# Tag Info

10

This is an area rife with potential misunderstanding, so we need to be absolutely clear what we mean. Suppose I take a ruler and a clock and I use rulers to mark out $x, y, z$ axes in space and the clock to note the positions of events in time. Assuming spacetime is flat, I now have a universal coordinate frame that everyone who is stationary relative to me ...

7

Here are some facts: As others have said, the evolution of a quantum state, including entanglement, can be simulated arbitrarily well classically with sufficient resources. Actually, modelling the evolution of a quantum system is not even (believed to be) NP hard- if it was, a quantum computer could solve NP problems! That said, it does generally require ...

5

No, entanglement is not a classical phenomenon. That is because its very definition is quantum: Let $H_1, H_2$ be quantum spaces of states of two system. A quantum state $\chi\in H_1\otimes H_2$ of the combined system is called separable if it is a simple tensor, i.e. if there are $\phi\in H_1,\psi\in H_2$ such that $\chi = \phi\otimes\psi$. A state is ...

3

It seems that you're confusing, perhaps rightfully, the difference between exhibiting entangled behavior and describing entangled behavior. It's no surprise that a classical system can describe entanglement. As you point out, a digital computer can simulate an entagled system. However, there is no meaningful entanglement happening in the computing process; ...

3

You can calculate the density matrix of the state $$\rho = |\varphi\rangle\langle\varphi|$$ And then, the reduced density matrix for one of the particles taking the partial trace $$\rho_A = \textrm{tr}_{B,C}(\rho)$$ The state is entangled if and only if the reduced matrix is a mixed state. This can be checked, for example, calculating the von Neumann ...

3

There can be no blanket answer about what properties are shared between two entangled particles because it depends entirely on the type of entanglement. Moreover, the entanglement of a single property does not necessitate the entanglement of the others (see this question). My understanding is that if particles A and B are say, spin-entangled, a kick applied ...

2

This is a very complex question you are asking and very much depends on the notion of "operational" you consider. This paper, for instance, shows that any entangled state can help to do some task better than without it. Regarding PPT vs. NPT, e.g. PPT states cannot be distilled, while most (?) NPT states can be distilled. (Whether all NPT states can be ...

2

It all goes back to EPR experiment. In a paper published by Einstein, Podolsky and Rosen the authors argued that quantum mechanics is "incomplete". They argued that principles (such as the principle of locality) needed to be restored in order for it be a complete theory. The problem is, ultimately quantum theory is a "nonlocal" theory. What this means is ...

2

$\newcommand{\ket}[1]{\left| #1 \right>}$A state $\ket \psi \in H_1 \otimes H_2 \otimes H_3$ is said to be entangled if there exist no coefficients $a_i,b_i,c_i$ such that: $$\ket \psi = \sum_{ijk} d_{ijk} \ket{e^1_i} \otimes \ket{e^2_j} \otimes \ket{e^3_j} = \sum_i a_i \ket{e^1_i} \otimes \sum_j b_j \ket{e^2_j} \sum_k c_k \ket{e^3_k} \tag{1}$$ ...

2

Quite inefficiently compared to "direct implementation" in quantum hardware, but yes, entanglement defintely can be simulated to arbitrary degree of precision give adequate memory and time resources in a conventional, entirely classical digital computer! Just go and solve multi-particle Schrodinger equation. And this fact has nothing to do with Bohmian ...

1

When someone says that spin measured about different axis can't both be known, they mean that whatever state you pick will have variability in at least one of the possible spin measurements you can do. So that is what you will get when measure the spin, you will get variable results. This happens even with entanglement with even just one particle. With ...

1

It's not uncommon that popular science articles will refer to paradoxes in physics. However it is extremely important to understand that there are no paradoxes in physics. Our current theories of physics are self consistent and do not contain paradoxes (though there are some conditions not covered by any of our existing theories). Non-physicists tend to use ...

1

As previously said (e.g. in this answer), you’re right, any possible physical theory can be described by hidden variables. The whole point of Bell inequality is to look at some properties of these hidden variables. The bell tests rule out local hidden variables. But non-local hidden variable are still allowed, and quantum physics can be seen as a non-local ...

1

You can use the teleportation protocol to teleport and part of a larger quantum state (which can be arbitrarily entangled), and it will work the way it should: I.e., if initially A+C hold $\vert\psi\rangle_{AC}$, after the protocol B+C hold $\vert\psi\rangle_{BC}$. The same is true if the initially shared state is mixed. This follows from the linearity of ...

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