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The paper is about quantum teleportation, which is a process enabled by entanglement. However, it does not provide faster-than-light communication. The complete process of teleportation always needs a conventional communication link without which the process cannot be completed. In the theoretical description of entanglement, the outcome of the joint ...
4
You are right. Bob can only determine the state of his particle (without measuring it) once he receives information from Alice about the outcome of her measurement - and this information can be transmitted no faster than the speed of light.
Similarly, quantum “teleportation” requires Bob to receive information from Alice, transmitted no faster than the speed ...
3
In the setting you describe, you cannot say much about the effect on the state on $\lvert+\rangle\langle+|$ (with $|+\rangle=(|0\rangle+|1\rangle)/\sqrt{2}$.
To see why, consider the two channels (acting on qubits)
$$
S_1(\rho) = \tfrac12(\rho+Z\rho Z)\ ,
$$
which sets the off-diagonal entries to zero, and $S_2(\rho)=\rho$ (the identity channel).
Then,
\...
2
When you prepare an entangled state you have just put some information into it. For example, in Bell state
$$
\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)
$$
you know that both qubits have to be in same state, either 0 or 1, after measurement.
Imagine that you prepared the Bell state, one qubit remains with you and second one is sent to your friend. When you ...
2
Let $|i\rangle$ be the eigenbasis of the Hilbert space of system B which diagonalizes $X$
$$
X = \sum_{i=1}^N x_i |i\rangle\langle i|
$$
where $x_i \in \mathbb{R}$ and define
$$
|\psi_k\rangle = \frac{1}{\sqrt{1 + x_k^2}} (|0\rangle + x_k |1\rangle)
$$
$$
\rho_k = |\psi_k\rangle\langle\psi_k| = \frac{1}{1 + x_k^2}\begin{pmatrix}
1 & x_k \\
x_k & x_k^...
2
The screen is a detector. The crucial point is that you can't just make a measurement in quantum mechanics. You have to choose what measurement to perform, and some alternatives are "incompatible" (noncommuting). The screen measures the particle in a way that doesn't reveal which slit it went through. You can instead measure the particle in a way ...
1
Yes, because $\mathrm{Tr}_B\circ\mathrm{Tr}_A=\mathrm{Tr}$, that is, the composition of the partial trace operations gives the standard trace.
1
Yes, you can make a separate state that is not entangled at all. Just measure two independent spins, let us say the results are up and down, then you have the separable state $|\psi>=|u d>$.
A change of base will not entangle them, let us say you rotate it some angle and use a different base, the new state will be:$|\psi'>=|a_i r+b_il,c_i r+d_il >...
1
The document cites "Experimental delayed-choice entanglement swapping" by Ma et al, which says:
In the entanglement swapping procedure, two pairs of entangled photons are produced, and one photon from each pair is sent to Victor. [...] If Victor projects his two photons onto an entangled state, [the other] photons are entangled although they have ...
1
No. This is precisely the definition of entanglement: What cannot be created with local operations and classical communication.
1
The screen is a detector, but instead of detecting the electron at one specific slit, it detects the electron after having been through both slits. The screen where you look at the interference is absolutely an example of a quantum measurement, but it measures after the interference, not before. By the time the electron reaches the screen, the interference ...
1
I think the answer boils down to this - for electrons that can either exist in single particle spatial wavefunctions $\phi_A(r)$ and $\phi_B(r)$, and single particle spin states $|\uparrow\rangle$ or $|\downarrow \rangle$. The 'final measured' state (where the maximum information is measured and known) will be something like 'you measure an electron in box A ...
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