The total Hamiltonian of a nuclear system must be rotational invariant because the state should not have a preferred direction. Thus, the total spin 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 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.
For an axial deform nucleus with , the ground state has to rotate so that the total spin 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 . And the spin is precessing around the body (symmetry) axis. Thus, the introduction of rotation 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 orbital, for example, the 3/2[101] is a pure p3/2. For such an orbital, the Nilsson orbital only contains one single . In that case, the state for need not be rotated.
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 is precessing around the body axis. And the rotation must change in both direction and magnitude with the spin to keep the total spin to be constant. So, the rotation of a deformed nucleus can be very complicated for . One interesting and simple case is that the , which is aligned on the body axis. In that case, the rotation direction is also rotating with a constant magnitude. Nevertheless, a Nilsson orbital usually contains many , and each of them is coupled to different rotations to form a constant . 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
where is the moment of inertia. For an ellipsoid with a long axis length is and a short axis length is is
So,
The term , substitute , the mass of the nucleus is .
For 180Hf, the estimated , 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.