# I Spinor rotation

1. Apr 17, 2016

### gentsagree

Hi,

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

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

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

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

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

2. Apr 17, 2016

### 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.

3. Apr 17, 2016

### gentsagree

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