If light is passed through two polarizing filters before arriving at a target, and both of the filters are oriented at 90° to each other, then no light will be received at the target. If a third filter is added between the first two, oriented at a 45° angle (as shown below), light will reach the target.

Why is this the case? As I understand it, a polarized filter does nothing except filter out light--it does not alter the light passing through in any way. If two filters exist that will eliminate all of the light, why does the presence of a third, which should serve only to filter out additional light, actually act to allow light through?

Image of three polarizers, target is at the right

  • $\begingroup$ A polarizer does alter the light!!! it! It causes a 90 degree rotation. If the light comes in vertical it leaves horizontal. Thus when you add the 3rd unit it will pass about 50% on to the next filter. $\endgroup$ Commented Mar 6, 2022 at 13:20

7 Answers 7


This link: http://alienryderflex.com/polarizer/ has an excellent explanation; much better than anything I could write here.

Essentially, it says that this occurs because the 45 degree filter outputs a projection of the vertical rays at 45 degrees. This, in turn, has a horizontal component, which the final filter projects in its output.

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    $\begingroup$ It might be worth pointing out that, although one can explain the effect classically, one can also do the experiment with single photons. In that case, a quantum mechanical interpretation is needed. $\endgroup$
    – hanno
    Commented Jan 25, 2014 at 0:37
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    $\begingroup$ @hanno very true! But, and forgive me if this sounds at all dismissive, I wish to leave my answer as is because it is more likely that a google searcher will be looking for a brief and succinct explanation to the classical effect. However, I do encourage someone to post a quantum mechanical explanation; that would definitely round out this question nicely. $\endgroup$
    – Jim
    Commented Jan 26, 2014 at 14:24
  • $\begingroup$ @Jimnosperm - a year and some later, I added that second explanation. Of sorts. $\endgroup$
    – Floris
    Commented Apr 10, 2015 at 22:27
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    $\begingroup$ I have read the link in this answer and must say I am not really happy with it. It just postulates that the filter can "take the projection of the polarization vector" without explaining how that happens. What does the filter actually do to the wavefront that could have this effect? I understand the QM explanation, but the classical explanation seems to me to be no explanation at all, at least as it is presented here. $\endgroup$
    – Thriveth
    Commented Apr 3, 2016 at 4:21
  • $\begingroup$ Very good, but Malus' law uses a SQUARED cosine - the idea is fine, but 50% is the magnitude reduction of a SINGLE 45 degree polarizer, not two. $\endgroup$
    – kaay
    Commented May 12, 2017 at 16:24

This answer was written for another question that was deleted a few minutes ago. I decided to post it here even though the effect it describes duplicates Floris' answer:

Photons passing through a medium don't just punch their way through like bullets. They are absorbed by the atoms of the medium and then re-emitted. (Incidentally, the reduction in speed for light passing through a medium is caused by each photon "orbiting" the nucleus of an atom before being re-emitted. They travel at c, but the distance travelled is greater.)

Light passing through a vertically oriented polarization filter emerges with a classical mechanical wave polarization in the vertical direction. If the light then passes through a horizontal filter, 100% of the classical mechanical wave action is eliminated.

But if you insert a filter oriented at 45 degrees between the vertical and horizontal filters, you introduce an element of quantum probability into the apparatus. It creates a quantum effect, and you can witness the quantum probabilistic transmission of light.

All the photons passing through the vertical filter are vertically oriented UNTIL they pass through the filter oriented 45 degrees. When they're absorbed and re-emitted by the 45 degree filter, 50% are vertically oriented, and 50% are horizontally oriented, as the quantum effect allows photons to be EITHER up or down, vertical or horizontal. 45 degrees is not allowed, but as 45 degrees is 50% of the angle between the vertical and horizontal filters, the emissions from that filter are half vertical and half horizontal.

The horizontal filter then emits only the horizontally oriented classical mechanical wave action that has passed through the 45-degree filter.

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    $\begingroup$ Note that the effect can be described purely classically. $\endgroup$
    – jinawee
    Commented Nov 15, 2018 at 7:19

A year later, here is a probabilistic (pseudo QM) explanation.

I am confused by the diagram that appears to show unpolarized laser light - I thought that most lasers by their nature produce polarized light; after the first polarizer that question is moot, so let's start there.

A polarized photon can be thought of as being in a mixture of states - when it approaches a polarizer, it's either parallel or perpendicular (it either passes, or it doesn't). When you polarize a photon and then immediately "test" it with another polarizer at right angles, you know the state it's in: it is "perpendicular" to the second polarizer and will be stopped.

