Why do you use $n_r = n -\ell -1$ as quantum number instead of $n$ for hydrogen atom? I got two different quantum numbers for the same problem: Hydrogen atom without any interaction.
Then, my energy is $$ E_n = -\frac{R_y}{n^2} $$ with the quantum number $n = 1, 2, 3, ....$
In another version it is $$ E_{n_r} = -\frac{R_y}{(n_r+\ell+1)^2} $$ with the quantum number $n_r = n - \ell -1$.
Unfortunately, there is no explanation why it uses $n_r$ in the second formula and what the difference physically is between $n$ and $n_r$. I guess the index stands for the quantum number due to the radial part of the wave function $R_{n\ell}$ but I am confused because it works totally fine with $n$.
Is there any explanation written somewhere or can anybody explain?
 A: The Hamiltonian of the hydrogen atom is a central potential
$$H_{\rm H}\propto\frac{1}{r}$$
this means that the angular-momentum is conserved. This translates to quantum mechanics as a decomposition in terms of spherical harmonics and thus you only need to quantum numbers $n_r$ and $l$ to describe the energies. However that is not all, the hydrogen atom has a hidden symmetry, that is the conservation of the Runge-Lenz vector, which translates into a spectrum that only needs one quantum number $n$. The link between symmetries and conservations is given by Noether's theorem.
A: The radial quantum number $n_r$ is related to the number of nodes in the radial part of the wavefunction since the associated Laguerre polynomial
$$
R_{n_r}^{2\ell+1}(r/a_0)
$$
is a polynomial of degree $n_r=n-\ell-1$.  Moreover, since $n\ge 1$, this also gives a bound on the largest possible $\ell$ for a given $n$ since obviously $n_r\ge 0$.  Thus, if $n=1$ then it must be that the largest (and only possible value of) $\ell$ is $\ell=0$.
A: Because an atom is three dimensional and the nucleus-electron interation is spherically symmetric - thus the resulting potential is spherically symmetric. The wavefunction can have angular components in addition to radial components in that case.
All this means is that in addition to $n$, you also get $l$ as a quantum number together defines the energy state of an electron. If you review from page 4 of this link: http://www.damtp.cam.ac.uk/user/tong/qm/qm4.pdf you may get a better picture of what I'm talking about.
