Undergrad An elementary question about rotations

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Applying a rotation operator by an angle θ around the z-axis on a positive spin-1/2 eigenstate results in a phase factor of e^{-iθ/2}, indicating that while the state appears changed, observable measurements remain unaffected. This phase factor is significant in quantum mechanics, as a rotation by 2π introduces a -1 factor, meaning a full rotation does not return the system to its original state until 4π is applied. This phenomenon is not exclusive to spin-1/2 particles but applies generally in quantum mechanics. Experimental verification, such as neutron interferometry, demonstrates these principles through observable interference patterns. The discussion also raises questions about the effects of rotations on systems of multiple spins, particularly regarding phase cancellations.
Silicon-Based
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Rotation of spin-1/2 state
Suppose I have a positive spin-##1/2## eigenstate pointing in the ##z##-direction. If I apply a rotation operator by an angle ##\theta## around the ##z##-axis the state should of course not change. However, if I write it out explicitly, I find something different:
$$R_z(\theta)|\uparrow\rangle =
\begin{pmatrix}
e^{-i\theta/2} & 0 \\
0 & e^{i\theta/2}
\end{pmatrix}
\begin{pmatrix}
1 \\
0
\end{pmatrix} =
e^{-i\theta/2}|\uparrow\rangle.
$$Is this factor of ##e^{-i\theta/2}## like a phase that can be added and removed at will because it's not an observable? I don't see a different way to rationalize this with the physical expectation above.
 
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Silicon-Based said:
Summary: Rotation of spin-1/2 state

Suppose I have a positive spin-##1/2## eigenstate pointing in the ##z##-direction. If I apply a rotation operator by an angle ##\theta## around the ##z##-axis the state should of course not change. However, if I write it out explicitly, I find something different:
$$R_z(\theta)|\uparrow\rangle =
\begin{pmatrix}
e^{-i\theta/2} & 0 \\
0 & e^{i\theta/2}
\end{pmatrix}
\begin{pmatrix}
1 \\
0
\end{pmatrix} =
e^{-i\theta/2}|\uparrow\rangle.
$$Is this factor of ##e^{-i\theta/2}## like a phase that can be added and removed at will because it's not an observable? I don't see a different way to rationalize this with the physical expectation above.

As you say, the value of all observables measured on that state will not change. However, rotating the system has changed the state. Note, for example, that rotating the system by ##2\pi## results in a phase factor of ##-1##. In QM you have to rotate a system by ##4\pi## to return to the original state.

Note that this is a general result for rotations of systems in QM. Not just spin 1/2 particles.

This prediction of QM can be tested. There is an experiment quoted in Sakurai, for example, using neutron interferometry. The idea is to split a beam of neutrons, with one path involving a rotation and then to recombine the paths and observe the interference pattern.

If one path involves a rotation of ##2\pi## you should see destructive interference, but not for a rotation of ##4 \pi##.
 
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If I have a system of two unentangled spins, with state vector like ##\left|\right.\uparrow\rangle\left|\right.\uparrow\rangle##, would a rotation of ##2\pi## add a factor of -1 or would the minus signs from the two spins cancel? Just a thought that came to my mind and it's not immediately clear to me what happens...
 
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Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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