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

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

A spinor rotates under a $2\pi$ rotation as

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

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.