Physical meaning behind the double rotation of spin 1/2 particles

Question

From the Bloch sphere, it is mathematically clear that a $720°$ rotation is necessary to bring a spin $1/2$ particle back to its initial state, as a full rotation changes the sign of the state.

However, what is the physical meaning behind that?
How can we understand the state of the particle after only one full rotation?

Answer 1

States are unit vectors up to phases. The representation of a full rotation is just a phase $-1$ in the Hilbert space of spin 1/2 particles. So this transformation is nothing but the identity on the states as physics requires. It is physically wrong saying that “it changes the sign of a state”. It just changes the sign of a state vector leaving the state invariant.

This property of the unitary representation of rotations for semi integer spin particles has however physical consequences. In particular, a super selection rule which states that no coherent superposition of states with integer and semi integer spin is physically possible.

Answer 2

Here is Feynman's illustration, without math:

Hold a coffee cup in the palm of your hand, in front of you. Now raise your elbow and pivot your forearm towards your body, and swing your hand under your arm the rest of the way through a full arc so it is once again in front of you. The cup has thus been rotated through 360 degrees.

Note the state of your arm: it has a twist in it that wasn't there before.

Now continue swinging through another arc, this time bringing your hand around above your arm so the cup is once again in front of you. Note that the twist is now gone and you are back to your original state- after 720 degrees of rotation.

Feynman's point was that in the spin-half case, there is a topological connection between the cup, your arm, and the 3-dimensional space they are embedded in which twists the connection in a 360 degree rotation, requiring a second 360 degree rotation to untangle the twisted connection.

Answer 3

It's a wrong question is ask "why a $720°$ rotation is necessary to bring a spin 1/2 particle back to its initial state".

Actually, a $360°$ rotation is necessary to bring a spin 1/2 particle back to its initial state. The proper question is "why a $360°$ rotation of a spin 1/2 particle is translated to $720°$ rotation of vector".

Let's look at the different ways spinor/fermion $\psi$ and vector $V$ transform under a rotation: $$ \quad \psi \rightarrow e^{iT\theta}\psi, \\ \quad V \rightarrow e^{iT\theta}Ve^{-iT\theta}= e^{2iT\theta}V, $$ where $T$ is a rotation generator of the Lie algebra which anti-commutes with $V$.

Because of the double-sided nature of $e^{iT\theta}Ve^{-iT\theta}$, a $360°$ rotation of a spinor/fermion is translated to a $720°$ rotation of a vector.

You may want to argue that since the rotation angle is perspective (spinor vs. vector) dependent, why do we take the spinor's perspective ($360°$), rather than the vector's perspective($720°$)? It has to do with the fact that spinor is more fundamental than vector. Figuratively speaking, vector is the square of spinor. In other words, a vector is a bi-spinor or spinor bilinear. Therefore, a lot things of vector are doubled, including the rotation angle. See more details here.