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I am confused as to what is being measured in the boxes in the example drawings shown on the Entanglement and Bell Tests section in the IBM Q Experience:

enter image description here

https://quantum-computing.ibm.com/docs/guide/mult-entang/entanglement-and-bell-tests

My understanding is that the $(|00 \rangle + |11 \rangle)/ \sqrt{2}$ represents a superposition of two states, $|00 \rangle$ and $|11 \rangle$. And that $|00 \rangle$ is a two qubit entangled state. If Alice measures one of the qubits to be a $0$, the other qubit must be a $0$ as well due to the entanglement. The same is true if Alice measures a $1$.

My question comes from not understanding why $A$ and $A'$ are present in the box. What are they measuring? Is $A$ measuring one qubit while $A'$ is measuring the second qubit? Or is $A$ measuring one qubit while $A'$ measures the same qubit. Same questions for box $B$.

MY BACKGROUND: I am working through the IBM Q Experience. I have some exposure to QM through College level Physical Chemistry. I have a basic understanding of linear algebra, superposition, and entanglement.

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Both $A$ and $A'$ measure the same qubit, while $B$ and $B'$ are applied to the second qubit. Alice chooses (randomly) which of the two measurement settings ($A$ or $A'$) she will choose for each run of the experiment, while Bob chooses between $B$ and $B'$.

You mention the perfect correlation in the computational basis. Note however that there's nothing "quantum" about correlation in a single basis. You can easily reproduce this in a "classical" theory. Replace the photons by balls with color and the measurement by determining the color. The source is just someone (repeatedly) giving each of Alice and Bob one ball. It would not surprise you if you find that whenever Alice has a blue ball also Bob has a blue ball. As a rule of thumb: If you do not find anything unusual about an experiment when you replace photons by balls then there's nothing quantum about it. ;-)

You need at least two measurement settings for both parties for a Bell inequality (that can be violated).

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  • $\begingroup$ In the first formulation of a Bell Game I saw, it was presented with each player either moving a joystick attached to their box either to the left or to the right. Here, is Alice picking between A and A', while Bob is picking between B and B'? $\endgroup$
    – Abstract Acumen
    Commented May 14, 2020 at 19:28
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    $\begingroup$ Yes, that's correct. $\endgroup$
    – M. Stern
    Commented May 15, 2020 at 16:59

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