PreposterousUniverse said:
The quantum number n determines the energy, and for each n the allowed values for the angular momentum quantum number are -(n-1),...,(n-1).
This doesn't seem resonable to me. Classically increasing the orbital angular momentum will result in an increase in the energy of the system. But why is it that quantum mechanically the energy doesn't depend on the orbital angular momentum. Could someone explain this?
This is a special symmetry of the (non-relativistic) hydrogen atom. The motion of two particles with a ##1/r## interaction potential has not only the usual space-time symmetries with its "10 conservation laws" from Galilei symmetry, but an additional "dynamical symmetry".
In the classical realm this is reflected in the fact that all bound motions are closed trajectories, i.e., "Kepler ellipses". That means that in addition to the 10 conservation laws what's also conserved is the Runge-Lenz vector, which points from the focus of the ellipse to the "perihelion" of the orbit. This additional symmetry makes the energy independent of the modulus of angular momentum, i.e., you can have the same energy with different values for ##\vec{L}^2##, i.e., the total energy for bound motion (##E<0##, i.e., elliptical/circular orbits) in the center-mass system is given by
$$E=-\frac{G m_1 m_2}{2 a},$$
where ##a## is the major semiaxis of the ellipse.
In quantum theory, the Runge-Lenz vector is an additional set of generators of symmetry transformations (due to Noether's theorem in the Hamiltonian formulation of classical mechanics as well as in quantum mechanics). It turns out that together with the angular-momentum operators it generates the symmetry groups for the different cases of the energy eigenvalues: for ##E<0## (the bound-state solutions) it's a SO(4) symmetry group, for ##E=0## (scattering states, referring to parabolic orbits in the classical theory) the symmetry group is that of the Euclidean plane ISO(3) (the usual semidirect product of rotations and translations in the 3D Euclidean plane), and finally for ##E>0## (scattering states, referring to hyperbolic orbits in the classical theory) it's the Lorentz group ##\text{SO}(1,3)^{\uparrow}##.
That's why the discrete energy eigenstates of the bound-state solutions depend only on the "main quantum number", ##n##, and not as well separately on the orbital-angular-momentum quantum number ##\ell##. Due to rotational symmetry for any central-potential problem there's always a degeneracy with respect to the "magnetic quantum number", ##m## (i.e., the eigenvalue ##m \hbar## with ##m \in \{-\ell,-\ell+1,\ldots,\ell-1,\ell \}##).
Because for any given ##n \in \mathbb{N}## the ##\ell \in \{0,1,\ldots,n-1\}## and for each ##\ell## there are ##(2 \ell+1)## possible ##m## -values for the bound-state energies you have thus the degneracy
$$\sum_{\ell=0}^{n-1} (2 \ell+1)=n^2$$
for each energy-eigenvalue ##E_n=-1 \text{Ry}/n^2##.