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The orbital angular momentum of a particle can be related to the revolution of that particle about some external axis. But in quantum mechanics, the spin angular momentum of a particle can't really be thought of as the rotation of the particle about its own axis. This is for a number of reasons. For one thing, you need to rotate the spin state of an electron 720 degrees, not 360, to get back the original spin state, which isn't how rotations work. For another thing, as I discuss here Goudsmit and Uhlenbeck showed that if the spin of an electron really was due to rotation about its own axis, then the a point on the equator would be moving with a speed greater than the speed of light. And in any case if the electron wasn't a point particle that would cause all sorts of problems. Finally, there isn't a definite "axis of rotation" for spin, because the three components of spin angular momentum don't commute with one another.

But my question is, can spin be related to a rate of change of anything at all with respect to time? Spin may not be related to rotation in $\mathbb{R}^3$, but can we relate it to a rotation or other kind of motion in some other space, possibly a non-Euclidean space? It may take 720 degrees to fully "turn" an electron, but is there actually a period of time in which it "turns" or does something else by 720 degrees?

To put it another way, if a particle has a fixed spin state, does it make any sense to say that the particle is "doing" anything, or does it simply "have" a property?

EDIT: Ehrenfest's theorem relates the expectation value of the linear momentum operator to the rate of change of the expectation value of the position operator with respect to time. Can the expectation value of the spin angular momentum operator be related to the rate of change of the expectation value of some operator?

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    $\begingroup$ Einstein–de Haas effect $\endgroup$ – Count Iblis Oct 31 '16 at 2:41
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    $\begingroup$ I always find it very odd when folks assert that spin has "no" connection with rotation, given that if you keep adding more half-units of it it eventually morphs into ordinary classical angular momentum and rotation. If someone has ever explained how that curious transition works, I surely have never seen it. Pauli, in his inimitable aggressive style, got frustrated with that little conundrum and so of course declared it unsolvable, and that everyone should therefore shut up about it. Too bad, that. $\endgroup$ – Terry Bollinger Oct 31 '16 at 5:33
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    $\begingroup$ @TerryBollinger What do you mean it "morphs into ordinary classical angular momentum"? How does one "add" half-units of spin? Spin is an intrinsic property of a quantum object, you can't "add" spin to something. $\endgroup$ – ACuriousMind Nov 5 '16 at 14:17
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    $\begingroup$ @TerryBollinger Yes, you can convert between spin and classical angular momentum by getting the photons absorbed. But a spin-2000 object still doesn't rotate classically (at least, I see no reason why it should). If you can demonstrate that quantum mechanics predicts it would, then that would constitute a good answer to this question. $\endgroup$ – ACuriousMind Nov 5 '16 at 15:29
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    $\begingroup$ @ACuriousMind, this is intriguing, I did not realize how different the mental mappings of this topic could be. Because particle spins (electrons in particular) due to Pauli exclusion are powerfully energetically inclined to pair up in antiparallel combinations, there's a break point at about spin 1 where most particle spins stop adding up in composite systems. But try this: What is the full set of possible angular momentum states of a large molecule, say maybe a C60 buckyball, in a vacuum? It will have a couple of quantized states at or near 0, subject to quantum rules and statistics. $\endgroup$ – Terry Bollinger Nov 5 '16 at 15:51
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I'll mainly address your last question:

Ehrenfest's theorem relates the expectation value of the linear momentum operator to the rate of change of the expectation value of the position operator with respect to time. Can the expectation value of the spin angular momentum operator be related to the rate of change of the expectation value of some operator?

Well, let's see what do we get if we apply the Ehrenfest theorem to a spin 1/2 particle in a magnetic field. The interaction energy between a magnetic dipole and a magnetic field B is

$$E = \mathbf µ· \mathbf B$$

where $\mathbf µ$, the magnetic moment, is a vector operator, and is given by

$$\mathbf µ = \gamma \mathbf S$$

Here $\gamma$ is the gyromagnetic ratio.

