Exercise 6.8
We’ll show that the triplet states are eigenvectors of \(\sigma \tau\), so the eigenvalue has the same value as the expectation value.
\[ \ket{T_2} = \frac{1}{\sqrt{2}} (\ket{uu} + \ket{dd} ) \]
\[ \begin{flalign*} \sigma_z \tau_z \ket{T_2} & = \frac{1}{\sqrt{2}} \sigma_z (\ket{uu} - \ket{dd} )&\\ & = \frac{1}{\sqrt{2}} ( \ket{uu} + \ket{dd})&\\ & = \ket{T_2} \end{flalign*} \]
\(\ket{T_2}\) is eigenvector of \(\sigma_z \tau_z\) with eigenvalue \(+1\), hence \(\braket{T_2 | \sigma_z \tau_z | T_2 } = +1\).
\[ \begin{flalign*} \sigma_x \tau_x \ket{T_2} & = \frac{1}{\sqrt{2}} \sigma_x (\ket{ud} + \ket{du} )&\\ & = \frac{1}{\sqrt{2}} ( \ket{dd} + \ket{uu})&\\ & = \ket{T_2} \end{flalign*} \]
\(\braket{T_2 | \sigma_x \tau_x | T_2 } = +1\)
\[ \begin{flalign*} \sigma_y \tau_y \ket{T_2} & = \frac{1}{\sqrt{2}} \sigma_y (\mathrm{i}\ket{ud} - \mathrm{i}\ket{du} )&\\ & = \frac{1}{\sqrt{2}} ( \mathrm{i}^2 \ket{dd} -\mathrm{i} (-\mathrm{i}) \ket{uu})&\\ & = - \ket{T_2} \end{flalign*} \]
\(\braket{T_2 | \sigma_y \tau_y | T_2 } = -1\)
\[ \ket{T_3} = \frac{1}{\sqrt{2}} (\ket{uu} - \ket{dd} ) \]
\[ \begin{flalign*} \sigma_z \tau_z \ket{T_3} & = \frac{1}{\sqrt{2}} \sigma_z (\ket{uu} + \ket{dd} )&\\ & = \frac{1}{\sqrt{2}} ( \ket{uu} - \ket{dd})&\\ & = \ket{T_3} \end{flalign*} \]
\(\braket{T_3 | \sigma_z \tau_z | T_3 } = +1\)
\[ \begin{flalign*} \sigma_x \tau_x \ket{T_3} & = \frac{1}{\sqrt{2}} \sigma_x (\ket{ud} - \ket{du} )&\\ & = \frac{1}{\sqrt{2}} ( \ket{dd} - \ket{uu})&\\ & = - \ket{T_3} \end{flalign*} \]
\(\braket{T_3 | \sigma_x \tau_x | T_3 } = -1\)
\[ \begin{flalign*} \sigma_y \tau_y \ket{T_3} & = \frac{1}{\sqrt{2}} \sigma_y (\mathrm{i}\ket{ud} - (-\mathrm{i})\ket{du} )&\\ & = \frac{1}{\sqrt{2}} ( \mathrm{i}^2 \ket{dd} + \mathrm{i} (-\mathrm{i}) \ket{uu})&\\ & = \ket{T_3} \end{flalign*} \]
\(\braket{T_3 | \sigma_y \tau_y | T_3 } = +1\)
An expectation value of \(+1\) means: if Alice and Bob both measure the same component of their spins, they will always get an equal result.
An expectation value of \(-1\) means: if Alice and Bob both measure the same component of their spins, they will always get an opposite result.