Why do you use $n_r = n -\ell -1$ as quantum number instead of $n$ for hydrogen atom?

Question

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?

Answer 1

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 $\ell=0$.

Answer 2

There is no difference. You can use either formula.

The radial quantum number $n_r$ is equal to the number of nodes in the radial part of the wave-function $\psi$ (the spheres where $\psi(\textbf r) = 0$, excluding the origin and infinity).

The principal quantum number $n$ is often defined as

$n := n_r + \ell + 1$

and it represents the energy levels of the electron in the gross-structure approximation.

Answer 3

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.

Answer 4

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.