# Operational definition of rotation of particle

...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:

- "Theoretical and conceptual analysis of the celebrated $4\pi$-symmetry neutron interferometry experiments", https://arxiv.org/abs/1601.07053.

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.