# Is the instantaneous polarisation direction of circularly polarised light observable?

In classical electromagnetism, a circularly polarised electromagnetic wave is described by an electric field $$\mathbf E(\mathbf r, t)=\mathbf E_0 e^{i(\mathbf k\cdot\mathbf x-\omega t)}$$, where $$\mathbf E_0\in\mathbb C^3$$ has the form (assuming without loss of generality that the propagation direction is $$z$$) $$\mathbf E_0=(E_0,i E_0,0),\qquad E_0\in\mathbb R.$$ Because only the real part of $$\mathbf E$$ is physical, this corresponds to the physical electric field $$\mathrm{Re}[\mathbf E(\mathbf r,t)]=E_0\big( \cos(kz-\omega t),\sin(kz-\omega t), 0 \big).$$ In other words, the direction of the electric field rotates in the $$xy$$ plane.

This should mean that, even if only in principle, it should be possible to measure the instantaneous direction of the polarisation of a coherent light beam. For example, I should be able to set up an experiment in which, by measuring the polarisation with an high enough frequency, I should observe the polarisation direction rotating in time.

In particular, I shold then also be able to find positions where to put linear polarisers in such a way to have the light always pass through the polarisers without attenuation (because if I know the frequency of the light, I also know the distance corresponding to a full rotation of the polarisation, and then if I put linear polarisers at these positions in a suitable direction, the light would always cross them without any attenuation).

What I find odd is that assuming the above is true, I'm confused as to how this situation would be described when switching to the formalism of quantum mechanics (QM). In QM, circular polarisation is a specific polarisation state, which I can write as $$|0\rangle+i|1\rangle$$, and there is no notion of a "rotation in time". This is at odds with the classical description, so how do we describe this classical situation from a QM point of view (if at all possible)?

• Do the calculation: 1550 micron (telecom wavelength) corresponds to an oscillation period of about $5\times 10^{-15}$ s, (5 femto-seconds). There does not exist any electronics that can sample an electromagnetic field at this rate. There wont' be in the foreseeable future. – flippiefanus Oct 11 '19 at 12:57