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3

Entanglement + a classical channels allows you to build a quantum channel using teleportation. Thus, adding a quantum channel does not give you additional power.


2

As @Rammus has mentioned in the comments one does not need a quantum channel to utilize entanglement as a resource. One can utilize quantum correlations to aptly perform tasks that are impossible to perform with classical correlations (i.e. shared randomness). For example in a nonlocal game, two players share entanglement (but don't need any kind of quantum ...


0

They did use them in the Vienna loophole-free Bell test, which is a pre-condition for implementing DIQKD. Their benefit is that you can do a loophole-free violation of the CHSH inequality efficiency down to $2/3$, whereas if you use maximally entangled states the required efficiency is much higher, $2\sqrt2-2$. Therefore I fully expect them to be used for ...


1

This is a good question, but one that probably doesn't have a satisfying answer, as of yet. One can say what entanglement is not; how you should not imagine it. In particular, it is not a causal link between observations/systems (as per no-communication theorem etc.). It is, instead, a form of correlation between observations made on different systems. You ...


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 $|\psi_m\rangle$ ...


3

Your circuit gets lucky on the probabilities for this one particular value, but doesn't work for any other. It also produces the wrong state, as you can confirm in this adjusted version of the quirk teleportation circuit. Compared to the original initialized with your value your phase is off. The difference becomes even clearer when you use the rotating ...


-4

The entanglement principle implies the use of some form of elementary building blocks, and of those blocks exist several versions. Most of them are just some forms of combinations from other blocks, but some of them are absolutely unique. Unique in such a sense, that there exists exactly only one of each of the unique blocks. Those are the prime blocks. They ...


3

Using shadow puppets. Imagine making shadow puppets. However in this setup, instead of one you have two screens and two torches (US: flashlights), pointing 90 degrees apart so that the image formed by torch 1 is projected onto screen 1 and the image formed by torch 2 is simultaneously projected onto screen 2. screen 1 screen 2 / \ / ...


6

What is entanglement? Briefly, entanglement is a type of dependence between subsystems of a composite system. Quantum mechanics can be viewed as a variant or extension of probability theory. In this view, state vectors are analogous to probability distributions, superpositions correspond to distributions that are non-deterministic (i.e. that do not assign $1$...


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 ...


3

As mentioned in the article, you can rewrite the GHZ state as \begin{align} \frac{1}{\sqrt{2}} (|000\rangle + |111)&= \frac{1}{2\sqrt{2}}(|000\rangle + |111 \rangle + \overbrace{|001\rangle + |110 \rangle - |001\rangle - |110\rangle}^{=0} + |000\rangle + |111\rangle ) \\ &= \frac{1}{2\sqrt{2}}\left[(|000\rangle + |111 \rangle + |001\rangle + |110 \...


0

I suspect that there is a bug in the code. Here is a short analysis following the steps of your scenario. First, diagonalize $M$ $$ M = |a\rangle\langle a| - |b\rangle\langle b| = P_a - P_b $$ where $|a\rangle = \alpha_0|0\rangle + \alpha_1|1\rangle$, $|b\rangle = \beta_0|0\rangle + \beta_1|1\rangle$ and $P_a = |a\rangle\langle a|$ and $P_b = |b\rangle\...


0

Your final bullet in the repo states, The books tell us that once Alice has performed her measurement, she can predict with certainty what will Bob measure... which is only true if Alice knows that Bob is measuring in either (i) the same basis (perfect anti-correlation), or (ii) a basis rotated by $\pi$ (perfect correlation). If Alice has no information ...


0

Contrary to what you have mentioned in the question, when you use a $u_3$ gate, Qiskit does not break it up into multiple gates. In fact $u_3$ gate is one of the basis gates for all Qiskit backends and all the single qubit gates are ultimately implemented as $u_1, u_2$ or $u_3$ gates (You can check more about those gates here). Since $u_3$ gate is the most ...


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