14

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 $\sigma_A^i$, $\sigma_B^i$ are states on systems $A$ and $B$ respectively, for all $i$. In other words, a state is entangled if it cannot be written as a convex ...


7

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 cannot be expressed as a tensor product, i.e. $$\nexists |x⟩ |y⟩: |\phi⟩ = |x⟩\otimes |y⟩$$ in the two qbits case. On the other hand, the Bell State is a ...


7

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.


6

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} \begin{pmatrix} 0 \\ 1 \end{pmatrix}= 0 $ and similarly, $ \langle 1|0\rangle = 0$. Also note that $|00\rangle = |0\rangle \otimes |0\rangle$, $|01\rangle = |0\...


6

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] &= |\psi_i\rangle \langle \psi_i | \psi_j\rangle \langle \psi_j | - | \psi_j\rangle \langle \psi_j |\psi_i\rangle \langle \psi_i | \\ &= 0 \end{align} So to get a ...


5

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 is to switch from Dirac notation to standard linear algebra by replacing the kets and bras with appropriate column and row vectors. After this conversion you ...


4

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: $$ |10\rangle \xrightarrow{I\otimes H} \frac{1}{\sqrt{2}}|1\rangle (|0\rangle + |1\rangle) \xrightarrow{CX} \frac{1}{\sqrt{2}}(|10\rangle + |01\rangle) $$ and $$ |01\rangle \...


4

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 state $|\psi \rangle = \dfrac{|01\rangle + |10\rangle}{\sqrt{2}} $ you need to apply another $X$ gate to any of the two qubits in addition to what you had before. ...


4

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 states. This corresponds to the fact that almost all pure two-qubit states are entangled. It is the product states that are rare (more precisely: they form a measure ...


4

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. However, do note that if Alice and Bob do not communicate, there is nothing Alice can do to influence the measurement she will get, since its qubit is in the ...


4

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 $|+\rangle$,$|-\rangle$ are in the $x$-axis. The state you chose is a sum of products of single states that are the same, and it turns out that the same is true when you convert it to ...


4

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 operations and classical communication Given that local operations and classical communication cannot increase entanglement, the possibility of performing $|\...


3

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^+ \rangle = \dfrac{1}{\sqrt{2}}\big(|01\rangle + |10 \rangle \big) $ $|\Psi^- \rangle = \dfrac{1}{\sqrt{2}}\big(|01\rangle - |10 \rangle \big) $ and place them as ...


3

A simpler answer with fewer mathematical symbols: The Bell states are examples of entangled states, but there's other examples of entangled states that are not Bell states, for example the W state: $$ \tag{1} \frac{1}{\sqrt{3}}\left(|001\rangle + |010\rangle + | 100\rangle\right), $$ the GHZ states for example: $$ \tag{2} \frac{1}{\sqrt{2}}\left(|000\rangle ...


3

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 $$ CNOT(H \otimes I) $$ As any operation on quantum computer (measurement and reset being exceptions), this operation is unitary. A unitary operation preserve angles among vectors and ...


3

The circuit you gave certainly generated an entangled state. In particular, it generates the state $|\psi\rangle = \dfrac{|00\rangle + |11\rangle}{\sqrt{2}}$ . But this is just one particular entangled state, there are many more. For instance, the following circuit prepares the state $|\psi \rangle = \dfrac{|01\rangle + |10\rangle}{\sqrt{2}} $ which is ...


3

Your calculation is correct. It is not true that QFT maps $\mathcal{B}$ to $\mathcal{B}$. It is also not true that QFT keeps $|\psi^-\rangle$ fixed. For a proof by contradiction assume that QFT maps $\mathcal{B}$ to $\mathcal{B}$. Then QFT2 = CNOT also maps $\mathcal{B}$ to $\mathcal{B}$. However, CNOT maps every Bell state to a product state. The ...


3

Yes, application of $U$ to Alice's half of $|\beta\rangle=(|00\rangle+|11\rangle)/\sqrt{2}$ and appliaction of $U^T$ to Bob's half are equivalent. In fact the identity $$ (A \otimes I)|\beta\rangle = (I \otimes A^T) |\beta\rangle $$ is true for any matrix (not necessarily unitary or real). Let $$ A = \begin{pmatrix} a_{00} & a_{01} \\ a_{10} & a_{11} ...


