# Rotation of spin operator

1. Oct 20, 2013

### jk22

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 $$Rspin=\left(\begin{array}{cc} cos(\theta/2) & -sin(\theta/2)\\sin(\theta/2) & cos(\theta/2)\end{array}\right)$$ and the new component of the spin operator is, let say for z : $$R_{spin}^{-1}\sigma_z R_{spin}$$

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

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

2. Oct 20, 2013

### Bill_K

What you've written as Rspin is only a special case. It represents a rotation through angle θ/2 about the y-axis: Rspin = exp(i θ/2 σy).

One way of parametrizing the 3-d rotations is:
$$\left(\begin{array}{cc}exp i(φ+ψ) cos θ/2 &exp i(-φ+ψ) sin θ/2 \\exp i(φ-ψ) sin θ/2&exp -i(φ+ψ) cos θ/2\end{array}\right)$$