15

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

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 ...


6

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 depends on the global$^1$ phase $e^{i\theta}$ of a state. In particular, $$|\psi\rangle=\frac{|00\rangle-|11\rangle}{\sqrt2}$$ and $$e^{i\pi}|\psi\rangle=\frac{|11\...


5

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 $|\...


5

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

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

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

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

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 $$ |\Psi\rangle=(|01\rangle-|10\rangle)/\sqrt{2}. $$ Trivially, $(U\otimes U)I(U\otimes U)^\dagger=I$ and it is well known that $$ U\otimes U|\Psi\rangle=|\Psi\rangle $$ (perhaps ...


3

If I understand the question correctly, you are asking, given a unitary $U$, what are the possible unitaries $V$ such that $V^\dagger UV=U\iff [U,V]=0$. In other words, you are asking what are the possible symmetries of $U$. These all come from the degeneracies of $U$. To see this, write the eigendecomposition of $U$ as $$U = \sum_k \lambda_k\Pi_k,$$ where $\...


3

Start with a two-qubit system: $$|00\rangle$$ An arbitrary single-qubit state can be achieved by applying the unitary $$ U = e^{i\alpha}R_z(\theta)R_y(\gamma)R_z(\delta) $$ to one of the qubits. Since we are starting in the $|0\rangle$ state it is sufficient to set $\alpha = \delta = 0$, yielding $$ U|0\rangle = \cos\frac{\gamma}{2}|0\rangle + e^{i\theta}\...


3

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 ...


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

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

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

(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

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

@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

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

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

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

The key observation is that swapping the first and third column yields $$ \frac{1}{\sqrt2}\begin{bmatrix} 0 & 1 & 1 & 0 \\ 0 & -1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & -1 \end{bmatrix} \to \frac{1}{\sqrt2}\begin{bmatrix} 1 & 1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & 1 &...


2

Bob does not know the outcome of Alice's measurement. All he knows is that Alice would have obtained $|\psi\rangle$ with probability $\frac12$ and $|\psi^\perp\rangle$ with probability $\frac12$ for some orthonormal basis $|\psi\rangle$, $|\psi^\perp\rangle$. Therefore, his state is a mixture $$ \rho_B = \frac12|\overline\psi\rangle\langle\overline\psi|+\...


1

Explicit indices The difficulty here arises from making indices implicit in tensor product expressions. For example, the unitary $U$ corresponding to controlled-NOT gate on qubits $1$ and $2$ and identity on qubit $3$ is often written down as $$ U=\text{CNOT}\otimes I\tag1 $$ but a similar unitary $U'$ corresponding to controlled-NOT on qubits $1$ and $3$ ...


1

Before you measure on the $X$ basis as you're doing, you need to write your statevector in terms of the eigenstates of that operator. We have that $$ \begin{align} |0\rangle &= \frac{1}{\sqrt{2}}(|+\rangle+|-\rangle) \\ |1\rangle &= \frac{1}{\sqrt{2}}(|+\rangle-|-\rangle) \end{align} $$ Thus, you're state would become $$ \begin{align} |\psi\rangle &...


1

When a measurement is performed in a certain basis, an orthogonal projection is performed on the state in question. So given a state $|\psi\rangle$, a measurement in the hadamard basis would result in the following: $$\frac{|+\rangle\langle+|\psi\rangle}{|\langle+|\psi\rangle|}$$ with probability $|\langle+|\psi\rangle|^{2}$ or $$\frac{|-\rangle\langle-|\psi\...


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