the potential is

V(r,\theta,\phi) = \begin{pmatrix} 0 & |r|<a \\ \infty & |r| \geq a \end{pmatrix}

The Laplacian in spherical coordinate is:

\nabla^2 = \frac{d^2}{dr^2} + \frac{2}{r}\frac{d}{dr} - \frac{L^2}{r^2}

since the L is the reduced angular momentum operator, if we set the solution be:

\psi(r,\theta,\phi) = R(r) Y_{lm}(\theta,\phi)

Then the angular part was solved and the radial part becomes:

L^2 Y_{lm} = l(l+1) Y_{ml}

\left(r^2 \frac{d^2}{dr^2} + 2 r \frac{d}{dr}+(k^2 r^2 - l(l+1))\right)R(r) = 0

k^2 = 2 m E/ \hbar^2

The radial equation is the spherical Bessel function.

The solution was common written as:

R(r) = j_l( k r) = \left( - \frac{r}{k} \right)^l \left(\frac{1}{ r} \frac{d}{dr}\right)^l \frac{sin(k r)}{kr}

The Boundary condition fixed the k and then the energy,

j_l ( k_{nl} a ) = 0

the all possible root are notated as n. thus the quantum numbers for this system are:

  • n , the order of root
  • l , the angular momentum

We can see in here, the different between Coulomb potential and spherical infinite well:

  • there is no restriction on n and l, therefore, there will be 1s, 1p, 1d, 1f orbit.
  • the energy level also depend on angular momentum, since it determined the order of spherical Bessel function.

we can realized the energy level by the graph of Bessel function. we set some constants be 1, the root are :

k_{nl} a = \frac{1}{\hbar} \sqrt{ 2 m a^2} \sqrt{E_{nl}} = \pi \sqrt{E_{nl}}

Thus, we plot

j_l( \pi \sqrt{E_nl})

 

Advertisements