should apply the Wigner D-Matrix to the intrinsic wave function. ( I am not sure the following equation is correct, but the idea is rotating the body-frame wavefunction with Wigner D-Matrix to get the Lab-frame wave function. In Lab frame the total angular momentum must be a good Quantum number as rotational symmetry restored, so as J_z = M. The problem of the following equation is that the J is not a good Q-number in Nilsson wavefunction )

\displaystyle |JMK\rangle = \sum_{M} D_{MK}^{J} |Nn_zm_lK\rangle

However, the Wigner D-Matrix does not conserve parity transform:

\displaystyle D_{MK}^J \rightarrow (-1)^{J+K} D_{M-K}^{J}

In order to restored the parity, we need to include \pm K in the Lab-frame wave function.

\displaystyle |JMK\rangle = \sum_{M} \left( D_{MK}^J \pm (-1)^{J+K} D_{M-K}^J \right) |Nn_zm_lK\rangle

where + for positive parity, – for negative parity.

From the above equation, for K^\pi = 0^+ (0^-), J must be even (odd). For K > 0 , J = K, K+1, K+2, ... .

rotaional Band of 205Fm.png

rotational band of 253No.png

rotational band of 19F.png

We can see for K = 1/2 , the J = 5/2, 9/2, 11/2 are lower to the main sequence. This was explained by adding an extra term in the rotation Hamiltonian that connect \Delta K = 1 .

\displaystyle \langle JMK | H'(\Delta K = 1) |JMK \rangle

\displaystyle \rightarrow \langle D_{MK}^J | H' | D_{MK}^J \rangle+ \langle D_{M-K}^J |H'| D_{MK}^J \rangle + \langle D_{MK}^J | H' | D_{M-K}^J \rangle+ \langle D_{M-K}^J |H'| D_{M-K}^J \rangle

The term with \Delta K = 0 vanished. And since $\latex K = 1$, the only non-zero case is K = 1/2 .

A possible form of the H' (\Delta K = 1) = \frac{1}{2} \omega (J_+ + J_-) . These are the ladder operator to rise or lower the m-component by 1. In 19F case, we can think it is a single proton on top of 18O core.  A rotation core affect the proton with an additional force, similar to Coriolis force on earth.