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I was just writing another thread when I stumbled upon something strange:

What if I now make a slight change and replace ##\bf{b}\rightarrow-\bf{b}##? The expectation values containing ##\bf{b}## change signs and I get greypilgrim said: ↑Hi,

I'm trying get a better understanding of Bell's inequality in the form

$$\left|E\left(\bf{a},\bf{b}\right) -E\left(\bf{a},\bf{c}\right)\right|\leq 1+E\left(\bf{b},\bf{c}\right)\enspace.$$

I'm considering the Bell state

$$\left|\psi\right\rangle= \frac{1}{\sqrt{2}}\left(\left|+\right\rangle_1\left|-\right\rangle_2- \left|-\right\rangle_1\left|+\right\rangle_2\right)\enspace.$$

and the expectation value ##E^{qt}## of the product of the result of a measurement of spin 1 in direction ##\bf{a}## and spin 2 in direction ##\bf{b}##

$$E^{qt}\left(\bf{a},\bf{b}\right)= \left\langle\psi\right|\vec{\sigma_1} \cdot\bf{a}\otimes \vec{\sigma_2} \cdot\bf{b} \left|\psi\right\rangle=-\bf{a}\cdot\bf{b}$$

which is a straigthforward calculation. I'm using the notation from 1.5.1 in

http://www.uibk.ac.at/exphys/photonik/people/gwdiss.pdf

which is, however, in German.

We can violate the inequality by choosing e.g. ##\bf{a}=e_x##, ##\bf{b}=\left(e_x+e_z\right)/\sqrt{2}##, ##\bf{c}=e_z## which yields

$$\left|-\frac{1}{\sqrt{2}}-0\right|=\frac{1}{\sqrt{2}}\leq 1-\frac{1}{\sqrt{2}}$$

which is obviously wrong. I think this choice of vectors also maximally violates the inequality.

$$\left|\frac{1}{\sqrt{2}}-0\right|=\frac{1}{\sqrt{2}}\leq 1+\frac{1}{\sqrt{2}}$$

which is correct! But I've only turned my measurement device 180°, i.e. exchanged the (+1) and (-1) results, or more mathematically, I've permuted my measurement basis. How can this lead to such a profound change, i.e. this would allow a local-realistic description for these measurement angles? Surely there has to be something wrong.

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# Bell experiment: Rotate measurement device 180°

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