How do particles get their spin?

In summary: I guess. In summary, according to Ballentine, the spin of a quantum particle type has everything to do with being a unitary representation of the (quantum) SO(3) group, corresponding to spatial rotations.
  • #1
BadgerBadger92
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TL;DR Summary
How do particles get their spin?
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  • #2
BadgerBadger92 said:
TL;DR Summary: How do particles get their spin?

https://physics.stackexchange.com/questions/418865/why-does-an-electron-have-spin

I’ve been trying to do research on this and haven’t gotten a clear answer. It sounds like they get it through particle interactions, but I may be misreading it. Is there an answer to this? And if there is, how did spin start?
One thing to understand: the spin of a fundamental particle has nothing to do any physical rotation. It is a property of an electron to have a spin of ##\pm \hbar/2## just as much as it is that it has a charge of ##-1.6 \times 10^{-19}## C. It's a physical property of the electron. It can only have two values, either plus or minus, otherwise it's not an electron. So we can't say "where" the particle gets its spin from... it has it simply because it's a fundamental particle. Possible spins and their particle classifications are:
Integer spin: boson
1/2 integer spin: fermion
Spin 0: scalar particle
Spin 1/2 (massless): Weyl spinor
Spin 1/2 (massive): Dirac particle (spin 1/2 spinor)
Spin 1 (massless): photon
Spin 1 (massive): Proca particle
Spin 3/2: Rarita-Schwinger spinor
Spin 2: graviton

There are, of course, subcategories for specific particle theories that use or generalize these spin categories. These are just the ones that I learned in class.

-Dan
 
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  • #3
topsquark said:
One thing to understand: the spin of a fundamental particle has nothing to do any physical rotation.
This is not correct. The spin of a quantum particle type has everything to do with being a unitary representation of the (quantum) SO(3) group, corresponding to spatial rotations. One can even demonstrate the half-integral nature of such spin in interferometry experiments, where the beam particles on one side are physically rotated before being merged with the beam from other side.

topsquark said:
[...] So we can't say "where" the particle gets its spin from... it has it simply because it's a fundamental particle. Possible spins and their particle classifications are:
Integer spin: boson
1/2 integer spin: fermion
Spin 0: scalar particle
Spin 1/2 (massless): Weyl spinor
Spin 1/2 (massive): Dirac particle (spin 1/2 spinor)
Spin 1 (massless): photon
Spin 1 (massive): Proca particle
Spin 3/2: Rarita-Schwinger spinor
Spin 2: graviton

There are, of course, subcategories for specific particle theories that use or generalize these spin categories. These are just the ones that I learned in class.
Take a look at Ballentine section 7.1 where he derives the half-integral constraint on quantum angular momentum numbers. He uses (almost) no other assumption beyond requiring that the SO(3) generators must be represented as Hermitian operators on a Hilbert space.

It's basically just using the so(3) commutation relations, the so(3) Casimir operators,finding the largest set of mutually commuting operators (in this case the Casimir and one other, say ##J_z##), decomposing the Hilbert state space in terms of simulataneous eigenstates of these operators, and the positive-definite nature of a Hilbert space inner product.

Edit: I will say that the answers given in the stackexchange link in the OP are disturbingly poor. Read my Ballentine reference above to begin to understand why. :oldfrown:
 
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  • #4
BadgerBadger92 said:
I’ve been trying to do research
Great. What did you read, and what parts were unclear?
 
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  • #5
strangerep said:
This is not correct. The spin of a quantum particle type has everything to do with being a unitary representation of the (quantum) SO(3) group, corresponding to spatial rotations. One can even demonstrate the half-integral nature of such spin in interferometry experiments, where the beam particles on one side are physically rotated before being merged with the beam from other side.Take a look at Ballentine section 7.1 where he derives the half-integral constraint on quantum angular momentum numbers. He uses (almost) no other assumption beyond requiring that the SO(3) generators must be represented as Hermitian operators on a Hilbert space.

It's basically just using the so(3) commutation relations, the so(3) Casimir operators,finding the largest set of mutually commuting operators (in this case the Casimir and one other, say ##J_z##), decomposing the Hilbert state space in terms of simulataneous eigenstates of these operators, and the positive-definite nature of a Hilbert space inner product.

