The question in brief: what does it mean, operationally, to rotate an electron?

Elaboration/background: I am trying to understand how representation theory applies to quantum mechanics. A stumbling block for many learners, myself included, is spin. As it happens I already know the story about how $\operatorname{SO}(n)$ is not simply connected, how its double cover $\operatorname{Spin}(n)$ is simply connected, and how some representations of the latter fail to factor through $\operatorname{SO}(n)$. While I am no expert in representation theory, I know the basics, and can learn more myself as I need it.

Rather, my problem is to understand exactly how experimental facts and procedures give us a representation. (In that sense, my question isn't really about electrons or even about quantum mechanics, since special relativity "is" representation theory too, but my lack of understanding becomes especially acute in the case of spin.)

The expositions I have looked at usually say something like "when you rotate an electron, the wavefunction transforms like a spinor", but this is only helpful if I know

  1. what it means to rotate a particle; and
  2. how to translate, at least in principle, from the physical situation to its mathematical description.

I have looked around at questions here, but they are either quite advanced (QFT stuff) or focus on the mathematical machinery as opposed to how that machinery lines up with actual operations on lab equipment. For example, in another thread, Eric Zaslow writes:

Imagine going to the rest frame of a massive particle. In this frame, there is rotational symmetry, which means that the Lie algebra of rotations acts on the wave function. So the wave function is a vector in a representation of Lie(SO(3)) = Lie(SU(2)). "Spin" is the label of precisely which representation this is. Note that while SO(3) and SU(2) share a Lie algebra, they are different as groups, and it is a fact of life ("the connection between spin and statistics") that some particles -- fermions, with half-integral spin -- transform under representations of SU(2) while others -- bosons, with integral spin -- transform under SO(3).

Zaslow's answer

This is clearly stated but does not attempt to address my concern here, which is how operations in the lab translate to Lie group/algebra actions on the state space.

To put my confusion in even starker terms: if I accept, and I do, that the state space of a particle is $\mathbb{C}^2$ or rather the projectivization $\mathbb{CP}^1$, then any Lie group which acts on $\mathbb{C}^2$ acts on the state space of the particle. So clearly the existence of an action is not the whole story. The remaining part of the story is the operational significance of the group...I think. :)

So again, but in different words: when the theoretician acts on an electron wave function with a particular element of $\operatorname{Spin}(3)$, what does the experimentalist do?

  • 1
    $\begingroup$ The experimentalist rotates the equipment and thus the coordinate system. For example, she might make a magnetic field point in a diffetent direction. $\endgroup$
    – G. Smith
    Dec 1, 2018 at 1:08
  • $\begingroup$ @G.Smith Thanks for your comment! If rotating the apparatus counts as rotating the particle, then I think this should be submitted as an answer. After all, rotating equipment has meaning in normal language, whereas rotating a point particle does not. Hence this is precisely the kind of operationalization I was hoping for. $\endgroup$
    – Melissa
    Dec 1, 2018 at 20:41
  • $\begingroup$ I don’t like to think about rotating the particle because I don’t know what rotating a point particle would mean. So I think about rotating the particle’s wave function (which in general is an extended, non-spherically-symmetric thing). This is just rotation of a function of three coordinates, and mathematically “actively” rotating a 3D function within a fixed coordinate system and “passively” rotating the coordinate system are equivalent. $\endgroup$
    – G. Smith
    Dec 2, 2018 at 18:46
  • $\begingroup$ Fair enough! :) I suppose the people who write about rotating the particles are the ones who should be explaining themselves. $\endgroup$
    – Melissa
    Dec 3, 2018 at 12:18

2 Answers 2


...when the theoretician acts on an electron wave function with a particular element of Spin($3$), what does the experimentalist do?

Here are three scenarios. The first two are boring, and the third one is interesting. In all three scenarios, let $R$ denote the element $-1$ in Spin($3$), which corresponds to the identity element in SO($3$).

  • First scenario: The theoretician is using a model in which the electron is the only thing in the universe. In this case, applying $R$ to the electron just changes the overall sign of the state-vector. Since the overall coefficient of the state-vector has no physical significance, there is no difference between what the experimentalist does to prepare the without-$R$ state and the with-$R$ state. (I'm trying to word this carefully while still being concise.)

