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I ask "After a slit, are photons polarized?" and later found two answers for the question "Why does the electric field dominate in light?". From this two answers I have concluded or summarized, that photons which are influenced by sharp edges, are polarised. And the reason for this phenomenon is the interaction between the electric field component of the photon and the surface electrons of the edge.

Now I'm interested in how this happens in detail. May be any quantization during the rotation of the light's electric field takes place? Perhaps the rotation led to a deflection or dissipation of the light?

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HolgerFiedler
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  • What do you mean by "quantization taking place"? Quantization is a theoretical procedure that takes a classical theory and süits out a quantum theory, not a physics process that can take place. – ACuriousMind Dec 20 '15 at 19:09
  • https://en.wikipedia.org/wiki/Polarizer#Wire-grid_polarizer –  Dec 20 '15 at 22:09
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    I don't understand. In a polarizer there is no rotation of the polarization (direction of the el.field) during propagation in a polarizer. One component (perpendicular to the slit) is not transmitted, so you have losses. A rotation can happen in a birefringent medium. – scrx2 Dec 20 '15 at 22:47
  • @fpdx Take a third polarizer between two crested with 90° polarisers and you will see, that some light is going through all polarizers. Hence the polarisers manipulating the light. And light with randomly distributed electric field components of the involved photons (unpolarized light) can go through one well designed polarizer by 50% and all this 50% is polarised then. – HolgerFiedler Dec 21 '15 at 05:15
  • @brucesmitherson From your link: "A wire-grid polarizer converts an unpolarized beam into one with a single linear polarization. Coloured arrows depict the electric field vector. The diagonally polarized waves also contribute to the transmitted polarization. Their vertical components are transmitted, while the horizontal components are absorbed and reflected. (This is not clearly shown.)" And my question is about how this happens in detail. – HolgerFiedler Dec 21 '15 at 05:19
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    @HolgerFiedler It seems to me that you are pointing out what is called the leaking of a real polarizer (usually less than 1%). No rotation happens, just absorption of one component with <100% efficiency. You always have some leaky transmission after two crossed polarizers. – scrx2 Dec 21 '15 at 11:11
  • @fpdx No, this has nothing to do with leaky transmission. You really get about much more than a few percents. – HolgerFiedler Dec 21 '15 at 11:23
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    @HolgerFiedler Ok. If you have: P1(0°) -> I1 -> P2(45°) -> I2 -> P3(90°) -> I3 , Malus law predicts: I2=I1/2, I3=I2/2=I1/4. Is this what you mean? But still, IMHO there is no polarization rotation, just classical decomposition and losses. – scrx2 Dec 21 '15 at 11:46
  • Or the polariser 0° and 180° both let through 50% of the light with polarisation from -45 to 45 and with 135 to 225° (this of course is possible for well designed polarizers and for some wavelength. Than one get 25% behind. And a third polariser under 45° does not change to much. Or both polarisers turn the light from above mentioned angles to zero and 180° and no light comes out from the second polarizer. In this case a polarizer under 45° has to rotate the light. – HolgerFiedler Dec 21 '15 at 15:59

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Polarization rotation due to a polarizer: in the comments I said that a polarizer does not rotate the polarization (a birefringent medium does). It blocks one component, the linear component parallel to its axis. I hold this, but I thought about the following system where a polarization rotation seems to happen due to the action of many polarizers. I never thought about it before, I hope (it's true and) it can help (but sorry, no quantization here, just classical).

Let's imagine a linear polarization at $0^o$ arriving on a series of $N$ polarizers as in this picture: series of polarizers

each polarizer is rotated by the same angle with respect to the previous one, so their axis spiral from horizontal up to the angle of the last one ($\theta(N)$, $=\pi/2$ in the figure, but you can choose it). Malus law says that the transmission of one polarizer is $I=I_o \cos^2(\alpha)$, where $\alpha$ is the angle between its axis and its input linear polarization. So the total transmission from the series of polarizers will be $T=(\cos^2(\theta(N)/N))^N$, where we can choose the angle of the last one ($\theta(N)$).

In the next figure I plot $T(N)$ for $\theta(N)=0, \pi/4, \pi/2$. When $\theta(N)=0$, no matter how many polarizers are there, they'll be all parallel and $T=1$ always. For $\theta(N)=45^o$, if there is only one polarizer (i.e. the last one, $N=1$) $T=0.5$. For $\theta(N)=90^o$, if there is only the last polarizer ($N=1$), $T=0$ because it's perpendicular to the polarization.

T(N)

But in any case, increasing $N$ the transmission goes to 1. So, the polarization is kind of guided by the polarizers, and actually rotates with little losses if $N$ is large. I think this could be what happens in the liquid crystal displays, where the twisted arrangement of the molecules act as the polarizers here. This is the closest I could think when you mention rotation by a polarizer.

scrx2
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  • "Therefore the charges accelerated parallel to the surface will re-emit light linearly polarized along a direction parallel to the edge." seems to be not correct. I'm sure that such re-emission should happens with different frequencies of the photons and the fringes have to have different colour edges.

     – HolgerFiedler Dec 25 '15 at 09:24
  • In the second point I'm with you. Birefringent mat rials rotate the light and/or divide it in two paths. As you pointed out, you (and I too) don't able to describe the quantised process (if it is possible at all). Happy holidays – HolgerFiedler Dec 25 '15 at 09:38
  • @HolgerFiedler: it was wrong... I removed that point – scrx2 Dec 25 '15 at 21:50