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 '19 at 10:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.