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.

  • $\begingroup$ 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 $\endgroup$ – xiang sun Apr 9 at 10:02

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