Looking at an hydrogen atom with the bohr model we get

$$L=n\hbar, \qquad n\in\mathbb{N}\tag{1}$$

for the angular momentum.

But by solving the Schrödinger equation, we get


How is it possible these two equations give different values in some cases?

I have already seen this post regarding this topic this post regarding this topic. But the answer didn't really satisfy me. Because if i have $\ell=1$, the the solution of the Schrödinger equation gives me $L=\sqrt 2\hbar$. Thus I have still a contradiction to the Bohr-solution.

What am I missing here?

  • 1
    $\begingroup$ why would you expect that two different, inequivalent models give rise to the same solution? $\endgroup$ Jul 21 '17 at 14:27
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/144066/2451 $\endgroup$
    – Qmechanic
    Jul 21 '17 at 14:33
  • $\begingroup$ Notice that you recover the angular momentum in Bohr's postulate for $n, l \gg 1$ of Schrödinger's result. See the linked duplicate post for more details $\endgroup$
    – Rol
    Oct 27 '21 at 3:52

The angular momentum values predicted by the Bohr model are plain incorrect. For instance, the ground state of hydrogen would have $\ell=1$ as per Bohr but $\ell=0$ as per Schrodinger. Moreover, Bohr predicts a single value of $\ell$ per energy level, whereas Schrodinger predicts many.

Experimental evidence (from the Zeeman effect) contradicted the Bohr model and reconciliation between observed and predicted values of $\ell$ was a triumph of Schrodinger's approach.


The Bohr model is wrong in this context. The solution of the angular part of the Shrodinger equation for the hydrogen atom gives you the spherical hatmonics $Y^m_\ell(\theta,\phi)$ which are the eigenfunctions of the angular momentum poerator $\hat L^2$ (and $\hat L_z$). Those functions are described by the two quantum numbers $\ell$ and $m$ which are different from $n$. That's why the eigenvalues of the angular momentum are given in terms of $\ell$ not $n$.
$$\hat L^2 Y^m_\ell(\theta,\phi)=\hbar \ell(\ell+1)Y^m_\ell(\theta,\phi)$$


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