Exercise 6.9

Using the results of Exercise 6.7 and Exercise 6.8 it is easy to show that

\[ \begin{flalign*} \vec{\sigma} \cdot \vec{\tau} \ket{sing} & = \sigma_x \tau_x \ket{sing} + \sigma_y \tau_y \ket{sing} + \sigma_z \tau_z \ket{sing}&\\ & = - \ket{sing} - \ket{sing} - \ket{sing}&\\ & = -3 \ket{sing} \end{flalign*} \]

\[ \begin{flalign*} \vec{\sigma} \cdot \vec{\tau} \ket{T_1} & = \sigma_x \tau_x \ket{T_1} + \sigma_y \tau_y \ket{T_1} + \sigma_z \tau_z \ket{T_1}&\\ & = \ket{T_1} + \ket{T_1} - \ket{T_1}&\\ & = \ket{T_1} \end{flalign*} \]

\[ \begin{flalign*} \vec{\sigma} \cdot \vec{\tau} \ket{T_2} & = \sigma_x \tau_x \ket{T_2} + \sigma_y \tau_y \ket{T_2} + \sigma_z \tau_z \ket{T_2}&\\ & = \ket{T_2} - \ket{T_2} + \ket{T_2}&\\ & = \ket{T_2} \end{flalign*} \]

\[ \begin{flalign*} \vec{\sigma} \cdot \vec{\tau} \ket{T_3} & = \sigma_x \tau_x \ket{T_3} + \sigma_y \tau_y \ket{T_3} + \sigma_z \tau_z \ket{T_3}&\\ & = - \ket{T_3} + \ket{T_3} + \ket{T_3}&\\ & = \ket{T_3} \end{flalign*} \]