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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

    Bill_K

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    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]
     
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