Questions tagged [bell-basis]

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What is the property of any entangled pair of qubits?

We know that any of the Bell states has the special property that the second qubit will be found in a state which is predictable based on measuring one of the qubits. But assuming any other pair of ...
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1answer
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Problem with quantum tomography on two qubits

With reference to question on how to do quantum tomography on two qubits, I would like to ask you for help again. I tried to do the tomography on state \begin{equation}\psi=\frac{1}{2}\begin{pmatrix}...
2
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1answer
144 views

Transforming the first Bell state into the other Bell states

As I understand it, you can transform the different Bell states into one another by applying various gates. Wikipedia has the Bell states written out as follows: And says that you can generate bell ...
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1answer
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Find the Bell States of A and B in the following scenario

Let A has quantum memories $M_1$ and $M_2$ and B has quantum memories $M_3$ and $M_4$ capable of holding one qubit. 2 Bell states are shared among A & B in the following way: First Bell state $|...
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0answers
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Determining the quantum secret

I earlier posted I question Representing a Bell measurement on non adjacent qubits for which I got an excellent answer. Now I want to build upon that and do further analysis which is where I am stuck. ...
1
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1answer
65 views

Representing a Bell measurement on non adjacent qubits

I have a state $${|\psi\rangle} = s {\Bigl(|1\rangle_1|1\rangle_2-|0\rangle_1|0\rangle_2\Bigr)}\otimes{\Bigl(|0\rangle_3|1\rangle_4-|1\rangle_3|0\rangle_4\Bigr)}\otimes{\Bigl(|0\rangle_5|1\rangle_6-|1\...
2
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1answer
61 views

If Alice and Bob share a Bell state, can Alice send her individual qubit to a third party?

Suppose Alice wants to send Bob information. Each of them has been sent each of the qubits of the Bell state in advance. Suppose Alice has the first qubit and Bob has the second. $|\Phi^+\rangle = \...
2
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2answers
51 views

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

I understand that, projecting $\lvert 00\rangle$ on the Bell states would produce $\lvert\Phi^+\rangle$. Because, $$ CNOT(H\lvert0\rangle \otimes \lvert0\rangle) = \frac{1}{\sqrt{2}}(\lvert00\rangle +...