Let’s consider an experiment. A spinless neucleus emits an electron and a positron, which are both spin-half particles. Alice observe the spin of electron in derection $ \textbf{a}$ , which is a unit vector. Bob measures at the same time the spin of positron in derection $ \textbf{b}$ at a similar distance from the neucleus. The result of measurement is either 1/2 or -1/2. Here are some results from Binney’s textbook.

And then Binney claims that the expectation value of the product of the result got by alice (which is $ \sigma_A \in \{1/2, -1/2\}$ ) and Bob satisfies

On the other hand, however, I got something unexpected from my own calculation. We know that

$ <\sigma_e(\textbf{a})\sigma_p(\textbf{b})>=\sum_{i,j \in \{1/2, -1/2\}}P(\sigma_e(\textbf{a})=i,\sigma_p(\textbf{b})=j)ij\=-\sum_{i,j \in \{1/2, -1/2\}}P(\sigma_e(\textbf{a})=i,\sigma_e(\textbf{b})=j)ij$

Therefore, $ <\sigma_e(\textbf{a})\sigma_p(\textbf{b})>-<\sigma_e(\textbf{a})\sigma_p(\textbf{b’})>\ =-\sum_{i,j,k \in \{1/2, -1/2\}}P(\sigma_e(\textbf{a})=i,\sigma_e(\textbf{b})=j, \sigma_e(\textbf{b’})=k)i(j-k)\ =-\sum_{i,j,k \in \{1/2, -1/2\}}P(\sigma_e(\textbf{a})=i,\sigma_e(\textbf{b})=j, \sigma_e(\textbf{b’})=k)ij(1-4jk)\ $

where I have use the fact that $ \sigma_e(\textbf{b})^2=1/4$

Take abosolute value on both sides yields

$ |<\sigma_e(\textbf{a})\sigma_p(\textbf{b})>-<\sigma_e(\textbf{a})\sigma_p(\textbf{b’})>| \ \leq \frac{1}{4}\sum_{j,k \in \{1/2, -1/2\}}P(\sigma_e(\sigma_e(\textbf{b})=j, \sigma_e(\textbf{b’})=k)(1-4jk)$

where I took the largest possible value of $ \sigma_e(\textbf{a})$ and $ \sigma_e(\textbf{b})$ . Expanding gives

$ |<\sigma_e(\textbf{a})\sigma_p(\textbf{b})>-<\sigma_e(\textbf{a})\sigma_p(\textbf{b’})>| \leq \frac{1}{4} + <\sigma_e(\textbf{b})\sigma_p(\textbf{b’})>$

**Question: Don’t you feel it’s Bell’s inequality? I did not introduce hidden variables at all! What’s wrong?**

And of course, it leads to contradictions.