# "Precession" of a free electron

I'd like to gain as deep of an understanding as possible of the following diagram from Introductory Quantum Mechanics, 4th ed., by Richard Liboff:

I'm not looking for help with solving the easy problem; it's very straightforward for example to calculate $$\left < \alpha_z|S_x|\alpha_z \right > =0$$ by simply plugging in the matrix / vector expressions for $$\left < \alpha_z \right |$$ , $$S_x$$ and $$\left | \alpha_z \right >$$ and doing the matrix multiplication and dot product. Instead, what I'm interested in is exactly how the diagram makes sense in the context in which it is presented.

Note that the diagram is not illustrating Larmor precession. The electron of course has a magnetic moment, but the external magnetic field here is zero, implying that the Larmor frequency is zero.

The diagram also isn't illustrating Thomas precession. The electron's velocity is nonrelativistic here, and indeed can be taken to be zero.

Since the book calls the diagram a "dynamical conception", and I can't find anything online about the precession of an electron's spin in the absence of an external electromagnetic field, I'm guessing that an electron's spin doesn't literally precess, i.e. change with time, in the absence of an external field, and the diagram is only intended to show how $$\left < S_x\right >=\left =0$$ is intuitively plausible in a state in which $$S_z=\hbar/2$$ and $$S^2=3\hbar^{2}/4$$. But if I'm wrong, what is the name of the precession involved, if there is one, and what are the details of how it works, e.g., with what frequency does the spin precess, and how is that frequency arrived at theoretically?

If the so-called "precession" can't be taken literally as something that happens over time, how can the diagram be validly interpreted, rigorously?

Although this came from a book on nonrelativistic quantum mechanics, please feel free to explain it at as advanced of a level as is needed to avoid "lies to children".

The talk about precession is misleading. The electron is in an eigenstate of the Hamiltonian, and is thus in a stationary state. The only time evolution is a global phase:

$$\exp{(-i\frac E {\hbar} t)}$$

There is no precession. The cone is a result of the Heisenberg Uncertainty Principle...that $$S_x, S_y, S_z$$ don't commute.

So $$\vec S$$ is in a superposition of all states lying on the cone with $$S_z=\hbar/2$$.

• "So $\vec S$ is in a superposition of all states..." That doesn't make sense if $\vec S$ is an operator... an operator acts on states it is not "in a superposition of... states." I assume the notation $\vec S$ denotes a vector of operators since in the previous paragraph its components are operators.
– hft
Apr 26, 2022 at 16:33
• @hft Position is also an operator, yet there is $\psi(x)$.
– JEB
Apr 26, 2022 at 17:06
• Yes, but so what? $X$ is an operator. $|x>$ is a state, not an operator. There are continuous position eigenbras $<x|$ and associated coefficients of expansion for any state $|\psi>$ given by $<x| \psi>=\psi(x)$. This does not mean that states and operators are the same thing. In fact, I'm not seeing how your example of "yet there is $\psi(x)$" addresses my comment in any way.
– hft
Apr 26, 2022 at 17:13

If the so-called "precession" can't be taken literally as something that happens over time... how can the diagram be validly interpreted, rigorously?

The diagram is pedagogical. The diagram is not "right" or "wrong," it is just a picture.

The diagram seems to be trying to illustrate how a vector can have all three components non-zero, generally, at any given time, but only one component non-zero on average. In this case the z-component would be non-zero on average, but the x- and y- components are zero on average (when averaging about one complete precession). The averaging in the diagram is a classical averaging over time due to a hypothetical "precession," which is not normally how one would describe the quantum physics of electron spin. But it is a way that a teacher might motivate a student to try and understand what is happening.

In addition, the diagram seems to illustrate in a classical way how a vector of length $$\sqrt{3/4}$$ and z-component $$1/2$$ could have zero expectation value in the x and y directions. How else, would you explain this classically/geometrically other than some type of classical "precession"? How else would you "explain this result geometrically," as the text asks?

We have no other intuition then classical intuition (because we are macroscopic objects). We try to understand things classically, and sometimes this leads us astray. Is the classical intuition in that case right or wrong? I'd say, it's mostly wrong, but it does not really matter.

The point is not that a classical explanation is correct, the point is to try and understand what is happening geometrically. The classical explanation will not be "rigorously" correct, but it may help you understand some of the pros and cons of a classical interpretation of quantum phenomena.