Is is possible to generate circularly polarized light using a rotating half-wave plate?

Question

Let's assume that I have a laser beam with a frequency $\omega$ that is linearly polarized (say coming out of a laser diode). If I pass it through a half-wave plate (HWP), the light that is transmitted will still be linearly polarized, although with its axis rotated with respect to the initial one depending on the angle between the axis of the HWP and that of the polarization. Then, if I keep the input laser fixed, and put the HWP on a motorized mount that is rotating with an angular frequency $\Omega$, the polarization axis of the output beam will also be co-rotating at a frequency $\Omega$. Let's assume first $\Omega \ll \omega$. What will happen if I start increasing the rotation frequency of the mount until $\Omega=\omega$? Will the output beam start behaving at that point like a circularly polarized beam?

From an EM point of view, I'm temped to say yes because that seems like the definition of a circularly polarized beam: the electric field vector will be tracing a circle in space as a function of time at the frequency of the beam.

However, from a QM, single photon point of view, this sounds wrong. If the photons are in the pure state $|R\rangle = \frac{1}{\sqrt{2}}(|H\rangle+i|V\rangle)$, then that state is clearly not $|H\rangle$,$|V\rangle$, $|+\rangle$ or $|-\rangle$, nor a mixed state. We can see that the density matrix $|R\rangle\langle R|$ is clearly different from that of any of the other states, namely it is an arrow pointing in the $y$-axis of the Bloch sphere. The output state of the beam after the rotating HWP will be an arrow that is tracing a circle along the $x-z$ plane of the Bloch sphere, but that is certainly not $|R\rangle$.

If I try to describe the polarization of a laser beam that is diagonally polarized ($|+\rangle = \frac{1}{\sqrt{2}}(|H\rangle+|V\rangle)$), then it would be clearly wrong to describe it as being a mixture of $|H\rangle$ and $|V\rangle$. It is also clearly not a quick cyclical succession of $|H\rangle \rightarrow |V\rangle \rightarrow -|V\rangle \rightarrow -|H\rangle$. Using the same reasoning it sounds wrong to describe $|R\rangle$ as a mixture of $|H\rangle$ and $|V\rangle$, or even a quick cyclical succession of $|H\rangle \rightarrow |V\rangle \rightarrow -|H\rangle \rightarrow -|V\rangle$, but to me the classical EM picture does suggest that (perhaps naively). There has been a related question asked about circularly polarized light as viewed from a rotating frame, although it seems like there wasn't a clear answer back then.

My initial scenario is a macroscopic EM wave, but I think I should be able to make sense of if down to the single-photon level. In principle, I could attenuate the laser beam a lot and then perform quantum state tomography on the beam using single-photon detectors (or, if really necessary, jump to an SPDC setup and carry out a similar experiment with a true single-photon source). I'm also very curious about the $\Omega>\omega$ regime.

I understand that it might be too hard to reach such high rotational frequencies to carry out the experiment using visible light, but I guess in principle we could switch to lower-frequencies such as micro- or radio-waves and try a similar experiment so I'm thinking the answer shouldn't depend on $\omega$ in general.

Answer 1

True circular polarized light is just a mixture of multiple normal/linear polarized light. According to QM if your laser emits one photon it will strike the plate either being reflected or transmitted .... when transmitted it will normal/linear polarized ..... which would just depend on angle of the plate.

Similarly you could just spin your laser and get the same effect.

Important point to realize here is all photons are linear/normal photons upon creation, if we mix 2 or more pure polarizations depending on the amount and angle of each one we can calculate a degree of circular polarization.

Answer 2

For simplicity, let's consider the two dimensional Hilbert space representing the polarization of the photon. Macroscopically, the polarization of light is represented by Poincare Sphere, which is the same thing of the Bloch sphere, so the last is enough to organize the argument.

The action of a linear polarizer is represented by a projector. When vertically oriented, it will be $P_V = |V\rangle\!\langle V|$. In Bloch sphere

$$ P_V = \frac 12 \left( \mathbb I + Z\right) $$

Where $Z$ is the z Pauli matrix. The state after the polarizer is (unormalized) $|\psi'\rangle\propto P_V |\psi\rangle$. Now let's consider any other direction on ZX plane. It could be parametrized by an angle $\theta$ and we could write

$$ P_\theta = \frac 12 \left ( \mathbb I \cos (\theta) Z + \sin (\theta) X \right ). $$

If $\theta =0$, we recover $P_V$, i.e., the polarization in the vertical direction. Directions in the Bloch sphere are not the same as directions in space. The state $|H\rangle$ is orthogonal to $|V\rangle$, but it is given by $\theta = \pi$ in the formula above, i.e., antiparallel to the Bloch vector of $|V\rangle$. If $|V\rangle$ and $H\rangle$ represent two perpendicular directions on the plane of polarization, any linear polarization could be written as a combination of these two, like $|P_\phi\rangle = \cos(\phi)|V\rangle + \sin(\phi)|H\rangle$. The angle $\phi$ are related to directions in the polarization plane, perpendicular of direction of propagation, so it represents direction in real space. To connect it with Bloch sphere, we could decompose the projector associated to the state above, and what we would find is that $\theta=2\phi$. From now and on, all the description could be done in Bloch sphere.

