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I often hear about subatomic particles having a property called "spin" but also that it doesn't actually relate to spinning about an axis like you would think. Which particles have spin? What does spin mean if not an actual spinning motion?

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Was there something in particular you didn't understand in the wikipedia article? en.wikipedia.org/wiki/Spin_%28physics%29 –  j.c. Nov 2 '10 at 19:11
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@j.c.: Does it matter if there was or wasn't? I was under the impression that we answered questions regardless of whether or not they have been answered in other sites, unless you're just asking to see if there's a specific property of spin the asker wants explained in detail. –  Mana Nov 2 '10 at 19:43
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@Mana, I agree, which is why I didn't vote to close. I tend to think my time answering questions is best spent if I'm writing the answer at a level the questioner might understand. –  j.c. Nov 3 '10 at 14:24
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I don't like the subatomic tag here. Is particle-physics more appropriate? –  Nick Nov 3 '10 at 16:34
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Tried understanding all the answers and they don't help. Maybe if you could say what spin does rather than what it is I might have an idea. Does it cause a particle to move on a curved rather than a straight path or does it affect another particle a certain way if they collide? It seems to have something to do with waves. Particles seem to vibrate in a certain space and the energy applied makes them respond in a sort of sequential way --one after another--to make a wave sequence. Is spin a quality that causes energy to be expressed this way? I'm kind of old so don't throw too much jargon my wa –  user9713 Jun 7 '12 at 2:10

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Spin is a technical term specifically referring to intrinsic angular momentum of particles. It means a very specific thing in quantum/particle physics. (Physicists often borrow loosely related everyday words and give them a very precise physical/mathematical definition.)

Since truly fundamental particles (e.g. electrons) are point entities, i.e. have no true size in space, it does not make sense to consider them 'spinning' in the common sense, yet they still possess their own angular momenta. Note however, that like many quantum states (fundamental variables of systems in quantum mechanics,) spin is quantised; i.e. it can only take one of a set of discrete values. Specifically, the allowed values of the spin quantum number s are non-negative multiples of 1/2. The actual spin momentum (denoted S) is a multiple of Planck's constant, and is given by $S = \sqrt{s (s + 1)}$.

When it comes to composite particles (e.g. nuclei, atoms), spin is actually fairly easy to deal with. Like normal (orbital) angular momentum, it adds up linearly. Hence a proton, made of three constituent quarks, has overall spin 1/2.

If you're curious as to how this (initially rather strange) concept of spin was discovered, I suggest reading about the Stern-Gerlach experiment of the 1920s. It was later put into the theoretical framework of quantum mechanics by Schrodinger and Pauli.

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@Albert: They most certainly are! There are fringe theories which consider them composite particles, but the standard model of particle physics and most extensions of the theory consider them fundamental point particles. –  Noldorin Nov 2 '10 at 19:54
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The only sense in that they are not point-like is that there position may be indefinite/fuzzy due to the Heisenberg principle. They are still however considered "point particles" since their position eigenstates are functions of a single position vector. –  Noldorin Nov 2 '10 at 19:55
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I think the flaw in your thinking comes from considering angular momentum as a consequence or property of spinning. Angular momentum really is the fundamental quantity here. It is a direct consequence of the rotational symmetry of the universe (see Noether's theorem). Realise, the maths is the basis of the physics here; nothing gives a more precise pictures. –  Noldorin Nov 4 '10 at 17:29
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Note that experimentally one can only say that election have no structure down to [length scale] and/or up to [energy scale]. Which makes them point-like for the purposes of all the experiments we've been able to do. I believe the limiting length scale is currently around $10^{-18}$ meters, which is pretty small. –  dmckee Nov 30 '10 at 4:06
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@حكيمالفيلسوفالضائع, yes they are for all intents and purposes. Certainly in traditional QM they are. In QFT, there are no particles at all really; particles are just conceptual approximations... but you're over-simplifying things. Something tells me you're just a reader of pop-science and think you understand it because of that. –  Noldorin Feb 20 at 18:19

Imagine going to the rest frame of a massive particle. In this frame, there is rotational symmetry, which means that the Lie algebra of rotations acts on the wave function. So the wave function is a vector in a representation of Lie(SO(3)) = Lie(SU(2)). "Spin" is the label of precisely which representation this is. Note that while SO(3) and SU(2) share a Lie algebra, they are different as groups, and it is a fact of life ("the connection between spin and statistics") that some particles -- fermions, with half-integral spin -- transform under representations of SU(2) while others -- bosons, with integral spin -- transform under SO(3).

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An accurate answer, but if the poster doesn't understand the actual concept of spin (not to mention group theory), this is all but useless. –  Noldorin Nov 2 '10 at 19:32
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I agree. This comment speaks to my confusion (see my Physics Meta question) over how this Physics site is conceived. Namely, what is the level (high school, undergrad, grad) of the intended audience? –  Eric Zaslow Nov 2 '10 at 22:05
    
Yeah, it's a point worth discussing. I think all three of those levels you point out should be acceptable. (You are clearly aiming at lower graduate level plus in this case.) Personally I would not like to see this site dominated by too many basic high school questions nor by many research-level ones. –  Noldorin Nov 3 '10 at 13:22

Spin is angular momentum of particles. The lowest possible spin is 1/2 h-bar. It's impossible for any particle with angular momentum to have a lower angular momentum than this, and whatever angular momentum a particle does have must be an integer multiple of this. Consider it the angular momentum building block. It's value is 340 dB below a kilogram meter squared radian per second.

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Yes, but one must keep in mind that while spin is indeed angular momentum, this momentum does not come from a spinning particle and does not have a classical analogy as such. –  Emilio Pisanty Oct 7 '12 at 12:38

I'm trying to give a less technical answer. It's not rigorous but should give you the idea how spin and the regular rotation related.

Maxwell's equations say in order to have magnetic field, you need a ring current.

This can be achieved by giving angular momentum to charged particles. This can be orbital or simply because the particle is spinning. This was the original thought hence the name 'spin'.

So in the classical picture, if you spin a tiny charged ball you'll have a spinning magnet. The axis of spinning and the north pole of the magnet pointing to the same direction.

If you put this spinning magnet into a magnetic field. The field will apply torque on it to turn it into the direction of the field (this is how compasses work).

But since the our magnet is spinning this torque cause the axis of spinning precess around the magnetic field. This means the component of the rotation axis that is parallel to the magnetic field (typically referred as the Z component) won't change while the other two components (X,Y) will circle around this axis.

On the other hand if the magnetic field inhomogeneous there will be a net force on the particle that will move it (that's why magnets can snap and repel each other). This force is proportional to the Z component. So the axis perpendicular to the magnetic field there will be no force, if it's parallel there will be maximum force (basically a dot product). This allows us measuring the Z component of the rotation axis.

That's the point of the Stern–Gerlach experiment. We would normally expect that particles will spin in a whole variety of random axes. So we would expect to measure random values for the Z component.

But in reality they have measured only two possible values corresponding to the Z angular momentum component: $ħ/2$ and $-ħ/2$ (for electrons). And not any other random values. Here the classical picture breaks down, angular momentum is also quantized. You can see spin is not the classical rotation vector. It's something you can dot multiply a vector to and you can only get two possible values. The positive component typically referred as the 'up' spin component while the negative is the 'down' spin.

Precession renders all axes other than the one being measured uncertain. This is how uncertainty principle plays role here: if you measure the Z component first, then measure the X component, then the Z again, you get random up/down results again, because the measurement of the X components precessed the Y and Z component. Also, you cannot cheat here: you may want to use weaker magnetic field to reduce the precession, the displacement will be too weak to distinguish between the up and down spins. If you try to use timing; you cannot cheat again because if you measure the time accurately, then the energy so the precession rate becomes uncertain.

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