# Angular momentum and Zeeman effect

Supposing an Alkali atom positioned in magnetic field in $$z$$-axis. I understand linearly polarized laser propagating in $$z$$-axis can induce $$\sigma$$ transmission. But I could not figure out how laser propagating in x/y axis can also induce $$\sigma$$ transmission, since the angular momentum of photon is parallel or antiparallel to the propagation direction.

I understand such transmission can be understood from the perspective of semiclassical theory. What I want to know is how to understand it when taking the fact laser is the ensemble of photons. Is it because photon directing in $$x$$-axis can be viewed by adding two virtual photons in $$z$$-axis with opposite propagation direction?

It's the photon's angular momentum about some chosen $$z$$ axis, not its linear momentum, that is important when considering whether the photon will induce a $$\sigma^{\pm}$$ or $$\pi$$ transition between states whose projections are referred to the $$z$$ axis. Linearly polarized light propagating along $$\hat{z}$$ can be decomposed into a sum of $$\sigma^{\pm}$$ components, which will induce $$\sigma^{\pm}$$ transitions.
Light propagating perpendicular to the quantization axis, for example along $$\hat{x}$$, might be polarized along $$\hat{z}$$. If this is the case, it carries zero angular momentum about the $$z$$ axis, and cannot induce $$\sigma^{\pm}$$ transitions. It can, however, induce $$\pi$$ transitions. A photon propagating along $$x$$ and polarized along $$y$$ can again be thought of as a superposition $$(\sigma^+ + \sigma^-)/\sqrt{2}$$, and will induce $$\sigma^{\pm}$$ transitions.
For an atom interacting with a laser, in the absence of any high-finesse optical cavity that alters the vacuum states of light, you are often better off just thinking of the electric field of the laser classically. In particular, since a laser is approximately a coherent state of the light, it's about the closest you can come to a classical light field in quantum mechanics. That being said, I would say that no, you should not think of it as a pair of photons counter-propagating along $$z$$. You just have to remember that photons have more degrees of freedom than just their propagation direction -- they also have polarization and therefore angular momentum.
• I missed an "angular" before one "momentum" as corrected before. So why polarized along y can be viewed as a superposition $(σ_{+}+σ_{_})/2$ while along z can not, when photon propagates along z axis – xiang sun Apr 9 '19 at 10:02