How an electron spin 1/2 needs two full rotations instead of one to return to its original state? In Quantum Mechanics an electron (or any fermion) wave function changes by a negative sign or -1 under a rotation of 360°. And if we
rotate the electron in physical space we would need to rotate it twice around or $2 \times 360° = 720°$ for
it to come back to its original state. Apart from the unphysical?
and bizare topology of such an object is there a mathematical proof of this in quantum mechanics?
 A: This certainly is a bizarre property of spin-half particles. Remember that angular momentum is generated by the operators of rotation, so that for a state say $| \psi \rangle$ if we consider a rotation about, say  the z-axis ($\hbar = 1$), then
$J_z | \psi \rangle =  -i \frac{\partial}{\partial \theta} | \psi \rangle$
where $J_z$ is the angular momentum operator. Let's say that $J_z$ has
a definite value so that a measurement of spin along the z-axis gives us a value $m$ and $m$ is the intrinsic spin ("magnetic") quantum number. That is,
$-i \frac{\partial}{\partial \theta} | \psi \rangle = m  | \psi \rangle$
and solving this, the state $| \psi \rangle$ can be represented by  $e^{i m \theta} | \psi (\theta) \rangle$. So if $m = \frac{1}{2}$ and we rotate
by $2 \pi$ then $| \psi \rangle \rightarrow - | \psi \rangle$. Doing this once again will retain the original state $| \psi \rangle$.

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*It might be pointed out that multiplication by a phase-factor $e^{i m \theta}$ is just that and does not change the physics of the system. This is true, but it should also be noted that the fact that the state changes sign after a rotation (rather rotation of the space around the particle represented by the state $| \psi \rangle$) is physically observable.

