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
$\mathrm{X}$ 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 state of qubit B i.e., $|0\rangle_B\to|1\rangle_B$ and $|1\rangle\to|0\rangle_B$, and does not depend on the state of qubit A.
5
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 $\frac{1}{\sqrt 2}\left(|0\rangle + |1\rangle \right)$,
you can work out that $H(|+\rangle) = |0\rangle$
So,
$$CNOT(H|+\rangle \otimes |+\rangle)$$
$$= CNOT(|0\rangle \otimes |+\...
5
If we measure states $|\Psi^-\rangle$ and $|\Psi^+\rangle$ in computational basis, both look identical: if one qubit is measured $|0\rangle$, the other is measured $|1\rangle$; if one qubit is measured $|1\rangle$, the other is measured $|0\rangle$.
Similarly, if we measure states $|\Phi^-\rangle$ and $|\Phi^+\rangle$ in computational basis, both look ...
5
In case you are interested in an alternative proof...
For a state $|\psi\rangle$ as given, the reduced density matrices of each of the three qubits must all be of the form $\sum_i\lambda_i|i_X\rangle\langle i_X|$, where the basis $|i_X\rangle$ is orthonormal. That means that all three density matrices have the same spectrum (eigenvalues $\lambda_i$).
Now, ...
5
The most basic but laborious way of checking that Bell states are orthonormal is to carry out the calculations for all sixteen inner products such as $\langle\Phi^+|\Psi^-\rangle$.
One way to do this is to switch from Dirac notation to standard linear algebra by replacing the kets and bras with appropriate column and row vectors. After this conversion you ...
4
The circuit you shown is to create the state $ \dfrac{|00\rangle + |11\rangle}{\sqrt{2}}$ have you started from the state $|00\rangle = |0\rangle \otimes |0\rangle$. Now to get this state to the state $|\psi \rangle = \dfrac{|01\rangle + |10\rangle}{\sqrt{2}} $ you need to apply another $X$ gate to any of the two qubits in addition to what you had before. ...
4
If you want to create $\frac{|01\rangle + |01\rangle }{\sqrt{2}} $, you actually need to start from the state $|10\rangle$, not $|01\rangle$. Quickly written, the calculation is:
$$ |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
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
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 example such as
$$
|\psi\rangle=\frac{3}{5\sqrt{2}}|00\rangle+\frac{3}{5\sqrt{2}}|01\rangle+\frac{4}{5\sqrt{2}}|10\rangle-\frac{4}{5\sqrt{2}}|11\rangle
$$
So, as ...
3
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 $|0\rangle$ (the $|00\rangle$ pair in your state obtained at 2.) and perform the tensor product, so you get the first quarter of the final expansion. Apply the same ...
3
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 entangled Bell state Alice has. The CNOT gate creates another entangled state whose measurement Alice will send to the second party, say it's Bob, to perform the ...
3
Perhaps the following diagram helps to show what acts where. I think you've got a little muddled.
3
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
$$\begin{align}
M|\phi^+\rangle =& \phantom{{}-{}}|\phi^+\rangle,\\
M|\phi^-\rangle =& -|\phi^-\rangle,\\
M|\psi^+\rangle =& -|\psi^+\rangle,\\
\text{and } M|\psi^-\rangle =& \phantom{{}-{}}|\psi^-\rangle
\end{...
3
No, you wouldn't find $0.9$ again. To make the partial trace calculation simpler you can note that the state $|\psi'\rangle$ is separable under the bipartition $a_1b_1 | a_2 b_2$, i.e. $|\psi'\rangle = |00\rangle \otimes (\sqrt{a} |00\rangle + \sqrt{1-a} |11\rangle)$. So irrespective of the value of $a$ we have $\operatorname{Tr}_{a_2b_2}[|\psi'\rangle\...
3
The most direct way to do this using normal notation is to simply write the Bell projections using the same convention for subscripts:$\def\ket#1{\lvert#1\rangle}\def\bra#1{\langle#1\rvert}\def\idop{\mathbf 1}$
$$\begin{aligned}
\bra{\Phi^+}_{1,5} \;&=\; \tfrac{1}{\sqrt 2}\Bigl(\,\bra{0}_1\bra{0}_5 \,+\, \bra{1}_1 \bra{1}_5\,\Bigr),
\\
\bra{\Phi^-}_{1,...
3
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 (mathematically this would be represented as tensor product of two single-qubit states). The best description of the individual qubits received by Alice and Bob is that ...
3
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 measure in the Bell basis, i.e. project onto one of these four states. If we started with the state $|00\rangle$, then we can write it as
$$
|00\rangle=\frac{1}{\sqrt{...
3
There is a more direct characteristic that makes the state of an entangled pair of qubits distinct from a non-entangled pair (which is also known as a separable state).
When two qubits are not entangled, the state of the one qubit can be described without any knowledge of the state of the other qubit. Mathematically we can write:
\begin{equation}
|\psi\...
3
Suppose you have 2-qubit state $|\psi\rangle_{AB}$. Entangled or not, you can always write it as
$$|\psi\rangle_{AB}=a_0|0\rangle_A|\psi_0\rangle_{B}+a_1|1\rangle_A|\psi_1\rangle_{B}$$
If you measure qubit $A$ in state $|0\rangle$, then the qubit $B$ is in state $|\psi_0\rangle$, and you can compute the probability of qubit $B$ being in state $|0\rangle$ as $...
3
I find it a little tough to understand your calculations directly. I am especially confused by the circuit diagrams in your question; why they are there and what you are using them for.
If you are performing calculations on theoretical data (without noise), then I feel you can make do with an easier approach for quantum state tomography. As per my answer on ...
3
Simply start by writing out everything
$$
|B_{00}\rangle_{13}|B_{00}\rangle_{24}=\frac12\left(|00\rangle_{13}|00\rangle_{24}+|00\rangle|11\rangle+|11\rangle|00\rangle+|11\rangle|11\rangle\right)
$$
Let me rearrange each of these terms
$$
\frac12\left(|00\rangle_{12}|00\rangle_{34}+|01\rangle|01\rangle+|10\rangle|10\rangle+|11\rangle|11\rangle\right).
$$
Now ...
3
To avoid long computation of all pair-wise inner products in Bell basis, you can think as follow.
Switching from computational basis to the Bell basis is done by transformation
$$
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 ...
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