6

When people talk about a loophole free Bell test, what they really mean is that the two loopholes that most concern the majority of people are closed simultaneously: the measurement loophole and the locality loophole. Let us briefly review the protocol: A Bell state $(|00\rangle+|11\rangle)/\sqrt{2}$ is produced, and two parties, Alice and Bob, each take ...


5

The difference is as follows: the original Bell inequality requires that outcomes from the same setting are always perfectly anti-correlated. It says nothing about the case where they are even marginally different. by contrast, in CHSH, the ideal (giving maximum violation) is that outcomes from the same setting would be anti-correlated, but it is not ...


5

I think that I can explain the definition through the following simple example: Suppose that you perform two experiments in the same house in two separate rooms. In the first you measure the observable $A$ and instantaneously in the second you measure the observable $B$. The measurements are afflicted with noise, so you do not get a definite answer every ...


5

I think there is a conceptual thing going on here that needs clarifying (I'll leave the experimental links to others). I presume the question is predicated on the idea that, well, measurements are made within a certain time of each other, which is compared to the distance between the places where the measurements are being made. The concern is that this only ...


5

The Hadamard gate is: $$\frac{1}{\sqrt 2} \left(|0\rangle \langle 0 | + |0\rangle\langle 1| + |1\rangle \langle 0| - |1\rangle \langle 1|\right)$$ And since $|+\rangle$ is $\frac{1}{\sqrt 2}\left(|0\rangle + |1\rangle \right)$, you can work out that $H(|+\rangle) = |0\rangle$ So, $$CNOT(H|+\rangle \otimes |+\rangle)$$ $$= CNOT(|0\rangle \otimes |+\...


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

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, ...


4

Measuring an observable does not mean applying the observable operator to a quantum state but rather measuring the state in the eigenbasis of the operator. A measurement will basically produce an eigenvalue of that observable operator and the system will collapse to a subspace corresponding to all states having that eigenvalue. For example, if you're ...


4

I think you're doing things a little bit backwards. You probably shouldn't be calculating $P(a|x)$ or $P(b|y)$ in advance, because you're simply trying to ask: Given a set of $\{P(ab|xy)\}$, do there exist assignments to $P(a|x)$ and $P(b|y)$ that satisfy $P(ab|xy)=P(a|x)P(b|y)$ for all $a,b,x,y$? So, how do you evaluate the probability of getting ...


3

The four Bell states are $$ |\Phi_{\pm}\rangle=(|00\rangle\pm|11\rangle)/\sqrt{2}\qquad |\Psi_{\pm}\rangle=(|01\rangle\pm|10\rangle)/\sqrt{2}. $$ So, let's consider what happens then we try and measure in the Bell basis, i.e. project onto one of these four states. If we started with the state $|00\rangle$, then we can write it as $$ |00\rangle=\frac{1}{\sqrt{...


3

Simply start by writing out everything $$ |B_{00}\rangle_{13}|B_{00}\rangle_{24}=\frac12\left(|00\rangle_{13}|00\rangle_{24}+|00\rangle|11\rangle+|11\rangle|00\rangle+|11\rangle|11\rangle\right) $$ Let me rearrange each of these terms $$ \frac12\left(|00\rangle_{12}|00\rangle_{34}+|01\rangle|01\rangle+|10\rangle|10\rangle+|11\rangle|11\rangle\right). $$ Now ...


3

Yes, there are. I just wrote a paper about it, actually. You need to define carefully what you mean by obtaining a Bell violation or ruling out local hidden variables. You can't demand to have a result which is impossible to explain with local hidden variables: the local bound of a Bell inequality is inherently probabilistic, so it is possible to obtain any ...


3

I believe you're thinking of the all-versus-nothing proofs based on GHZ states. You start with a state such as $$ |\Psi\rangle=\frac{1}{\sqrt{2}}(|000\rangle+|111\rangle) $$ and you select at random one of the four measurements to implement: $X\otimes X\otimes X$, $X\otimes Y\otimes Y$, $Y\otimes X\otimes Y$ or $Y\otimes Y\otimes X$. Assuming every one of ...


3

Are there scenarios in which Bell nonlocality can be observed without such averages, that is, in a single-shot scenario? No, you have to collect statistics. Any single result you see could have been due to classical players picking completely at random and getting lucky. Making the chance of classical luck arbitrarily close to zero requires repetition (or ...


