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I have read this question (in the comments):

Photons can be polarized which somehow affect their physical shape considering photons polarized parallel to a slit can make it through but do not make it through when polarized perpendicularly to the slit.

What do photons look like?

And this one:

That's why photons of microwave frequencies can't escape from the microwave, either. But the interpretation is different: the wave function isn't directly measurable like the electric fields. Instead, it encodes (after squaring) the probability density that the photon is here or there.

If photons move linearly, what's actually stopping them from passing through a microwave oven mesh?

enter image description here

Now as far as I understand, photons are defined as elementary point particles in the Standard Model. I do understand that a large number of photons build up the classical EM waves in a very nice way.

Nowadays we do have single photon emitters, and if such a device emits a single photon towards a slit so that the photon is perpendicularly polarized (relative to the slit), then will the photon pass or not?

Now the image above explains the filter (polarizer) as where only parallel components get through, and perpendicularly polarized components do not get through.

I do not really understand this explanation in case of a single photon, since it is an elementary particle, and, even though it shows wave properties, it should not have parallel or perpendicular components (or constituents).

So basically, if a photon is an elementary particle, and does not have any components or constituents, then how can we state that it will not pass through the slit (if it is perpendicularly polarized)?

I do understand that slits can stop photons based on their wavelength, that is, for example in the case of a microwave mesh. But this example is not about wavelength, but polarization. How can a slit stop a photon just based on polarization, if the photon is elementary, and does not have for example vertical and horizontal components?

The last one says that single photons are described by the wavefunction, that isn't directly measurable like the EM fields (so it does not have vertical or horizontal components), but rather it describes the probability of finding the photon in space. But if the single photon is described by the wavefunction (and does not have vertical or horizontal components), then the slit should not be able to stop the photon based on polarization.

Question:

  1. Why can't a single photon pass through a slit if it is perpendicularly (relative to the slit) polarized?
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    $\begingroup$ I think I've told you this many times before, but you're always going to be misled if you keep using these online resources for kids... You've asked hundreds of questions based on them, but all of them would have been answered automatically if you just used a regular textbook. $\endgroup$
    – knzhou
    Oct 15, 2021 at 0:15
  • $\begingroup$ Even a point particle occupies a nonzero volume of space, due to quantum uncertainty. Also the illustration is wrong --- a vertical slot aperture will pass horizontally polarized waves, not vertical. $\endgroup$
    – The Photon
    Oct 15, 2021 at 0:46
  • $\begingroup$ Whether you call it horizontal or vertical it’s obvious the OP means photons that Have gone through a vertical polarizer, are polarized vertically. Then his question is why do they not go through Slits the are polarized perpendicular to that vertical polarization. Also who cares if it’s resources for kids, can you answer the question? $\endgroup$ Oct 15, 2021 at 2:21
  • $\begingroup$ @knzhou Thank you, I appreciate the comment, can you please direct me towards a online source that deals with this topic, single photon going through slits based on polarization? $\endgroup$ Oct 15, 2021 at 4:10
  • $\begingroup$ @ÁrpádSzendrei Unfortunately no one wants to discuss Single photons doing anything but being detected. Questions like yours tend to be ignored if you even imply that a photon may possibly have other properties other than a speed in a direction. You can check out my paper at billalsept.com for a starter then reach out to discuss. Thanks $\endgroup$ Oct 15, 2021 at 22:09

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"I do not really understand this explanation in case of a single photon, since it is an elementary particle, and, even though it shows wave properties, it should not have parallel or perpendicular components (or constituents)."

This is false. A single photon is an excitation of the electromagnetic field, which certainly has parallel and perpendicular components. It is absolutely not a point particle. Famously, photons have spin 1, which makes them a vector particle that has orientation in space.

So, a priori, there is no reason why an experiment like this, which is designed to detect the spatial orientation of a photon, wouldn't work. Photons have orientation, whether it is one or a whole beam of them. They are made out of electromagnetic field, after all.

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  • $\begingroup$ Thank you so much! "Famously, photons have spin 1, which makes them a vector particle that has orientation in space.", can you please elaborate on this? $\endgroup$ Oct 18, 2021 at 5:13
  • $\begingroup$ @ÁrpádSzendrei: a proper treatment is an entire chapter of a quantum field theory textbook, so I don't know if I can "elaborate", but for a taste of it, I'll point out that the thing that gets quantized when defining the photon is literally a vector, called the "electromagnetic 4-potential" $A^a$, from which one can get the electric and magnetic fields by taking various derivatives. (and again, if you want to know more, you have to go to a textbook, the explanation is several pages of an E&M textbook) $\endgroup$ Oct 18, 2021 at 16:09
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Although the image is drawn as a slit, the captioning more properly calls it a "filter". Polarizing filters are not slits.

Just because it is an elementary particle does not imply that it doesn't have perpendicular components. In this case there is an electric field associated with the photon, and that field has an orientation.

Polarizing filters preferentially interact with electric fields in certain directions and not others. The orientation of the photon's electric field therefore describes the probability that the filter will absorb it.

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  • $\begingroup$ Thank you so much! "In this case there is an electric field associated with the photon, and that field has an orientation.", can you please elaborate on this? $\endgroup$ Oct 15, 2021 at 1:54
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All EM radiation begins with the emission of photons. Without the emission the EM field is empty.

