When writing the time independent Schr\"odinger equation in spherical coordinates, we need to plug the Laplacian in Spherical Coordinates into the time independent Schr\"odinger equation. The Laplacian was found to be
![{\displaystyle \nabla _{sph}^{2}={\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial }{\partial r}}\right)+{\frac {1}{r^{2}sin\theta }}{\frac {\partial }{\partial \theta }}\left(sin\theta {\frac {\partial }{\partial \theta }}\right)+{\frac {1}{r^{2}sin^{2}\theta }}{\frac {\partial ^{2}}{\partial \phi ^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d00f5e8083478d396ec8b097c43b35f40ec17b3)
Using the three dimensional Schr\"odinger equation we then have
![{\displaystyle {\hat {H}}\psi (r,\theta ,\phi )=-{\frac {\hbar ^{2}}{2m}}\left[{\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial \psi (r,\theta ,\phi )}{\partial r}}\right)+{\frac {1}{r^{2}sin\theta }}{\frac {\partial }{\partial \theta }}\left(sin\theta {\frac {\partial \psi (r,\theta ,\phi )}{\partial \theta }}\right)+{\frac {1}{r^{2}sin^{2}\theta }}{\frac {\partial ^{2}\psi (r,\theta ,\phi )}{\partial \phi ^{2}}}\right]+V(r,\theta ,\phi )\psi (r,\theta ,\phi )=E\psi (r,\theta ,\phi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/33a960585e3b935651e3b8ed57f845dc287718a0)
We can gain insight into this somewhat ugly equation by rewriting it using the square of the angular momentum operator in spherical polar coordinates:
![{\displaystyle {\hat {L}}^{2}={1 \over \sin \theta }{\partial \over \partial \theta }\left(\sin \theta {\partial \over \partial \theta }\right)+{\frac {1}{\sin ^{2}\theta }}{\frac {\partial ^{2}}{\partial \phi ^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ce0f26a6ecbde85a70b94bc9b7ead0bf36e9492f)
This leads to
![{\displaystyle \left(-{\frac {\hbar ^{2}}{2m}}\left({1 \over r^{2}}{\partial \over \partial r}\left(r^{2}{\partial \over \partial r}\right)\right)+{\frac {1}{2m}}{\frac {{\hat {L}}^{2}}{r^{2}}}+V(r,\theta ,\phi )\right)\psi (r,\theta ,\phi )=E\psi (r,\theta ,\phi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/346d58976ea834380c74eebfae40b438b4dcc473)
This equation is only exactly solvable if
, a function without angular dependence. We then write
leading to the following equation:
![{\displaystyle {\begin{matrix}\left(-{\frac {\hbar ^{2}}{2m}}\left({1 \over r^{2}}{\partial \over \partial r}\left(r^{2}{\frac {\partial }{\partial r}}\right)\right)+{\frac {1}{2m}}{\frac {{\hat {L}}^{2}}{r^{2}}}+V(r,\theta ,\phi )\right)\psi (r,\theta ,\phi )R(r)Y(\theta ,\phi )&=ER(r)Y(\theta ,\phi )\\-{\frac {\hbar ^{2}}{2m}}\left(Y(\theta ,\phi )\left({\frac {1}{r^{2}}}{\partial \over \partial r}\left(r^{2}{\frac {\partial }{\partial r}}\right)\right)R(r)\right)+{\frac {R(r)}{2m}}{\frac {{\hat {L}}^{2}\,Y(\theta ,\phi )}{r^{2}}}+V(r)R(r)Y(\theta ,\phi )&=ER(r)(Y(\theta ,\phi )\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8aa9687200f8a8264b7812fb519fa2315970202c)
To solve this equation we need to remove the angular dependence. This is simply done by substituting the eigenfunctions of
into the equation. These are known to be the spherical harmonics,
. We also know that these have eigenvalues
, i.e.
![{\displaystyle {\hat {L}}^{2}\,Y_{l}^{m}(\theta ,\phi )=\hbar ^{2}l(l+1)Y_{l}^{m}(\theta ,\phi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3169fb5b59bbf6ff9f2be991ba8c491f2ef999eb)
We now substitute this result into the Schr\"odinger equation and divide through by a common factor of
![{\displaystyle \left(-{\frac {\hbar ^{2}}{2m}}\left({\frac {1}{r^{2}}}{\partial \over \partial r}\left(r^{2}{\frac {\partial }{\partial r}}\right)+{\frac {\hbar ^{2}l(l+1)}{r^{2}}}\right)+V(r)\right)R(r)=ER(r)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e73a3598942f6b297fbbc526983c732f5c87996a)
This is the radial equation.\\
{\mathbf References}
[1] Griffiths, D. "Introduction to Quantum Mechanics" Prentice Hall, New Jersey, 1995.