# Tag Info

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

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 ...

4

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, $$CNOT(H|+\rangle \otimes |+\rangle)$$ $$= CNOT(|0\rangle \otimes |+\... 4 Measuring an observable does not mean applying the observable operator to a quantum state but rather measuring the state in the eigenbasis of the operator. A measurement will basically produce an eigenvalue of that observable operator and the system will collapse to a subspace corresponding to all states having that eigenvalue. For example, if you're ... 4 I think you're doing things a little bit backwards. You probably shouldn't be calculating P(a|x) or P(b|y) in advance, because you're simply trying to ask: Given a set of \{P(ab|xy)\}, do there exist assignments to P(a|x) and P(b|y) that satisfy P(ab|xy)=P(a|x)P(b|y) for all a,b,x,y? So, how do you evaluate the probability of getting ... 3 The four Bell states are$$ |\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

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 ...

2

... 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 $\pm 1$ ...

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. ...

1

The key to figuring out the probability of any measurement result is Born's rule, which says that if you have a state $\left|\psi\right\rangle$ the probability of observing measurement outcome $\left|\phi\right\rangle$ is given by \begin{align} \Pr(\phi | \psi) = \left|\left\langle \phi | \psi \right\rangle \right|^2. \end{align} In the example you ...

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