But when you have another angle, you have a certain probability of passing or not passing. In particular, when you're coming in polarized at 45°, there is an equal chance of passing or not passing (it is in a mixture of two states, if you like). So half the photons will pass the second polarizer - and they will come out "rotated".

These photons now hit the third polarizer - again, at 45 degrees. Again, you have a 50-50 chance that such a photon is parallel, and is passed on.

We therefore have a 1 in 4 chance of passing the two polarizers, where before we had none.

  • $\begingroup$ After reading „So half the photons will pass the second polarizer - and they will come out "rotated".“, shouldn't this rotation be quantized? $\endgroup$ Commented Nov 1, 2018 at 18:21
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    $\begingroup$ @HolgerFiedler I don’t understand your question - yes they will have a new polarization state but what do you mean by “quantized”? It is in a pure state relative to one set of axes but a mixture compared to another. $\endgroup$
    – Floris
    Commented Nov 1, 2018 at 18:30
  • $\begingroup$ Are the photons rotated only to some angles or is any angle possible? $\endgroup$ Commented Nov 1, 2018 at 18:40
  • $\begingroup$ @HolgerFiedler I believe any angle is possible $\endgroup$
    – Floris
    Commented Nov 1, 2018 at 22:02
  • $\begingroup$ @HolgerFiedler Space itself is not quantized. There is neither a smallest translation nor a smallest rotation. In fact, given that we live on a rotating sphere, it would be incredibly obvious to us if there were a smallest unit of either $\endgroup$
    – Jim
    Commented Aug 19, 2022 at 13:41
  • quantum experiments are probabilistic (weird non-classical thing):

    • they take a mixture of states
    • the result of the experiment is one of the states
    • each result occurs with a probability proportional to the wave function amplitude of the states squared
    • after the experiment, the system assumes the state that was observed
    • well known example: probability that an electron hits a given part of a wall in the double slit experiment
  • when a photon hits a polarizer, that is an experiment, and two outcomes are possible:

    • it passes through, and its new orientation is the same as that of the polarizer
    • it does not pass through
  • a physical property of polarizers that we take for granted: the probability that a photon passes is given by the angle between the polarization and the filter:

    • parallel: 100%
    • perpendicular: 0%
  • but what about 45 degree polarized photon? We can conclude that from the horizontal and parallel ones:

    • every photon polarized vertically is equivalent to a superposition of a 45 and -45 degree states with wave function amplitude $\frac{1}{\sqrt{2}}$ each
    • the 45 state always passes (parallel)
    • the -45 degree state never passes (perpendicular)
    • so the probability that the vertical photon will pass is $\frac{1}{\sqrt{2}}^2 = \frac{1}{2}$ (probability is wave function amplitude squared).
  • when light passes the vertical filter, all photons that pass are now forced to be in the vertical polarized state due to wave function collapse (or the universe splitting up, or whichever interpretation you subscribe to)

    • if those photons hit the horizontal filter immediately, 0% of them pass (perpendicular)

    • if they hit a 45 degree filter first, $\frac{1}{2}$ pass, and assume the 45 degree state.

    • Then they hit the horizontal filter, that one is also at 45 degrees from the previous state, so again $\frac{1}{2}$ pass.

    • So the net number that passes is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

    • If the incident light is unpolarized and a vertical polarizer is present, this will add a factor of $\frac{1}{2}$ in front of this answer to get a transmission of $\frac{I_0}{8}$ ($I_0$ being the incident intensity on the first of the 3 polarizers). This is because 1/2 is the average probability over $cos(\theta)^2$ for all angles.

  • this is all very similar to multiple chained Stern Gerlach experiments

If this particular example is done with large amounts of light, not just single photons, then it does not clarify if there are truly probabilistic things happening at the atomic level, or if light is just a classical continuous wave with intensity = amplitude squared, and the polarizes are simply deterministically reducing light intensity based on the incoming polarization.

However, if we use single photons, e.g. generated with spontaneous parametric down-conversion, we observe that single photons either pass and exit with same incoming frequency (and thus the same energy since $E = hf$), or don't pass at all, then the fundamental probabilistic nature of things becomes more apparent.

One may then ask if there is some other not yet observed photon state besides polarization which is randomly generated in our experiment, and which if observed could deterministically decide if a given photon would pass or not if we were able to observe it. But the Bell tests suggest that this is not the case, unless there can be faster than light action.

