5

I want to emphasize that you are wasting your time with this historical stuff. But since you insist I'll answer anyway. First of all, note that the inequality \begin{equation} \left|\int d\lambda p(\lambda)(1-A(\vec a', \lambda)B(\vec b',\lambda))\right|+ \left|\int d\lambda p(\lambda)(1-A(\vec a', \lambda)B(\vec b,\lambda))\right| \\ \le 2-| E(\vec a',\vec ...


5

I'll describe the case of two party correlations but this can straightforwardly extended to more parties. Let's give a box to Alice and a box to Bob. Alice and Bob can interact with their boxes by providing them classical inputs $x \in \mathcal{X}$ and $y \in \mathcal{Y}$ respectively. They can also receive outputs from their boxes, $a \in \mathcal{A}$ and $...


5

It's easy to generate such a model for specific cases. For example, take the maximally entangled state $|\phi^+\rangle = \frac1{\sqrt2} (|00\rangle + |11\rangle)$, and let the observables in the CHSH inequality be $A_0=B_1=Z$, and $A_1=B_0=X$. Now if Alice and Bob measure in the same basis, they'll get results that are random but equal with probability 1, ...


3

This, in a nutshell, is the whole issue of quantum. You say "of course an electron will have either spin up or spin down". It might seem like it should, but this is founded purely on your intuition about what the world around you is like. You have no a priori law of the universe that rules out other options. And hence, when an experiment shows you ...


2

I tried to prepare Bell state with circuit described by $\text{CNOT} (H \otimes I)$ and the result is state $\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$ as expected. Here is my circuit in the composer (including the code): After measurement on simulator I get these results Please make sure that you measure correctly, i.e. each qubit to separate classical bit: ...


2

(1) The best description that I can give is a mixed state, $\rho=I/2$. (2) If you apply a unitary operation on your half, it does not change my description of my state. If you apply a measurement on your half, my best description only updates if you tell me the measurement result. (3) The combined state looks like $(|01\rangle+|10\rangle)/\sqrt{2}$.


2

The other answer already covered most of the bases. I'd just add an explicit example of a no-signalling, non-quantum distribution, because I think it's useful to have some in mind when discussing these things. Consider a two-two behaviour, that is, a conditional probability distribution $p(ab|xy)$ with $a,b,x,y\in\{0,1\}$, such that $p(a,a|x,y)=1/2$ for $xy=...


1

Short answer: Yes, it is the complementary angle because you have grabbed the wrong probability amplitudes from $|\psi\rangle$. \begin{align} \text{Pr}(A=\updownarrow, B=\searrow) &= \langle \psi | \left(|\updownarrow\rangle\langle\updownarrow| \otimes |\searrow\rangle\langle\searrow|\right) |\psi \rangle\\ &= \langle \psi | (\mathbb{I}_A \otimes R_B^...


1

One way to think about this is to realise that the "physical properties" in QM (but also more generally in physics) are not just a function of the system under examination, but also of the measurement apparatus that you are using to probe said system. In the case of QM, you cannot observe the system without significantly perturbing it, as you do in ...


1

The "realism" assumption basically amounts to saying "the system's state can be described by a probability distribution, operations on the system can be represented by probabilistic transitions, and here are some specific distinguishable states that I know the system can be in". An example of a system that doesn't have this property is ...


1

Actually, an electron will only have spin up or spin down when we decide to look at it. Generally, quantum mechanics postulates that the state of the spin exists in superposition between up and down, only collapsing to a definite value--and becoming an element of reality--when we perform a measurement. In the first assumption you cite, when the physical ...


1

You probably did something like this: Since you are only measuring the top qubit, and a Bell state $|\psi \rangle = \dfrac{|00\rangle + |11\rangle}{\sqrt{2}}$ has $1/2$ probability of the first qubit in the state $|0\rangle$ and $1/2$ probability of it being in the state $|1\rangle$. Hence, you got the above probabilities readout plot. If you also put in ...


1

As far as I know the distance record was obtained in the Canary islands experiment, where Alice and Bob were 144 km apart. There exists a paper that directly does what you want, lowerbound the "speed of quantum information". They obtain the bound $1.5 \times 10^4 c$. I read it as a work of comedy. Nobody thinks that quantum information is actually ...


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