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The part that still needs explaining is how the magnitude can be reduced. IIRC single photons can be polarized by these things, and AFAIK their wavelength is no

The part that still needs explaining is how the magnitude can be reduced. IIRC single photons can be polarized by these things, and AFAIK their wavelength is not changed so their energy is unchanged as well. I have always thought (how I got there I don't know) that the polarizer did something weird like rotate the photon to the correct phase angle AND passed it through with probability based on the angle / or didn't let it pass. This would give a similar reduction in intensity for a desktop experiment while having similar but different details when looking at the photon level. Is this correct?

Sharlin 4y ago
The polarization of a single photon is a quantum property, so it's essentially a probability distribution. Passing a photon through a polarizer modifies the probability distribution such that "more perpendicular" polarizations are now less likely and "more parallel" ones more likely. (Polarization is a superposition (ie. a linear combination) of two orthogonal basis vectors, and a polarizer projects a polarization vector onto one of the basis vectors.)
abdullahkhalids 4y ago
The explanation given is at the level of classical electromagnetism, and is sufficient to explain and predict experiments with regular light. If you want an explanation and prediction of what happens at the level of single photons, you need more structure from the theory of quantum optics. But briefly the filter at angle T does a measurement on the photon in the basis {T, T+pi/2}, and you end up seeing the photon on the other side of the filter only with whatever probability the photon has for being in state |T>, as opposed to state |T+pi/2>. So, filters are inherently destructive and fewer and fewer photons pass through each subsequent filter. And a photon that makes it through a filter at angle T, now has a new state |T>.
- jiggawatts 4y ago
  The thing is that there is no actual evidence that “single photons” exist in the electromagnetic field. That last bit is important: photons are a mathematical shorthand for dealing with emission and absorption by atomic orbitals. They’re not an explanation for continuous waves in between, but the mathematics largely works anyway because photons are very similar to how one would do a Monte Carlo numerical simulations of continuous wave phenomena. This has resulted in an unbelievable amount of confusion…
  - oh_my_goodness 4y ago
    There is lots of evidence for photons in the EM field, beginning with Planck's invention of photons. http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Planck%20(1900),%20Distribution%20Law.pdf http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Planck%20(... If you make the energy in a mode of the EM field continuous, you get the famous 'ultraviolet catastrophe.' This holds whether the matter involved has a continuous emission spectrum or not.
    - renox 4y ago
      I'm not so sure: 'black body' are made of atoms, atoms emits lights only on certain frequency due to the way electrons orbitals are structured, with discrete energy levels, but I don't see how this is related to EM fields themselves..
      - wyager 4y ago
        Free electrons also emit blackbody radiation. Photon quantization has nothing to do with the energy levels of atoms. It is a very fundamental property of the universe - you can tell because the only relevant constant is planck's constant.
        renox 4y ago
        I didn't say that light isn't quantized, I just said that for me blackbody radiation of ordinary matter isn't a convincing proof that light is quantized.. As for free electrons emitting blackbody radiations can you give me a reference? If you're talking about a wiggle (https://en.wikipedia.org/wiki/Wiggler_(synchrotron) https://en.wikipedia.org/wiki/Wiggler_(synchrotron)) I don't see how this is related to the issue.
        oh_my_goodness 4y ago
        The sun is full of free charged particles.
        4y ago
gnramires 4y ago
I think that although light is a quantum phenomenon and we can detect single photons, many people overlook that photons still behave similarly to Maxwell's (wave) equations. In particular, the average behavior is that of the electromagnetic wave (up to some extremes like extremely high energies). The 'bullet' model people think of when the word photon (particle) is mentioned is inadequate. This becomes clear in field theory (QED and QFT), where there's a more complete description based of this phenomena solely based on field (wavelike) behavior. It's believe any (small) system follows QFT exactly (there's still uncertainty around gravity). The exact nature of the relationship of quantumness and fields (i.e. how the single-particle behavior arises from QFT) is still unclear, which is why there are many competing interpretations of quantum mechanics. In the Copenhagen interpretation, which is the most "easy" one, the behavior of photons is just (almost) that of Maxwell's equations, on average, s.t. a single photon will be measured with probability equal to the average light intensity anywhere (they are said to "collapse" at the moment of measurement, which is surely a simplification of a more complete underlying theory).
roesel 4y ago
The magnitue drop is reasonably simple to understand in terms of fields. The oscillating optical field might be less effective at exciting material oscillations in the middle filter due to a mismatch in polarization, but it still does so at the same frequency. You can think of it as multiple photons (incoming field) collectively exciting the same electron on the same frequency but with reduced efficiency. The electron then re-emits fewer photons (outgoing field) of the same wavelength, leading to a lower light intensity detected after the filter.
- Arwill 4y ago
  I don't think photons are absorbed and re-emitted by electrons. At least that argument does not hold when discussing light slowing down in glass or water. Light is affected by the electromagnetic field of the material it is going trough, is slowed down, or absorbed based on some of its property, but photons that go trough are going trough without collision. Photons that get absorbed and re-emitted are scattered in all directions, and are mostly lost. You would not see a consistent image trough a polarising sunglass, if the photons you were seeing were re-emitted photons.
  - roesel 4y ago
    I think using the term "re-emit" is a mistake on my part. I fully agree that the photons are not absorbed and re-emitted. However, their speed of propagation in the material, and their properties can be altered by the interaction with dipole moments of the material. Therefore, photons do indeed "go trough without collision" but not without interaction. The dipoles in a dielectric material such as glass need to take the energy to oscillate from the field, and then they give it back to the field by radiating it out again. I agree that this is not "re-emission" in the absorption/emission diagram sense, and I understand that the field in a material cannot be separated into the material portion and the vacuum portion, since it is both at any given time, but what I have tried to outline in the previous comment is still a useful representation of the role of the material which the original article chose to leave out. I should have used "radiate" and "dipole" instead of "emit" and "electron" to make it a bit clearer.
gorkish 4y ago
At the single photon level, the photon that goes into the polarizer and the photon that goes out of the polarizer are not the same photon, so it's not right to say that a photon is changed or transformed. For a photon coming in at 45 degrees to the polarization angle, the probability that another photon will be emitted is sin(45deg) =~ 70% and the probability that it will be absorbed is 1/sin(45deg) =~ 30%. (This is also a simplification; polarization angle is similarly quantum in nature, and I have assumed it to be collapsed here)
wyager 4y ago
Easier to think about the polarizer's effect on a non-quantum EM wave, and then understand that our probability of observing a photon is based on the magnitude of the resultant EM wave.
cratermoon 4y ago
> single photons can be polarized by these things, and AFAIK their wavelength is not changed so their energy is unchanged as well. This is correct. The linked article doesn't explain how the single-photon behaves. It's the same issue that the two-slit experiment has if you try to ignore quantum behavior: You still can't say which slit a single photon passes through.

