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Solving the Schrodinger equation for hydrogen atom we arrive to the conclusion that quantum numbers $n$ and $\ell$ have the relation $$\ell=0,1,...,n-1.$$

Now,since we can not solve the Schrodinger equation for many electrons atoms we can approximate the potential $V(r_i)$ to be central than Hamiltonian becomes $$ \hat H= \sum_i \left( -\frac{\hbar^2}{2m} \nabla_i^2 +V(r_i) \right).$$

So for each electron we have the following equation $$-\frac{\hbar^2}{2m} \nabla_i^2\psi_i +V(r_i)\psi_i=E\psi_i.$$

And solution is of the form

$$\psi_i=R_i^{n,\ell}(r)Y^m_{\ell}(\theta, \phi).$$

My question is why quantum numbers $n$ and $\ell$ have the relation $$\ell=0,1,...,n-1.$$

Is this an assumption or can we proof it?

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marked as duplicate by ZeroTheHero, Aaron Stevens, Buzz, Kyle Kanos, Qmechanic quantum-mechanics Dec 29 '18 at 6:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/238180/2451 and links therein. $\endgroup$ – Qmechanic Dec 26 '18 at 10:41
  • $\begingroup$ Their not the same question since what i am just concerned about the relation between $\ell$ and $n$ with the assumption that the potential is central $\endgroup$ – amilton moreira Dec 26 '18 at 10:57
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    $\begingroup$ Do you understand why $\ell$ cannot be larger than $n-1$ for the hydrogen atom? $\endgroup$ – Aaron Stevens Dec 26 '18 at 13:28
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    $\begingroup$ The OP already asked the same question physics.stackexchange.com/q/450330 for which there is a perfectly valid answer: the range $0 \le \ell \le n-1$ is specific to the Coulomb potential. $\endgroup$ – ZeroTheHero Dec 26 '18 at 14:38
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    $\begingroup$ @BruceGreetham not sure I understand your comment. The rangle of $\ell$ depends on the specific potential, and we typically assume it’s smooth. If you solve for the various central potentials you will get different ranges, an in many cases there no relation between $\ell$ and $n$. $n$ just labels the order of appearance of an $\ell$ value. $\endgroup$ – ZeroTheHero Dec 26 '18 at 14:51