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

In QM, that becomes

For axial symmetry,

Remember, in deformed nuclei, the projection of along the symmetry axis in the body-frame is . The expected value of the Hamiltonian with state in the body-frame is proportional to for and for . i.e.

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 . The problem of the following equation is that the J is not a good Q-number in Nilsson wavefunction )

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

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

where + for positive parity, – for negative parity.

From the above equation, for (), must be even (odd). For , .

We can see for , the are lower to the main sequence. This was explained by adding an extra term in the rotation Hamiltonian that connect .

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

A possible form of the . 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.