# Why does calcium emit circularly polarized light?

One page 31 of Quantum Metrology, Imaging, and Communication by Simon, Jaeger and Sergienko, the authors state that:

... calcium has two electrons in its outermost, partially filled shell; these electrons are in $$\mathrm{s}$$ states, with oppositely aligned spins, so that the total angular momentum is $$J = 0$$ (a singlet state). Consider then an electron excited to a higher energy singlet state. Intermediate between the ground and excited singlet states, there is a $$J = 1$$ triplet state. The excited electron can decay in two steps, via this intermediate state. Two photons are emitted in the process. These photons may be emitted in any direction, but if we look only at photons emitted back-to-back, i.e. look at coincidence counts from detectors separated by $$180^\circ$$, then angular momentum conservation requires the two photons to have the same circular polarization (left-handed or right-handed).

Why should the two photons have circular polarization and not say horizontal or vertical? The angular momentum would still be conserved if the two photons have vertical polarization along say $$+z$$ and $$-z$$ directions.

• The angular momentum must also be conserved on each step of the two-photon process. Nov 1, 2021 at 12:50
• The angular momentum must be conserved upon the emission of the first photon, and then also upon the emission of the second photon. Nov 1, 2021 at 13:31
• I am probably missing something, but if circularly polarized light enters a linear polarizer, the exiting light is linearly polarized and a tiny torque is exerted on the polarizer. Not so for linearly polarized light entering the linear polarizer. I assume the torque was due to the requirement of conservation of angular momentum.
– Ed V
Nov 1, 2021 at 14:55
• There are various papers like this open access one around: J.-W. Liaw, Y.-S. Chen and M.-K. Kuo, Rotating Au nanorod and nanowire driven by circularly polarized light, Optics Express, Vol. 22, Issue 21, pp. 26005-26015 (2014) doi.org/10.1364/OE.22.026005
– Ed V
Nov 1, 2021 at 16:48
• The transitions are from $m=0$ in the singlet to $m=\pm 1$ in the triplet so will be circularly polarised. (usually these are observed as $\sigma^+\;\sigma^-$ in the Zeeman effect.) Why the polarisation is the same when detected 180 apart is perhaps due to a phase change imposed by the geometry necessary to keep the total angular momentum conserved- but this is just a handwaving guess. Nov 1, 2021 at 20:17

Photons have a spin of magnitude $$1$$. If we project the corresponding spin state vectors onto the momentum vector of the photon, we get what's called "helicity" with values $$\pm1$$ which are left and right circular polarization. Helicity 0 is forbidden for massless particles like photons. (You'll have to take it up with the physicists to really explain why. We'll just leave it at that.)

Since the process you describe involves two separate photon emissions, and each one changes the angular momentum of the atom by a magnitude of $$1$$, the photons that are emitted must also each carry a defined angular momentum to compensate the angular momentum change in the atom. The question is now: Which way are those angular momentum vectors (i.e., spin states) oriented?

Let's say that we place the calcium atom at the center of a Cartesian laboratory coordinate system. Let's further assume that the first photon is emitted along the $$+z$$ direction with a helicity of $$+1$$, i.e., the spin vector also points along the $$+z$$ axis. This would be right circularly polarized light. The atom, meanwhile, gains an angular momentum of opposite sense, described by a vector pointing along the $$-z$$ direction. Now, the second photon is emitted along the $$-z$$ axis, carrying away the angular momentum of the atom so that the latter returns to its $$J=0$$ state. This means that the second photon has its spin vector in the $$-z$$ direction; again, the helicity (projection of the spin onto its momentum) is $$+1$$, which is right circular polarization.

Now, let's talk about linear polarization some more, because it gets a little tricky.

A general photon whizzing though free space will have the same energy and momentum no matter whether it's in a $$+1$$ or $$-1$$ helicity state. Thus, we can put the photon into any superposition of the helicities and it will still have the same energy and momentum. (In a chemist's mind, think how combinations of complex-valued atomic orbital wavefunctions to the real-valued Cartesian ones still represent stationary solutions.)

This means that, in the general case, describing linear polarization as a superposition of circular polarization states is just as valid as the other way around.1 In the calcium experiment, we are perfectly able to describe the photons as being in a superposition state of horizontal and vertical linear polarization. But if we were to send the through linear polarizers to measure which polarization they have, each one would have a 50% chance for either result.

Note also that a single photon can actually have linear polarization if it is in a equally-weighted superposition state of $$+1$$ and $$-1$$ helicity. This does not mean that it has a helicity of $$0$$; instead, it means that when we measure the helicity, we have a 50% chance to get either result.2

• Opposite from the point of view at the atom, but the same from the point of view of each photon in its own direction of travel. Nov 2, 2021 at 1:10
• but left circular and right circular are not superimposable by just changing direction of observation ? Nov 2, 2021 at 8:39
• @Micheal Linearly polarized light has a total spin of zero, but the photons it consists of do have - positive and negative, just cancelling out at the proper phase. And consider it differently: When you say "horizontal/vertical polarization", which plane would that be horizontal/vertical to? There's no plane of preference in e.g. an isolated Ca atom, so photons can only be circular polarized and linear polarization is an emergent feature of multiple photons forming "the" photonic/electromagnetic field Nov 2, 2021 at 11:48
• @EdV I guess I should have worded it more carefully. Let's say we are an observer in the fixed laboratory frame. We are looking at the atom so that the first emitted photon comes straight at us. Let's say that it has a circular polarization that is clockwise from our point of view; for the photon itself, that's LCP. Then the next photon flies straight away from us with counterclockwise polarization from our view, which is again LCP for the photon. They key point here is that our perspective is stationary; we're not turning around to look in the travel direction of the photons. Nov 2, 2021 at 15:40
• @Antimon Good point! Yes, if you "look" exactly, the circular polarization corresponds to the pure two possible spin projections, but like everything in the quantum realm you can of course also have a superposition of both even for a single photon. Sorry for oversimplifying Nov 3, 2021 at 10:32

This is by way of a long comment.

The Ca atoms can take all sorts of orientations in space and presumably they suffer no collisions when in the excited singlet or triplet. The excited singlet must be produced quickly in a time much less that its lifetime. This singlet then decays to a triplet and then this decays to a singlet ground state, in both transitions a photon is emitted. We are told that both are both either right or left circular polarised and detection is at 180 degrees. This means that two identical detection sets are used and at 180 to one another. The fact that detection is at 180 effectively photo-selects the atoms, i.e. only some atoms will have a dipole so aligned to be detected at either position. Once the first photon is detected the second must come from an atom with the same initial dipole direction, provided no collisions or other interaction occurs.

The singlet has $$L=1, m=0$$, and the triplet $$L=1, m=-1,0,1$$. As the initial state is a singlet suppose it generates a photon that is right circular polarised (rcp) with momentum projection labeled $$m_p=+1$$, and leaves the atom in the $$m=-1$$ triplet sub-level and the total change in angular momentum will be zero. ($$L$$ changes by 1 singlet to triplet and a photon has 1 unit of angular momentum ($$L=1$$), plus projection quantum numbers $$m_p=\pm 1$$ but no $$m_p=0$$)

The second transition is independent of the first as there is a decay time for both singlet and triplet states, i.e overall not a two photon transition. The triplet decays to the ground state where $$L=0, m=0$$ from $$L=1, m=-1$$ so has the same polarisation state as the first transition but can only be detected at 180 (or 0) because of the photo-selection by virtue of where the detectors are placed.

(As an aside presumably detection of both photons at 0 degree needs v faster pd's so it is easier to use two and perform coincidence that way).