I'm trying to understand this derivation of Bell's Theorem on Wikipedia.
I've never encountered tensor products before, but from my reading I gather that the first bit of the tensor product acts on the first particle, and the second bit acts on the second particle. I tried using the Pauli matrices to work through the problem explicitly but I can't seem to get the $\frac{1}{\sqrt{2}}$ result. This is what I did:
$$\left|+x\right\rangle = \left( \begin{matrix}1\\ 1 \end{matrix} \right),\ \left|-x\right\rangle = \left( \begin{matrix}1\\ -1 \end{matrix} \right)$$
\begin{align} |\psi\rangle & = \frac{1}{\sqrt{2}}\left( \left|+x\right\rangle \otimes\left|-x\right\rangle -\left|-x\right\rangle \otimes\left|+x\right\rangle \right) \\ & = \frac{1}{\sqrt{2}}\left( \left( \begin{matrix}1\\ 1 \end{matrix} \right) \otimes \left( \begin{matrix}1\\ -1 \end{matrix} \right) - \left( \begin{matrix}1\\ -1 \end{matrix} \right) \otimes \left( \begin{matrix}1\\ 1 \end{matrix} \right) \right) \\ & = \frac{1}{\sqrt{2}}\left( \left( \begin{matrix}1\\ 1 \end{matrix} \right) \otimes \left( \begin{matrix}1\\ -1 \end{matrix} \right) + \left( \begin{matrix}-1\\ 1 \end{matrix} \right) \otimes \left( \begin{matrix}-1\\ -1 \end{matrix} \right) \right) \end{align}
The expected value of operators $AB$ is
$$\langle AB\rangle = \langle \psi |AB| \psi\rangle$$
In Wikipedia's notation I'm considering the case $(a,b)$, with operators $A=A(a)=S_z\otimes I$ and $B=B(b)=-\frac{1}{\sqrt{2}}I \otimes (S_X+S_Z)$. The operator B isthen
$$B = -\frac{1}{\sqrt{2}}I \otimes (S_Z + S_X) = -\frac{1}{\sqrt{2}} \left( \left( \begin{matrix}1 & 0\\ 0 & 1 \end{matrix} \right)\otimes \left( \begin{matrix}1 & 1\\ 1 & -1 \end{matrix} \right) \right)$$
so when operated on $\psi$ I get $$B|\psi \rangle = -\frac{1}{2}\left[\left( \begin{matrix}1\\ 1 \end{matrix} \right) \otimes \left( \begin{matrix}0\\ 2 \end{matrix} \right) + \left( \begin{matrix}-1\\ 1 \end{matrix} \right) \otimes \left( \begin{matrix}-2\\ 0 \end{matrix} \right)\right]$$
As for $A$,
$$A = S_Z \otimes I = \left( \begin{matrix}1 & 0\\ 0 & -1 \end{matrix} \right) \otimes \left( \begin{matrix}1 & 0\\ 0 & 1 \end{matrix} \right)$$
so
$$AB|\psi \rangle = -\frac{1}{2}\left[ \left( \begin{matrix}1\\ -1 \end{matrix} \right) \otimes \left( \begin{matrix}0\\ 2 \end{matrix} \right) +\left( \begin{matrix}-1\\ -1 \end{matrix} \right) \otimes \left( \begin{matrix}-2\\ 0 \end{matrix} \right)\right].$$
Then to do the inner product I need to make the bra $\langle \psi|$ by flipping everything inside into row vectors:
$$\langle \psi| = \frac{1}{\sqrt{2}} \left[ (1\ \ \ 1)\ \otimes\ (1\ \ -1)\ + (-1\ \ \ 1)\ \otimes\ (-1\ \ -1) \right]$$
Finally, I get $$\langle \psi | AB | \psi \rangle = -\frac{1}{2\sqrt{2}}\left[ 0 \otimes -2 + 0 \otimes 2 \right] = ?$$
At this point I don't know where I've gone wrong. I don't know what the value in the square brackets is (do they just multiply like normal numbers now?), but I don't see any way for it to be $-2,$ which is what it needs to be in order to cancel properly and make the thing equal $\frac{1}{\sqrt2}$. Am I not doing the inner product properly here? Have I completely misunderstood how tensors work?
\begin{pmatrix}
is way easier than lugging around a bunch of\left(
and\right)
s. $\endgroup$