Is spin of elementary particles same as the rotation of a planet?

Question

By the word spin of elementary particle, one would imagine the particle to be rotating around its own axis, just like a planet rotates, but is it actually true? While spinning does an elementary particle actually rotate around its own axis just like a planet?

Answer 1

No, for several reasons. For instance, it is possible to stop the rotation around its axis or, more precisely, to decrease or stop the corresponding intrinsic angular momentum of a macroscopic object by exerting a torsion. This is not possible for an elementary particle: the value of the spin is constant.

Answer 2

No. The spin of a quantum particle is not the same as the rotation of a planet around its axis.

There are deep reasons, sometimes referred to with the statement that "spin is a quantum property without a classical counterpart." That statement is partly true, although angular momentum is a concept born with classical mechanics. What is true is that, like in the case of velocity, we have to adapt classical ideas to the conceptual constraints of Quantum Mechanics (QM).

To maintain the parallel with the concept of velocity, we know that in QM, we have to get rid of the classical trajectories. Therefore we have no possibility of keeping the classical meaning of instantaneous velocity as $$ \lim_{\Delta t \rightarrow 0} \frac{{\bf r}(t+\Delta t)-{\bf r}(t)}{\Delta t}. $$ We have to introduce a QM analog of the classical velocity as $\frac{{\bf \hat p}}{m}$, where ${\bf \hat p}$ is the momentum operator. This way, we can translate many classical statements about velocity to QM without relying on the classical definition. Of course, in the process of such translation, the classically-based intuition should be abandoned, and a new intuition, based on what can and what cannot be said on the basis of QM axioms and definition, has to be built.

Going back to the spin, the first classical intuition we have to get rid of is the connection between angular momentum and rotation. At the level of orbital motion, this fact should be pretty obvious: there is no trajectory, and it is impossible to assign a meaning to position and momentum simultaneously. Still, angular momentum properties are there: conservation, in the case of a central interaction, the existence of a magnetic dipole moment, if the particle is charged, and so on.

In the case of extended classical objects, like planets, the intrinsic angular momentum is connected to the object's rotation. However, the QM spin is also meaningful in the case of true point-like particles. And even if we deal with "extended" QM objects (like, for example, a proton or a neutron), we are not allowed to consider them as small rigid bodies. Missing that, the connection between angular momentum and an intrinsic rotation is lost precisely as in the case of the orbital angular momentum.

Here again, some properties of angular momenta are kept. In my experience, the best intuition about spin can be obtained by considering that the presence of non-zero spin requires a multicomponent wavefunction. A multicomponent wavefunction may have an associated probability current density from which a non-zero angular momentum (and a non-zero magnetic momentum, in the charged case) may originate. More on this topic can be found in the literature ( Ohanian, H. C. (1986). What is spin? American Journal of Physics, 54(6), 500-505; Mita, K. (2000) Virtual probability current associated with the spin American Journal of Physics, 68(3), 259-264; C.T. Sebens, How electrons spin? ).

Answer 3

The question asks about elementary particles, but we can't necessarily tell what is an elementary particle and what is composite. We used to think the proton was an elementary particle, but now we know it's a composite of quarks and gluons. It's possible that we'll find out tomorrow that the electron is a composite particle. And in fact a "dressed" electron in quantum field theory really isn't pointlike -- there is a cloud of virtual particles around it. A composite system does have contributions to its angular momentum from the motion through space of its constituent particles. We refer to this as "orbital" angular momentum, as opposed to intrinsic angular momentum.

The definitive way to tell that spin can't just be orbital angular momentum in all cases is that we have particles with spin 1/2. Orbital angular momentum only comes in integer multiples of h-bar. When you have a particle, say a proton, with spin 1/2, that spin 1/2 can actually have contributions in it from orbital angular momenta of quarks and gluons. However, it can't all be made out of those orbital angular momenta, because then it would be an integer.

We can actually probe how much of the proton's spin is orbital by measuring things like the magnetic dipole moment and comparing with theory. This is a little iffy in the case of the proton, because we don't have a good, tractable way of doing QCD calculations. IIRC there are rough estimates, though.

An example where we get a more definitive answer is a heavy nucleus. So say, for example, you have the ground state of 175Hf, and its spin is some odd multiple of 1/2. We actually do have pretty decent models that allow us to say how this arises as the vector sum of different contributions: the spin-1/2 of an odd neutron, the orbital angular momentum of the odd neutron, and the collective angular momentum of the football-shaped nucleus rotating end over end. (You would think that the collective angular momentum would be zero in the ground state, but it's not.) The description of this collective angular momentum as the rotation of a rigid body or a fluid droplet is of course a classical description. However, the classical approximation can be quite good when the number of particles is large. You might say that this is getting pretty far afield from the original question, which was about "elementary" particles, but even a dressed electron is not really a simple pointlike object.

Answer 4

Just as linear momentum is associated with the number of wavelengths you can fit into a given distance of a wave repeating in space, so angular momentum is associated with the number of 'wavelengths' you can fit into the 360 degrees of a wave repeating in angle, around a point.

If you look at diagrams of the orbitals of an electron around an atom of defined angular momentum, you find spherical harmonics, which fit sinusoidal waves around a sphere. These are waves in angular space, just as the states of defined linear momentum are complex sinusoidal waves in linear space.

We can see how the orbital angular momentum is related to that of a macroscopic body if we compare the wave patterns of set of six particles moving linearly around a circle, and a single particle at the centre with spherical harmonic waves arranged around it. The linear waves in the particles of the macroscopic rigid body extend the radial pattern of the waves surrounding the central particle.

Particles with intrinsic angular momentum localised to a point have to have such a wave around them. A particle of spin half is represented by a spinor, and has to flip sign on rotating 360 degrees, and fit a whole number of wavelengths into 720 degrees. It can't be constant over all angles, and still manage to flip sign, unless it has zero amplitude. However, it's still just an angular wave, like orbital angular momentum.

Whether having an angular wavelength implies it is 'rotating' is a tricky question, like whether having a linear wavelength means a particle is 'moving' is. With linear waves, the phase velocity (the motion of the wave peaks) is completely different to the group velocity (the motion of a wave packet), which is what we normally think of as the particle 'moving'. A particle in a pure momentum state has no well-defined position, no observable distinguishing features anywhere which could be said to 'move' around. We can sum different pure momentum states to get a confined wavepacket that can 'move', and we can do the same sort of thing by combining orbital angular momentum states, but the minimal intrinsic angular momentum is fixed, so we probably don't have the freedom to play around with it in the same way.