# Assignment of $m_\ell$ values to $p_x, p_y$ and $p_z$ states

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.

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.
• I think he didn't mean momentum but the chemists notation of the three states with $l=1$ – Ofek Gillon Nov 2 '18 at 13:48