6 votes

How would a Quantum Computer (network) perform loophole-free Bell tests?

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 ...
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What is an example of an entangled state whose correlations are describable with a local hidden variable model?

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 ...
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5 votes
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Difference between Bell's inequality and CHSH

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 ...
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Definition of locality in Bell experiments

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 ...
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5 votes

References examining Bell inequality violations at large distances

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 ...
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Projecting $\lvert ++ \rangle$ on Bell Basis

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\...
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Formal proof of Bell's inequality in CHSH form

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(\...
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What is the no-signaling set and how can it be related to other types of correlations?

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 ...
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4 votes
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Simultaneous eigenstate of commuting observables and their tensor product

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 ...
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4 votes

Determining whether $P(ab|xy)$ factorizes in Bell experiments

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 ...
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3 votes

Projecting $\lvert ++ \rangle$ on Bell Basis

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 ...
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Find the Bell States of A and B in the following scenario

Simply start by writing out everything $$ |B_{00}\rangle_{13}|B_{00}\rangle_{24}=\frac12\left(|00\rangle_{13}|00\rangle_{24}+|00\rangle|11\rangle+|11\rangle|00\rangle+|11\rangle|11\rangle\right) $$ ...
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3 votes

Are there Bell-like violations that can be observed without collecting statistics?

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 ...
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3 votes

Are there Bell-like violations that can be observed without collecting statistics?

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 ...
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Are there Bell-like violations that can be observed without collecting statistics?

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 ...
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3 votes

References examining Bell inequality violations at large distances

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 ...
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3 votes

How is Bell’s Inequality converted to the CHSH inequality?

... 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 ...
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3 votes

In Bell's inequalities, what is the meaning of assuming that the physical properties $P_Q,P_R,P_S,P_T$ have definite values?

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 ...
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2 votes
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Why is $P(1,2)_{\text{same}} = \frac{1}{4}$ and not $\frac{1}{2}$ in Preskill's Bell experiment?

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 ...
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2 votes
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Problem about entanglement swapping

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^+\...
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2 votes

How to prepare an arbitrary two-qubit state?

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 ...
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2 votes

How to prepare an arbitrary two-qubit state?

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 ...
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What would the CHSH inequality be equal to if the two qubits were separable?

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 $|\...
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2 votes
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What are the two measurements per box being done in the Bell Tests in the IBM Quantum Experience "Entanglement and Bell tests" Section

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 ...
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Why is the CHSH inequality defined with a minus sign?

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 $QS\in\{\pm 1\}$...
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Where does $xy = a + b \pmod 2$ come from in the context of CHSH inequalities?

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 ...
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2 votes

Output of Bell State Measurement

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 (...
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2 votes
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Does the CHSH inequality fully characterise the local polytope?

Not quite. Consider the following no-signalling distribution $PR_1$ which I will write in the form $$ \begin{pmatrix} p(00|00) & p(01|00) & p(00|01) & p(01|01) \\ p(10|00) & p(11|00) ...
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2 votes
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EPR pair and individual operations

(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 ...
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2 votes

What is the no-signaling set and how can it be related to other types of correlations?

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 ...
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