Talk:Spin (physics)

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Help with this template Comments: edit – history – watch – refresh Lacks inline citations, very important topic! Snailwalker \| talk 00:34, 21 October 2006 (UTC)

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1 I just rewrote the section "Measurement of the spin along an arbitrary axis"
2 This article makes no sense to non-physicists

[edit] I just rewrote the section "Measurement of the spin along an arbitrary axis"

It was unnecessarily complicated in that it brought in spherical coordinates(!!) which have nothing to do with spin and vastly complicate calculations in this case. I also provided the general, normalized solution for spin-1/2 in an arbitrary direction.

I suspect that it is possible to write down a general solution for higher spin in arbitrary directions, given that an appropriate representation of SU(2) is provided. If someone thinks this is worth writing up, give me a ring at carl@brannenworks.com and I'll look around for a reference.

—Preceding unsigned comment added by 24.19.189.240 (talk) 07:32, 3 March 2008 (UTC)

OK, the operators describing spin observables do form an operator valued vector space defined by $[J_i,J_j]=i\epsilon_{ijk}\hbar J_k$ . Why did you revert? And how about the part on SO(3) representations and Clebsch-Gordon coefficients?

When one talks about an observable such as spin or energy, one is talking about the eigenvalue/eigenstate, not the corresponding operator(s). Spin does not transform like a vector, i.e. like the coordinates, under rotations. For example, if you have a spin-1/2 particle with a spin "up" in the x direction (i.e. it is an eigenstate of σ_x with eigenvalue +1, if you rotate your coordinate system by 360 degrees, you do not get back the same state; rather, it picks up a π phase shift. Alternatively, if you rotate your measurement apparatus by 90 degrees to measure spin along the y direction, you do not measure zero, you get a 50% chance of spin up or down. It is only as the angular momentum becomes large that you asymptotically recover the classical behavior.
- (Of course, strictly speaking, angular momentum is not a vector either; it is a pseudovector.)
- You are correct that the operators ( $σ 1$ , $σ 2$ , $σ 3$ ) for spin-1/2, or J's for higher spins, do form the basis of a 3-vector space. i.e. for the operator $σ = v 1 σ 1 + v 2 σ 2 + v 3 σ 3$ , the quantities $(v 1, v 2, v 3)$ transform like a 3-vector (or maybe a pseudovector?) as the coordinate system is rotated. This is just a reflection of the fact that these operators are defined relative to the coordinate system, and not on the transformation properties of the underlying observable quantity.
- (Also, please do not confuse 3-vectors with general abstract vector spaces, which of course include all quantum states.)

—Steven G. Johnson

[edit] An alternative calculation of spin

We have all noticed that spin is described as being a multiple of hbar/2. I thought that it would be better to set this value to a constant giving,

hdot = hbar/2 = 5.2728584118222738157569629987554e-35 J.s

But now the equations for spin did not work with hdot, so I had to correct them.

Here are the corrected equations,

|sv| = sqrt(s(s + 2)) * hdot

and

Sz = ms.hdot

where,

sv is the quantized spin vector,

|sv| is the norm of the spin vector,

s is the spin quantum number, which can be any non negative integer,

Sz is the spin z projection,

ms is the secondary spin quantum number, ranging from -s to +s in steps of two integers

For spin 1 particles this gives:

|sv| = sqrt(3).hdot and Sz = -hdot, +hdot

For spin 2 particles this gives:

|sv| = sqrt(8).hdot and Sz = -hdot, 0, +hdot

Now that the spin equations have been corrected, the definitions for fermions and bosons are incorrect, and must be redefined as follows.

Fermions are particles that that have an odd integer spin.

Bosons are particles that have an even integer spin.

Would these redefinitions have any other effects on the Standard Model?

Can these redefinitions explain any currently unexplained phenomena?

Are there any experiments that could confirm or refute these claims?

I would like eveyone to have a good think about this, and give me your objections to it, or even data to support it.

Giving a name to $\hbar/2$ has absolutely no affect on anything physical, just the way we describe it. Thus there are no experiments that could confirm or refute your claims as you claim nothing different from the status quo. However, you have made a massive change to notational conventions that would require most of the world's physics books to be rewritten, and therefore I think it is a bad idea. Uberdude85 01:40, 21 November 2006 (UTC)

I had a radical thought I'd just like to put out there for Visionaries. What if gravity is just the combined effect(similar to centrifugal force) of the Quantum Spin of large masses of matter Dark or Common. And what if a series of Quantum Gyroscopes could be created to Nullify this effect.... —Preceding unsigned comment added by 99.252.195.32 (talk) 15:42, 2 January 2008 (UTC)

[edit] spin

i think it would be really good if someone actually explains what is spin. fact is i dont know what it is

Spin is the property shown by quantum objects which causes them to show the behavior we associate with the word 'spin'. There really is no simpler answer than that - it is what it is. What I can tell you that it is [i]not[/i] is the rotation of a particle around its own axis. It's merely named with such an analogy in mind - but it is just a name. 81.156.75.42 14:32, 2 April 2007 (UTC)

mass could be a product of an objects contained energy ie. speed of its quantum spin more energy....faster spin....more gravity —Preceding unsigned comment added by 99.252.195.32 (talk) 15:51, 2 January 2008 (UTC)

[edit] purely quantum phenomenon

I'd like to dispute the claim that spin is a purely quantum phenomenon. For example, the classical electromagnetic field has intrinsic angular momentum. It is only half integer spin that is purely quantum mechanical. -Lethe | Talk 01:59, August 9, 2005 (UTC)

It's not just that. If you model the electron as a classical spinning object, then, given the experimental upper limit on the electron radius, the surface of the electron would have to be moving faster than light. -- CYD

Sure, so intrinsic angular momentum isn't due to a rotating sphere. It's intrinsic, just like the charge. but what does QM have to do with it? -Lethe | Talk 03:57, August 9, 2005 (UTC)

from Baez on spr:

Spin angular momentum is the angular momentum intrinsic to a particle. Most people are unfamiliar with spinning point particles in the context of classical mechanics, so they think of spin as a purely quantum business. This isn't quite right: there's a perfectly nice notion of a spinning point particle in classical mechanics. However, nobody thought of it until after quantum mechanics came along.

I think that you could just write a classical Hamiltonian like

$H=\frac{p^2}{2m}+\mathbf{S}\cdot\mathbf{B}$

or something, and do classical mechanics with it. no? -Lethe | Talk 09:46, August 9, 2005 (UTC)

Wait. Why would the speed of rotation have to be greater than that of light under relativity? I can see how in Newtonian mechanics a rotating body might need to have a rotational speed greater than that of light, but FTL isn't forbidden in Newtonian mechanics. Under relativistic mechanics FTL is forbidden, but mass increases with velocity (and thus momentum increases faster than predicted by Newtonian mechanics), so no object should have to have an FTL velocity of rotation to have a given amount of angular momentum. 12.37.33.3 04:49, 18 March 2006 (UTC)

Well, I don't like to say "mass increases with velocity". In modern usage, "mass" always refers to invariant mass. Nevertheless, you may be right; the formula l=Iω is a nonrelativistic formula, and with a relativistic version of this formula, I guess we may find that we can have a very high angular momentum without rω being greater than c. But what is the relativistic formula for angular momentum of a rigid body? I don't know. We have to be a little careful, because there is actually no such thing as a rigid body in relativistic mechanics, but still we should be able to come up with a relativistic version of that formula. I'm going to think about that. It's a good point. -lethe ^{talk +} 05:08, 18 March 2006 (UTC)

OK, relativistic mass increases with velocity. This, I believe, is because of Energy/Mass equivalence (in fact, I believe that all of relativity can be derived from adding En/Ma equivalence to Newtonian mechanics, though I don't have the math skill to prove it, my knowledge being more of the theory). As I understand it, kinetic energy has mass, so it "weighs down" a particle. To get to a certain speed, you need to add a certain amount of energy, to go beyond that speed, you need to overcome the inertia of the rest mass *plus* the inertia of the kinetic energy already on the object. The speed of light is the point where the KE of an object is infinitely greater than its rest mass (this is why it is impossible for a massive object to reach the speed of light. It would take infinite energy). Photons travel at exactly c because any positive, non-zero, real number is infinitely greater than 0 (the rest mass of a photon). Another way of looking at it might be to say that the speed of a particle is a function of what fraction of its mass is KE, with 50% KE giving a speed of ~.86c and 100% KE giving a speed of c. Anyways, since one of the determining factors of momentum is mass, and since angular momentum is a kind of momentum, a rotational speed greater than c should not be required to give ANY finite, positive, real angular momentum. 12.37.33.3 05:35, 18 March 2006 (UTC)

Well, I'm sick to death of conversations about relativistic mass versus invariant mass. Let me just tell you that no practicing physicist uses a concept of relativistic mass. Please read this. Mass does not increase with velocity. Nevertheless, it may be the case that angular momentum increases without bound as radial velocity of rotation approaches c, so your question might still be a good one. But please, I don't want to talk any more about what relativistic mass means. -lethe ^{talk +} 06:00, 18 March 2006 (UTC)

Sorry. I'm a recent high school graduate, not a proffesional physicist (and as my new username suggests, I'm likely to go into a totally different field). Was my error in not properly specifying invariant vs. relativistic mass, or in even treading on the topic of relativistic mass in the first place? If it was not being specific enough, is there a way of specifying relativistic mass in less than 6 syllables? If it was in treading on the topic at all I'll shut up. Linguofreak 06:27, 18 March 2006 (UTC)

You didn't make an error. Sorry if I was terse before, but talking about relativistic mass makes me edgy; I've been involved in countless battles about that matter. The point is, relativistic mass, while not wrong, is rather inconvenient. It's not invariant, meaning it's observer dependent, two different people see different relativistic masses if they're moving. This isn't so bad, the same holds true of energy and momentum, but we also need to have a word for invariant quantities. Rest mass is a bad word, since some things are never at rest (photons). And "relativistic mass" is already redundant, since it's the same as energy. So let's use "mass" for rest mass, and "energy" for relativistic mass. Relativistic mass isn't really a mass anyway, since it's only inertia in one direction. Transverse inertia (inertia in a direction perpendicular to the direction of motion) is always equal to rest mass.

The way to specify "relativistic mass" in fewer than 6 syllables is to say "energy". Energy = mγc². If you prefer units of mass, then say E/c². And m is always rest mass.

But this issue is purely linguistic. It doesn't affect the physics, and you raised a good point. Your point was that moment of inertia should increase with the rate of rotation (since mass does, and moment of inertia depends on mass). The better way of saying that would just be that the linear formula for angular momentum must be corrected at relativistic speeds. I think it's a good point, one I hadn't thought of, but it's probably true, though I've never studied relativistic tops, so I don't know the answer. I'm researching it now, and as soon as I find something, I'll post back. Anyway, thank you for raising an interesting question. -lethe ^{talk +} 06:59, 18 March 2006 (UTC)

Ok, yeah, "energy" works. I'll have to try and remember to use E or E/c². It's kinda weird though, because since I don't have any formal training in relativistic physics I'm used to calling it "energy" when it exhibits the classical charachteristics of energy, and mass when it exhibits the classical characteristics of mass (such as inertia).

I do have a couple more questions. Wouldn't a point mass elementary particle (as modern physics assumes), be a black hole? I think I've heard something about the ratio of charge to mass for such particles excluding them being black holes, but what about neutral particles? But a sphere of non-zero size doesn't really fit either because of the point you made earlier of there being no such thing as a truly rigid body. So neither assumption really seems to work.

No, a point particle need not be a black hole. For example, electrons are not black holes. I suppose you think that should happen because the classical energy density (which goes as 1/r²) becomes infinite as the radius goes to zero. This is true, but it's not the whole story. Mostly, that is a relic of the classical point particle theory. The problem persists in a different form in the quantum theory, where renormalization is the answer. We can't answer this question with a point particle anyway, because point particles can't rotate. We have to answer it with a body of finite size. As we both agree, rigid bodies are tricky. Our choices are to either figure out how to deal with rigid bodies (and there are ways), or else to try to deal with nonrigid bodies. I'm looking at both ways to answer the question. A third way is to just assume perfect rigidness, but this is unphysical at relativistic velocities (a good approximation at slow speeds though). -lethe ^{talk +} 23:05, 26 March 2006 (UTC)

While you're researching and calculating, could you tell me if relativity can be derived by adding Energy/mass equivalency to Newtonian mechanics, or am I missing something?

Linguofreak 18:29, 18 March 2006 (UTC)

No, relativity cannot be derived by adding mass energy equivalence to Newtonian mechanics. The finiteness of the speed of light is an experimental factor which has to be put in by hand. If you don't add that to your theory, you won't get relativity. However, it works nicely the other way; the equivalence of mass and energy can be derived from the finiteness of the speed of light (along with the conservation of momentum), so that is definitely the way you should try to understand it. -lethe ^{talk +} 22:53, 26 March 2006 (UTC)

I was going to write something else, but then I found in the article on Momentum what I was looking to show. Under relativistic mechanics $\mathbf{p} = \gamma m\mathbf{v}$ . So when v=c, and gamma is infinite, shouldn't momentum be infinite? (forgot to sign... again...) Linguofreak 21:27, 26 March 2006 (UTC)

Right, this was your original point: momentum and energy become unbounded as the speed approaches c, which implies that for a sufficiently large velocity less than c, the particle can attain any amount of energy or momentum. It's natural to assume that the same phenomenon occurs with angular momentum. If that is the case, then the statement "a rotating body with the angular momentum of an electron must be rotating with a radial speed greater than c" should be incorrect; you should be able to attain any angular momentum before your radial velocity surpasses c. It sounds reasonable. The question is, when the rω goes to c, does the angular momentum L become infinite? To answer that, we need a formula for angular momentum. One such formula is L=mγrv, and this expression certainly does become infinite. However, that formula is only valid for a point particle moving in a straight line. For a rotating disk, the answer becomes much more subtle. I've already done a calculation that shows that a rigid rotating disk has an angular momentum that stays finite, but rigid bodies are quite tricky in relativity. I'll say more about this later. Finally, let me just say that I think that while there are certainly difficulties with the classical theories of point particles, I don't think it's as bad as Xerxes makes it seem below. -lethe ^{talk +} 22:49, 26 March 2006 (UTC)

Assorted answers: A classical point mass particle would be a black hole. Ignoring quantum mechanics at that scale is foolish, however. Relativity is a broad unification of scalar and vector concepts into 4-vectors with Minkowski metric. Mass-energy-momentum is just one such unification. Infinite momentum is the equation's way of telling you "This is impossible." -- Xerxes 21:37, 26 March 2006 (UTC)

Did you mean finiteness of the speed of light, or constancy in all reference frames?

Finiteness of lightspeed without constancy does not give relativistic theory. However, if you knew that light carried energy, and were to somehow observe that E and m were related by a finite, non-zero ratio without knowing anything about the constancy of the speed of light or its finiteness, you should be able to deduce both, along with all of relativity from this. The thing is that it's easier to measure the speed of light directly (and find that it is finite and the same in all reference frames) than to figure out mass/energy equivalency directly.

This point is actually worth emphasizing. It is the finiteness of the speed of light, not the invariance, which should be viewed as the fundamental concept. The invariance of the speed of light can be derived mathematically. However, the math doesn't tell you whether this invariant speed is finite or infinite, that's something you have to choose for yourself, based on physical experiment. If you choose infinite invariant speed, you get Newtonian mechanics, while if you choose finite invariant speed, you get relativity. This point is not well-known, it's not taught in most of the books, I suppose because Einstein himself chose to adopt invariance of the speed of light as a postulate. But that's not fundamental; it can be derived. To sum up: yes indeed, finiteness alone without invariance of the speed of light alone is enough to get you the entire theory of special relativity. (I guess I should add a proviso here. When I say "speed of light", I don't necessarily mean the speed at which light travels. I just mean some universal invariant speed. Whether or not electromagnetism actually travels at that speed is a question of the details of electromagnetics, independent of the geometry of spacetime. It would be better if that speed had a more independent name, like say "the universal invariant speed" or something. Who knows? Maybe one day we'll find out the photon has a nonzero mass!)

Now about your proposal to adopt the equivalence of mass and energy as fundamental instead. Despite what I said earlier, I think probably this can work. The thing is, when you use the invariance/finiteness of c, you can use a purely geometric argument about distances between points. Flash a light inside a rocket, that sort of thing. If you want to argue with energy and mass instead, then you have to move to a less geometric picture, using variables that are a bit more abstract (energy versus distance). Also, I think you'll have to use the conservation of energy and/or the conservation of momentum. The argument is probably much less obvious, but maybe it still works. We should try it.

First question about your rigid rotating disk: Have you tried it for cases where the rate of rotation is greater than c? I have a hunch that it may work at c, but not faster. But as you said, relativity doesn't like rigid bodies. What about if you took it to be a rotation in a non-rigid body? The whole thing about point particles was more of an aside, because I didn't think a point particle or a rigid sphere would work. But now that I think you've explained quantum waveforms to me in a way that I can grasp on the uncertainty talk page, they seem like a viable alternative. I think I saw that you had left another response to me there, so I'll go check it out. Linguofreak 03:08, 27 March 2006 (UTC)

Yes, for the rigid rotating disk, the angular momentum is finite for all finite rω, below c and beyond it. But like I've said, rigid objects at relativistic speeds are quite unphysical. At the very least, there would need to be an enormous amount of internal energy holding the atoms together, to counteract the centrifugal force, and this ought to contribute to the angular momentum, so I don't put too much stock in the calculation. Now, Born provided a way to approximate a sort of relativistic form of rigidity (it's not a true rigidity, but rather some constraint about internal tensions), but I don't really know how to use that method. So I also have this citation for an old paper by Regge where he does such a calculation for a rigid sphere, but I haven't tracked down that paper yet. The other options are, as you mention, to consider non-rigid objects. Those are more physical anyway, but of course also more complicated. For that approach, there are books about relativistic neutron stars and such. A nice astrophysics book ought to have something about the angular momentum of a rapidly rotating star. The extreme example of this is the rotating black hole. A rotating black hole is described by the Kerr metric, and there is a formula in the classic GR textbook by Misner, Wheeler and Thorne (known affectionately as "the phonebook") which says that the angular velocity looks like this:

$\omega=\frac{a}{a^2+r^2}$

with r the horizon radius given by

$r=m+\sqrt{m^2-a^2}$

and the angular momentum given by L=ma. This isn't too hard to solve for L, and you get L=2m²β, where β is the ratio of the radial velocity at the horizon radius to the speed of light. Clearly this expression is finite for all β, so once again, it looks like L will not diverge. However, it's still not entirely satisfactory to me, because well, there is no mass at the horizon of a black hole, or indeed anywhere outside the singularity. Therefore we cannot conclude that this solution actually entails stuff going faster than light.

To sum up: the rigid disk doesn't diverge if you ignore rigidity issues, and I haven't found the calculation that takes them into effect. I haven't found the formula for the fast neutron star either, and I'm not satisfied with the black hole case. Thus, I don't have an answer to your seemingly simple question yet. We'll see what we can do though. 144.92.241.204 05:00, 27 March 2006 (UTC)

When Uhlenbeck and Goudsmith had the idea of spin, their professor, Ehrenfest, asked Lorentz what did he think about it. Lorentz sent him some papers covered with formulas, showing that it is impossible to have an electron spinning with the predicted angular momentum. Now, does anyone here know relativity better than Lorentz? Omsharan 09:44, 25 June 2007 (UTC)Om Sharan

[edit] points?

"even though they are believed to be point particles possessing no internal structure"

Are they really believed to be infinitesmal points, or are they just modeled as points? — Omegatron 20:57, 30 September 2005 (UTC)

[edit] Half-integers?

This page says the spin is a "half-integer (0, 1/2, 1, 3/2, ...)". But the page about Half-integers says integers are no half-integers, i.e. only 1/2, 3/2, ... Which one is it? If both pages are right, at least the link is misleading. Or "half-integer" has two meanings that should be distinguished.

The listed numbers are correct: N/2. Integer spin means boson and half-odd-integer spin means fermion. I'm not sure about the meanings of half-integer. --MarSch 12:53, 3 November 2005 (UTC)

I think you have been confused by the slightly unclear phrasing; it says "integer or half-integer (0, 1/2, 1, 3/2, 2, etc.)". 0, 1 and 2 are integers; 1/2, 3/2, 5/2 are half integers. So both pages are correct. Uberdude85 01:40, 21 November 2006 (UTC)

[edit] What Spin Is

Spin exists because physics is invariant under Lorentz Transformations, instead of being invariant under Galilean transformations. Quantum mechanics doesn't create spin, it merely quantizes it.

There is a very close relationship between physical laws being invariant under groups of transformations, and conserved quantities. Translation invariance implies momentum conservation, time invariance implies energy conservation, rotational invariance implies angular momentum conservation, and so on.

Now Lorentz Transformations have this funny little property that Galilean Transformations don't have. Two velocity changes at angles to each other may be equivalent to a velocity change PLUS A ROTATION. Because of this, the total conserved quantity under Lorentz spacial rotations isn't exactly classical angular momentum, but it's classical angular momentum plus "a little extra." We call this extra quantity, which we identify with the intrinsic rotational properties of different types of fields under Lorentz Transformations, spin. When we quantize the fields, we get quanta that each carry a little bit of this spin, which depends on the type of geometric object we quantized. Scalars - spin 0, vectors - spin 1, rank 2 tensors - spin 2, and of course, for various flavors of spinors, which carry an half-odd-integral spin representation of the Poincare group, 1/2, 3/2, 5/2, etc...

So in a nutshell, spin is not a quantum mechanical property. Spin is a relativistic property. Spin getting quantized is a quantum mechanical property. Classical spin, and even classical spinors, are perfectly well defined concepts.

Note: A huge number of undergraduate texts get this completely wrong, and present spin as some magical incomprehensible quantum mechanical property. It becomes apparent later, when you employ Langrangian densities and Poincare invariance to directly calculate the equations of motion and conserved quantities for various flavors of fields, how spin is fabricated out of the odd way Lorentz rotations compose.

It would be nice if the article explained what spin actually is, in terms comprehensible to an ordinary person.Hermitian 09:27, 12 January 2006 (UTC)

I disagree with you: spin arises naturally even in nonrelativistic settings. The noncommutativity of boosts in relativistic mechanics is not really relevant. -lethe ^talk 10:50, 12 January 2006 (UTC)

It's not noncommutativity. It's the composition of two pure velocity changes yielding a velocity change plus a rotation. Velocity changes are just rotations in a plane which includes "t" as a coordinate, whereas spacial rotations do not. If there was no mixing between purely spacial rotations and velocity changes, then total angular momentum would be the usual classical thing. Spin arises naturally in the mechanics of fields when you calculate the conserved total angular momentum for Poincare invariance, and doesn't when you calculate it for Galilean invariance. All spin in the standard model is created in this manner.

I'm not sure what you mean by "spin arises naturally even in nonrelativistic settings." Spin is a consequence of Poincare invariance of physical laws. That doesn't mean that the particles in question are moving at relativistic speeds.

A user in an infinitesimally displaced frame will see different coordinates and field variables. If the laws of physics are invariant, the variation in the action will be zero, and we will get a conservation law, of the quantity canonically conjugate to what was perturbed.

When we do this for a field and infinitesimal spacial rotations under the Galilean group, the conserved total angular momentum is a single term, which we can quickly identify as classical angular momentum. When we do this for a field and infinitesimal spacial rotations under the Poincare group, the conserved total angular momentum is the sum of two terms, because of the mixing between pure spacial rotations and velocity changes. We can quickly identify one of the terms as the same classical angular momentum we saw in the Galilean scenario, and we call the new term "spin angular momentum." The total angular momentum is the sum of these two.

All quantum field theory based on the Lagrangian Formalism and Schwinger's Variational Principle derives spin through this mechanism, with the spins of the quanta corresponding to what variety of geometric object is being employed as a field variable.Hermitian 11:40, 12 January 2006 (UTC)

Spin is a consequence of Poincare invariance of physical laws. No, spin is a consequence of rotational invariance. The fact that Poincaré invariance also implies spin simply speaks to the fact that the rotation group is a subgroup of the Poincaré group. You have the same spin representations of the nonrelativistic Galileo group as you do of the Poincaré group, therefore it cannot be correct to classify spin as a relativistic phenomenon. -lethe ^{talk +} 23:50, 21 February 2006 (UTC)

You are both wrong. So Let's clear some things up. The Galilean group is generated by the rotation group SO(3,R) and Galilean boosts. In the Galilean group, the set of boosts K forms an Abelian subgroup and thus an invariant subgroup. So the Galilean group is a "semidirect" product of SO(3) rotations J with the Abelian group of boosts K which exist on a flat affine velocity space. So J^2 is a Casimir operator. It commutes with everything and is preserved under Galilean transformations In contrast, in the Lorentz group the set of boosts exists on the curved Lobachevski velocity space and does not form a subgroup, but a groupoid. So the Lorentz group is a "quasidirect" product of the rotation group (as a subgroup of the automorphism group of the velocity groupoid) with the weakly associative groupoid of velocities. Thus the Lorentz group is IN NO WAY a semidirect product like the Galilean group, hence the Lorentz group is a simple group (ie noncommutative and having no nontrivial invariant subgroups, such as the SO(3) group as you claim). NOW, why this often gets confused as being a result of mysterious quantum mechanics has to do with Lie algebras and irreducible representations and the complexification of your original representation space, which are mathematical concepts that are more advanced than you are when you first learn quantum mechanics. So instead of dealing with angular momentum, as a more direct illustration take the electric and magnetic fields. Under SO(3,R) rotations, the sixtors E and B transform separately, giving a representation of the rotation subgroup that forms a direct sum, and hence an irreducible representation. This irreducibility does not hold if we allow for complex numbers as the number field for the representation spaces. Thus we can form representations under SO(3,C) where F+=E+iB, F-=E-iB. Notice this looks a lot like the forms J+ and J- in quantum mechanics. So really, in the Lorentz group, we must do the same thing in order to get a compact representation. So we complexify and find that our full rotations under the Lorentz group are J+=J+iK, J-=J-iK. This DOES arise from the fact that Lorentz boosts are non-Abelian. Hence we arrive at the Dirac or Weyl representation of spinors for a relativistic theory, which reproduces our 4D Lorentz algebra. But notice J^2 is no longer a Casimir operator. Now our equivalent Casimir operator is formed from the Pauli-Lubanski 4-vector, W. In the timelike case for a massive spin 1/2 particle, no one cares about the boosts K and the commutator for W reduces SO(3), and the Casimir now looks like m^2 J^2, which is the product of our 2 Casimirs P^2 and W^2. However if we boost, our theory is no longer SO(3) invariant, and spin is no longer a good quantum number. When we boost we have to add a term that looks like the helicity operator, which couples on momentum and spin. So using the fact that SO(3)~SU(2), the Lorentz group turns into a helicity representation SU(2)L x SU(2)R, which are coupled by the mass of the particle. Each transform under different representations since it is a direct product. —Preceding unsigned comment added by 71.105.84.231 (talk • contribs)

You say it yourself: for a massive spin 1/2 particle, no one cares about the boosts K and the commutator for W reduces SO(3), and the Casimir now looks like m^2 J^2. It's not that no one cares about boosts, it's that Wigner taught us that the right way to find representations of a group is to look at the isotropy subgroup. The isotropy subgroup for a massive particle is the rotation group. The rotation group is a subgroup of any spacetime symmetry group of physics, thus any system must live in a rep of the rotation group, which we call spin. All the other stuff you mention is certainly relevant to the representation theory in general, but does not change this simple fact. -lethe ^{talk +} 20:41, 17 March 2006 (UTC)

[edit] Forward referencing

It would be nice if this article didn't use the phrases, "which will be described in the remainder of this article", "A later section covers", "as we will see", etc. in describing information presented later in the article. I guess I'm saying the grammar could be improved.

[edit] Cleaning up the Intro

The intro goes from an explanation of Spin in a classical mechanics to an explanation of spin in a quantum mechanics wihout significant emphasis that the two are not physical analogs. Though my familiarity with the subject is elementary, I am under the impression that Spin is only mathematically analogous (and loosely at that, since half-integers are allowed), and the use of the term arose from a lack of understanding at the time that electrons et al. are in fact point particles. I'm going to try to rewrite the intro to make it more clear that Spin in quantum mechanics is a consequence of Quantum Mechanics -- i.e., schrodinger equation, heisenberg's formulation, etc. -- and is not understandable with the original Newtonian Mechanics (F = dp/dt) and its later but equally classical formulations (Lagrange's equation, Hamilton's canonical momentum, etc.). Ryanluck 16:59, 27 February 2006 (UTC)

More specifically, it is not the "atomic length scales" which give rise to Quantum Mechanical spin -- it's our formulation of quantum mechanics itself. As is well said in the article, classically a spinning electron at the classical radius would have surface speeds greater than c -- so clearly it's not the smallness but our understanding of the behavior of such small objects. Ryanluck 17:02, 27 February 2006 (UTC)

I disagree with your assessment of spin. I think of spin in the following way: spin is our word to describe how a system behaves under rotation. This has nothing to do with either classical mechanics or quantum mechanics, though of course both kinds of systems do have to have a specific behaviour under rotation. Angular momentum is the generator of rotations, so angular momentum is related to spin, and for classical objects, the angular momentum of a system is determined by how fast the system is revolving, which is where the term got its name. The only difference between classical spin and quantum spin is that quantum systems (whose space of states is a projective space) allow projective representations of the rotation group, while classical systems (whose space of states is a vector space) do not. Thus one sees half-integer spin in quantum mechanics, but not in classical mechanics. Certainly I agree with you that the smallness associated with quantum mechanics is not relevant, but I don't think Schrödinger's equation is relevant either. -lethe ^{talk +} 17:17, 27 February 2006 (UTC)

Ah, ok, that sheds some light. The rotation matrix as applied to spin states is where I first encountered a rigorous description of the weirdness -- the half-integer spins rotated 2*pi give you e^(pi*i) = -1 rather than the usual e^(2*pi*i) = 1, so your conception of spin as behavior under rotation makes sense. My intuition with the nature of the Hilbert space is still pretty weak, but your projective representations argument helps with that to. Thanks for the corrections and clarifications. Feel free to alter any of my edits as you see fit (not that such permission needs to be given or received in wikipedia). Ryanluck 17:58, 27 February 2006 (UTC)

Yes yes, that rotation matrix that acts on two component spinors is the projective representation of the rotation group. This means that it acts not on vectors in the Hilbert space of two component spinors, but rather on rays of that Hilbert space (the space of rays in a vector space is called a projective space). The reason for this is that phase is unmeasurable, because the measurable stuff in QM all uses magnitude squared. Some heavy duty mathematics can tell you that the projective representations of a group are just the representations of the universal cover (but only if the group is finite dimensional). So quantum mechanics requires you to consider the group SU(2) instead of SO(3) (SU(2) is the universal cover of SO(3)). -lethe ^{talk +} 18:11, 27 February 2006 (UTC)

[edit] Why is spin so important?

A particle may exist in many different quantum states. However, it seems that spin is somehow more fundamental to various other properties of particles than other things. E.g., spin states add and give bosons or fermions which have dramatically different properties, but principle quantum numbers don't. Also, most fundamental particles and even larger quantum mechanical systems are grouped most readily by their spin states (spin 1/2, sprin 0, spin 1, spin 2 (graviton?)). Can someone explain this in the article? Ryanluck 17:11, 27 February 2006 (UTC)

Well, the spin quantum number of particles is more important because it is intrinsic, i.e. it is a property of an electron, whereas the principle quantum number depends on how the electron is moving relative to a nucleus. The same can be said of mass and charge. This fact is mentioned in the article already. -lethe ^{talk +} 17:21, 27 February 2006 (UTC)

Hmm, I see that this mention was added by you. Well I agree with your edits. -lethe ^{talk +} 17:32, 27 February 2006 (UTC)

Upon looking closer, I've decided that I actually do not agree with your edits. Specifically, the use of the word "analogous". I've reverted. -lethe ^{talk +} 18:00, 27 February 2006 (UTC)

[edit] spin's relevance

I was hoping to find a correlation/link between spin/rotation and e=mc2. The other correlation/link is whether sound waves are generated and measureable by particles and large scale structures. Does the mass of an object include these energy outputs?

Sound waves are not generated by particle spin. Sound is a phenomenon meaningless at much smaller scales than that of mid-sized groups (a few thousand) of particles. It involves the transmission of vibrations between particles. Unless you are dealing with densities on the order of those in a neutron star or higher, these generally be atoms or molecules (molecules being made up of 2 or more atoms, and atoms being made up of 4 or more elementary particles (3 quarks and an electron for basic hydrogen, more for more complex atoms (not counting gluons, etc.))). Particle spin is an intrinsic property of individual elementary particles, and as far as I know, the spin of one particle does not induce vibrations in its neighbors. 12.37.33.3 05:09, 18 March 2006 (UTC)

[edit] Split?

Would it be useful to the reader to split this article? It seems like the discussion of spin in quantum mechanics would make a good article all by its self. Is it sufficently different from spin in the everyday world to warrant such a split? Or do they really deserve to be on the same page? Also, the physics bits of Rotation seem to share lot with he non-quantum bits of Spin (physics). Would it be useful to reader to merge those bits? (See also Talk:Rotation#Split?) Ewlyahoocom 12:57, 2 April 2006 (UTC)

But right now, almost the entire article is solely about quantum spin... -lethe ^{talk +} 14:27, 2 April 2006 (UTC)

It is a single concept. There is no justification for such a split. Removing. -- Xerxes 22:04, 2 April 2006 (UTC)

[edit] Clarify the First Sentence to make definition easier to understand

I am guessing that this sentence would be easier to read if the clauses were re-ordered to keep all of the definition together and put the anti-definition at the end of the sentence:

INSTEAD OF : In physics, spin refers to the angular momentum intrinsic to a body, as opposed to orbital angular momentum, which is generated by the motion of its center of mass about an external point.

REVISE SENTENCE: In physics, spin refers to the angular momentum intrinsic to a body, which is the rotation of the body around its own center of mass; as opposed to orbital angular momentum, which is generated by the motion of its center of mass about an external point.

OR, TO MAKE IT EVEN MORE OBVIOUS: In physics, spin is the rotation of a body around its own center of mass. Spin refers to the angular momentum intrinsic to a body. This should not be confused with orbital angular momentum, which is generated by the motion of a center of mass about an external point.

Regards: Greg Chalmers

Such a revision changes the meaning of the sentence and would be incorrect. -- Xerxes 23:16, 5 April 2006 (UTC)

[edit] Move the page to make it easier to understand

OK, I think I understand that this article is about quantum mechanics. But this is not clear from the lede or the title of the article. The phrase "intrinsic angular momentum" is presented as if that is supposed to clarify things, but it doesn't. People looking for the meaning of "spinning top" or "spinning wheel" or "spins round and round" would be better served by the information at Rotation or Angular momentum or even Billiards#English. Perhaps this page could be moved to e.g. Spin (quantum mechanics) or Spin (particle physics) or something like that. Ewlyahoocom 06:44, 7 April 2006 (UTC)

I'm against it. Exactly what sort of information did you have in mind to put in an article on spin? -lethe ^{talk +} 07:39, 7 April 2006 (UTC)

Against. People looking for the meaning of any of the phrases you mentioned would be more likely either to search Wiktionary for "spin," or to look up "top" or "wheel." The only field in which the term "spin" is used and is important enough to justify it's own article is physics. Linguofreak 17:58, 8 April 2006 (UTC)

Really? Then I guess most of the articles linked from Spin (disambiguation) derive their names from this quantum physics meaning, yes? Spin (public relations) actually refers to a spin-1/2 particle with non-zero magnetic moment? Figure-skating spin ("Pair spins", even!) has something in common with Pauli exclusion? Furthermore, I haven't suggested doing anything to Spin (physics) other than leaving it as the (automatically generated) redirect to the new name. With a new name, it would be more clear that the article is about quantum mechanics, and anyone arriving here by mistake would immediately know to click on the dablink to Rotation before wading through the article (for examples, see the above comments #Cleaning up the Intro, #Split?, #Clarify the First Sentence to make definition easier to understand). Ewlyahoocom 03:03, 9 April 2006 (UTC)

You've totally lost me. How will changing spin (physics) to spin (particle physics) make it easier for people looking for spin (public relations) to realize they're in the wrong place? -lethe ^{talk +} 17:23, 9 April 2006 (UTC)

Those comments are in response to Linguofreak's preceding comment that "The only field in which the term "spin" is used and is important enough to justify it's own article is physics." Ewlyahoocom 03:10, 10 April 2006 (UTC)

Ok, I admit, I made that point rather badly. Let me rephrase: The only field of physics where the term "spin" is used and is important enough to justify its own article is particle physics. Linguofreak 04:05, 10 April 2006 (UTC)

What are you talking about? Spin is important to all physics, not just particle physics. Atomic physics, molecular and chemical physics. Not to mention all of chemistry, in particular NMR, MRI, Open shell systems and radical chemistry, transition metal chemistry etc. It's a common and general physics concept, and it really does belong under "Spin (physics)". Also, spin is not a quantum mechanical phenomenon. So if someone comes to "Spin (physics)" expecting to read something else than what the entire physics world calls "spin", then that's their problem. --BluePlatypus 22:22, 13 April 2006 (UTC)

[edit] What is spin?

