Rotation of spin operator

  • Thread starter jk22
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  • #1
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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 ?
 

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  • #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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