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.