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Sorry that this isn't a quick question but I didn't know how to make it shorter. I am struggling with this for quite a long time and I would appreciate every help that I can get. I could not find a derivation of the standard deviation for Bell's inequality (CHSH's inequality) so I tried to make my own. Consider the following setup: Alice and Bob share an entangled state \begin{equation} |\Psi\rangle_{AB} = \frac{1}{\sqrt{2}}(|0\rangle_A\otimes|0\rangle_B+|1\rangle_A\otimes|1\rangle_B) \end{equation} Alice can either measure X or Z operators, where we use shorthand notation X and Z instead of $\sigma_x$ and $\sigma_z$. Bob can measure the operators $W = \frac{1}{\sqrt{2}}(X+Z)$ and $V = \frac{1}{\sqrt{2}}(Z-X)$. In general, for two arbitrary observables $\mathcal{A}=(\vec{a}\cdot\vec{\sigma}^{(1)})$ and $\mathcal{B}=(\vec{b}\cdot\vec{\sigma}^{(2)})$ where $\sigma^{(1)} = \vec{\sigma}\otimes\mathbb{1}$ and $\sigma^{(2)} = \mathbb{1}\otimes\vec{\sigma}$ we can prove that the mean value of $\langle \mathcal{A}\mathcal{B} \rangle$ is given by: \begin{equation} \langle \mathcal{A}\mathcal{B} \rangle = -\cos(\theta_{ab}) \end{equation} where $\theta_{ab}$ is the angle between $\vec{a}$ and $\vec{b}$. The proof goes as follows:

Since \begin{align*} \vec{\sigma}^{(tot)}|\Psi\rangle &= (\vec{\sigma}^{(1)}+\vec{\sigma}^{(2)})|\Psi\rangle\\ &= \frac{1}{\sqrt{2}}(\vec{\sigma}^{(1)}+\vec{\sigma}^{(2)}) (|00\rangle + |11\rangle)\\ &= 0 \end{align*} It follows that \begin{align*} \vec{\sigma}^{(1)}|\Psi\rangle = -\vec{\sigma}^{(2)}|\Psi\rangle \end{align*} With this in mind we find: \begin{align*} \langle \mathcal{A}\mathcal{B} \rangle &= \langle\Psi|(\vec{a}\cdot\vec{\sigma}^{(1)})(\vec{b}\cdot\vec{\sigma}^{(2)})|\Psi\rangle\nonumber\\ &= -\langle\Psi|(\vec{a}\cdot\vec{\sigma}^{(1)})(\vec{b}\cdot\vec{\sigma}^{(1)})|\Psi\rangle\nonumber\\ &= -a_ib_j\frac{1}{2}(\langle00|+\langle11|)(i\epsilon_{ijk}\sigma_k^{(1)}+\delta_{ij})(|00\rangle+|11\rangle)\nonumber \\ &= -a_ib_j\frac{\delta_{ij}}{2}(\langle00|+\langle11|)(|00\rangle+|11\rangle)\nonumber\\ &= -\vec{a}\cdot\vec{b} = -\cos(\theta_{ab}) \end{align*} where we used the identity $\sigma_i \sigma_j = \delta_{i j}\mathbb{1} + i\varepsilon_{i j k}\,\sigma_k$

If we now plug in the angles which are described above, we can find that \begin{align*} \nonumber \langle X \otimes W \rangle &=\langle Z \otimes W \rangle =\langle Z \otimes V \rangle = \frac{1}{\sqrt{2}}\\ \langle X \otimes V \rangle &= -\frac{1}{\sqrt{2}}\\ \Rightarrow |S| = |\langle Q \otimes W \rangle &+\langle R \otimes W \rangle +\langle R \otimes V \rangle -\langle Q \otimes V \rangle| = 2\sqrt{2} \end{align*} So far so good. If I wanted to calculate the uncertainty of $|S|$, I would have to find $(\Delta\mathcal{AB})^2 = \langle (\mathcal{A}\mathcal{B})^2 \rangle - \langle \mathcal{A}\mathcal{B} \rangle^2$ for each of the observables mentioned above. $\langle \mathcal{A}\mathcal{B} \rangle^2$ is easy since it is equal to \begin{align*} \langle \mathcal{A}\mathcal{B} \rangle^2=(\vec{a}\cdot\vec{b})^2 = a_1^2b_1^2+a_2^2b_2^2+a_3^2b_3^2+2a_1^2b_1^2a_2^2b_2^2+2a_1^2b_1^2a_3^2b_3^2+2a_2^2b_2^2a_3^2b_3^2. \end{align*}

On the other hand, $\langle (\mathcal{A}\mathcal{B})^2 \rangle$ can be found by using the exact same method as before:

\begin{align*} \langle (\mathcal{A}\mathcal{B})^2 \rangle &= a_i^2b_j^2\frac{1}{2}(\langle00|+\langle11|)(i\epsilon_{ijk}\sigma_k^{(1)}+\delta_{ij})^2(|00\rangle+|11\rangle)\nonumber \\ &= a_i^2b_j^2\frac{\delta_{ii}}{2}(\langle00|+\langle11|)(|00\rangle+|11\rangle)\nonumber\\ &= a_1^2b_1^2+a_1^2b_2^2+a_1^2b_3^2+a_2^2b_1^2+a_2^2b_2^2\\&+a_2^2b_3^2+a_3^2b_1^2+a_3^2b_2^2+a_3^2b_3^2 \end{align*}

Taking both together, this leads to the final result:

\begin{align*} (\Delta(\mathcal{A}\otimes \mathcal{B}))^2 &= (a_1b_2-a_2b_1)^2+(a_3b_1-a_1b_3)^2\\&+(a_2b_3-a_3b_2)^2 = (\vec{a}\times\vec{b})\cdot(\vec{a}\times\vec{b})\\ \Rightarrow \Delta(\mathcal{A}\otimes \mathcal{B}) &= |\vec{a}\times\vec{b}| = |\sin(\theta_{ab})| \end{align*}

Now the problem is the following: if I plug in the angles as described above (and those which saturate Bell's inequality), I find that the standard deviation becomes 1. Considering that $|S| = 2\sqrt{2}$, an error of $\Delta|S| = 1$ seems way too high. Is there anything I did wrong?

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    $\begingroup$ I would swear a very very similar question was posted earlier today, although it appears to have been deleted. $\endgroup$ – ZeroTheHero Jan 17 at 22:17
  • $\begingroup$ I asked that question earlier and deleted it afterwards because it was not very well structured. I thought I might put a little more effort into making it clear what I am looking for $\endgroup$ – xabdax Jan 17 at 22:25
  • $\begingroup$ I don't think you can compute the variance by simply summing the variances of the individual terms. If you have an observable written as $\mathcal O\equiv\hat A+\hat B+\hat C-\hat D$. Computing the variance of $\mathcal O$ amounts to computing $\langle \mathcal O^2\rangle-\langle\mathcal O\rangle^2$, which is different than the sum of the variances when the operators do not commute. $\endgroup$ – glS Feb 19 at 10:29
  • $\begingroup$ I tried that as well, but the result was not much better. In fact, I got an uncertainty of sqrt(8)... $\endgroup$ – xabdax Feb 21 at 23:48

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