# Intuitive argument for why $\Delta m=0$ for transitions with light polarised in $z$-direciton?

The selection rules have $$\Delta m =0$$ for light polarised in the z-direciton trying to excite an atom. Mathematically it is clear that $$\langle n, l, m_l| z |n, l, m_l \rangle = 0$$ unless $$\Delta m=0$$. Therefore I understand the mathematical origin of this fact, I'm just looking for the most intuitive physical description of why this is true.

I have thought of the fact that there are no $$x$$,$$y$$ terms in the electric dipole radiation term in the Hamiltonian means that $$m_l$$ must be conserved through these transitions, but again this seems more 'mathematical' than intuitive.

I guess maybe it could just be geometrically obvious that any direction that is distorted solely by its z-coordinate cannot have its direction with respect to the z-axis affected?

• The classical explanation is just that you need a torque about the z-axis in order to change the angular momentum about that axis, but a force parallel to the z-axis can't produce a torque about the z-axis. Feb 22, 2021 at 1:29
• Yeah that makes sense to me, thanks! Feb 22, 2021 at 11:55