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15 votes

Are entangled and Bell states the same thing?

A bipartite state $\rho_{AB}$ is called entangled if it cannot be written as $$ \rho_{AB} = \sum_i \lambda_i \sigma_A^i \otimes \sigma_B^i, $$ for some $\lambda_i\geq 0$, $\sum_i \lambda_i = 1$ and $\...
Rammus's user avatar
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8 votes
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How to show that Bell states are orthonormal

First, note that $|0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $|1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ and therefore $\langle 0 |1\rangle = \begin{pmatrix} 1 & 0 \end{pmatrix} \...
KAJ226's user avatar
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7 votes

What does the minus sign in the four bell states represent?

If we measure states $|\Psi^-\rangle$ and $|\Psi^+\rangle$ in computational basis, both look identical: if one qubit is measured $|0\rangle$, the other is measured $|1\rangle$; if one qubit is ...
kludg's user avatar
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7 votes

Are entangled and Bell states the same thing?

Put in simply words you could say that $$bell \implies entangled$$ but not the opposite. Namely the Bell state is a precise case of the entangled state. We have that an entangled state is such that it ...
simonegiancola09's user avatar
7 votes

What happens to the Bell state qubits after the Quantum Teleportation?

They get measured as part of the teleportation. It doesn't matter what happens to them after that. You can throw them away, reset them back to $|0\rangle$ and re-use them for something else, whatever.
Craig Gidney's user avatar
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6 votes

What gate combinations create entangled two-qubit states?

Yes, there are many other quantum circuits that prepare entangled states. In fact, in certain sense that can be made precise, almost all possible quantum circuits on two qubits produce entangled ...
Adam Zalcman's user avatar
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6 votes
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Does a basis of maximally entangled states exist for two-qubit or two-qutrit system so that the density matrices of the basis states don't commute?

No such (orthonormal) basis can exist. An orthonormal basis $\{|\psi_i\rangle\}$ requires $\langle \psi_i | \psi_j \rangle = 0$ for $i\neq j$, and so clearly \begin{align} [\rho_i, \rho_j] &= |\...
forky40's user avatar
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6 votes

Why are Bell states the maximally entangled ones?

It depends on what you want to take as your definition of maximally entangled. But, here's a good one: Given a Bell state, I can convert it into any other two-qubit entangled state using only local ...
DaftWullie's user avatar
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6 votes

Why are Bell states the maximally entangled ones?

Entanglement is analogous to correlation. Correlation over a property simply exists where simultaneous determination is not equivalent to individual determination of a property. What does it mean? Say ...
Harshit Gupta's user avatar
6 votes

Are $(|00\rangle-|11\rangle)/\sqrt2$ and $(|11\rangle-|00\rangle)/\sqrt2$ the same quantum state?

The kets $|\psi\rangle$ and $e^{i\theta}|\psi\rangle$ represent the same quantum state for any $|\psi\rangle$ and any phase factor $e^{i\theta}$ with $\theta\in[0,2\pi)$. No observable quantity ...
Adam Zalcman's user avatar
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6 votes
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Why is a Bell state involved in quantum teleportation?

The motivation for introducing the Bell state is simple: it gets the job done! This of course raises the question: what properties of the Bell state make it suitable for use as a resource in quantum ...
Adam Zalcman's user avatar
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5 votes
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Transforming the first Bell state into the other Bell states

$\mathrm{X}$ is not equivalent to a $\mathrm{CNOT}$ gate. The former is a 1-qubit gate whereas the 2nd is a 2-qubit gate (in essence, a controlled-$\mathrm{X}$). The $\mathrm{X}$ basically flips the ...
Sanchayan Dutta's user avatar
5 votes
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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\...
Mahathi Vempati's user avatar
5 votes
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Prove that there exist tripartite $|\psi\rangle$ which cannot be written as $|\psi\rangle=\sum_i\lambda_i|i_A\rangle|i_B\rangle|i_C\rangle$

In case you are interested in an alternative proof... For a state $|\psi\rangle$ as given, the reduced density matrices of each of the three qubits must all be of the form $\sum_i\lambda_i|i_X\rangle\...
DaftWullie's user avatar
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5 votes

What is the quantum circuit to prepare a Bell state?