  • Second scenario: The theoretician is using a more realistic model that includes many particles. Let $c^\dagger(\mathbf{x})$ denote the operator that creates an electron at the location $\mathbf{x}$. (This is a nonrelativistic model.) The theoretician considers a state of the form $$ c^\dagger(\mathbf{x})|\psi\rangle, \tag{1} $$ where $|\psi\rangle$ is the state of everything else. Once again, applying $R$ to the electron just changes the overall sign of the state-vector, because it just changes the overall sign of $c^\dagger(\mathbf{x})$. So again there is no difference between what the experimentalist does to prepare the without-$R$ version and the with-$R$ version of the state.

  • Third scenario (this is the interesting one): The theoretician considers a state of the form $$ \big(c^\dagger(\mathbf{x})+ c^\dagger(\mathbf{y})\big)|\psi\rangle, \tag{2} $$ where the points $\mathbf{x}$ and $\mathbf{y}$ are far away from each other, and then applies $R$ only to any electron that happens to be located near $\mathbf{y}$. The result of applying $R$ is $$ \big(c^\dagger(\mathbf{x})- c^\dagger(\mathbf{y})\big)|\psi\rangle. \tag{3} $$ The state-vectors (2) and (3) are not proportional to each other, so they represent different physical states. This time, the experimentalist must do something different to prepare the state (3) instead of the state (2).

The question is, what must the experimentalist do differently to prepare (3) instead of preparing (2)? If we change "electron" to "neutron", then this has actually been done in the neutron interference experiments reviewed in this paper:

These are basically two-slit experiments with a macroscopic distance between the two paths in the interferometer. Diffraction in a crystal was used as a substitute for "slits." Magnets were arranged in a way that would cause precession of any neutron that passes through one of the paths, and the effect on the resulting two-slit interference pattern displays the effect of the sign-change under $2\pi$ rotations that characterizes spin-1/2 particles. In this experiment, the difference between (2) and (3) corresponds to turning on the magnetic field — or, more generally, adjusting the strength of the magnetic field to interpolate between (2) and (3).

(By the way, in the neutron interference experiments cited above, there was typically only one neutron in the interferometer at any given time — so these experiments are good examples of single-particle self-interference.)

If we want our model to handle experiments like this, we need a way to construct things that change sign under a $2\pi$ rotation. Representations of the rotation group $O(3)$ don't do this. The fundamental representation of the covering group does.

  • $\begingroup$ Thank you, this is helpful and addresses my question. I will look at the paper. In the meantime, would you agree to the statement "rotating a particle has no simple, immediate operational meaning (unlike translating it)", or is that a mis-reading on my part? $\endgroup$
    – Melissa
    Dec 1, 2018 at 20:38
  • $\begingroup$ @Melissa I don't have a clear yes or no answer -- this is exceeding the limits of my intuition! The best I've been able to do (so far, at least) is to think about how a given transformation affects a state-vector (like applying a rotation to just one of the two terms in a superposition) and then thinking about how the resulting state-vector would be prepared experimentally. That probably doesn't qualify as a "simple, immediate operational meaning"; but just because I haven't thought of one doesn't mean there isn't one. This is a great question! $\endgroup$ Dec 1, 2018 at 21:34

So again, but in different words: when the theoretician acts on an electron wave function with a particular element of Spin(3), what does the experimentalist do?

(what is Spin(3)?)

In my opinion you have the cart in front of the horse, to start with.

A theory in physics models acts, does not act. A validated theory has a calculational correspondence with the experiment, but not all "acts" in the theory lead to an observable effect in an experiment.

Experimenters recorded the behavior of atoms of silver in magnetic fields in 1922, and this was correlated with that of single electrons. Then the mathematical modeling related the behavior to angular momentum with the group structure of SU(2) and the behavior of the wavefunction under this algebra.

This model, when used to predict new setups , was validated. and is continuously validated up to now by all experimental data.

In addition, the spin hypothesis kept angular momentum conservation (Noether's theorem) possible in particle interactions.

  • $\begingroup$ Thanks for your answer, and for the references! It is a fact that many expositions contain formulations like "when you rotate an electron...". I am asking what that means. As for Spin(3), that's the double cover of SO(3). It's also isomorphic to SU(2) but that's exceptional. $\endgroup$
    – Melissa
    Dec 1, 2018 at 20:32

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