For the orbital angular momentum quantum number $\ell=1$, there are three possible $m_\ell$ values, namely, $-1,0$ and $+1$. Which $m_\ell$ value is associated with which of the three p spates, namely, $p_x,p_y$ and $p_z$ and why?


$m=0$ is the case where the wave function behaves like $\cos \theta$, which is what is commonly referred as $p_z$.

For $p_x,p_y$, you need a superposition of $m=1,-1$: $$ Y_1^{1} + Y_1^{-1} \propto \cos \varphi \sin \theta $$ so one can see that $p_x = \frac{1}{\sqrt{2}} (|1\rangle + |-1 \rangle) $ and in the same matter, $p_y$ will be a superposition with a minus sign.

Hope it helped.

edit: I found a wikipedia page explaining the issue. Here: https://en.wikipedia.org/wiki/Atomic_orbital#Real_orbitals


None of them. The traditional$^*$ approach is to use the operators $\hat H$, $\hat L^2$, and $\hat L_z$ which corresponds to the Hamiltonian (energy), the magnitude of angular momentum, and the z-component of angular momentum. We can specify these three at once because it turns out that these operators commute, so there exists a common eigenbasis we can work in such that the basis vectors are eigenvectors of all three operators.

So the $\ell=1$ you state relates to the eigenvalues of the $\hat L^2$ operator, and the three $m_{\ell}$ values you site deal with three different eigenvalues of the $\hat L_z$ operator. They do not pertain to any single linear component of momentum (since $L_z=xp_y-yp_x$).

$^*$ We technically don't have to work with $\hat L_z$. We could choose any single component of $L$ to work with. By tradition we specify $L_z$, and we can always adjust our coordinates so that the z-axis is in a place to do this.

  • $\begingroup$ I think he didn't mean momentum but the chemists notation of the three states with $l=1$ $\endgroup$ – Ofek Gillon Nov 2 '18 at 13:48
  • $\begingroup$ @OfekGillon Yes. $\endgroup$ – mithusengupta123 Nov 2 '18 at 13:50
  • $\begingroup$ @mithusengupta123 Oh ok sorry. I am not a chemist :) $\endgroup$ – Aaron Stevens Nov 2 '18 at 13:57
  • $\begingroup$ @AaronStevens , it's ok I also got confused at first. You are welcome to see what are these p orbitals here: en.wikipedia.org/wiki/Atomic_orbital#Real_orbitals . This is also relevant to the OP $\endgroup$ – Ofek Gillon Nov 2 '18 at 14:57

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