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I have posted here regarding this before. I always get conflicting information when I research about this stuff.

Some people say that a photon spins in or against it’s direction of motion…meaning if it’s going along the Z axis, from point a to b, it’ll either spin forwards or backwards with respect to it’s direction of propagation. Others say that it spins circularly(as in left or right with respect to it’s propagation direction).

Some people also say that polarization of the electric field is the same thing as spin and that the electric field is perpendicular to it’s direction of propagation.

You can see why I’m confused, especially because I’m finding so much conflicting information. And when it comes to polarizing lenses for a movie theater for example, I’m curious how it works on a quantum level.

I’d really appreciate if someone could tell me in plain English how this works, without using conventional terminology (some people say that left and right handed spin is just conventional and it really spins along it’s direction of propagation). I also see that angular momentum and orbital momentum are factors in spin as well, which I can’t figure out if it’s true or not.

Right now I feel like Homer Simpson when he’s clueless(derrr).

The way I learned it originally was that polarization and spin are not the same thing. Polarization is the direction of the electric field in the x and y planes(so horizontal or vertical). But I keep seeing that spin is important with a polarizer(like lenses or an lcd screen pixel). So it almost seems like spin and polarization are the same thing or at least 2 sides of the same coin.

So to sum it up: which direction is a photon actually spinning relative to it’s direction of propagation along the Z axis? And is polarization really the same as spin? I keep finding conflicting info after hours of googling. So when I think of photons going through a polarizer, some people say they spin left or right due to the electric field, some say that they don’t spin at all..my head hurts!

Any help would be greatly appreciated. My last post on here helped me some, but I was u able to get the info I was looking for in full. Please explain it to me as if I’m a dumb Homer. How does this actually work, in a literal sense(not conventional graph sense)?

Sorry for the long post. I know there are lots of questions. But I just need a basic reply explaining this as a whole. You don’t have to put too much effort into replying. Im thankful for any help here. So, Thank you!

Edit: if there’s a video that you know of and you think would explain it well, please share it! I’d appreciate it. I’ve found nothing so far. So I’m not being lazy, just not smart enough to fathom it. I feel so stupid haha

theguineapigking
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  • See: https://physics.stackexchange.com/a/232600/50583, https://physics.stackexchange.com/a/291180/50583 – ACuriousMind Sep 19 '22 at 10:05
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    I don't understand how this differs from your previous question. The second paragraph of this one seems to be a rewording of the third paragraph of that one. I answered that one, you asked for a clarification of that paragraph specifically in a comment, I responded in another comment, then you accepted my answer... and now you're asking the same thing again? – benrg Sep 19 '22 at 15:12

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It's a really great question. I regularly get confused about this topic and questions around it, and I have to use this topic near every day in my research. Don't be discouraged.

Spin and polarization are not the same. Roger Vadim's answer has the right idea, which I'm just going to expand on and hopefully make clearer.

Polarization is a classical quantity. Notably, what this means is that, for a particular photon, it's defined: you know it, in all relevant axes. For example, if you know the polarization is linear in the $+x$ direction, you know you won't have electric field in the $y$-direction. If the light is circular, you know you have equal field in both directions and you know they're offset by a particular phase ($\pi/2$).

Spin is not like this. Spin is purely quantum. If you know $s_z$ you don't know $s_y$ or $s_x$; in fact you cannot know them.

That said, spin and polarization are intimately connected. As an atomic experimentalist, I have to think about this all the time. We send light to our atoms and expect the electrons to go to certain states. Those transitions have to follow conservation of energy but also conservation of angular momentum. So how do we know where they'll go?

To know it we have to specify a "quantization axis," which is a direction where the angular momentum states of the atom are well-defined. (This corresponds to choosing the "$z$"-axis in the common QM notation.) We typically do this by putting a small magnetic field on the atoms, which splits states of differing angular momentum by small amounts of energy. Then if you "tilt your head" and try to look at the states along a different axis, they'll be ill-defined, in the sense that if you know your atoms to be in, say, $|F=1, m_F=+1\rangle$ ($F$ is some angular momentum) when the magnetic field is in the $+z$ direction, then you won't know the state of your atoms if you suddenly change this magnetic field to a different direction (suddenly so that the electrons' states don't change). This is the same statement that we don't know $s_x$ or $s_y$ if we know $s_z$.