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    $\begingroup$ A polarizer does alter the light!!! it! It causes a 90 degree rotation. If the light comes in vertical it leaves horizontal. Thus when you add the 3rd unit it will pass about 50% on to the next filter. Blindly applying probability and super position is a mistake many scientists make. $\endgroup$ Commented Mar 6, 2022 at 13:23
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    $\begingroup$ Best explanation by far. Needs more upvotes! $\endgroup$ Commented Oct 27, 2023 at 15:29

Another way to think of this is to see the polarized light as truly polarized, no super position of states but that the polarizer acts on a probability basis when at 45 degrees. So a nicely polarized photon arrives at the filter and can't happen to find a nice electron waiting for it in the perfect state or position to cause transmission, so it gets reflected. This happens about 50% of the time for the 45 degree polarizer.


Filter 1 allows the component of the light vibrating in the same direction of the grating though. If the angle of the first polarizer is 90 degrees (parallel to the y-axis), then the percent of the photon’s energy passing through will equal the sine of the photon’s angle of vibration (polarization mode). With an equal distribution of polarization modes, that component will average out to 50% of the total energy passing through the first filter, and all of it will have the same 90 degree polarization mode.

With just Filter 3 added, all of the energy is vibrating at 90 degrees has no component along the x-axis.

Adding Filter 2 at 45 degrees results in a percent of the 50% equal to the sine of 45 degrees: (sqrt 2)/2. Passing through another 45 degree filter will reduce it a further (sqrt 2)/2.

Filter 1: 1/2 Filter 2: (sqrt 2)/2 Filter 3: (sqrt 2)/2

Summary: (1/2)•((sqrt 2)/2)•((sqrt 2)/2) = (1/2)•(2/4) = 1/4

25% of the original optical energy will exit Filter 3.

  • $\begingroup$ As a side note, polarization mode becomes critical for long haul high speed optical circuits. If one mode takes too much longer than another mode to reach a receiver, a single pulse of light can end up “smearing” into the next pulse making ‘on’ and ‘off’ more difficult to distinguish. This occurs due to birefringence, non-circular core profiles, differently manufactured fusion spliced fibers, and excessive curvature in the fiber. $\endgroup$
    – Digcoal
    Commented Apr 14, 2021 at 23:10

I think the 'quantum eraser' will interest you (wikipedia.org/wiki/Quantum_eraser_experiment#Introduction, see long answer below). It's weird that every light transmission/reflection/absorption event is probabilistic (Feynman's rules) and that a polarizer can 'label' a photon's polarization state on output (see below)...

"Although the nonlocality of quantum mechanics is most apparent in tests of Bell’s inequalities, it also plays a central role in experiments exploring complementarity. One such, the “quantum eraser,” was discussed by Scully, Englert and Walther [20] in connection with the micromaser. ...This particular version of the quantum eraser has a straightforward classical-wave explanation when the light source is describable in terms of coherent states. Thus it could be argued that there is nothing particularly quantum about this quantum eraser. Nevertheless, Jordan has proposed a similar Mach-Zehnder version of this experiment [23], in which he has argued on the basis of the correspondence principle that despite the existence of a classical explanation, such first-order interference experiments can be interpreted as true quantum erasers. ...We stress that it is the mere possibility of obtaining which-path information that destroys the interference; no actual polarization measurements need to be made." - https://arxiv.org/pdf/quant-ph/9501016.pdf

My understanding: They say that the first two polarizers are considered "labelers", since they determine polarized state output. And as stated, this 'which path test' quantum mechanically entangles the photons, which is evident as a wave-particle-duality when the 'test' involves interference.

Here's a cool description by Scientific American: arturekert .org/miscellaneous/quantum-eraser.pdf

"Wave-particle duality, which represents the complementary nature of the wavelike and particlelike behaviors of a quantum system, is perhaps the example that has garnered the greatest share of attention. The wavelike behavior is manifested in interference experiments. However, when a welcher-weg ~which-path! measurement is carried out on an interfering system, the system becomes entangled with the measuring apparatus, so that the paths of the system become distinguishable, and the fringe visibility vanishes." - people .bu.edu/alexserg/PRA32106.pdf

So yeah, every time QM is impossible to intuit. Frankly, I always end up confused interpreting real life examples. I was tripped up because it did not seem that the expected two slit pattern appeared and the classical explanation seemed a suspicious coincidence. When considering this for quantum cryptography, I think it is fair game to consider the light subsequent the eraser and between the 3m as impossible to observe without interfering with communication (ie. unhackable). See: optics .rochester.edu/workgroups/lukishova/QuantumOpticsLab/2010/OPT253_reports/Justin_Essay.pdf


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