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This may seem like a slightly trite question, but it is one that has long intrigued me.

Since I formally learned classical (Newtonian) mechanics, it has often struck me that angular momentum (and generally rotational dynamics) can be fully derived from normal (linear) momentum and dynamics. Simply by considering circular motion of a point mass and introducing new quantities, it seems one can describe and explain angular momentum fully without any new postulates. In this sense, I am lead to believe only ordinary momentum and dynamics are fundamental to mechanics, with rotational stuff effectively being a corollary.

Then at a later point I learned quantum mechanics. Alright, so orbital angular momentum does not really disturb my picture of the origin/fundamentality, but when we consider the concept of spin, this introduces a problem in this proposed (philosophical) understanding. Spin is apparently intrinsic angular momentum; that is, it applies to a point particle. Something can possess angular momentum that is not actually moving/rotating - a concept that does not exist in classical mechanics! Does this imply that angular momentum is in fact a fundamental quantity, intrinsic to the universe in some sense?

It somewhat bothers me that that fundamental particles such as electrons and quarks can possess their own angular momentum (spin), when otherwise angular momentum/rotational dynamics would fall out quite naturally from normal (linear) mechanics. There are of course some fringe theories that propose that even these so-called fundamental particles are composite, but at the moment physicists widely accept the concept of intrinsic angular momentum. In any case, can this dilemma be resolved, or do we simply have to extend our framework of fundamental quantities?

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No, the concept of “intrinsic angular momentum” doesn’t imply the system must be a point particle. The system only has to admit so-called small symmetry group; see… for details. Such composite particles as baryons, mesons, atomic nuclei, and helium atoms (in the 1s² state) have certain value of spin as well. – Incnis Mrsi Aug 20 '14 at 11:14
@IncnisMrsi: I know composite particles also have spin, but the point is, one could envisage that arising from angular momentum by motion, and not "intrinsically". – Noldorin Aug 20 '14 at 13:21
Not necessarily. Does the spin 1 of ortho-positronium (at 1s) arise from angular momentum by motion? An s orbital doesn’t rotate. The same for triplet state of ¹H. – Incnis Mrsi Aug 20 '14 at 13:37
Sure, but then you could just say the spin is carried over from the intrinsic spin of the fundamental particles... – Noldorin Aug 20 '14 at 13:59 – Qmechanic Feb 16 '15 at 20:09

10 Answers 10

Note As David pointed out, it's better to distinguish between generic angular momentum and orbital angular momentum. The first concept is more general and includes spin while the second one is (as the name suggests) just about orbiting. There is also the concept of total angular momentum which is the quantity that is really conserved in systems with rotational symmetry. But in the absence of spin it coincides with orbital angular momentum. This is the situation I analyze in the first paragraph.

Angular momentum is fundamental. Why? Noether's theorem tells us that the symmetry of the system (in this case space-time) leads to the conservation of some quantity (momentum for translation, orbital angular momentum for rotation). Now, as it happens, Euclidean space is both translation and rotation invariant in compatible manner, so these concepts are related and it can appear that you can derive one from the other. But there might exist space-time that is translation but not rotation invariant and vice versa. In such a space-time you wouldn't get a relation between orbital angular momentum and momentum.

Now, to address the spin. Again, it is a result of some symmetry. But in this case the symmetry arises because of Wigner's correspondence between particles and irreducible representations of the Poincaré group which is the symmetry group of the Minkowski space-time. This correspondence tells us that massive particles are classified by their mass and spin. But spin is not orbital angular momentum! The spin corresponds to group $Spin(3) \cong SU(2)$ which is a double cover of $SO(3)$ (rotational symmetry of three-dimensional Euclidean space). So this is a completely different concept that is only superficially similar and can't really be directly compared with orbital angular momentum. One way to see this is that spin can be a half-integer, but orbital angular momentum must always be an integer.

So to summarize:

  • orbital angular momentum is a classical concept that arises in any space-time with rotational symmetry.
  • spin is a concept that comes from quantum field theory built on the Minkowski space-time. The same concept also works for classical field theory, but there we don't have a clear correspondence with particles, so I omitted this case.

Addition for the curious

As Eric has pointed out, there is more than just a superficial similarity between orbital angular momentum and spin. To illustrate the connection, it's useful to consider the question of how particle's properties transform under the change of coordinates (recall that conservation of total angular momentum arises because of the invariance to the change of coordinates that corresponds to rotation). Let us proceed in a little bit more generality and consider any transformation $\Lambda$ from the Lorentz group. Let us have a field $V^a(x^{\mu})$ that transforms in matrix representation ${S^a}_b (\Lambda)$ of the Lorentz group. Thanks to Wigner we know this corresponds to some particle; e.g. it could be scalar (like Higgs), bispinor (like electron) or vector (like Z boson). Its transformation properties under the element ${\Lambda^{\mu}}_{\nu}$ are then determined by (using Einstein summation convention)

$$ V'^a ({\Lambda^{\mu}}_{\nu} x^{\nu}) = {S^a}_b(\Lambda) V^b (x^{\mu}) $$

From this one can at least intuitively see the relation between the properties of the space-time ($\Lambda$) and the particle ($S$). To return to the original question: $\Lambda$ contains information about the orbital angular momentum and $S$ contains information about the spin. So the two are connected but not in a trivial way. In particular, I don't think it's very useful to imagine spin as the actual spinning of the particle (contrary to the terminology). But of course anyone is free to imagine whatever they feel helps them grasp the theory better.

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The second statement is not really correct. Spin is a natural concept for nonrelativistic QM. Moreover, spin variables is not a good way to classify the representations of Poincare, the correct way is to use helicity and total angular momentum. – Grisha Kirilin Nov 15 '10 at 11:48
@Grisha: spin is not natural in QM. It is inserted by hand. If you want to understand its origin, you have to study QFT (or at least Dirac equation). As for the latter part: I am talking about massive particles. There is really no need to talk about helicity there. You only need it for massless particles. – Marek Nov 15 '10 at 12:43
@Marek. There are Poisson brackets for the angular momentum in classical mechanics. If you use Heisenberg's canonical quantization, you obtain the algebra of the angular momentum operators. Only from this algebra you can easily show that 2j+1 should be an integer number, where "j" is the maximal projection of the momentum. Therefore, "j" can be either integer or half-integer. Of course, it is because Lie algebra accounts for local properties of a group which is the same for SO(3) and SU(2). It is a simple QM, where is no Poincare. – Grisha Kirilin Nov 16 '10 at 10:06
Moreover, if you use the method of induced representations to construct the representations of Poincare, then, of course, the little group (aka stabilizer subgroup) would be SU(2) for a massive state. But this little group classifies the total angular momentum of the state in the rest frame. In QFT you can not construct operator only for the spin for any state - it exists only for the total angular momentum. Consider a Dirac particle in the Coulomb field, there aren't states with a definite electron spin - just because it is not conserved. Spin is an essentially nonrelativistic concept. – Grisha Kirilin Nov 16 '10 at 10:14
@Marek. It seems you skip my arguments finished by the phrase "Therefore, j can be either integer or half-integer", where I said nothing about orbital momentum. I was talking about the operator, because you mentioned the representations of Poincare. The physical way to classify representations is to choose operators which are separately commute with the Hamiltonian. There is no such operator as "spin". You can check the explicit form of spherical bispinor (Landau&Lifshitz Vol.4 Eq.(24.13)) - they mix different projections of spin and orbital momentum - only total angular momentum is definite. – Grisha Kirilin Nov 16 '10 at 13:21

In the field of classical mechanics angular momentum is almost always derived form linear momemtum. This actually might be the problem because it is also possible to do it the other way around: linear momentum is a limiting case of angular momentum where the radius of rotation becomes infinite. In this view the split between rotational and linear vanishes - the new concept that is introduced is: infinity.

This is not a new idea of mine, it has been established since the 19th century. By using projective geometry one can integrate linear and angular kinematics and dynamics in one framework (i.e. a translation is a rotation around an infinite axis; a pure moment is a force along an infinite line of action). Keywords: Felix Klein, linear complexes.

Another issue is intrinsic angular momentum. I could say: study the fundamentals, the principles, and the maths, and eventually you'll get a holistic picture, but that's not what I believe. I think we are in need of some kind of geometrical electron model that permits us to depict intrinsical angular momentum.

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Interesting thoughts there, I agree we're in need of a more geometrical model. May have a look at that framework you mention. – Noldorin Nov 15 '10 at 12:12
Do you have some references for seeing linear momentum as a limiting case of angular momentum via projective geometry? – student May 28 '12 at 17:48
I googeld and saw that a book of Portmann/Wallner is online: Part 3.4. – Gerard Sep 6 '12 at 20:52
Basically a translation can be seen as a rotation around a line at infinity at 0 degrees. Likewise a pure moment (i.e. zero net force) can be seen as force of 0 along a line at infinity. – Gerard Sep 6 '12 at 20:55

whether you call a similar concept "fundamental" is a matter of taste - and the proposition is just a meaningless emotional slogan. The angular momentum is surely an important quantity that is, in a very well-defined sense, as important as the normal momentum. Incidentally, both of them are conserved if the physical laws are symmetric with respect to translations and rotations, respectively.

So the real question is why the spin in quantum mechanics can't be reduced to the orbital motion - i.e. to the "linear motion" and ordinary "momentum". It's because the objects in quantum mechanics are described not just by their shape in space but by wave functions, and wave functions may be said to transform nontrivially (into something else) under rotations.

In particular, if the wave function (or a field) is a vector or a tensor or, most typically, a spinor, then it means that in a different coordinate system, the values of the components of the wave function will be different. This is possible even in the case when the wave function (or field) is fully localized at one point, i.e. nothing is rotating "orbitally".

The angular momentum is defined by the change of the phase of the wave function under rotations, which may come from the dependence of the wave function on space, but also from the transformations of the components of the wave function among each other, which is possible even if everything is localized at a point. So even point-like objects may carry an angular momentum in quantum mechanics, the spin.

Note that the spin is a multiple of $\hbar/2$ and $\hbar$ is sent to zero in the classical limit, so in the classical limit, the spin as the internal angular momentum becomes zero and disappears, anyway.

Another new feature of the spin is that unlike the angular momentum, it may be half-integer, not just a multiple of $\hbar$: also $\hbar/2$ is possible. That's because the wave functions (and fields) may transform as spinors that change the sign if they're rotated by 360 degrees. Only a rotation by 720 degrees is topologically indistinguishable from "no rotation", so wave functions are obliged to return to their original values under a 720 degree rotation. But fermions change their signs under rotations by 360 degrees which corresponds to their half-integral spin.

If the word "fundamental" means that it can't be reduced to some other things such as a classical intuition about motion and rotation, then be sure that the spin is damn fundamental, much like the rest of quantum mechanics.

Best wishes Lubos

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Thanks for your reply. I think your reasoning is right. Physicists like to use the term "fundamental" a lot, but it probably isn't very well-defined. – Noldorin Jan 14 '11 at 17:48
Dear Noldorin, I actually use it often, too - but just not for random quantities such as the angular momentum. I use it for important principles and universal laws - anything that is not just an approximation; anything that is unique and doesn't have lots of "sibling concepts"; anything that matters in the whole Universe. In particular, the fundamental scale is probably the Planck scale - more generally, it's the place where the most accurate, not approximate laws of the Universe show their physical consequences directly. – Luboš Motl Jan 17 '11 at 10:49

In classical mechanics, the fundamental entities change according to the framework you opt for. If you do classical Newtonian mechanics, I'd say the fundamental entities are postions and velocities. All the others can be derived from them and the dynamics of particles are described in terms of functions of these (forces are functions of time, positions and velocities).

But if you go to Hamiltonian mechanics, then positions and momenta become fundamental. And the Hamiltonian can be expressed as a function of these and possibly time.

Clearly, in classical mechanics, angular momentum is always a derived quantity, because it is always an orbital angular momentum, never an intrinsic angular momentum. Even when you have an object rotating on an own axis, this can be understood as the particles constituting the object executing an orbital motion. Of course, you can write Hamiltonians which depend on the angular momentum of the top, but these are higher level descriptions, the angular momentum of the top could still be decomposed in principle into the orbital angular momenta of its constituents. This would not be a very practical approach to problem-solving of course.

Therefore, as you say, a fundamental intrinsic angular momentum is a novelty in quantum mechanics. The way it enters the equations is usually through the multiple-valuedness of the wave function. Say a spin 1/2 particle has to be described by two independent component wave functions (there could be more components, but these would not be independent). I don't know of any way around this. This is a basic fact of how nature works and it is related to the representations of the symmetry group of space-time.

Since the symmetry group of space-time is basically the same in quantum and in classical physics, I don't see however why it should not be possible to describe particles with intrinsic momenta in classical mechanics. I think it is certainly possible in principle. The question is, is it useful? Since all our elementary particles have to be described at the quantum level, what use is a classical theory of particles with intrinsic momenta? Except in the sense of tackling problems like the top by simplification or something?

EDIT: As a matter of fact, classical field theories do have spin. Think of the Maxwell equations for instance.

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Thanks for your reply. It confirms a few of my views for sure. I wasn't aware classical field theories predict spin. Ordinary quantum mechanics is not a field theory and predicts spin, however? – Noldorin Nov 15 '10 at 12:13
@Noldorin: it doesn't predict it. You can work in QM without spin as well. Also, in QM mechanics you can have spin 1/2 bosons, which isn't really consistent with reality. That is why Dirac equation was such a huge success: it indeed did predict spin! But it wasn't until later that people understood where spin is really coming from. For that you need to consider fields. – Marek Nov 15 '10 at 15:02
@Raskolnikov: classical field theory and quantum particles are deeply related. The bridge is through quantum field theory. This is obtained by quantization of classical field theory. Once you have it quantized, you can notice that there exists something called "particle approximation" (this is about Feynman diagrams). So in the end you'll arrive at particles. So it's morally correct to say their spin comes from classical field theory. – Marek Nov 15 '10 at 15:08
Stackexchange sites are great: my usage was correct. – Marek Nov 15 '10 at 18:39
Really? I've never heard it used by anyone in Britain, least of all in public. Physicists are known for corrupting language, however! I can concede it's used in some areas though, so fair enough. :) Just a warning: the chance you are understood outside of the physics/science community is about zero. – Noldorin Nov 16 '10 at 0:09

A hint of the special role of angular momentum happens when you look for its conjugate variable. It is angular position, which is adimensional. And then you have that any product of a variable times its conjugate has units of action, which are the same units that angular momentum. So classical mechanics already tell us that something is going on. (Caveat: that you can have the same units with scalar product and with cross product, and the physical meaning is different. If you have checked the pamphlets from German car and enginemakers, you could have noticed the unit "Nm", newton times meter, and the unit "Joule", used differently.)

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In QM it is impossible and is not necessary to impose R = 0 (see my blog) to have a system at rest. On the contrary, one has to put P = 0. It does not mean point-likeness but ubiquity instead.

There is an article by R. Ohanian on spin. But I am afraid it is finally a tautology or so.

I think the angular momentum is fundamental. I think that even in Classical Mechanics a description of anything with help of solely three coordinates R(t) is too primitive. Generally everything is not point-like and rotates, roughly speaking. So the intrinsic angular momentum J is as fundamental as the linear momentum P (as well as color, charge, and flavor ;-).

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Vlad, you are in a catch-22 situation. In most cases, you don't want to answer, but only to comment an answer. So, it is not an answer and you get negative scores. But you will not be able to comment until you accummulate 50 points of reputation. Break the loop, look for some questions you can answer in a useful way, and/or do questions of general interest. – arivero Jan 18 '11 at 0:12
@Vladimir: I'm not sure I agree with your answer, but I'm not sure why you got the down-votes either. (People should indeed leave reasons!) – Noldorin Jan 18 '11 at 23:16
@Noldorin: many think of elementary particles in QM as of stable point-like objects whereas there is no stable solution localized at a point all the time. Wide wave packets may be more or less "stable" but they are not point-like objects. The latter case is much more realistic due to necessity of stability while preparing and measuring the spin projections. – Vladimir Kalitvianski Jan 18 '11 at 23:50
Interesting. I'm not very familiar with QFT, but you saying that all particles (field wave packets) are unstable to some degree? Are there any solitons in QFT? – Noldorin Jan 19 '11 at 0:19
@Noldorin: Yes, they (the wave packets) are unstable and the extent of their instability is determined by the preparation device (source, diaphragms, etc.). In addition, if we speak of charge scattering, in the final state you always have many (soft) photons. You cannot scatter without radiation (elastically). It means the initial system is always "broken apart" in some way (inelastic scattering). It is a strict QED result. The system is "large and soft", easy to deformate inelastically. It is incompatible with a soliton-like construction. – Vladimir Kalitvianski Jan 19 '11 at 10:21

