Which quantum numbers does the orbital angular momentum depend on? [closed]








**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
 
 






---








 **Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)




Closed last year.








[Improve this question](/posts/165950/edit) 



I learned from a question that I recently solved that the orbital angular momentum depends on both the azimuthal and magnetic quantum numbers.


I did not expect this because the formula for the orbital angular momentum is $ \sqrt{l(l+1)} \frac {h}{2π} $ and so I expected it to only depend on the azimuthal quantum number.


So, why does orbital angular momentum depend on the magnetic quantum number? A detailed explanation will be appreciated.




*Orbital angular momentum*, like any other form of angular momentum, is a vector quantity.\* Thus, it has both a magnitude as well as a direction.


The *magnitude* of the orbital angular momentum is given by the formula you wrote, and is thus only dependent on the quantum number $l$:


$$|L| = \hbar\sqrt{l(l+1)}.$$


However, the *orientation* of the angular momentum vector is partly described by the quantum number $m$. In particular, the $z$-component of the orbital angular momentum is given by


$$L\_z = m\hbar.$$


It's a quirk of quantum mechanics, however, that you can't simultaneously know all three components $(L\_x, L\_y, L\_z)$, so this is the best we can really do. Ultimately, this is because the operators $L\_x, L\_y, L\_z$ don't commute with one another.


So, given that we know the magnitude $|L|$ as well as the $z$-component $L\_z$, we can say that the orbital angular momentum lies somewhere along a *cone*, as the following picture from Atkins' *Molecular Quantum Mechanics* (4th ed.) shows. Atkins uses the general symbol $I$ to refer to some kind of angular momentum; the fact that it's orbital angular momentum doesn't change anything. Indeed, the [Wikipedia page on magnetic quantum number](https://en.wikipedia.org/wiki/Magnetic_quantum_number) has a very similar picture, as pointed out in the comments.


[![angular momentum cone](https://i.stack.imgur.com/wgcme.png)](https://i.stack.imgur.com/wgcme.png)




---


\* Pseudovector, for the picky.


 5