As you wrote at point 3, $CNOT$ is the sum of two tensor products, each one involving two matrices. Consider the first tensor product and apply the two matrices respectively to $|0\rangle$ and to $|0\...
Michele Amoretti's user avatar
5 votes

How to show that Bell states are orthonormal

The most basic but laborious way of checking that Bell states are orthonormal is to carry out the calculations for all sixteen inner products such as $\langle\Phi^+|\Psi^-\rangle$. One way to do this ...
Adam Zalcman's user avatar
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4 votes

What is the quantum circuit to prepare a Bell state?

I will answer the question in a different way. Let's assume your two qubits are represented as below. Consider your $|00\rangle$, represented by $|q_1q_0\rangle$, is the state of above circuit before ...
Abbyss's user avatar
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4 votes
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What is the quantum circuit to prepare a Bell state?

I would recommend to use a direct matrix representation. An input state $|00\rangle$ can be writen as vector $$x= \begin{pmatrix} 1\\0\\0\\0 \end{pmatrix} $$ First step, i.e. Hadamard gate on first ...
Martin Vesely's user avatar
4 votes
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How to apply the Schmidt Decomposition to a Bell state?

First suggestion: don't try it on a Bell state! This is already in the Schmidt basis, and lots of things that you do to the state will keep in Schmidt decomposed as well! Instead, why not try an ...
DaftWullie's user avatar
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4 votes

How does Bell measurement work in the teleportation?

Perhaps the following diagram helps to show what acts where. I think you've got a little muddled.
DaftWullie's user avatar
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4 votes

How to show that Bell states are orthonormal

To avoid long computation of all pair-wise inner products in Bell basis, you can think as follow. Switching from computational basis to the Bell basis is done by transformation $$ CNOT(H \otimes I) $$ ...
Martin Vesely's user avatar
4 votes

Q-Sphere representation of Bell States

The circuit you shown is to create the state $ \dfrac{|00\rangle + |11\rangle}{\sqrt{2}}$ have you started from the state $|00\rangle = |0\rangle \otimes |0\rangle$. Now to get this state to the ...
KAJ226's user avatar
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4 votes
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Q-Sphere representation of Bell States

If you want to create $\frac{|01\rangle + |01\rangle }{\sqrt{2}} $, you actually need to start from the state $|10\rangle$, not $|01\rangle$. Quickly written, the calculation is: $$ |10\rangle \...
Lena's user avatar
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4 votes
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Show that the Bell states form a basis

If you are taking the four Bell states $|\Phi^+ \rangle = \dfrac{1}{\sqrt{2}}\big(|00\rangle + |11\rangle \big) $ $|\Phi^- \rangle = \dfrac{1}{\sqrt{2}}\big(|00\rangle - |11\rangle \big) $ $|\Psi^+ \...
KAJ226's user avatar
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4 votes
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What happens if a Pauli $X$ gate is applied to part of a Bell state?

Your reasoning is correct, this is indeed the resulting state they would get. More generally, you can think about it as applying a $X\otimes I$ gate on the whole system, where $I$ is the identity gate....
Tristan Nemoz's user avatar
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4 votes

Intuition for why $\frac{|00\rangle+|11\rangle}{\sqrt{2}}$ can be written as $\frac{|++\rangle+|--\rangle}{\sqrt{2}}$

Notice that it is somewhat a coincidence of that particular Bell state and choice of basis. The states $|0\rangle$ and $|1\rangle$ are in the $z$ axis of the Bloch sphere and $|+\rangle$,$|-\rangle$ ...
Mauricio's user avatar
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4 votes

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

The reason why swap works in this way can be seen from its eigenbasis. We have $$ \text{SWAP}=P_+-P_- $$ where $P_+=I-P_-$ and $P_-$ is the projector onto the anti-symmetric state $$ |\Psi\rangle=(|01\...
DaftWullie's user avatar
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4 votes
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State tomography on a subsystem of the GHZ state

Quantum state tomography returns the expected result. State of a subsystem In quantum mechanics the state of a subsystem is given by the partial trace of the state of the full system over the ...
Adam Zalcman's user avatar
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3 votes
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How does Bell measurement work in the teleportation?

In the teleportation protocol, the two parties share the entangled Bell State and it is implemented via a CNOT gate between the state to be sent (suppose Alice is sending the state) and the part of ...
aditikatoch's user avatar
3 votes
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What is the representative matrix for a measurement in the Bell-state basis?

I think your reasoning is just fine and I checked that the Bell states are indeed eigenvectors of $$M=U^\dagger(Z\otimes Z)U,$$ as $$\begin{align} M|\phi^+\rangle =& \phantom{{}-{}}|\phi^+\rangle,\...
M. Stern's user avatar
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