New answers tagged teleportation
0
I think what you could do is measure the bits, and then possibly flip the answer based on whether a drawn random number is less than the error rate associated with the outcome, i.e the error rates of measuring 0 but really given a 1, and measuring 1 but really given a 0. However, doing this on actual HW is a bit more tricky. Namely all of the logic needs ...
1
If Alice and Bob share a maximally entangled state wherein each controls one qubit: $$|\Phi^{+}\rangle=\frac{|0\rangle^{A}\otimes|0\rangle^{B}+|1\rangle^{A}\otimes|1\rangle^{B}}{\sqrt{2}}$$ and Alice wants to teleport one qubit $|\psi\rangle^{S}=a|0\rangle^{s}+b|1 \rangle^{s}$ Now, as you can see, these two states are product with one another, so the total ...
2
There are lots of different situations one can talk about from a cryptography perspective. But here's one that has a huge practical relevance:
There are physical realisations of quantum computers which, individually, are limited in the number of qubits they can use. For example, ion traps. For the sake of argument, assume you can hold 10 qubits in each ion ...
3
In a certain sense, local measurements of entangled states already can be viewed as a kind of "teleportation".
For example, suppose Alice and Bob share the Bell state $B=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$. You can check that for any $\theta$ for states $|\phi\rangle = \text{cos}(\theta)|0\rangle+\text{sin}(\theta)|1\rangle$ and $|\psi\...
2
The two most evocative types of proofs, to me, are proofs based on circuit identities and proofs based on ZX calculus diagrams.
For circuit identities, you start with a swap and then just keep doing obviously correct things like "CZ = CNOT with target conjugated by hadamards" until you get to teleportation.
For the ZX calculus, the proof is almost ...
1
I will try to give a high level interpretation of what is going on during Quantum Teleportation.
We first begin with Alice and Bob each holding 1 bit from a Bell State, otherwise known as a maximally entangled state. At a high level we can think of the entanglement "connecting" the two bits of information and correspondingly the information ...
2
The recovery operator $R_{xz}$ is the same one as you would apply when doing quantum state teleporation
$R'^\dagger_{xz}$ will be some combination of Clifford gates and Pauli gates, the latter classically conditioned on the Bell basis measurement outcome.
I think that a single qubit ($n=1$) will be sufficient to demonstrate whats going on here. Let the ...
Top 50 recent answers are included