5-3

Show

$$\bm{\epsilon}_{+}\cdot \bm{\epsilon}_{+}^* =1,\quad \bm{\epsilon}... ...silon}_{+}^* =0,\quad \vn \times \bm{\epsilon}_{-}^* =-i \bm{\epsilon}_{-}^*\,.$$ (27)

5-3解答

$$\bm{\epsilon}_{+}\cdot \bm{\epsilon}_{+}^* = \frac{1}{2} \left(\bm{\epsilon}_1+i\bm{\epsilon}_2\right) \cdot \left(\bm{\epsilon}_1-i\bm{\epsilon}_2\right) =\frac{1}{2}\cdot 2 = 1$$

$$\bm{\epsilon}_{-}\cdot \bm{\epsilon}_{-}^* = \frac{1}{2} \left(\bm{\epsilon}_1-i\bm{\epsilon}_2\right) \cdot \left(\bm{\epsilon}_1+i\bm{\epsilon}_2\right) =\frac{1}{2}\cdot 2 = 1$$

$$\vn \times \bm{\epsilon}_{+}^* = \frac{1}{\sqrt{2}} \vn\times \left( \bm{\epsilon}_1-i\bm{\epsil... ...i}{\sqrt{2}}\left(\bm{\epsilon}_1-i\bm{\epsilon}_2\right) =i\bm{\epsilon}_{+}^*$$

$$\bm{\epsilon}_{+}\cdot \bm{\epsilon}_{-}^* = \frac{1}{2} \left(\bm{\epsilon}_1+i\bm{\epsilon}_2\right) \cdot \left(\bm{\epsilon}_1+i\bm{\epsilon}_2\right) = 0$$

$$\bm{\epsilon}_{-}\cdot \bm{\epsilon}_{+}^* = \frac{1}{2} \left(\bm{\epsilon}_1-i\bm{\epsilon}_2\right) \cdot \left(\bm{\epsilon}_1-i\bm{\epsilon}_2\right) = 0$$

$$\vn \times \bm{\epsilon}_{-}^* = \frac{1}{\sqrt{2}} \vn\times \left( \bm{\epsilon}_1+i\bm{\epsil... ...}{\sqrt{2}}\left(\bm{\epsilon}_1+i\bm{\epsilon}_2\right) =-i\bm{\epsilon}_{-}^*$$

著者: 茅根裕司 chinone_at_astr.tohoku.ac.jp