# Why does circularly polarized light break time-reversal symmetry?

I've encountered some interesting paper on 2D materials where authors use circularly polarized light to break time-reversal symmetry to split energy levels.

Here you can find the paper:

Valley-selective optical Stark effect in monolayer WS2. E.J. Sie et al. Nature Materials 14, 290 (2015), arXiv:1407.1825.

My first question is, why is the degeneracy of levels mentioned in the paper locked by time-reversal symmetry?

The second question is, why does circularly polarized light break time-reversal symmetry?

I would like to have some brief physical explanation, maybe, with some brief mathematics involved.

• Well, what does circularly polarized light become after time-reversal? – ACuriousMind Aug 14 '15 at 2:42
• Well, I guess it would become the same circularly polarized light travelling in the opposite direction, right? How can it help to address my question though? – Capo Pavel Mestre Aug 14 '15 at 2:47
• I'm more thinking that it switches from clockwise to counter-clockwise polarized light, which might interact differently with some states, so a system that interacts with light that is only polarized in one circular direction is not time-reversal symmetric. – ACuriousMind Aug 14 '15 at 2:49
• yeah, but in the paper it is said that the system is initially time-reversal symmetrical and circularly polarized light breaks this symmetry in material. So when the circularly polirized light interacts with the system described in the paper then the system is no longer symmetrical under time-reversal procedure. The question is why? – Capo Pavel Mestre Aug 14 '15 at 3:09
• Circularly polarized light is an eigenstate of angular momentum. Angular momentum is odd under time reversal (because $\mathbf x \times \mathbf p$ is). – Robin Ekman Aug 14 '15 at 3:18

Circularly polarized light is an eigenstate of angular momentum. Angular momentum is odd under time reversal ($T$), since $\mathbf x\times\mathbf p$ is an angular momentum and $\mathbf p \mapsto -\mathbf p$ under $T$.
Therefore $T$ makes left-circular polarized light into right-circular polarized light and vice versa. Thus a Hamiltonian containing an external circular-polarized field is not $T$-invariant, and Kramers theorem does not apply.