I dug up a few good articles on the subject, namely "What is spin?", Ohanian, H.C., Am. J. Phys., 54(6), 500. And "Understanding electron spin", Boeyens, J. C. A., J. Chem. Ed., 72(5), 412. Both do a fairly good job of explaining how spin is not a quantum mechanical and/or relativistic effect. Perhaps something for the reference section? --BluePlatypus 22:29, 13 April 2006 (UTC)

I've been planning a rewrite of this article for a while which would eradicate from head to toe the notion that spin is either relativistic or a quantum phenomenon. Of course, it means that classical spin has to get some press time, which isn't really a concept that's used much, but well, I think it's worth it. My planned rewrite makes use of concepts from differential geometry, so it may not be as accessible to physicists as an article on spin deserves, so I'm a bit hesitant. Anyway, feel free to post references. Maybe I'd like to read them too. Are they online? -lethe ^{talk +} 23:57, 13 April 2006 (UTC)

[edit] Cottage industry?

The study of the behavior of such "spin models" is a thriving cottage industry in condensed matter physics.

Cottage industry? I don't think that's the right phrase. —Keenan Pepper 15:45, 6 June 2006 (UTC)

You know, I saw that phrase a few weeks ago, and it seemed weird to me too, but my roommate who is in condensed matter physics says that it is used. -lethe ^{talk +} 16:08, 6 June 2006 (UTC)

[edit] even and odd spin

When it turned up that spin 1/2 exists, one talks about half odd integral spin versus integral spin. This ugly language should be changed. In the unit of h-bar/2 the fermions have odd spin and the bosons have even spin. That is easier.

When it turned up that the quark charges are integer multiples of 1/3 proton charge, the decent thing is to introduce a new elementary unit so that the proton charge is 3 and the electron charge is −3. This is not what the physicists did, however, for they had grown accustomed to the Millikan unit, but for the teaching of young people it should be done right.

Am I wrong? Bo Jacoby 14:25, 19 July 2006 (UTC)

You are not wrong, however fixing notation is not within the purview of Wikipedia. The right approach is probably to write some physics textbooks. After a few decades, if your proposal has caught on in physics education, then Wikipedia may follow suit. -lethe ^{talk +} 17:17, 19 July 2006 (UTC)

Thanks for your respons. A short explanation might save future readers from some of the confusion which I felt by the mystery of half odd integer spin. The subject matter is sufficiently complicated already. This extra complication may be the straw that breaks the camel's back. Bo Jacoby 14:24, 20 July 2006 (UTC)

[edit] Contradiction on definition of spin?

Can someone explain the following contradiction in the article, under the heading "Spin of elementary and composite particles"?

First it states:

Theoretical and experimental studies have shown that the spin possessed by these particles cannot be explained by postulating that they are made up of even smaller particles rotating about a common center of mass...

But a little further on it says:

The spin of composite particles, such as protons, neutrons, atomic nuclei, and atoms, is made up of the spins of the constituent particles, and their total angular momentum is the sum of their spin and the orbital angular momentum of their motions around one another.

Which statement is correct here? Mehrshad123 00:02, 14 December 2006 (UTC)

[edit] Not really a contradiction...

Think of it this way: Just as a figure skater speeds up by bringing in her or his arms, thereby conserving angular momentum, the smaller a particle is, the faster its components must whirl about each other to maintain the same overall angular momentum.

While protons, neutrons, and other composite particles are certainly tiny, they have well-defined diameters that are sufficiently large to allow their constituents (quarks and gluons) to whirl about each other with enough separation to add significantly to the overall angular momentum of the particle. This is the "orbital angular momentum of their [quark and gluon] motions about one another" mentioned in the article.

An electron, in contrast, has never been found to have a measurable diameter no matter how closely one looks (where in particle physics "looking closely" translates into "no matter how high an energy one uses"). While very small components of electrons can (and have; see Preons) been proposed, the trouble is that no matter how one slice and dices and whirls such hypothetical pieces, they all must move within a region so incredibly tiny that they would have to exceed the speed of light many times over to provide significant additional angular momentum. This again is in sharp contrast to particles such as protons with diameters large enough to add angular momentum while whirling about at less than the speed of light.

Incidentally, the reason why the net angular momentum of a proton is a bit of a mystery is that from a theoretical perspective no one can quite pin down why all that whirling and twirling of the quarks and gluons seems to add up to... well... zero, more or less. That is, if you do nothing more than assume that two quarks have opposing 1/2 spins and the third one acts as a sort of tie breaker, you end up with a pretty good first guess at what the spin of a proton is: 1/2. Such unanticipated simplicity is a bit like looking down over a huge skating rink chock full of twirling dancers, some with arms locked together, some forming long chains moving at frightening speeds at the tips ... and yet finding that the only ones that really count when all is said and done were three smaller dancers dressed in red, yellow, and blue, all of whom were twirling at only half the speed of the very slowest of the other dancers. How, exactly, did all that other whirling and twirling by the other (mostly gluon) dancers manage to cancel itself out?

Terry Bollinger 02:01, 19 December 2006 (UTC)

[edit] Spin quantum number

Spin quantum number has some nice technical info about spin, and it does not really make sense to have these two as separate articles. --Itinerant1 21:25, 16 February 2007 (UTC)

I agree, I was looking for something in the article spin but the actual info was in Spin quantum number. Carnelain 07:41, 17 April 2007 (UTC)

The split is a mess. It would make sense to turn Spin quantum number into an specific, more general / chemistry - oriented article on electron spin and move the general features of spin physics into this article. 84.92.241.186 22:44, 27 April 2007 (UTC)

[edit] Exchange of identical particles

Mention must be made of the connection between spin and the sign change of wavefunctions upon an exchange of identical particles. References should include the Dirac chapter on permutation and the Feynman chapter on the Stern-Gerlach apparatus. —The preceding unsigned comment was added by 59.92.152.114 (talk) 17:12, 25 February 2007 (UTC).