Edit: I will say that the answers given in the stackexchange link in the OP are disturbingly poor. Read my Ballentine reference above to begin to understand why. :oldfrown:
Okay, I guess this is another flaw in my Quantum armor. What you say does make sense: in retrospect J = L + S, so J, L, and S have to live in the same space and J and L are spatial. Ergo so is S. For some reason I had never put that together.

Thanks for the catch!

-Dan
 
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  • #6
Vanadium 50 said:
Great. What did you read, and what parts were unclear?
I posted the link.

It says it comes from particle interactions but my engineer friend says it hasn’t been discovered yet.
 
  • #7
strangerep said:
This is not correct. The spin of a quantum particle type has everything to do with being a unitary representation of the (quantum) SO(3) group, corresponding to spatial rotations. One can even demonstrate the half-integral nature of such spin in interferometry experiments, where the beam particles on one side are physically rotated before being merged with the beam from other side.Take a look at Ballentine section 7.1 where he derives the half-integral constraint on quantum angular momentum numbers. He uses (almost) no other assumption beyond requiring that the SO(3) generators must be represented as Hermitian operators on a Hilbert space.

It's basically just using the so(3) commutation relations, the so(3) Casimir operators,finding the largest set of mutually commuting operators (in this case the Casimir and one other, say ##J_z##), decomposing the Hilbert state space in terms of simulataneous eigenstates of these operators, and the positive-definite nature of a Hilbert space inner product.

Edit: I will say that the answers given in the stackexchange link in the OP are disturbingly poor. Read my Ballentine reference above to begin to understand why. :oldfrown:
What websites would you recommend?
 
  • #8
BadgerBadger92 said:
[Stackexchange answer(s)] says it comes from particle interactions
Rubbish. Read the Ballentine reference I gave above.

BadgerBadger92 said:
but my engineer friend says it hasn’t been discovered yet.
Absolute rubbish! Read the Ballentine reference I gave above.

[Sorry if I sound a bit harsh, but this question comes up frequently, often followed by ill-informed answers. I guess I need to take a Bex and have a good lie down...]
 
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  • #9
strangerep said:
Rubbish. Read the Ballentine reference I gave above.Absolute rubbish! Read the Ballentine reference I gave above.

[Sorry if I sound a bit harsh, but this question comes up frequently, often followed by ill-informed answers. I guess I need to take a Bex and have a good lie down...]
A lot of people seem to get the wrong answers here too. People argue about it all the time.
 
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  • #10
strangerep said:
Rubbish. Read the Ballentine reference I gave above.Absolute rubbish! Read the Ballentine reference I gave above.

[Sorry if I sound a bit harsh, but this question comes up frequently, often followed by ill-informed answers. I guess I need to take a Bex and have a good lie down...]
Do you have websites you can recommend?
 
  • #11
BadgerBadger92 said:
What websites would you recommend?
I learned this stuff from textbook(s), not websites. Used copies of Ballentine are available quite cheaply. If you *really* can't afford maybe $30 (Ballentine is a really good investment), then PM me and I might be able to suggest other alternatives.

Btw, when asking this sort of question, it would help to include a brief statement of what level of formal education you have reached in physics and mathematics, and what textbook(s), if any, you have studied (even if they're quite basic).
 
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  • #12
I may shoot you a PM tomorrow.

I only have a high school level education but I like reading about this subject. I’m going back to school next year, I’m going to get a student loan.
 
  • #13
BadgerBadger92 said:
I only have a high school level education [...]
OK. In that case, Ballentine is likely to be far too advanced. It's a graduate (or upper-undergrad) text. One needs at least basic proficiency in linear algebra, classical mechanics, multivariate calculus (including complex numbers), and Lie group theory -- which are all university-level.

I did a quick google for videos that might be on a suitable level, but everything I found was I either too basic (and often subtly wrong), or too advanced (and didn't do quantum angular momentum from the simplest perspective that Ballentine uses in his section 7.1).

Walter Greiner's series of physics textbooks, many of which are "introductory" level, might therefore be a good investment. There should be lots of used copies available cheaply. Greiner's teaching style is good for self-study because he doesn't skip steps in nontrivial calculations.

BadgerBadger92 said:
I’m going back to school next year, I’m going to get a student loan.
Good for youI Although,... I hope you get extremely good advice on what courses to take so you can eventually pay back that loan without too much pain. :oldfrown:
 
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  • #14
It's a bit difficult to describe "spin", which is a generic quantum property of quanta, which have no good analogue in classical point-particle mechanics. The reason is that "spin" on the classical level is a notion of fields rather than point particles.

