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If we consider a spin-$\frac12$ particle at rest in the absence of any potentials, we can use the Pauli spin operators and an associated basis to describe the observable.

Let's arbitrarily choose the $Z$ basis and consider the particle to initially be in state $\newcommand{\k}[1]{\left|\;#1\;\right\rangle}$ $\k{\downarrow}$.

If we then move into an accelerating reference frame, do we require a new spin operator? Secondly, what is the spin state of the particle in this new reference frame?

To make this somewhat more intuitive, this question arose from considering a particle in free fall in a uniform gravitational field. Does the spin state change depending on whether we measure the particle in the laboratory frame or the particle own frame of reference?

EDIT:

The particle and its measurement apparatus are at rest in a laboratory. They are technically in an accelerating reference frame subject to gravitational forces. The spin is measured and found to be in the $\k{\downarrow}$ eigenstate. If we then release both the particle and the apparatus, they will be in an inertial (freely falling) frame. Is there a way to quantify the state of the particle now? Will it still be in the $\k{\downarrow}$ eigenstate or do the eigenstate have to be redefined altogether?

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    $\begingroup$ The spin tensor, like all tensors, transforms under coordinate change. The spin operator is just a restriction to the spatial parts of the spin tensor integrated over all space. $\endgroup$ – Slereah Aug 6 '15 at 9:02
  • $\begingroup$ I'm not quite clear what you're asking here Hedra. Can I say this though: if you drop an electron, its spin changes. If it didn't, it wouldn't fall down. If you were falling alongside it you might claim that the spin state hadn't changed, but you'd be wrong. When it comes to relativity, remember this: when your measurement changes it's because the thing you're measuring has changed, or because you have changed. And when your measurement doesn't change, it might be because there is no change, or because both you and the thing you're measuring have changed. $\endgroup$ – John Duffield Aug 6 '15 at 9:15
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    $\begingroup$ I have zero knowledge of this area, but if it's of any interest chapter 24 of this book discusses transformations in accelerated frames and identifies a Wu operator as (and I quote) the spin operator in linearly accelerated frames. $\endgroup$ – John Rennie Aug 6 '15 at 10:46
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    $\begingroup$ Thanks for the responses. They are all relevant to what I am asking. I have edited my question to try to be more clear. John Duffield, would you be able clarify what you mean by "If it's spin doesn't change then it wouldn't fall down." $\endgroup$ – Hedra Aug 7 '15 at 4:33
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    $\begingroup$ Unfortunately we measure spin by exerting forces on it, so it will not be "in a frame" if we measure the spin. You can easily talk about the spin in an instantaneously comoving frame. $\endgroup$ – Timaeus Sep 10 '15 at 11:33
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A series of lorentz boosts can generate a rotation therefore an observer in a reference frame that is moving with respect to the rest reference frame where the particle is at rest will measure a value of spin that depends on the momentum of the particle in the transformed frame and the lorentz transformation needed to relate the two reference frames. This is all true in special relativity. It is hard for me to imagine that things would be simpler in accelerating reference frame i.e GR. So yes, the spin of the particle should change since it can change in special relativity.

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I think such experimental evidence of spin-flips due to frame velocity or intense gravitational field have not been seen or reported in literature; however looking at nature of spin states of particles which are devoid of any mechanical rotations as such should not be affected by gravity generated by space time curvature or such forces generated by motion of the frames (mechanical in nature); the spin/color and other quantum mechanical attributes are distinguishing labels and has nothing in common with usual spinning tops....

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