Since we only have knowledge of the electrons' state in this one special direction, whatever light we send in we have to think about its effects relative to that axis. The photons' angular momentum is along its propagation direction, so we have to decompose that direction relative to the quantization axis to figure out the transitions that will happen.

Suppose we send light in along the magnetic field, for example. Then LHC and RHC light induce $\Delta m_F = \pm 1$ or $\sigma^{\pm}$ transitions, so that $s_z = \pm 1$, as you might hope. For linear polarization, we must decompose it into LHC and RHC components, so we get both $\sigma^+$ and $\sigma^-$ transitions.

But what happens if we put in light perpendicular to the magnetic field, say in $+x$? Then linear polarization induces $\pi$ transitions, ones for which $\Delta m_F=0$. And circular polarizations must be decomposed into two linear out-of-phase components, one of which must in turn be both RHC and LHC. So you can actually get $\sigma^+$, $\sigma^-$, and $\pi$ transitions in this case!

These statements correspond to these mathematical ones:

enter image description here

So you see that the spin of the photon, the thing that constraints what transitions you can make, depends on the quantization axis, while the polarization of the light doesn't. So although they're closely connected, they're not the same.

Please ask questions if you have them! I can always get better at understanding this.

flevinBombastus
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  • Hi. Thank you for your helpful and respectful response :) I really struggled to understand the equations(my fault since I’m dumb). When you mention circular polarization, are you referring to multiple photons being out of phase, so it looks circular on a graph? Or do you mean a single photon is literally spinning in a circular motion? Also, is the spin in the direction of propagation or against(parallel or antiparallel)? If so, how does spin affect anything? Since it’d be spinning up or down? Does that make sense? – theguineapigking Sep 19 '22 at 12:57
  • Also, can a single photon have a linear polarization of 45 degrees? Or is it a superposition of vertical and horizontal, which acts like 45 degrees when averaged in a polarizer? Because people only ever talk about 90degree polarization, but if I tilt my head, it could appear to be 45degrees. So I don’t know if it’s just a convention or not. If you need more clarification, please let me know. I have Asperger’s and struggle to no end to learn. So thank you so much. – theguineapigking Sep 19 '22 at 13:00
  • @theguineapigking Not dumb at all, this is hard stuff. When I talk about circular polarization, what I mean is a single photon. But not a single photon "spinning." If you think about linear polarization, you know this is a field that waves in a single direction. I can always decompose this field along any two axes I choose, not just the "actual" direction and the direction perpendicular to it. – flevinBombastus Sep 19 '22 at 13:01
  • @theguineapigking So, for example, I can pick a physical x and y axis to decompose the polarization. Then the actual field can be waving in the same direction as the x-axis I chose, or it could be in the same direction as the y-axis I chose, or it could have equal components in the x- and y-axes I chose, which makes it 45°-polarized as you asked. – flevinBombastus Sep 19 '22 at 13:03
  • @theguineapigking This is EXACTLY what you said about tilting your head --- you can decompose along any direction you want and that changes your perspective about whether something is "horizontally" polarized or +45°. – flevinBombastus Sep 19 '22 at 13:04
  • Thank you. So as for spin, I read on Wikipedia that it means it’s spinning in or against the direction of propagation, and that the whole right hand left hand rule is just for convention purposes.. Is that true? So when you refer to circular polarization, in this case, are you referring to spin? If not, do you mean quantum superposition, where it has equal horizontal and vertical counterparts? Hopefully that makes sense. – theguineapigking Sep 19 '22 at 13:05
  • @theguineapigking In all of this though, there's a secret extra degree of freedom: the relative phase between the two components. For all linear polarizations the relative phase is zero and they're "in phase." But for circular light, we have equal field amplitudes in the x- and y-directions and those components are out of phase by 90° or $\pi/2$. Then at any instant there's a direction the field is pointing, and that direction appears to be rotating about the propagation direction, so it's called "circular." – flevinBombastus Sep 19 '22 at 13:06
  • @theguineapigking Sorry, I'm trying to keep up with your comments with the limited character space :) I don't like to think of "spin" as "spinning." There's no physical object spinning if you have quantum spin, and thinking otherwise usually leads to trouble (though not always). What is true is that, if you have a photon, then linear or circular polarization corresponds to angular momentum along the direction of propagation. – flevinBombastus Sep 19 '22 at 13:09
  • Would you mind if I chat with you a bit more about this? If not it’s okay. But you seem to be very knowledgeable and helpful. – theguineapigking Sep 19 '22 at 13:09
  • Let us continue this discussion in chat. – flevinBombastus Sep 19 '22 at 13:10
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    +1 good job explaining! – Roger V. Sep 19 '22 at 15:00
  • @flevinBombastus it is somewhat misleading to use $s_x, s_y, s_z$ when talking about photon spin. A photon doesn't have a rest frame, so you cannot sensibly define three projections of its spin because most rotations would also change the momentum. Only the projection on the direction of the momentum is well-defined, and it behaves exactly like a polarization of a transverse plane wave. In fact, there is a mathematical correspondence between the two. A related fact is that, at one spacetime point, $B_x, B_y, B_z$ all commute. – Peter Kravchuk Sep 25 '22 at 10:54
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Polarization is a classical notion, that is, it can be defined already in classical electrodynamics as a direction of the electric field (or as a manner of this field changing in time, e.g., when talking about circular polarization.)