The best intuitive semi-classical description is to consider what's observable related to spin. Take an electron, which is an elementary particle (to the best of our present knowledge). As any elementary particle an electron has "intrinsic properties", which are defined for an electron at rest. To simplify things I only consider the electromagnetic interaction. Then these intrinsic properties are the mass of the electron (##m_e \simeq 0.51 \; \text{MeV}/c^2##), its electric charge (one negative elementary charge, ##q_e=-e##), and finally spin ##s=1/2##. This tells you that an electron at rest has a kind of intrinsic angular momentum. The closest thing we know from everyday life is the classical spin of an extended (rigid) object: Its center of mass can be at rest and the particle is spinning around this center of mass, but that's not really explaining what spin of an elementary quantum particle like an electron is, because the electron as a generic "elementary quantum object" has no extension in the usual sense.

A consequence of having spin 1/2 for the electron also is that it has also a magnetic moment, i.e., in addition to having an electric charge it also a bit like a point-like permanent magnet. Again, it's also not to be taken too literally, because it must be described with quantum theory, which gives quite different meaning to the quantities we know from classical objects.

To really understand all this, you indeed need a pretty abstract mathematical description, and the only link with the meaning of the observables in classical point-particle physics can be given in terms of the symmetries of the classical space-time description (BTW we don't have a generic quantum description of space and time yet, but that's another story). In Newtonian physics the space is homogeneous (i.e., the physical laws are the same at any place) and isotropic (i.e., the physical laws don't depend on the orientation), and there are mathematical ways to describe these symmetries in quantum mechanics. If you work all this out you come to the conclusion that in addition to angular momentum due to the motion of a particle around a center (e.g., the planets around the Sun) there can also be this generic quantum type of angular momentum, which we call spin.
 
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  • #15
BadgerBadger92 said:
A lot of people seem to get the wrong answers here too. People argue about it all the time.
If you are referring to PF threads, some specific references would be helpful.
 
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  • #16
PeterDonis said:
If you are referring to PF threads, some specific references would be helpful.
🦗🦗🦗

(those are supposed to be crickets)
 
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  • #17
Vanadium 50 said:
🦗🦗🦗

(those are supposed to be crickets)
PeterDonis said:
If you are referring to PF threads, some specific references would be helpful.
Check out the first two replies
 
  • #18
BadgerBadger92 said:
Check out the first two replies
Both of those replies are correct. And neither one says the things in the Stack Exchange thread that @strangerep correctly labeled as rubbish.
 
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  • #19
topsquark said:
the spin of a fundamental particle has nothing to do any physical rotation.
strangerep said:
The spin of a quantum particle type has everything to do with being a unitary representation of the (quantum) SO(3) group, corresponding to spatial rotations.
It's a floor wax! No, it's a dessert topping! Stop, you're both right--Shimmer is a floor wax and a dessert topping!

When it comes to QM, seemingly outlandish things like that actually turn out to be true. It is true both that the spin of a quantum particle comes from being a unitary representation of a rotation group (though I would say the appropriate group is SU(2), not SO(3), since we have to include spin-1/2), and that the spin of a quantum particle does not mean there is anything actually rotating, as would be the case for a classical object with spin.
 
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  • #20
PeterDonis said:
Both of those replies are correct. And neither one says the things in the Stack Exchange thread that @strangerep correctly labeled as rubbish.
I see
 
  • #21
BadgerBadger92 said:
Check out the first two replies
Just for reference, the problem with my post was that I said that spin was not a physical rotation. That isn't quite correct. What I should have said is that it is not an angular rotation. Quantum spin does not imply any sort of angular rotation, though it is related to the group of angular rotations... an electron would have to go around twice to get back to its starting point, which is ridiculous. This does happen, but not in Physical space; it happens in a Hilbert space. Except for that one word I was correct.

And, for the record, strangerep and I were not arguing. He pointed out a flaw in my post and I accepted it. Mistakes happen.