3

The largest scale Bell test done thus far is the "Cosmic Bell Test" of 2017. It ruled out hidden variables within a distance of 600 light years from Earth. The 16 significant Bell test experiments performed between 1972 and 2018 are listed here with references to the original papers.


3

... and the question changes again. Re Update 4: I'm not sure what you intended the variable $a$ to be in the question (I think part of the problem is that you've got muddled between the chosen questions and the given answers), but I have used it very carefully, to always be the $\pm1$ answer given. In a quantum setting, $a$ is the answer, a value $\pm 1$ ...


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 do understand that the sum of these three probabilities is greater than one because there are some constraints already involved; like if we uncover all three coins at least two have to be the same. So naturally, there's some redundancy leading to a sum of probabilities that is greater than one! I would say the explanation is simpler than that. ...


2

I guess the way that I'd start (aside from just getting a computer to do it!) is to remember that the Bell states form an orthonormal basis. So, you can ask, for example, about what the $|\Phi^+\rangle^{AD}$ component is: $$ \langle\Phi^+|^{AD}|\Phi\rangle^{ABCD}=-\frac12|\Phi^+\rangle^{BC}. $$ You do this for each of the four states, and you can use that to ...


2

GiannisKol is correct in an abstract sense -- you simply want to specify any unitary matrix with the first column containing elements $(\alpha,\beta,\gamma,\eta)$. You then complete the other columns using any valid assignment (columns must be orthonormal, so you might use the Gram Schmidt procedure). However, there are obviously many options, and you are ...


2

First of all, I'm assuming you mean the state : $\alpha \left|00\right> + \beta \left|01\right> + \gamma\left|10\right> + \eta \left|11\right>$. What you actually want to do is to act with a gate $U$ on the initial state $\left|00\right>$ so that $U\left|00\right> = \alpha \left|00\right> + \beta \left|01\right> + \gamma\left|10\right&...


2

First, is this value correct? Yes, it is. If you expand out the calculation you're doing, this is the same as $$ \sqrt{2}\langle\psi|X\otimes X+Z\otimes Z|\psi\rangle $$ for any two-qubit state $|\psi\rangle$. In the particular case $|\psi\rangle=|00\rangle$, it's easy to extract that this is $\sqrt{2}$. Indeed, for any separable state of the form $$ |\psi\...


2

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 "...


2

Imagine you had a general formula $$ C=a_1QS+a_2RS+a_3RT+a_4QT. $$ Algebraically, we know that if $Q$, $S$, $R$ and $T$ are random variables with values $\pm 1$, then each term such as $QS\in\{\pm 1\}$. Hence, there is a trivial bound $$ C\leq |a_1|+|a_2|+|a_3|+|a_4| =C_\max. $$ This can never be beaten by any model, be it local hidden variable, quantum, ...


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

Alice has a choice of two bases in which to measure, we call her choice $x$. Bob also has a choice of two bases in which to measure, we call his choice $y$. When they measure we call Alice's measurement outcome $a$ and Bob's measurement outcome $b$. All $a,b,x,y \in \{0,1\}$. Then Alice and Bob win if $xy = a + b \mod 2$. For example if Alice chooses to ...


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 $(|01\rangle+|10\rangle)/\sqrt{2}$.


1

No, this isn't a Bell test. Testing for a violation of Bell's inequality requires testing over several distinct measurement bases. For a good example of Bell tests run on IBM hardware, see section II(3) here. To run the circuit in the OP, just move the second $H$-gate on $q_0$ in front of the measurement on $q_2$. This will have no impact on your outcomes ...


1

The part you're overlooking is these lines: # Players do a sqrt(X) based on their referee's coin. circuit.append([ cirq.CNOT(alice_referee, alice)**0.5, cirq.CNOT(bob_referee, bob)**0.5, ]) The players rotate their qubits conditioned on what the local referee said. This is equivalent to changing the measurement basis conditioned ...


1

I figured out the answer while writing the question, but figured I'd still post it for future reference. The problem with the calculation is that it was not taking into account the locality constraint. Namely, the fact that local deterministic behaviours also have to satisfy $$p(ab|xy)=p(a|x)p(b|y).$$ Taking this into account, we notice that there are $\...


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