 Solution

For the nonrelativistic Coulomb Hamiltonian with reduced electron-proton mass $\mu$, separation in Spherical coordinate system gives ${\psi }_{n\ell m}(r,\theta ,\varphi )=R_{n\ell }(r)Y_{\ell m}(\theta ,\varphi )$. The quantum numbers obey $n\ge 1$, $0\le \ell \le n-1$ and $-\ell \le m\le \ell$. These are joint eigenstates of energy, $L^2$ and $L_z$, with the latter two eigenvalues $\ell (\ell +1){\hslash }^2$ and $m\hslash$.

The Coulomb bound-state energies and characteristic length areFor example, ${\psi }_{100}=(\pi a_{\mu }^3{)}^{-1/2}{e}^{-r/a_{\mu }}$ is spherically symmetric and has no radial node. In general the radial part contains an exponential times a Generalized Laguerre polynomial, with $n-\ell -1$ radial nodes. The spherical harmonic specifies angular structure and azimuthal phase; real linear combinations of degenerate $m$ states give familiar directional orbitals.

The leading Coulomb energy depends on $n$ alone, giving $n^2$ spatial states for each principal shell, or $2n^2$ including electron spin. Fine structure, Lamb shift and Hyperfine structure lift parts of this ideal degeneracy. An atomic orbital is a probability amplitude, not a classical planetary path.^[ref]