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Theoretically, after a rotation of $2\pi$, a fermion wavefunction picks up a minus sign, and it is after a rotation through $4\pi$ that it returns to its initial quantum state. Now, the wave-functions or the quantum state is not a directly measuable quantity. Then how will the fact that "the wavefunction picks up a minus sign" be reflected in measurements?

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The main experiment to mention here is neutron interference. While you cannot detect the total phase of a state - $\lvert \psi \rangle$ and $-\lvert \psi\rangle$ represent the same quantum state and cannot be distinguished - you can detect relative phases, i.e. the $\phi$ in $$ \lvert \psi_1\rangle + \mathrm{e}^{\mathrm{i}\phi}\lvert \psi_2\rangle.$$ So the trick here is to "rotate" $\lvert \psi_2\rangle$ but not $\lvert \psi_1\rangle$. For this, we take advantage of the coupling of spin to magnetic fields, described for instance here. The reason we use neutrons is that they are uncharged and will therefore not be deflected by a magnetic field in the manner electrons would.

We split a neutron beam in two and subject one of the two resulting beams to a constant magnetic field $B_0$ in $z$-direction. The Hamiltonian (generically $\propto \vec S\cdot \vec B$) becomes $$ H = k B_0 S_z,$$ where $k = \frac{g_n e}{m_e c}$ is a bunch of constants determining the magnetic moment of the neutron. With $\omega := k B_0$ this now means that the phase $\phi$ is $\omega t/2$ (the half comes from the half-spin of the neutron and is precisely what we want to detect!), where $t$ is the time travelled, which you can relate to the distance travelled and thereby to the interference pattern on your screen.

That's basically it, once you see that your pattern is consistent with $\omega t/2$, you've shown that a $2\pi$ rotation, corresponding to $\omega t = 2\pi$ since the time evolution is then $\mathrm{e}^{\mathrm{i}2\pi S_z}$, aka a "full rotation", just acts as a half-rotation on the neutron state.

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    $\begingroup$ "Main experiment" sounds a bit objective for something that's entirely as matter of taste and familiarity. $\endgroup$ – DanielSank Mar 28 '17 at 22:29

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