Exercise 6.7

Consider the Triplet state \[ \ket{T_1} = \frac{1}{\sqrt{2}}(\ket{ud} + \ket{du}) \]

Expectation value of \(\sigma_z \tau_z\)

\[ \begin{flalign*} \sigma_z \tau_z \ket{T_1} & = \sigma_z \tau_z \frac{1}{\sqrt{2}}(\ket{ud} + \ket{du}) &\\ & = \frac{1}{\sqrt{2}} \sigma_z (-\ket{ud} + \ket{du})&\\ & = - \frac{1}{\sqrt{2}} (\ket{ud} + \ket{du})&\\ & = - \ket{T_1}& \end{flalign*} \]

\(T_1\) is an eigenvector of \(\sigma_z \tau_z\) with eigenvalue \(-1\). Thus the expectation value \(\braket{T_1 | \sigma_z \tau_z | T_1 } = -1\).

If Alice and Bob both measure the z-component of their spins, they will always get an opposite result.

Expectation value of \(\sigma_x \tau_x\)

\[ \begin{flalign*} \sigma_x \tau_x \ket{T_1} & = \sigma_x \tau_x \frac{1}{\sqrt{2}}(\ket{ud} + \ket{du}) &\\ & = \frac{1}{\sqrt{2}} \sigma_x (-\ket{uu} + \ket{du})&\\ & = \frac{1}{\sqrt{2}} (\ket{du} + \ket{ud})&\\ & = \ket{T_1}& \end{flalign*} \]

\(T_1\) is an eigenvector of \(\sigma_x \tau_x\) with eigenvalue \(+1\). Thus the expectation value \(\braket{T_1 | \sigma_x \tau_x | T_1 } = +1\).

If Alice and Bob both measure the x-component of their spins, they will always get an equal result.

Expectation value of \(\sigma_y \tau_y\)

\[ \begin{flalign*} \sigma_y \tau_y \ket{T_1} & = \sigma_y \tau_y \frac{1}{\sqrt{2}}(\ket{ud} + \ket{du}) &\\ & = \frac{1}{\sqrt{2}} \sigma_y (-\mathrm{i}\ket{uu} + \mathrm{i}\ket{du})&\\ & = \frac{1}{\sqrt{2}} ((-\mathrm{i}) \mathrm{i} \ket{du} + \mathrm{i} (-\mathrm{i}) \ket{ud})&\\ & = \frac{1}{\sqrt{2}} (\ket{du} + \ket{ud})&\\ & = \ket{T_1}& \end{flalign*} \]

\(T_1\) is an eigenvector of \(\sigma_y \tau_y\) with eigenvalue \(+1\). Thus the expectation value \(\braket{T_1 | \sigma_y \tau_y | T_1 } = +1\).

If Alice and Bob both measure the y-component of their spins, they will always get an equal result.