-Dan
 
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  • #22
topsquark said:
I said that spin was not a physical rotation. That isn't quite correct.
topsquark said:
an electron would have to go around twice to get back to its starting point, which is ridiculous. This does happen, but not in Physical space; it happens in a Hilbert space.
Sort of. The "rotation" in question is a change of phase--but physical states in Hilbert space are represented by rays, i.e., just multiplying an element of the Hilbert space by a phase does not change what physical state it represents. So what happens to the state vector in Hilbert space isn't really a "rotation", except in a fairly abstract sense; the state vector is not "going around twice to get back to its starting point" in any physical sense in the Hilbert space, since which physical state it represents stays the same.
 
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  • #23
topsquark said:
[...] What I should have said is that it is not an angular rotation. Quantum spin does not imply any sort of angular rotation, though it is related to the group of angular rotations... an electron would have to go around twice to get back to its starting point, which is ridiculous. This does happen, but not in Physical space; it happens in a Hilbert space.
Well,... in fact does happen also in physical space. There exist devices for rotating the neutrons in the beam on one side of an interferometer. When the rotated and unrotated beams are merged back together, the results of the rotation can be observed via the interference patterns generated. It turns out that fermions really do need to be rotated by 720deg (not merely 360deg) to return to their original state. IIRC, Ballentine talks about this in his textbook, but I don't recall the exact chapter/section. (If you can't find it, let me know and I'll try to be a bit more helpful.)

Btw, if you haven't heard of the (in)famous "Dirac Belt Trick", I'm sure you'd find it as interesting/fascinating/maddening as I did when I first read about it in Penrose's "Spinors & Spacetime" book. Nowadays it's easy to find youtube videos demonstrating it. The "trick" shows that there are ordinary situations in everyday life where a 360deg rotation is not equivalent to a 720deg rotation, thus revealing that our human intuition about physical 3D space is incomplete.

[Edit: you (and @PeterDonis) might also be interested in Misner, Thorne & Wheeler, Box 5.6 Part D, which goes through the math about how the notion of "intrinsic angular momentum" also has a classical counterpart. I.e., it's not just a quantum concept.]
 
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  • #25
PeterDonis said:
Is this paper the type of experiment you are talking about?

https://arxiv.org/abs/1904.07085#
On a very quick skim-read, yes, I think so. But I was actually thinking of some of the older ones (circa 1975) which are referenced in that paper, e.g., ref[18] which is Werner, Colella and Overhauser 1975. I think this was among those mentioned by Penrose & Rindler in their now-quite-old textbook.
 
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  • #26
PeterDonis said:
Sort of. The "rotation" in question is a change of phase--but physical states in Hilbert space are represented by rays, i.e., just multiplying an element of the Hilbert space by a phase does not change what physical state it represents. So what happens to the state vector in Hilbert space isn't really a "rotation", except in a fairly abstract sense; the state vector is not "going around twice to get back to its starting point" in any physical sense in the Hilbert space, since which physical state it represents stays the same.
Gah! Honestly, this time I knew that. I was trying to simplify it a bit, but I said it badly again. I'm just going to give up on this one! :H

-Dan
 
  • #27
strangerep said:
Misner, Thorne & Wheeler, Box 5.6 Part D, which goes through the math about how the notion of "intrinsic angular momentum" also has a classical counterpart. I.e., it's not just a quantum concept.
This is true, but it should be noted that the classical concept described there relies on the object having a well-defined center of mass worldline (and a well-defined stress-energy tensor, which means a well-defined congruence of worldlines, occupying a "world tube" surrounding the CoM worldline, that describes the matter in the object). In QM these things do not exist, so "intrinsic spin" cannot be defined exactly in this classical way. This is one way of stating the fact that in QM intrinsic spin cannot be interpreted as an actual physical object spinning around an axis, as it can in classical physics.
 
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  • #28
PeterDonis said:
This is true, but it should be noted that the classical concept described there relies on the object having a well-defined center of mass worldline (and a well-defined stress-energy tensor, which means a well-defined congruence of worldlines, occupying a "world tube" surrounding the CoM worldline, that describes the matter in the object). In QM these things do not exist, so "intrinsic spin" cannot be defined exactly in this classical way. This is one way of stating the fact that in QM intrinsic spin cannot be interpreted as an actual physical object spinning around an axis, as it can in classical physics.
The part that I've emboldened above might be overly simplistic. There are versions of Ehrenfest's theorem relating to angular momentum which may be applicable here, (but I need to do more study on that).

IoW, if one can sensibly relate the quantum case to the classical case by use of expectation values, that's all one can reasonably expect from a quantum-classical correspondence.
 
