The Hamiltonian is

Use cylindrical coordinate, the Schrodinger equation is

Where the energy is

Note that . One of the reason is there are 2 degree of freedom, it cannot be solely expressed into 1 parameter.

Set ,

The angular part, we can set

The solution is

The z-part is usual 1D Harmonic oscillator

Thus the rest is

rearrange

Set a dimensionless constant , and

Using the normalisation formula, , set

Because of the long-range and short-range behavior, set

Set

Define .

This is our friend again! The complete solution is

where

The energy is

The quantum number has same meaning as in spherical case.

The notation for the state is

with the spin, the only good quantum number is the z-component of the total angular momentum long the body axis, i.e. , thus, the state is

Note that the total angular momentum is not a good quantum number in deformation, as the rotational symmetry is lost. However, the quantum number is linked with the angular momentum of the Nilsson single particle orbit. It is because when a particle has , the angular momentum must at least .

The above is a general solution for the harmonic oscillator in cylindrical coordinate. When , it reduce to spherical case.

According to P. Ring & P. Schuck (2004) (The Nuclear Many-Body Problem, P.68), the Hamiltonian can be expressed as a quadruple deform field by setting

This Hamiltonian has similarity with the deformation

take the first quadruple term, and calculate

Compare, we have

The ratio

We need volume conservation

Thus,

The energy is

Here is some plots with various

Next time, I will add the LS coupling and L-square term to recreate the Nilsson diagram. Also I will expand the solution from cylindrical coordinate into spherical coordinate. This unitary transform is the key to understand the single particle-ness.

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