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 $\...
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  • 4,192
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 ...
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  • 3,084
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 ...
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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.
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  • 23.2k
6 votes
Accepted

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} \...
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  • 12.6k
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] &= |\...
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  • 4,877
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 ...
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  • 14.3k
5 votes
Accepted

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 ...
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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\...
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5 votes
Accepted

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\...
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  • 47.5k
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 ...
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  • 14.3k
5 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 ...
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  • 14.3k
5 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 ...
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  • 47.5k
5 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 ...
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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.
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  • 47.5k
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 ...
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  • 12.6k
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 \...
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  • 2,394
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....
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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 $|+\rangle$,$|-\rangle$ ...
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  • 1,213
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\...
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  • 47.5k
3 votes
Accepted

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 ...
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  • 47.5k
3 votes

Bell state preparation

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\...
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3 votes
Accepted

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 ...
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3 votes
Accepted

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,\...
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  • 2,363
3 votes
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Fidelity With Bell State Calculation

No, you wouldn't find $0.9$ again. To make the partial trace calculation simpler you can note that the state $|\psi'\rangle$ is separable under the bipartition $a_1b_1 | a_2 b_2$, i.e. $|\psi'\rangle =...
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  • 4,192
3 votes
Accepted

Representing a Bell measurement on non adjacent qubits

The most direct way to do this using normal notation is to simply write the Bell projections using the same convention for subscripts:$\def\ket#1{\lvert#1\rangle}\def\bra#1{\langle#1\rvert}\def\idop{\...
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3 votes

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

Q1) The qubits in the $|\Phi^+\rangle$ state are entangled - this means that (by definition) you can not represent the state of one of them individually without talking about the second one (...
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3 votes

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

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 ...
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  • 47.5k
3 votes

What is the property of any entangled pair of qubits?

There is a more direct characteristic that makes the state of an entangled pair of qubits distinct from a non-entangled pair (which is also known as a separable state). When two qubits are not ...
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  • 4,748
3 votes

What is the property of any entangled pair of qubits?

Suppose you have 2-qubit state $|\psi\rangle_{AB}$. Entangled or not, you can always write it as $$|\psi\rangle_{AB}=a_0|0\rangle_A|\psi_0\rangle_{B}+a_1|1\rangle_A|\psi_1\rangle_{B}$$ If you measure ...
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  • 3,084

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