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  • #29
PeterDonis said:
This is true, but it should be noted that the classical concept described there relies on the object having a well-defined center of mass worldline (and a well-defined stress-energy tensor, which means a well-defined congruence of worldlines, occupying a "world tube" surrounding the CoM worldline, that describes the matter in the object). In QM these things do not exist, so "intrinsic spin" cannot be defined exactly in this classical way. This is one way of stating the fact that in QM intrinsic spin cannot be interpreted as an actual physical object spinning around an axis, as it can in classical physics.
I THINK I understand what you guys are saying. I’m going to read more about this to make sure.

I have one more question. I may have asked this before but ECT has really harmed my memory.

What do particles do with their spin?
 
  • #30
strangerep said:
if one can sensibly relate the quantum case to the classical case by use of expectation values, that's all one can reasonably expect from a quantum-classical correspondence.
For quantum-classical correspondence, I agree, but that still does not ground a claim that, in the quantum case, a particle having spin means something is actually spinning.
 
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  • #31
BadgerBadger92 said:
What do particles do with their spin?
The total set of a particle's properties , e.g., energy, momentum, parity, charge(s), etc, etc., characterizes the particle. Some properties (the invariants such mass, total angular momentum and charge) determine what type of particle it is. Others, such as energy, momentum, etc, determine its state relative to other particles.

As to what a particle might "do" with any of these properties, well, it could participate in interactions with other particles, possibly producing new particle(s) with new values of these properties.
 
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  • #32
BadgerBadger92 said:
What do particles do with their spin?
I have no idea what you mean by this question.
 
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  • #33
PeterDonis said:
For quantum-classical correspondence, I agree, but that still does not ground a claim that, in the quantum case, a particle having spin means something is actually spinning.
Umm,... I hope you don't think I actually said that, or think that, else we're about to have another major run of mutual misunderstandings... :oldfrown:
 
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  • #34
PeterDonis said:
I have no idea what you mean by this question.
I mean that particles have their spin, and they don’t spin with it. So what are they doing with it? Is it just a property of the particle and doesn’t do anything with it?
 
  • #35
strangerep said:
I hope you don't think I actually said that
I'm clarifying for the OP's sake.
 
<h2>1. What is spin in particle physics?</h2><p>Spin is a fundamental property of particles in quantum mechanics. It refers to the intrinsic angular momentum of a particle, which is not related to its physical rotation but is a quantum mechanical property that determines how the particle interacts with other particles.</p><h2>2. How do particles get their spin?</h2><p>Particles do not "get" their spin in the traditional sense. Spin is an inherent property of particles that cannot be created or destroyed. It is a fundamental characteristic of particles, similar to mass and charge.</p><h2>3. Can the spin of a particle change?</h2><p>Yes, the spin of a particle can change through interactions with other particles. For example, when two particles collide, their spins can combine or cancel out, resulting in a change in the total spin of the system.</p><h2>4. What is the significance of spin in particle physics?</h2><p>Spin is a crucial property in particle physics as it helps to classify particles and understand their behavior. It also plays a role in determining the stability and interactions of particles.</p><h2>5. How is spin measured in particle physics experiments?</h2><p>Spin is measured indirectly through the observation of the particle's interactions with other particles. This can be done using detectors and sophisticated mathematical analysis to determine the spin of a particle based on its decay products.</p>

1. What is spin in particle physics?

Spin is a fundamental property of particles in quantum mechanics. It refers to the intrinsic angular momentum of a particle, which is not related to its physical rotation but is a quantum mechanical property that determines how the particle interacts with other particles.

2. How do particles get their spin?

Particles do not "get" their spin in the traditional sense. Spin is an inherent property of particles that cannot be created or destroyed. It is a fundamental characteristic of particles, similar to mass and charge.

3. Can the spin of a particle change?

Yes, the spin of a particle can change through interactions with other particles. For example, when two particles collide, their spins can combine or cancel out, resulting in a change in the total spin of the system.

4. What is the significance of spin in particle physics?

Spin is a crucial property in particle physics as it helps to classify particles and understand their behavior. It also plays a role in determining the stability and interactions of particles.

5. How is spin measured in particle physics experiments?

Spin is measured indirectly through the observation of the particle's interactions with other particles. This can be done using detectors and sophisticated mathematical analysis to determine the spin of a particle based on its decay products.

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