When electric field is quantized, it can be shown that photons possess angular momentum (spin.) This angular momentum possess the general properties of quantum angular momentum - notably, one can measure simulatenously its magnitude and projection on a selected quantization axis, but not simultaneous projections on different axes. As the quantization axis it is often convenient to chose the direction of the photon propagation, in which case it is an example of helicity, more generally defined in particle physics for all particles.

Thus, when we say that the photon spin projection on the direction of its propagation is $\pm 1$, we are thinking of it as an angular momentum, with obvious classical analogy of a top, rotating clockwise or counterclockwise, which is where it joins with our thinking about circular polarization.

What may add to the confusion is that terms polarization, spin and helicity are often use dinterchangeably.

Roger V.
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  • Thank you. So are people referring to helicity when they talk about it spinning forward or backwards with respect to it’s propagation direction? And spin is left or right? If so, is this literal, or is it conventional(meaning not what it’s doing in real life)?

    The issue is that I keep seeing that spin, helicity, and polarization are the same thing, or not the same thing. Would you mind specifying the difference? My small mind can’t exactly grasp your answer very well. Thank you so much

     – theguineapigking Sep 19 '22 at 10:10
  • Polarization may originate as a classical notion, but the fact remains that you can describe any photon as a superposition of left and right circular polarization α|L> + β|R>. Any combination of α and β with α = β exp(2φi) yields linearly polarized light with a precisely defined polarization angle φ. Thus, all the lingo about linear/elliptical/circular polarized light and polarization angles is "just" a phenomenological description of quantum mechanical phenomenon from a time before quantum mechanics was a thing. – cmaster - reinstate monica Sep 19 '22 at 22:03
  • @cmaster-reinstatemonica What you said doesn't contradict the notion of polarization as classical. The difference is that, when you specify the polarization, you know the alpha and beta and therefore know the state of the polarization exactly. In comparison, for spin, even if you know the $z$-component, you don't know the $x$- and $y$-components. – flevinBombastus Sep 20 '22 at 00:07
  • @flevinBombastus I would not say that I don't know the x and y components of the spin, I would say that they are simply undefined. Because the word "to know" seems to imply that the thing is actually knowable. However, in quantum mechanics, the spin simply has no defined x and y components until I try to measure them. That's when, depending on your favorite interpretation, either the wave function collapses, worlds split, or you get entangled with the particles state. Point is, photons only have a single basis with defined spin (|L> and |R>), the rest is superposition of the two. – cmaster - reinstate monica Sep 20 '22 at 06:24
  • @flevinBombastus So, when I measure linear polarization, I'm measuring in exactly one of the α/β superposition bases. After the measurement, I know the exact phase angle φ, but since I now have a superposition of equal amounts of |L> and |R>, it is meaningless to ask for the spin of the photon. Measurement of the spin simply leads to collapse / world split / entanglement, it is undefined due to superposition. Likewise, when I have a photon with defined spin, the phase angle φ is undefined. That's exactly what the polarization lingo describes without mentioning "superposition". – cmaster - reinstate monica Sep 20 '22 at 06:37
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I am forced to copy the figure from the link I gave you in your duplicate question:

photpol

Left and right circular polarization and their associated angular momenta

When a light beam is circularly polarized, each of its photons carries a spin angular momentum (SAM) of $+-h/{2π}$ the ± sign is positive for left and negative for right circular polarizations (this is adopting the convention from the point of view of the receiver most commonly used in optics). This SAM is directed along the beam axis (parallel if positive, antiparallel if negative). The above figure shows the instantaneous structure of the electric field of left (L) and right ( R) circularly polarized light in space. The green arrows indicate the propagation direction.

