The sun sends out unpolarized light. There are infinite degrees in which these photons are oriented. A polarizer only lets in light of one specific orientation.

In statistics, the infinitesimal area/slice of a single value in a continuous probability distribution is virtually meaningless, but I'll call it zero.

How can there be any light coming through a polarizer if only a 'slice' of the continuous circle of orientations is being allowed through? Is it that the polarizers allow light within a certain degree of deviation of that orientation?

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    $\begingroup$ I'm curious: why did this post receive downvotes? It seems like a thoughtful and well-framed question. It's based on a misconception, yes, but if anything that makes it a better question? $\endgroup$
    – Myridium
    Apr 12 at 4:30
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    $\begingroup$ If you have three filters, things start to get weird, as shown at the beginning of this video on quantum entanglement youtu.be/zcqZHYo7ONs $\endgroup$
    – Rodney
    Apr 12 at 9:37
  • $\begingroup$ @FlatterMann - I'm wary of cluttering the comments. Happy to chat here if you like: chat.stackexchange.com/rooms/145291/… $\endgroup$
    – Myridium
    Apr 12 at 19:49
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    $\begingroup$ I’ve hidden a number of comments that should have been posted as answers, and replies to them. $\endgroup$
    – rob
    Apr 13 at 6:04
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    $\begingroup$ @Myridium I agree, it's a perfectly good question. It's just that downvoting and patronizing "quality control" are sadly the national sports of SE. $\endgroup$
    – Tfovid
    Apr 13 at 7:45

5 Answers 5


A polarizer is not a gatekeeper that lets correctly polarized photons through and says "you shall not pass" to incorrectly, or halfway-correctly polarized ones. It interacts with each and every one of them.

Any passive optical device (such as a polarizer) has an effect on the light by absorbing and re-emitting photons. Incoming photons interact with the polarizer (which may be an anisotropic crystal or a wire grid or whatever) by "jiggling" the charged particles (mostly free and bound electrons) it is made of back and forth.

If we have a perfect polarizer, photons that are polarized 90 degrees to the polarizer will always be absorbed and, possibly, reflected or deflected. But when struck by any other photon there's a non-zero chance of re-emitting a photon travelling in the same direction, but polarized parallel to the polarizer.

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    $\begingroup$ Ah! This explains one of my long-standing problems with polarisers: if you place two at 90° to each other, they let very little light through — but place a third between them, at 45° to the others, then you get a lot more light through. That seems to make no sense if a polariser is a simple filter (which is how they're usually described). $\endgroup$
    – gidds
    Apr 12 at 22:14
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    $\begingroup$ @gidds Yes, that's exactly the point I was trying to make! Glad, I managed to get it across. $\endgroup$
    – LLlAMnYP
    Apr 13 at 7:38
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    $\begingroup$ Even accounting for polarizers being more than simple filters, they still let more light through than they "should". This being a proof of quantum mechanics: youtu.be/zcqZHYo7ONs $\endgroup$ Apr 14 at 1:43
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    $\begingroup$ @Tyrannosaur yes, I agree. It seems strange that many people still believe that the three filters problem can be simply explained by saying "the filters mess with the photons". This is not how quantum mechanics works. $\endgroup$
    – Guilherme
    Apr 14 at 5:42
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    $\begingroup$ @Tyrannosaur the video you linked insists on the same premise I am rejecting. A photon that "passed" through a polarizer is not the same photon that was incident on the polarizer. It is, at the very least, in a different state of polarization. If you will, the photon $|\psi\rangle$ was incident on the polarizer $A$. After interacting with the polarizer we will have a different photon that can be describe as $\hat{A}|\psi\rangle$. Of course the interaction of photon $\hat{A}|\psi\rangle$ with polarizer $B$ is different to the interaction of $|\psi\rangle$ with $B$. $\endgroup$
    – LLlAMnYP
    Apr 14 at 6:25

A polarizer only lets in light of one specific orientation.

A polarizer passes the component of the incident light that is aligned with its axis of polarization.

Any polarization that is not perfectly perpendicular to the axis of polarization has a non-zero component parallel to the axis.

Net result, the intensity of the transmitted beam, for an ideal polarizer and a linearly polarized input, is $I_0\cos^2\theta$ where $I_0$ is the intensity of the input beam and $\theta$ is the angle between the axis of the polarizer and the E-field polarization of the beam.

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    $\begingroup$ A polarizer will pass 50 of 100 photons of random polarization .... A photon has a probability of going thru the polarizer ..... the probability is close to 1 when perfectly aligned and 0 when perfectly unaligned. Your equation above is a probability equation. $\endgroup$ Apr 11 at 23:07
  • $\begingroup$ A practical, realisable polarizer must allow more polarizations that perfectly aligned. $\endgroup$
    – D Duck
    Apr 12 at 8:22
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    $\begingroup$ @DDuck, Even an ideal one does, remember that $\cos^2\theta$ doesn't drop off particularly fast (i.e. the derivative is ~0) for $\theta$ near 0, and it's only 0 when $\theta$ is exactly $\pm\pi/2$. $\endgroup$
    – The Photon
    Apr 12 at 15:17
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    $\begingroup$ It may help to think of the polarization of the light as a vector quantity. $\endgroup$
    – Izzy
    Apr 13 at 8:45

Are you familiar with the rope in a slit analogy for polarization?

