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I Spinor rotation

  1. Apr 17, 2016 #1

    I am confused on a very basic fact. I can write [itex]\xi = (\xi_{1}, \xi_{2}) [/itex] and a spin rotation matrix as

    U =
    \left( \begin{array}{ccc}
    e^{-\frac{i}{2}\phi} & 0 \\
    0 & e^{\frac{i}{2}\phi}
    \end{array} \right)

    A spinor rotates under a [itex]2\pi[/itex] rotation as

    \xi ' =
    \left( \begin{array}{ccc}
    e^{-i\pi} & 0 \\
    0 & e^{i\pi}
    \end{array} \right)
    \left( \begin{array}{c}
    \xi_{1} \\
    \end{array} \right)
    \left( \begin{array}{ccc}
    -\xi_{1} \\
    \end{array} \right)

    which is [itex](-\xi_{1}, \xi_{2})[/itex], and not [itex]-\xi[/itex], so only one component changes sign. Is this correct?
  2. jcsd
  3. Apr 17, 2016 #2


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    Science Advisor
    Gold Member
    2017 Award

    How do you come to this conclusion? Since ##\exp(\mathrm{i} \pi)=\exp(-\mathrm{i} \pi)=-1## your rotation by ##2 \pi## leads to ##\hat{U} \xi=-\xi## as it should be.
  4. Apr 17, 2016 #3
    Ahah, ok, I should not try to do physics today.
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