Appendix A: Detailed Calculations of the First Examination
For the spin-state of the particle along the x or y directions, we have:
$$\begin{aligned} \vert x+ \rangle =&\frac{1}{\sqrt{2}} \bigl( \vert z+ \rangle +\vert z- \rangle \bigr) \end{aligned}$$
(A.1)
$$\begin{aligned} \vert x- \rangle =&\frac{1}{\sqrt{2}} \bigl( \vert z+ \rangle -\vert z- \rangle \bigr) \end{aligned}$$
(A.2)
$$\begin{aligned} \vert y+ \rangle =&\frac{1}{\sqrt{2}} \bigl( \vert z+ \rangle +i\vert z- \rangle \bigr) \end{aligned}$$
(A.3)
$$\begin{aligned} \vert y- \rangle =&\frac{1}{\sqrt{2}} \bigl( \vert z+ \rangle -i\vert z- \rangle \bigr) \end{aligned}$$
(A.4)
with single probabilities:
$$ P_{x}^{+}=\frac{1}{2}, \quad P_{x}^{-}=\frac{1}{2}, \quad P_{y}^{+}=\frac{1}{2}, \quad P_{y}^{-}=\frac{1}{2} $$
(A.5)
Now, where, e.g., \(P_{x}^{+}\) is the probability of finding the result \(\frac{\hbar}{2}\) for the spin component along the x-direction of particle, the other probabilities are defined similarly.
For left-hand side of inequality (7), one finds:
$$ -\frac{1}{2}\ln\frac{1}{2}-\frac{1}{2}\ln\frac{1}{2}- \frac{1}{2}\ln\frac{1}{2}-\frac{1}{2}\ln \frac{1}{2}=2\ln2=1.386 $$
(A.6)
For the right hand side of (7), one gets:
where, 1.386>0.317 as was expected.
Appendix B: Detailed Calculations of the Second Examination
Here we define the single probabilities at t
1 and t
2 as:
$$\begin{aligned} P_{+}^{t_{1}} =&\frac{1}{2} ( 1+\sin\theta_{a} ) \end{aligned}$$
(B.1)
$$\begin{aligned} P_{-}^{t_{1}} =&\frac{1}{2} ( 1-\sin\theta_{a} ) \end{aligned}$$
(B.2)
$$\begin{aligned} P_{+}^{t_{2}} =&\frac{1}{2} ( 1+\sin\theta_{a} \cos\theta_{ab} ) \end{aligned}$$
(B.3)
$$\begin{aligned} P_{-}^{t_{2}} =&\frac{1}{2} ( 1-\sin\theta_{a} \cos\theta_{ab} ) \end{aligned}$$
(B.4)
where, e.g., \(P_{+}^{t_{1}}(P_{+}^{t_{2}})\) is the probability of finding the result \(+\frac{\hbar}{2}\) for the spin component of the particle along the direction \(\hat{a}(\hat{b})\) at t
1(t
2). Other probabilities are defined similarly. For spin-states of the particle at t
1 and t
2, we have:
$$\begin{aligned} \bigl \vert \varphi_{+}^{t_{1}} \bigr\rangle =&\cos\frac{\theta_{a}}{2}\vert z+ \rangle +\sin\frac{\theta_{a}}{2}\vert z- \rangle \end{aligned}$$
(B.5)
$$\begin{aligned} \bigl \vert \varphi_{+}^{t_{2}} \bigr\rangle =&\cos\frac{\theta_{b}}{2}\vert z+ \rangle +\sin\frac{\theta_{b}}{2}\vert z- \rangle \end{aligned}$$
(B.6)
$$\begin{aligned} \bigl \vert \varphi_{-}^{t_{1}} \bigr\rangle =&-\sin\frac{\theta_{a}}{2}\vert z+ \rangle +\cos\frac{\theta_{a}}{2}\vert z- \rangle \end{aligned}$$
(B.7)
$$\begin{aligned} \bigl \vert \varphi_{-}^{t_{2}} \bigr\rangle =&-\sin\frac{\theta_{b}}{2}\vert z+ \rangle +\cos\frac{\theta_{b}}{2}\vert z- \rangle \end{aligned}$$
(B.8)
Now, the probability amplitudes for the same or different results at successive times can be obtained as:
$$\begin{aligned} \bigl\langle \varphi_{+}^{t_{1}}\bigm|\varphi_{+}^{t_{2}} \bigr\rangle =&\cos\frac{\theta_{a}}{2}\cos\frac{\theta_{b}}{2}+\sin \frac{\theta_{a}}{2} \sin\frac{\theta_{b}}{2}=\cos\frac{\theta_{ab}}{2} \end{aligned}$$
(B.9)
$$\begin{aligned} \bigl\langle \varphi_{+}^{t_{1}}\bigm|\varphi_{-}^{t_{2}} \bigr\rangle =&\sin\frac{\theta_{a}}{2}\cos\frac{\theta_{b}}{2}-\cos \frac{\theta_{a}}{2} \sin\frac{\theta_{b}}{2}=\sin\frac{\theta_{ab}}{2} \end{aligned}$$
(B.10)
$$\begin{aligned} \bigl\langle \varphi_{-}^{t_{1}}\bigm|\varphi_{+}^{t_{2}} \bigr\rangle =&- \biggl[ \sin\frac{\theta_{a}}{2}\cos\frac{\theta_{b}}{2}-\cos \frac{\theta_{a}}{2} \sin\frac{\theta_{b}}{2} \biggr] =-\sin \frac{\theta_{ab}}{2} \end{aligned}$$
(B.11)
$$\begin{aligned} \bigl\langle \varphi_{-}^{t_{1}}\bigm|\varphi_{-}^{t_{2}} \bigr\rangle =&\cos\frac{\theta_{a}}{2}\cos\frac{\theta_{b}}{2}+\sin \frac{\theta_{a}}{2}\sin\frac{\theta_{b}}{2}=\cos\frac{\theta_{ab}}{2} \end{aligned}$$
(B.12)
So, for the left-hand side of the inequality (18) we have,
Similarly, for the right hand side of (18), one obtains: