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Questions tagged [bell-basis]

For questions related to Bell basis - i.e. converting a state to the Bell basis, working with such states and measuring in the basis.

Bell basis is composed of so-called Bell states:

  • $\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$
  • $\frac{1}{\sqrt{2}}(|00\rangle - |11\rangle)$
  • $\frac{1}{\sqrt{2}}(|01\rangle + |10\rangle)$
  • $\frac{1}{\sqrt{2}}(|01\rangle - |10\rangle)$

This means that it described two qubit states. Or in other words, any two qubit state can be expressed as a linear combination of Bell states above.

A state can be transformed to the Bell basis by operation $\mathrm{CNOT}(H \otimes I)$. An inverse operation, i.e. $(H \otimes I)\mathrm{CNOT}$, is used for a measuring in the Bell basis

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What quantum gates admit a basis-independent interpretation of their action?

The SWAP gate swaps the state of the two qubits so that in the computational basis $|01\rangle \rightarrow |10 \rangle$ with a matrix representation given by: \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1…
user97154
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Prove that the singlet is invariant under $U\otimes U$

I have a Bell state ${\Psi}^{-}= \frac{1}{\sqrt2} (|01\rangle - |10\rangle).$ How can I prove that this state is invariant (up to a global phase), when doing the same unitary $U$ on each qubit? That is, how can I show that, for all $2\times 2$…
arsene stein
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How does quantum teleportation work with mixed shared states?

I am given the scenario that instead of the two parties (A & B) sharing the bell state $|\phi_+\rangle$ they share the mixture $\rho_\lambda = \lambda|\phi_+\rangle\langle\phi_+|+(1-\lambda)\frac{\mathbb{1}}{2}\otimes \frac{\mathbb{1}}{2}$ in the…
Luca Ion
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Why are Bell states the maximally entangled ones?

I just want to know why actually Bell states are examples of maximally entangled states and significance of that "maximal" term. Is there anything for proving that?
user18115
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Are entangled and Bell states the same thing?

I am a little confused whether the entangled state and Bell state are the same thing? If they have a bit of contrast, what is the difference between them?
P.Chian
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What is the quantum circuit to prepare a Bell state?

I was watching some lectures on qubits. They were talking about how to generate a Bell state. They described it as follows: Prepare state 00: $$\left |0 \right> \otimes \left |0 \right>$$ Apply the Hadamard: $$ (H \otimes I)(\left |0 \right>…
STOI
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What gate combinations create entangled two-qubit states?

I know that in order to make a two quit entangled state, this quantum circuit is used: But I was wondering if there are any other gate combinations which also create entangled two quit states. If there are other quantum circuits which achieve this,…
Will
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How to apply the Schmidt Decomposition to a Bell state?

I am trying to understand the Schmidt Decomposition, currently in my QC class. We had a tutorial where we were told if $|\psi\rangle$ is a pure state of a composite system A then there exists $|i_A\rangle$ and $|i_B\rangle$ which are orthonomral…
Yiğit Aras Tunalı
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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 just reversing back the transformation would give me…
Bidon
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Prove that there exist tripartite $|\psi\rangle$ which cannot be written as $|\psi\rangle=\sum_i\lambda_i|i_A\rangle|i_B\rangle|i_C\rangle$

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 $$| \psi \rangle = \sum_i \lambda_i | i_A \rangle | i_B \rangle | i_C \rangle .$$ In this unofficial…
MBolin
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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}1 \\ i \\-1 \\-i\end{pmatrix}\end{equation} This…
Martin Vesely
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Simultaneous measurements and Bell basis measurements to estimate $\lvert\text{Tr}(\sigma \rho)\rvert^2$ in Huang et al. paper

Theorem 2 of this paper says if one is able to prepare $\rho^{\otimes k}$ then it is possible to predict expectation values of all $n$-qubit Pauli observables using $O(n)$ number of copies of $\rho$. It then gives an explicit procedure in Appendix…
user8183310
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General representation for GHZ states in any orthonormal basis

We know that if we consider a Bell state for example $$ |\Phi^+\rangle = \frac{|00\rangle + |11\rangle}{\sqrt{2}} $$ Then we can write this state in some other orthonormal basis in the same form. Like: $$ |\Phi^+\rangle = \frac{|00\rangle +…
Adam Warlock
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Are mixtures of pairs of Bell states perfectly distinguishable by local operations?

Consider the four Bell states $$ |\psi^{00}\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle), \hspace{2mm} |\psi^{01}\rangle = \frac{1}{\sqrt{2}}(|00\rangle - |11\rangle),\hspace{2mm} |\psi^{10}\rangle = \frac{1}{\sqrt{2}}(|01\rangle +…
user16106
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Intuition for why $\frac{|00\rangle+|11\rangle}{\sqrt{2}}$ can be written as $\frac{|++\rangle+|--\rangle}{\sqrt{2}}$

In analyzing measurement of $\frac{|00\rangle+|11\rangle}{\sqrt{2}}$ in the local $|+\rangle$, $|−\rangle$ basis, through algebra manipulation, the initial state is first written as $\frac{|++\rangle+|--\rangle}{\sqrt{2}}$. I understand how to get…
Claire
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