Hello, everyone, in order to calculate deuteron by Hartree-Fock method, I need a basis. The basis of infinite spherical square well is too “rigid”, that it has to “extension” to non-classical region. Beside of the basis of Wood-Saxon potential. The finite spherical square well is a good alternative. The radial equation is basically the same as infinite spherical square well.

The potential is

Within the well, the wave vector is

, outside the well, the wave vector is

The solution is spherical Bessel function. Since the Bessel function of the first and second kind are oscillating like sin or cosine function. To form a decay function when , we need the Hankel function with complex argument.

To make it real, we need a factor .

The boundary conditions are continuity and differential continuity.

These two conditions solved for two parameters and . However, I cannot find an analytical solution to the energy .

In mathematica, the spherical Hankel function is a build in function.

SphericalHankelH1[n, r]

If the potential depth is 60 MeV for proton. Radius is 1 unit for a light nuclei. Set . (This is equivalent to radius of

The result is follow,

The 1st column is 1s, 1p, 1d, 1f, and 1g. The 2nd column is 2s, 2p, 2d, 2f, and 2g. The 3rd column is 3s, 3p, and 3d.

By compare with infinite square well,

The energies are lower in finite well, because the wave functions can spread-out to non-classical region, so that the wave length is longer and energy is lower.

In this example, the 3d and 2g orbits are bounded (of course, all orbits in infinite well are bound). This is not because of the depth of the well, but the boundary of the well. In other word, to bring down an orbit, the wave function has to spread out, that is connected with the neutron-halo.