Sharlin 4y ago

The polarization of a single photon is a quantum property, so it's essentially a probability distribution. Passing a photon through a polarizer modifies the probability distribution such that "more perpendicular" polarizations are now less likely and "more parallel" ones more likely. (Polarization is a superposition (ie. a linear combination) of two orthogonal basis vectors, and a polarizer projects a polarization vector onto one of the basis vectors.)

abdullahkhalids 4y ago

The explanation given is at the level of classical electromagnetism, and is sufficient to explain and predict experiments with regular light. If you want an explanation and prediction of what happens at the level of single photons, you need more structure from the theory of quantum optics. But briefly the filter at angle T does a measurement on the photon in the basis {T, T+pi/2}, and you end up seeing the photon on the other side of the filter only with whatever probability the photon has for being in state |T>, as opposed to state |T+pi/2>. So, filters are inherently destructive and fewer and fewer photons pass through each subsequent filter. And a photon that makes it through a filter at angle T, now has a new state |T>.

jiggawatts 4y ago
The thing is that there is no actual evidence that “single photons” exist in the electromagnetic field. That last bit is important: photons are a mathematical shorthand for dealing with emission and absorption by atomic orbitals. They’re not an explanation for continuous waves in between, but the mathematics largely works anyway because photons are very similar to how one would do a Monte Carlo numerical simulations of continuous wave phenomena. This has resulted in an unbelievable amount of confusion…
- oh_my_goodness 4y ago
  There is lots of evidence for photons in the EM field, beginning with Planck's invention of photons. http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Planck%20(1900),%20Distribution%20Law.pdf http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Planck%20(... If you make the energy in a mode of the EM field continuous, you get the famous 'ultraviolet catastrophe.' This holds whether the matter involved has a continuous emission spectrum or not.
  - renox 4y ago
    I'm not so sure: 'black body' are made of atoms, atoms emits lights only on certain frequency due to the way electrons orbitals are structured, with discrete energy levels, but I don't see how this is related to EM fields themselves..
    - wyager 4y ago
      Free electrons also emit blackbody radiation. Photon quantization has nothing to do with the energy levels of atoms. It is a very fundamental property of the universe - you can tell because the only relevant constant is planck's constant.
      - renox 4y ago
        I didn't say that light isn't quantized, I just said that for me blackbody radiation of ordinary matter isn't a convincing proof that light is quantized.. As for free electrons emitting blackbody radiations can you give me a reference? If you're talking about a wiggle (https://en.wikipedia.org/wiki/Wiggler_(synchrotron) https://en.wikipedia.org/wiki/Wiggler_(synchrotron)) I don't see how this is related to the issue.
        oh_my_goodness 4y ago
        The sun is full of free charged particles.
        4y ago

