Why are all the orbitals that have the same principal number in Hydrogen degenerate?

In hydrogen, all orbitals with the same principal quantum number 'n' (1,2,3...) are degenerate, regardless of the orbital angular momentum quantum number'l' (0,1...n-1 or s,p,d..). However, in atoms with more than one electron, orbitals with different values of l for a given value for n are not degenerate. Why is this? Surely the radial distribution functions are similar for hydrogen (in that there's still penetration of orbitals and so on). Or is it that orbitals with different values of l are degenerate for a value of n greater than would be occupied in that particular atom's ground state? 




The answer is right there in your question. It's only the interaction of ***multiple*** electrons in an atom like He, Li, Be, etc. that makes the different angular momenta wave functions differ in energy.


Consider this.. For the one electron system, why should a $p$ or $d$ orbital differ in energy from an $s$. What makes them differ?


In the multi-electron case, the $p$ orbitals have different spatial extent, different angular components, so the electron density caused by an electron in that orbital will interact differently with the other electrons. In other words, you need to have more than one electron for the "shape" of the $p$ and $d$ and $f$ orbitals to matter to the other electrons.


In the H atom, there's only one electron, so there's no electron-electron repulsion to differentiate the $s$, $p$, and $d$ orbitals.


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This was asked a long time ago, but I think I'll add an answer. 


The answer is penetration and shielding. For a hydrogen atom, if its electron is promoted to 2s or 2p, the only thing that determines its energy is its radial distance, determined by quantum number n. There is no electron-electron repulsion, obviously. 


For Helium, however, if one of its electrons is promoted to 2s or 2p, there's still an electron remaining in the core 1s shell, so the other electron in the 2s or 2p orbitals will be shielded because of electron-electron repulsion. Since 2s is shielded less by the core than 2p, it experiences a higher Zeff and is able to penetrate the nucleus more and hence has a lower potential than 2p, which is shielded more by the core and has a lower Zeff than 2s. Therefore, 2s is lower in energy than 2p because of penetration and shielding.


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