[edit] Outside Link: origin of spin

My website contains a mathematical derivation that shows clearly how the phenomenon of spin comes from a relativistic treatment of the Schrodinger equation for the electron. It also clearly shows how spin is hard to conceptualize by virtue of the fact that spin terms in the relativistic Schrodinger equation are imaginary. The site also contains a link to a .pdf file of all my handwritten notes detailing every tedious detail of the derivation. I think readers would benefit greatly from exposure to this. Please check it out, and if anyone thinks it is a good idea please link to it. The url is http://www.teraproofs.com/spin.html --Leiding 23:36, 18 June 2007 (UTC)

[edit] Spin IS rotation

And I thought elementary particles DO rotate. That's what "spin" means! How can anything have angular momentum if it doesn't revolve or rotate? By the way, what letter stands for spin in particle physics?--Mathexpressions 18:53, 12 July 2007 (UTC)

Elementary particles may spin, but this is not what quantum spin describes. Quantum spin terms in the relativistic Hamiltonian are imaginary terms, and this implies that we cannot easily conceptualize exactly what quantum spin is.--Leiding 16:28, 19 July 2007 (UTC)

The magnetic moment we observe as 'spin' isn't from the electron spinning. It'd have to be rotating with a surface speed faster than c to create the moment we observe. It's simply an intrinsic property of fermions. (I say 'simply'. Of course, I've no doubt there are fuller models for it at a level I don't understand.) The opening part of the article contradicts this at present, implying it's due to a rotation of the body. I'll have to dig up a few citations and correct this. 314159 00:13, 20 July 2007 (UTC)

[edit] SU(2) symmetry

Could someone please tell me what SU(2) symmetry is and its relation to spin? I've had an entire mathematical course on group theory and to them all SU(2) is is a group. I understand you can generate any eigenstate of spin with an SU(2) transformation (a rotation in spin space), but what does it mean to have SU(2) symmetry?--Loodog 15:17, 20 August 2007 (UTC)

[edit] (In)accurate prediction of the electron g-factor

In the section "Spin and magnetic moments" there is a discussion of the prediction of the electron g-factor, and it's experimentally determined value is listed as "2.0023193043768(86)." The whole number portion and fracational portion of this value are then considered. It is stated that the whole number portion, two (2), arises from "the Dirac equation." It is further stated that there is a "correction of 0.00231456893..." which arises from its own and surrounding electromagnetic fields.

I might be confused but when I do the math that is implied in that section:

2.0023193043768(86) − 2 - 0.00231456893... = 0.0000047354435(53)

2.0023193043768(86)≠ 2.00231456893...

Where did this correction of "0.00231456893..." come from? To what number is the correction made? Maybe it would be better to state:

  the predicted value of the electron g-factor,
  the experimentally determined value of the electron g-factor,
  then expand upon or reference a discussion of any correction(s)

Cbobo01 (talk) 02:21, 18 January 2008 (UTC)

[edit] This article makes no sense to non-physicists

I'm not a physicist and it doesn't make sense, which just proves my statement. The amount of confusing jargon here is ridiculous.

It even seems to contradict itself by implying that "spin" is not the same as it is in classical mechanics.

The introductory section talks more about what spin is not than what it is. After you tell me about classical rotation, you fail to finish your thought and tell me what quantum spin actually is. The introductory paragraph says that it doesn't mean "rotation" yet then you later mention how you have to rotate a spin-1/2 particle 720 degrees to return it to its original quantum state. But you don't mention what that "quantum state" really means.

The very first paragraph in the body of the article has a completely meaningless formula: $[S_i, S_j ] = i \hbar \epsilon_{ijk} S_k$ - which could maybe be more useful if you said what those variables are, but probably not.

"Spin in quantum mechanics" is also a silly section name given that, as far as I know, the entire article concerns itself with that subject.

How about you start with introducing spin for a layman and then get to the formulas that could only possibly help a physicist later on?

njaard (talk) 01:07, 15 April 2008 (UTC)

unfortunately, the quantum world is a very confusing place, it takes a litte faith to understand that 'spin' is just a name that is given to a property of quantum particles, it is not actually comparable to classical spin, i.e a ball spinning, and unfortunately by trying to compare it to classical spin to make it more understandable, it usually just ends up more confusing, just think of it as a property of the particle that happens to be called spin, rather than imagining an electron spinning like the earth does 193.60.83.241 (talk) 14:40, 14 May 2008 (UTC)

I recognize that. So why the comparison to classical spin at all? Just say "it has nothing to do with rotating bodies" and leave it at that. njaard (talk) 06:18, 15 May 2008 (UTC)

I agree, but have finals soon so do not currently have the time to amend 193.60.83.241 (talk) 20:11, 15 May 2008 (UTC)

I hate to disrespect, but non-physicists are not supposed to be able to understand quantum theory. Richard Feynman is quoted as saying that "No-one understands quantum mechanics", is he not? The reason the article is littered with jargon (seemingly incomprehensible to us novices) is that quantum theory is designed around very particular jargon, and none of it is particularly comprehensible at all, unless you take the years to study in a university wherein you could acquire mastery of this jargon.

As humans, we generally aren't good enough to write up everything in an equation as simple as "E=mc^2", our world doesn't work so "cleanly", so to speak. (and rumbles in the scientific community are starting to indicate that relativity may be due for revision)

Should anyone figure out a way to explain quantum physics to novices, their effort would be hailed as a breakthrough unparalleled in human history. To use a favored turn of phrase, that's just the way it goes. Quantum physics is confusing. The sky is blue, grass is green, and politicians lie. Feel free to eviscerate me over my opinions. CameronB (talk) 11:43, 23 May 2008 (UTC)

I agree it is very hard to understand for those not familiar with physics at a degree level, but by no means should that mean we should no try, and certainly not discourage people from trying to understant, if we go around telling people 'im not going to explain this because youll never understand it' then physics in the public eye will not improve, we must strike a compromise between how likely people are to understand a given concept, and the importance of not excluding the public from the world of physics193.60.83.241 (talk) 14:08, 24 May 2008 (UTC)