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Rather predictably, I'm having a hard time understanding quantum spin. I know that it has units of angular momentum and relates to rotation, but that it can't be thought of as the actual spin of a particle with real size. Also, I understand that it's quantized.

What I'm wondering is, what happens to quantum spin in a rotating frame of reference?

Specifically, are there (rotating) frames of reference in which (for instance) an electron has spin that isn't half-integer? Is there a frame of reference rotating about a single electron such that that one electron acts like a spin-0 particle?

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The spin of a particle is a characteristic of it, so the answer regarding a possible change of the spin qunatum number for an elementary particle by applying a physical coordinate change (may it be considered continuous in time or not) is a somewhat tautological "no".

That number, say $\tfrac{1}{2}$ for the electron, will stay the same. But there are three dimensions $x^i$ and a spin operator $\sigma_{x^i}$ associated to each of them and that observable will naturally change if you turn around and measure in the new direction you look at. Of course, if the spinor-state doesn't change in time and you measure in the old direction, not a thing will have changed. Also, keep in mind that you compute expectation values and when you make a measurement then the knowledge or input regarding the state you're working with changes.

The state of the quantum system is represented by a vector $|\psi\rangle$, transforming like $$|\psi\rangle\rightarrow U(\alpha_t)|\psi\rangle,$$ and for every observable $A$ there is the associated expectation value $$\langle A\rangle_{\psi}:=\langle\psi|A|\psi\rangle.$$ In the Heisenberg picture the rotation looks like $$A\rightarrow U(\alpha_t) A U(\alpha_t)^\dagger$$ and the last interesting thing to note here is that "by the Noether theorem", the observables are the same objects that generate the transformation of the associated transformation. By the linearity of the expectation value $$\langle -2\text{i}\ A+\text e^{\pi \sqrt{163}}\ B\rangle_{\psi}=-2\text{i}\ \langle A\rangle_{\psi} +\text e^{\pi \sqrt{163}}\ \langle B\rangle_{\psi}$$ and because there are commutation relations which enable you to express products and powers of $\sigma_{x^i}$ as linear expressions, the measurement expectations is a function of the previous expectation values $\langle \sigma_{x^i}\rangle_{\psi}$ und the relative turning angle $\alpha$ or $\alpha_t$.

As a handwaving guess, since the eigenvalues of the spin matrices are something like $O(\tfrac{\hbar}{2})$ and the transformation is unitary, the prediction for up/down etc. in the rotated direction of choice will change mildly, a short sum involving $\cos{(\alpha_t)}$-like coefficients. And e.g. measuring in one direction $\sigma_{z}$ and rotating w.r.t the very same direction will change nothing as the associated transformation operator is only a function of $\sigma_{z}$ itself/commutes with the observable.

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