# Is the spin state of an atom related to the polarization of the photon it spontaneously emits?

From literature I've been reading, I find that scientists are able to "map" atomic states onto photon states. Are they talking about spin states and corresponding photon polarization states? Can somebody explain how these two things are related? I am having a tough time keeping track of the orientation of the measurement basis.

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Can you add some references? There are many ways how to map quantum states between light and matter so it's hard to guess which one is the right one. –  Ondřej Černotík Jan 2 '13 at 10:01
Here is a link to the actual paper: I can't find a free version though. :( prl.aps.org/abstract/PRL/v85/i26/p5639_1 –  QEntanglement Jan 3 '13 at 5:44

In the experiment mentioned in the paper, the authors use continuous variables (CV) of light and atomic ensembles. In the light modes, the CV used are the amplitude and phase quadratures $\hat{X}$, $\hat{Y}$, corresponding to Stokes vector components $\hat{S}_z$, $\hat{S}_y$, respectively. In the atomic ensembles, phase shifts between ground and excited states, in the paper denoted by $\hat{F}_y$, $\hat{F}_z$, are used.
The Hamiltonian governing the light-atom interaction is $\hat{H}\propto \hat{S}_z\hat{F}_z$, which means that these CVs are left unchanged and the atom-light system undergoes a non-demolition interaction where the operators $\hat{S}_z$, $\hat{F}_z$ are imprinted onto $\hat{F}_y$, $\hat{S}_y$ [Eq. (3) in the paper].