2

Firstly, note that every Bell state $|\psi_{ij}\rangle=(|0i\rangle+(-1)^j|1\bar i\rangle)/\sqrt{2}$ is an eigenstate of $E_i\otimes E_i$ for all $i$ (the eigenvalues are either $\pm 1$). Hence, a Bell-diagonal state remains Bell-diagonal under the action of the map. This already suggests that a Bell-diagonal state is likely to be the ultimate destination of ...


2

Starting with the state $|\psi \rangle = \dfrac{|01\rangle + |10 \rangle }{\sqrt{2}} = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} $. If you want to find the probability of measuring $+1$ in observable $X = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} $ for the first qubit, and $+1$ in the observable $Z = \begin{pmatrix} ...


2

I tried to prepare Bell state with circuit described by $\text{CNOT} (H \otimes I)$ and the result is state $\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$ as expected. Here is my circuit in the composer (including the code): After measurement on simulator I get these results Please make sure that you measure correctly, i.e. each qubit to separate classical bit: ...


2

(1) The best description that I can give is a mixed state, $\rho=I/2$. (2) If you apply a unitary operation on your half, it does not change my description of my state. If you apply a measurement on your half, my best description only updates if you tell me the measurement result. (3) The combined state looks like $(|01\rangle+|10\rangle)/\sqrt{2}$.


2

General procedure First, write down the eigendecomposition of your observable $A$ $$ A = \sum_{m} \lambda_m P_m $$ where $\lambda_m$ is the $m$th eigenvalue of $A$ and $P_m$ is the associated projector. Next, compute the projections of the input state $|\psi\rangle$ on the eigenspaces of $A$ $$ |\psi_m\rangle = P_m|\psi\rangle. $$ Note that $|\psi_m\rangle$ ...


2

What you should take from your previous question is a basic protocol. If you measure the ends of two Bell pairs using a Bell measurement, and if you apply the appropriate connection, what you are left with is a Bell state between the two unmeasured qubits. I could draw this diagrammatically as: So, this is a repeating trick: So far, this appears to ignore ...


2

@Kenneth comment is right. Note that: $$ XZ |\Phi^+\rangle = XZ\bigg(\dfrac{|00\rangle + |11\rangle}{\sqrt{2}} \bigg) = \dfrac{|10\rangle - |01\rangle}{\sqrt{2}} = - \bigg( \dfrac{|01\rangle - |10\rangle}{\sqrt{2}} \bigg) = -|\Psi^-\rangle $$ But there is no distinction between the state $-|\Psi^-\rangle $ and $|\Psi^-\rangle $ quantum mechanically. That is, ...


2

The Bell states form an orthonormal basis of 2-qubit Hilbert space. The way to show it is to come back to the definition of what an orthonormal basis is: All vectors have length 1 They are orthogonal to each other. The 2 qubit Hilbert space is 4 dimensional and you have 4 (orthonormal) vectors which implies linear independence. So the only thing you need to ...


2

Definitions for same starting point: Quantum teleportation: We know that quantum teleportation is the transfer of quantum state over a distance. Bell States: The Bell states are four specific maximally entangled quantum states of two qubits. They are in a superposition of 0 and 1 – a linear combination of the two states. Evaluate your question: From your ...


2

Yet another derivation Applying a local unitary $U^A$ on the first subsystem of a bipartite maximally entangled state $|\psi^{AB}\rangle$ is equivalent to applying a possibly different unitary $V^B$ on the second subsystem $$ (U^A\otimes I)|\psi^{AB}\rangle = (I\otimes V^B)|\psi^{AB}\rangle\tag1. $$ In the specific case of the Bell state $(|00\rangle+|11\...


2

One strong element of the intuition is related to the fact that it is maximally entangled. One definition of a pure state $|\psi\rangle$ being maximally entangled is that the individual systems have density matrices $$ \rho_A=\text{Tr}_B(|\psi\rangle\langle\psi|)=\frac{I}{d} $$ where $d$ is the dimension of $A$'s Hilbert space. Now, one thing that the ...


2

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 you have 2 coins - one black and one white. Let's say if you have a black coin you get a $+1$ colour value whereas with white you get a $-1$ colour value. If ...


Only top voted, non community-wiki answers of a minimum length are eligible