Understanding Spinor Rotations

[gentsagree]

Hi,

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

<br /> U =<br /> \left( \begin{array}{ccc}<br /> e^{-\frac{i}{2}\phi} &amp; 0 \\<br /> 0 &amp; e^{\frac{i}{2}\phi} <br /> \end{array} \right)<br />

A spinor rotates under a 2\pi rotation as

<br /> \xi &#039; = <br /> \left( \begin{array}{ccc}<br /> e^{-i\pi} &amp; 0 \\<br /> 0 &amp; e^{i\pi} <br /> \end{array} \right)<br /> \left( \begin{array}{c}<br /> \xi_{1} \\<br /> \xi_{2}<br /> \end{array} \right)<br /> =<br /> \left( \begin{array}{ccc}<br /> -\xi_{1} \\<br /> \xi_{2} <br /> \end{array} \right)<br />

which is (-\xi_{1}, \xi_{2}), and not -\xi, so only one component changes sign. Is this correct?

 

[vanhees71]

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.

 

[gentsagree]

Ahah, ok, I should not try to do physics today.