Questions tagged [bell-basis]

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Creating Bell states (a question from tutorial)

I was going through a Quantum Computation lecture series on YouTube given by Umesh Vazirani. There in Lecture 6.3 he talks about creating a Bell state using a Hadamard and CNOT gate. He seems to be ...
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Constructing and Measuring in an Arbitrary 3-qubit basis

As part of a Quantum Theory project I have "constructed" an arbitrary 3-qubit basis: $\left|B_0\right> =\left|000\right>$ $\left|B_1\right> = \frac{1}{\sqrt{2}}\cos(x)(\left|100\right> +...
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Bell states from EPR pair of two-level atoms [closed]

Considering a case in which we prepare an entangled EPR pair from two two-level atoms interacting with a cavity, one prepared in ground state and the other in excited state but moving in at a later ...
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2answers
63 views

How to measure in the bell basis?

The question is quite straightforward, I have some two qubit state and I want to measure it in the Bell basis. This question answers me partly, but I still have some doubts because, I thought that ...
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31 views

Do Bell state measurements satisfy the completeness condition from the no-signalling theorem?

Do Bell state measurements satisfy the completeness condition from the no-signalling theorem? In a quantum experiment involving unitary evolution and projective measurements,  do Bell state ...
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3answers
116 views

No Schmidt decomposition for tripartite states

Exercise 2.77 in Nielsen and Chuang asks to show by example that there exist tripartite states $| \psi \rangle_{ABC} $ which cannot be written as $$| \psi \rangle = \sum_i \lambda_i | i_A \rangle | ...
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1answer
128 views

Is back in time (classical) information transfer possible (based on delayed choice entanglement swapping)?

Delayed choice entanglement swapping. Two pairs of entangled photons are produced, and one photon from each pair is sent to a party called Victor. Of the two remaining photons, one photon is sent to ...
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2answers
112 views

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
67 views

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}...
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1answer
178 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. ...
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1answer
66 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\...
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1answer
73 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 = \...
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2answers
54 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 +...