Why is the orbital angular momentum of a pi electron along the axis of two atoms' molecule one? I'm reading quantum chemistry. The book says that the orbital angular momentum of a $\pi$ electron along the symmetry axis of a molecule made up of two atoms is $\pm 1$. I think this is a primary question, but I do not know why (I'm a student studying physics)



I currently have a preliminary understanding of this:

In a molecule inculding two atoms, potential energy axisymmetric about z' axis(the line connecting the two atoms). So, the angular momentum along z axis is quantized. That is to say $m_z$ is good quantum number. Let's consider $\pi$ orbit made up of two $p_z$. The orbital angular momentum along z axis(the symmetry axis of $p_z$) of an electron in $p_z$ is 0. So considering orbital angular momentum along z' axis of this electron, the electron is in $\frac 1 {\sqrt 2}(|+&gt;-|-&gt;)$. So the orbital angular momentum along z' axis is either 1 or -1. So the orbital angular momentum along z' axis of a $\pi$ electron is either 1 or -1.

Is my understanding correct?
Let the molecular axis be the $z$ axis, A and B the two atoms forming the bond. The $2p_z$ AO has atomic quantum number $m_l = 0$ and so has $L_z =0$. The MOs formed from
2pz AOs are therefore $\sigma$ MOs, bonding and antibonding. 
The $p_x$ AO is a linear combination of $m =+ 1$ and $m = -1$ AOs and has $|m| =1$.

$2p_x \equiv \frac{(2p_1 + 2p_{-1}) } {\sqrt 2}$

$2p_y \equiv \frac{(2p_1 - 2p_{-1}) } {i\sqrt 2}$

Therefore the MOs made from the $2p_x$ and $2p_y$ AOs have $|m| = 1$  and are $\pi$ MOs.