From a thermic source photons get emitted in a unpredictical way. The only thing you observe is an energy transfer to another body or your measurement instrument.

From photons, emitted from an antenna, are polarized. All electrons synchronous get accelerated forth and back and the emitted photons have an electric field component parallel and also periodically up and down orientated. And you are able to measure thom this bunches of photons their common magnetic field. It is oriented perpendicular to the electric field and periodically with and again the clock.

Since the photons are the constituents of the EM field (see the first paragraph), the only explanation for the field components of the EM wave is that the photon carries these field components.

So basically, if a photon is an elementary particle, and does not have any components or constituents,…“ does not give the photon the properties it has. The photon is an elementary particle and contains an electric and a magnetic field component that travels with c from emission to absorption of the photon.

Why can't a single photon pass through a slit if it is perpendicularly (relative to the slit) polarized?

The photon has an observable radius of action. If the slit is to small no photons of a given energy content can pass the slit. If the slit is an opening, the photons in the middle of the opening are passing through without any disturbance and only at the edges of the opening the diffraction of photons takes place.

The interaction is explainable by the electric fields as was pointed out in another answer. So a single photon does not pass through the slit if its electric field is oriented in a way, it interacts with the walls of the slit and get blocked.

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  • $\begingroup$ Thank you so much! "The photon is an elementary particle and contains an electric and a magnetic field component that travels with c from emission to absorption of the photon.", can you please elaborate on this? Are you saying that photons have electric and magnetic field components? $\endgroup$ Oct 18, 2021 at 5:12
  • $\begingroup$ A radio wave is made from polarised photons and the common E an M fields from all these photons are measurable. Isn’t this a strong indication that the photon is a propagating oscillation of these two field components? $\endgroup$ Oct 18, 2021 at 14:06
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The explanation in the picture uses a mechanical wave on a string as an easy to understand analogy, not as a description of an EM wave. A photon polarized perpendicularly to the slit will pass easily there. It's more likely to be attenuated if it's polarized along the slit (if the slit is conducting), due to the interface conditions requiring the $\vec E$-field to vanish at the sides of the slit. This effect is used in wire grid polarizers.

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  • $\begingroup$ Thank you so much! "A photon polarized perpendicularly to the slit will pass easily there.", can you please elaborate on this? Are you saying that a single photon will pass if it is perpendicularly polarized (relative to the slit)? $\endgroup$ Oct 15, 2021 at 4:12
  • $\begingroup$ @ÁrpádSzendrei I'm saying that for a perpendicularly polarized photon there's nothing except simple geometry that prevents its passing, while for a parallel polarization there's an additional "squeezing" factor that reduces transmission. $\endgroup$
    – Ruslan
    Oct 15, 2021 at 7:43
  • $\begingroup$ I do hope some of the downvoters here are preparing their own answers... $\endgroup$
    – Ruslan
    Oct 16, 2021 at 8:40
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Everyone talks about duality but they never seriously consider a photon as a particle. Maybe on impact but never during propagation. If you truly consider a single photon propagating through space with a frequency then you can begin to picture what’s happening. Imagine the photons energy oscillating through maximum and minimum amplitude’s, at a certain frequency, as it propagates. Now imagine those maximum amplitudes expanding along a certain polarization. This model will answer both questions about being perpendicular to the slit or transmitting through a mesh. With the microwave oven you need to think linearly, where the oscillating max amplitude of a photon, polarized in any direction is stretched out (longer wavelength) so that it always makes contact with the Mesh. See “Single Edge Certainty” at billalsept.com

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  • $\begingroup$ Thank you so much! "Now imagine those maximum amplitudes expanding along a certain polarization.", can you please elaborate on this? Are you saying that the EM energy oscillates so that it fits through the slits? $\endgroup$ Oct 15, 2021 at 3:47
  • $\begingroup$ A single photon has no frequency (or wavelength). The wave characteristic of light emerges from a large number of photons. Think about the double slit experiment where single photons are sent and initially there is no pattern but fairly random dots. After time however you see a wave (interference) pattern emerge. I think that’s why someone downvoted this answer. $\endgroup$
    – joseph h
    Oct 15, 2021 at 4:02
  • $\begingroup$ A photon That produces green light has a wavelength around 500nm. Which really is a frequency of more than 600 trillion oscillations per second. As the photon propagates, picture it expanding and collapsing in a spherical shape every oscillation. At the same time imagine the sphere spinning and at higher frequencies the spin is even faster. Imagine the sphere stretching outward while still oscillating. The faster the photon spins the more it is polarized along that plan. Photons polarized along a vertical plane will travel through a vertical slit. Polarized horizontally and they will not. $\endgroup$ Oct 15, 2021 at 4:14
  • $\begingroup$ @josephh you need to seriously consider photons as Particles. You can’t have it both ways. $\endgroup$ Oct 15, 2021 at 4:17
  • $\begingroup$ "you need to seriously consider photons as Particles" That was my whole point. Also, on a separate point, wave/particle duality means you can. That is the essence of QM. $\endgroup$
    – joseph h
    Oct 15, 2021 at 4:27

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