If we study atomic structure, we will find the principle quantum number must larger than azimuthal quantum number (the number of angular momentum), i.e.

But most text book give this result using mathematics that there is no solution for the radial equation when . Some text book solves the radial equation using power series, and argue that the series has to be terminated at power of because the number of node is . The exactly solution is related to Laguerre polynomial, and the polynomial is only defined for .

Also, in nuclear physics, there is no restriction and and can take any integers. Why these two cases are different? of course, the key point is the potential difference. Coulomb potential is used in atomic orbit. Wood-Saxon potential (or finite square well ) is used in nuclear orbit.

The Schrodinger equation for potential is

Separate the radial and spherical part

The radial equation further reduce by using

we can see, the effective potential is

For Coulomb potential,

Introduce dimensionless quantities, or Express everything using the fine structure constant. is Bohr radius,

The equation becomes

The simplified effective potential,

Now, we related the principle quantum number and azimuthal quantum number in one simply formula.

Since the repulsive part due to the rotation is getting larger and larger, the minimum point of the effective potential is

When , .

I plot the case for

The bottom line is . We can see, when , the minimize point of the potential is at . The potential is vary shallow, and the bound state is very diffuses. Since the 2nd derivative of the wave function around the minimum point is

For , is negative, that give oscillating wave function.

For , is positive, that gives explosive wave function that cannot be normalized.

Therefore, .

In fact, if we plot the effective potential,

we can find that when , the potential is also very shallow. For such a picture, we can see the wave function must be oscillating that the number of node must be larger then 1, so the principle quantum number should be larger than 1.

In nuclear orbit, the effective potential using Wood-Saxon potential is

where $R \approx 1.25 A^{1/3} [fm]$ and . For simplicity, we use

The position of minimum potential is no simple formula. I plot R=10, a=1

We can see, for any there is a “well” that can bound a nucleon.

I haven’t solve the finite spherical well, but the infinite spherical well is solved. The solution is the spherical Bessel function of first kind. that function only has 1 parameter, which is the angular momentum. To determine the energy is simply find the node position, and the number of node is related to the principle quantum number. Therefore, the principle quantum number and angular momentum is unrelated. A similar result should be applied well in the case of finite spherical well like Wood-Saxon type, that, for a fixed angular momentum, there can be many solution with different number of nodes.

Wood-Saxon potential, the effective potential is always sloping, especially when outside of the well, that forces that wave function to be decay faster. However, the Coulomb potential is a long range force, the effective potential can be slowly changing for long distance, that allow the wave function to be oscillating for long range.