If we consider a spin 1/2 particle, then, the rotation of the spinor for each direction is given by a rotation matrix of half the angle let say theta [tex]Rspin=\left(\begin{array}{cc} cos(\theta/2) & -sin(\theta/2)\\sin(\theta/2) & cos(\theta/2)\end{array}\right)[/tex] and the new component of the spin operator is, let say for z : [tex]R_{spin}^{-1}\sigma_z R_{spin}[/tex](adsbygoogle = window.adsbygoogle || []).push({});

On the other hand one could consider the rotation of the spin vector operator : [tex]R\vec{\sigma}[/tex] where R is a 3x3 rotation matrix.

I don't understand what the angle [tex]\theta[/tex] represents when compared to the rotation in 3d space of the spin vector, where we have 3 angles ?

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Rotation of spin operator

**Physics Forums | Science Articles, Homework Help, Discussion**