Hot answers tagged

19

I'm assuming that you have a finite-dimensional base Hilbert space $\mathcal H_0$ and that you're building your full Hilbert space as $\mathcal H=\mathcal H_0\otimes \mathcal H_0$. In these conditions, the set of separable states has measure zero. (It gets a bit more complicated if you have $\mathcal H_0^{\otimes 4}$ and you're allowed to split it any way ...


12

The answer is no: whether or not the state can be written as a product state does not depend on the basis. And you are precisely correct: there is indeed a basis-independent invariant that characterizes the entanglement. It is called the "entanglement spectrum": the eigenvalue spectrum of the reduced density matrix produced by taking the partial trace over ...


8

No, in fact, "most states" are entangled. (This is meant to be a heuristic; I freely admit this is probably a sticky thing to get into formally as far as randomly picking a state.) My intuition/reason for saying this is that there exist approximation methods which work by restricting themselves to a low-entanglement subspace, such as Matrix Product States (...


6

I know from this talk by M Horodecki that entangled states are more abundant than separable states. Also see this paper on how separable states form a set of measure zero in pure state space. As for your argument, I think it can also be seen this way. $|\psi\rangle = |\psi_1\rangle \otimes|\psi_2\rangle$ can be viewed as a constraint on the structure of ...


5

The set of entangled states is open and dense in the space of all states (for a given system). In that sense, almost every state is entangled.


5

No, not really. The amount of entanglement and the amount of energy in a state are completely independent: entanglement (together with discord) is a property of the state itself and its relationship to a bipartite (or multipartite) structure of the Hilbert space in which it lives, whereas energy is a joint property of the state and the system's Hamiltonian....


4

No, the entanglement (yes/no) doesn't depend on the basis of the two subsystems, only on the way how the two subsystems are separated from one another. A non-entangled state is a state of the form $|j\rangle \otimes |\alpha\rangle$ for some states $|j\rangle,|\alpha\rangle$ of the two subsystems; all other states in the composite Hilbert space are entangled....


3

when I open the box and see the cat dead, there should be a timeline split, creating a universe where the cat is alive instead of dead. Is this possible? I know we can't really know for sure, but is it possible? One has to keep in mind clearly that physics is about mathematical models that fit data and predict new observations. One can have an infinity of ...


3

There is something called quantum steering. The correlation between two EPR pairs means that if one pair is in some dynamic evolution due to some interaction that the other part of the pair will execute the same evolution. An example is with ESR/NMR for two charged spin particles in an entangled state. If one particle enters region A with a magnetic field it ...


2

The outcome is determined (probabilistically) by the state, and is either $|00\rangle$ or $11\rangle$ (assuming that's the observation you're making) equiprobably.


2

When we create an entangled state in the lab, really what we are doing is creating a 'known' entangled state that we can then experiment with. As ACuriousMind said in the comment, most of the time things ARE entangled. What is entangled depends on whether the observer has interacted with it or not. If an observer interacts with a dynamic system and then ...


2

Note that you can use the entanglement entropy to calculate the amount of entanglement in a bipartite pure state, but this is not a good measure for a general bipartite (mixed) state. In the general case there are several different entanglement measures currently used, which have certain desiderata: https://quantiki.org/wiki/axiomatic-approach. Invariance ...


2

Your initial state is entangled, which means it exhibits a correlation between the spin of the two particles. Each particle is spin up or spin down with equal probability, but the spins are (anti)correlated so that they cannot have the same spin. The states $\left|\uparrow \downarrow\right>$ and $\left|\downarrow \uparrow\right>$ are in agreement ...


2

Yes - these experiments have been conducted, most famously by Aspect et al., but also by others, see Wikipedia. They all observed violations of Bell's inequalities - Our world is therefore not local-realistic in the sense of Einstein. An extension of Bell's inequalities by Legett (Legett inequalities) holds for non-local realistic theories. Their violation ...


2

The canonical example for MPS (in fact, the first MPS ever) is the AKLT model (http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.59.799, https://projecteuclid.org/euclid.cmp/1104161001). The 2nd reference also discusses the 2D (=PEPS) version of the state. Another example of an exact MPS/PEPS model are (nearest-neighbor) RVB states (https://arxiv.org/...


1

Interaction between a physical object in a superposition, and another object, usually causes the second object to also go into a superposed state, that will be correlated in some way with the original superposed state. For example, this is true of the quantum-mechanical description of a measurement interaction - some part of the measurement device has to ...


1

I will turn my comment to an answer, in reality answering your statement in the comments: Thus I am asking "what everyday processes require quantum entanglement rather than just classical correlation?" Classical probabilistic analysis depends on classical dynamics, the emergence of entropy from statistical mechanics is a good example. There have ...


1

The question assumes that quantum entanglement actually happens the way it has been described by quantum mathematics. Which may or may not be true. All is not set yet. Leaving that aside - In any case, entanglement is not a law in itself, it is a phenomenon. It is a consequence of basic underlying laws at quantum level. For example, anti correlation of spin ...


1

To add on to WillO's answer, I think the key in entanglement is not "which measurement happens first" or "which measurement affects the other", but rather that $|00\rangle$ and $11\rangle$ are the only possible outcomes! No matter what frame you switch to, you will NEVER get $|01\rangle$ or $|10\rangle$. Even if you consider the different frames in which ...


1

First an important clarification about simultaneity you need to be aware of: In special relativity we learn that there is no such thing as two spatially separated events A and B happening 'at the same time', at least not in any absolute sense. If one inertial observer sees the events as simultaneous, another perfectly legitimate inertial observer sees A ...



Only top voted, non community-wiki answers of a minimum length are eligible