All this is classical physics, but I'd say we can extend the equations to quantum mechanics in a straightforward way. If we take the spin to be a matrix, then its Hamiltonian is (source of the derivation)

$$H=-\gamma \mathbf S·\mathbf B$$

As an example, if choose our coordinate system so that $\mathbf B = B\mathbf k$, then

$$H=-\gamma BS_z=-\gamma B\frac{ℏ}{2}\sigma_z$$

where $$\sigma_z=\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)$$

Applying the theorem then gives the rate of change of $\left\langle S_x \right\rangle$,

$$\displaystyle\dfrac{d\left\langle S_x \right\rangle}{dt}= \frac{1}{i\hbar}\langle\left[ S_x,H \right]\rangle = \omega \left\langle S_y \right\rangle$$

where $\omega=-\gamma B$ is the Larmor frequency. Larmor precession is the precession of the magnetic moment of any object with a magnetic moment about an external magnetic field. According to Wikipedia, the phenomenon is similar to the precession of a tilted classical gyroscope in an external gravitational field (the torque produced by the magnetic moment being here analogous to the external gravitational torque in the case of the gyroscope).

For the rates of change of $\left\langle S_y \right\rangle$ and $\left\langle S_z \right\rangle$, we obtain:

$$\displaystyle\dfrac{d\left\langle S_y \right\rangle}{dt}= - \omega \left\langle S_z \right\rangle, \displaystyle\dfrac{d\left\langle S_z \right\rangle}{dt}= 0 $$

Using the properties of Pauli matrices, we can write the preceding equations in a more compact manner:

$$\displaystyle\dfrac{d\left\langle \mathbf S \right\rangle}{dt}= \gamma \left\langle \mathbf S \right\rangle \times \mathbf B$$

According to Peter H. Holland's The Quantum Theory of Motion, a classical analog for this precessional equation of motion of the spin vector in a magnetic field is possible (in fact, the first equation he derives is more complicated, as it includes a "quantum torque"). In general, he states (section 9.3.3., Is there a classical analog of spin?):

We conclude that the classical analogue of the systems governed by the Pauli equation is an ensemble of charged dipoles and one passes continuously between the two regimes by varying the effectiveness of the quantum potential and torque. The "spinning" object does not disappear in the limit, it simply evolves differently.

My way to see it is that the time dependence of spin expectation values follows the classical equation of motion for angular momentum vector. This conclusion is also to be found in this paper called Significance of Ehrenfest theorem in quantum–classical relationship, which in addition claims:

In the measurement of magnetic moment of neutron and other nuclei by the nuclear induction method, Bloch* has used essentially these classical equations dispensing with the Schrödinger equation from the simple argument based on ET.

(*reference)

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  • $\begingroup$ I'm not talking about the time-dependence of the expectation value of the spin angular momentum operas, and I'm not talking about a specific system like a particle in a magnetic field. I'm asking, for any arbitrary system, even when the spin state of the particle is constant, can we relate the expectation value of the spin angular momentum operator to the time derivative of the expectation value of some operator? $\endgroup$ – Keshav Srinivasan Nov 6 '16 at 19:47
  • $\begingroup$ @KeshavSrinivasan I don't think so. As you said, those relations between operators were given by the application of Ehrenfest's theorem. My point was to demonstrate that, for this particular example, Ehrenfest's theorem gives a result analogous to classical theory. The rates of change of each spin component are related to the values of the other components, that's all you can get. The same is appliable to the angular momentum operator. Maybe someone will come up with a more general answer, but I'd say there isn't the kind of relationship you're thinking of. $\endgroup$ – David Herrero Martí Nov 6 '16 at 19:56
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When you study the reprepesentation theory of the poincare group, one of the things you learn is the so called little groups or isotropy groups. These classify the representations of the poincare group by a method called induced representations. When you do all that you eventually find that each massive particle has a quantum number that makes it either bosonic or fermionic. Now since the poincare group simply encodes the transformations of space-time we, I think, are free to conclude that spin is a property that a particle has. Since after all we could have assumed that there was only one particle in the universe removing the possibility that spin was connected to any thing happening to it or it doing something.

