How an electron spin 1/2 needs two full rotations instead of one to return to its original state?

Question

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?

Answer 1

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 $| \psi \rangle$.

• 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.