Alpha cluster and alpha separation energy

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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

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The total Hamiltonian of a nuclear system must be rotational invariant, because the state should not have prefer 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 prefer direction, it rotates around the “z-axis”. Since this “z-axis” is arbitrary, the wave function (or the spherical harmonic) still have no spacial preference and can point to any direction. i.e., the energy is the same for all direction. The nuclear spin 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 magnetic field.

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

For 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 Nilsson model, the intrinsic angular momentum is not a good quantum number, in other words, a Nilsson orbital is a superposition of many difference 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 previous paragraph, there is some exceptional cases for the maximum \Omega orbital, for example, the 3/2[101] is a pure p3/2. For such orbital, the Nilsson orbital only contains one single j. In that case, the state for J = j need not be rotate.

Annotation 2020-07-08 222954.png

In above picture, the left illustrates the relation between difference 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 align 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 coupled to different rotation to form a constant J. Therefore, when a deformed nucleus start to rotate, many rotation 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 that the Coriolis force will be there. This will be the next topic.

Decoupling factor of 19F

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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.

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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

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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

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N=4.png

N=3.png

N=2.png

Rotational Band

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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.