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Let's consider that our linear polarizer is rotating in a plane, parallel to that generated by $|V\rangle,|H\rangle$. For simplicity, the angle starts at $\phi(0)=0$ (vertical) and it rotates with constant angular velocity so $\phi(t)= \Omega t$. The projector at time t would be

$$ P_\Omega = \left (\mathbb I + \cos{(\Omega t)}Z + \sin{(\Omega t)} X \right) $$

And the state after the polarizer $P_\Omega |\psi\rangle \cos{(\Omega t)} \propto |V\rangle + \sin{(\Omega t)} |H\rangle$.

Look that all states of linear polarization in the polarization plane could and would be accessed, at some time t. But we will never access any state of circular polarization, like $|V\rangle + i|H\rangle$. In the Bloch sphere, this is more evident: we never leave the zx plane, so we couldn't access the whole Bloch sphere. A circular polarizer would be

$$ P_R = \frac 12 \left ( \mathbb I + Y\right ), $$

Never achieved by $P_\Omega$, no matter what frequency we set neither how many time we wait.

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It seems that we could generate a rotating polarized wave with rotating linear polarizer, but why it shouldn't be a circular polarized wave? By the reason that in circular polarization we have dephasing between the components of the electric field, something that we could not produce with rotations of the linear polarizer. You'll produce a rotating polarization, but where both components of $E_x$ and $E_y$ are always in phase, and it could not be called circular polarization, simple because it doesn't fit its definition and doesn't share its properties.

If you want to connect it with classical electromagnetism, you should work with the Stokes parameters, defined as

$$ \begin{aligned} S_0 &= E_x^2 + E_y^2\\ S_1 &= E_x^2 - E_y^2\\ S_2 &= 2E_xE_y\cos\delta\\ S_3 &= 2E_xE_y\sin\delta \end{aligned} $$

where $\delta$ is the dephasing aforementioned. Look that if you produce linear polarized light, $\delta = 0$, doesn't matter if you rotate the direction of polarization, which in this case is to change $E_x$ and $E_y$ in phase.

To match Bloch and Poincare sphere, look that $S_1$ corresponds to $-Z$, $S_2$ corresponds to $X$ and $S_3$ to $Y$.

Answer 3

1. How does a polarizer work?

Designed for the frequency range of the photons to be influenced, ideally 50 % of the photons pass through the polarizer (for photons with frequencies above and below the ideal frequency, either the transmission rate decreases or the photons are not all fully polarized). How do I influence a beam of photons at a polarizer?

To get a clearer idea here, I remind you of the experimental setup in which all photons are blocked behind two crossed polarizers. If you move a third polarizer at 45° between the crossed polarizers, light comes out again.

This only leads to the conclusion that the edges of the slits of the polarizers - or more precisely their surface electrons - interact with the electric field component of the photon and align it parallel to the slits. Although I am not familiar with this and do not rule out the interaction of the magnetic field component or a 90° alignment to the slit - but this does not change the fact that unpolarized photons are subject to a rotation of their E and B field components at the slit.

2. How do I obtain a torque for the field components of the photon?

In my opinion, there are - at least - two possibilities:

• When photons are emitted from a rotating molecule, the emission should carry some of the torque of the molecule - the E and B components of the photon rotate. In the case of the molecule, this is most likely to be elliptical circular.
• When passing through a crystal structure, there are structures that are characterized by a periodicity of their structure that exactly matches the resonance of the E and B field components of photons of excellent frequencies. And among these structures, there are some whose crystal lattice is not completely cubic, but is oriented in such a way that the photons receive a twist.

Danyel: Is is possible to generate circularly polarized light using a rotating half-wave plate?

If you follow the above, then clearly yes. A rotation of the influencing medium will give the transmitting photons a twist.

TL;DR
I deliberately write about a twist, not a spin. The spin of the subatomic particles is clearly to be seen in relation to the orientation of the magnetic field component of the electron and proton. The entire spintronics is based on magnetic phenomains. Talking about a spin in the photon is very misleading. There is a chirality of the sequence of the field components of the photon, and this is opposite for photons made from electrons and positrons.
TL;DR