gnramires 4y ago

I think that although light is a quantum phenomenon and we can detect single photons, many people overlook that photons still behave similarly to Maxwell's (wave) equations. In particular, the average behavior is that of the electromagnetic wave (up to some extremes like extremely high energies). The 'bullet' model people think of when the word photon (particle) is mentioned is inadequate. This becomes clear in field theory (QED and QFT), where there's a more complete description based of this phenomena solely based on field (wavelike) behavior. It's believe any (small) system follows QFT exactly (there's still uncertainty around gravity). The exact nature of the relationship of quantumness and fields (i.e. how the single-particle behavior arises from QFT) is still unclear, which is why there are many competing interpretations of quantum mechanics. In the Copenhagen interpretation, which is the most "easy" one, the behavior of photons is just (almost) that of Maxwell's equations, on average, s.t. a single photon will be measured with probability equal to the average light intensity anywhere (they are said to "collapse" at the moment of measurement, which is surely a simplification of a more complete underlying theory).

roesel 4y ago

The magnitue drop is reasonably simple to understand in terms of fields. The oscillating optical field might be less effective at exciting material oscillations in the middle filter due to a mismatch in polarization, but it still does so at the same frequency. You can think of it as multiple photons (incoming field) collectively exciting the same electron on the same frequency but with reduced efficiency. The electron then re-emits fewer photons (outgoing field) of the same wavelength, leading to a lower light intensity detected after the filter.

Arwill 4y ago
I don't think photons are absorbed and re-emitted by electrons. At least that argument does not hold when discussing light slowing down in glass or water. Light is affected by the electromagnetic field of the material it is going trough, is slowed down, or absorbed based on some of its property, but photons that go trough are going trough without collision. Photons that get absorbed and re-emitted are scattered in all directions, and are mostly lost. You would not see a consistent image trough a polarising sunglass, if the photons you were seeing were re-emitted photons.
- roesel 4y ago
  I think using the term "re-emit" is a mistake on my part. I fully agree that the photons are not absorbed and re-emitted. However, their speed of propagation in the material, and their properties can be altered by the interaction with dipole moments of the material. Therefore, photons do indeed "go trough without collision" but not without interaction. The dipoles in a dielectric material such as glass need to take the energy to oscillate from the field, and then they give it back to the field by radiating it out again. I agree that this is not "re-emission" in the absorption/emission diagram sense, and I understand that the field in a material cannot be separated into the material portion and the vacuum portion, since it is both at any given time, but what I have tried to outline in the previous comment is still a useful representation of the role of the material which the original article chose to leave out. I should have used "radiate" and "dipole" instead of "emit" and "electron" to make it a bit clearer.

gorkish 4y ago

At the single photon level, the photon that goes into the polarizer and the photon that goes out of the polarizer are not the same photon, so it's not right to say that a photon is changed or transformed. For a photon coming in at 45 degrees to the polarization angle, the probability that another photon will be emitted is sin(45deg) =~ 70% and the probability that it will be absorbed is 1/sin(45deg) =~ 30%. (This is also a simplification; polarization angle is similarly quantum in nature, and I have assumed it to be collapsed here)

wyager 4y ago

Easier to think about the polarizer's effect on a non-quantum EM wave, and then understand that our probability of observing a photon is based on the magnitude of the resultant EM wave.

cratermoon 4y ago

> single photons can be polarized by these things, and AFAIK their wavelength is not changed so their energy is unchanged as well. This is correct. The linked article doesn't explain how the single-photon behaves. It's the same issue that the two-slit experiment has if you try to ignore quantum behavior: You still can't say which slit a single photon passes through.