If you want something more physical then imagine putting a magnetic field near an electron in some direction. The spin couples to the magnetic field. Then on the bloch sphere this causes a precession about the axis that couples to the magnetic field. So concretely take the hamiltonian to be $H =\sigma_z B_z $ where $B_z$ is some constant related to strength of magnetic field then there will precession about the spatial z direction. This can can be seen by calculating $ \langle\sigma_x(t)\rangle \text{ and } \langle\sigma_y(t)\rangle $.

Also remember change in angular momentum is torque, but torque requires a notion of force which can't exist in quantum mechanics. Position and momentum operators' expectation values are piggy backing off their classical limit.Quantum spin has no classical limit that is the point of the stern-gerlach experiment.

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    $\begingroup$ "Since after all we could have assumed that there was only one particle in the universe removing the possibility that spin was connected to any thing happening to it or it doing something." Consider the case of momentum. Even if there was only one particle in the Universe, momentum would still involve "doing something"; the expectation value of the momentum operator is related to the rate of change of the expectation value of the position operator. So why couldn't the expectation of the spin angular momentum operator be related to the rate of change of the expectation of some operator? $\endgroup$ – Keshav Srinivasan Nov 1 '16 at 18:31
  • $\begingroup$ The momentum operator can be defined as the generator of translations. All this requires is the existence of space. We could then do an infinite number of experiments and find out that momentum can be related to change in expectation value of position operator but that requires introduction of particles and observers to do this. Plus you are using momentum and position operators which have analogues in classical mechanics, there is no reason to expect spin which is purely quantum mechanics to have a classical mechanical like relation with some other operator. $\endgroup$ – Amara Nov 1 '16 at 21:20
  • $\begingroup$ Bringing in the Poincare group is right, but it's not true that quantum spin has no classical analogue. $\endgroup$ – Andrew Steane Jun 10 at 19:56
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Without any interaction, I would say the spin property is meaningless. It manifests itself as a flip operator (having discrete negative/positive eigenvalues) through interactions, such as the one with a Stern-Gerlach apparatus. But time dependence of spin becomes relevant if an electron continuously interacts with external electromagnetic field. Look up for the Larmor precession. When you put electron in a uniform magnetic field spin starts to precesses around the direction, determined by the magnetic field. For the more general case where we have uniform electric and magnetic fields, there is Bargmann-Michel-Telegdi (BMT) equation for the spin precession:

$$ \frac{ds^{\mu}}{d\tau}=\frac{e}{m}\left[\frac{g}{2} F^{\mu\nu}s_{\nu} +\left(\frac{g}{2}+1\right)u^{\mu}\left(S_{\lambda}F^{\lambda\nu} u_{\nu}\right)\right] $$ where $\tau$ is the proper time. This equation is supposed to be solved in accordance with classical Lorentz force equation: $$ \frac{du^{\mu}}{d\tau}=\frac{e}{m} F^{\mu\nu}u_{\nu} $$ BMT equation is a classical equation. In the operational sense, $s^{\mu}$ corresponds to the electron's spin polarization and BMT equation gives the rate of change at which transverse polarization transforms into longitudinal one and vice versa. You may look into this paper for the details.

As far as I know, we do not have quantum mechanical analog of BMT equation. But in the Heisenberg equations of motion for the operators, the spin shows up:

$$\begin{align} \frac{d x_{\mu}}{d\tau}=\Pi_{\mu} ,\qquad \Pi_{\mu}=p_{\mu}-\frac{e}{c}A_{\mu}(x)\\ \frac{d\Pi_{\mu}}{d\tau}=\frac{e}{c}F_{\mu\nu} -\frac{i e}{2c}\partial_{\nu} F_{\mu\nu} + \frac{e}{4c}\sigma_{\lambda\nu}\,\partial_{\nu}F_{\lambda\nu},\qquad \sigma_{\lambda\nu}=\frac{i}{2}\left[\gamma_{\lambda},\, \gamma_{\nu} \right] \end{align} $$ These equations reduce to Lorentz force equation in the classical limit. So electron's spin have no effect on the classical trajectory. But in intense electromagnetic fields that show very rapid oscillations, spin affects the expectation values of the operators.