Rope sits in a vertical slit. If you wobble it left-to-right, the wave gets stuck at what seems like a narrow hole. If you wobble it top-to-bottom, the wave can pass through as if it were empty space (which it is). If you wobble it diagonally, something interesting happens. Beyond the slit, the rope will behave as if you gave it a (smaller) vertical wobble. Such of your wave as was horizontal gets suppressed, but because there was some motion in the vertical axis there's still something to propagate. That is, the polarizer changes the wave you put in.

You might say, instead of "A polarizer only lets in light of one specific orientation" that "A polarizer only lets out light of one specific orientation." It can satisfy that constraint either by blocking the light or by changing the orientation, and it does both in some mixture based on the input orientation. You can check this with two polarizers at 45 degrees to each other. Since there is only vertically polarized light coming out of the first, if the second could only accept 45-degree light it would be black. Instead it accepts some of the vertical light but lets it through with the new polarization.

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    $\begingroup$ +1 Rope in a slit is a good analogy. Pushing a card through a slit is a bad (yet common) analogy, as it passes only under very narrow range of angles. That might have confused the OP. $\endgroup$
    – dominecf
    Apr 13 at 8:28

There are infinite degrees in which these photons are oriented.

If you mean to say there there are infinite degrees of freedom in the photon polarization, this is not true. As a transverse vector field, electromagnetic waves have only two degrees of freedom. Since the EM wave equation is linear, it obeys the principle of superposition; i.e. any linear combination of solutions is another solution, so while there are infinitely many different polarizations, as you've noted, there are only three linearly independent polarizations for a vector field, one of which corresponds to longitudinal waves—which are forbidden for EM waves—yielding two degrees of freedom.

Now that we've established that there are really only two, rather than infinite, properly distinct (i.e. orthogonal) polarizations (subject to a choice of basis), we can see that it does make sense to say that a polarizer simply blocks one of them and allows the other to pass. Therefore, since unpolarized light contains a mix of both the transmitted and blocked polarizations, some nonzero amount will be let through the polarizer.

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    $\begingroup$ I strongly suspect that "There are infinite degrees in which these photons are oriented." means that it could be $1^\circ$, it could be $43^\circ$, and any of an infinitude of other degrees. And that it does not refer to degrees of freedom. $\endgroup$
    – Arthur
    Apr 12 at 10:07
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    $\begingroup$ @Arthur I agree that that is probably what OP meant, but that also seems to be precisely the point of confusion. Though there are infinitely many polarization axes, the polarizer prescribes exactly one 2D basis and only cares about the projection to those axes. $\endgroup$ Apr 12 at 11:17
  • $\begingroup$ "electromagnetic waves have only two degrees of freedom" There's four degrees of freedom, see the Stokes parameters. The phase offset between the two linear directions leads to the components of circular polarization. $\endgroup$
    – user71659
    Apr 12 at 18:29
  • $\begingroup$ @user71659 Polarization is a vector in $\mathbb{C}^2$, a 2 dimensional vector space, although I suppose you could say it's 4 if you insist on everything being real. $\endgroup$
    – Sandejo
    Apr 12 at 21:59

A polariser causes incoming light to decohere into a classical probabalistic distribution of orientations. The polarisation space of light has two (complex) dimensions. A polariser produces very different outcomes for light which is aligned with the polariser, and light which is aligned orthogonally to it. The interaction of the polariser with the light and with the broader environment causes quantum decoherence, which is the phenomenon where a quantum superposition `collapses' into a classical distribution. (Instead of a coin being in a quantum superposition of Heads and Tails, the universe splits into two parallel universe, where in one it's 100% Heads and in the other it's 100% Tails).

Let's assume monochromatic light with a fixed direction, shining straight into the polariser at an orthogonal angle, and an infinitely thin polariser located at $z=0$ for concreteness. The polarisation of any wavepackets of light at $z=0$ can be decomposed into a linear combination of two vectors (with complex components). To anthropomorphise the situation and make it relatable, a polariser takes the polarisation vector, decomposes it into 2 vectors in a particular basis (one that aligns with the polariser orientation, and the orthogonal one) and then throws out the half that's aligned the wrong way.

So you see, rather than thinking about light having just the right polarisation to pass through, you should be thinking about what the length is of the vector that projects onto the polarisation axis. (Slightly rough picture but it conveys the correct mathematical description, without overcomplicating with complex numbers etc.)

We would call something a 'polariser' when we observe that it causes light to behave as described above. At a more technical level, what's happening is that a polariser is a system which causes decoherence of light into the component that aligns with the polariser direction, and the orthogonal direction. Then, classical probability theory takes over: in rough terms, each photon has a % chance to be aligned correctly, and if so then it passes through.


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