If I rotate a system by $2\pi$, will I invert it? In this post, I saw the following

...if you have a two-fermion system, interchanging them is equivalent to rotating the system by an angle $2\pi$.

But in my view, rotating the system by $2\pi$ does not interchange the particles, but rather results in the same system. By interchange here, I mean $(p_1,p_2)\to(p_2,p_1)$.
Why rotating a system by 2π the same as reversing the particles?
 A: The operator for rotating a field with spin $\sigma$ through an angle $\theta$ has the form
$ e^{i\sigma\theta}$, so a “complete” rotation of $\theta=2\pi$ radians will cause a spin-half particle to accumulate a phase factor of $e^{i\frac12 2\pi} = -1$.
Warning, warning: the previous sentence is oversimplified enough that anyone who’s worked through the details will have complaints about it — me included. As I remember, the “simple” treatment in Streater and Wightman is ten or fifteen pages long. But the gist of my oversimplification is correct: spinors “double cover” the rotation group, and must rotate through $2\times2\pi$ radians to return to their initial state.
The double-cover property of spinor rotations is closely related to the spin-statistics theorem, which says a state with two identical particles of spin $\sigma$ must accumulate a phase $(-1)^{2\sigma}$ when the particles are exchanged: that is, two-boson wavefunctions are symmetric under exchange, and two-fermion wavefunctions are antisymmetric.
In the early decades of quantum field theory it was believed this sign change under “complete” rotations was a mathematical curiosity, because the absolute phase of a single fermion isn’t an observable. However, interferometry permits observation of phase differences, and the phase change of single neutrons under a $2\pi$ rotation was first observed in 1975.
Your linked answer (v2) seems to be saying

*

*Rotating a spinor field by $2\pi$ introduces a phase factor of $-1$.


*Exchanging two identical fermions introduces a phase change of $-1$.


*Therefore, rotating a two-fermion state by $2\pi$ has the same effect as exchanging the particles.
This is different from how I understand the relationship among these effects. I am under the impression that you accumulate the rotation-phase for each particle in the state. So rotating your two-fermion system about its center by a half-turn, $\frac{2\pi}{2}$ radians, would give a phase factor $e^{i\pi/2} = i$ on each particle, or a total phase $-1$ for the system, at the same time as putting particle $A$ in the position of particle $B$ and vice-versa.
That is to say, I think you’re right: a half-turn is the same as an exchange for a two-fermion system. If the linked answer is actually correct, that’s very interesting.
A: Yes, classically turning around by $2\pi$ is the trivial transformation. However the whole new thing about spinors is that for them this transformation is not trivial - the spinor instead changes sign. But because all provabilities involve spinor bilinears they don't change and this allows such novel objects in the quantum theory.
A: In my humble opinion, we do not rotate a physical system at some angle; instead, we recalculate the observational results from one still reference system to another one turned at this angle with respect to the original reference frame. So, observation of a QM system at the angle $0+\varepsilon$ may not be different from observation at the angle $2\pi - \varepsilon$, if $\varepsilon\to 0$.
The difference between spin and a classical angular momentum  is in quantizing the spin projections $S_z$ and in discretenes of the total spin $S$. Any observation implies exchange of the energy with the system, so the system never stays the same if observed - we observe transitions. In Classical Mechanics, where the space is 3D and the maximum "rotation" angle is $2\pi$, there is another implicit hypothesis, namely, our observation does not change the system, contrary to QM. If you forget observation influence, then you reason in terms of non observable constructs and you have paradoxes everywhere.
Interchange of identical particles (their mutual transitions) may not lead to any physical effect by definition of identical particles; thus the wave function is (anty) symmetric and may only acquire an inessential sign factor (whatever it is) disappearing from observable quantities.