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  • $\begingroup$ I'm not really talking about the time-dependence of the spin-state of a particle. Even if we have a free particle so that particle is stuck in a given momentum state, it is still the case that the expectation value of the momentum operator can be related to the time derivative of the expectation value of the position operator. Just like that, why can't the expectation value of the spin angular momentum operator be related to the time derivative of the expectation value of some operator? $\endgroup$ – Keshav Srinivasan Nov 6 '16 at 18:07
  • $\begingroup$ This is a deep question. The main reason in my opinion is that spin has no classical analog. Position and momentum are defined as conjugate phase space variables in the classical Hamiltonian formulation. Classical flow equations can be transformed into operator equations via quantization. But with no predefined notion of spin in the classical phase space, there is no conjugate variable to spin to define the Poisson brackets that may generate a flow. $\endgroup$ – user91411 Nov 6 '16 at 18:21
  • $\begingroup$ @user91411 "The main reason in my opinion is that spin has no classical analog." Well, the orbital angular momentum operator does have a classical analog, but its change of rate can't be related to the time derivate of another operator, either. $\endgroup$ – David Herrero Martí Nov 6 '16 at 20:10
  • $\begingroup$ @DavidHerreroMartí The expectation of the orbital angular momentum operator can be related to a time derivative; see my question and answer here: physics.stackexchange.com/a/292323/27396 $\endgroup$ – Keshav Srinivasan Dec 23 '16 at 15:02
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The most helpful way to begin here is with some group theory which applies equally to both classical and quantum physics. Whether in classical or in quantum physics, we are interested in properties that might be preserved under translations, rotations and boosts (changes between inertial frames), or behave in simple ways under such basic transformations. Thus we end up defining things like mass and energy and momentum, and also ordinary (or 'orbital') angular momentum.

Now it happens that spin angular momentum was discovered in physics at about the same time as quantum theory, which resulted in people assuming it was something to do with quantum mechanics. It is, but no more so than energy and momentum and other more familiar properties. You can have a classical physics of spin.

The group of Lorentz transformations and translations is called the Poincare group. When you study what sorts of things can undergo such transformations, an angular-momentum-like property that can be associated with point particles naturally arises. Another way to see this is to learn about spinors, and to learn that they behave in sensible ways under Lorentz transformations, and this is all classical physics (not quantum).

What this means is that the property we call spin can be associated with point particles in the same way that mass and charge can be. The difference, of course, is that it is a vector property (an axial vector to be precise). It is not a property of their motion through space, but its presence can be manifested when it influences that motion. Certain types of interaction that result in a torque on the particle can change the direction of its spin, and it turns out that when this happens the orbital angular momentum of an isolated system is not conserved, but the sum of spin and orbital angular momentum is conserved.

The classical limit for spin can in principle be observed. For example, one might take a set of many atoms, and form a total spin state which consists of a sum over spin states such that all three components of the total spin are reasonably well-defined. This is a bit like taking states of the harmonic oscillator and combining them into the combinations (superpositions) called Glauber coherent states, where the position and momentum have Gaussian distributions. In the spin case, each of the spin components has a large mean and a Gaussian distribution about that mean with a smallish standard deviation. What distinguishes classical from quantum here is that the classical system can be observed without significantly disturbing it.

Such a classical system would have plenty of spin angular momentum, but it would not have any motion through space associated with that spin, including rotation. There need not be any motion at all. Zilch. Zippo. In an interaction with another thing, it would be possible for the other thing to acquire orbital angular momentum while the spin state of our classical spin changed, for example by getting smaller or pointing in another direction.

I noted that some comments on the question seem to suggest that some people think spin cannot add up in this way. This may be because they have in mind the solid state, where there are substantial spin-orbit interactions. But in a gas what I have written would apply, and there are some experiments in atomic physics using laser-cooled gases where this sort of thing is explored.

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This is not an answer per se, because why would it be? There isnt any clear cut idea of what a spin of a particle should be thought of akin to. This question has been beaten to death and its been recycled and rephrased in so many different ways, frankly its getting tiresome. Or may be i am too negative; either way.

Excerpt from Uhlenbeck and Goudsmith's description of the account of where they propose the spin of an electron to have $\pm \frac{1}{2}$.

