Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rotation of spin operator

  1. Oct 20, 2013 #1
    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]

    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 ?
  2. jcsd
  3. Oct 20, 2013 #2


    User Avatar
    Science Advisor

    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:
    (Rotate through angle φ about z-axis)(Rotate through angle θ about y-axis)(Rotate through angle ψ about z-axis)

    What you get instead of Rspin is

    [tex]\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)[/tex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook