# 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)?

## 1 Answer

Not sure I understand what exactly you mean by timing the measurement. If we reduce the frequency enough so that we are dealing with low frequency radio waves, then one can measure the direction of the field based on the force it would exert on a charged particle. However, for light, the frequency is too high to do that. The integration time of a photo detector is typically much longer than the period of a light field. So although one can argue about this on a theoretical level, one would not be able to perform such an experiment in practice.

In the quantum case, remember that one can always perform a unitary transformation to convert linear polarization basis into a circular polarization basis. In effect, the measurement setup performs such a unitary transformation to bring the state into the eigenbasis of the measurement setup.

Another issue with the measurement is the phase. (Perhaps this is the core of your problem.) At very low frequencies, we can measure the phase of the field, because the time it takes for the measurement is much shorter than the period of the wave. We cannot do this at light frequencies. Therefore, we can only measure phase as a relative concept using interference effects. It is for this reason that the quantum measurement becomes probabilistic. We simply don't know what the absolute phase is, because we cannot measure it.

• Thanks. If by "light frequencies" you mean "visible light frequencies", let me point out that I don't particularly care about visible light here, any frequency is fine. By timing I mean that, knowing the frequency of the incoming (assuming it is monochromatic) beam, I could think of just "looking" at the measurement outcome at fixed intervals, in order to always catch the wave on a fixed direction. But it seems like you are saying that even actively measuring the rotation of the way is too much for current day technology? That's a bit surprising honestly – glS Oct 11 at 9:33
• Isn't there a way to use very long-wavelength to do this sort of thing? Of course one would need to also have detectors and polarizers that work at that wavelength, which I don't if they're available. I do know of quantum optics experiments done at telecom wavelengths though, so I though it wouldn't be to hard to do something like this. Regardless of the practicality of the experiment though, you do agree that in principle you should be able to measure this, yes? – glS Oct 11 at 9:35
• 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 at 12:57
• good point. Still, my main question stands: should this be possible in principle? And if so, how is this consistent with the fact that in the corresponding QM description it is not? – glS Oct 13 at 20:49
• QM only addresses the probabilistic nature, because in practice it is all that one can observe. So even if it would be in principle possible to resolve a single oscillation of an optical field, one would then not expect the result to correspond to what QM tells us. Remember, physics is a science. As such, we (should) only make statements about what we can observe. To say that we should be able to say if something can happen in principle, assumes that we have knowledge of something that cannot be confirmed scientifically. – flippiefanus Oct 14 at 4:15