[...] Lorentz received us with his well known great kindness, and he was very much interested, although, I feel, somewhat skeptical too. He promised to ,think it over. And in fact, already next week he gave us a manuscript, written in his beautiful handwriting, containing long calculations on the electromagnetic properties of rotating electrons. We could not fully understand it, but it was quite clear that the picture of the rotating electron, if taken seriously, would give rise to serious difficulties. For one thing, the magnetic energy would be so large that by the equivalence of mass and energy the electron would have a larger mass than the proton, or, if one sticks to the known mass, the electron would be bigger than the whole atom! In any case, it seemed to be nonsense.

You may have read this most probably but this is just to show people that any attempt to 'describe' the spin as 'something' would lead to it being reduced to ashes by people engaging in reductio ad absurdum. We just raise hands and say for better or for worse that spin is just something.....like you anticipate as being like mass or charge that a particle 'just has'.

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    $\begingroup$ Goudsmit and Uhlenbeck only showed that the specific assumption that spin is due to a electron rotating about its own axis leads to a contradiction, they most definitely did not show that all possible descriptions of spin as a process a particle does lead to a contradiction. $\endgroup$ – Keshav Srinivasan Oct 31 '16 at 7:44
  • $\begingroup$ @KeshavSrinivasan.....i am sorry if i gave you that impression that i was saying that all possible descriptions of spin as a process a particle does lead to a contradiction......this was just one of the many attempts to try and think of electron spin as something even remotely conceivable and being shot down immediately by the authorities aka LORENTZ and such. I did not want to mention the speed of equator being greater than $c$ ......it is an ad nauseam ....i am sure there might be many more such examples $\endgroup$ – Prasad Mani Oct 31 '16 at 9:02
  • $\begingroup$ @PrasadMani I gave the speed of the equator reason and other reasons in the beginning of my question. But all of that is specifically about reasons why spin cannot be due to rotation. None of these reasons address anything about whether spin is a process that's completely different from rotation. $\endgroup$ – Keshav Srinivasan Oct 31 '16 at 22:46
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(...) my question is, can spin be related to a rate of change of anything at all with respect to time? Spin may not be related to rotation in $\mathbb{R}_3$, but can we relate it to a rotation or other kind of motion in some other space, possibly a non-Euclidean space?

It really depends on what you mean by "change of anything with respect to time", but the general answer is no, not in the way you are probably thinking of it.

No, because it is not correct to think of the spin angular momentum (SAM) of a particle as due to any kind of "rotation" or other "movement" in any (Hilbert) space. An electron with spin $+1/2$ is not rotating, nor changing, for the sole fact of having a SAM. In fact, if no external influences act on the electron, such spin state (or any other spin state for that matter) will be an eigenstate of the system, meaning that by definition it is stationary: nothing has to "move" for that electron to possess such SAM.

If that were the case, it would mean that spin is not really a "fundamental" quantum number, but only a "feature" of some more fundamental property. It is important to notice that a similar argument also holds for orbital angular momentum: a particle (or composite particle, or any other type of quantum state) with a definite orbital angular momentum is not necessarily something that can be thought of as "rotating around": a single particle with an appropriately structured spatial wavefunction can have a definite orbital angular momentum, yet not be rotating in any sense of the word (on the other hand, it is true that the orbital angular momentum is generally a feature of the spatial profile of the wavefunction, and is therefore in this sense not a fundamental property of the system).

We like to picture the electrons in atoms as "rotating" around the nucleus, but that picture is really as wrong as the picture of a rotating spin: the electrons (or better said, the nucleus+electron systems) are in a stationary state, meaning that nothing is changing with respect to time. The difference in such cases is however that the orbital angular momentum quantum number is completely written in the structure of the spatial wavefunction, and is therefore by no means an intrinsic property of the particles.

Yes, in the sense that of course the spin state can be changing with respect to time, or other properties of the particle be changing with respect to time depending on their spin state. Put an electron in an appropriate magnetic field and you will see it move in one direction or the other depending on its spin state. Or the spin direction itself can be rotating with a fixed angular frequency, like what happens to nuclei in an NMR machine.

