So say you have a vertically polarized single photon impinging on an atom. The atom absorbs the photon and re-emits it. Does the re-emitted photon have the same polarization (vertical) as the incident photon? My gut tells me it does not, but I do not know.

I remember reading in a QFT book that the free photon has an inherent circular polarization to it, so does that mean the emitted photon will be circularly polarized? (Is this even correct?) (I am not super knowledgeable in QFT, but I have had a good amount of QM courses)

  • $\begingroup$ That depends on the atomic state, see e.g. the Faraday effect or birefringence. $\endgroup$
    – CuriousOne
    Mar 15, 2016 at 20:44
  • $\begingroup$ If a photon of a given polarization goes in to the experiment, and a photon of the same polarization comes out, how would you know if it were absorbed and re-emitted? (Well, conceivably by a time delay.) $\endgroup$
    – Jon Custer
    Mar 15, 2016 at 20:58
  • $\begingroup$ Jon, what if the photon was absorbed and re-emitted with the same polarization but in a completely different direction. Wouldn't the angle confirm that it was a different photon? $\endgroup$ Mar 15, 2016 at 23:23
  • $\begingroup$ Here it's about circular polarisation, and it's relation with spin (that is supposed to be conserved... somewhere :-) ). $\endgroup$ Mar 16, 2016 at 20:37
  • $\begingroup$ @BillAlsept, do you mean like a mirror? $\endgroup$
    – S. McGrew
    Dec 3, 2020 at 1:07

3 Answers 3


If an atom first absorbs a photon and then emits a photon of the same (or very close) frequency, the two photons do not have to have similar wave vector or polarization.

Absorption and re-emission
If the photon is actually in resonance with an atomic transition, we are most likely talking about two separate processes: the absorption of the incident photon and an emission of a new photon, and the two are independent (there is some degree of decoherence between the two). That the initial photon is not identical with th emitted one is used in laser cooling of atoms.

Rayleigh scattering
But such a re-emission may also happen as a single process of Rayleigh scattering, in which case the photon does not have to be in resonance with the atomic transition. The fact that the re-emitted photons differs from the incident one has long-reaching implications -such as the diffusion of the solar light in the atmosphere, resulting in the blue color of the sky.

Stimulated emission
What might be the source of confusion here is the phenomenon of stimulated emission, when an (already excited) atom emits a photon identical to the incident one - identical in all aspects: frequency, wave vector, polarization.

  • $\begingroup$ I wouldn't call Rayleigh scattering reemission, but I guess that the distinction between scattering and absorption followed by reemission is not 100% sharp. $\endgroup$
    – my2cts
    Sep 15, 2021 at 8:52
  • $\begingroup$ @my2cts I agree that reemission is not a very precise term here, since we don't really have well-defined absorption and emission. But I don't think the OP was aware that the same outcome may come from different processes,a nd I tried to explain it in their language. $\endgroup$ Sep 15, 2021 at 9:01

This is not a full answer, as I only know about hydrogen, but here the answer is "Not necessarily".

When photons close to the resonance of the 1→2 transition, i.e. the so-called Ly$\alpha$ photons, scatter on neutral hydrogen, they may become polarized. The amount of polarization depends on whether the energy of the photon is very close to the line center ("core scattering"), or farther away ("wing scattering"), as well as on the angle $\theta$ between the incident and outgoing photon.

The phase function — i.e. the probability distribution that determines the angle of scattering — also determines the degree of polarization. For instance, for core scattering this is $$ W(\theta) \propto 1 + \frac{R}{Q} \,\cos^2\!\theta, $$ where $R/Q$ is the degree of polarization for 90º scattering.

I won't go into too much detail if this is not what you had in mind, but it turns out that this can result in either unpolarized or partially polarized photons, which in general is different from the polarization before the scattering event.

If this is of interest to you problem, I can elaborate a bit on this.

  • $\begingroup$ Thanks for the reply. This is very interesting and IS of interest to my problem. My research deals with scattering of a single photon from a collection of atoms. These "atoms" aren't strictly defined in terms of what atoms they actually are, they are being treated more like "scatterers." I am wondering how similar they would behave in relation to the hydrogen atom for very first order approximations in terms of polarization changes. $\endgroup$
    – user41178
    Mar 15, 2016 at 21:16

So say you have a vertically polarized single photon impinging on an atom.

You cannot vertically polarize a photon. The photon has spin 1 which, because the mass of the photon is zero, will be either +1, i.e. in its direction of motion, or -1, against its direction of motion. The photon does not have an electric and magnetic field defining it so as to talk of classical polarization. Only momentum and $E=hν$

This can be seen pictorially in how photons build a polarized beam :

phot pol

Note how the photon spin is + or - to its direction of motion, while the light has circular polarization.

The atom absorbs the photon and re-emits it. Does the re-emitted photon have the same polarization (vertical) as the incident photon? My gut tells me it does not, but I do not know.

Cannot be vertical for the individual photon, so the answer is no. If a photon with spin +1 is absorbed by an atom can give a photon of spin +1 to its direction of motion when it deexcites, will depend on the energy levels involved, i.e. the angular momenta for the individual energy levels and atoms.

Light, the quantum mechanical superposition of zillions of photons can have vertical (to its direction of motion) and circular polarization, depending how the individual wavefunctions of the photons $Ψ$ are added into one wavefunction for the ensemble. This link shows how the classical field is built up from the quantum mechanical fields.


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