As for spin and extended particles, I'd say the contrary: it is not contrary to intuition that point particles have some intrinsic angular momentum, because a point looks as if it had some rotation invariance built-in. The surprising thing is that extended objects have this angular momentum, without a point for the rotational symmetry to pivot in.

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Quantum physics requires a symmetry of the space-time, not of an “extended object” as you see it with physical intuition. You will give different answers on the “is this thing rotationally symmetric?” question depending on exact formulation. Can such molecules as water (H₂O) or methane (CH₄) be rotationally symmetric? Geometric intuition says: no, their molecular geometry denies it. But corresponding composite wave functions (of all nuclei and electrons, but with translational symmetry removed) for the ground state are rotationally symmetric. – Incnis Mrsi Aug 20 '14 at 12:05

Existence of a spin of a particle is of course an indication that the particle is in fact composed of space-separated parts. This does not mean though that the particle is composed of other particles.

For example, at least a part of the electron's spin is currently known to be in fact the orbital momentum of the quantum vacuum fluctuations which are involved by electron's core into rotation. This part is known as anomalous angular momentum of electron.

Another example is photon where the spin can be explained as an order in which the energy contained in electric and magnetic fields rotates around the axis laid along the direction of the photon's propagation.

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-1: this answer is wrong. There is no anomalous angular momentum of the electron. There is an anomlous magnetic moment, but this is not angular momentum, it is current. – Ron Maimon Sep 19 '11 at 0:36

There exists a very simple and concise semi-classical explanation of the electron's spin angular momentum, without the notion of rotation of any material object: Qualitatively speaking, the electron’s spin angular momentum is the electromagnetic field's angular momentum resulting from the combined electromagnetic field surrounding an electron just in such a way that a nonzero Poynting vector circulating around the electron’s dipole axis is created, which also means that a permanent electromagnetic energy flux circulates around the electron’s dipole axis. Relativistic electrodynamics demonstrates that any kind of energy flux is associated with momentum flux (parallel to the Poynting vector) which in itself can be associated with angular momentum relative to a given point or axis of reference. Hence energy circulation around the electron’s dipole axis is equivalent to momentum circulation. If integrated over the whole space around an electron the result is a substantial fraction if not all of an electron’s spin angular momentum being distributed in that space. (See e.g. Feynman Vol. II)

A quantitative assessment of the electromagnetic field angular momentum of an electron is given in:
S.M. Blinder: Singularity-free electrodynamics for point charges and dipoles: a classical model for electron self-energy and spin, Eur. J. Phys. 24 (2003)271-275 (arxiv preprint).

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There is more to it than Spin being intrinsic angular momentum. An electron has an "internal degree of freedom" - to be left handed or right handed, and it can leave point A with RH spin and arrive at B with LH spin. Thus Pauli needs two complex components in his equation. (unlike a photon which arrives with the same spin although it too has LH and RH, so there is no internal degree of freedom ). This is distinct from the spin-vector which defines a direction in space. The two-valuedness comes from rotation being about a bivector which can point up or down along the axis of rotation. One can do spatial rotations either way - and electrons seem to make the distinction - as if there are two kinds, but everything else is the same mass and charge, so we say it is the same particle, with opposite spins. So it seems that there is no necessary connection to either relativity (except for fixing up the Thomas factor in the Pauli eq ) or QFT. Hamilton had the algebra to make the classical distinction between left and right - it is built into quaternion algebra, but he did not see it as a mechanical property of particles - but heck, he did not see the Maxwell equation either.

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With this “two-valuedness comes from rotation being about a bivector which can point up or down along the axis of rotation” you put the cart in front of the horse. How many “components” is needed depends on representations, see… for details. – Incnis Mrsi Aug 20 '14 at 20:08

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