But SAM seems so much like a regular angular momentum!

Does it? Do you think of a single photon with a definite value of the polarization as rotating? Probably not, but you should! The "polarization" of light is nothing but a different name for the spin of photons. Yet in the case of photons most people seem to not have so many difficulties in seeing the polarization as something unrelated to a kind of rotation.

Can the expectation value of the SAM operator be related to the rate of change of the expectation value of some "classically interpretable" operator?

No! The only way for this to make sense is the existence of a "classical equivalent" of SAM. There is no such thing. Why? Plainly said, because there isn't: classical objects are really just only characterized by position and momentum.

But what about Ehrenfest theorem?

What about it? Sure you can apply it, in its general form stating the relation between the variation of the expectation value of an operator and the expectation value of the commutator of the operator with the Hamiltonian: $$ \frac{d}{dt}\langle A \rangle = \frac{1}{i\hbar}\langle [ A, H] \rangle, $$ holding for any operator $A$ measuring an intrinsic (that is, time-independent) property of a particle. If you take $A=S_z$, or some other component of the spin, you will get an expression for the rate of variation of the average value of the SAM in the particular system under consideration. If it is an electron in an electromagnetic field you will get an expression for the time variation of $\langle S_z \rangle$ with respect to the particular electromagnetic field applied, which will tell how the spin of the particle evolves, but not much more. If you instead want to find some operator $A$ such that $$ \frac{d}{dt}\langle A \rangle \simeq \langle S_z \rangle, $$ you can probably find it: you need to find a system described by an Hamiltonian $H$ such that there is an operator $B$ such that $[A,H] \simeq S_z$. I see no reasons for this to not be possible, but it will hardly give you any insight into the nature of SAM, as it will be highly dependent on the particular system you are considering. In fact, you shouldn't think of Ehrenfest theorem as something giving general connection between classical and quantum mechanics, as it is nothing but an alternative and equivalent way to state Schrödinger's equation.

But why is spin angular momentum so strictly related to the orbital angular momentum then?

Because the quantum object "corresponding" to the classical angular momentum is the total angular momentum operator, $\boldsymbol J = \boldsymbol L + \boldsymbol S$. This is a consequence of the fact that the Lagrangian describing the interactions between the various particles only preserves the total angular momentum, not singularly the spin or orbital components of it. This is not so unnatural as it may at first appear. Think for example at the 3D wavefunction of a photon or electron. While we are used to think of one such particle as having some value of SAM, in general you will have a probability amplitude of the particle having a value of SAM for every point of the wavefunction (or equivalently said, entanglement between SAM and position). The SAM operator $\boldsymbol S$ acts on such wavefunction rotating the SAM degrees of freedom at every point in space, leaving unchanged the spatial amplitude distribution. The orbital angular momentum operator $\boldsymbol L$ on the other hand rotates the spatial distribution of amplitudes, carrying around untouched the spin degree of freedom associated to each point. Now, why should this whole complicated mess be invariant when you apply only one of these two operations? Indeed in won't in the general case: you will need to rotate the spatial distribution and the internal degrees of freedom accordingly to obtain the same structure.

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  • $\begingroup$ to the downvoter: care to explain where you disagree? $\endgroup$ – glS Nov 6 '16 at 14:39
  • $\begingroup$ Well, here's the thing: a momentum eigenstates of a free particle is an eigenstate of the Hamiltonian. And yet momentum can still be related to a rate of change; the expectation value of the momentum operator can be related to the rate of change of the expectation value of the position operator with respect to time. And I'm sure you could relate the expectation value of the orbital angular momentum operator to the rate of change of the expectation value of some operator. So why can't you do likewise for the expectation value of the spin angular momentum operator? $\endgroup$ – Keshav Srinivasan Nov 6 '16 at 15:03
  • $\begingroup$ @KeshavSrinivasan I added some points, see the edit $\endgroup$ – glS Nov 6 '16 at 16:49
  • $\begingroup$ I'm not looking for a particular situation in which the expectation value of the spin angular momentum is related to the rate of change of the expectation value of some operator, I'm looking for a relationship that holds in general, just as the relationship between the expectation value of the momentum operator can be related to the rate of the change of the expectation value of the position operator. And the operator doesn't need to be "classically interpretable", I'm fine if we just find any operator whose rate of change of expectation value can be related to spin. $\endgroup$ – Keshav Srinivasan Nov 6 '16 at 16:57
  • $\begingroup$ @KeshavSrinivasan I know, and my point was in fact just that: there isn't one. There cannot be because the fundamental interactions between particles depend on the total angular momentum, not only its spin component. This means that when you take the commutator of some operator with the Hamiltonian you will at best find yourself with some function of $\boldsymbol J$, not $\boldsymbol S$. $\endgroup$ – glS Nov 6 '16 at 17:00
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Does spin have anything to do with a rate of change?

