Can a single photon be elliptically polarised ? or is elliptically polarised EM-radiation a mixture of circular and linear polarised ?
2 Answers
A single photon can indeed be elliptically polarized. A photon is, ultimately, a (quantized) excitation of a specific mode of the electromagnetic field, and while it is usually convenient to use modes with specific properties (like a definite frequency, wavevector, or linear or circular polarization), this is in general not necessary, and there are plenty of ways to define modes with elliptical polarization, or even with a spatially-dependent polarization.
On the other hand, even if you have a single photon that's elliptically polarized, in general it will also be possible to describe it as a superposition of a horizontally polarized photon and a vertically polarized one, as well as a superposition of circularly polarized photons. However, this is a red herring, because the statement "the system is in a superposition state" is always true, unless you're working in a trivial space of dimension 1.
To give just one example of this, a wavefunction $\psi(x)=Ae^{-x^22/\sigma^2}$ with a tight width $\sigma$ might be considered to just be the state "the particle is very close to the origin", and one would be justified in thinking this is just "one state". However, a look at its momentum-space representation shows that this state is also the superposition of very many momentum states with a very wide span in momentum. So is the system in a superposition or not? It's a meaningless question!
So: you can indeed have states of the EM field which contain a single photon with a well-defined elliptical polarization. These states are also superpositions (but note that this is different from a mixed state!) of linearly polarized photons and of circularly polarized photons. And the distinction is meaningless otherwise.
- 137,480
Sure it can. Using a combination of polarizers and waveplates you can produce any pure polarization you want, from linear to circular to elliptical. And you can verify this by further combinations of polarizers and waveplates, undoing the polarization change and setting up a situation where a photon would only get through 100% of the time if it were, say, elliptically polarized at some particular point in the beam path.
There's nothing fundamentally more "real" about any of these states than any others. All of these are still pure states. There's no rule that says a pure state has to be in any sense "nice" or symmetrical or anything. Remember that a pure state in one basis is a superposition in another; there's no fact of the matter as to whether it "really" is a superposition or not.
This is distinct from the situation in, say, an unpolarized beam. The polarization state of such a beam cannot be described with any pure state. Instead you have to use a density operator, which in effect describes a classical-ish probability distribution of quantum mechanical pure states.
- 759
- 3
- 8