[Physics] Does spin have anything to do with a rate of change

operatorsquantum mechanicsquantum-spinrotationspinors

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?

Best Answer

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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