Yes. Spin is real.

The orbital angular momentum of a particle can be related to the revolution of that particle about some external axis.

Yes, but don't think of the particle as something like a planet. See atomic orbitals on Wikipedia: "The electrons do not orbit the nucleus in the manner of a planet orbiting the sun, but instead exist as standing waves". A better analogy for orbital angular momentum is playing hula hoop.

But in quantum mechanics, the spin angular momentum of a particle can't really be thought of as the rotation of the particle about its own axis. This is for a number of reasons. For one thing, you need to rotate the spin state of an electron 720 degrees, not 360, to get back the original spin state, which isn't how rotations work.

Intrinsic spin is a real rotation. The Einstein-de Haas effect demonstrates that "spin angular momentum is indeed of the same nature as the angular momentum of rotating bodies as conceived in classical mechanics". The 720 degrees is there because the electron isn't just a "spinor", it's a "bispinor". I think the best way to grasp this is by thinking of a wave going round like this:

enter image description here Image courtesy of Adrian Rossiter's torus animations

For another thing, as I discuss here Goudsmit and Uhlenbeck showed that if the spin of an electron really was due to rotation about its own axis, then the a point on the equator would be moving with a speed greater than the speed of light.

It's a non-sequitur. You can see it in an old version of the Wikipedia Stern-Gerlach article: "If this value arises as a result of the particles rotating the way a planet rotates, then the individual particles would have to be spinning impossibly fast. Even if the electron radius were as large as 2.8 nm (the classical electron radius), its surface would have to be rotating at 2.3×10$^{11}$ m/s. The speed of rotation at the surface would be in excess of the speed of light, 2.998×10$^8$ m/s, and is thus impossible. Instead, the spin angular momentum is a purely quantum mechanical phenomenon". Of course the electron doesn't spin like a planet, it's a spin ½ particle. It's a straw man to say it can't be spinning like a planet so it can't be spinning at all. It's fatuous to say electron intrinsic spin is some magical mystic thing that surpasseth all human understanding. It's like the spin of a cyclone. Take the spin away from a cyclone using an anticyclone, and all you've got is wind. Take the spin away from an electron using a positron, and all you've got is light.

And in any case if the electron wasn't a point particle that would cause all sorts of problems.

There is no evidence that the evidence is a point particle. The electron's field is what it is. Saying it's a point-particle causes all sorts of problems, such as renormalisation.

Finally, there isn't a definite "axis of rotation" for spin, because the three components of spin angular momentum don't commute with one another.

That's right, there is no definite axis of rotation. But that doesn't mean there is no rotation. If there really was no rotation, the electron wouldn't have a magnetic dipole moment.

But my question is, can spin be related to a rate of change of anything at all with respect to time? Spin may not be related to rotation in R3, but can we relate it to a rotation or other kind of motion in some other space, possibly a non-Euclidean space? It may take 720 degrees to fully "turn" an electron, but is there actually a period of time in which it "turns" or does something else by 720 degrees?

It's just a double rotation. It isn't something magical and mysterious. See this paper, and this image by Martin van der Mark:

enter image description here

To put it another way, if a particle has a fixed spin state, does it make any sense to say that the particle is "doing" anything, or does it simply "have" a property?

The former. Spin is real. Electron and positrons don't go round in opposite circles in a magnetic field because of some kind of juju magic. But because each is a dynamical spinor. The positron has the opposite chirality to the electron.

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