...

The language of physics is mathematics, and physics cannot be described by words only since the time of Newton, who introduced calculus.

anna v
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  • Hi Anna. Thanks for your reply. So if a photon spins in or against it’s direction of propagation, then would that affect it passing through a polarizer, as some people have stated? Because in this case it wouldn’t affect the angle of the electric field passing through a horizontal or vertical slit…ie it wouldn’t matter if it’s spinning forwards or backwards right? If what I said above is correct, does this mean that the spin of a photon doesn’t have anything to do with polarization?

    Thank you again for your reply.

     – theguineapigking Sep 19 '22 at 11:40
  • I know you probably think I’m dumb. I have Asperger’s and struggle to learn. That’s why I posted again. It’s really hard for me to understand people sometimes. So I’m sorry if it’s annoying. – theguineapigking Sep 19 '22 at 11:42
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    I'm going to downvote because I don't think it really answers the question and is a tad snotty in the process. – flevinBombastus Sep 19 '22 at 12:15
  • "if a photon spins in or against it’s direction of propagation, then would that affect it passing through a polarizer" Look at the picture. the left diagram,of the rotating electric field is built up from a huge number of photons aligned with the corresponding direction of spin as in the middle . It is the quantum addition of photon spin, as shown on the right, that becomes the polarization of the classical wave . – anna v Sep 19 '22 at 15:49
  • @theguineapigking The thing about linear polarization is that it is a superposition of |L> and |R> with a known phase angle between them. The phase angle is what determines the angle of the polarization. When you pass light through a linear polarizer, you are forcing the photons into one of two such superpositions of |L> and |R> where the allowable phase difference is determined by the angle of the polarizer. – cmaster - reinstate monica Sep 19 '22 at 21:09
  • @theguineapigking So, if you want to know the spin of the photons, transform whatever polarization it has into the |L>/|R> basis. However, if you do this for linearly polarized photons, you get equal amplitudes for |L> and |R>. Thus, linearly polarized photons can behave either as spin +/-1 (circular polarization measurement for the |L>/|R> basis), or as spin 0 (linear polarization measurement for some superposition basis). – cmaster - reinstate monica Sep 19 '22 at 21:43
  • @cmaster-reinstatemonica the problem boundary conditions will define what the solutions is at the photon level. – anna v Sep 20 '22 at 03:35
  • @anna v we can only associate circular polarization to photon spin. Linear polarization only gives that either spin state is equally likely and both are in superposition, right? – Duke William Jun 01 '23 at 08:36
  • @DukeWilliam The electric and magnetic fields are transverse to the direction of the beam. at most + , which is part of the photon directions. – anna v Jun 01 '23 at 10:42
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Nobody is actually saying that a photon spins in or against it’s direction of motion. It just sounds like they are saying that, because of the use of conventional terminology.

When we describe a photon there is an arrow that rotates around, like clock-hand of a clock that was thrown flat side first.

When the length of the arrow stays constant, then we have a circularly polarized photon. This photon is non-polarized in such sense that there is no special direction that the arrow has all the time, so orientation of one polarizing lens has no effect on how the photon goes through.

If we have many of these circularly polarized photons, then there appears to be an constant strength electric field that rotates around the same way as the aforementioned arrow does.

If the length of the arrow changes as the arrow rotates, and the length is the longest when the arrow points up and when it points down, then we have a vertically polarized photon.

That vertically polarized photon can pass through a vertical long opening on a screen, because its arrow can fit through nicely, so to speak, although that may be a bad wording.

If we have many of these photons, then there appears to be an non-constant strength electric field that also rotates around like the aforementioned arrow does.

The photon arrow is a "photon absorption probability arrow".

If we have many photons with identical arrows, then a similar arrow as that arrow can be said to describe an electric field's strength and direction.

stuffu
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  • -1 for all the talk about many photons. Polarization (linear, circular, elliptical) and spin are phenomena that can be attributed to single photons. With the slight complication that polarization can be measured in any of an infinite number of bases on top of the circular basis. – cmaster - reinstate monica Sep 19 '22 at 21:32